Class Note: Physics

(3rd and 4th Semester)

Astronomy Special Paper

Department of Biological Sciences

MIDNAPORE CITY COLLEGE
Kuturiya, P.O. Bhadutala, Pin-721129
Paschim Medinipur
West Bengal, India
 Astronomical Scales
UNIT 1 ASTRONOMICAL SCALES
Structure

1.1 Introduction
Objectives
1.2 Astronomical Distance, Mass and Time Scales
1.3 Brightness, Radiant Flux and Luminosity
1.4 Measurement of Astronomical Quantities
Astronomical Distances
Stellar Radii
Masses of Stars
Stellar Temperature
1.5 Summary
1.6 Terminal Questions
1.7 Solutions and Answers

1.1 INTRODUCTION
You have studied in Units 9 to 11 of the Foundation Course in Science and
Technology (FST-1) that the universe is vast. You know that the Sun is one amongst
billions of stars situated in as many galaxies. You have also learnt that the distances
between planets and stars are huge, and so are their masses. For example, the distance
between the Sun and the Earth is of the order of 1.5×1011m. The radius of the Sun
itself is about 7×108m, which is almost 100 times the Earth’s radius. The mass of the
Earth is of the order of 1024kg and the Sun’s mass is a million times larger. The time
scales involved are also huge. For example, the estimated age of the Sun is about
5 billion years, compared to the lifetime of a human being, which is less than
100 years in most cases. All these numbers are very large compared to the lengths,
masses and time scales we encounter everyday. Obviously, we need special methods
to measure them and represent them.

The distances and masses of celestial objects are of fundamental interest to astronomy.
Does a star in the night sky seem bright to us because it is closer, or is it so because it
is intrinsically bright? The answer can be obtained if we know the distance to a star.
You have also learnt in Unit 10 of FST-1 that the mass of a star determines how it will
evolve.

In this unit we introduce you to some important physical quantities of interest in

astronomy, such as distance, size, mass, time, brightness, radiant flux, luminosity,
temperature and their scales. You will also learn about some simple methods of
measuring these quantities. In the next unit, you will learn about the various
coordinate systems used to locate the positions of celestial objects.

Objectives

After studying this unit, you should be able to:

• describe the distance, mass, time and temperature scales used in astronomy and
astrophysics;
• compare the brightness and luminosity of astronomical objects; and
• determine the distance, size and mass of astronomical objects from given data.
5
Basics of Astronomy
1.2 ASTRONOMICAL DISTANCE, MASS AND TIME
SCALES
In astronomy, we are interested in measuring various physical quantities, such as
mass, distance, radius, brightness and luminosity of celestial objects. You have just
learnt that the scales at which these quantities occur in astronomy are very different
from the ones we encounter in our day-to-day lives.
Therefore, we first need to understand these scales and define the units of
measurement for important astrophysical quantities.
We begin with astronomical distances.
Astronomical Distances
You have studied in your school textbooks that the Sun is at a distance of about
1.5 × 1011m from the Earth. The mean distance between the Sun and the Earth is
called one astronomical unit. Distances in the solar system are measured in this unit.
Another unit is the light year, used for measuring distances to stars and galaxies.
The parsec is a third unit of length measurement in astronomy.
We now define them.

Units of measurement of distances

• 1 Astronomical Unit (AU) is the mean distance between the Sun and the
Earth.

1 AU = 1.496 × 1011m

• 1 Light Year (ly) is the distance travelled by light in one year.

1 ly = 9.460 × 1015m = 6.323 × 104 AU

• 1 Parsec (pc) is defined as the distance at which the radius of Earth’s orbit
subtends an angle of 1″ (see Fig.1.1).

1 pc = 3.262 ly = 2.062 × 105AU = 3.085 × 1016m

EARTH
angle = 1 arc second

1 AU

SUN
STAR
distance = 1 parsec

Fig.1.1: Schematic diagram showing the definition of 1 parsec. Note that 1°° ≡ 60′′ and 1′′ = 60″″.
Thus, 1″″ = 1/3600 degree
6
Dimensions of Astronomical Objects Astronomical Scales

The sizes of stars or stellar dimensions are usually measured in units of solar radius
RΘ. For example, Sirius (yqC/kd), the brightest star in the sky, has radius 2RΘ. The
radius of the star Aldebaran (jksfg.kh) in Taurus is 40RΘ and that of Antares (T;s"Bk) in
Scorpius is 700 RΘ.

Unit of measurement of size

1 solar radius, RΘ = 7 × 108m

Mass
Stellar masses are usually measured in units of solar mass MΘ. We know that
MΘ = 2 × 1030 kg. For example, the mass of our galaxy is ~ 1011 MΘ. The mass of a
globular cluster is of the order of 105 − 106 MΘ. S. Chandrasekhar showed (Unit 11)
that the mass of a white dwarf star cannot exceed 1.4 MΘ. This is called the
Chandrasekhar limit.

Unit of measurement of mass

1 solar mass MΘ = 2 × 1030 kg

Time Scales
The present age of the Sun is about 5 billion years. It has been estimated that it would
live for another 5 billion years in its present form. The age of our galaxy may be
around 10 billion years. Various estimates of the age of the universe itself give a
figure between 12 and 16 billion years. On the other hand, if the pressure inside a star
is insufficient to support it against gravity, then it may collapse in a time, which may
be measured in seconds, rather than in millions of years.
In Table 1.1, we list the distances, sizes and masses of some astronomical objects.
Table 1.1: Distance, radii and masses of astronomical objects

Distance Radius Mass Remarks

Sun 1 AU 1 RΘ 1 MΘ −
−6
Earth − 0.01 RΘ 10 MΘ −
−3
Jupiter 4 AU 0.1 RΘ 10 MΘ Largest planet
(5 AU from
the Sun)
Proxima 1.3 pc 0.15 RΘ 0.12 MΘ Nearest star
Centauri
Sirius A 2.6 pc 2 RΘ 3 MΘ Brightest star
Sirius B 2.6 pc 0.02 RΘ 1 MΘ First star identified
as white dwarf
Antares 150 pc 700 RΘ 15 MΘ Super giant star

7
Basics of Astronomy You may like to express the distances and sizes of some astronomical objects in
various units introduced here.

Spend SAQ 1
10 min.
a) Express the distance between Jupiter and Sun in parsecs, and the distance between
the Earth and the Sun in light years.

b) Express the radius of the Earth in units of RΘ.

Next time when you look at the familiar stars in the night sky, you will have some
idea of how far these are from us, and also how massive they are.
An important problem in astronomy is to find out how much energy is emitted by
celestial objects. It is expressed in terms of the luminosity and is related to the radiant
flux and brightness of the object. You may have noticed that some stars in the night
sky appear bright to us, some less bright and others appear quite faint. How do we
estimate their real brightness? Let us find out.

1.3 BRIGHTNESS, RADIANT FLUX AND LUMINOSITY

It is a common experience that if we view a street lamp from nearby, it may seem
quite bright. But if we see it from afar, it appears faint. Similarly, a star might look
bright because it is closer to us. And a really brighter star might appear faint because it
is too far. We can estimate the apparent brightness of astronomical objects easily, but,
if we want to measure their real or intrinsic brightness, we must take their distance
into account. The apparent brightness of a star is defined in terms of what is called
the apparent magnitude of a star.
Apparent Magnitude
In the second century B.C., the Greek astronomer Hipparchus was the first astronomer
to catalogue stars visible to the naked eye. He divided stars into six classes, or
apparent magnitudes, by their relative brightness as seen from Earth. He numbered
the apparent magnitude (m) of a star on a scale of 1 (the brightest) to 6 (the least
bright). This is the scale on which the apparent brightness of stars, planets and other
objects is expressed as they appear from the Earth. The brightest stars are assigned
the first magnitude (m = 1) and the faintest stars visible to the naked eye are
assigned the sixth magnitude (m = 6).

Apparent Magnitude
Apparent magnitude of an astronomical object is a measure of how
bright it appears. According to the magnitude scale, a smaller
magnitude means a brighter star.

The magnitude scale is actually a non-linear scale. What this means is that a star, two
magnitudes fainter than another, is not twice as faint. Actually it is about 6.3 times
fainter. Let us explain this further.
The response of the eye to increasing brightness is nearly logarithmic. We, therefore,
need to define a logarithmic scale for magnitudes in which a difference of 5
magnitudes is equal to a factor of 100 in brightness. On this scale, the brightness
ratio corresponding to 1 magnitude difference is 1001/5 or 2.512.
Therefore, a star of magnitude 1 is 2.512 times brighter than a star of magnitude 2.
8
It is (2.512)2 = 6.3 times brighter than a star of magnitude 3. Astronomical Scales

How bright is it compared to stars of magnitude 4 and 5?

3
It is (2.512) = 16 times brighter than a star of magnitude 4.
4
And (2.512) = 40 times brighter than a star of magnitude 5.
5
As expected, it is 2.512 = 100 times brighter than a star of magnitude 6.
For example, the pole star (Polaris, Dhruva) has an apparent magnitude +2.3 and the
star Altair has apparent magnitude 0.8. Altair is about 4 times brighter than Polaris.
Mathematically, the brightness b1 and b2 of two stars with corresponding magnitudes
m1 and m2 are given by the following relations.

Relationship between brightness and apparent magnitude

b 
m1 − m2 = 2.5 log10  2  (1.1)
 b1 
b2 b1
= 100 ( m1 − m2 ) / 5 = 100 −(m1 −m2 ) / 5 (1.2)
b1 b2

In Table 1.2, we give the brightness ratio for some magnitude differences.
Table 1.2: Brightness ratio corresponding to given magnitude difference

Magnitude Difference Brightness Ratio

0.0 1.0
0.2 1.2
1.0 2.5
1.5 4.0
2.0 6.3
2.5 10.0
3.0 16.0
4.0 40.0
5.0 100.0
7.5 1000.0
10.0 10000.0

Modern astronomers use a similar scale for apparent magnitude. With the help of
telescopes, a larger number of stars could be seen in the sky. Many stars fainter than
the 6th magnitude were also observed. Moreover, stars brighter than the first
magnitude have also been observed.
Thus a magnitude of zero or even negative magnitudes have been assigned to extend
the scale. A star of −1 magnitude is 2.512 times brighter than the star of zero
magnitude. The brightest star in the sky other than the Sun, Sirius A, has an apparent
magnitude of − 1.47.
9
Basics of Astronomy The larger magnitude on negative scale indicates higher brightness while the
larger positive magnitudes indicate the faintness of an object.
The faintest object detectable with a large modern telescope in the sky currently is of
magnitude m = 29.
22
Therefore, the Sun having the apparent magnitude m = − 26.81, is 10 times brighter
than the faintest object detectable in the sky.
In the following table we list the apparent magnitudes of some objects in the night
sky.
Table 1.3: Apparent magnitudes of some celestial objects

Object Indian Name Apparent Magnitude

Sun Surya −26.81
Full Moon Chandra −12.73
Venus Shukra −4.22
Jupiter Guru −2.60
Sirius A Vyadha −1.47
Canopus Agastya −0.73
α-Centauri −0.10
Betelgeuse Ardra +0.80
Spica Chitra +0.96
Polaris Dhruva +2.3
Uranus Varuna +5.5
Sirius B +8.68
Pluto +14.9
Faintest Star +29
(detected by a modern
telescope)

Let us now apply these ideas to a concrete example.

Example 1: Comparison of Brightness

Compare the brightness of the Sun and α-Centauri using the apparent magnitudes
listed in the Table 1.3.
Solution
From Table 1.3, mSun − mαC = − 26.81 − (− 0.10) = −26.71. Therefore, using
Eq. (1.2), we obtain
bαC
= 100 −( 26.71) / 5 = 10 −10.7
bSun
or
bSun
= 1010.7 , i.e. the Sun is about 10 times brighter than α-Centauri.
11
bαC

10 You may now like to solve a problem based on what you have studied so far.
SAQ 2 Astronomical Scales
Spend
a) The apparent magnitude of the Sun is − 26.81 and that of the star Sirius is − 1.47. 10 min.
Which one of them is brighter and by how much?
b) The apparent magnitudes of the stars Arcturus and Aldebaran are 0.06 and 0.86,
respectively. Calculate the ratio of their brightness.

The apparent magnitude and brightness of a star do not give us any idea of the total
energy emitted per second by the star. This is obtained from radiant flux and the
luminosity of a star.

Luminosity and Radiant Flux

The luminosity of a body is defined as the total energy radiated by it per unit
time.
Radiant flux at a given point is the total amount of energy flowing through per
unit time per unit area of a surface oriented normal to the direction of
propagation of radiation.

−1 −2 −1
The unit of radiant flux is erg s cm and that of luminosity is erg s .
In astronomy, it is common to use the cgs system of units. However, if you wish to
convert to SI units, you can use appropriate conversion factors.
Note that here the radiated energy refers to not just visible light, but includes all
wavelengths.
The radiant flux of a source depends on two factors:
(i) the radiant energy emitted by it, and
(ii) the distance of the source from the point of observation.
Suppose a star is at a distance r from us. Let us draw an imaginary sphere of radius r
round the star. The surface area of this sphere is 4π r2. Then the radiant flux F of the
star, is related to its luminosity L as follows:

L
F=
4πr 2 (1.3)
The luminosity of a stellar object is a measure of the intrinsic brightness of a star. It is
expressed generally in the units of the solar luminosity, LΘ, where

LΘ = 4 × 10 26 W = 4 × 10 33 erg s −1

For example, the luminosity of our galaxy is about 1011 LΘ.

Now, the energy from a source received at any place, determines the brightness of the
source. This implies that F is related to the brightness b of the source: the brighter the
source, the larger would be the radiant flux at a place. Therefore, the ratio of
brightness in Eq. (1.2) can be replaced by the ratio of radiant flux from two objects at
the same place and we have
F2
= 100 ( m1 − m2 ) / 5 (1.4)
F1
11
Basics of Astronomy You know from Eq. (1.3) that the flux received at a place also depends on its distance
from the source. Therefore, two stars of the same apparent magnitude may not be
equally luminous, as they may be located at different distances from the observer: A
star’s apparent brightness does not tell us anything about the luminosity of the star.
We need a measure of the true or intrinsic brightness of a star. Now, we could easily
compare the true brightness of stars if we could line them all up at the same distance
from us (see Fig. 1.2). With this idea, we define the absolute magnitude of a star as
follows:

Absolute Magnitude

The absolute magnitude, M, of an astronomical object is defined as its

apparent magnitude if it were at a distance of 10 pc from us.

Fig.1.2: Absolute magnitude of astronomical objects

Let us now relate the absolute magnitude of a star to its apparent magnitude. Let us
consider a star at a distance r pc with apparent magnitude m, intrinsic brightness or
luminosity L and radiant flux F1. Now when the same star is placed at a distance of
10 pc from the place of observation, then its magnitude would be M and the
corresponding radiant flux would be F2. From Eq. (1.4), we have

F2
= 100 ( m − M ) / 5 (1.5)
F1

Since the luminosity is constant for the star, we use Eq. (1.3) to write
2
F2  r pc 
=   (1.6)
F1  10 pc 

Using Eq. (1.6) in Eq. (1.5), we get the difference between the apparent magnitude
(m) and absolute magnitude (M ).
12
It is a measure of distance and is called the distance modulus (see Fig. 1.3). Astronomical Scales

Distance modulus
 r pc 
m − M = 5 log10   = 5 log10 r − 5 (1.7)
 10 pc 

Farther away

Closer

Fig.1.3: Star cluster showing distance modulus as a measure of distance. For the star farther away,
m = 12.3, M = 2.6, r = 871 pc. For the closer star, m = 8.0, M = 5.8, r = 28 pc

We can also relate the absolute magnitudes of stars to their luminosities. From
Eq. (1.3), we know that the ratio of radiant flux of two stars at the same distance
from the point of observation is equal to the ratio of their luminosities. Thus, if M1
and M2 are the absolute magnitudes of two stars, using Eq. (1.5), we can relate their
luminosities to M1 and M2.
Relationship between Luminosity and Absolute Magnitude

L2
= 100 ( M1 − M 2 ) / 5 (1.8)
L1
or
L 
M 1 − M 2 = 2.5 log10  2  (1.9)
 L1 

Thus, the absolute magnitude of a star is a measure of its luminosity, or intrinsic

brightness.
Often if we know what kind of star it is, we can estimate its absolute magnitude. We
can measure its apparent magnitude (m) directly and solve for distance using
Eq. (1.7). For example, the apparent magnitude of Polaris (pole star) is +2.3. Its
absolute magnitude is −4.6 and it is 240 pc away. The apparent magnitude of Sirius A
is −1.47, its absolute magnitude is +1.4 and it is at a distance of 2.7 pc.
You may now like to stop for a while and solve a problem to fix these ideas.

SAQ 3 Spend
5 min.
a) The distance modulus of the star Vega is −0.5. At what distance is it from us?
b) If a star at 40 pc is brought closer to 10 pc, i.e., 4 times closer, how bright will it
appear in terms of the magnitude?
13
Basics of Astronomy We now discuss some simple methods of measuring astronomical distances, sizes,
masses and temperatures.

1.4 MEASUREMENT OF ASTRONOMICAL

QUANTITIES
Since the brightness of heavenly objects depends on their distances from us, the
measurement of distance is very important in astronomy. You must have measured the
lengths of several objects in your school and college laboratories. But how do we
measure astronomical distances? Obviously, traditional devices like the metre stick or
measuring tapes are inadequate for such measurements. Other less direct ways need to
be used. We now discuss some common methods of measuring astronomical
distances. Since stars have been studied most extensively, we will focus largely on
them in our discussion.
1.4.1 Astronomical Distances
You may be familiar with the method of trigonometric parallax. To get an idea of
what it is, perform the following activity.

Activity 1: Trigonometric Parallax

Extend your arm and hold your thumb at about one foot or so in front of your eyes.
Close your right eye and look at your thumb with your left eye. Note its position
against a distant background. Now close your left eye and look at your thumb with
your right eye. Do you notice that the position of the thumb has shifted with respect to
the background? Your thumb has not moved. However, since you have looked at it
from different point (left and right eyes), it seems to have shifted. The shift in the
apparent position of the thumb can be represented by an angle θ (Fig. 1.4).
Parallax is the apparent change in the position of an object due to a change in the
location of the observer.

Fig.1.4: Parallax angle and baseline

We call θ/2, the parallax angle. The distance b between the points of observation (in
this case your eyes), is called the baseline. From simple geometry, for small
θ b
angles, = , where d is the distance from the eyes to the thumb.
2 d

14
The parallax method can be used to measure the distances of stars and other objects in Astronomical Scales
the sky. The principle of the method is similar to the one used in finding the height of
mountain peaks, tall buildings, etc.
Let us now find out how this method can be used to measure astronomical distances.
Stellar Parallax
For measuring the distance of a star, we must use a very long baseline. Even for
measuring the distance to the nearest star, we require a baseline length greater than the
Earth’s diameter. This is because the distance of the star is so large that the angle
measured from two diametrically opposite points on the Earth will differ by an
amount which cannot be measured. Therefore, we take the diameter of the Earth’s
orbit as the baseline, and make two observations at an interval of six months (see
Fig. 1.5).
One half of the maximum change in angular position (Fig. 1.5) of the star is defined as
its annual parallax. From Fig. 1.5, the distance r of the star is given by

d SE
= tan θ (1.10a)
r

where dSE is the average distance between the Sun and the Earth. Since the angle θ is
very small, tan θ ≅ θ, and we can write

d SE
r= (1.10b)
θ
Remember that this relation holds only when the parallax angle θ is expressed in
radians.

dSE

Fig.1.5: Stellar parallax

Since, dSE = 1 AU, we have

1AU
r= (1.10c)
θ 15
Basics of Astronomy If we measure θ in arc seconds, then the distance is said to be in parsecs.

One parsec is the distance of an object that has a parallax of one second of an
arc (1″).

The nearest star Proxima Centauri has a parallax angle 0.77 ″. Thus its distance is
1.3 pc. Since the distance is proportional to 1/θ, the more distant a star is, the
smaller is its parallax.
In Table 1.4 we give the parallax angles and distances of some stars.
Table 1.4: Parallaxes and distances of some bright stars
Star θ (in arc-seconds) distance (r pc)
α-CMa 0.375 2.67
αCMi 0.287 3.48
αAquila 0.198 5.05
αTauri 0.048 20.8
αVirginis 0.014 71.4
αScorpii 0.008 125

Note that the angle θ cannot be measured precisely when the stellar object is at a large
distance. Therefore, alternative methods are used to determine distances of stellar
objects.
You could now try an exercise to make sure you have grasped the concept of parallax.

Spend SAQ 4
10 min.
a) The parallax angles of the Sun’s neighbouring stars (in arc-seconds) are given
below. Calculate their distances.
Star Parallax
Alpha Centauri 0.745
Barnard’s star 0.552
Altair 0.197
Alpha Draco 0.176

b) A satellite measures the parallax angle of a star as 0.002 arc-second. What is the
distance of the star?

You have just learnt that the parallax method helps us in finding the distances to
nearby stars. But how can we find out which stars are nearby? We can do this by
observing the motion of stars in the sky over a period of time.
Proper Motion
All celestial objects, the Sun, the Moon, stars, galaxies and other bodies are in relative
motion with respect to one another. Part of their relative motion is also due to the
Earth’s own motion. However, the rate of change in the position of a star is very slow.
It is not appreciable in one year or even in a decade. For example, if we photograph a
small area of the sky at an interval of 10 years, we will find that some of the stars in
the photograph have moved very slightly against the background objects (Fig.1.6).
16
 Astronomical Scales

Fig.1.6: Motion of a star with respect to distant background objects

The motion of a star can be resolved along two directions:

i) Motion along the line of sight of the observer, (either towards or away from the
observer) is called the radial motion.
ii) Motion perpendicular to the line of sight of the observer is called proper motion
(see Fig. 1.7).

Radial
motion Space
motion

Fig.1.7: Radial and proper motion of a star

Radial motion causes the spectral lines of a star to shift towards red (if the motion is
away from the observer) or towards blue (if the motion is towards the observer). This
shift is the well-known Doppler shift. The proper motion is very slow. It is measured
over an interval of 20 to 30 years. It is expressed in arc seconds per year. The average
proper motion for all naked eye stars is less than 0.1 arc second/yr.
The proper motion is denoted by µ. For a star at a distance r from the Earth it is
related to its transverse velocity as follows:

transverse velocity
proper motion =
distance of the star

v
or µ= θ , (1.11a)
r
where vθ is the transverse velocity.

Hence,

vθ = µr (1.11b)
17
Basics of Astronomy If µ is measured in units of arc-seconds per year and r in pc, the transverse velocity is
given by

vθ ( km s −1 ) = 4.74µr (1.11c)

If we add the radial velocity vector and the proper motion vector, we obtain the space
velocity of a star (Fig.1.7).

We can locate stars that are probably nearby by looking for stars with large proper
motions (see Fig. 1.8a). Proper motion of a star gives us statistical clues to its
distance. If we see a star with a small proper motion, it is most likely to be a distant
star. However, we cannot be absolutely certain since it could also be a nearby star
moving directly away from us or toward us (see Fig. 1.8b).

(a) (b)

Fig.1.8: a) If two stars have the same space velocity and move perpendicular to the line of sight, the
one with the larger proper motion will be nearer; b) Two stars at the same distance with
the same velocity may have different proper motions, if one moves perpendicular to the
line of sight and the other is nearly parallel to the line of sight

We know that the Sun itself is not stationary. The space velocity vector of a star must
be corrected by subtracting from it the velocity vector of the Sun.
The space velocity of a star corrected for the motion of the Sun is termed as the
peculiar velocity of the star.
The peculiar velocities of stars are essentially random and their typical magnitude is
such that in a time of about 106 years the shape of the present constellations will
change completely and they would not be recognisable (Fig. 1.9).

100,000 years ago Present 100,000 years from now

Fig.1.9: Change in the shape of the Big Dipper due to peculiar velocities

You may now like to solve a problem based on these concepts.

18
SAQ 5 Spend Astronomical Scales
2 min.
The star η CMa is at a distance of 800 pc. If the proper motion of the star is 0.008″/yr,
calculate its transverse velocity in km s−1.

So far you have learnt how we can find distances of stars. In astronomy, it is equally
important to know the sizes of stars. Are they all the same size, or are some of them
smaller or larger than the others? Let us now find out how stellar radii may be
measured.

1.4.2 Stellar Radii

There are several ways of measuring the radii of stars. Here we describe two methods:
• the direct method, and
• the indirect method.
Direct Method
We use this method to measure the radius of an object that is in the form of a disc. In
this method, we measure the angular diameter and the distance of the object from the
place of observation (see Fig. 1.10).

D
r θ

Fig.1.10: Direct method

If θ (rad) is the angular diameter and r is the distance of the object from the observer
then the diameter of the stellar object will be
D=θ×r (1.12)
This method is useful for determining the radii of the Sun, the planets and their
satellites. Since stars are so far that they cannot be seen as discs even with the largest
telescopes, this method cannot be used to find their radii. For this we use other
methods.
In Table 1.5 we give the radius of some stars.
Table 1.5: Radius of some stars

Star θ (in arc seconds) Radius (in RΘ)

α Tau 0.020 48
α Ori 0.034 214
α Sco 0.028 187

The luminosity of a star can also reveal its size since it depends on the surface area
and temperature of star. This provides a basis for the indirect method of determining
stellar radii.
19
Basics of Astronomy Indirect Method
To obtain stellar radii, we can also use Stefan-Boltzmann law of radiation
F = σ T4 (1.13)

where F is the radiant flux from the surface of the object, σ, Stefan’s constant and T,
the surface temperature of the star. You have learnt in Sec. 1.3 that the luminosity L of
a star is defined as the total energy radiated by the star per second. Since 4π R2 is the
surface area, we can write
L = 4π R 2 F
where R is the radius of the star. If the star’s surface temperature is T, using
Eq. (1.13), we obtain

L = 4πR 2 σT 4 (1.14)
The knowledge of L and T gives R.

Now let us consider two stars of radii R1 and R2 and surface temperatures T1 and T2,
respectively. The ratio of luminosities of these two stars will be
2 4
L1  R1   T1 
=    (1.15)
L2  R2  T 
 2

But from Eq. (1.8), we have

L2
= 100 ( M1 − M 2 ) / 5 (1.16)
L1

where M1 and M2 are the absolute magnitudes. Therefore, from Eqs. (1.15) and
(1.16), we get

R22T24
= 100.4(M 1 − M 2 ) (1.17)
R12T14

Using Eq. (1.17) let us now determine the ratio of radii of Sirius A and Sirius B.

Example 2: Determining stellar radii

The surface temperatures of Sirius A and Sirius B are found to be equal. The absolute
magnitude of Sirius B is larger than that of Sirius A by 10. Thus, M1 − M2 = −10 and
we have
R2 = 0.01 × R1
Thus the radius of Sirius A is 100 times that of Sirius B.

You may like to attempt an exercise now.

Spend SAQ 6
5 min.
The luminosity of a star is 40 times that of the Sun and its temperature is twice as
much. Determine the radius of the star.

Mass is also a fundamental property of a star, like its luminosity and its radius.
Unfortunately, mass of a single star cannot be found directly. If, however, two stars
revolve round each other, it is possible to estimate their masses by the application of
Kepler’s laws.

20
1.4.3 Masses of Stars Astronomical Scales

Two stars revolving around each other form a binary system. Fortunately, a large
fraction of stars are in binary systems and therefore their masses can be determined.
Binary stars can be of three kinds:
1. Visual binary stars: These stars can be seen moving around each other with the
help of a telescope. If both the stars have comparable masses, then both revolve
around their common centre of mass in elliptic orbits. If, however, one is much
more massive than the other, then the less massive star executes an elliptic orbit
around the more massive star (Fig.1.11).

Fig.1.11: Orbit of visual binary stars

2. Spectroscopic binary stars: The nature of these stars being binary is revealed by
the oscillating lines in their spectra. Consider the situation in Fig.1.12a. Here star
1 is moving towards the observer and star 2 is moving away from the observer.
The spectral lines of star 1 are, therefore, shifted towards blue region from their
original position due to Doppler Effect. The lines of star 2 are shifted towards red.
Half a period later, star 1 is moving away from the observer and star 2 is moving
towards the observer (Fig.1.12b). Now the spectral lines of the two stars are
shifted in the directions opposite to the earlier case. In this way the spectral lines
oscillate.

1 2 2 1
d

To Earth To Earth

(a) (b)

Fig.1.12: Spectroscopic binary stars

21
Basics of Astronomy Observations of oscillating lines indicate that the stars are binary stars. If only one
of the stars is bright, then only one set of oscillating lines is observed. If both the
stars are bright, then two sets of oscillating lines are seen.
3. Eclipsing binary stars: If the orbits of two stars are such that the stars pass in
front of each other as seen by an observer (Fig.1.13), then the light from the group
dips periodically. The periodic dips reveal not only the binary nature of the stars,
but also give information about their luminosities and sizes.

d
a c
b

c
Total light output

a a

b b
Due to limb
darkening
d
Time

Fig.1.13: Eclipsing binary stars

Now suppose M1 and M2 are the masses of the two stars and a is the distance between
them, then we can write Kepler’s third law as

GP 2
2
(M 1 + M 2 ) = a 3 (1.18)
4π

where P is the period of the binary system and G is the constant of gravitation. This
relation gives us the combined mass of the two stars. However, if the motion of both
the stars around the common centre of mass can be observed, then we have
M 1a1 = M 2 a 2 (1.19)
where a1 and a 2 are distances from the centre of mass. Then both these equations
allow us to estimate the masses of both the stars.
Masses of stars are expressed in units of the solar mass, MΘ = 2 × 1030 kg. Most stars
have masses between 0.1 MΘ and 10 MΘ. A small fraction of stars may have masses of
50 MΘ or 100 MΘ.
So far we have discussed the ways of measuring stellar parameters such as distance,
luminosity, radii and mass. Stellar temperature is another important property of a star.
1.4.4 Stellar Temperature
The temperature of a star can be determined by looking at its spectrum or colour. The
radiant flux (Fλ) at various wavelengths (λ) is shown in Fig.1.14. This figure is quite
similar to the one obtained for a black body at a certain temperature. Assuming the
star to be radiating as a black body, it is possible to fit in a Planck’s curve to the
observed data at temperature T. This temperature determines the colour of the star.

22
 Astronomical Scales

Eλ
λmax λ

Fig.1.14: Total energy flux at different wavelengths.

The temperature of a star (corresponding to a black body) may be estimated using

Wien’s law:
λmaxT = 0.29 cm K (1.20)
Such a temperature is termed as surface temperature, Ts. In general it is difficult to
define the temperature of a star. For instance the temperature obtained from line
emission is indicative of temperature from a region of a star where these lines are
formed. Similarly the effective temperature of a star corresponds to the one obtained
using Stefan-Boltzman law, i.e., F = σTe4 .
In Table 1.6 we give the range of values of stellar parameters of interest in astronomy
such as mass, radius, luminosity and stellar temperature.

Table 1.6: Range of Stellar Parameters

Stellar Parameters Range

Mass 0.1 − 100 MΘ
Radius 0.01* − 1000 RΘ
Luminosity 10−5 − 105 LΘ
Surface Temperature 3000 − 50,000 K

*It is difficult to put any lower limit on the radii of stars. As you will learn later, a neutron star has a
radius of only 10 km. The radius of a black hole cannot be defined in the usual sense.

We can find various empirical relationships among different stellar parameters, e.g.,
mass, radius, luminosity, effective temperatures, etc. Observations show that the
luminosity of stars depends on their mass. We find that the larger the mass of a star,
the more luminous it is. For most stars, the mass and luminosity are related as
3.5
L  M 
=  
 (1.21)
LΘ  M Θ 

In this unit we have introduced you to a number of astronomical quantities and

described some simple ways of measuring them. We now summarise the contents of
this unit.

1.5 SUMMARY
• The astronomical units of distance, size, mass and luminosity are defined as
follows:
− 1 astronomical unit (AU) is the mean distance between the Sun and the
Earth. 1 AU = 1.496 × 1011m.

− 1 light year (ly) is the distance travelled by light in one year.

1 ly = 9.4605 × 1015m = 6.32 × 104 AU. 23
Basics of Astronomy − 1 parsec (pc) is defined as the distance at which the radius of Earth’s orbit
subtends an angle of 1″. 1 pc = 3.262 ly = 2.06 × 105AU = 3.084 × 1016m.

− 1 solar radius RΘ = 7 × 108 m.

− 1 solar mass MΘ = 2 × 1030 kg.

− 1 solar luminosity LΘ = 4 × 1026 W.

• Apparent magnitude of an astronomical object is a measure of how bright it

appears. Its absolute magnitude is defined as its apparent magnitude if it were at
a distance of 10 pc from us.
• The difference in apparent and absolute magnitude is called the distance modulus
and is a measure of the distance of an astronomical object:

 rpc 
m − M = 5 log10  
 10pc 

• Radiant flux is the total amount of energy flowing per unit time per unit area
oriented normal to the direction of its propagation. The luminosity of a body is
defined as the total energy radiated per unit time by it.
• Brightness and radiant flux of an object are related to its apparent magnitude as
follows:

b2
= 100 ( m1 − m2 ) / 5
b1
F2
= 100 ( m1 − m2 ) / 5
F1

• The absolute magnitude and luminosity are related as follows:

L 
M 1 − M 2 = 2.5 log10  2 
 L1 

• If θ is the parallax of an object in arc seconds, then its distance in parsecs is given
by
1 AU
r=
θ
• The motion of an object can be resolved into two components: radial motion and
proper motion. The proper motion µ of a star is related to its transverse velocity
vθ as follows:

v
µ= θ
r
where r is its distance.
• Stellar radii are related to the absolute magnitudes and temperatures of stars:

R22T24
= antilog{+ 0.4(M 1 − M 2 )}
R12T14
24
• The masses M1 and M2 of stars in a binary system can be estimated from the Astronomical Scales
following relations:

GP 2
2
(M 1 + M 2 ) = a 3
4π

M 1a1 = M 2 a 2

where P is the time period of the binary system, G the constant of gravitation, a
the distance between them and a1, a2, their distances from the centre of mass,
respectively.
• The temperature of a star can be estimated by fitting observed data to Planck’s
black body radiation curve or using Wein’s law: λ max T = 0.29 cm
The temperature T of a star can also be estimated from Stefan-Boltzmann law:

F = σT 4 .

1.6 TERMINAL QUESTIONS Spend 30 min.

1. The apparent magnitude of full moon is − 12.5 and that of Venus at its brightest is
− 4.0. Which is brighter and by how much?
2. The apparent magnitude of the Sun is − 26.8. Find its absolute magnitude.
Remember that the distance between the Sun & the Earth is 1.5 × 1013 cm.
3. After about 5 billion years the Sun is expected to swell to 200 times its present
size. If its temperature becomes half of what it is today, find the change in its
absolute magnitude.
4. The mass of star Sirius is thrice that of the Sun. Find the ratio of their luminosities
and the difference in their absolute magnitudes. Taking the absolute magnitude of
the Sun as 5, find the absolute magnitude of Sirius.

1.7 SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs)

1. a) Jupiter is 5 AU from the Sun.

1 pc = 2.06 × 105 AU
5 AU = 2.43 × 10−5 pc

Distance between Earth and the Sun = 1 AU.

1 ly = 6240 AU

1 AU = 1.6 × 10−4 ly

b) The radius of the Earth is 0.01 RΘ .

2. a) The Sun is brighter

mSun − mSi = − 26.81 − (− 1.47) = − 25.34

25
Basics of Astronomy b1
= 100 25.34 / 5 = 100 5.07 = 1.38 × 1010
b2

The Sun is about 1010 times brighter than Sirius.

b1
= 100 −(0.06−0.86) / 5 = 100
+.80/5
b) = (10)0.32 = 2.09
b2

 r 
3. a) m − M = − 0.5 = 5 log10  
 10 pc 

 r   0 .5 
log10   =  −  = −0.1
 10 pc   5 

r
= (10 )−0.1
10 pc

r = 7.9 pc

2
F2  40 
b) =   = 16
F1  10 

It will appear 16 times brighter, which corresponds to m = 3 from Table 1.2.

1 AU
4. a) r= pc ;
θ
Alpha Centauri 1.34 pc;
Barnard’s star 1.81 pc;
Altair 5.07 pc;
Alpha Draco 5.68 pc

b) Distance = 500 pc

5. vθ = 4.74 µ r = 30.34 km s−1

6. M1 − M2 = 40

4
R22 T4 L 1
= 1 . 2 =   × 40
2 4 L
R1 T2 1  2

( )
R22 = R12 × 2.5

R2 = 1.58RΘ .

26
Terminal Questions Astronomical Scales

1. Apparent magnitude of moon is lower (larger negative number), than that of

Venus. Therefore, moon is brighter than Venus.
Moreover,

bmoon
= 10 −0.4( mmoon − mvenus )
bvenus

= 10 −0.4( −12.5+ 4.0) = 10 −0.4−8.5 = 10 3.4 = 2.5 × 10 3.

2. The relation between apparent magnitude m and absolute magnitude M is

M = m − 5 log r + 5

where the distance r is in parsec. Distance of the Sun in parsec is

1.5 × 1013/3 × 1018 = 5 × 10−6. So,

M = − 26.8 − 5 (log 5 − 6) + 5 = − 26.8 − 5 log 5 + 30 + 5

= 8.4 − 3.5 = 4.9

3. According to Eq. (1.17)

(200) 2
= 10 0.4( M1 − M 2 )
4
( 2)

where M1 is the present absolute magnitude of the Sun. Therefore,

 200 × 200 
M 1 − M 2 = 2.5 log 
 16 

= 2.5 log (2500) = 2.5 × 3.4 = 8.5

So, the absolute magnitude of the Sun will decrease by 8.5 and it will, therefore,
become much more luminous.

4. Using Eq. (1.21)

LSirius
= (3) 3.5
LΘ

27
Basics of Astronomy Now using Eq. (1.16),

(3) 3.5 = 10 0.4( M Θ −M Sirius )

where MΘ and MSirius are absolute magnitudes of the Sun and Sirius.

So, (MΘ − MSirius) = 2.5 log (46.8)

= 2.5 × 1.7 = 4.25.

∴ MSirius = 5 − 4.25 = 0.75.

28
 Basic Concepts of
UNIT 2 BASIC CONCEPTS OF POSITIONAL Positional Astronomy

ASTRONOMY
Structure
2.1 Introduction
Objectives
2.2 Celestial Sphere
Geometry of a Sphere
Spherical Triangle
2.3 Astronomical Coordinate Systems
Geographical Coordinates
Horizon System
Equatorial System
Diurnal Motion of the Stars
Conversion of Coordinates
2.4 Measurement of Time
Sidereal Time
Apparent Solar Time
Mean Solar Time
Equation of Time
Calendar
2.5 Summary
2.6 Terminal Questions
2.7 Solutions and Answers

2.1 INTRODUCTION

In Unit 1, you have studied about the physical parameters and measurements that are
relevant in astronomy and astrophysics. You have learnt that astronomical scales are
very different from the ones that we encounter in our day-to-day lives.

In astronomy, we are also interested in the motion and structure of planets, stars,
galaxies and other celestial objects. For this purpose, it is essential that the position of
these objects is precisely defined. In this unit, we describe some coordinate systems
(horizon, local equatorial and universal equatorial) used to define the positions of
these objects. We also discuss the effect of Earth’s daily and annual motion on the
positions of these objects.

Finally, we explain how time is measured in astronomy. In the next unit, you will
learn about various techniques and instruments used to make astronomical
measurements.

Study Guide

In this unit, you will encounter the coordinate systems used in astronomy for the first
time. In order to understand them, it would do you good to draw each diagram
yourself.

Try to visualise the fundamental circles, reference points and coordinates as you make
each drawing.

You could use clay/plastecene balls or spherical potatoes to model various planes on
the celestial sphere. You could also mark the meridians and the coordinates on them
for a visual understanding of the coordinates we describe in this unit.
29
Basics of Astronomy Objectives
After studying this unit, you should be able to:
• identify the fundamental great circles such as horizon, celestial equator, ecliptic,
observer’s/local meridian;
• assign the horizon and equatorial coordinates of a celestial object; and
• calculate the apparent solar time at a given mean solar time on any day.

2.2 CELESTIAL SPHERE

When we look at the clear sky at night, the stars appear to be distributed on the inside
surface of a vast sphere centred on the observer. This sphere is called the celestial
sphere (Fig. 2.1).

Fig.2.1: The celestial sphere centred on the observer. Note that the celestial sphere is extremely
large compared with the Earth; for all practical purposes, the observer and the centre of
the Earth coincide

Actually, the stars are at different distances from us. But since the distances of the
stars are very large, they appear to us as if they are at the same distance from us.
So it is sufficient if only the directions of stars are defined on the sphere. Since the
distance is not involved, it is usual to take the radius of the celestial sphere as
unity.
You know that on the surface of a sphere, we need just two coordinates to describe
the position of a point. We shall study below some ways used to specify these two
coordinates for stars and other celestial objects. But before that you should familiarise
yourself with the geometry of a sphere.
30
2.2.1 Geometry of a Sphere Basic Concepts of
Positional Astronomy
Let us consider a sphere of radius R = 1 unit (Fig. 2.2). We know that a plane
intersects the sphere in a circle. If the plane passes through the centre of the sphere we
get a great circle, and if it does not pass through the centre then it is a small circle.

Fig.2.2: Great and small circles on a sphere. A great circle is the intersection of a sphere and a
plane passing through its centre.

In Fig.2.3, FG is a great circle whose plane passes through the centre O, and EH is a
small circle. Points P and Q are called the poles of the great circle FG. Notice that PQ
is the diameter of the sphere perpendicular to the plane of the great circle FG. The
distance between two points on a sphere is measured by the length of the arc of the
great circle passing through them. It is the shortest path on the surface of the sphere
between these points. Thus, in Fig. 2.3, the distance between points C and D on the
great circle FG is the arc length CD. It is equal to R (=OD) times the ∠COD in
radians. Since R = 1, arc CD = ∠COD. In astronomy, it is convenient to denote
distances on a sphere in angles (radians or degrees) in this way.
An angle between two great circles is called a spherical angle. It is equal to the angle
between the tangents to the great circles at the points of their intersection. It is also the
angle between the planes of the two great circles. In Fig. 2.3, PACQ and PBDQ are
two great circles intersecting at the points P and Q. ∠APB and ∠CPD are spherical
angles between these great circles. These angles are equal. They are also equal to
∠COD, since the arc length CD is common to both great circles.

E H

A B

F O G

C D

Fig.2.3: Spherical angle is the angle between two great circles

31
Basics of Astronomy 2.2.2 Spherical Triangle
A closed figure formed by three mutually intersecting great circles on a sphere is
called a spherical triangle. For example, in Fig. 2.3, PCD is a spherical triangle. All
the 6 elements of the triangle, namely, 3 sides and 3 angles are expressed in angular
units. The elements of a spherical triangle ABC in Fig. 2.4 are the 3 spherical angles
denoted by A, B and C and the 3 sides denoted by a, b and c.

B
a

c
C
b
A

Fig.2.4: Spherical triangle is bounded by three arcs of great circles: AB, BC and CA. The spherical
angles are A, B, C and the corresponding sides are a, b and c

Note that a spherical triangle is not just any three-cornered figure lying on a sphere.
Its sides must be arcs of great circles.
The essential properties of a spherical triangle are:
i) The sum of the three angles is greater than 180°,
ii) Any angle or side is less than 180°,
iii) Sum of any two sides/angles is greater than the third side/angle.
Note: Remember that the sides are being measured in degrees or radians.
The four basic formulae relating the angles and sides of a spherical triangle (Fig. 2.4)
are given below. These are for reference only. You need not memorise them.

Sine formula

sin A sin B sin C

= = (2.1)
sin a sin b sin c

Cosine formula

cos a = cos b cos c + sin b sin c cos A

cos b = cos c cos a + sin c sin a cos B (2.2)
cos c = cos a cos b + sin a sin b cos C

Analogous cosine formulas are:

sin a cos B = cos b sin c − sin b cos c cos A

32 sin a cos C = cos c sin b − sin c cos b cos A (2.3)
Four parts formula Basic Concepts of
Positional Astronomy
cos a cos B = sin a cot c − sin B cot C (2.4)

Now that you have become familiar with the geometry of the celestial sphere, you can
learn about the different coordinate systems used to locate a celestial object.

2.3 ASTRONOMICAL COORDINATE SYSTEMS

The method of fixing the position of a star or celestial object on the celestial sphere is
the same as that of fixing the position of a place on the surface of the Earth. We shall,
therefore, first consider geographical coordinates of a place on the surface of the
Earth.
2.3.1 Geographical Coordinates
As you know, any place on the Earth is specified by two coordinates: geographical
latitude and geographical longitude (see Fig.2.5).

Longitude
Prime meridian
Latitude

Fig.2.5: The geographical coordinates (longitude and latitude) uniquely define any
position on the Earth’s surface

Note that the line joining the poles is always perpendicular to the equator.

The circles parallel to the equator are the circles of latitude. The great circles drawn
through the north and South Poles are called the circles of longitude.

How do we specify the coordinates of the place P on the Earth’s surface? The great
circle passing through P is called its meridian.

By international agreement, the geographical coordinates (longitude and latitude)

of the place P are specified with respect to the equator and the meridian through
Greenwich, which is also called the prime meridian.

33
Basics of Astronomy The general definitions are given in the box ahead.

Definitions of longitude and latitude

The longitude of a place is defined by the angle between Greenwich
meridian and meridian of the place. It is measured from 0 to 180°° east or
west of Greenwich.
The latitude is defined by the angular distance of the place from the
equator. It is measured from 0 to 90°° along the meridian of the place
north or south of the equator.

In Table 2.1, we give the longitudes and latitudes of some cities in India.
Table 2.1: Longitudes and latitudes of some cities in India

Cities Longitude Latitude

Allahabad 81° 50′E 25° 26′N

Ahmedabad 72° 44′E 23° 08′N

Chennai 80° 10′E 13° 06′N

Kanyakumari 77° 32′E 8° 05′N

Kolkata 88° 21′E 22° 34′N

Mumbai 72° 51′E 19° 07′N

New Delhi 77° 12′E 28° 46′N

Shillong 91° 52′E 25° 34′N

Srinagar 74° 48′E 34° 05′N

Ujjain 75° 46′E 23° 10′N

We shall now extend the concept of geographical coordinates on the Earth to define
the coordinates of a celestial body on the celestial sphere. But before that you may
like to work out a problem.

Spend SAQ 1
3 min.
Using Table 2.1, find the difference between a) the longitudes of Ahmedabad and
Shillong, and b) the latitudes of Srinagar and Kanyakumari.

The longitude-latitude system illustrates the principle of an astronomical coordinate

system. What we require for reference is:
i) A fundamental great circle (such as the equator), and
ii) A reference point, or origin on the chosen great circle.
Depending on the choice of these two references, we describe the following
coordinate systems, which are most often used in astronomy:
1. Horizon system
2. Equatorial system
34
2.3.2 Horizon System Basic Concepts of
Positional Astronomy
Suppose the observer is at the point O at latitude α (see Fig. 2.6). The point vertically
overhead the observer is called the zenith. It is denoted by Z in Fig. 2.6. The point Note that the term horizon
vertically below the observer is called the nadir. It is denoted by Z′. used here does not have its
common meaning as the
The great circle on the celestial sphere perpendicular to the vertical line ZO Z′ is line where the sky meets the
land.
called the celestial horizon or just the horizon. This is the fundamental great circle
chosen for reference in the horizon system.
The great circle ZNZ′SZ containing the zenith of the observer is called the observer’s
meridian or the local meridian. N and S are the points of intersection of the horizon
and the observer’s meridian. These points indicate geographical north and south
directions. Either N or S can be taken as the reference point in this system.

Zenith Z

Observer′′s meridian

O
S N

horizon Celestial equator

Nadir Z′′

Fig.2.6: The celestial horizon and the observer’s meridian

Let X be the celestial body, say a star. You can now fix the coordinates of X in this
system.
Draw the great circle ZXZ′ through the star X in Fig.2.6 (see Fig.2.7a). Let it intersect
the horizon at Y.
The position of X in the horizon system is defined with respect to the horizon and the
reference point N or S.
One coordinate of X is the arc YX, called the altitude (a). Mark it on Fig. 2.6.
To fix the other coordinate, choose N as the origin. Then the arc length NY is the
second coordinate, called the azimuth (A). Mark it on the above figure.
The position of X from the zenith, the arc ZX = 90−a, is called the zenith distance and
is denoted by z. Mark z on Fig. 2.6.
Thus, in the horizon system the position of X can be specified by the coordinates
(A, a) or (A, z). 35
 Basics of Astronomy Compare your drawing with Fig. 2.7a.

Zenith Z

zenith distance z
X
X
altitude
W W

horizon a altitude horizon a

O O
S NS N
A azimuth A

Y azimuth
E E Y

Nadir Z′′

(a) (b)

Fig.2.7: a) Horizon coordinate system; b) the azimuth gives the direction in which to look for X and
the altitude gives the angle by which a telescope must be raised from the horizon

The altitude (a) varies from 0° to 90° from the horizon to the zenith or 0° to −90°
from the horizon to the nadir. The azimuth (A) of the star is measured from N
eastward or westward from 0 to 180°. Thus, in Fig.2.7a, the great circle arc NY along
the horizon is the azimuth (east) of X. If S is chosen as origin, the azimuth is measured
from 0 to 360° through west.

Horizon coordinate system

Horizon Coordinates: (A, a) or (A, z)

The azimuth, A, is the arc length NY along the horizon if N is taken as the
origin.
The altitude, a, is the arc length YX along the great circle ZXYZ′ containing the
zenith and the star.
The zenith distance, z, is the arc length ZX from the zenith to the star on the
great circle ZXYZ′.
Fundamental great circle: Horizon
Reference point: The points of intersection N or S of the horizon and the
observer’s meridian.

The horizon system is a very convenient system and most small telescopes use it to
locate a celestial object. The azimuth gives the direction in which to look for an
object. The altitude then gives the angle by which the telescope must be raised from
the horizon to locate the object (Fig. 2.7b).
36
The pole star is situated in the geographical north direction. Let us locate its Basic Concepts of
Positional Astronomy
coordinates in the horizon system.

Example 1: Coordinates of the pole star in the horizon system

RECALL
What are the horizon coordinates of the pole star?
Solution
Study Fig. 2.8. The pole star is at P. Notice that the angle α (∠BCO) is the latitude of P
the observer. Simple geometry shows that this angle is equal to ∠PON. Do you
recognise that the arc PN or ∠PON is the altitude of P? O λ

Zenith Z

Celestial sphere

P The latitude λ of a point P

on the surface of the Earth
D 90°° − α is given by the angle between
W the line OP and a line in the
α plane of the equator. Here O
α is the centre of the Earth.
S N
B αO North Pole
C
Earth Equator

Nadir Z′′
The longitude tells us our
Fig.2.8: Horizon coordinates of the pole star. BC and OD are parallel. Hence ∠ZOD = α, east-west position. It is
∠POD = 90°° and ∠ZON = 90°°. ∠POZ = 90°° − α and ∠PON = α measured from the 0º line of
the prime meridian of
Thus, the altitude of the pole star is equal to the latitude of the observer. Greenwich. Lines of
longitude run from the
What is its azimuth? With respect to N, it is 0° and with respect to S, it is 180° west. North Pole to the South Pole.
Thus the horizon coordinates of the pole star are (0°,α).

You may now like to attempt a problem.

SAQ 2 Spend
Suppose you wish to point a small telescope to a star whose azimuth is 30° and 3 min.
altitude is 45°. Describe the procedure you will adopt.

The horizon system, though simple, has two shortcomings. Firstly, different observers
have different horizons. Therefore, at a given time, the horizon coordinates of an
object will be different for different observers. Secondly, as the Earth rotates from
west to east, celestial objects move from east to west across the sky. Looking from far
above the North Pole, say from a spacecraft, the motion of the Earth is anticlockwise.
Thus, even for the same observer, the horizon coordinates keep changing with time.
So, the horizon system of coordinates is not very useful for practical purposes. These 37
 Basics of Astronomy drawbacks are removed in the equatorial system, a system that gives the same
coordinates for an object for all observers on the Earth.
2.3.3 Equatorial System
Consider Fig. 2.9 showing the celestial sphere for an observer O. The great circle
whose plane is parallel to the equatorial plane of the Earth, and contains the centre O
of the celestial sphere is called the celestial equator. P and Q are the poles of the
celestial equator: P is the north celestial pole and Q, the south celestial pole. These
poles are directly above the north and south terrestrial poles. As you know, the point P
also points to the pole star.
In Fig. 2.9, the great circle NESW is the observer’s horizon with zenith Z as its pole
and RWTE is the celestial equator for which P and Q are poles.
The celestial equator and the observer’s horizon intersect in two points, E and W,
called the East and West points.
Notice that the observer’s
meridian is a full great Any semi-great circle through north and south celestial poles P and Q is called a
circle, unlike the meridian. meridian. But as you have learnt in Sec. 2.3.2, the full great circle through the
observer’s zenith (PZRQT) is called the observer’s meridian or the local meridian.

U Zenith Z

R Celestial P
Equator Diurnal
Celestial
X circle
sphere

O
S horizon N
C
D
E V

Observer’s
meridian

Q
T

Nadir Z′′

Fig.2.9: Schematic diagram of the celestial equator and diurnal circle UVX, upper (U) and lower
REMEMBER (V) transits. In the segment DUC the star is above the horizon. It is below the horizon in the
segment CVD
THE DIURNAL
CIRCLE OF A STAR As a result of the Earth’s rotation from west to east, the celestial bodies appear to
IS ALWAYS move across the sky in the opposite direction, i.e., from east to west. This daily or
PARALLEL TO THE diurnal motion is common to all celestial objects. The path of a star X will be along
CELESTIAL the small circle XUV parallel to the celestial equator. It is called the diurnal circle
EQUATOR. (Fig. 2.9). It intersects the observer’s meridian at the points U and V.
The star completes one circuit of its diurnal circle in 24 hours, which is the Earth’s
rotation period. As you can see from the Fig. 2.9, the star will cross (transit) the
observer’s meridian at the points U and V during its diurnal motion. These are called
the upper and lower transits. A star is said to rise when it rises above the observer’s
38
horizon. It is said to set when it goes below the horizon. In Fig. 2.9, in the segment Basic Concepts of
Positional Astronomy
DUC, the star is above the horizon. It is below the horizon in the segment CVD.
Now, suppose we want to define the equatorial coordinates of a star X (see Fig. 2.10).
The celestial equator (RWTE) is the fundamental great circle in this system. The
reference point is R, the point of intersection of the observer’s meridian (ZRQP) and
the celestial equator above the horizon.
The semi-great circle PXQ is known as the hour circle of the star X. It is called hour
circle because it indicates the time elapsed since the star was on the upper transit on
the observer’s meridian. Let the hour circle of X intersect the celestial equator at J
(Fig. 2.10).

U Z

R
P
Diurnal
X circle
J

W
Celestial
Equator O N
S C
D
E V

Q
T

Z′′

Fig.2.10: Schematic diagram showing local equatorial coordinates (H = RJ, δ = JX). Note that as
the diurnal circle is parallel to the celestial equator, JX does not change due to the star’s
diurnal motion. But RJ changes as the star moves in the diurnal circle

One coordinate of the star X is given by the great circle arc RJ along the celestial
equator. It is called the hour angle (H) of the star. It is measured towards west from 0
to 24 hours.
The other coordinate required to define the position of the star is the circle arc JX
from the equator along the star’s hour circle. It is called declination (δ).
As the path of the diurnal motion of the star is parallel to the celestial equator, the
declination, δ, of a given star does not change during the diurnal motion.
The hour angle and the declination specify the position of the star in the local
equatorial system.

In astronomy, angles such as hour angle are frequently expressed in time units
according to the relation:

24 hours = 360°, lh =15°, 1m = 15′ and 1s = 15″.

39
Basics of Astronomy The declination of a star varies from 0 to 90° from the celestial equator to the celestial
poles. It is taken as positive if the star is situated north of the equator and negative if it
is south of the equator.
Notice that in this system, one of the coordinates, the declination, does not change
during the diurnal motion of the star while the other coordinate, the hour angle, varies
from 0 to 24 hours due to the rotation of the star. That is why this system is called the
local equatorial system.

Local Equatorial System

Local Equatorial Coordinates: (H, δ)
The hour angle, H, is the arc length RJ along the celestial equator.
The declination, δ, is the arc length JX along the star’s hour circle.
Fundamental great circle: Celestial equator.
Reference point: Intersection of observer’s meridian and celestial equator.

We shall now define a coordinate system in which both the coordinates of a star
remain unchanged during its diurnal motion. For this, we choose the celestial equator
as the fundamental great circle and the vernal equinox (ϒ) as the reference point. Let
us understand what ϒ is.
Vernal equinox
You know that the Earth moves around the Sun in a nearly circular orbit. The plane of
the Earth’s motion around the Sun intersects the celestial sphere in a great circle
called the ecliptic (Fig. 2.11a).
As seen from the Earth, the Sun appears to move around the Earth in this plane. Thus
the ecliptic is the annual path of the Sun against the background stars. In Fig. 2.11a, R
ϒ T is the celestial equator and M ϒ M′ is the ecliptic. The ecliptic is inclined to the
equator at an angle of 23°27'. Points K and K′ are the poles of the ecliptic.
The ecliptic and the celestial equator intersect at two points called vernal equinox (ϒ)
and autumnal equinox (Ω).
During the annual apparent motion, the Sun is north of the equator on ϒ M Ω and
south of the equator on Ω M′ ϒ. Thus, the Sun’s declination changes from south to
north at vernal equinox (ϒ) and from north to south at autumnal equinox (Ω).
We can define the position of a star X with respect to the celestial equator and the
vernal equinox (Fig. 2.11b). Let the hour circle of X intersect the celestial equator at J.
One coordinate of X is the arc length ϒJ along the celestial equator measured from
vernal equinox eastward. It is known as the right ascension of the star and is
denoted by α.
The other coordinate is the declination δ (arc JX ) as already defined.
As the Earth rotates, the points ϒ and J on the celestial equator rotate together.
Thus, the separation between ϒ and J does not change and the right ascension of the
star remains fixed.
This system of coordinates (α, δ) is called the universal equatorial system. The α
and δ of stars do not change appreciably for centuries.
40
 North Basic Concepts of
Positional Astronomy
celestial pole

K
P

R Autumnal equinox
Sun
Celestial Ω
equator δ
Earth
M' M
Ecliptic
α
ϒ
Vernal equinox
ϒ
T
Ecliptic
(b)

South K'
celestial pole

(a)

Fig.2.11: a) The ecliptic, vernal equinox and autumnal equinox; b) universal equatorial coordinates
α, δ)
(α

Universal Equatorial System

Universal Equatorial Coordinates: (α, δ)
The right ascension, α, is the arc length ϒJ eastward along the celestial equator.
The declination, δ, is the arc JX along the star’s hour circle.
Fundamental great circle: Celestial equator.

Reference point: Vernal equinox.

Table 2.2 gives the universal equatorial coordinates of some prominent objects.
Table 2.2: Universal equatorial coordinates of prominent astronomical objects

Object Universal equatorial coordinates

Crab nebula RA 5: 35 (h,m) Declination 22°:01m
Andromeda RA 00:42.7 (h,m) Declination 41°:16m
Ring nebula RA 18:53:35.16 (h,m,s) Declination 33°:01m
Orion nebula RA 05:35.4 (h,m) Declination −05°:27m
41
Basics of Astronomy Let us now find the coordinates of a star.

Example 2: Equatorial Coordinates

Calculate at latitude of 50°N, the zenith distance of a star of declination 20°N when it
is on the observer’s meridian.
Solution
Look at Fig. 2.12.

Z
40°°
20°° N X

Celestial P
R equator

20°° S X' 50°°

S O horizon N

Fig.2.12

Since the latitude of the place is 50°N, the altitude of the pole star is also 50°. So, arc
PN = 50°. Therefore, arc PZ = 40°, since arc NZ is 90°. As the declination of X is
20°N, RX = 20°. Thus, arc ZX, which is the zenith distance of X, is 30°, since
RP = 90°. If the declination of the star were 20°S, it would have been at X′, then the
zenith distance ZX′ would have been 50° + 20° = 70°.

You should now practice calculating the equatorial coordinates of a few celestial
objects.

Spend SAQ 3
10 min.
a) Determine the declination (δ) of (i) celestial North Pole, (ii) celestial South Pole
(iii) zenith. (Hint: Look at Fig. 2.12 again).
b) Determine the right ascension (α) and declination (δ) of (i) vernal equinox, (ii)
ecliptic north pole, (iii) ecliptic South Pole. (Hint: Look at Fig. 2.13. Notice that
arc ϒR = 6h.)

42
 Basic Concepts of
Positional Astronomy
K
P
δ

R
Celestial Sun
equator
Earth
M' M
Ecliptic

ϒ
Vernal equinox T

δ
Q

K'

Fig.2.13

We shall now briefly describe the diurnal motion of the stars and the Sun as seen from
different places on the surface of the Earth.

2.3.4 Diurnal Motion of the Stars

See Fig. 2.14, drawn for an observer in northern latitudes. The diurnal circles of stars
north of the equator are more than half above the horizon (arc JKL). The diurnal
circles for stars south of the equator are less than half above the horizon (arc MCM′).

Z
K
R Celestial
Celestial P Z
equator
equator

C W
S M O J N
L
M' E
Dec. 21
Horizon N
S
Mar. 21
Q T Sept. 22 June 21

Z'
(a) (b)

Fig.2.14: a) Diurnal circles for stars north and south of the equator; b) diurnal circles on particular days
43
Basics of Astronomy Thus, an observer in northern latitudes will see more northern stars for longer duration
above the horizon compared to the southern stars. For an observer on the equator, the
zenith is on the celestial equator. Hence, both northern and southern stars are equally
visible for 12 hours as their diurnal circles are perpendicular to the horizon (Fig. 2.15).

Z Celestial
equator

Horizon
Q O P

δmax = 90°° δmax = 90°°

Z'

Fig.2.15: Diurnal circles of stars seen from the equator

If we observe stars to the north, we find that some stars never set, i.e., they never go
below the horizon. Instead they trace complete circles above it (Fig. 2.16). Such stars
are called circumpolar stars. (The Egyptians called such stars as the ones that know
no destruction.) Such stars are also there near the South Pole.

Zenith Z

R Celestial P
Equator
Celestial
X Star
sphere never
sets
W

S Horizon O N
C
D
E

Star Observer’s
never meridian
rises
Q
T

Nadir Z′′

Fig.2.16: Circumpolar stars

44
For an observer at latitude φ in the northern hemisphere, the declination of Basic Concepts of
Positional Astronomy
circumpolar stars is δ ≥ (90 − φ) (Fig. 2.17). The southern stars with δ < (φ − 90) are
never seen above the horizon. They never rise. The situation is exactly the opposite in
the southern hemisphere. For an observer on the equator, φ = 0, and the maximum
value of δ is 90° at the north and South Poles. Hence, there are no circumpolar stars,
as the declination of stars can never exceed 90° (see Fig. 2.15).

R
P

W
Celestial X
equator O N
S
90°°− φ
φ δ
E α

Q T

Z′′

Fig.2.17: Declination of circumpolar stars in the northern hemisphere for an observer at latitude φ
in the northern hemisphere

At North Pole (φ = 90°), all the stars lying north of the equator, that is, δ ≥ 0 are
circumpolar, while the southern stars (δ< 0) are below the horizon. These will be
invisible (Fig. 2.18).
Z, P

Celestial
equator Horizon
O

Z', Q
Fig.2.18: Circumpolar stars at the North Pole

Motion of the Sun

You have studied in Sec. 2.3.3 that because of the yearly revolution of the Earth round
the Sun, the Sun appears to move around the Earth along the ecliptic. It moves at a
45
Basics of Astronomy rate of about 1° per day. As a result, the right ascension and declination of the Sun
change continuously. The Sun is at ϒ (Fig. 2.19) on or around March 21 every year.
At that time its coordinates are α = 0, δ = 0.
At M, around June 21, the Sun is at α = 6h, δ = +23.5°. At Ω around September 23,
α = 12h, δ = 0° for the Sun. At M' around December 22, the Sun is at α = 18h,
δ = −23.5° (see Table 2.3).

The point M with δ = δmax = + 23.5° is the summer solstice which marks the
beginning of Dakshinayan, the south-ward motion of the Sun. The opposite point M'
with δ = δmin = − 23.5° is the winter solstice which is the beginning of Uttarayan,
the north-ward journey of the Sun. Note that the arc lengths ϒM and ϒM′ are equal
Table 2.3 even though it may not seem so in Fig. 2.19.
~ Day (α, δ)
North
March 21 (0, 0) celestial pole
h
June 21 (6 , + 23.5°)
K
September 23
h
(12 , 0°) P
h
December 22 (18 , -23.5°)
Celestial
equator
Ω (Sept.
( 23)
R
α = 12h, δ = 0°° Sun

Earth
(Dec. 22) M' M (June 21)
Ecliptic
α = 18h, δ = −23.5°° α = 6h, δ = +23.5°°

ϒ (Mar 21)
α = 0, δ = 0 T

Q
South
celestial pole
K'

Fig.2.19: Motion of the Sun

The Sun as a Circumpolar Star

You have learnt that for an observer at latitude φ, all stars with δ ≥ 90° − φ are
circumpolar. Thus, at North Pole φ = 90° and for a star to be circumpolar, δ ≥ 0. You
have also learnt that the declination of the Sun varies from −23.5° to +23.5° annually.
Hence, for an observer on the North Pole, the Sun is circumpolar from March 21 to
September 23 when the Sun is north of the equator since δ ≥ 0 for the Sun during this
period.
Further, from September 23 to March 21 the Sun is invisible at the North Pole, since
its declination is negative during this period. On March 21, the Sun appears above the
horizon and continues to remain above the horizon for 6 months till September 23.
46
Thus, it is seen that the Sun is above the horizon for six months and below the horizon Basic Concepts of
Positional Astronomy
for 6 months for observers at the North Pole.
Can you say how an observer at the South Pole sees the Sun?
The Sun is visible from September 23 to March 21 at the South Pole and invisible for
the next six months.
During the ‘day’ of 6 months, that is, from March 21 to September 23 at the North
Pole, the Sun moves around the sky in small circles parallel to the celestial equator,
which coincides with the horizon (see Fig. 2.20).

Z, P

Celestial
equator
Horizon

Z', Q

Fig.2.20: Sun’s motion for an observer on the North Pole for whom the celestial equator coincides
with the horizon.

On a given date, the Sun is circumpolar at all those places whose latitude φ ≥ (90 − δ),
where δ is the declination of the Sun on the date under consideration. On June 21/22,
for example, when the Sun's declination is +23.5°, the Sun is circumpolar at all the
places north of 66.5° latitude.
You may now like to stop for a while and concretise these ideas. Try this SAQ.

SAQ 4 Spend
10 min.
a) Find the approximate zenith distance of the Sun on June 22 and December 23 at
Delhi (φ = 28°46′N).
b) At what latitudes is the star Procyon (δ = 05°18′S) circumpolar?

2.3.5 Conversion of Coordinates

We now state the relevant formulae for conversion of coordinates from local
equatorial system to horizon system. Their mathematical derivation can be found in
any standard book dealing with positional astronomy such as Smart (1960). You
need not memorise these formulae.
47
Basics of Astronomy Local Equatorial and Horizon System:
Conversion of (H, δ) to (A, z)
sin z sin A = cos δ sin H
sin z cos A = sin δ cos φ − cos δ sin φ cos H (2.5)
cos z = sin δ sin φ + cos δ cos H cos φ

Conversion of (A, z) to (H, δ):

cos δ sin H = sin z sin A
cos δ cos H = cos z cos φ − sin z sin φ cos A (2.6)
sin δ = cos z sin φ + sin z cos φ cos A
So far we have discussed how to locate the position of a celestial object in horizon,
local equatorial and universal equatorial coordinate systems. However, you know
that along with the space coordinate, we also need to specify the time to be able to
completely describe the motion of an object. Let us now learn about how time is
measured in astronomy.

2.4 MEASUREMENT OF TIME

The rotating Earth is our natural clock. The ancient people were familiar with rotation
and revolution of the Earth around the Sun and the revolution of the moon around the
Earth. It was natural for them to base time measurement on these motions. Their basic
unit of time was from one Sunrise to the next. They called it the day. Revolution of
the Earth around the Sun provided another unit called the year. A third unit, the
month, is approximately the period of revolution of moon around the Earth.
The basic unit of time, the day, is defined as the time required for the Earth to rotate
once on its axis. If this rotation is considered with respect to the Sun then the time for
one rotation relative to the Sun is the solar day, that is, the time from one Sunrise to
the next Sunrise. On the other hand, rotation period of the Earth relative to the stars
is called a sidereal day. The sidereal day is different from a solar day. This difference
arises due to the Earth’s revolution around the Sun (see Fig. 2.21).

F B

F F'

Fig.2.21: The difference between one sidereal day and one solar day

In Fig. 2.21, the Earth is initially at A in its orbit around the Sun. Suppose at a given
place F, the Sun and a distant star are overhead. Thus, it is noon for an observer at F
(the foot of the solid arrow) and midnight at F′ (the foot of the dotted arrow). One
sidereal day later (say at time t), the Earth is at B but now the solid arrow does not
point to the Sun, though it would still point towards the distant star as it is very far
48
away compared to the Sun. So at time t, the Sun is not overhead at F, and it is not Basic Concepts of
Positional Astronomy
noon at F. The solid arrow points towards the Sun only a little later at C, when the
Earth has turned upon its axis and moved a bit in its orbit.
The time from A to C is equal to one solar day, while the time from A to B is equal to
one sidereal day.
The Earth has to rotate by about 1° more to bring the Sun overhead and complete the
solar day. It takes roughly 4 minutes for the Earth to rotate through 1° since it rotates
through 360° in 24 hours. Thus, the sidereal day is shorter than the average solar day
by about 4 minutes. In a month (30 days), the sidereal time falls behind the mean solar
time by two hours, and in six months the difference becomes 12 hours. Then at solar
midnight we have sidereal noon.
Since the Sun regulates human activity on the Earth, we use solar time in our daily
life. The sidereal time is not useful for civil purposes but is used in astronomical
observations.

2.4.1 Sidereal Time

Due to the rotation of the Earth, the vernal equinox ϒ, moves along the equator once
in 24 hours like any other star. The hour angle of ϒ for an observer is called the
sidereal time for that observer or the local sidereal time (LST).

Z
P
X

R
J
W

S N
ϒ
E

Q
Z'

Fig.2.22: Sidereal Time

In Fig. 2.22, NWSE is the horizon, P and Z are the north celestial pole and the zenith,
respectively. The hour angle (HA) of vernal equinox is the great circle arc Rϒ
measured towards the west from the observer’s meridian. When ϒ is at R, at upper
transit on the observer’s meridian, its HA is 0h and consequently the local sidereal
time (LST) is 0h at this instant.

The interval of time between two successive transits of vernal equinox over the same
meridian is defined as one sidereal day. One sidereal day is divided into 24 sidereal
hours, one sidereal hour into 60 sidereal minutes and one sidereal minute into 60
sidereal seconds.
Let X be the position of a star in Fig. 2.22. The arc RJ (towards west) is the HA of X
(HAX) and the arc ϒJ (towards east) is the right ascension of X (RAX). We have
Rϒ = RJ + ϒJ. Hence
49
Basics of Astronomy LST = HAX + RAX (2.7)
This is an important relation in the measurement of time. If X is the Sun, denoted by

, then we have the relation

LST = HA
+ RA
(2.8)

2.4.2 Apparent Solar Time

The hour angle of the Sun at any instant is the apparent solar time. The interval
between two consecutive transits of the Sun over the same meridian is called one
apparent solar day (see Fig. 2.23).

Local celestial meridian

Celestial Equator
(a)

Ecliptic

EAST SOUTH WEST

Local celestial meridian

Celestial Equator
(b)

Ecliptic

EAST SOUTH WEST

Fig.2.23: Apparent solar day

In Fig. 2.23a, we show the Sun at noon on March 21, when the Sun and the vernal
equinox are on the local meridian. The celestial sphere rotates and some time during
the next day, the vernal equinox is again on the meridian. That interval of time is one
sidereal day; a fixed point on the celestial sphere has gone around once.
But in that time the Sun has moved eastward along the ecliptic (see Fig. 2.23b). But
the Sun is not yet on the meridian and it is not yet noon. The celestial sphere has to
rotate a bit more to bring the Sun up to the meridian. This time interval represents the
apparent solar day.
The apparent solar day is found to vary throughout the year. Thus, the Sun is an
irregular timekeeper and consequently the time based on the motion of the Sun is not
uniform. The non-uniformity in the solar time is due to two reasons: the first is that
the path of apparent motion of the Sun is elliptical and not circular. This means that
the Sun moves at a non-uniform rate. The second reason is that the Sun's apparent
motion is in the plane of the ecliptic and not in the plane of the equator; and the
measurement of time, that is, the measurement of hour angle is made along the
50 equator.
2.4.3 Mean Solar Time Basic Concepts of
Positional Astronomy
In order to have a uniform solar time, a fictitious body called the mean Sun is
introduced. The mean Sun is supposed to move uniformly (that is, in a circular path)
along the equator completing one revolution in the same time as the actual Sun does
round the ecliptic. The hour angle of the mean Sun at any instant is defined as the
Mean Solar Time (MST), or simply, the Mean Time.
The time interval between the successive transits of the mean Sun over the same
meridian is defined as the mean solar day.
The mean solar day is subdivided into hours, minutes and seconds. The civil day
begins at mean midnight for reasons of convenience. Our clocks are designed to
display the mean solar time.
By international agreement, the time of the Greenwich meridian is called the
Greenwich mean time (GMT). It can be related to the mean time of any other place
through the longitude of that place.
Since the Earth completes one rotation on its axis, that is 360 degrees of longitude in
24 hours, each degree of longitude corresponds to 4 minutes of time. Therefore, GMT
at a given instant is related to the mean time of any other place by a simple relation:
Local Mean Time = GMT + λ (in units of time) (2.9)
where λ is the geographical longitude of the place expressed in units of time. In the
above relation, λ is to be taken as positive if the longitude of the place is east of
Greenwich, and negative if it is west of Greenwich. The above relationship holds good
for both the sidereal as well as the solar time.
It is obvious that if each town and city in a country were to use its own local time, the
life of citizens would become very difficult. To overcome this problem, a whole
country uses the local time of a place as its standard time.
For example, the Indian Standard Time is the local time of a place on longitude 82.5°
east. The IST is, therefore, ahead of the GMT by 5½ hours. Very large countries such
as United States have more than one standard time.

Example 3
What will the local time of Shillong be when the local time of Ahmedabad is 6 p.m.?

Solution
Refer to Table 2.1. The difference between longitudes of Ahmedabad and Shillong is
approximately 19°. Since each degree of longitude is equal to 4 minutes of time, the
difference in the local time of these two places is 19 × 4 = 76 minutes. Since Shillong
is east of Ahmedabad, local time of Shillong will be ahead of the local time of
Ahmedabad by one hour and sixteen minutes.

And now you can attempt an exercise.

SAQ 5 Spend
5 min.
Refer to Table 2.1. If the local time of Mumbai is 8:00 p.m., what would be the local
time at Kolkata at that time?

We will now discuss the Equation of Time, which relates the apparent solar time and
the mean solar time.
51
Basics of Astronomy 2.4.4 Equation of Time
NOTE
HA
is the symbol of As we have seen in Eq. (2.8)
hour angle for the Sun.
LST = HA
+ RA
(2.10)
RA
is the symbol of
right ascension of the Using Eq. (2.10) for the Sun and the mean Sun, we can write
Sun.
RAMS − RA
= HA
− HAMS (2.11)
HAMS is the symbol The difference in right ascensions of the mean Sun and true Sun at a given instant (see
of hour angle for the
Fig. 2.24a) is called the Equation of Time (ET).
mean Sun.
Celestial equator
RAMS is the symbol non-uniform
of right ascension of rate
the mean Sun.

Ecliptic

uniform rate

uniform rate

(a)

Real Sun
time of
perihelion time of equinox Fictitious Sun

Not to scale Equation of time Mean Sun

RAMS − RA

18
16
14
12
10
8
Sundial-clock time

6
4
2
(b) 0
−2
−4
−6
−8
−10
−12
−14
−16
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

Fig.2.24: Equation of time

52
Thus, we have Basic Concepts of
Positional Astronomy
ET = HA
− HAMS (2.12)
We see that the ET at any instant is the difference between the apparent solar time
given by HA
and the mean solar time given by HAMS.
In Fig. 2.24b, you can see that during the year the ET varies between −14 1/4 to + 16
1/4 minutes and it vanishes 4 times during the year on or about April 16, June 14,
September 1 and December 25.
Any discussion on time in astronomy would perhaps be incomplete if we do not
mention the calendar.

2.4.5 Calendar
The civil year contains an integral number of 365 mean solar days. As the actual
period of Earth’s revolution, called a tropical year contains 365.2422 mean solar
days, a fraction 0.2422 of a day is omitted each year. This resulted in the loss of a
number of days over several centuries and the civil year would get out of step with the
seasons.
To overcome this confusion, Julius Caesar introduced Julian calendar, named after
him, in which the year was made of 365.25 mean solar days. Consequently, every 4th
year was made to contain 366 mean solar days. This was called a leap year. The extra
day was added in February. Thus, in this calendar, a year, which is divisible by 4, is a
leap year in which the month of February has 29 days.
In Julian calendar, the tropical year was assumed to be of 365.25 mean solar days, but
its actual length is 365.2422 mean solar days. The small difference between assumed
and actual lengths of the year, created serious problem by the 16th century and the
civil year was out of step in relation to the seasons. To overcome this problem, Pope
Gregory, introduced in 1582 the calendar known as Gregorian calendar, which we
are using now.
In this calendar, a leap year is defined as before, with the exception that when a year
ends in two zeros, it will be a leap year only if it is divisible by 400. Thus in a cycle of
400 years, there are 100 leap years according to Julian Calendar while there are only
97 in Gregorian Calendar, since years 100, 200 & 300 are not leap years. This makes
the average civil year to consist of 365.2425 mean solar days, which is very close to
the true length of the tropical year. No serious discrepancy can arise in the Gregorian
calendar for many centuries.
With this we come to the end of Unit 2. In this unit, we have studied various aspects
of positional astronomy. We now summarise the main points of the unit.

2.5 SUMMARY
• In positional astronomy, we are concerned with the relative directions of the
heavenly bodies, which appear to be distributed on the surface of a vast sphere
called the celestial sphere.
• Any plane passing through the centre of the sphere intersects it in a circle called
the great circle. The extremities of the diameter of the sphere perpendicular to the
plane of the great circle are called poles of the great circle. Arcs of three great
circles form a spherical triangle in which any angle / side is less than 180°. The
sum of the three angles is greater than 180°.
• Due to rotation of the Earth on its axis, the heavenly bodies appear to move across
the sky from east to west in small circles parallel to the equator. 53
Basics of Astronomy • In the horizon coordinate system, the altitude and azimuth specify a star’s
position. The altitude of the north celestial pole at any place is equal to the latitude
of the place.
• In the local equatorial system, the coordinates of a heavenly body are given by the
hour angle and the declination. In the universal equatorial system, the
declination and the right ascension specify the position of a celestial object.
• Circumpolar stars are those, which never set, i.e., they are always above the
horizon. At a given place of latitude φ, a star becomes circumpolar if its
declination δ ≥ (90 − φ). It follows that at geographical North Pole (where
φ = 90°) the Sun is circumpolar for 6 months from March 21 to September 23.
Thus, at poles the day will last for 6 months and night for 6 months during the
year. At equator (φ = 0) there is no circumpolar star and all the stars move in
circles perpendicular to the horizon. The day and night are of 12-hour duration
each.
• Hour angle of vernal equinox is defined as the sidereal time. The sidereal day is a
measure of the rotation period of the Earth with respect to the distant stars.
• The hour angle of the mean Sun is defined as mean solar time and hour angle of
the Sun is defined as apparent solar time.
• The difference between the apparent solar time and the mean solar time is known
as the equation of time. The difference between the two systems of time varies
between −14 1/4 to +16 1/4 minutes and becomes zero 4 times during the year.

2.6 TERMINAL QUESTIONS Spend 30 min.

1. What is the latitude of the place at which celestial horizon and equator coincide?
At what latitude is the celestial equator perpendicular to the horizon?
2. Show that a star attains its maximum altitude when it is on the observer’s
meridian.
3. A and B are two places in north latitude on the surface of the Earth; their latitudes
are 24° 18′N and 36° 47′N, respectively; and their longitudes are 133° 36′E and
125° 24′W, respectively. Find the difference in their local time.
4. Show that at a place of latitude φ (North) at upper transit, the zenith distance of a
star equals φ − δ .

2.7 SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs)
1. a) Difference in the longitudes of Ahmedabad and Shillong
= 91° 52′ − 72° 44′ = 19° 08′
b) Difference in latitude between Srinagar and Kanyakumari
= 34° 05′ − 8° 05′ = 26° 00′
2. Turn the telescope 30° degrees from north to west, keeping it horizontal. Then
raise it so that it makes an angle of 45° with the horizontal.
3. a) See Fig. 2.25. δ of celestial north pole = 90°.
δ of celestial South Pole = − 90°.
δ of zenith = φ, latitude of the place.
54
 Basic Concepts of
Z Positional Astronomy
40°°
20°° N X

Celestial P
R equator

20°° S X' 50°°

S O horizon N

Fig.2.25

b) See Fig. 2.26. (α,δ) of vernal equinox = (0,0). (α,δ) of ecliptic north pole
= (18h, 66½°) since ϒJ is measured eastward. (α,δ) of ecliptic South Pole
= (6h, − 66½°).

K
P

R
Celestial Sun
equator
Earth
M' M
Ecliptic

ϒ
Vernal equinox T

K'

Fig.2.26 55
Basics of Astronomy 4. a) At Delhi, zenith makes an angle of 28° 46′ with the equator. The Sun on
June 22 and December 22 makes an angle of 23° 27′ with the equator.
Therefore, zenith distance of the Sun on that day = 28° 46′ − 23° 27′ = 5° 19′
b) φ ≥ 90 − δ = 84° 42′

5. The difference in longitudes of Mumbai and Kolkata is around 15 ½°. Thus, the
time difference is 62 minutes. Since Kolkata is to the east, the local time there
will be 2 minutes past 9 p.m. when it is 8 p.m. in Mumbai.

Terminal Questions
1. φ = 90°, φ = 0°.
2. See Fig. 2.27. At the transit point M on the diurnal circle, the altitude is maximum.

Z
M
R
Celestial P
equator

W
S O N

Q T

Z'

Fig.2.27

3. Difference in longitude = 133° 36′ E − 125° 24′ W = 259°.

∴ Time at A will be ahead by 259° × 4 min = 1036 min = 17 h 16 min.

4. See Fig. 2.28. The star is at U. In general, z = φ − δ.

φ
z
δ Z
U

Horizon

S O N

Celestial
equator

Z'

Fig.2.28
56
 Astronomical Techniques
UNIT 3 ASTRONOMICAL TECHNIQUES
Structure
3.1 Introduction
Objectives
3.2 Basic Optical Definitions for Astronomy
Magnification
Light Gathering Power
Resolving Power and Diffraction Limit
Atmospheric Windows
3.3 Optical Telescopes
Types of Reflecting Telescopes
Telescope Mountings
Space Telescopes
3.4 Detectors and Their Use with Telescopes
Types of Detectors
Detection Limits with Telescopes
3.5 Summary
3.6 Terminal Questions
3.7 Solutions and Answers

3.1 INTRODUCTION
In Unit 1 you have learnt about various measurable astronomical quantities and the
methods of measuring some of them such as distance, size and mass. You have also
studied about the radiant flux or brightness of different astronomical sources like the
planets, the sun, stars and galaxies. A great deal of information in astronomy is
gathered with the help of ground-based instruments. Therefore, in this unit we
describe some instruments as well as the techniques used to measure radiant flux and
to analyse the radiation emitted by astronomical objects. We shall first revisit the
basic optical definitions, viz. magnification, light gathering power and resolving
power, relevant to astronomy. You may recall having studied them in Unit 11 of the
Physics elective PHE-09 entitled Optics.
In Sections 3.3 and 3.4, we shall describe instruments like optical telescopes and
detectors that help us measure radiant flux from stars and other objects. A telescope is
like a camera. It is a light-focusing instrument, while a detector, like a photographic
film, is the medium on which the photons leave an impression, which can be
measured later.
You know the poem ‘Twinkle-twinkle little star’ from your early days. In this unit,
you will study how the Earth’s atmosphere, that makes the stars twinkle, affects the
observations made with telescopes on the ground. Finally, you will learn about
techniques to detect faint astronomical light, which originates from very distant
objects.
In the next unit we discuss the basic principles of physics applicable in astronomy and
astrophysics.
Objectives
After studying this unit, you should be able to:
• define angular magnification, light gathering power and resolving power of a
telescope;
• describe different types of reflecting telescopes and telescope mountings; 57
Basics of Astronomy • explain how atmosphere affects the incoming starlight;
• describe the different detectors and techniques used for observations; and

• estimate detectability limit of a telescope.

3.2 BASIC OPTICAL DEFINITIONS FOR ASTRONOMY

The telescope is the most common instrument used in ground-based astronomy. Most
people think of a telescope only in terms of its magnifying power. However,
magnification is only one of three powers of a telescope, the other two being its
resolving power and light gathering power. Before we describe various telescopes,
we would like to discuss these concepts as applicable in astronomy.
3.2.1 Magnification
As you know from the course in Optics, the magnifying power of a telescope refers to
its ability to make the image appear bigger. If the angle subtended by a distant object
at the objective of a telescope is β and that subtended by the virtual image at the eye is
α (Fig. 3.1), then the angular magnification (A.M.) of the telescope is given by

α
A.M. = (3.1)
β

fo fe

α
Fo Fe
β
Fe

To ‘top’ point of
virtual image at infinity

Fig.3.1: The image of a distant object is formed at the focal plane of the objective. The rays
entering the eye are rendered parallel by the eye-piece. The eye sees the image at infinity

We calculate the A.M. of a telescope by dividing the objective’s focal length by the
focal length of the eyepiece.
f
A.M. = o (3.2)
fe

For example, if a telescope has an objective with a focal length of 60 cm and an eye-
piece of focal length 0.5 cm, its angular magnification is 60/0.5, or 120 times. We say
that the magnification is 120 X.
Further, the A.M. is also equal to the ratio of entrance pupil (e.g., the objective lens
diameter) to the human eye’s pupil diameter. If we assume the diameter of a normal
human eye pupil to be about 8 mm for the eye adapted to perfect darkness, then the
lowest possible magnification will be equal to the ratio of objective diameter (in mm)
to 8. Similarly, if we assume a normal pupil diameter of 3 mm, the highest possible
58 magnification will be the diameter of the objective divided by 3.
The highest possible magnification of a telescope is limited by its optics which Astronomical Techniques
includes the quality of lenses and mirrors and the thermal insulation of the telescope
tube so that the exchange of heat does not disturb the air inside the tube. The
magnification is also limited by the disturbance of light rays suffered in the Earth’s
atmosphere. Also, the higher the magnification, the smaller is the field of view, i.e., the
area of the sky which can be observed by the telescope becomes smaller (see Fig. 3.2).

(a) (b) (c)

Fig.3.2: Increasing the magnification decreases the field of view and makes the image dimmer.
Pictures show simulated views of the Orion Nebula at magnifications of approximately 50x,
80x, and 120x

Notice that the images in Figs. 3.2 b and c are dimmer compared with the one in Fig. 3.2a.
Is there a way to make it brighter? We can do so by gathering more light from the object.
This brings us to the concept of light gathering power of a telescope.

3.2.2 Light Gathering Power

The light gathering power of a telescope refers to its ability to collect light from an
object. Most interesting celestial objects are faint sources of light and in order to get
an image we need to capture as much light as possible from them. This is somewhat
similar to catching rainwater in a bucket, the bigger the bucket, the more rainwater it
catches.

Similarly, the light gathering power of a telescope is proportional to the area (i.e.,
diameter squared) of the telescope objective. Now, the area of a circular lens or mirror
of diameter D is π (D/2)2. Thus, the ratio of light gathering powers of two telescopes
is given by
2
LGP1 D 
=  1  (3.3)
LGP2  D2 

For example, a telescope of 100mm diameter can gather (100/8)2 = 156.25 times more
light than a human eye with a typical pupil diameter of 8 mm.

Similarly, if we compare telescopes of 24 cm and 4 cm diameters, the former gathers

(24/4)2 = 36 times more light than the latter. Thus, even a small increase in telescopes
diameter produces a large increase in its light gathering power and allows astronomers
to study much fainter objects.

The ability of a telescope to reveal fine detail of an object is determined by its

resolving power, which we now discuss.
59
Basics of Astronomy 3.2.3 Resolving Power and Diffraction Limit
You have learnt in Sec. 11.2 of Unit 11 of Optics that the image of a point object
through an optical instrument is not a sharp point-like image but a bright circular disc
called the Airy disc. The disc is surrounded by a number of alternate bright and dark
fringes produced due to diffraction. The central bright disc represents the image of the
object.
Now suppose we wish to observe two close stars that appear equally bright. We
should be able to see two Airy discs. However, whether we see them as distinct discs
or overlapping each other, depends on the resolving power of the telescope. Fig. 3.3
shows three situations where two equally bright stars are closely placed. The pattern
of rings seen at each star image is called the Airy pattern. The two stars are said to be
just resolved when we can just infer their images as two distinct Airy discs (Fig. 3.3b).
To resolve these Airy discs, we use the Rayleigh criterion.

Rayleigh criterion
Two equally bright stars are said to be resolved when the central maximum
of one diffraction pattern coincides with the first minimum of the other.

(a) (b) (c)

Fig.3.3: Resolving a pair of two equally bright stars. In a) the stars are easily resolved; in b) the
stars are JUST resolved, and in c) the stars are too close and not resolved.

You may like to ask: What factors determine a telescope’s resolving power?
Naturally, the quality of lenses is a major factor. But even with perfect optics, the
resolving power of a telescope is limited by diffraction.
The diffraction limit of resolution (R) of a telescope is defined as
R (in radians) = (1.22 λ /D), (3.4)
where λ is the wavelength of light and D is the telescope diameter. Both λ and D have
to be expressed in the same units. [Note that 1 degree = 60 arc-minutes (60′) = 3600
arc-seconds (3600″)]. R can be expressed in arc-seconds as

1.22λ
R (arc-sec) = (206265) (3.5)
D
For human eye

60 R ~ 1′ for absolute sharp vision,

 R ~ 2′ for clear vision and Astronomical Techniques

R ~ 4′ for comfortable vision.

Remember that the smaller the value of R, the better is the instrument’s ability to
resolve nearby objects.

Example 1: Diffraction limit of resolution of telescopes of different sizes for a

given wavelength
Table 3.1 shows the diffraction limit of resolution of telescopes of different sizes for
λ = 457 nm.
Table 3.1: Diffraction limit of resolution of various telescopes

Telescope Diameter (mm) R (″″)

50 2.3

100 1.15

200 0.58

400 0.29

500 0.23

You may like to calculate R for a telescope.

SAQ 1 Spend
10 min.
Calculate the diffraction limit of resolution of Mount Palomar telescope of 200 inch
diameter for λ = 457 nm. Compare its light gathering power with a telescope of
200 mm diameter.

Based on size alone, the largest telescopes should have large resolving powers. But
the resolution of large telescopes is limited by the passage of light through the Earth’s
atmosphere. When we look through a telescope, we are looking through several
kilometres of turbulent air, which blurs the image. The Earth’s atmosphere does not
allow ground-based telescopes to resolve better than 1-2 arc-seconds in the sky (for
even the best astronomical sites).

The major limitation for ground-based astronomy is the Earth’s atmosphere and it can
affect the observations in many ways. Moreover, all wavelengths cannot pass through
the atmosphere. This brings us to the concept of atmospheric windows.

3.2.4 Atmospheric Windows

Some of the effects of the Earth’s atmosphere on electromagnetic radiation are:

Absorption, scintillation, scattering and turbulence. Atmospheric molecules such as
carbon dioxide and water vapour give rise to absorption. Thus, only certain bands of
frequencies in the electromagnetic spectrum pass through the atmosphere. These
regions of the electromagnetic spectrum are called atmospheric windows. Fig. 3.4
shows the absorption properties of the Earth’s atmosphere. On the x-axis is the
wavelength in cm and on the y-axis is the altitude in km at which the intensity of the
radiation entering the atmosphere is reduced to half.
61
Basics of Astronomy Opaque
Visual Radio

Earth’s atmosphere
window window

Transparency of
Altitude
Infrared
and
Gamma rays, X-rays microwaves
and UV rays are are blocked
blocked by the by the
atmosphere. atmosphere. Transparent
Gamma X ray UV Micro-
Infrared wave
ray UHF VHF FM AM
−12 −10 −8 −4 −2 2 4
10 10 10 10 10 1 10 10
Visual
Wavelength (metres)
7
4 × 10− m 7
7 × 10− m

Fig.3.4: The spectrum of electromagnetic waves from gamma rays to long radio waves showing the
optical and radio ‘windows’ and the regions of atmospheric transparency

The atmosphere allows only visible radiation and radio waves to come through to the
surface of the Earth. For observations at other wavelengths we have to fly detecting
instruments to altitudes at which these wavelengths are not completely absorbed.
Before the advent of artificial satellites, the instruments were flown in balloons and
rockets. Now-a-days observations are carried at various wavelengths by instruments
on board the space satellites. It is important to remember that astronomers like to
observe objects in as many wavelengths as possible. This helps them to understand
astronomical objects better.
Spend SAQ 2
5 min.
For the same diameter, compare the resolving power of an optical telescope operating
at λ 457 nm and a radio telescope operating at λ 1 cm.

You have learnt in SAQ 2 that the resolving power of a radio telescope is quite poor.
Therefore, radio telescopes need to have very large apertures.
So far we have discussed some optical definitions relevant to astronomy. We now turn
our attention to the instruments and techniques used for making astronomical
observations from the Earth. Let us now learn about optical telescopes.

3.3 OPTICAL TELESCOPES

Telescopes have undergone tremendous evolution from Galileo’s times when only
refracting telescopes were used. The most recent telescopes have mirrors that can be
actively shaped according to the observer’s need. The refracting telescopes became
outdated very soon because
i) it was difficult to make good large lenses required for large gathering area,
ii) large lenses were very heavy and balancing the telescope became difficult, and
iii) lenses suffered from optical aberrations.
Reflecting telescopes with mirrors (parabolic or hyperbolic) have been around for
more than a century. In the last few decades, advanced technologies have been
developed which allow the observer to effectively control the optical system so that it
can be adapted to the needs of the observations.
Such active and adaptive optics have completely changed this basic tool of
observational astronomy. We now describe various types of reflecting telescopes.
62
3.3.1 Types of Reflecting Telescopes Astronomical Techniques

James Gregory of Scotland (1638-75) proposed the first design for a reflecting
telescope in which he used a paraboloidal mirror as the objective (primary mirror) to
minimize chromatic and spherical aberrations (Fig. 3.5).
In this type of telescope, light from a distant object hits the primary mirror and is
reflected to a secondary mirror which reflects it down to the telescope tube again to a
secondary focus. The light emerges from the telescope through a small central hole in
the primary mirror and is observed with an eyepiece or a detector.

Focal
point
Parallel light rays
from the star

Secondary Paraboloidal
mirror primary mirror

Fig.3.5: Gregorian reflecting telescope with equatorial mounting; ray diagram for the telescope

The performance of the Gregorian telescope was not satisfactory and Newton (1642-
1727) built a working reflector.
A Newtonian telescope (Fig. 3.6) uses a parabolic primary but a flat mirror as a
secondary, which is set at 45° to the axis of the tube. The light is brought to a focus at
the side of the telescope where the eyepiece is placed. This system is widely used
even now.

Secondary
mirror

Primary
Eye-piece mirror

Fig.3.6: Newtonian reflecting telescope

A little later, a French optician, G. Cassegrain (who lived at the time of Newton),
designed another reflector (Fig. 3.7).
Now-a-days the most commonly used telescope is the Cassegrain type with a central
hole in the primary mirror which allows the light to come out for placing an eyepiece
or any other general detector or instruments. This design also allows the folding of the
focussed beam and thus provides a very compact telescope.

63
Basics of Astronomy

Fig.3.7: Cassegrain reflecting telescope

All telescopes need to be pointed at the desired part of the sky, and then they have to
follow or track the objects in the sky as their direction changes due to Earth’s
rotation. For example, suppose you are viewing Jupiter and its moons with a
telescope. Due to the Earth’s rotation, you will see Jupiter moving across your view
until it is gone. You will have to move your telescope to follow Jupiter if you want to
keep viewing it. However, to track it, you will need a mounting for the telescope so
that it can be turned in the desired direction. We will now briefly describe telescope
mountings.

3.3.2 Telescope Mountings

Most of the small telescopes (less than 1 metre diameter) use the equatorial mount
(Fig. 3.8). In an equatorial mounting, the pier or the base on which the telescope is
mounted is set so that its axis points to the North Pole. This is done by raising the axis
by an angle equal to the latitude of the place. This axis is called the polar axis. A
rotation about the polar axis is used for adjustment in right ascension. The telescope is
also provided with motion about an axis perpendicular to the polar axis, called the
declination axis, for adjustment in declination. Thus, a combination of these two
motions allows the telescope to point to any object whose equatorial coordinates are
known.

Fig.3.8: The equatorial mounting

Since the telescope is fixed on the Earth, it moves with the Earth. In order that the
object remains in the field of the telescope, the mounting is made to rotate in the
direction opposite to that of the Earth with the same speed as that of the Earth. All
64
these adjustments make this type of mounting heavier than the altitude-azimuth Astronomical Techniques
mounting described ahead. That is why it is not used with very big telescopes.
All modern telescopes (larger that 2 m diameter) use Alt-Azimuth type mount shown
in Fig. 3.9. You have learnt in the last unit that altitudes and azimuths of objects
change with the location of the observer and with time for the same observer. This
was a major limitation of this type of mounting in earlier times but with advances in
the technology using computers in the past few decades, it is no more an issue.

AZIMUTH AXIS

ALTITUDE AXIS

Fig.3.9: The Alt-Azimuth type mount

In Fig. 3.10a we show the largest telescope in India set up by the Indian Institute of
Astrophysics at Kavalur Observatory in Karnataka.

(a) (b)

Fig.3.10: a) The Vainu Bappu telescope in Kavalur Observatory; b) domes housing twin Keck
telescopes

At present, the largest operating telescopes are the twin Keck 10 m telescopes at
Mauna Kea, in Hawaii, placed at about 4200 m height above sea-level (Fig. 3.10b).
There are plans now for making very large aperture (up to 20 − 30 m) telescopes! 65
Basics of Astronomy 3.3.3 Space Telescopes
You have learnt in Sec. 3.2 that the main advantages of a telescope over direct
observation with the human eye, are magnification, light collection and resolution.
For light collection, one could fabricate telescopes of increasingly large diameters.
But these will be limited by the effect of the Earth’s atmosphere. However, if we
could put a telescope in space (high above the Earth’s atmosphere, in a balloon or a
satellite), then the Earth’s atmosphere would not be a limiting factor, and we could
achieve diffraction-limited images. In such situations the size of the telescopes that
can be built for such platforms is the only limitation due to the costs involved and
limitations of available technology.
The Hubble Telescope of about 2 m diameter is the best example of this type of
telescope (Fig. 3.11). Over the past decade it has made very high quality observations
of stars, nebulae, galaxies, supernovae and other objects. Some of these objects belong
to a very early phase of the universe. These observations have led to improved
understanding of these objects.

Fig.3.11: The Hubble Space Telescope

Both the types of telescope − ground-based and space-based − have somewhat

different roles to play and complement each other. That is why there is always a
demand to build large diameter ground-based telescopes.
The active and adaptive telescopes are the latest instruments in use today. In the
active telescopes, the shape of the primary mirror can be changed (although it retains
its parabolic form) according to the observer’s need. This is done with the help of
small peizo driven pistons placed at the back of the mirror support. This is now
routinely done for large mirror based telescopes. In the adaptive telescope the
changes in the shape of the primary mirror are done in a controlled manner depending
upon the changes in the Earth’s atmosphere during night. In some sense, the adaptive
telescopes can beat the effect of the atmosphere and render images as good as the
space telescopes for some duration in the night!
Telescopes must use detectors to help us obtain useful information about the universe.
We now discuss various detectors and how these are used with telescopes.

3.4 DETECTORS AND THEIR USE WITH TELESCOPES

Detectors are used for measuring the light output from a telescope and play a major
role in obtaining information about the stars, galaxies, etc. The actual light available
66
from an astronomical object is very small as will be clear from the following example.
 Astronomical Techniques
Example 2: Estimate of Available Light from a Star
Vega is one of the bright stars in the sky and is about 26 light years away. The
radiation emitted in visible band at the star’s surface is 175,000 Wcm−2. This reduces
by 10−16 times on its arrival outside Earth’s atmosphere. Another 20% is lost in the
atmosphere due to absorption and scattering. About 30% is lost in the telescope optics.
Thus, if we use a 25 cm (10″) telescope to collect this light, then only about
0.5 × 10−9 W is available at its Cassegrain focus. Eventually only a fraction of this is
actually detected since the detector is never an ideal one and has an efficiency of at
best 80%. It is amazing that even with such a small amount of light detected, we can
study many things about a star, e.g., its light variation, brightness, composition. We
can even take a spectrum (which actually divides this small available light into smaller
bits of wavelengths)!

Detectors are used with telescopes in the following two modes of operation:
• Imaging: This involves taking direct pictures of star fields and extended objects
like gas clouds or galaxies. Since sharp images are required over a wide field
which may extend up to several square degrees, careful optical design is a natural
requirement.
• Photometry: This involves measuring total brightness, spectrum etc. of single
objects. Compared to imaging mode, poorer images are acceptable in this case but
the stellar image has still to be small enough to enter an aperture or slit of a
spectrograph.
We now briefly describe various types of detectors.
3.4.1 Types of Detectors
Detectors used in the imaging mode are mainly 2-dimensional (2D type) since we are
trying to form images of objects in a given area. Examples of such detectors are the
photographic emulsion, human eye and the most modern detector, the charge-
coupled device (CCD). Detectors used for photometry of single objects are 1D type
(one dimensional), since they receive photons from one object only. The photometer
is a 1D detector.
Photometer
Before the advent of CCDs, the measurements of light intensity and colour were made
using a photometer, a highly sensitive light meter attached to a telescope (Fig. 3.12a).
A photometer is still used in the photometry of single stars. It is used more commonly
for stars whose light output varies with time, called variable stars. The most
important component of a photometer is a photomultiplier tube that is based on the
photoelectric effect about which you have studied at +2 level. A photon when
incident on a photocathode emits an electron. The electric current thus generated is
amplified further and can be measured directly. The calibration of intensity or colour
is done by observing a comparison star. Today, however, most photometric
measurements are made on CCD images.
Charge-coupled device
A charge-coupled device (CCD) is a special computer chip of the size of a postage
stamp (Fig. 3.12b). It contains a large number (~ millions) of microscopic light
detectors arranged in an array. A CCD can be used like a small photographic plate,
though it is much more sensitive. CCDs detect both bright and faint objects in a single
exposure. The image from a CCD is stored in a digitised form in a computer.
Therefore, brightness and colour can be measured to high precision. Moreover, it is 67
Basics of Astronomy easy to manipulate the image to bring out details. At present, the only major drawback
of CCD is that its maximum size is limited (about 70 mm square) as compared to the
most basic 2D detector, i.e., photographic plates which can be as large as
300 mm square. This disadvantage of CCD is also being overcome by combining a
large number of CCDs.

(a) (b)

Fig.3.12: a) Photometer attached to a telescope; b) a CCD

Efficiency of a Detector
A basic parameter which defines the efficiency of any detector is its Quantum
Efficiency (Q.E.). It is the ratio of number of photons actually detected (or recorded)
by it to the number of photons recorded by an ideal and perfect detector. Since ideal
detector by definition would detect all photons incident on it with 100% efficiency,
this ratio is nothing but the ratio of actually detected photons by the detector versus
the number of photons incident on it. Fig. 3.13 gives the efficiency curves for various
detectors.

100
QUANTUM EFFICIENCY (PERCENT)

CCD

10
PHOTOMULTIPLIER TUBE

PHOTOGRAPH
1

EYE

0.1
300 500 700 900 1100
λ (nm)

Fig.3.13: Quantum efficiency of various detectors in terms of wavelength of light

68
Note that the human eye and photographic emulsion are detectors with the lowest Astronomical Techniques
sensitivity and photomultiplier tubes are only marginally better. The CCD works over
a large wavelength region in the visible band with a Q.E. of the order of 60-80%. You
must remember that the y-axis in Fig. 3.13 is in log scale. In spectroscopy or
spectrophotometry, we analyze
As you have read in Unit 1, the limiting magnitude of the naked eye is about +6. This the light in great detail by
means that the faintest object visible under normal observing conditions, with an eye using a spectrograph which
spreads the light into a
adapted to darkness has a magnitude of +6. It takes 25-30 minutes for the eye to get spectrum according to
completely dark-adapted, otherwise we can see only up to fourth or fifth magnitude. wavelength. You are familiar
with a prism dispersing white
Experiments have, however, shown that a dark-adapted eye, looking at a patch of dark light into its component
sky through a modest telescope can see stars as faint as mv = +8.5. This corresponds to colours. Nearly all modern
a flux of about 200 photons per second. spectrographs use a grating
instead of a prism and the
We would also like to find out how far the limits of detection are extended when we spectrum is recorded directly
use a telescope with various detectors. on a CCD camera. Since we
understand how light is
emitted, scattered, or absorbed
3.4.2 Detection Limits with Telescopes by matter, a spectrum carries a
lot of information about the
Modern detectors like CCDs can detect individual photons, and the limiting source of light and the medium
magnitude is normally governed by the background noise. through which it has passed.
We shall discuss this in detail
For a typical photographic plate, only about 0.1% of the incident photons are recorded in Unit 7.
depending on the type of grains on the film. On such a plate, an image is detectable
only after about 5 × 104 photons have been received.
At 200 photons per second, the photographic plate can match eye’s limit with an
exposure of about 4 minutes.
Obviously, the plate can detect much fainter objects with longer duration exposures.
The limit to the flux of visible radiation which is detectable is given by

vis 1
F ∝ (3.6)
lim D 2t

where t is the exposure time and D is the telescope aperture.

For a telescope, the magnitude limit in the visible range, in a very approximate
manner without considering any efficiency factors etc., is given by

mvlim ~ 2 + 5 log10 D (3.7)

where D is in mm.
Thus using the human eye as the detector along with a telescope we can have the
limiting magnitudes as mv ~ 12.9 for a 6 inch and mv ~ 15.3 for an 18 inch telescope.

SAQ 3 Spend
5 min.
Find the magnitude of the faintest object that the 3.5 metre New Technology
Telescope at the European Southern Observatory in Chile can detect.

With this we come to the end of the unit. We now summarise its contents.

3.5 SUMMARY
• If a telescope has an objective of local length f0 and the eye-piece has a focal
length fe, then angular magnification is equal to the ratio f0/fe.

69
Basics of Astronomy • The light gathering power of a telescope refers to its ability to collect light from
an object.

• The resolving power of a telescope is its ability to reveal fine detail. The
resolving power of a telescope is given by

11.6
α=
D

where D is the diameter of the telescope’s objective. A point light source observed
through a telescope would not appear as a point source. Diffraction would cause
the image to appear as a round disk of light, called Airy’s disk. The diffraction
limit of resolution in radians is given by

λ
R = 1.22
D

where λ is the wavelength of light. Both λ and D must have the same units.

• At the Earth’s surface, electromagnetic radiation can be detected only in the

radio, infrared and optical windows.

• Telescopes can be of refracting or reflecting types.

• All modern telescopes are reflecting type. There are three kinds of reflectors:
Gregorian, Newtonian and Cassegrain. The Cassegrain reflectors are the most
popular.

• Telescopes can be operated in imaging or photometry mode.

• The larger is the diameter of the objective, the fainter is the source that a telescope
can detect.

• The human eye, photographic emulsion, photometer and charge-coupled devices

are various types of detectors.

• Of these the CCDs are used most by modern astronomers. These are used for
recording images, measuring brightness and colour of celestial objects.

3.6 TERMINAL QUESTIONS Spend 30 min.

1. Why can infrared observations be made from high mountains while X-ray
observations can be made only from space?

2. Nocturnal animals have large pupils in their eyes. Can you relate that to
astronomical telescopes?

3. The moon has no atmosphere. If we had an observatory on the moon, at what

wavelengths could we observe the astronomical objects?

4. What is the resolving power of a 20 cm telescope if observations are made at λ

550nm?

5. What do two stars 1.5 arc-seconds apart, look like through a 25 cm telescope ?

6. Compare the light gathering powers of the 5 m telescope and a 0.5 m telescope.

70
 Astronomical Techniques
3.7 SOLUTIONS AND ANSWERS
Self Assessment Questions (SAQs)

1.22λ
1. R= × 206265 arc - sec
D

1.22.457 × 2.06265
= × 10 − 2 arc - sec
2 × 2.54
= .023 arc - sec

2
LGPMP  200 × 2.54 × 10 
=   = (2.54)2 × 10 2 = 645.2
LGP200mm  200 

1.22λ1 D 1
2. Ratio of resolving powers = × 206265 × ×
D 1.22λ 2 206265

λ 4.57 ×10 − 9 × 10 2
= 1 =
λ2 1

= 4.57 × 10 − 7

The resolving power of the radio telescope is 4.57 × 10 − 7 times the resolving
power of an optical telescope operating at 4.57 nm.
3. Limiting magnitude

mν = 2 + 5 log10 D

= 2 + 5 log10 (3.5 × 100 × 10 )

= 2 + 15 + 5 log10 (3.5)

= 19.7
Terminal Questions
1. At the top of a high mountain, we can receive only IR. X-rays are available only
much above the top of the highest mountains. Therefore, to observe in X-ray band,
we will need to observe from a space-based telescope.
2. The larger the diameter of the pupil, larger is the light gathering power. Nocturnal
animals are adapted to night vision by virtue of this. Astronomical telescopes also
require bigger and bigger objectives to see fainter and fainter objects.
3. At all wavelengths, provided we have detectors sensitive in those regions.
4. 0.58 arc-seconds.
5. Resolved.
6. The 5 m telescope has 100 times the light gathering power of a 0.5 m telescope.

71
Basics of Astronomy
UNIT 4 PHYSICAL PRINCIPLES
Structure
4.1 Introduction
Objectives
4.2 Gravitation in Astrophysics
Virial Theorem
Newton versus Einstein
4.3 Systems in Thermodynamic Equilibrium
4.4 Theory of Radiative Transfer
Radiation Field
Radiative Transfer Equation
Optical Depth; Solution of Radiative Transfer Equation
Local Thermodynamic Equilibrium
4.5 Summary
4.6 Terminal Questions
4.7 Solutions and Answers

4.1 INTRODUCTION
So far in this block we have provided the basic information which is useful in
astronomy. You have learnt about astronomical quantities of interest, various
coordinate systems, astronomical instruments and techniques. We now turn our
attention to astrophysics. The aim of astrophysics is to apply principles of physics to
understand and explain the behaviour of various astronomical systems.
There are certain physical principles and concepts which are used in astrophysics so
universally that it is worthwhile to discuss them before we begin studying specific
astronomical systems. We have decided to put together some such basic physical
principles in this Unit. You must have already learnt many of these principles in your
other physics courses. Now you will learn how these can be applied to astrophysical
systems.
Let us consider some issues which are of universal concern in astrophysics. We know
that gravitation is the dominant force in virtually any astrophysical setting. Since
gravitation is always attractive, it must be balanced in some way in a system which is
not shrinking. We shall discuss a very general and powerful principle called the virial
theorem, which helps us understand how gravitation is balanced in astrophysical
systems.
Another topic of importance in astrophysics is the interaction of radiation with
matter. Astrophysics is a very special science in which we cannot do experiments
with our systems (stars, galaxies, etc.) in our laboratories. Virtually everything we
know about these systems is learnt by analysing the radiation reaching us from these
systems. If we want to make inferences about the systems which emitted the radiation
or through which the radiation passed, then we need to understand how matter and
radiation interact with each other. Many astronomical systems like stars emit radiation
simply because they are hot. So we often need to apply various principles of thermal
physics to understand how matter in these systems behaves. Therefore, we plan to
recapitulate some of the important results of systems in thermodynamic equilibrium
and then develop the theory of transfer of radiation through matter.
Objectives
After studying this unit, you should be able to:
72 • apply virial theorem to simple astrophysical systems;
• identify the situations in astrophysics to which Newton’s theory of gravitation or Physical Principles
general theory of relativity can be applied;
• determine the specific intensity, energy density, radiant flux and radiation,
pressure for a given radiation field; and
• solve the radiative transfer equation for simple cases and interpret the results.
Study Guide
In this unit, we will be using certain concepts discussed in various units of the physics
electives PHE-01 entitled ‘Elementary Mechanics’ (Unit 10), PHE-06 entitled
‘Thermodynamics and Statistical Mechanics (Unit 9), and PHE-11 entitled ‘Modern
Physics’ (Unit 9). Please keep these units handy for ready reference.

4.2 GRAVITATION IN ASTROPHYSICS

In your elementary physics courses, you must have learnt about two important long-
range forces, whose range of influence extends to infinity. These are the gravitational
force and the electromagnetic force. The electromagnetic force can be attractive or
repulsive depending on the nature of charges. So, if a system has equal amounts of
positive and negative charges, and if there are no relative motions between these two
types of charges, then electromagnetic forces are screened off. That is, the system
does not produce a large-scale electromagnetic field. On the other hand, gravitation is
always attractive and cannot be screened off. So it is the dominant force acting over
the entire universe.
Since gravitation is always attractive, a natural question to ask is: Why do the celestial
objects not shrink in size? Obviously, it has to be balanced in a system which is not
shrinking in size (such as a star or a galaxy). We shall now use Newtonian theory of
gravitation to discuss an important theorem (virial theorem) which tells us how this
balancing takes place. Then we shall briefly consider whether Newtonian theory of
gravitation is adequate in astrophysics or whether we have to apply a more complete
theory of gravitation − the general theory of relativity due to Einstein.

4.2.1 Virial Theorem

Why is our solar system not shrinking in size? We shall analyse this problem in the
non-inertial frame of reference attached to a planet. From Unit 10 of the course
PHE-01 entitled ‘Elementary Mechanics’, you know the answer: the Sun’s
gravitational attraction on a planet is balanced by the centrifugal force due to the
orbital motion of the planet. Let M and m be the masses of the Sun and a planet in the
solar system. For simplicity, if we assume the planet to go in a circular orbit of radius
r with speed v, then the force balance equation in the given non-inertial frame of
reference is

mv 2 GMm
=
r r2
We can rewrite this equation in the form

1  GMm
2 mv 2  − =0 (4.1)
2  r

Now note that the gravitational potential energy of the system (the Sun and the planet)
is
GMm
EG = − ,
r
73
Basics of Astronomy whereas the kinetic energy of the system is

1 2
EK = mv
2
if the Sun is assumed to be at rest. We can now rewrite Eq. (4.1) in the following form

2 E K + EG = 0 (4.2)

This is the virial theorem for a planet going around the Sun.
What is the significance of this result? This tells us that the gravitational potential
energy and the kinetic energy of a system will have to be of the same order if
gravitation is to be balanced by motion. We have proved virial theorem for the simple
case of a planet going around the Sun in a circular orbit. However, Eq. (4.2) can be
proved quite generally for a system in which gravitation is balanced by motions such
that the system is not shrinking in size. The motions needs not be circular, but can be
of any type. For example, inside a star, gravity is balanced by thermal motions of its
particles (atoms, electrons, ions). Even in this situation, Eq. (4.2) can be shown to
hold provided we take the total kinetic energy of all the particles in the star for EK. If a
galaxy or a star cluster is not shrinking in size, the total kinetic energy EK of the stars
in it should be related to the total gravitational potential energy EG by Eq. (4.2).

Fig.4.1: Spiral galaxies

In a type of galaxy known as spiral galaxy, stars seem to be moving in nearly circular
orbits. However, in typical star clusters and in galaxies known as elliptical galaxies,
stars move randomly. Due to these random motions, the stars do not fall to the centre
due to gravitation. Do you feel puzzled by the idea that random motions can balance
gravitation? To understand this concept, consider the air around you. Earth’s gravity is
pulling all the molecules of the air. Then why are not all molecules settling on the
floor of the room due to this attraction? It is the random motion of the molecules
which prevents this from happening.
Deriving the virial theorem Eq. (4.2) for a completely general situation is a very
mathematically involved problem. It is beyond the scope of this elementary course. So
let us apply the virial theorem in a simple situation so that you feel comfortable with
it.

Example 1: Estimating the average temperature in stellar interior

A star (we have the Sun in mind) has a mass of 1033g and a radius of 1011cm. Make an
Fig.4.2: The Sun order-of-magnitude estimate of the average temperature in the interior of the star.
Solution
Since this is an order of magnitude estimate, we shall take the virial theorem to imply
74 that the potential and the kinetic energies are approximately equal. In c.g.s. units, we
take the following approximate values of various physical constants: gravitational Physical Principles
−7 −16 −24
constant G ≈ 10 , Boltzmann constant kB ≈ 10 , mass of hydrogen atom mH ≈ 10 .
The gravitational potential at the surface of the star has the magnitude GM/R. The
approximate gravitational potential energy of the star is given by multiplying this by
M, which is

( )2 ≈ 1048 erg
GM 2 10− 7 × 1033
=
R 1011

Now we need to equal the total kinetic energy to this. Now, suppose that the star is
made up of hydrogen. Then, the star consists of about M/mH particles. Each of them
has kinetic energy of order kBT. Therefore, the total kinetic energy of the star is

M 1033 × 10 −16 T
k BT ≈ − 24
≈ 1041T
mH 10

If this is equated to 1048 erg, then we obtain the temperature of the star as

T ≈ 107 K

Note that detailed calculations suggest a temperature of about 15 million degrees at

the centre of the Sun. The above order of magnitude estimate thus provides a fairly
good estimate of the average temperature inside a star.

You may now like to solve a problem to fix these ideas.

SAQ 1 Spend
10 min.
A globular cluster of stars has about a million stars. The stars inside such a cluster of
radius 1020 cm have random velocities of order 106 cm s−1. Estimate the mass of the
33
star cluster and the number of stars in it. Take the mass of a star to be about 10 g.
Hint: The mass of the cluster is Nm, where N is the number of stars in the cluster.
Now, equate the total K.E. of the cluster to its P.E.

After Newton formulated his theory of universal gravitation, for more than two
centuries it was regarded as a supreme example of a successful physical theory.
However, in 1915, Einstein showed that this theory was incomplete and formulated
his new theory of gravitation known as general relativity. In this section, we
investigate the situations in which Newton’s theory may have to be replaced by the
general theory of relativity.

4.2.2 Newton versus Einstein

We now know that Newton’s theory is only an approximation. But it is such an
exceptionally good approximation in most circumstances that we do not need general
relativity at all. Only when the gravitational field is sufficiently strong, we have to
apply general relativity. Although we shall not discuss general relativity in this
elementary course, we would like to point out when you can safely use Newtonian
theory and when general relativity is needed.
Even people without any technical knowledge of general relativity now-a-days have
heard of black holes. These are objects with gravitational fields so strong that even
light cannot escape. Let us try to find out when this happens. Newtonian theory does
not tell us how to calculate the effect of gravitation on light. So let us figure out when
a particle moving with speed of light c will get trapped, according to Newtonian
theory. Suppose we have a spherical mass M of radius r and a particle of mass m is 75
Basics of Astronomy ejected from its surface with speed c. The gravitational potential energy of the particle
is

GMm
−
r
If we use the non-relativistic expression for kinetic energy for a crude estimate (we
should actually use special relativity for a particle moving with c!), then the total
energy of the particle is

1 2 GMm
E= mc −
2 r
Newtonian theory tells us that the particle will escape from the gravitational field if E
is positive and will get trapped if E is negative. In other words, the condition of
trapping is

1 2 GMm
mc − <0
2 r
or
2GM
>1 (4.3)
c 2r

It turns out that more accurate calculations using general relativity give exactly the
same condition (4.3) for light trapping, which we have obtained here by crude
assumptions.

General relativity is needed when the factor

2GM
f = (4.4)
c2r

is of the order unity.

Newtonian theory is quite adequate if f is much smaller than 1.

Let us investigate the case of the Sun.

Example 2
Let us examine if the Newtonian theory is adequate for the Sun.
Solution
The Sun has mass 1.99 × 1033 g and radius 6.96 × 1010 cm. Substituting these values in
Eq. (4.4), we get

f = 4.24 × 10−6 << 1

Hence, Newtonian theory is quite adequate for all phenomena in the solar system.
Only if we want to calculate very accurate orbits of planets close to the Sun (such as
Mercury), we have to take into consideration general relativity.

When is general relativity applicable in the case of the Sun? We can use Eq. (4.3) to
calculate the radius to which the solar mass has to be shrunk such that f is of order
unity. Then the light emitted at its surface gets trapped. Why don’t you do this
76 calculation yourself?
SAQ 2 Spend Physical Principles
2 min.
Calculate r for the Sun such that f ~ 1.

You would have calculated the radius to be 3 km. Therefore, general relativity will
apply to the Sun, once it shrinks to this size. As we shall discuss in more detail in
Block 3, when the energy source of a star is exhausted, the star can collapse to very
compact configurations like neutron stars or black holes. General relativity is needed
to study such objects.
If matter is distributed uniformly with density ρ inside radius r, then we can write

4 3
M= πr ρ
3
and Eq. (4.4) becomes

8π Gr 2ρ
f = (4.5)
3 c2

We note that f is large when ρ is large or r is large (for given ρ). The density ρ is very
high inside objects like neutron stars. You may ask: Can there be situations where
general relativity is important due to large r? We know of one object with very large
size − our universe itself. The distance to farthest galaxies is of the order 1028 cm. It is
very difficult to estimate the average density of the universe accurately. Probably it is
of the order 10−29g cm−3. You may like to substitute these values in Eq. (4.5), and
calculate f.

SAQ 3 Spend
2 min.
Calculate the value of f from Eq. (4.5) using the data given above for the Universe.

The result of SAQ 3 tells us that we should use general relativity to study the
dynamics of the whole universe. This subject is called cosmology.
Thus, in astrophysics, we have two clear situations in which general relativity is
very important:
• the study of collapsed stars and
• the study of the whole universe (or cosmology).
In most other circumstances, we can get good results by applying Newtonian theory of
gravitation.
As we mentioned in the introduction, basic principles of thermal physics apply to
many astronomical systems with high temperatures. Let us briefly revisit these
principles.

4.3 SYSTEMS IN THERMODYNAMIC EQUILIBRIUM

You have learnt about thermodynamic equilibrium in the course PHE-06 entitled
Thermodynamics and Statistical Mechanics. If a system is in thermodynamic
equilibrium, then several important principles of physics can be applied to that
system. Let us first recapitulate some important laws and equations, relevant for such
systems, namely, Maxwellian velocity distribution, Boltzmann distribution law,
Saha’s equation and Planck’s law of black body radiation. Afterwards, we shall
discuss whether we can assume astrophysical systems to be in thermodynamic
equilibrium and whether these principles can be applied to them.
77
Basics of Astronomy Maxwellian velocity distribution
You have studied about Maxwellian velocity distribution in Unit 9 of PHE-06.
Different particles in a gas move around with different velocities. Recall that if the gas
is in thermodynamic equilibrium at temperature T, the number of particles per unit
volume having speeds between v and v + dv is given by

 m 
3/ 2  mv 2 
dN v = 4πN   v exp −
2  dv (4.6)
 2πk BT   2 k BT 
 

where N is the number of particles per unit volume and m is the mass of each particle.
Boltzmann and Saha Equations
You have studied in Unit 9 of the physics elective PHE-11 entitled Modern Physics
that a hydrogen atom has several different energy levels. It is also possible to break
the hydrogen atom into a proton and an electron. This process of removing an electron
from the atom is called ionisation. If a gas of hydrogen atoms is kept in
thermodynamic equilibrium, then we shall find that a certain fraction of the atoms will
occupy a particular energy state and also a certain fraction will be ionised. The same
considerations hold for other gases.

If N0 is the number density of atoms in the ground state, then the number density Ne of
atoms in an excited state with energy E above the ground state is given by

Ne  E 
∝ exp  −  (4.7)
N0  k BT 

This is the Boltzmann distribution law.

In 1919, the famous Indian physicist M.N. Saha derived an equation which tells us
what fraction of a gas will be ionised at a certain temperature T and pressure p. The
derivation of this equation involves some statistical mechanics. Here we merely quote
the result without derivation. If nI is the number of hydrogen atoms out of which nII
are ionised at temperature T, then Saha’s equation gives
3/2
n II  2 πm e   I 
p e =   (k B T )5/2 exp −  (4.8)
nI  h2   k BT 

where I is the ionisation potential of hydrogen, pe is the partial pressure of electrons,

h is Planck’s constant and me is the mass of electron. A form convenient for
calculation is

log (n II p e /n I ) = 2.5logT − (5040/T ) I − 0.48

Spend SAQ 4
5 min.
Assuming pe to be 100 dyne/cm2, calculate the fraction of hydrogen atoms ionised at
T = 10,000 K. The ionisation potential of hydrogen is 13.6 eV.

Planck’s law of blackbody radiation

You have studied this law and its consequences in Unit 15 of PHE-06. You know that
when radiation is in thermodynamic equilibrium with matter, it is called blackbody
radiation. The spectral distribution of energy in blackbody radiation is given by the
famous law derived by Planck in 1900 (see Fig. 4.3). The energy density uνdν lying in
the frequency range between ν and ν + dν is given by

78
 Physical Principles
8πh ν 3dν
u ν dν = (4.9)
c3  hν 
exp  − 1
 k BT 

1.0
T3
Energy intensity (arbitrary units)

T3 > T2 > T1

0.5
T2

T1

0
0 2,000 4,000 6,000 8,000 10,000 12,000 14,000 16,000 18,000 20,000

wavelength (angstrom)

Fig.4.3: Blackbody radiation curve

We can now use these results to understand the interaction of matter with radiation.

4.4 THEORY OF RADIATIVE TRANSFER

Matter can both emit and absorb radiation. It is possible to use quantum mechanics to
calculate the rates at which atoms emit or absorb energy. Here, however, we shall not
do it. We would study the processes of emission and absorption by matter, by
introducing suitable coefficients of emission and absorption. Radiative transfer is the
name of the subject in which we study the interaction of radiation with matter having
prescribed emission and absorption coefficients.
Let us first consider how we can provide the mathematical description of radiation at a
given point in space.

4.4.1 Radiation Field

You know that it is particularly easy to give a mathematical description of blackbody
radiation, which is homogeneous and isotropic inside a container. Specifying the
energy density uν associated with the frequency ν, which is given by Planck’s law
(Eq. 4.9), more or less provides us complete information about blackbody radiation. In
general, however, the radiation is not isotropic. When we have sunlight streaming into
a room, we obviously have a non-isotropic situation involving the flow of radiation
from a preferred direction.

We now define the radiation field for a non-isotropic situation. Let us consider a small
area dA at a point in space (Fig. 4.4). Let dEν dν be the energy of radiation passing
through this area in time dt from the solid angle dΩ centred at θ and lying in the
frequency range ν, ν + dν. The energy dEνdν is proportional to the area dA cosθ
projected perpendicular to the direction of radiation, time interval dt, solid angle dΩ
and frequency range dν. Hence we can write

dE ν dν = I ν ( r , t , nˆ )cosθ dA dt dΩ dν (4.10) 79
Basics of Astronomy where n̂ is the unit vector indicating the direction from which the radiation is coming
and N̂ is the unit vector normal to the area dA. The quantity I ν (r , t , n̂ ) is called the
specific intensity. As you can see, it is a function of position r, time t and direction
n̂ .

n̂

dΩ

dA

Fig.4.4: Illustration of specific intensity

Radiation field
If I ν (r , t , n̂ ) is specified for all directions at every point of a region at a
time, then we say that the radiation field in that region is completely
specified.

In this elementary treatment, we shall restrict ourselves only to radiation fields which
are independent of time.
If we know the radiation field at a point in space we can calculate various quantities
like radiant flux, energy density and pressure of radiation. For example, radiant
flux is simply the total energy of radiation coming from all directions at a point per
unit area per unit time. Hence, we simply have to divide Eq. (4.10) by dA dt and
integrate over all solid angles to get the flux. Thus, we can define the radiant flux
associated with frequency ν, and the total radiant flux as follows:

Radiant flux in terms of specific intensity

The radiant flux associated with frequency ν is given by

∫
Fν = I ν cosθ dΩ (4.11)

The total radiant flux is obtained by integrating over all frequencies

∫
F = Fν dν (4.12)

The pressure of the radiation field over a surface is given by the momentum
exchanged per unit area per unit time, or momentum flux, perpendicular to that
surface. Let us obtain an expression for momentum flux.
You know from Unit 3 of PHE-11 that the momentum associated with a photon of
energy dEν is dEν /c.

Its component normal to the surface dA is dEν cosθ/c. On dividing this by dA dt, we
get the momentum flux associated with dEν ;
80
 dE ν cos θ Physical Principles
Momentum flux =
c dA dt
Using Eq. (4.10), we get the expression for momentum flux in terms of specific
intensity:
dEν cos θ 1 I
= ν cos 2θ dΩ (4.13)
c dAdt c
The radiation pressure pν is obtained by integrating the momentum flux over all
directions.

Radiation pressure

1
pν =
c ∫
I ν cos 2θ dΩ. (4.14)

If the radiation field is isotropic, i.e., it is independent of θ and φ, then

dΩ = sinθ dθ dϕ and dΩ = ∫ 4 π . Hence, we get,

Iν 4π I ν
c ∫
2
pν = cos θdΩ = (4.15)
3 c
We now apply these results to calculate energy density and radiation pressure.

SAQ 5 Spend
5 min.
Perform the integration in Eq. (4.15) and verify the result.

Example 3: Calculating energy density, specific intensity and radiation pressure

Calculate the energy density uν of a radiation field at a point and use that expression
of energy density to write down the specific intensity of a blackbody radiation. Show
that the pressure due to isotropic radiation is given by 1/3 of the energy density.

Solution

Let us consider energy dEν of radiation associated with frequency ν as given by

Eq. (4.10). This energy passes through area dA in time dt in the direction n̂ . Since the
radiation traverses a distance cdt in time dt, we expect this radiation dEν to fill up a
cylinder with base dA and length cdt in the direction n̂ during this time. Now the
volume of this cylinder is cosθ dAcdt. Therefore, from Eq. (4.10), the energy density
of this radiation in the solid angle dΩ is
dE ν I
= ν dΩ
cosθ dAcdt c
To get the total energy density of radiation at a point associated with frequency ν, we
have to integrate over all directions, so that
Iν
uν = ∫ c
dΩ

4π I ν
For isotropic radiation u ν =
c
81
Basics of Astronomy Since blackbody radiation is isotropic, the specific intensity of blackbody radiation
usually denoted by Bν (T) should be independent of direction. Hence the energy
density of blackbody radiation is simply given by

Bν (T )
uν = ∫ c
dΩ

Since Bv(T) is independent of direction, integration over all solid angles gives 4π.
4πBν (T )
∴ uν =
c
Therefore, making use of Eq. (4.9), we now get the specific intensity of blackbody
radiation as

2 hν 3 1
Bν (T ) =
c 2  hν 
exp  − 1
 k BT 

Using Eq. (4.15), we get the radiation pressure for isotropic radiation as
1
pν = u ν
3
4π Bν (T )
For black body radiation pν = .
3 c

In astrophysics, we need to understand the interaction of matter and radiation to

explain spectra of objects such as stars, interstellar gas clouds and galaxies. We now
discuss the effect of matter on radiation field.

4.4.2 Radiative Transfer Equation

If matter is present, then in general the specific intensity of the radiation field keeps
changing as we move along a ray path. Before we consider the effect of matter, first
let us find out what happens to specific intensity in empty space as we move along a
ray path.
See Fig. 4.5. Let dA1 and dA2 be two area elements separated by a distance R and
perpendicular to a ray path. Let Iν1 and Iν2 be the specific intensities of radiation in the
direction of the ray path at dA1 and dA2.

dA1 dA2

Fig.4.5: Two area elements perpendicular to a ray path

We want to determine the amount of radiation passing through both dA1 and dA2 in
82 time dt in the frequency range ν, ν + dν. If dΩ2 is the solid angle subtended by dA1 at
dA2, then according to Eq. (4.10), the radiation falling on dA2 in time dt after passing Physical Principles
through dA1 is

I ν2dA2 dtdΩ 2 dν

From considerations of symmetry, this should also be equal to

I ν1dA1dtdΩ1dν

where dΩ1 is the solid angle subtended by dA2 at dA1. Equating these two expressions
and noting that

dA2 dA1
dΩ1 = , d Ω 2 =
R2 R2

we get

Iν1 = Iν2 (4.16)

In other words, in empty space the specific intensity along a ray path does not change.
If s is the distance measured along the ray path, then we can write

dI ν
=0 (4.17)
ds
in empty space.
At first sight, this may appear like a surprising result. We know that the intensity falls
off as we move further and further away from a source of radiation. Can the specific
intensity remain constant? The mystery is cleared when we keep in mind that the
specific intensity due to a source is essentially its intensity divided by the solid angle
it subtends, a quantity called the surface brightness of an object. This means that the
specific intensity is a measure of the surface brightness. As we move further away
from a source of radiation, both its intensity and angular size falls as (distance)2.
Hence the surface brightness, which is the ratio of these two, does not change.
Let us now consider what happens if matter is present along the ray path. If matter
emits, we expect that it will add to the specific intensity. This can be taken care of by
adding an emission coefficient jν on the right hand side of Eq. (4.17). On the other
hand, absorption by matter would lead to a diminution of specific intensity and the
diminution rate must be proportional to the specific intensity itself. In other words, the
stronger the beam, the more energy there is for absorption. Hence the absorption term
on the right hand side of Eq. (4.17) should be negative and proportional to Iν. Thus,
we obtain the radiative transfer equation which gives the value of specific intensity in
the presence of matter:

Radiative transfer equation

dI ν
= jν − α ν I ν (4.18)
ds
where αν is the absorption coefficient

The radiative transfer equation provides the basis for our understanding of interaction
between radiation and matter.
It is fairly trivial to solve this equation if either the emission coefficient or the
absorption coefficient is zero. Let us consider the case of jν = 0, i.e., matter is assumed
to absorb only but not to emit. Then Eq. (4.18) becomes
83
Basics of Astronomy dI ν
= −α ν I ν (4.19)
ds
On integrating this equation over the ray path from s0 to s, we get
 s 
 s0
∫
I ν ( s ) = I ν ( s0 ) exp − α ν (s′) ds′

(4.20)

We will discuss below more general solutions of the radiative transfer equation. These
solutions will provide us answers to questions such as: Why is the radiation emitted
from nebula usually in spectral lines? Why do we see absorption lines in stellar
spectra?

4.4.3 Optical depth; Solution of Radiative Transfer Equation

To obtain a general solution of the radiative transfer equation, we need to define two
quantities, namely, the optical depth and the source function. Let us first define the
optical depth τν through the following relation:

dτ ν = α ν ds (4.21)

such that the optical depth along the ray path between s0 and s becomes
s
τν = ∫s α ν (s′)ds′
0
(4.22)

If matter does not emit radiation, i.e., jν = 0, it follows from Eqs. (4.20) to (4.22) that
the specific intensity along the ray path falls as

I ν ( τ ν ) = I ν ( 0) e − τ ν (4.23)

Based on the values of τν we can define objects as optically thick or optically thin.

Optically thick and optically thin objects

If the optical depth τν >> 1 along a ray path through an object, then the
object is known as optically thick.
An object is known as optically thin if τν << 1 for a ray path through it.

It follows from Eq. (4.23) that for an optically thick object Iν (τν) = 0 and it
extinguishes the light of a source behind it. What about an optically thin object? It
does not decrease the light much. Hence the terms optically thick and optically thin
roughly mean opaque and transparent at the frequency of electromagnetic radiation
we are considering.

We now define the source function

j
Sν = ν (4.24)
αν

Dividing the radiative transfer equation Eq. (4.18) by αν and using Eqs. (4.21) and
(4.24), we get

dI ν
= −I ν + Sν (4.25)
dτ ν
84
 Physical Principles
Multiplying this equation by e τ ν , we can write it as:

d
dτ ν
( )
I νe τ ν = S νe τ ν

Integrating this equation from optical path 0 to τν (i.e., from s0 to s along the ray
path), we get the general solution of the radiative transfer equation:

General solution of the radiative transfer equation

τν
− (τ ν − τ′ν )
I ν (τ ν ) = I ν (0) e − τ ν + ∫0 e S ν (τ′ν ) dτ′ν (4.26)

If matter through which the radiation is passing has constant properties, then we can
take Sν to be constant and solve the integral in Eq. (4.26). This gives

I ν (τ ν ) = I ν (0) e − τ ν + S ν (1 − e − τ ν )

We are now interested in studying the emission and absorption properties of an object
itself without a source behind it. Then we take Iν(0) = 0 and write

I ν (τ ν ) = S ν (1 − e − τ ν ) (4.27)

Let us consider the cases of optically thin and thick objects.

• Optically thin object

For an optically thin object, τν << 1, and e − τ ν may be approximated to 1 − τν.

Thus Eq. (4.27) becomes

I ν (τ ν ) = S ν τ ν

For matter with constant properties, we take τν = ανL, where L is the total length
of the ray path. Making use of Eq. (4.24), we get the following result

Iν = jνL (4.28)
• Optically thick object

If the object is optically thick, we can neglect e − τν compared to 1 in Eq. (4.27).

Then we get the result

Iν = Sν (4.29)

Let us put these results together for ready reference.

Specific Intensity of

Optically thin object: Iν = jνL

Optically thick object: Iν = Sν

We have derived two tremendously important results in Eqs. (4.28) and (4.29). To
understand their physical significance, we have to look at some thermodynamic
considerations.

85
Basics of Astronomy 4.4.4 Local Thermodynamic Equilibrium
Suppose we have a box kept in thermodynamic equilibrium. If we make a small hole
on its side, we know that the radiation coming out of the hole will be blackbody
radiation. We have already derived the specific intensity of blackbody radiation as

2 hν 3 1
Bν (T ) = (4.30)
c 2  hν 
exp  − 1
 k BT 

The specific intensity of radiation coming out of the hole is simply

I ν = Bν (T ) (4.31)

We now keep an optically thick object behind the hole inside the box as shown in
Fig. 4.6. If this object is in thermodynamic equilibrium with the surroundings, then it
will not disturb the environment and the radiation coming out of the hole will still be
blackbody radiation, with specific intensity given by Eq. (4.31). On the other hand, we
have seen in Eq. (4.29) that the radiation coming out of an optically thick object has
the specific intensity equal to the source function. From Eq. (4.29) and (4.31), we
conclude

S ν = Bν (T ) (4.32)

when matter is in thermodynamic equilibrium.

Bν (T)

Fig.4.6: Blackbody radiation coming out of a box with an optically thick obstacle placed behind
the hole inside the box

On using Eq. (4.24), we finally obtain the famous result known as Kirchoff’s law.

Kirchoff’s law

jν = α ν Bν (T ) (4.33)

Let us now stop and try to understand what we have derived. Very often matter tends
to emit and absorb more at specific frequencies corresponding to spectral lines. Hence
both jν and αν are expected to have peaks at spectral lines. But, according to
Eq. (4.33), the ratio of these coefficients should be the smooth blackbody function
Bν(T).

We now look at the results of Eq. (4.28) and (4.29).

Eq. (4.28) tells us that the radiation from an optically thin source is essentially
determined by its emission coefficient. Since the emission coefficient is expected to
have peaks at spectral lines, we find that the emission from an optically thin system
86 like a hot transparent gas is mainly in spectral lines.
On the other hand, from Eq. (4.29), the specific intensity of radiation from an Physical Principles
optically thick source is its source function. This has been shown to be equal to the
blackbody function Bν(T) in Eq. (4.31). Hence, we expect an optically thick object
like a hot piece of iron to emit roughly like a blackbody.
The nature of radiation from an astrophysical source crucially depends on whether the
source is optically thin or optically thick. Emission from a tenuous nebula is usually in
spectral lines. On the other hand, a star emits almost like a blackbody.
Why is the radiation from a star not exactly blackbody radiation? Why do we see
absorption lines? Recall that we have derived Eq. (4.29) by assuming the source to
have constant properties. This is certainly not true for a star. As we go down from the
star’s surface, temperature keeps increasing. Hence Eq. (4.29) should be only
approximately true. It is the temperature gradient near the star’s surface which gives
rise to the absorption lines.
By assuming thermodynamic equilibrium, we have derived the tremendously
important result Eq. (4.32) that the source function should be equal to the blackbody
function Bν(T). In a realistic situation, we rarely have strict thermodynamic
equilibrium. The temperature inside a star is not constant, but varies with its radius. In
such a situation, will Eq. (4.32) hold?
We have already mentioned in section 4.3 that the Maxwellian velocity distribution,
the Boltzmann law and the Saha equation hold if the system is in thermodynamic
equilibrium. This generally means that the temperature does not vary much over it.
For Planck’s law also to hold, the radiation has to interact with matter efficiently.

We note from the radiative transfer equation Eq. (4.18) that αν has the dimension of
inverse length. Its inverse α ν−1 gives the distance over which a significant part of a
beam of radiation would get absorbed by matter. Often this distance α ν−1 is referred to
as the mean free path of photons, since this is the typical distance a photon is
expected to traverse freely before interacting with an atom.

The smaller the value of α ν−1 , the more efficient is the interaction between matter and
radiation. If α ν−1 is sufficiently small such that the temperature can be taken as
constant over such distances, then we expect Planck’s law of blackbody radiation to
hold. In other words, if both α ν−1 and the mean free path of particles are small
compared to the length over which the temperature varies appreciably, then all the
important laws of thermodynamic equilibrium hold.

Such a situation is known as Local Thermodynamic Equilibrium, which is abbreviated

as LTE. Inside a star, we expect LTE to be a very good approximation and we can
assume Eq. (4.32) to hold when we solve radiative transfer equation inside the star. In
the outer atmosphere of a star, LTE may fail and it often becomes necessary to
consider departures from LTE when studying the transfer of radiation there.
In this unit, we have discussed some basic principles of physics applicable to
astrophysical systems. We now summarise the contents of this unit.

4.5 SUMMARY
• The virial theorem, states that the gravitational potential energy EG and kinetic
energy EK of a system are of the same order
2 E K + EG = 0

87
Basics of Astronomy • Newton’s theory of gravitation is generally adequate. However, if the quantity
2GM
f = becomes of order unity in a system, then the general theory of
c2r
relativity has to be used instead of Newton’s theory.

• For astrophysical systems in thermodynamic equilibrium, the following results

hold:
Maxwellian velocity distribution
 m 
3/2  mv 2 
dN ν = 4πN   v 2exp −  dv
 2πk BT   2 k BT 
 

Boltzmann’s law

Ne  E 
= exp  − 
N0  k BT 

Saha’s equation
3/2
nII  2πm   I 
pe =  2 e  (k BT )5/2 exp− 
nI  h   k BT 

Planck’s law of black body radiation

8πh ν 3 dν
u ν dν =
c3  hν 
exp  − 1
 k BT 

• The energy of radiation lying in the frequency range ν and ν + dν passing

through an area dA cosθ in time dt from the solid angle dΩ centred at θ is given
by

dE ν dν = I ν ( r , t , nˆ )cosθ dA dt dΩ dν

where Iν is called the specific intensity.

• The radiant flux of a time independent radiation field is defined in terms of
specific intensity Iν as

∫
Fν = I ν cosθdΩ

• The radiation pressure is given by

1
pν =
c ∫
I ν cos 2θ dΩ

• The interaction of matter with radiation is given by the radiative transfer

equation

dI ν
= jν − α ν I ν
ds
The general solution of this equation is given as
tν
∫0 e − (τ ν − τ ν )S ν (τ′ν ) dτ′ν
′
I ν (τ ν ) = I ν (0) e − τ ν +
88
• The optical depth of an object along a ray path between points s0 and s is given Physical Principles
by

s
τν = ∫s α ν (s′)ds′
0

• Kirchoff’s law for a system in thermodynamic equilibrium is given as

jν = α ν Bν (T )

4.6 TERMINAL QUESTIONS Spend 30 min.

1. Suppose the Sun contracts at a uniform rate to half its present size in 107 yr.
Suppose that all its energy is radiated from its surface. Calculate the luminosity of
the Sun during the contracting phase.

2. Determine the size to which the Earth must shrink so that the use of Einstein’s
theory of gravitation becomes necessary.

3. Calculate the optical depth at which the specific intensity reduces to one-
hundredth of its original value in a system in which no emission of radiation is
taking place. Perform your calculation at a given frequency. Would the system be
optically thick or thin?

4.7 SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs)

1. We need to equate the total kinetic energy of the globular star cluster with its
gravitational potential energy. If the cluster has N stars with masses of individual
stars of order m, and velocities of order v, the total kinetic of the star cluster is
about

Nmv2.

The total gravitational potential energy is

G ( Nm) 2
,
R

On equating these two, we have

GNm
v2 ≈ ,
R

from which

Nm ≈ ≈
( )2
v 2 R 106 × 1020
≈ 1039 g
G −7
10

This is the mass of the cluster. Taking the masses of stars to be of order 1033g
(~ mass of the Sun), the cluster has about million stars in it.

89
Basics of Astronomy 2GM
2. f = =1
c 2r

2GM 2 × 6.673 ×10 −11 ×1.99 × 1030

⇒ r= = m
c2 (3 ×10 )8 2

2 × 6.673 × 1.99
= ×10 3 m
9
= 2951 m

8π Gr 2 ρ 8π 6.673 × 10 −8 × 10 28 × 10 28 × 10 −29
3. f = = .
3 c2 3 3 × 1010 × 3 ×1010

8π × 6.673
= ×10 −1
27
~ 0.6, which is of the order 1.

n 5040
4. log pe II = 2.5 log T − . I − 0.48
nI T

n 5040
log pe + log II = 2.5 log T − . I − 0.48
nI T

n 5040
log II = − 2 + 2.5 log T − . I − 0.48
nI T

( )
= − 2.48 + 2.5 log 10 4 −
5040
T
× 13.6

= 0.67

I
5.
c ∫
pν = ν cos 2θ . 2π . sin θ dθ

I − cos 3 θ π 4π I ν
= 2π ν . =
c 3 0 3 c

Terminal Questions

GM 2
1. Gravitational potential energy of a star of radius r =
r

2GM 2
For the Sun of radius RΘ , gravitational P. E. =
2 RΘ

GM 2
Present gravitational potential energy of the Sun =
RΘ

2GM 2 GM 2 GM 2
∴ Energy radiated from its surface = − =
RΘ RΘ RΘ

(Remember that when the star contacts, gravitational energy is released)

90
 Physical Principles
GM 2
∴ L= (T is the time of contraction)
RΘT

GM 2
=
RΘ .107 × 365 × 24 × 3600

=
(
6.673 × 10 −8 × 2 × 1033 )2
7 × 1010 × 107 × 3.65 × 10 2 × 2.4 × 10 × 3.6 × 103

6.673 × 4 6.673 × 4 ×1035 1

= 1035 = .
7 × 3.65 × 2.4 × 3.6 7 × 3.65 × 2.4 × 3.6 4 × 1033

(in terms of present solar luminosity)

6.673
= ×10 2
7 × 3.65 × 2.4 × 3.6
= 3.02 (in terms of present solar luminosity)

2GM Earth
2. f ≈1=
c 2r

2GM Earth
∴ r=
c2

2 × 6.673 × 10 −11 × 6 × 10 24
= m (SI Units )
(3 × 108 )2
2 × 6.673 × 6
= ×10 −3 m
9
= 8.9 × 10−3 m

Iν
3. Eq. (4.23) ⇒ = e− τ ν
I ν (0 )

1
= e− τ ν ⇒ τ ν = log e ( 100 ) = 4.61
100

Since τ ν >>1 , the system is optically thick.

91
 The Sun
UNIT 5 THE SUN
Structure
5.1 Introduction
Objectives
5.2 Solar Parameters
5.3 Solar Photosphere
5.4 Solar Atmosphere
Chromosphere
Corona
5.5 Solar Activity
5.6 Basics of Solar Magnetohydrodynamics
5.7 Helioseismology
5.8 Summary
5.9 Terminal Questions
5.10 Solutions and Answers

5.1 INTRODUCTION
From Unit 4, you know that the principles of different branches of physics such as
mechanics, thermodynamics and quantum mechanics are used in astronomy and
astrophysics. In the present and subsequent Units, you will use these principles to
investigate the behaviour and properties of the universe and its constituents.

On the cosmic scale, the Sun is just another star; there are bigger and brighter stars in
the universe. The Sun is, however, very important to us because i) it is the nearest
and the only star in our planetary system and ii) it provides almost all of our
energy. Do you know that a slight variation in the energy received from the Sun can
threaten life on the earth! Further, the Sun being the nearest star, we can study its
structure, atmosphere, and other physical characteristics in greater detail. The
information/data so obtained can be used to test the theories of stellar structure and
evolution. In this way we can improve our theories and have a better understanding of
other stars. In the present Unit, you will study about the Sun.

Due to the efforts of astronomers, today we have detailed information regarding the
Sun. In Sec. 5.2, you will learn to arrive at the estimates of the basic solar parameters
such as mass, radius and effective surface temperature. As far as we are concerned, all
the visible radiation from the Sun comes from its surface layer called the photosphere.
Above the photosphere is the atmosphere of the Sun consisting of two distinct layers
namely the chromosphere and the corona. In Sec. 5.3, you will study the characteristic
features of these layers. Interaction of the Sun’s magnetic field with highly mobile
charged particles in it gives rise to a variety of observable events. These events,
collectively known as solar activity, have been discussed in Sec. 5.4. The theoretical
analysis of the interaction of magnetic field with conducting matter in motion is
known as magnetohydrodynamics and it provides a basis to understand solar activity
and related features of the Sun. Solar magnetohydrodynamics has been discussed in
Sec. 5.5. In Sec. 5.6, you will learn, in brief, about helioseismology which provides
valuable information about the Sun’s internal structure.

Objectives
After studying this unit, you should be able to:

• estimate values of the basic parameters of the Sun;

5
The Solar System • describe different layers of the Sun’s atmosphere;
and Stars
• describe some of the observed features such as sunspot, prominence and solar
flare associated with solar activity;
• explain the role of the Sun’s magnetic field in solar activity;
• derive the basic results of solar magnetohydrodynamics; and
• explain the seismology of the Sun.

5.2 SOLAR PARAMETERS

The basic parameters of the Sun are its mass (MΘ), radius (RΘ), luminosity (LΘ) and
effective surface temperature (Teff). In the following discussion, you will learn to
estimate these parameters.
11
Mass: You know that the mean distance of the Sun from the Earth is 1.5 × 10 m. It is
called the mean solar distance, a. In astronomy we measure mean distances in terms
of a and it defines the Astronomical Unit (1 AU = mean solar distance). To obtain an
expression for the mass of the Sun, we may use Kepler’s third law under the
assumption that the mass of the planet can be neglected in comparison with the mass
of the Sun. This assumption is valid for the Sun-Earth system and we can write:

4π 2 a 3
= GM Θ (5.1)
You may recall from Unit 6 of P2
the physics elective course
PHE-01 entitled Elementary where a, P, G and MΘ are the mean solar distance, orbital period (~ 365 days) of the
Mechanics that Kepler’s third Earth, gravitational constant and mass of the Sun, respectively. With the values of the
law is given by:
orbital period and the mean solar distance available at present, the value of GMΘ is
4π2 a 3 estimated to be 132712438 × 1012 m3s−2. Since the laboratory measurements for G
T2 =
GM gives a value equal to 6.672 × 10−11 m3kg−1s−2, we obtain:
where T, a, G and M are
MΘ ≈ 2 × 1030kg.
respectively the time period,
semi-major axis, gravitational This value is taken as the mass of the Sun as it exists today. In fact, solar mass
constant and mass of the Sun.
decreases continuously since the Sun continuously emits radiation and particles which
carry with them some mass. However, the total mass loss during the Sun’s estimated
life time (~ 1010 yrs) is found to be less than 1027kg. This value is much less than the
error in measurement of the solar mass and is, therefore, negligible. This method can
also be used to estimate the masses of the satellites/Moons of the planets in our solar
system. How about solving an SAQ of this nature?

Spend SAQ 1
5 min.
One of the four Galilean satellites of the planet Jupiter is Io. Its orbital period is
In astronomy, we encounter
typically small angles (of the
1.77 days. The semi-major axis of its orbit is 4.22 × 1010cm. Calculate the mass of
order of minute (′) and second Jupiter under the assumption that the Jupiter is too massive in comparison to Io.
(″)), for which simple
approximations are used for
Radius: The radius of the Sun can be estimated if we know the values of its angular
trigonometric functions. Thus,
for small values of the angle θ, diameter, θ and the mean solar distance, a (Fig. 5.1). The angular diameter of the Sun
is 32′ and mean solar distance is 1.5 × 1011m. Thus, with the help of Fig. 5.1, we
tanθ = θ = sinθ
obtain the value of the solar radius RΘ as:
and
cosθ = 1 RΘ =
1
2
[ (1.5 × 10
11
m) × ( 32 × 2.9 × 10 − 4 rad) ]
However, as you know, angles
must be measured in radian if
these formulae are to be valid. = 6.7 × 108m.
6
Astronomical observations indicate that the solar radius is not constant; rather, its The Sun
9
value changes slowly. Over a period of ~ 10 years, the average change is about
2.4 cm per yr. Further, radius of the Earth is 6.4 × 106m. Thus, the Sun’s radius is

1 AU
Sun

32´

Earth

Fig.5.1: When viewed from the Earth, the angular diameter of the Sun is approximately 32′′

almost 100 times larger than that of the Earth. To get an idea of the relative sizes of
the Sun and the Earth, refer to Fig. 5.2.
Sun

Moon’s orbit

Earth

Fig.5.2: The size of the Sun is so big that it can contain the Earth as well as the orbit of the Moon!

Luminosity: The solar luminosity, LΘ, is defined as the total energy radiated by the
Sun per unit time in the form of electromagnetic radiation. To estimate the value of
luminosity of the Sun, let us imagine a sphere with the Sun at its centre (Fig. 5.3). The
radius of this imaginary sphere is a, the mean distance between the Sun and the Earth.
Now, each unit area A of the sphere receives energy equal to S, called the solar
constant. Therefore, luminosity can be expressed as:

LΘ = 4πa2S (5.2)

a
1m
Earth
Sun 1m
A

1 AU

Fig.5.3: Imaginary sphere of radius a surrounding the Sun where a is its mean solar distance from
the Earth 7
The Solar System Since the solar radiation is absorbed in the Earth’s atmosphere, it is obvious that S
and Stars should be measured above the atmosphere. S has now been measured accurately using
satellites and its value is 1370 Wm−2. Substituting the value of S and the mean solar
11
distance, a = 1.5 × 10 m in Eq. (5.2), we get:
26
LΘ = 3.86 × 10 W.

Temperature: The temperature of the Sun at its surface and its interior regions are
different. The surface temperature can be estimated using Stephan-Boltzmann law
which you studied in our course on Thermodynamics and Statistical Mechanics
(PHE-06). We leave this as an exercise for you in the form of an SAQ.

Spend SAQ 2
5 min.
Assume that the Sun radiates like a black body at temperature T. Calculate T using
−8 −2 −4
Stephan-Boltzmann law. Take Stephan constant σ = 5.67 × 10 Wm K .

On solving SAQ 2, you would have found that the temperature of the Sun is 6000 K.
This estimated temperature is called the effective surface temperature because it is
the temperature of a black body whose surface emits the same flux as the Sun. This is
the temperature of the surface layer of the Sun called the photosphere from which all
the radiation is emitted.
To appreciate the validity of the approximation that the Sun radiates like a black body,
refer to Fig. 5.4. It shows the observed solar radiation in the ultraviolet, visible and
infrared regions of electromagnetic spectrum. Also plotted in this figure is the energy
curve of a black body at 6000 K. In view of the similarity of the two curves, it is fair
to assume that the Sun radiates like a black body at temperature 6000 K.

You learnt that the Sun

radiates like a black body at
approximately 6000 K. You
may think that if all the
energy is radiated outward
from the Sun, its surface
should gradually cool down.
This does not happen
because there is a constant
flow of energy from the
interior of the Sun towards
its surface.

Fig.5.4: Solar energy curve

The Sun is a hot, bright gaseous ball and it does not have a well defined surface like
the Earth. The visible surface of the Sun is called photosphere. Let us learn about it
now.

5.3 SOLAR PHOTOSPHERE

The photosphere (Fig. 5.5) is the visible surface of the Sun. All the light received from
the Sun, in fact, comes from the photosphere. You may ask: Why do not we receive
the radiation in the same form as generated in the interior of the Sun? At the
centre of the Sun, the energy is generated in the form of high energy photons called γ-
Fig.5.5: The Sun’s photosphere
rays. As these photons travel outwards, they collide with particles of matter and lose
energy continuously. By the time these photons reach the surface − the photosphere −
they are reduced to photons of visible region of electromagnetic spectrum. So, visible
8 radiation is emitted from the photosphere.
The density of photosphere is 3400 times less than the density of the air we breathe. The Sun
The thickness of the photosphere is about 500 km and the temperature at its base is
~ 6500 K. The temperature decreases upward and reaches a minimum value of
~ 4400 K at the top. This assumption is corroborated by the Sun’s absorption
spectrum which indicates that the light we receive must be passing through a cool gas
in which photons get absorbed.
The photosphere is not a quiet region (see Fig. 5.6). It shows a granular structure. If
you look at Fig. 5.6a carefully, you can see that the photosphere consists of bright and
irregularly shaped granules; each granule surrounded by dark edges. It has been found
that these granules are very hot and their typical size is ~ 1500 km. The hot gas in the
granules rises up with a speed of the order of 500 ms−1 and bursts apart by releasing
energy. The cool material subsequently sinks downward along the dark edges or lanes
between granules. The rising hot granules are seen only for a very short time (~ 10
minutes) before they dissolve.

Granule boundary

Granule
Granule

(a) (b)
Fig.5.6: a) Photograph of the photosphere showing granular structure; and b) schematic diagram
showing granules and their boundaries

The question is: What causes granulation of the photosphere? It is caused due to
convection (a mode of energy transport by matter, about which you will study in As they look inwards into the
solar atmosphere,
Unit 8). The granulation can be visualised (Fig. 5.6b) as the top layer of a region astronomers have discovered
where, due to convection, hot gas from below the photosphere moves upward. Thus, that, within a distance of
the centre of the granule is hotter and it emits more radiation and looks brighter in about 500 km, the solar
comparison to the edges which are relatively cooler and emit less radiation. atmosphere changes from
being optically thin to
Convection based explanation seems valid because the spectra of granules indicate optically thick. This distance
that their centres are much hotter than the edges. Further, the solar granulation is only ~ 0.07% of the Sun’s
provides observable evidence supporting the idea that there exists a convection zone radius. This region gives the
below the photosphere. impression that the Sun has a
sharp edge as viewed from
the earth.
You may now like to know: What is the chemical composition of the photosphere?
It consists of 79 percent hydrogen and the remaining 21 percent consists of nearly 60
other chemical elements. Interestingly, all the elements of the photosphere are known
elements and their proportion in the earth is more or less the same as that in the As you have learnt in Unit 4,
photosphere. This similarity in the chemical compositions of the photosphere and the optically thin/thick medium
refers to the medium
earth is of utmost importance for understanding the formation of the solar system. characterised by low/high
absorption of electromagnetic
Though the photographs of the Sun give the impression that it has a clear edge, such radiation.
clear and distinct edge does not exist. Outside the apparent edge are the Sun’s outer
layers, collectively known as the Sun’s atmosphere. These layers can be seen and
probed and valuable information about their physical characteristics can be obtained.
Let us learn about the various layers of the solar atmosphere. 9
The Solar System
and Stars 5.4 SOLAR ATMOSPHERE
The Sun’s atmosphere is divided into two layers namely, the chromosphere and the
corona. A schematic diagram of these layers is shown in Fig. 5.7.

Corona

2600 km
Transition zone
2300 km

Chromosphere

500 km

Photosphere

0 km

Sun’s interior

Fig.5.7: Schematic diagram showing the layers of solar atmosphere

5.4.1 Chromosphere
Chromosphere lies above the photosphere (Fig. 5.7) and extends up to ~ 2000 km.
This layer of the solar atmosphere is normally not visible from the Earth because of its
faintness. However, it can be seen during a solar eclipse. The name chromosphere is
derived from the fact that a few seconds before and after a total solar eclipse, a bright,
pink flash appears above the photosphere (Fig. 5.8). The spectrum obtained at that
Fig.5.8: Chromosphere just time is called a flash spectrum (Fig. 5.9).
before a total solar
eclipse

You may be aware that each

chemical element
emits/absorbs electromagnetic
radiations of characteristic
wavelengths. These radiations
of different wavelengths (also
called lines) constitute the
spectrum of the element.
Balmer lines, which fall in the
visible region, are the spectral
lines of hydrogen atom. The
emission/absorption of the
spectral lines is predominantly
dependent on the temperature Fig.5.9: Emission lines observed at the time of total solar eclipse
and density of the material.
You will study about the origin The appearance of pink colour is due to the emission of the first Balmer line (Hα)
of spectral lines in Unit 7. which occurs in the red region. The temperature, density and pressure in the
chromosphere determine the intensities of various emission lines. In the
10 chromosphere, the density decreases by a factor of ~ 104 from that of the photosphere
while the temperature rises to ~ 25000 K within a short distance of ~ 2000 km. The Sun
Therefore, the spectral lines that are not produced at relatively higher density and
lower temperature of the photosphere are formed in the chromosphere as emission
lines (see margin remarks).

At this point, it is logical to ask: Why does the temperature in the chromosphere
increase with height? The clue to the answer of this question lies in observing the
chromosphere just before the total solar eclipse. Hot gas, in the form of jets called
spicules, is observed throughout the chromosphere (Fig. 5.10). These spicules extend
upward in the chromosphere up to a height of ~ 10000 km and last for as long as 15
minutes. This implies that the lower part of the chromosphere is highly turbulent and
the spicules transport energy and matter from the photosphere to the chromosphere.
This causes heating of the chromosphere.

Fig.5.10: Spicules in the Sun’s chromosphere

Now, your next logical question could be: What causes spicules? The origin of
spicules is not yet understood completely. However, it appears to be caused by the
Sun’s magnetic field. Further, just above the chromosphere, there exists a transition
region extending up to ~ 3000 km. In this region, the temperature rises sharply
to ~ 106 K (Fig. 5.11). The transition region links the chromosphere with corona, the
outermost part of the solar atmosphere.

107 1014

1012
106
1010
Temperature (K)

Density (cm─3)

105
108

104 106

103 2
104
4 6
10 10 10
Height from the Sun's surface (km)

Fig.5.11: The variation of temperature and density in the Sun’s atmosphere with distance

11
 The Solar System 5.4.2 Corona
and Stars
Corona, the outermost layer of the Sun’s atmosphere is named after the Greek word
for Crown. Like the chromosphere, the corona can be observed only during total solar
eclipse − when the Moon completely covers the solar disc (Fig. 5.12). You may
wonder why we cannot see corona at normal times! The fact is that the density of
matter in both the chromosphere and corona is very low (see Fig. 5.11). They emit
very little light and, as a result, they are very faint. In the bright light of the
photosphere, they are not visible.
Chromosphere

40000 Corona
Si X
Si VII
Temperature (103 K)

20000 Mg X
O VI
10000 OV

O II
200 Si IV
O II
40
Si II

10
1 2 4 10 20 40
Height above the surface(103 km) Fig.5.12: Two photographs of the solar corona

Fig.5.13: Height versus The spectrum of corona consists of bright lines superimposed on a continuous
temperature plot
showing emission
spectrum. When these lines were first discovered, they were thought to be due to a
lines of different new element, coronium, not found on the earth. Later, it was realised that these lines
ionised atoms were due to highly ionised atoms and not due to the so-called ‘new’ element
coronium. Fig. 5.13 shows the temperature and height in the corona at which emission
lines of various ionised elements are formed. You may note here that to excite the
emission lines from highly ionised elements, say spectral line of SiX ( read margin
remarks), a temperature greater than 2 × 107K is required. The observed emission
lines of highly ionised atoms of iron, nickel, neon, calcium etc., in the spectrum of
corona clearly indicate that the temperature prevailing in corona is very high (more
than 106 K). Now, before proceeding further, how about testing yourself?

Spend SAQ 3
5 min. a) The temperature of chromosphere and corona is very, very high in comparison to
that of the photosphere. Still, we observe that the photosphere is the brightest of
the three. Why?

b) Calculate the temperature at which a particle will have sufficient energy to ionise
a hydrogen atom.

Due to high temperature, electrons in the corona region have high energies. These
In astronomy, it is common
to denote a neutral atom,
electrons interact with ionised atoms and give rise to emission of X-rays. The coronal
such as silicon, as SiI. By X-ray emission is much larger than that of the photosphere. Remember that the
this convention, singly temperature of the photosphere is only 6000 K. So, it emits very little energy in the
ionised silicon is denoted by X-ray region. The Sun, as observed in X-rays is shown in Fig. 5.14 which clearly
SiII. Hence, SiX denotes a
silicon atom whose nine
indicates the existence of very high (~ 106 K or more) temperature in the corona.
electrons have been
removed due to repeated You have already learnt that the temperature of the photosphere is lower than that of
ionisation. the chromosphere and as one goes further up in the corona, temperature rises to more
than a million degree K. This gives rise to a very simple but important question:
Despite being closer to solar interior, why is the photosphere far cooler than the
12 corona? You know that the second law of thermodynamics precludes such a scenario
as heat cannot flow from a cooler region to a hotter region on its own. We also know The Sun
that the radiation from the photosphere passes through corona almost freely because
of its (corona’s) low density. Since hardly any absorption of radiation takes place in
the corona, the existence of such high (~ million degree) temperature in the corona
presents a paradoxical situation. Several mechanisms have been proposed to resolve
the paradox. It is now generally believed that the magnetic field of the Sun might, in
some way, be responsible for coronal heating. You will study the basics of this
mechanism in Sec. 5.5 of this Unit. The observed overlapping of regions of intense X-
ray emission and strong magnetic fields lend support to this idea.

Solar Wind Fig.5.14: X-ray picture of the

Sun
Unlike its visual appearance, the solar corona extends much beyond into the space.
The outer layer of the solar atmosphere, in fact, continuously emits charged particles
which fill the entire solar system. This emission is called the solar wind. It comprises
streams of charged particles (mainly protons and electrons) and causes continuous loss
of mass from the Sun. The phenomenon of solar wind was predicted much before its
detection. Its characteristics can be investigated using rockets and satellites. For
instance, the solar wind velocities range from 200-700 km s−1 at the distance of the
Earth from the Sun. The number density of the solar wind at this distance
is ~ 7 particles per cm3.
You may like to know: What gives rise to the solar wind? In view of the high
temperature prevailing in the corona, the gas contained therein exerts tremendous
pressure outward. In fact, the pressure is much higher than the inward pressure due to
the Sun’s gravity. The gas, therefore, streams outward from the Sun and fills the One of the manifestations of
interplanetary space. In 1962, Mariner II spacecraft detected the solar wind by its on- solar wind is observed in the
board instruments. shape of comets. You know
that the tail of a comet points
away from the Sun. It is
because the solar wind sweeps
along the material of the comet.
Outer radiation belt
Magnetic axis of
the Earth

You know that when a charged

particle passes through a
magnetic field, it experiences a
force which changes its
direction of motion. The force
experienced by a moving
Inner radiation belt charged particle is known as
Rotational axis of Lorentz force.
the Earth

Fig.5.15: van Allen radiation belts

As you know, satellites and
space missions comprise
The electrically charged particles carried by the solar wind cannot cross the lines of very sophisticated integrated
force of the Earth’s magnetic field. These particles are deflected by the Earth’s circuits, solar cells and other
magnetic field, spiral around the field lines and move back and forth between the electronic gadgets.
Therefore, care is taken to
magnetic poles of the Earth. As a result, two doughnut-shaped zones of highly minimise the damaging
energetic charged particles are created around the Earth and they are collectively effects of the radiation belts
called the van Allen radiation belts. These radiation belts are shown in Fig. 5.15. on the satellites and space
missions.
Astronomers have observed a variety of short-lived events, collectively known as
solar activity, occurring on or near the surface of the Sun. The root cause of all these 13
The Solar System activities is the existence of strong and localised magnetic field in the photosphere.
and Stars Studies of these events/activities provide valuable information about the Sun and the
nature of its magnetic field. You will now learn about some of these short-lived
events.

5.5 SOLAR ACTIVITY

Sunspots

If you look at the photographs (Fig. 5.16) of the Sun, you see dark spots on its visible
surface. These dark spots are called sunspots. Sunspots can be seen sometimes even
with unaided eye at sunrise or sunset. (But you should not attempt to see the Sun with
unaided eye as it may cause irrepairable damage to your eyes because of its intense
brightness.) Naked eye observations of sunspots date back to ~ 2000 years in China. It
was in the seventeenth century that Galileo, using the telescope which he himself had
fabricated, observed sunspots and found that these dark spots were in motion. This led
him to suggest that the Sun was spinning in space. Galileo also observed that the sizes
and shapes of the sunspots kept changing as they rotated with the Sun.

Fig.5.16: Photographs of a sunspot group

The sunspot temperature is ~ 4000 K. With such high temperature, you may
wonder, why they appear dark! Sunspots appear darker because they are cooler
than their surrounding areas in the photosphere that have an effective temperature of
6000 K. A typical white light picture of a large sunspot is shown in Fig. 5.17. Note
that it consists of a dark central region, called umbra, surrounded by a less dark
region, called penumbra. We do not see such details in the picture of smaller
sunspots.

Fig.5.17: Sunspot structure

At this stage, a logical question is: Why is the temperature of the sunspots lower
than their surroundings? It is due to the existence of strong magnetic fields in the
sunspots. In the presence of a magnetic field, a spectral line emitted by an atom at a
14
single wavelength is split into three lines. This is called the Zeeman Effect. Such The Sun
Zeeman splitting is observed in the spectrum of sunspots. Since the line separation, ∆λ
is proportional to the applied magnetic field, a magnetic field up to ~ 3000 Gauss has
been estimated in sunspots. Fig. 5.18 shows the mechanism of Zeeman splitting of a
spectral line and the Zeeman splitting of a spectral line of a sunspot.

Energy
levels
Transition

Spectrum

(a) (b)

Fig.5.18: Zeeman splitting of a) a spectral line; and b) a spectral line of the sunspot

The presence of strong magnetic fields in sunspots restrains the flow of hot material
from layers below the photosphere. Therefore, within a sunspot, less heat comes up
and they (sunspots) are cooler/darker than the surrounding region. Within a sunspot,
the umbral magnetic field is quite intense ~ 3000 Gauss. It spreads like an umbrella
and weakens in the penumbral region. The field strength in the penumbra is estimated
to be ~ 1000 Gauss.

Sunspots can last for weeks. The question is: How do these cooler regions survive
for so long amidst the hotter regions? This could happen due to the magnetic field.
You know that magnetic field exerts pressure (equal to B2/2µ) across the lines of
force. This pressure, along with the pressure of matter inside a sunspot balances the
material pressure outside and the sunspots can exist in equilibrium.

Sunspot Cycle

The observed motion of the sunspots indicates that the Sun is spinning in space. In
1843, Heinrich Schwabe, a German who observed the sky for fun, discovered a
periodic variation in the numbers of visible sunspots. He found an interval of 5.5 yrs
between the time when maximum number of sunspots (sunspot maxima) were
observed and the time when the minimum number of sunspots (sunspot minima) were
observed. Over the last two centuries, sunspot observations clearly suggest a periodic
variation of about 11 years between two successive sunspot maxima (Fig. 5.19a).

Another important observation pertaining to sunspot is that the sunspot zones migrate
along solar latitude. It is observed that the first sunspot zone appears at latitude
of ~ 35° in, say, the northern hemisphere and it migrates to lower latitudes. It lasts
till it reaches a latitude of ~ 10°. The latitude migration of sunspot zones is shown in
Fig. 5.19b. This is the famous butterfly diagram which shows a period of ~ 11 years
between the successive occurrences of a sunspot at a given latitude. It is believed that
the sunspot cycle is caused due to differential rotation of the Sun; it rotates faster at
the equator compared to higher latitudes.
15
The Solar System
and Stars Number of
Sunspots

Year
(a)

Year
(b)
Fig.5.19: a) Sunspot cycle; and b) butterfly diagram which shows the migration of sunspots from
higher to lower latitudes

Solar Prominences

Refer to Fig. 5.20 which depicts loop like structures surging up into the corona when
the Sun is viewed along the edge of the solar disc. These structures are called
prominences which are intimately connected with and formed due to the Sun’s
magnetic field. The material in prominences comprises hot ionised gases trapped in
magnetic fields associated with the active regions. Since these gases came from
deeper layers of the solar atmosphere, they are cooler and therefore denser than the
coronal gas. It is for this reason that prominences appear as bright structures.
However, when viewed against the photosphere (solar disk), prominences appear as
dark snake like objects, called filament.

Fig.5.20: Solar prominences

Solar Flares

Yet another form of solar activity is called solar flare. Solar flares are sudden eruptive
16 events which occur on the Sun (Fig. 5.21). Each event may involve energy in the
 The Sun

Approximate size of the Earth

Fig.5.21: Solar flare

range of 1022 to 1025 Joules. Usually the flares last anywhere between a few minutes
to more than an hour. A large flare may have linear dimension as large as 105 km and
may be seen as a short-lived storm on the Sun. Such energetic eruptions are usually
linked to sunspots because these quite often occur at the top of magnetic loops that
have their feet in sunspots. Thus, the most likely places of occurrence of solar flare are
the regions of closely packed sunspots. The tremendous amount of energy carried in
solar flare is released in the form of X-rays, ultraviolet and visible radiation, high
speed electrons and protons. You may ask: What is the source of energy in solar
flares? This question can be answered on the basis of a model for solar flare shown in
Fig. 5.22.

Fig.5.22: A solar flare model

17
The Solar System A Model of Solar Flare
and Stars
All kinds of solar activities i.e., the sunspots, prominences, flares etc., are possibly
linked to the release of stored magnetic energy. It is believed that the energetic solar
eruptions are caused due to coming together and merging of magnetic fields in the
active regions (the phenomenon is known as magnetic field reconnection) and
thereby releasing the stored magnetic energy. To appreciate this phenomenon,
magnetic lines of forces can be considered as stretched springs with certain amount of
energy associated with each unit length. If the length of the lines of forces gets
reduced by certain mechanism, energy is released. This is what happens when lines of
force pointing in opposite directions meet and merge with each other (Fig. 5.22).

In order to fix these ideas, you should answer the following SAQ.

Spend SAQ 4
3 min.
a) What is the basis to conclude that the Sun is rotating in space?
b) What is the difference between spicules and solar prominences?

So far, you have studied about the photosphere, solar atmosphere and solar activity.
You must have noted that the Sun’s magnetic field plays an important role in solar
activity. Further, gaseous matter in the Sun is in the ionised form, that is, it is a
conducting matter. We will now try to understand the nature of interaction between
the Sun’s magnetic field and the conducting matter (fluid) in motion. This
understanding is of utmost importance in astronomy because, everywhere in the
universe, we find conducting matter moving in the presence of magnetic fields. The
study of the motion of conducting fluid in the presence of magnetic field is called
magnetohydrodynamics. Let us now turn to this subject.

5.6 BASICS OF SOLAR MAGNETOHYDRODYNAMICS

We begin our discussion of solar magnetohydrodynamics with Maxwell’s equations.
You may recall from the physics course entitled Electric and Magnetic Phenomena
(PHE-07) that three of the Maxwell’s equations can be written as:

∇ . B = 0, (5.3)

∇ × B = µj, (5.4)

and

You are familiar with the ∂B

Ohm’s law expressed
∇×E= − (5.5)
∂t
mathematically as:
j=σE
However, when a conducting where µ, B, E and j are the magnetic permeability of the medium, magnetic field
fluid is moving with velocity intensity, electric field intensity and electric current density respectively. Note that in
v in a magnetic field B, there Eq. (5.4), we have not written the displacement current term. This is because,
is an induced electric field
magneto-hydrodynamic phenomena in the Sun are usually slow whereas the
given as v × B. Thus, in the
presence of two kinds of displacement current gives rise to fast phenomena such as electromagnetic radiation.
electric fields − one due to
charged particles and another Further, if the fluid (conducting matter) velocity is v and its electrical conductivity is
due to their motion − the σ, then Ohm’s law gives (see the margin remark):
Ohm’s law takes the form:
J = σ (E + v × B)
j = σ (E + v × B) (5.6)

18
where (v × B) term represents the electric field induced due to the motion of The Sun
conducting fluid in the presence of magnetic field. Using Eqs. (5.3) to (5.6), you can
readily obtain:

∂B
= ∇ × ( v × B ) + η∇ 2 B (5.7)
∂t

where η = (µσ)−1 is called the magnetic diffusivity. Its value is generally constant in
solar conditions.

SAQ 5 Spend
3 min.
Derive Eq. (5.7).

The first term on the right hand side of Eq. (5.7) gives the change in B due to the fluid
motion. The second term represents the change in B due to conductivity σ. It is
generally called the Ohmic decay of the field. Note that for v = 0, Eq. (5.7) reduces to:

∂B
= η∇ 2 B (5.8)
∂t

Eq. (5.8) is the diffusion equation. It gives the rate at which magnetic field diffuses
out due to conductivity. In the limit of infinite conductivity, we have η → 0, and
Eq. (5.7) becomes

∂B
= ∇ × ( v × B) (5.9)
∂t

Without going into the mathematical details, it is possible to understand the relative
importance of the two terms on the right hand side of Eq. (5.7). To do so, let us obtain
the orders of magnitude, of the values of the two terms. The order of magnitude of the
VB
first term is , where V, B and L are the typical values of the fluid velocity,
L
magnetic field and the dimension of the system. Similarly, the magnitude of the
ηB
second term is . The ratio of the two terms is called magnetic Reynold number,
L2
Rm, and is given by:

VL
Rm = (5.10)
η

If Rm >> 1, the second term in Eq. (5.7), i.e. the diffusion term, is negligible. The
condition Rm >>1 is obtained in two situations: when the conductivity is very high
because η appears in the denominator, and secondly when the dimension of the
system, L is very large. In astrophysical systems, we have Rm >> 1 because of the
second situation as their dimensions are very large. In any case, Rm >> 1 implies that
there is no decay of the magnetic field as if the conductivity of the medium is infinite.
Actually, the conductivity is finite, but the large dimensions ensure that Rm >> 1 and
so there is no decay of the field.

If we drop the diffusion term in Eq. (5.7), the remaining equation (Eq. (5.9)), implies
that the magnetic flux linked to a cross-section of the fluid remains unchanged as the
fluid moves about. In other words, the magnetic field is frozen in the fluid. The
idea of frozen field means that the magnetic flux is transported along with the material
motion. We can show this formally in the following manner: 19
 The Solar System Let us consider a cross-sectional area A placed in a magnetic field B (Fig. 5.23). The
and Stars magnetic flux linked with area A may be written as:
l
Φ= ∫ B.d a (5.11)
A

where da is an element of area on the surface A. Let l be the curve enclosing the area
dl v × dldt A. Let us further assume that, in the time interval dt the area changes from A to A′ as
the fluid moves around. The magnetic flux Φ′ linked with A′ may be different from Φ
because (i) magnetic field B may have changed and/or (ii) some flux might have been
A A' exchanged through the surface of the volume generated between A and A′. Now, the
rate of change of magnetic field is ∂B/∂t. The area of the curved surface surrounding
Fig.5.23: Motion of a closed the volume between A and A′ is v × dl dt where dl is an element of length of the
curve l with fluid with contour of A. Therefore, the difference ∆Φ between fluxes through A′ and A is given
velocity v by:

 ∂B 
 
∆Φ = 
 ∂t ∫ l ∫
.da + o B.. ( v × dl )  dt.

(5.12)
A 

You may recall that area is a Further, using Stoke’s theorem, we can write the second term on the right hand side of
vector quantity having direction
along the normal to the area, Eq. (5.12) as:
whose sense is defined by the

∫ol B.. (v × dl) = −∫ ∇ × ( v × B) . da

right hand rule.
(5.13)
A

So, we can write the rate of change of flux (Eq. (5.12)) as:

dΦ  ∂B 
dt
= ∫ 
 ∂t
− ∇ × ( v × B )  . da

(5.14)
A

∂B
From Eq. (5.9), we have = ∇ × ( v × B) . Thus, from Eq. (5.14), we find that the
∂t
rate of change of magnetic flux is zero. It implies that Φ remains constant. We may,
therefore, conclude that in astrophysical systems the lines of forces are
completely attached or glued to the moving fluid when Rm >> 1.

Now, let us pause for a moment and think about the significance of the above
conclusion for solar activity. Recall that the active regions consisting of sunspots have
strong magnetic fields and solar activities such as prominence and flare occur in these
regions. The structures associated with solar prominences and solar flares are very
similar to the magnetic lines of force. It is, therefore, believed that these activities are
caused due to frozen magnetic lines of force in the conducting fluid.

You have learnt earlier in this unit that, in almost all the events associated with the
Sun’s magnetic field, energy is also transported. The energy transported by the
magnetic field glued to the conducting matter is responsible for heating the
chromosphere and corona. Would not you like to know how it happens? To
understand this process, we begin with the fact that an electric current exists in the
solar atmosphere due to the drift of electrons with respect to ions carrying opposite
charges. If a magnetic field B is also present in the plasma, then a volume force, also
1
called the Lorentz force (= j × B), acts on the material. Since j = (∇ × B ) (from
µ
20
Eq. (5.4)) we may write the expression for Lorentz force as:
 The Sun
1
j× B = (∇ × B ) × B
µ

1  B2 
= (B.∇ )B − ∇   (5.15)
µ  2µ 

The first term on the RHS of Eq. (5.15) denotes magnetic tension and the second term
denotes the gradient of magnetic pressure. Therefore, in the presence of a magnetic
field, a lateral pressure (due to the second term in Eq. (5.15)) acts on a conducting gas.
To maintain equilibrium, the lateral pressure due to magnetic field is balanced by the
gas pressure. The magnetic tension term is similar to the tension of a stretched string.
When a stretched string is distorted, its tension provides the restoring force. Therefore,
when the magnetic field lines (frozen in the conducting fluid) are disturbed we have a
similar restoring force (due to the magnetic tension). Thus, a disturbance may
propagate as a transverse wave along the magnetic field lines. Such waves are called
Alfven waves. The speed of these waves is given by:

B
vm = .
4πρ

where ρ is the density of the plasma. Since Alfven waves propagate along the
magnetic field lines, it is possible to transport energy outward along magnetic field
lines which are threading the outer solar atmosphere. This energy is believed to be
responsible for heating the chromosphere and corona.

SAQ 6 Spend
3 min.
A pressure of 103Pa (Pascal) prevails in the solar atmosphere. What should be the
strength of the magnetic field required to balance such a pressure?

Now, before we close our discussion about the Sun, we will briefly discuss a new area
of solar research called helioseismology.

5.7 HELIOSEISMOLOGY The name helioseismology

derives from the fact that, in
In 1962, Leighton, Noyes and Simon noticed wiggling back and forth of some of the Greek, helios means the Sun
absorption lines in the solar spectrum with periods ~ 5 min. They conjectured that the and the word seismos is
used for an earthquake.
movement of the surface of the Sun is responsible for such observations.

Subsequently, careful observations of the solar surface confirmed the idea of waves As you know, seismology
refers to the study of seismic
rising up and down on the surface of the Sun (Fig. 5.24). Astronomers now use these waves moving inside the
waves to probe the solar interior much the same way as seismologists probe the Earth. Their arrival at
earth’s interior using vibrations caused by an earthquake. various points on the Earth’s
surface enables us to find
the point of origin of these
So far, millions of different vibrational modes also called acoustic modes or p-modes waves. This way we are able
have been observed on the Sun’s surface. All of them have different frequencies and to construct the internal
surface patterns. Thus, the Sun appears to have a rhythmic surface motion similar to structure of the Earth.
that of a beating heart. The observed rise and fall of the surface of the Sun is, in fact,
due to superposition of many different acoustic modes. These modes are now believed
to be driven by irregular motions in the convective envelope under the solar surface.

21
The Solar System
and Stars
Disk of Sun
5 min

Position
Time

Fig.5.24: Waves on the surface of the Sun

The combined effect, or the superposition of millions of these acoustic waves, results
in the observed up and down motion of the photosphere with a period of the order of 5
minutes. The extent or size of a wave is called its horizontal wavelength. The relation
between sizes and periods of waves, obtained theoretically, suggested that only
specific combination of periods and sizes can resonate inside the Sun.

Fossat and Grec observed the solar oscillation from the South pole for around 120
hours. The analysis of this continuous record showed that the entire surface of the Sun
is ringing like a bell with periods in the range of 5 minutes and the vibrations may last
for days and weeks.

The natural frequencies of oscillations can be computed for any solar model. In view
of the precision now possible for determining the frequencies, we may compare these
with those computed for a given solar model. In case there is lack of agreement
Since we cannot go inside a
star, we build theoretical
between the predicted and observed frequencies, the model is slightly modified to
models which tabulate improve agreement. The improvement in the model brings it closer to reality. It has
pressure, temperature and been found that the depth of the solar convection zone is at a radius of 71.3 percent of
density at various depths radius of the Sun. It has also been confirmed now that the proportion of helium in the
inside the star.
Sun lies between 0.23 − 0.26. This is quite consistent with the value of 0.25 for
helium believed to have been formed in the early phase of the universe after the big-
bang.
Now, let us summarise what you have learnt in this unit.

5.8 SUMMARY

• The mass, radius and effective surface temperature of the Sun are 2 × 1030kg,
7 × 108m and 6000 K, respectively.
• The Sun’s atmosphere comprises of the photosphere, chromosphere and
corona.
• The photosphere is the visible surface of the Sun and all the radiation we receive
from the Sun is emitted by this layer.
• The photosphere has a granular structure which is caused by convection of hot
gas from below the photosphere.
• The chromosphere, which lies above the photosphere, extends up to ~ 2000 km
and is normally not observable from the earth except during total solar eclipse.
The temperature of the chromosphere is much higher than the photosphere.
• Relatively higher temperature of the chromosphere is perhaps caused due to
spicules − jets of hot gas − which extend upward in the chromosphere.

22
• The outermost layer of the solar atmosphere, corona is also not visible at normal The Sun
times due to its low material density. The temperature of the corona is of the order
of 106K, much more higher than that of the photosphere.
• The sunspots are the dark spots on the solar disk. They appear dark because they
are cooler than their surrounding areas in the photosphere. The movement of
sunspots indicate that the Sun is spinning in space.
• Solar prominences are the loop like structures surging up into the corona. They
are caused due to the Sun’s magnetic field.
• Solar flares are sudden eruptive events involving energy in the range of 1022 to
1025 joules. The most likely places of occurrence of solar flares are the regions of
closely placed sunspots and they are possible caused due to the release of stoned
magnetic energy.
• The study of the motion of conducting matter in the presence of magnetic field is
called magnetohydrodynamics.

• The magnetic Reynold number is given by

VL
Rm =
η

which indicates that for infinite conductivity (i.e. η → 0), there is no decay of the
magnetic field. In other words, magnetic field is frozen in the conducting matter
in motion.
• Transportation of magnetic flux with conducting matter explains some of the solar
activities like prominence and solar flare.
• Helioseismology has its origin in the observed wiggling back and forth of some of
the absorption lines in the solar spectrum. The movement of the surface of the
Sun, causing these back and forth motion of absorption lines, gives rise to waves
on it. Investigation of the nature of these waves provides valuable information
regarding the internal structure of the Sun.

5.9 TERMINAL QUESTIONS Spend 20 min.

1. In a sunspot, magnetic diffusivity, linear dimension and velocity of conducting

fluid respectively is 103m2s−1, 104km, and 103ms−1. Estimate the magnetic
Reynold number, Rm. Is it possible to assume that the conductivity in the sunspot
is virtually infinite?

2. The temperature inside a sunspot is 4000 K and that of its surface is 6000 K.
Calculate the strength of the magnetic field inside the sunspot which will balance
the pressure inside and outside.
[Hint: Remember that the magnetic pressure is B2/2µ where µ is magnetic
permeability of the medium and its value for the present case can be taken as
4π × 10−7 NA−2].

3. The number density of particles (assume hydrogen) in the photosphere is 1020

particles per cm−3 and the strength of the magnetic field of the Sun is 1 G.
Calculate the velocity of the Alfven waves in the photosphere.

23
The Solar System
and Stars 5.10 SOLUTIONS AND ANSWERS
Self Assessment Questions (SAQs)

1. Since the mass of the planet Jupiter is very large compared to its satellite Io, we
can use Kepler’s third law for Jupiter and Io system. Thus, we can write:
4π 2 a 3
= GM J (i)
P2

where MJ is the mass of Jupiter.

From the problem, we have

10 8
a = 4.22 × 10 cm = 4.22 × 10 m

P = 1.77 days = 1.77 × 24 × 60 × 60 s

G = 6.672 × 10−11 m3 kg−1 s−2

Substituting these values in Eq. (i) above, we get

4π 2 a 3
MJ =
GP 2
4 × (3.14) 2 × (4.22 × 10 8 m) 3
=
(6.672 × 10 −11 m 3 kg −1s −2 ) × (1.77 × 24 × 60 × 60 s) 2

= 1.97 × 1027 kg

2. According to the Stephan-Boltzmann law, the amount of energy radiated by a

black-body per unit time per unit area at temperature T is given by

E=σT4 (i)

where σ is Stephan constant.

The energy radiated by the Sun can also be expressed in terms of its luminosity LΘ
as

LΘ
E= (ii)
4πR 2

where R is the radius of the Sun.

Comparing the above two expressions, we can write

LΘ
= σT 4
2
4πR

Rearranging the terms and substituting the values of R (= 6.7 × 108m),

σ (= 5.67 × 10−8 W m−2 K−4) and L = (3.86 × 1026W), we get,

1/ 4
 LΘ 
T =  
24  4πR 2 σ 
 1/ 4 The Sun
 3.86 × 10 26 W 
= 
 4 × (3.14) × (6.7 × 10 8 m) 2 × (5.67 × 10 −8 Wm − 2 K − 4 ) 
 
1/ 4
 3.86 × 10 2 
=  × 1016 K 4 
2
 4 × 3.14 × (6.7) × 5.67 

= 0.5895 × 104K

≈ 6000 K.

3. a) It is because the density of matter in photosphere is much higher than the

density of matter in chromosphere and corona.

b) We know that the ionisation energy of the hydrogen atom is 13.6 eV. Further,
energy acquired by a particle at temperature T is kBT. If this energy is equal to
the ionisation energy of the hydrogen atom, it will be ionised. Thus, we must
have
k BT = (13.6 × 1.6 × 10 −19 ) J

(13.6 × 1.6 × 10 −19 J)

T=
(1.38 × 10 −23 JK −1 )

4
≈ 13.6 × 10 K.

4. a) Motion of sunspots.

b) Spicules are the jet like structures comprising hot gas and is observed
throughout the chromosphere. Prominences look like structures surging up
into the corona.

5. Ohm’s law can be written as:

J = σ (E + v × B)

Taking curl of both sides, we get:

∇ × J = σ (∇ × E + ∇ × (v × B))

Using Eqs. (5.4) and Eq. (5.5) we can write:

1 ∂B
(∇ × ∇ × B) = − + ∇ × ( v × B)
σµ ∂t

1 ∂B
[∇(∇. B) − (∇.∇) B] = − + ∇ × ( v × B)
σµ ∂t

From Eq. (5.3), we have ∇ . B = 0. Thus, we get

∂B 1
= ∇ × ( v × B) + ∇2 B
∂t σµ

25
The Solar System
= ∇ × ( v × B) + η∇ 2 B
and Stars

which is Eq. (5.7).

2
6. We know that the pressure generated by magnetic field B is equal to B /2µ. Thus,
3
for generating pressure equal to 10 Pa, we must have

B2
≈ 103 Pa
2µ
−7 −2
Substituting µ = 4π × 10 NA in the above expression, we get:

B ≈ 5 × 10 −2 T = 5 × 10 2 G.

Terminal Questions

1. The magnetic Reynold number is given by (Eq. (5.10)):

VL
Rm =
η

Substituting the values of V, L and η from the problem, we get

(10 3 ms −1 ) × (10 7 m )
Rm =
10 3 m 2 s −1

≈ 107

Yes, conductivity can be taken to be virtually infinite because Rm >> 1.

2. On the basis of the equation of state for the sunspot and using the fact that
magnetic pressure is equal to B2/2µ, we can write:

B2
= Nk B (T2 − T1 )
2µ
where N is number density. Since µ = 4π × 10−7 NA−2 , T2 = 6000 K and
T1 = 4000 K, we can write:

B2 = 2 × (4π × 10−7 NA−2) × (1023 m−3) × (1.38 × 10−23J K−1) × (2000 K)

B = 0.08 T = 800 G

3. Velocity of Alfven waves is given by

B
vm =
4πρ

where ρ is density. Substituting the values of B and ρ, we get:

1G
v =
m 4π × (10 × 1.6 × 10 − 24 g cm −3 )
20

≈ 25 cm s−1.
26
 The Solar Family
UNIT 6 THE SOLAR FAMILY
Structure
6.1 Introduction
Objectives
6.2 Solar System: Facts and Figures
6.3 Origin of the Solar System: The Nebular Model
6.4 Tidal Forces and Planetary Rings
6.5 Extra-Solar Planets
6.6 Summary
6.7 Terminal Questions
6.8 Solutions and Answers

6.1 INTRODUCTION
In Unit 5, you have learnt about various features of the Sun such as solar atmosphere,
solar activity and energy transportation. You know that the Sun is the only star in our
solar system. It is the most massive object in the system and all the planets revolve
around it. You have already studied about the solar system in your school science
course as well as in Foundation in Science and Technology (FST-1) course.
Therefore, you know most of the facts and figures such as mass, density, distance
from the Sun, and surface temperature of all the nine planets of the solar system.
However, being a student of physics and astronomy, you may not be satisfied only
with the facts and figures and would like to know: How did the solar system come
into being? Is it possible to explain the observations pertaining to the solar system on
the basis of the principles of physics? Do all the stars have planetary systems similar
to our solar system? Does life exist on any other planet? In the present unit, we shall
address some of these issues.

In Sec. 6.2, you will recapitulate some facts and figures about the solar system. This is
necessary because any model for the formation of the solar system must be consistent
with these observations. In Sec. 6.3, you will learn the nebular model which is at the
core of all the contemporary theories of formation and evolution of the solar system.
In this section you will also discover that the nebular model explains most of the
dynamic properties of the solar system. Further, it has been argued that the
gravitational force gives rise to tides in the oceans on the Earth and planetary rings
around the outer planets such as Jupiter and Saturn. The genesis of tidal forces and
planetary rings has been discussed in Sec. 6.4. And, in Sec. 6.5, you will learn about
the efforts made by astronomers to investigate the existence of extra-solar planets.

Objectives
After studying this unit, you should be able to:

• describe the planets of the solar system;

• understand how the terrestrial planets differ from the jovian planets;
• explain the nebular model of the solar system;
• discuss the role of gravitational forces in generating tides and in the formation of
Earth’s tidal bulge and planetary rings; and
• discuss the possible existence of extra-solar planets.

27
The Solar System
and Stars 6.2 SOLAR SYSTEM: FACTS AND FIGURES
The solar system consists of the Sun, nine planets (see the margin remarks), satellites
of planets, asteroids and comets. The nine planets, arranged according to their
increasing distances from the Sun, are: Mercury (Buddha), Venus (Shukra), Earth
In a recent development,
the International
(Prithvi), Mars (Mangal), Jupiter (Brihaspati), Saturn (Shani), Uranus (Arun),
Astronomical Union Neptune (Varun) and Pluto (Yama). The sizes of these planets with respect to the
(IAU) has decided to Earth are shown in Fig. (6.1).
remove Pluto from the list
of planets in the solar
system. So, the solar
Note in Fig. 6.1a that the sizes of the first four planets are similar to that of the Earth
system now has only and they are called terrestrial planets. On the other hand, from Fig. 6.1b, it is
eight planets. Pluto has obvious that the sizes of the next four planets are bigger than the Earth. They are
now been categorised as called jovian planets. The status of the ninth planet, Pluto, is somewhere in-between.
an object of the Kuiper
belt, found in the outer
region of the solar
system. Kuiper belt Relative sizes of the terrestrial planets Relative sizes of the jovian planets
contains many objects of (Earth = 1) (Earth = 1)
the size of Pluto. In fact,
it was the realisation that
R = 11.3
many more objects of the R = 0.38 R = 4.0
size of Pluto could be
discovered in future that
prompted IAU not to
consider Pluto as planet.
Mercury
Uranus

R = 1.0
R = 0.95 R = 1.0 Jupiter

Earth
R = 9.4

Venus R = 3.9
Earth

R = 0.53
R = 0.26 Neptune
Saturn

R = 0.18
Moon
Mars
Pluto

(a) (b)

Fig.6.1: Sizes of a) terrestrial; and b) jovian planets relative to the Earth

The properties of the terrestrial and jovian planets are different. For example,
terrestrial planets are mainly made of rocks and metals having an average density of
4 or 5 g cm−3 whereas jovian planets consist mainly of gas and ice with an average
density of 1 or 2 g cm−3.

All these planets revolve around the Sun in elliptical orbits. The planetary orbits are
almost in the same plane except that of Pluto which is inclined at an angle of ~ 17° to
the common plane (Fig.6.2). Asteroids are believed to be the captured objects which
were wandering in the solar system. Their orbits are mostly located in between the
orbits of Mars and Jupiter. Since all the planets revolve round the Sun, it is considered
the ‘head’ of the solar family. In addition, the Sun contains almost 99.87% of the total
mass of the solar system. Among the planets, Jupiter and Saturn are the most massive,
accounting for 92% of the mass of all the planets.
28
 The Solar Family

Earth
Mercury
Venus Sun

Mars

(a)

Jupiter
Uranus
Saturn

Neptune
Pluto

(b)

Fig.6.2: Orbits of the a) terrestrial; and b) jovian planets in the solar system

Table 6.1 summarizes some basic data about these planets and the Moon. Many of
these features could directly be attributed to the distance of a planet from the Sun.
Table 6.1: Some basic data of planets and the Moon

Planets Mass [kg] Density Mean Rotation Period Revolution Inclination of Percentage of
[g cm−3] distance (hours or days) Period (days) orbit to the light reflected
from the plane of the
Sun (km) Solar System
(deg)
23 7
Mercury 3.30×10 5.4 5.8×10 58 d 88 7.0 7
24 8
Venus 4.87×10 5.2 1.1×10 243 d 245 3.4 76
24 8
Earth 5.97×10 5.5 1.5×10 23 h 56 min 365.25 0.0 39
23 8
Mars 6.42×10 3.9 2.3×10 24 h 37 min 687 1.8 18
27 8
Jupiter 1.90×10 1.3 7.7×10 9 h 55 min 4333 1.3 51
26 9
Saturn 5.69×10 0.7 1.4×10 10 h 30 min 10743 2.5 50
25 9
Uranus 8.70×10 1.1 2.8×10 17 h 14 min 30700 0.8 66
26 9
Neptune 1.03×10 1.7 4.5×10 18 h 60280 1.8 62
22 9
Pluto 1.00×10 2.1 5.9×10 6 d 09 h 17 min 90130 17.1 15
22 5
Moon 7.35×10 3.3 3.8×10 * 27.33 d 27.33 5.0
* This is Moon’s mean distance from the Earth.

SAQ 1 Spend
5 min.
Using the data given in Table 6.1, verify Kepler’s third law:

T2
= Const .
r3
where T is the orbital period of a planet and r is its mean distance from the Sun.

Further, as mentioned above, the division of planets into two groups namely,
terrestrial and jovian, is based on the similarity of some major characteristics of the
planets in a particular group. Table 6.2 lists these characteristics of terrestrial and 29
The Solar System jovian planets. Such classification helps in developing theoretical models for their
and Stars formation and evolution.
Table 6.2: Characteristics of the terrestrial and jovian planets

Characteristics Terrestrial Jovian

Basic form Rocky Gas/Liquid/Ice
Mean orbital distance (AU) 0.39−1.52 5.2−30.1
Mean surface temperature (K) 200−700 75−170
Mass (relative to the Earth) 0.055−1.0 14.5−318
Equatorial radius (relative to the Earth) 0.38−1.0 3.88−11.2
Mean density (g cm−3) 3.96−5.42 0.68−2.30
Sidereal rotation period (equator) 23.9 h − 243 h 9.8 h − 19.2 h
Number of known Moons 0−2 8−20
Ring systems No Yes

We present some salient features of the individual planets in Table 6.3.

Table 6.3a: Salient features of the terrestrial planets

Mercury Venus Earth Mars

• It can be seen near • It appears as the third • Its crust, extending to • Though half in size
the horizon at brightest object in the 10 km deep under the of the Earth, this
sunset or sunrise sky after the Sun and oceans and up to 40 planet has various
with unaided eyes. the Moon, as it is km under the similarities with the
nearest to the Earth. continents, consists Earth
• Like the Moon, it mainly of silicon
has no atmosphere • Its surface is dry, hot (27.7%), and oxygen i) A day on Mars is
and its surface is and volcanic. Its (47.3%). Elements 24h 40 minutes
full of craters. atmosphere contains like aluminium, iron, and a year lasts for
about 96 percent carbon calcium, sodium etc. 1.88 Earth year,
• It is, at the same dioxide, 3.5 percent make the bulk of its
time, the coldest nitrogen and remaining matter and less than ii) its rotation axis is
and the hottest half percent is water 2% is made of all tipped at 25° and
planet because its vapours, argon, other elements. thus has seasons
periods of sulphuric acid, and polar caps.
revolution and hydrochloric acid etc. • Its rotation axis is
rotation are almost tipped by ~ 23° • This is the most
equal which keeps • The planet is covered causing various extensively probed
its same surface by a thick cloud mainly seasons and polar planet and a few
face the Sun all consisting of sulphuric caps. automated
the time. acid droplets. laboratories have
• Its atmosphere also been landed.
• Its surface • Its surface temperature consists of distinct The atmosphere and
temperature varies is very high (~ 470°C) layers called geology of this
between + which is perhaps troposphere, planet has many
340°C to − caused due to the stratosphere and features similar to
270°C. greenhouse effect: the ionosphere. the Earth.
infrared radiation Troposphere
emitted by the planet is comprises mainly of • Martian surface has
not allowed to escape 78% nitrogen, 21% craters of all sizes
due to the presence of oxygen; stratosphere and enormous
carbon dioxide in its contains ozone which volcanoes.
atmosphere thus absorbs the harmful
causing the heat ultraviolet radiation • Martian soil, like the
received from the Sun from the Sun. Earth, is mostly
to be trapped and raise made of silicates.
its temperature. However, due to the
presence of 16
percent iron oxide in
its soil, it has the
characteristic red
colour. It is also
known as the Red
30 Planet.
 Table 6.3b: Salient features of the Jovian planets The Solar Family

Jupiter Saturn Uranus Neptune Pluto

• It is the largest • It is the second • This planet is • This planet was • This planet was
(having diameter largest planet smaller than discovered in predicted to exist
11.2 times that exceeded in Jupiter and four 1846 and it is so theoretically to
of the Earth) and size and mass times farther far away that, account for the
the most massive only by Jupiter. from the Sun. In seen from this observed
planet (contains Like Jupiter, it a telescope, it planet, the Sun irregularities in
71% of all the consists mainly appears as a would look like the orbits of
planetary mass). of hydrogen green disc with a bright spot! Uranus and
and helium. vague markings. Neptune.
• It is like a • Its colour is
spinning ball of • It is the last • Its axis of faint blue which • It is a very small
gas and liquids (sixth) planet rotation is tipped is caused due to (about one-fifth
with no solid visible from 97.9° from the larger the size of the
surface. In this the Earth. It perpendicular to percentage of Earth), cold and
regard, it is has beautiful its orbit. This methane present dark planet.
similar to the rings (which causes its poles in it.
Sun. It has a can be seen to nearly point • Unlike most
large number of through a towards the Sun • Its cloud planetary orbits,
satellites. telescope). The and it would temperature is Pluto’s orbit is
rings are, in seem that the about − 237°C. quite elliptical
• It is covered by a fact, a planet is rolling and therefore it
turbulent, thousand along the orbit • Like other can come closer
gaseous tightly packed like a wheel. jovian planets, to the Sun than
atmosphere individual this planet also Neptune.
comprising of ringlets. • The rings of this has rings.
hydrogen, planet were • Being so far away
helium and small • The discovered as • Two Moons of from the Sun, this
traces of water temperature at late as 1977 this planet are planet is cold
vapour, its cloud tops is (Voyager 2) and visible from the enough to freeze
ammonia, − 180°C. It is they comprise of Earth. most compounds.
methane etc. colder than very dark
Jupiter. material, as black • Triton is the • Its mass is only
• It has a Great as coal. largest Moon of 0.002 times the
Red Spot on its • This planet is this planet Earth mass!
surface which less dense than • It has a large which is
has an oval water and number of orbiting it in the
shape. It is big being mostly Moons. clockwise
enough to liquid and direction i.e.
accommodate rotating opposite to the
two Earths! It is rapidly, its rotation of the
presumed to be shape is planet. Triton
due to huge flattened. also has
cyclonic atmosphere of
disturbance in its • It has a large its own
atmosphere. number of comprising of
satellites nitrogen and
• In view of its orbiting at the methane.
composition, edge of the
size and the rings. These
number of satellites
Moons, Jupiter (Moons) are
looks like a star composed of
having its own rock and ice
‘solar system’! and have
craters.
• It has rings and
it emits radio • Its main
waves. satellite, Titan,
is very large
(diameter
5800 km) and
has atmosphere
of its own as
dense as ours.

31
The Solar System At this stage, you may pause for a moment and think about what have you learnt till
and Stars now. You have basically recapitulated what you learnt in school about the solar
system and have acquainted yourself with the characteristic features of each of its
planets. In the next section, you will learn about the origin of the solar system. A
model for the origin of the solar system must explain its characteristic features listed
below:

i) Most of the mass of the solar system is contained in the Sun.

ii) Except for Mercury and Pluto, the orbital planes of all the planets are more or less
in the same plane.

iii) When viewed from above, the planets are found to revolve around the Sun in the
anticlockwise direction; the direction of rotation of the Sun is also the same.

iv) Except for Venus, Uranus and Pluto, the direction of rotation of planets is the
same as their direction of revolution.

v) The direction of revolution of the satellites of each planet is the same as the
direction of rotation of the planet itself.

vi) Total angular momentum of all the planets is more than the angular momentum of
the Sun.

vii) Terrestrial planets comprise mainly of rocky material whereas jovian planets
comprise mainly of gaseous material.

Spend SAQ 2
3 min.
List some common features of terrestrial and jovian planets.

6.3 ORIGIN OF THE SOLAR SYSTEM: THE NEBULAR

MODEL

The formation of the Earth or, in fact, the entire solar system has been of considerable
interest to human being for ages. A variety of speculative ideas were proposed which
gave rise to two kinds of theories: catastrophic and gradualistic. According to the
catastrophic theories, the planets were formed out of the material ejected from the Sun
when a giant comet collided with it or the planets came into being due to ripping-off
of material from the Sun caused by the tides generated by a close by passing star.
However, these theories do not explain many features of the solar system. Further,
during the 20th century, astronomical evidences have been found which support the
gradualistic theories. According to these theories, formation of planets is a gradual
process and is a natural by-product of the formation of stars like the Sun.

Contemporary gradualistic theories of the solar system are based on the nebular
hypothesis. According to this hypothesis, the Sun as well as the planets formed from
an interstellar cloud of gas and dust. A model of the formation of the solar system
based on this hypothesis is called a nebular model. The basic principle of physics
underlying the nebular model is the Newton’s law of gravitation. According to the
nebular model, formation of stars including the Sun, begins when the interstellar
cloud with enough mass and low internal pressure, contracts due to its own gravity.

Can you guess the consequences of gravitational contraction of an interstellar

cloud? You are on the right track if you think that it leads to increase in density of the
32 cloud. Simultaneously, the kinetic energy of the particles in the cloud increases which
causes increase in its temperature. As a result, the internal pressure of the collapsing The Solar Family
cloud increases. Eventually, the gravitational contraction is balanced by the internal
pressure of the cloud. The contracting cloud is called a protostar and when its
internal temperature is high enough to initiate thermonuclear reactions, a full-fledged
star (such as the Sun) is born. This is the general picture of the formation of a star
according to the nebular model.

Well, you may ask: How did our solar system come into existence? According to
the nebular model, our solar system formed due to gravitational contraction of the
rotating interstellar cloud called the solar nebula. Due to rotation, the solar nebula
takes the form of a disc (Fig. 6.3). When the Sun formed at the centre of this disc and
became luminous enough, the remaining gas and dust was pushed away due to the
Sun’s radiation pressure. The blown away gas and dust condensed into various planets
orbiting the Sun. One of the natural consequences of the solar nebula model is that
most stars in the galaxies should have planetary systems!

(a) (d)

Venus (e) Uranus Mercury

(b) Jupiter Earth

Saturn
Neptune
Pluto Asteroids
Mars (f)
(c)

Fig.6.3: Schematic diagram showing different stages (a to f) of the formation of the solar system

Formation of Solar Nebular Disk

You must have noted that one of the significant features of the solar nebular model is
the formation of a disk of interstellar cloud around the Sun. You may ask: Is there
any evidence supporting the formation of the disk? Two facts support this
assumption. Firstly, as you know, the orbits of planets, except that of Mercury and
Pluto, lie in nearly the same plane. Such coplanar planetary orbits are consistent with
the formation of nebular disk. Secondly, the rotation of the Sun and revolution of the
planets is along the same direction and their equatorial planes are very close to the
plane of the solar system. Thus, the motion of the Sun and the planets are consistent
with the disk hypothesis.

Your next logical question could be: Can we explain the formation of nebular disk
on the basis of physical principles? It can indeed be done provided we assume that
the interstellar cloud, giving rise to the solar system, had some initial rotational
motion. Having assumed this, we can invoke the principle of conservation of angular
momentum to explain the formation of a disk. As the cloud contracted due to gravity,
each gas and dust particle would come closer to the axis of rotation. To conserve the
angular momentum during the contraction, the particles coming nearer to the axis of
rotation must revolve faster. At some point, due to the increased speed of revolution, 33
The Solar System the centripetal acceleration of the particles of the cloud balance the gravitational
and Stars contraction and equilibrium is attained.

To understand the flattening of the rotating cloud, refer to Fig. 6.4. Particles at points
A (near the pole) and B (near the equator) have the same angular momentum because
points A and B are equidistant from the axis of rotation. Thus, it is energetically more
favourable that particle at A falls under gravity along the line AB and reaches the point
B (closer to the centre) than say a particle at point C to come to point B. It is so
because, in the first situation, no change in angular momentum is involved. Thus,
rotating cloud near the poles contracts more than those near the equator giving rise to
the formation of disk.

(a) (b) (c)

Fig.6.4: a) Interstellar cloud with initial rotational motion; b) due to gravitational attraction,
material at the poles contracts along the rotation axis; and c) the final disk shaped solar
nebula

The principle of the conservation of angular momentum also explains why most of the
planets rotate along the same direction. It is because each planet has retained some of
the angular momentum of the solar nebula. Let us discuss the angular momentum of
the solar system to appreciate this point better.

Angular Momentum of the Solar System

The present distribution of angular momentum in the solar system seems inconsistent
with the basics of the nebular model. The argument goes like this. Since the Sun and
the planets are formed from the same spinning interstellar cloud, the angular
momentum per unit mass of each of them must be same. The facts are otherwise: the
Sun possesses approximately 99.9% of the total mass of the solar system but only 1%
of its total angular momentum; orbital angular momentum of Jupiter exceeds the
rotational angular momentum of the Sun by a factor of 20; among the planets having
99% of the total angular momentum of the solar system, Jupiter possesses the most of
it! To appreciate these facts and figures, you should solve an SAQ.

Spend SAQ 3
5 min.
Calculate the total angular momentum of the Sun-Jupiter system assuming that Jupiter
has a circular orbit of radius 5.2 AU, and its orbital period is 11.86 yr. Assume that the
Sun interacts only with Jupiter.

You may ask: How does the nebular model explain the inconsistency in the
distribution of angular momentum? It is proposed that during the formation of the
34 solar system, the angular momentum is transported from the central part of the nebula
to the outer regions. Two possible mechanisms for such a transfer have been The Solar Family
suggested. According to the first mechanism, the interaction of the charged particles
and the magnetic field of the evolving Sun causes transfer of angular momentum. You
know that charged particles spiral along the magnetic field. Thus, the charged
particles created by ionisation of solar nebula by the Sun are dragged along by the
magnetic field of the rotating Sun (Fig. 6.5). The magnetic field links the outer
nebular matter (charged particles) with Sun’s rotation. As a result, the nebular
material in the outer region gains angular momentum at the expense of the Sun. The
other mechanism suggested for transfer of angular momentum is based on viscosity.
You know that due to viscosity, motion of fluid in one part is affected by the motion
of the fluid in the adjoining part. Thus, it is quite possible that the slow moving
particles located at outer edges of the nebula gain in velocity due to their interaction
with the fast moving particles in the smaller orbits of the nebula and vice-versa. This
may cause transfer of angular momentum from the Sun to the outer planets of the
solar system.

Now, you will learn about formation and evolution of planets according to the nebular
model.

Formation of Planets

You know that the solar nebula consisted of gas and dust. As the solar nebula Fig.6.5: Magnetic lines of force
of the rotating Sun
contracted, it became hot enough and most of the dust particles evaporated. So, the
solar nebula consisted mainly of gaseous matter. The question is: How did the
planets form from the solar nebula? The formation of planets is a two-stage
process: firstly, small solid particles are formed from the gaseous matter and then
these particles stick together and grow into planets.

Depending upon the temperature of the nebular region, the nebular gas condensed into
solid matter of different types. In the inner region, the temperature was very high and
materials with very high melting points were formed. The sequence of condensation
of gas into different types of materials, from the centre of the nebula (the Sun) to its
periphery, is called the condensation sequence and is given in Table 6.4.

Table 6.4: The condensation sequence

Temperature Material(s) formed Planet

(K) (Temperature of formation)
With the help of computers, it
is possible to study the
1500 Metal oxides Mercury (1400 K) behaviour of a large number of
particles when they interact in
1300 Metallic Iron and the manner as in the process of
Nickel accretion. Such computer
experiments are called
1200 Silicates
simulation. The results of such
simulation studies suggest that
1000 Feldspars Venus (900 K) there could be as many as 100
planetesimals of the size of the
Moon along with a large
680 Troilites (FeS) Earth (600 K)
number of smaller objects in
Mars (450 K) the region of inner planets. In
addition, there could be about
175 H2O, Ice Jovian (175 K) 10 bodies with masses
comparable with the planet
Mercury and many other
65 Argon-neon Ice Pluto (65 K) objects of the size of the planet
Mars.

From Table 6.4, it is evident that terrestrial planets formed from high density materials
and jovian planets formed from low density materials.
35
The Solar System Having discussed the general chemical composition of planets, we now focus on the
and Stars evolution of planets. The evolution of planets involves three stages:

a) The first stage involves the growth of macroscopic grains of solid matter from the
interstellar cloud. The size of these grains range from a few cm to a few km and
they are called planetesimals. Planetesimals can grow through two processes:
condensation and accretion. In the condensation process, grains grow by adding
one atom at a time to a ‘nucleus’ atom, from the surrounding gaseous cloud. This
is similar to the growth of snowflakes in the Earth’s atmosphere. In the accretion
process, solid particles stick together. Further, the planetesimals would tend to
rotate in the plane of the solar nebula.

b) In the second stage, planetesimals coalesce and form protoplanets − objects

having planetary sizes and masses. You may ask: How do the protoplanets
form? Since all the planetesimals are moving along the same direction in the
nebula, they collide with each other at a low relative velocity and stick together to
form protoplanets. Further, growth of protoplanets is helped by gravity because
the nebular matter is attracted by the protoplanets.

c) At the third stage, when a protoplanet grows into a stable planet, a large amount
of heat is generated in its core due to the decay of short-lived radioactive
elements. Heat is also generated due to collision of these planets with other
objects. Due to high temperature, the planets melt and facilitate the process of
gravitational separation in which materials in the planet segregate themselves
according to their density. Therefore, the inner regions of the planets hold heavier
elements and compounds and lighter elements are pushed to the surface.

This, in the nutshell, is the ‘story’ of planet building! You must have noted that
The creation of a massive there are many unanswered questions and many ifs and buts in the mechanism
object, like Jupiter, described above. To understand the details of formation of planets is still an active
influences the orbits of
nearby planetesimals. Most area of research in astronomy and astrophysics. We will, however, not go into
of the objects present in the those details and close our discussion with a few words about the success of the
asteroid belt today had their nebular model.
orbits changed gradually into
more eccentric orbits till they The solar nebular model successfully accounts for the following three important
were sucked in by Jupiter. It
is also likely that they either features of the solar system:
left the solar system or
crashed into the Sun. This i) Disk shape of the solar system: The model suggests that the rotating solar
process might in fact have nebula ultimately evolves into disk shape due to gravitational contraction and
resulted in the smaller mass conservation of angular momentum.
left in the asteroid belt and
also a smaller planet Mars. ii) The orbits of most of the planets are coplanar: As per the model, this situation
exists because of the disk shape of the solar nebula from which planets are
formed.
iii) The direction of rotation of the Sun and the directions of revolution of most of
As we go further from Jupiter the planets are the same: It is so because the Sun and the planets formed from
and beyond in the nebula, the the same rotating nebula.
material density becomes
very low. So, the accretion
process for the formation of a You have also learnt that the condensation sequence of the solar nebula explains why
planet-like object takes much terrestrial planets comprise of compounds having high melting point and jovian
longer time. Saturn perhaps planets comprise of ices and gases.
took twice as long to form as
Jupiter, while the planet The nebular model has some very obvious limitations. You know that the Moon’s
Uranus took an even longer
time. The planet Neptune is surface is not smooth; it has craters of varied sizes, small and large. Also, its surface
estimated to have taken twice composition is extremely poor in hydrogen, helium etc. These observations suggest
the time it took to form the continued collisions of bigger planetesimals with the Moon’s surface even after its
Uranus. formation. It is quite likely that other objects (planets) in the solar system experienced
similar collisions with planetesimals. Thus, the theory of solar system formation must
36 account for massive encounters endured by planets in the early stages of their
formation. These massive collisions or encounters are possibly responsible for The Solar Family
different orientation of the spin axis of the planets in the solar system. It is now known
that Venus, Uranus and Pluto have retrograde rotations.

The cause for tipping of rotation axis, retrograde rotation etc. of planets cannot be
understood as such by gradualistic models like the nebular model. We need to
consider catastrophic events. During the formation of the solar system, the wandering
planetesimals did perhaps collide with other planets such as Mercury, Venus, Earth
etc. The collision with Mercury ripped off its low density mantle while such a
collision with Venus flipped its rotation axis. A collision with Earth led to the
formation of the Earth-Moon system. Further, massive planetesimals crashed into
Mars as well as other outer planets changing orientation of their rotation axes. It is
also possible that some of these planetesimals were captured by planets as their
Moons.

Now, before you proceed further, how about answering an SAQ?

SAQ 4 Spend
7 min.
a) List two evidences supporting the assumption that a disk shaped solar nebula
existed during the evolution of the solar system.

b) Why do terrestrial planets comprise mainly of materials having high melting

points?

c) What is the difference between the condensation and accretion processes?

6.4 TIDAL FORCES AND PLANETARY RINGS

You must be aware of tides occurring on the Earth’s surface. In fact, those of you who
live in the coastal areas must be quite familiar with tides. Two high tides occur once in
every ~ 24 hours depending on the local features of the coastal area and its latitude.
However, generally people are not familiar with the tidal bulge of the Earth along the
equator which measures around 10 cm. Do you know what causes tides or tidal
bulge of the Earth? It is caused due to the difference between gravitational force of
the Moon at different locations, say points A and B, on the Earth (Fig. 6.6). Due to
similar reasons, tidal bulge (as large as 20 m) caused by the Earth has also been
observed on the Moon. The larger value of the bulge of the Moon is because the Earth
is more massive than the Moon.

Tidal effects play an important role in astronomy, particularly in understanding the

creation of large number of satellites and ring systems in the jovian planets. We now
derive an expression for the tidal force for the Earth-Moon system. Then we shall use
this concept for a qualitative explanation of the formation of planetary rings.

A B

Moon
Earth

Fig.6.6: Tidal bulge of the Earth is due to the difference in gravitational force of the Moon
experienced by mass elements at, say points A and B, on the Earth.

37
The Solar System Let a mass m on the Earth be located at a distance r from the centre of mass of the
and Stars Moon (Fig. 6.7). If the mass of the Moon be M, the magnitude of the gravitational
force acting on m, in the direction shown, is given by:

Mm
F =G , (6.1)
r2

where G is the gravitational constant.

r dr
M
m m
Moon
Earth

Fig.6.7: Gravitational force on an element of mass m located on the Earth at a distance r from the
centre of mass of the Moon

Now consider a similar mass located at a distance dr from the earlier mass element
along the same line. The difference between the gravitational forces experienced by
these two mass elements can be obtained by differentiating F in Eq. (6.1):

Mm
dF = −2G dr (6.2)
r3

Note that the difference in the gravitational forces on the two equal masses separated
by a distance dr will cause them to move with different accelerations. The differential
force given by Eq. (6.2) is called the tidal force. The r3 term in the denominator of
Eq. (6.2) clearly shows that the tidal force has a stronger dependence on distance
compared to the gravitational force (which varies as 1/r2). Further, in case of the
Earth-Moon system, the tidal force becomes more pronounced if the mass element is
closer to the Moon.

Now to appreciate the effect of the tidal force of the Moon on mass elements located
at different points on the Earth, e.g., at the equator and at the poles, let us obtain a
general expression for the tidal force considering the Earth as a two-dimensional
object. Let us consider an element of mass m located at the centre (C) of the Earth and
another element of mass m located at an arbitrary point (P) at latitude φ on the surface
of the Earth (Fig. 6.8). By assuming that the Moon lies along the x-direction, the
components of gravitational force at points C and P can be written as:

Mm
FC , x = G ,
r2

FC , y = 0,

and

Mm
FP, x = G cos θ ,
s2

Mm
FP , y = −G sin θ. (6.3)
38 s2
where s is the distance between points P and the centre of mass of the Moon. The Solar Family

x
P
θ
R
s
φ
θ
Earth
C
r
Moon

Fig.6.8: The schematic diagram of the tidal force on an arbitrary point on the Earth due to Moon

Let the unit vectors along the x- and y-directions be î and ĵ , respectively. Thus, from
Eq. (6.3), the difference between the magnitudes of gravitational forces at the points P
and C can be written as:

∆F = FP − FC

 cos θ 1   sin θ 
= GMm  2 − 2  ˆi − GMm  2  ˆj (6.4)
 s r   s 

Further, s can be expressed in terms of r, R and φ as:

s 2 = (r − R cos φ) 2 + R 2 sin 2 φ

 2R 
≈ r 2 1 − cos φ 
 r 

Substituting for s2 in Eq. (6.4) using the small angle approximations, cosθ ≈ 1 and
R
sinθ ≈ sin φ, and using binomial expansion we obtain:
r

GMmR
∆F ≈ (2 cos φ ˆi − sin φ ˆj) (6.5)
r3

SAQ 5 Spend
6 min.
Derive Eq. (6.5).

You may ask: What does Eq. (6.5) signify physically? This expression clearly
indicates that the magnitude of the tidal force is dependent on the latitude. For mass
element located at the equator of the Earth, that is, for φ = 0, the magnitude of tidal
force is maximum. And, at the poles, where φ = π/2, the magnitude of the tidal force is
minimum. This is the cause of tidal bulge around the equator and also causes tides in
the oceans.
39
The Solar System In the above discussion, we have considered the Earth-Moon system to be an isolated
and Stars one. In fact, we must also consider the tidal forces due to the Sun. When the Sun,
Earth and the Moon are all aligned in a straight line, the differential forces discussed
above add up to produce large tides on the Earth. These tides are called spring tides.
When the Sun, Earth and the Moon form a right angle, the differential forces due to
the Sun and the Moon are directed opposite to each other and the tides on the Earth
(known as neap tides) are very small.

Yet another effect of tidal force is that the Earth’s speed of rotation is slowed down.
This results in longer days at present compared to many years ago. Further, just as the
Moon causes tides on the Earth, the Earth also gives rise to tides on the Moon. You
know that on the Earth we see the same side of the Moon which implies that its
rotation and revolution periods are the same. It is quite likely that earlier the Moon’s
rotation period was shorter than its orbital period. As time progressed, because of the
tidal friction, the rotation period of the Moon increased and has become equal to its
orbital period (also called one-to-one synchronous rotation). We find such
synchronous motion to be quite a common phenomenon in the solar system. For
instance, the two Moons of Mars and four Moons of Jupiter and majority of Saturn’s
Moons are in synchronous rotation. Many moons of the outer planets behave in a
similar manner.

You know that most outer planets of the solar system have rings around them
comprising of small (~ 10 µm to 10−5m) particles. The concept of tidal forces can be
invoked to understand these ring systems. There are two possible scenarios. Since the
jovian planets are massive, tidal gravitational force of the planets on their satellites
must be very strong. Thus, if a satellite comes very close to, say, Jupiter, it
experiences a strong tidal force and breaks up into pieces. The particles formed during
such a process revolve around the planet in ring formation. According to the second
scenario, during the formation of jovian planets, the tidal forces restrained the
particles from condensing into a satellite.

You may argue: The tidal forces must ultimately disrupt the ring system; what
maintains the ring system? It is suggested that the Moons associated with these
planets play an important role in preserving the rings. The particles in the rings are
restrained from moving out of their orbit due to the combined gravitational forces of
these Moons.

Human beings have always wondered about the existence of planets around other
stars. Only in recent times there has been a confirmation that planets do exist outside
the solar system. These planets are called extra-solar planets. You will learn about
them now.

6.5 EXTRA SOLAR PLANETS

On August 4, 1997, the Hubble Space telescope took the first image of an extra-solar
planet around another star (Fig. 6.9). The picture shows a double star located at about
450 light years away towards the constellation of Taurus. It was the first direct look at
a planet outside our solar system. It is now believed that this planet is a “runaway”
object, thrown out of the binary system, as indicated by the filament tracing. The
planet is located at about 1500 times the Earth-Sun distance from its parent. The
Fig.6.9: Photograph of an binary system is believed to be 300,000 years old and the planet seems to be quite
extra solar planet similar to Jupiter with a mass of around 2-3 times that of Jupiter. This discovery led to
taken by Hubble
telescope careful investigation by several groups of astrophysicists and by now more than 100
such extra solar planets have been found.

Young planets in the new systems are difficult to detect as the parent stars
40 overshadow their feeble glow. Since it is difficult to detect a planet orbiting a distant
star, we look for alternative methods of detection, such as the influence of the planets The Solar Family
on their parent stars. You may like to know: What are these influences? It is easy to
think of gravitational influence. As the planet orbits the star, it will tug at it from
different sides causing it to wobble back and forth. The gravitational influence gives
rise to the following methods for detection of the planet:

a) Astrometric Detection
b) Radial Velocity Detection

The first method is based on measuring the position of a star relative to its background
stars. If the star is accompanied by an orbiting planet, it gets a tug from the planet and
its position changes a little. This change in the position of the star is measured which
would show a periodic change (back and forth) indicating the presence of an orbiting
object.

The second method is based on Doppler effect. As an orbiting planet tugs on to its
companion star, the light from the star experiences a Doppler shift. If the planet pulls
the star slightly away from us, the light emitted by star would shift towards red end of
the electromagnetic spectrum while if it tends to pull the star towards the Earth, the
light would be shifted towards the blue end of the spectrum. To measure the Doppler
shift, we choose a particular spectral line and observe its shift from red to blue and
back.

In 1995, Michel Mayor and Didier Qucloz of the Geneva Observatory observed that
the Sun-like star, 51 Pegasi is wobbling back and forth at the rate of 56 ms−1. The only
valid explanation for this observation was the presence of a planet like object orbiting
the parent star. The mass of the planet was estimated to be half the mass of Jupiter
with a radius of about 0.05 AU. Subsequently, Mayor also detected a planet of about
0.16 times the mass of Jupiter orbiting the star HD 83443 in the constellation of Vela
about 141 light years away from the Earth.

Recently David Charbonneau, Timothy Brown and Robert Gilliland used the Hubble
spectrometer and made the first direct detection and chemical analysis of the
atmosphere of the planet HD 209458b orbiting a yellow, Sun-like star, HD 209458, in
the constellation of Pegasus. The mass of the planet is estimated to be 70% of the
mass of Jupiter. It passes in front of its star every 3.5 days and contains sodium in its
outer layers. The extremely short period of the planet suggests its very close proximity
to the star and therefore its atmosphere getting heated to around 1100 degrees Celsius.

Many new results are pouring in and are compelling astronomers to have a re-look on
the theories of solar system formation. The search for extra-solar planets is going hand
in hand with the search for life in the Universe. Exploration of the solar system has so
far not revealed any signs of even primitive life on any of the planets or their
satellites. Scientists strongly believe that there must be a large number of extra-solar
planets where conditions are suitable for the existence of life. Indeed, some of these
planets could be hosting life more advanced than our own. It is possible that some of
these intelligent beings are trying to contact us just as we are looking for them. It must
be remembered that any message from any of these beings would be coded in radio
waves. That is why SETI, an organisation searching for Extra Terrestrial Intelligence,
is looking for such signals in the radiation at radio frequencies coming from outside
the solar system. Scientists are hopeful that someday they would be able to detect
extra-solar intelligence.

Let us now summarise what you have learnt in this Unit:

41
The Solar System
and Stars
6.6 SUMMARY
• The first four planets of the solar system namely Mercury, Venus, Earth and Mars
are called terrestrial planets and they mainly comprise of solid matter. The next
five planets, namely Jupiter, Saturn, Uranus, Neptune and Pluto are called jovian
planets and they mainly comprise of ices and gases.

• One of the significant features of jovian planets is the ring system.

• Some of the characteristic properties of the solar system are:

i) Most of the mass of the solar system is contained in the Sun.
ii) Orbital planes of all the planets except Mercury and Pluto are coplanar.

iii) The direction of rotation of the Sun and direction of revolution of the planets
is the same.
iv) Total angular momentum of all the planets is more than the angular
momentum of the Sun.
• According to the nebular hypothesis of the origin of solar system, the Sun as
well as the planets formed from an interstellar cloud of gas and dust. A model
based on this hypothesis is called nebular model.
• According to the nebular model, formation of stars including the Sun takes place
when the interstellar cloud contracts due to its own gravity.
• The contracting interstellar cloud takes a disk shape due to its rotational motion.
Formation of nebular disk causes coplanar planetary orbits.
• The present distribution of angular momentum in the solar system seems
inconsistent with the basics of the nebular model.
• The angular momentum problem is addressed by proposing that there is a transfer
of angular momentum from the inner to outer regions of the solar system during
its evolution.
• The formation of planets results due to condensation of nebular gas as per the
condensation sequence.

• The formation of planets involves three stages:

i) Formation of planetesimals,
ii) Formation of protoplanets, and
iii) Stabilisation of the planet.

• Tidal forces result due to the difference in gravitational force at two points on the
Earth. For the Earth, tidal forces due to the Moon have a significant effect. Tidal
force at the equator is maximum and at the poles, its value is minimum. The
general expression for the tidal force is given by

GMmR
∆F ≅ (2 cos φ ˆi − sin φ ˆj)
r3

• In the recent past, a few extra-solar planets have been detected.

6.7 TERMINAL QUESTIONS Spend 20 min.

42
1. Explain, in your own words, the theory of the solar system formation based on The Solar Family
nebular hypothesis.
2. Explain how angular momentum can be transferred from the Sun to the outer
planets of the solar system.

3. Is it possible that the Earth also suffered collisions with other bodies of the solar
system and its surface also had craters like those on the surface of Mercury? Can
you guess what happened to those craters?

4. Explain how extra-solar planets can be detected. Why can they not be seen
directly?

6.8 SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs)

1. To verify Kepler’s third law, let us calculate (T 2 r 3 ) for three representative

planets, say the Mercury, the Earth and the Saturn. From Table 6.1, we have

TMercury = 88 days = 88 × 24 × 60 × 60 s = 7.6 × 106s

and
rMercury = 5.8 ×10 7 km = 5.8 ×1010 m
So,
(TMercury ) 2 (7.6 ×10 6 s) 2
=
(rMercury ) 3 (5.8 ×1010 m) 3
= 2.9 ×1019 m − 3s 2

Similarly, using data from Table 6.1, you can easily show that:

(TEarth ) 2
= 2.9 ×1019 m − 3 s 2
3
(rEarth )
and
(TSaturn ) 2
= 2.9 ×1019 m − 3s 2
3
(rSaturn )

Since the ratio (T 2 r 3 ) for all the three planets has the same value
2.9 ×1019 m − 3s 2 , Kepler’s third law is true.

2. See text.

3. For two bodies of mass M1 and M2 moving around their centre of mass, the
reduced mass is given by:

M 1M 2
µ=
M1 + M 2

So, for the Sun-Jupiter system, we can write:

43
The Solar System MΘ M J
and Stars µ=
MΘ + M J
≈ MJ

Using Kepler’s third law, it can be readily shown that the angular momentum L
and the period T of the orbiting mass are related as:

2πabµ
L= .
T

where a, and b respectively are the semi-major and semi-minor axes of the
elliptical orbit of the planet. As per the problem, the orbit of Jupiter is to be
considered circular. Thus, we can write:

a = b = 5.2 AU

= 5.2 × 2.279 × 1011m

Substituting the values of MJ = 2 × 1027 kg and T = 11.86 yr, we get:

L=
(
2 × 3.14 × (5.2 × 2.279 × 1011 m) 2 × 2 × 10 27 kg )
(11.86 × 365 × 24 × 3600 s)

17.63 × 10 51 kg m 2
=
4.44 × 10 9 s

= 3.9 × 10 42 kg m 2 s −1

4. See text.

5. Refer to Fig. 6.8. From Eq. (6.4), we have the difference between the magnitudes
of gravitational force at points P and C as:

∆ F = FP − FC

 cos θ 1 ˆ  sin θ  ˆ
= GMm  −  i − GMm   j
 s2 r2   s2 
s 2 = r 2 + R′ 2
= r 2 + R 2 sin 2 φ
= (r − R cos φ) 2 + R 2 sin 2 φ
2 2 2 2 2
≅ r − 2r R cos φ + R cos φ + R sin φ
= r 2 − 2r R cos φ + R 2
 2R 
≈ r 2 1 − cos φ  (neglecting higher order terms in R)
 r 

Substituting for s2 and making small angle approximations, cos θ ≈ 1 and

R
sin θ ≈ sin φ, we can write:
r

44
 The Solar Family
   R 
   sin φ 
 1 1 ˆ  r ˆ
∆ F = GMm  −  i − GMm  j

 r 1 − 2 R  2
 
 r 1 − 2 R 
cos φ  
2 r 2
  cos φ    
  r     r 

GMm  r   R sin φ  ˆj

 − 1 ˆi −  
2  r − 2 R cos φ
=
r    r − 2 R cos φ  

GMm  2 R cos φ  ˆ  R sin φ  ˆj

=   i −  
r 2  r − 2 R cos φ   r − 2 R cos φ  

≅
GMm R
r3
(
2 cos φ ˆi − sin φ ˆj ) (because r>> 2 R cos φ).

Terminal Questions

1. See text.
2. See text.
3. See text.
4. See text.

45
The Solar System
and Stars UNIT 7 STELLAR SPECTRA AND
CLASSIFICATION
Structure
7.1 Introduction
Objectives
7.2 Atomic Spectra Revisited
7.3 Stellar Spectra
Spectral Types and Their Temperature Dependence
Black Body Approximation
7.4 H-R Diagram
7.5 Luminosity Classification
7.6 Summary
7.7 Terminal Questions
7.8 Solutions and Answers

7.1 INTRODUCTION
In Unit 5, you have studied about the Sun and in Unit 6, you have learnt about the
solar system. You also learnt the characteristic features of the solar atmosphere and
solar activity. You now know that the Sun is the nearest and the only star in our solar
system. All the stars, including the Sun are located at very large distances from the
Earth. Thus, a logical question is: How do we obtain information about the stars? All
the information about stars is obtained by analysing their spectra and this is the
subject matter of the present Unit.

You may recall from Thermodynamics and Statistical Mechanics (PHE-06) course
that the radiation emitted by an object at a given temperature covers a range
(spectrum) of wavelengths with a characteristic peak wavelength. The value of the
characteristic wavelength depends on temperature of the object. Therefore, by a
careful analysis of the radiation emitted by a star, we can estimate its temperature. In
addition, we can also obtain useful information regarding composition, pressure,
density, age etc. of a star on the basis of its spectrum. In Sec. 7.2, we briefly
recapitulate the atomic origin of emission and absorption spectra and explain how the
temperatures and luminosities of stars can be inferred from their spectra.
One of the earliest uses of stellar spectra was to classify stars on the basis of strength
of certain spectral lines. Later on, it was discovered that the relative strengths of
spectral lines depend basically on the star’s temperature. The spectral classification
has been discussed in Sec. 7.3. The most comprehensive classification of stars was
done on the basis of the correlation between their luminosities (an observable
parameter of stars) and temperatures. This classification gave rise to Hertzsprung-
Russell (H-R) diagram about which you will learn in Sec. 7.4. This diagram is of
utmost importance in astronomy. In Sec. 7.5, you will learn yet another classification
of stars, called luminosity classification, which tells us about the size of a star.
Objectives
After studying this unit, you should be able to:
• explain the atomic origin of emission and absorption spectra;
• discuss the correlation between the strength of spectral lines of an element and the
temperature of the region containing the elements;
• list the spectral types of stars and their characteristic features;
• discuss the salient features of different groups of stars on the H-R diagram;
• describe the need for luminosity classification of stars; and
46 • estimate the size of a star on the basis of its luminosity class.
 Stellar Spectra and
7.2 ATOMIC SPECTRA REVISITED Classification

To appreciate the relation between the physical parameters of an object and the
radiation emitted by it, let us imagine heating an iron bar. First, the iron bar begins to
glow with a dull red colour. As the bar is heated further, the dull red colour changes to
bright red and then to yellowish-white. If we can prevent the bar from melting and
vaporising with further increase of temperature, it would glow with a brilliant bluish
colour. This simple experiment reveals how the intensity and colour of light emitted
by a hot object varies with temperature. Similar is the situation with stars. You must
have noticed, while looking at night sky, that all stars do not look the same in terms of
their brightness or colour: the differences are determined by their surface
temperatures.

If you happen to look at Orion

nebula (Fig. 7.1), it has not only
blue stars but also some red
stars. Similarly the brightest
visible star to the naked eye −
Sirius (Vyadh) − looks whitish;
Canopus (Bramhahrudaya) is
yellowish and Aldebran
(Rohini) is reddish.

Fig.7.1: The Orion nebula

From Unit 3, you know that spectroscopy refers to the analysis of light in terms of its
wavelength and it is extensively used to obtain information regarding temperature,
composition etc. of stars. Such analyses provide valuable information. But, the
question is: How do we analyse the light from a star? To answer this, we need to
recapitulate the atomic origin of emission and absorption of light. Recall from the
Modern Physics (PHE-11) course that:

i) Electrons in an atom exist in certain allowed energy states. Each

element has a characteristic set of energy levels (see Fig. 7.2).

0 0 0

−0.85 eV n=4
−1.50 eV n=3

−3.40 eV n=2

−13.60 eV −25eV −25eV

n=1
(a) (b)

Fig.7.2: Energy level diagram of a) hydrogen atom; and b) helium atom

47
The Solar System ii) When an electron makes a transition from a higher energy level to a lower
and Stars
energy level, radiation is emitted. An electron goes from a lower energy level
to a higher energy level when it absorbs energy and vice-versa (Fig. 7.3). Such
transitions follow certain selection rules and are always accompanied by the
emission/absorption of radiation. You know from Unit 9 of the PHE-11 course
that the wavelength (or frequency) of the emitted/absorbed radiation is
determined by the difference in the energies of the two atomic energy levels:

∆ E = E2 − E1 = hv (7.1)
where E1 and E2 are the energies of the levels involved in transition, h is Planck’s
constant , v is the frequency of the emitted/absorbed radiation and ∆ E is the
energy of the corresponding photon.

etc…
E4 Level 4

E3 Level 3

E2 Level 2
Energy

Bright emission line

E1 Level 1

(a)

etc…
Level 4
E4
Level 3
E3

E2 Level 2
Energy

Dark absorption line

E1 Level 1

(b)
Fig.7.3: The atomic electron a) emits radiation when it makes a transition from a higher energy
level to a lower energy level giving rise to emission spectrum; and b) makes a transition to
a higher energy level on absorbing radiation giving rise to absorption spectrum

To understand the origin of atomic spectra, refer once again to Fig. 7.3 which shows
the energy level diagram and transition of electron between these levels for a
hydrogen atom. In Fig. 7.3(a), a hydrogen atom makes a transition from the 2nd
energy level to the 1st, giving off light with an energy equal to the difference of
energy between levels 2 and 1. This energy corresponds to a specific colour or
wavelength of light -- and thus we see a bright line at that exact wavelength! This is
an emission spectrum.

On the other hand, what would happen if we tried to reverse this process? That is,
what would happen if a photon of the same energy was incident on a hydrogen atom
in the ground state? The atom could absorb this photon and make a transition from the
ground state to a higher energy level. This process gives rise to a dark absorption
line in the spectrum as shown in the figure. This is an absorption spectrum.
48
If light from a star with a continuous spectrum is incident upon the atoms in its Stellar Spectra and
Classification
surrounding atmosphere, the wavelengths corresponding to possible energy transitions
within the atoms are absorbed. Thus, an observer will see an absorption spectrum.

−0.54eV
n=∞ 0 eV
−0.85eV n=4
−1.51eV n=3
−3.39eV n=2

Lyman series

−13.6eV
n=1 1215 1025 972 950 937
λ (Å)
(a)

n=4
n=3

n=2
Balmer series

4000 5000 6000 7000

λ (Å)

n=1
(b)
Fig.7.4: Emission of radiation by hydrogen atoms corresponding to Lyman and Balmer series:
a) Transitions to level n=1 gives Lyman series; and b) Transitions to n=2 level gives Balmer
series

On the basis of above, we can understand the genesis of atomic spectra. Let us take
the simplest example of the spectrum of hydrogen atom. In Fig. 7.4 we show the
various spectral series for the hydrogen atom. The transition of electrons from higher
energy levels to the lowest (n = 1) energy level gives rise to emission of a series of
characteristic wavelengths known as Lyman series. Similarly, transition of electrons
from higher energy levels to the first excited (n = 2) energy level results into the
emission of another series of characteristic wavelengths called the Balmer series.
You may note that some of the spectral lines in Balmer series fall in the visible region
of the electromagnetic spectrum (Fig. 7.4b). This makes them very useful spectral
lines for spectroscopic analysis. Further, transitions to the second excited (n = 3)
energy level give rise to the Paschen series.
Emission Spectrum
Now, let us consider the nature of emission of radiation by hydrogen gas (a collection
of many hydrogen atoms) kept at a high temperature. Each atom in the gas can
49
The Solar System absorb thermal energy as well as the energy due to collisions with other atoms. As a
and Stars
result, each atom would absorb a different amount of energy. Therefore, the electrons
in the lowest energy level of various atoms get excited to different higher energy
levels (Fig. 7.5). The question is: What kind of emission spectrum would we obtain
in such a situation?

etc… etc…
Level 4 Level 4
Level 3 Level 3

Level 2 Level 2
Energy Energy

Level 1 Level 1

Fig.7.5: At high temperatures, electrons in hydrogen atoms get excited to various higher energy
levels; transitions to levels 2 and 3 are shown here as examples

To answer this, let us consider a hydrogen atom in this hot gas whose electron has
been excited to the second excited energy level (Fig. 7.6). This electron can return to
the lowest energy level either directly (Fig. 7.6a) or in steps. For example, it can go to
the first excited energy level and then to the lowest level (Fig.7.6b). In the two cases,
the energies of the emitted photons would be different. Thus, in a gas of hydrogen
atoms at high temperature, electrons can be excited to many possible levels and can
make transitions to any of the lower levels emitting radiations of the corresponding
frequencies. This implies that photons of different frequencies will be emitted and we
will observe more than one emission line in the emission spectrum of a hot gas.

etc… etc…
Level 4 Level 4
Level 3 Level 3

Level 2 Level 2
Energy Energy

Level 1 Level 1

(a) (b)

Fig.7.6: Different ways of transition of electron in a hydrogen atom from an excited energy level to
the lowest energy level

However, you must remember that the spectrum of hydrogen gas is not continuous,
that is, not all frequencies will be emitted; radiations of only those frequencies will be
emitted whose energies are equal to the difference in energy between the allowed
energy levels. Therefore, the emission spectrum of hydrogen gas contains bright lines
of certain definite frequencies.

The frequencies contained in the emission spectra of each element are unique:
no two elements have the same type of emission spectra because each one of
them has a characteristic set of allowed energy levels.
50
Absorption Spectrum Stellar Spectra and
Classification
Let us assume now that the hydrogen gas is not at a high enough temperature to excite
the electrons to higher energy levels. In such a situation, there is no emission of
radiation. If white light (continuous spectrum) is passed through this cool hydrogen
gas, it will absorb light of only those frequencies which give electrons just enough
energy to make a transition to one of the excited energy levels (see Fig. 7.3b).
Therefore, just as photons of certain energies are emitted by hydrogen gas at a high
temperature, photons of the same energies are absorbed by the cool hydrogen gas.
You can argue that atoms absorbing photons and giving rise to absorption spectrum
will eventually re-emit photons of the same frequencies. Thus, it should cancel out the
effect of absorption and we could not observe any absorption spectrum! The question,
therefore, is: How to explain the observed absorption spectrum? The re-emitted
photons are repeatedly absorbed by the atoms of the cool gas before they finally
escape it. These re-emitted photons escaping from the gas travel along different
directions. Thus, the photons of certain frequencies, which were originally coming
towards the spectroscope as a part of white light, are scattered by the atoms of the
cool gas. As a result, the intensities of the light corresponding to these frequencies are
very low and we observe an absorption spectrum.
Thus when white light passes through a cool hydrogen gas, we obtain a series of dark
lines and such a spectrum is called the absorption spectrum.

Do remember that for any element, the dark lines of absorption spectrum occur
at exactly the same frequencies (or wavelengths) where we observe bright lines
in its emission spectrum.

Now, before proceeding further, you should answer an SAQ to fix these ideas.

SAQ 1 Spend
5 min.
a) Why do the spectra of different elements have different sets of lines?
b) What do you understand by a continuous spectrum? Under what condition do we
observe it?

Let us pause for a moment and think: How does the knowledge about the origin of
spectral lines help in analysing stellar spectra? Firstly, you have learnt that each
element produces a unique set of spectral lines. Therefore, observing the spectrum of a
star, we can tell which element(s) are present in its atmosphere. Secondly, the nature Strictly speaking, the stellar
spectra tell us about the
of spectrum  emission or absorption  depends on the temperature prevailing in the composition of the star’s
atmosphere through which star light passes. Thus, the nature of spectrum gives us an atmosphere. However, it is
idea about the temperature of the star’s atmosphere. generally believed that the
composition of the
In addition, stellar spectra also provide information about the density of the star’s atmosphere reflects the
composition of the star itself.
atmosphere and the motion of stars with respect to the Earth. Whether a star is moving
towards or away from the Earth can be inferred on the basis of Doppler effect:

According to the Doppler effect, there is an apparent change in the wavelength

of radiation emitted by a star due to its (star’s) motion; if the star is moving away
from the Earth, the wavelengths are shifted towards longer wavelengths (the
phenomenon is known as red shift) and if the star is moving towards the Earth,
the wavelengths are shifted towards shorter wavelengths (the phenomenon is
known as blue shift).

With this background information, you are now ready to study stellar spectra. In this
context, it is useful to remember a set of empirical rules of spectroscopic analysis
given below:
51
The Solar System i) A hot and opaque solid, liquid or highly dense gas emits a continuous spectrum;
and Stars
Fig. 7.7a (you have learnt about optically thin and thick media in Unit 4).
ii) A hot gas produces emission spectrum and the number and position of the
emission lines depends on the composition of the gas (Fig. 7.7b).
iii) If light having continuous spectrum is passed through a gas at low temperature, an
absorption spectrum consisting of dark lines is produced (Fig. 7.7c). Again, the
position and number of dark lines are characteristic of the elements present in the
gas.
iv) When light with continuous spectrum passes through a very hot, transparent gas, a
continuous spectrum with additional bright lines is produced.

Fig. 7.7 shows some of these situations.

Hot light source

Spectroscope
Cloud of gas

Continuous spectrum
with dark lines

Spectroscope Spectroscope (c)

Continuous spectrum Bright-line spectrum

(a) (b)

Fig.7.7: Diagrammatic representation of empirical rules for spectroscopic analysis of stellar spectra

7.3 STELLAR SPECTRA

Before discussing stellar spectra, you may like to know how they are obtained using a
spectroscope. Refer to Fig. 7.8 which shows how to obtain a spectrum using a
spectroscope. Light from the telescope (pointing towards the star) is passed through a
slit, a collimating lens, through the prism of the spectroscope to obtain the spectrum.
The resulting spectrum is focussed by another lens and can either be viewed directly
or photographed.

Prism
Red
Light from
telescope
Source slit Violet
Collimating lenses

Fig.7.8: Schematic diagram of a spectroscope

For a casual observer, all stars look alike. But, if observed carefully, some stars appear
bright and some faint. It is usually difficult to make out the colours of stars if they are
52 fainter than the second magnitude. This is due to poor colour sensitivity of the human
eye to fainter light. Stellar spectra, however, show that individual stars differ widely Stellar Spectra and
Classification
from one another in brightness and spectral details. While spectra of some stars
contain lines due to gases like hydrogen and helium, others show lines produced by In earlier times, the spectrum was
exposed on a photographic film.
metals. Some stars have spectra dominated by broad bands of molecules such as The film was then developed in
titanium oxide. the usual way and scanned for
intensity at each point of the film.
A typical spectrum of a star consists of a continuum, on which are superposed dark The data so obtained was plotted
absorption lines. Sometimes emission lines are also present. An example of a typical to obtain the spectrum. Now-a-
stellar spectrum is the solar spectrum (Fig. 7.9). Solar spectrum can be easily obtained days, with the advent of modern
detectors like the CCDs (about
by passing a narrow beam of sun light through a prism. It consists of a continuum which you have studied in Unit
background superposed by dark lines. 3), the spectrum can be directly
stored in a computer and plotted.

Fig.7.9: Solar spectrum

The dark lines in the solar spectrum are called Fraunhofer lines. You know that the You have learnt in Unit 1 that
the brightness of stars is
intensities of spectral lines indicate the abundance of various elements to which the expressed in magnitudes.
lines belong. The important lines in stellar spectra are due to hydrogen, helium,
carbon, oxygen, neutral and ionised metal atoms. Bands in the spectra are caused by
the molecules such as titanium oxide, zirconium oxide, CH, CN, C3 and SiS2.
Similarities in stellar spectra provided the basis for classification of stars into certain
categories. The earliest classification was done by Annie J. Cannon. She classified
more than 2, 50,000 stars by observing the strength of absorption lines, particularly,
the hydrogen Balmer lines. In this way stars have been classified into seven major
spectral types, namely, O, B, A, F, G, K and M.
For greater precision, astronomers have divided each of the main spectral types into
10 sub-spectral types. For example, spectral type A consists of sub-spectral types A0,
A1, A2.... A8, A9. Next come F0, F1, and so on. Thus there are 70 sub-spectral types
possible. However, in practice, all 70 types have not been observed. The advantage of
the finer division is to estimate the star's temperature to accuracy within about
5 percent. The Sun, for example, is not just a G star, but a G2 star, with a surface
temperature of about 5800 K.

7.3.1 Spectral Types and Their Temperature Dependence

Fig. 7.10 shows a set of spectra of seven main types of stars: O, B, A, F, G, K and M
(hottest O-type to coolest M-type). It was shown by M.N. Saha, an Indian scientist,
that Cannon’s spectral classification can be explained primarily on the basis of the M.N. Saha’s conclusion that
temperatures of stars because the intensities of spectral lines depend on the surface Cannon’s spectral classification
temperature of a star. To elaborate this argument, let us take the case of spectral lines is based on surface temperature
of stars is based on the fact that
of hydrogen Balmer series as an example. ionisation of atoms is a process
similar to chemical reaction.
You know that 50 to 80 percent of the mass of a typical star is made of hydrogen. This means that at a given
However, only a small fraction of the stars show Balmer lines in their spectra! It is so temperature, relative number of
atoms in various stages of
because most of the stars do not have sufficient surface temperature to raise the ionisation which are in
electrons in hydrogen atoms to the second and higher energy levels. And, as you equilibrium, is fixed. Further,
know, for production of Balmer lines, electrons in the hydrogen atoms make the intensity of an absorption
transitions from the higher energy levels to the first excited energy level. For example, line is a function of the number
a M type star whose surface temperature is about 3000 K, does not have enough of atoms which can absorb
radiation corresponding to this
energy for production of Balmer lines. Therefore, even though there is an abundance line and this number itself is a
of hydrogen in the stars, Balmer lines may not be observed in the spectra of all of function of temperature.
them: the surface temperature prohibits the formation of Balmer lines of hydrogen.
53
The Solar System H Hδ Hγ Hβ
and Stars Hydrogen Balmer series lines

350 nm 400 nm
Various 450 nm 500 nm
KH metals
lonized Ca Calcium

Fig.7.10: Examples of spectra of various types of stars (O to M type)

On the other hand, very hot stars with surface temperatures higher than 25000 K also
do not contain Balmer lines in their spectra. Can you guess why is it so? It is because
such stars are so hot that electrons of the hydrogen atoms are ripped off and the atoms
are ionised. Such ionised atoms cannot produce spectral lines. It is only when the
surface temperature of a star is between 7500 K and 11000 K that the conditions are
favorable for the production of Balmer lines. Therefore, surface temperature of a star
determines which spectral lines would be formed and what their intensities would be.
After hydrogen, the second most abundant element in stars is helium. A helium atom
has two electrons which are held very tightly together in their lowest states. Hence,
helium absorption lines are seen in the spectra of relatively hot stars with surface
temperatures in the range 11000 K to 25000 K. In stars hotter than 25000 K, one of
the two electrons in helium atoms is torn away. The question is: Are the spectral
lines produced by singly ionised helium atoms similar to those produced by un-
ionised helium atom?

The spectral lines produced by singly ionised helium are different from those
produced by neutral or un-ionised helium. Further, stars with temperatures in excess
of 40000 K are so hot that helium is completely ionized which cannot produce any
spectral lines.
In some stars, conditions are favourable for a molecule to produce spectral lines. Very
cool stars with temperatures less than 3500 K show very strong, broad bands of
titanium oxide (TiO). Table 7.1 shows the values of the temperature and some other
important parameters of seven main spectral types.
Table 7.1: Spectral types and their parameters
Spectral Approx. Hydrogen Other spectral Naked-eye Color
class temp. (K) Balmer features example (Star)
lines
O 40,000 Weak Ionised He Meissa (O8) Blue
B 20,000 Medium Ionised and Achenar (B3) Blue/White
Neutral He
A 10,000 Strong Ionised Ca weak Sirius (A1) White
F 7,500 Medium Ionised Ca weak Canopus (F0) Yellow/White
G 5,500 Weak Ionised Ca Sun (G2) Yellow
medium
K 4,500 Very weak Ionised Ca Arcturus (K2) Orange
strong
M 3,000 Very weak TiO strong Betelgeus (M2) Orange/Red
54
There is an alternative method to determine the temperature of a star on the basis of its Stellar Spectra and
Classification
spectrum. This method is based on the principle of black body radiation and you will
learn it now.

7.3.2 Black Body Approximation

As you know, the colour and brightness of a star is different from other stars. You also
know that these parameters depend on temperature. Does it, therefore, mean that we
can estimate the temperature of stars on the basis of their observed colour and
brightness?
It can indeed be done if we consider a star as an ideal object called black body. Refer
to Fig. 7.11 which shows a set of black body spectra at various temperatures. Note
that hotter bodies radiate most of their total energy in the shorter wavelength part of
the spectrum. On the other hand, the cooler bodies have the peak of their radiation at
the longer wavelength side of the spectrum and the total energy radiated by them is
relatively low.
It has been observed that the outer envelope of a star’s spectrum is quite similar to a
black body spectrum at a certain temperature. Thus, a stellar spectrum can be
approximated to a black body spectrum. You should, however, note that the
absorption features of stellar spectra distinguish it from the black body spectrum.
Therefore, the main spectral types, namely O to M, can be considered as referring to
different temperatures. In other words, we can say that spectral type directly refers to
the effective temperature of the star (see Table 7.1).

0
log10 Fλ

-2

-4

-5 -4.5 -4 -3.5
log10 λ (cm)

Fig.7.11: Black body radiation curves for various temperatures

To estimate the temperature of an astronomical object such as a star on the basis of

black body approximation, go through the following Example carefully.

55
The Solar System
and Stars Example 1
An astronomical object named Cygnus X-1, a strong X-ray source, is found to radiate
like a black body with peak wavelength at 1.45 nm. Calculate its temperature. Assume
that the constant for Wien’s displacement law is equal to 2.9×10−3 mK.
Solution
You may recall from Thermodynamics and Statistical Mechanics (PHE-06) course
that the Wein’s displacement law is given by:
λpeak T = constant
where λpeak is the value of wavelength corresponding to the peak of the black body
spectrum and T is the temperature of the black body. For Cygnus X-1, we have

λpeak = 1.45 × 10−9m

and
value of the constant = 2.9 × 10−3mK.
Thus, temperature of Cygnus X-1,

2.9 × 10 −3 mK
T= ≅ 2 × 10 6 K .
−9
1.4 × 10 m

As you know, the temperature of a star is not a directly measured quantity; rather, we
infer this parameter on the basis of black body approximation. Would it not be better
to classify stars on the basis of a parameter, say luminosity, which is more easily
measurable? This is what was attempted by Ejnar Hertzsprung and Henry Norris
Russell and their classification of stars resulted in a graph known as H-R diagram.
You will study about it now.

7.4 H-R DIAGRAM

H-R diagram is a graph which enables us to simultaneously classify stars on the basis
of their temperatures and their luminosities. Since H-R diagram involves parameters
such as temperature, luminosity and radius, it would be advisable to first recapitulate
these terms and their interdependence in the context of stars.
The luminosity (L) of a star is defined as the total energy radiated by it, in one second,
consisting of radiations of all wavelengths. On the other hand, you may recall from
Unit 1 that the absolute visual magnitude refers only to the visible range of the
electromagnetic radiation. Therefore, corrections need to be applied for the radiations
emitted at other wavelengths. The required extent of correction depends upon the
temperature of the star. This correction is called Bolometric Correction (B.C.) and
for medium temperature stars like our Sun, its value is small. After making necessary
correction, we obtain the absolute bolometric magnitude: for the Sun it is + 4.7 and
for Arcturus, it is − 0.3.
Thus, the expressions for the apparent magnitude as well as the absolute magnitudes
of a star are written as:

mbol = mv + B.C. (7.2a)

and
M bol = M v + B . C . (7.2b)

where mbol, mv, M bol and M v are respectively the apparent bolometric magnitude
apparent visual magnitude, absolute bolometric magnitude and absolute visual
magnitude of a star. Further, the absolute bolometric magnitude of a star can be
56 calculated from the following relation:
 Stellar Spectra and
bol bol  L  Classification
M star − MΘ = −2.5 log  star  (7.3)
 LΘ 
You have already learnt that a difference of 5 in absolute magnitude implies a ratio of
100 in luminosity. So, the luminosity of Arcturus is 100 LΘ . The most luminous stars
can have luminosities of the order of 105 LΘ , whereas the least luminous ones have
luminosities ~ 10−4 LΘ .
Thus, if we can measure the parallax of a star, we can find its distance, calculate its
absolute visual magnitude, calculate the bolometric correction and obtain absolute You have learnt in Unit 1
bolometric magnitude. As a result, we can estimate the luminosity of a star in terms of how the parallax of a star is
the Sun’s luminosity. How about solving an SAQ to check your understanding of the measured.
terms discussed above?
SAQ 2 Spend
5 min.
The absolute visual magnitude of a star is 8.7 and for its temperature, the bolometric
correction is − 0.5. Calculate the absolute bolometric magnitude and the luminosity of
the star.

You know that one of the fundamental parameters of a star is its diameter and it is
related to the star’s luminosity and temperature. To appreciate the relation amongst
these parameters, consider a normal candle flame which has a low surface area. The
candle flame, despite being very hot, cannot radiate much heat and its luminosity is
low. However, if the candle flame were 25 cm long, it would have larger surface area.
In this case, despite being at the same temperature, it would radiate more heat and
would have high value of luminosity.
In the same way, a star’s luminosity is affected by its surface area and temperature. To
obtain a relation between the luminosity and diameter of a star, we can assume the star
to be spherical in shape having surface area 4πR2, where R is the radius. Further, from
basic thermodynamics (PHE-06 course), you know that, the total energy radiated by a
black body per second per unit area is σT 4 (Stefan’s law). Assuming that the star
radiates like a black body, we can express its luminosity (L) as the product of its
surface area and the energy radiated by it per unit area per second, i.e.,
L = 4πR 2σT 4 (7.4)
Thus, by comparing the luminosity L of a star, with LΘ , the luminosity of the Sun, we
get:
2 4
L  R  T 
=    (7.5)
LΘ  RΘ   TΘ 

where RΘ and TΘ are the radius and temperature of the Sun.

Let us now refer to Fig. 7.12 which shows a H-R diagram for all known stars in our
solar neighbourhood. Note that the H-R diagram is a plot between absolute magnitude
or luminosity (along y-axis) and temperature (along x-axis).
The H-R diagram contains quite a lot of information about stars. Since the absolute
magnitude or luminosity refers to the intrinsic brightness of a star, H-R diagram
relates the intrinsic brightness of a star with its temperature. Moreover, the H-R
diagram separates the effects of temperature and surface area on the luminosity of the
stars because the brightness of two stars at the same temperature is proportional to
their radii. This feature enables us to classify stars in terms of their diameters.
Further, you must remember that the location of a star on the H-R diagram is in no
way related to its location in space: a star located near the bottom of the diagram
simply means that its luminosity is low and similarly, a star in the right indicates that
its temperature is low (because the temperature decreases away from the origin) and
so on. Another interesting feature of the H-R diagram is that the position of a star on 57
The Solar System it changes with time. This implies that the star’s luminosity and temperature change
and Stars
with its age. Again, this change in position of a star has nothing to do with the star’s
actual motion.
The stars have been divided into different types/groups on the basis of their location
on the H-R diagram. In Fig. 7.12, you may note that the distribution of stars follows a
pattern such that a majority of stars fall along a central diagonal called the main
sequence. The main sequence stars account for nearly 90 per cent of all stars. The
other types of stars such as giants, supergiants, white dwarfs populate other regions.
The giant stars (named so because of their big size) located at the top right of the H-R
diagram have low temperature but high luminosity.
Spectral class
O B A F G K M

Supergiants
-20 17

6
Red giants
3.2
Absolute magnitude

-10 Main
sequence 1.8
1.5
1.3 (a)
1
5 0.7
White dwarfs 0.5

0.3
10

20000 10000 7000 6000 5000 3000

Temperature (K)

Spectral Class
O B A F G K M
−10

−−−−5

0
Absolute Magnitude

(b)
+5

+10

+15

+20
20,000 14,000 10,000 7000 5000 3500 2500

Temperature (K)

Fig.7.12: a) A schematic H-R diagram (Note that on the main sequence, masses of the stars are
indicated in units of solar mass.); and b) magnified version of the H-R diagram
58
When we move further up in the H-R diagram (Fig. 7.12), we find super giants. Stellar Spectra and
Classification
These high luminosity and low temperature stars must be extraordinarily big in size. It
has been estimated that the diameters of super giants are roughly 100 to 1000 times
the diameter of the Sun! Further, when we come to the bottom left (below the main
sequence), we come across stars known as white dwarfs  stars which are very hot
but their luminosity is very low. Obviously, white dwarf stars must be very small in
size compared to the Sun to have such low luminosities.

Example 2

A bright star in Orion constellation, Betelgeuse, has a surface temperature of 3500K

and is 105 times more luminous than the Sun. Calculate its radius in terms of RΘ , the
radius of the Sun. What kind of star could it be on the basis of the H-R diagram?

Solution

We have

LB
TB = 3500K; TΘ = 5800K; = 105
LΘ

From Eq. (7.5), we have

LB R B2 TB4
=
LΘ 2 4
RΘ TΘ
or,
10 5 = [3500 5800]4 (R B /RΘ )2
or,
R B ≈ 1000RΘ

The luminosity and surface area of Betelgeuse is much higher than the Sun whereas
its temperature is lower than the Sun. These characteristics, as per the H-R diagram,
indicate that Betelgeuse is a super giant.

To fix the ideas expressed above, you should answer the following SAQ.

SAQ 3 Spend
8 min.
a) Suppose that the surface temperature of two stars A and B is the same and the
luminosity of A is higher than B. Which of the two stars is bigger in size? Why?

b) Choose a typical giant star in Fig. 7.12 and estimate its radius.

Although the basic H-R diagram depicts stars in terms of their temperatures and
absolute magnitudes, we can have its many other representations as well depending
upon the extent of information we wish to incorporate in it. Fig. 7.13 shows a
composite H-R diagram incorporating several parameters such as luminosity, absolute
magnitude, temperature, and spectral type.

59
The Solar System
and Stars

Fig.7.13: A composite H-R diagram showing various parameters of the stars

Spend SAQ 4
5min.
a) In which part of the main sequence are the less massive stars located?

b) A main sequence star has a luminosity of 400 LΘ. What is its spectral type?

Thus, on the basis of above discussion, you will agree that the H-R diagram provides
information about the following parameters of stars:

a) size
b) luminosity
c) mass
d) spectral type, and
e) absolute magnitude

It is because of this reason that the H-R diagram is so important in astronomy. So far,
you have studied about stellar spectra and classification of stars on the basis of their
temperature. You have also learnt that, on the basis of H-R diagram, one can get an
idea about the size of a star. However, to get a better idea about the size of a star, we
can use the correlation between the sharpness of spectral lines and luminosity. This
gives rise to luminosity classification. You will learn about it now.

7.5 LUMINOSITY CLASSIFICATION

We have discussed earlier in this unit that the luminosity of a star predominantly
depends on its size: the bigger the star is, the higher is the value of its luminosity.
Detailed study of the stellar spectra helps in ascertaining the density and size of a star
which give a fairly good idea of its luminosity. You may ask: Which features of the
stellar spectra provide such information? Recall that the spectrum of an isolated
atom differs from that of the gas of such atoms. The difference is manifested in the
form of broadening of spectral lines of the atom in the latter case due to collision
between atoms.

The next logical question is: Under what conditions can collisions readily take
place? For frequent collisions, a denser gas provides a better environment than a rarer
gas. Therefore, if the spectral lines in stellar spectra are broadened, we can safely
conclude that the density of the star is high. Examples of such stars are the main
sequence stars. On the other hand, giant stars have very low densities. As a result,
their spectral lines are fairly narrow compared to the main sequence stars.

60
Thus, sharp lines in the stellar spectra clearly indicate that the size of the star is large Stellar Spectra and
Classification
and hence its luminosity is high. The converse is also true. Thus, looking at the width
of spectral (absorption) lines, we can obtain a fairly good idea about its luminosity.
Table 7.2: Luminosity classes
As such, stars have more or less continuous range of luminosities. Still, on the basis of
the width of their spectral lines, they are categorized into various luminosity classes. Ia Bright supergiant
Table 7.2 gives the various luminosity classes denoted by Roman numerals I to V
Ib Supergiant
with the supergiants further subdivided into classes Ia and Ib.
II Bright giant
The star Rigel (β Orionis) is a bright supergiant (class Ia) and Polaris, the North star is
a regular supergiant (class Ib).The star Adhara (ε Canis Majoris) is a bright giant III Giant
(Class II); Capella (α Aurigae) is a Giant (III) and Altair (α Aquilae) is a subgiant
(IV). The Sun is a main-sequence star (V). Thus, the complete spectral classification IV Subgiant
for the Sun is G2V. This complete classification is also called the spectro-luminosity
V Main-sequence star
classification. The spectro-luminosity class of star Vega is A0V.

Refer to Fig. 7.14 which shows the position of luminosity classes on the H-R diagram.
You must remember that the lines corresponding to each class is just an
approximation; star of a particular class may lie just above or below the line
corresponding to that class.

Fig.7.14: H-R diagram depicting the location of luminosity classes

Now, let us sum up what you have learnt in this unit.

7.6 SUMMARY
• Emission and absorption of radiation is caused due to transition of electrons in
atoms between allowed energy levels. Analysis of stellar spectra provides
information about the temperature and size of a star.
• On the basis of the strength of spectral lines, particularly Balmer lines, in stellar
spectra, stars were classified into seven main spectral types namely O, B, A, F,
G, K and M. It was shown by M N Saha that this classification essentially refers
to the temperatures of stars.
• H-R diagram enables us to classify stars on the basis of their temperatures and
luminosities.

61
The Solar System • Luminosity of a star is defined as the total energy rediated by it in one second
and Stars
consisting of radiations of all wavelengths. The relation between luminosity,
radius and temperature of a star is given by:
2 4
L  R   T 
=  


T


LΘ  RΘ   Θ 

• On the basis of H-R diagram, stars are grouped into four categories namely main
sequence, giants, super giants and white dwarfs. Stars of each group have
characteristic temperatures, sizes and luminosities.
• The density of a star affects the sharpness of its spectral lines. This fact is used for
luminosity classification of stars.
• According to the luminosity classification, stars are classified into five classes: I,
II, III, IV and V with class I further sub-divided into Ia and Ib.

7.7 TERMINAL QUESTIONS Spend 25 min.

1. Assign spectral class to each of the objects whose characteristics are given below:
a) Temperature ~ 40,000K
b) Weak Balmer lines but moderately strong Ca lines
c) Strongest hydrogen lines
d) Molecular bands of TiO
e) Neutral helium lines
bol
2. A star has luminosity ~ 100 LΘ and apparent bolometric magnitude, m star = 9.7.
bol
If Sun has M Θ = + 4.7, calculate the distance of the star.

3. Calculate the radius of a star which has the same effective temperature as the Sun
but luminosity 10,000 times larger.
4. In the following table, which star is a) the brightest; b) the most luminous; c) the
largest; and d) the smallest?

Star Spectro-luminosity type mv

1. G2V 5
2. B1V 8
3. G2Ib 10
4. M5III 19
5. White Dwarf 15

7.8 SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs)

1. a) Spectra of different elements have different lines because their atomic energy
levels are different.
b) A continuous spectrum contains all wavelengths. We get such a spectrum
when the density of a system is high, as in a solid.
2. From Eq. (7.2b), we know that
M bol = M v + B . C .
Substituting the values of M v and B.C., we get, M bol = 8.7 − 0.5 = 8.2
62
 Further, from Eq. (7.3), we have: Stellar Spectra and
Classification
Lstar (M bol − M bol )
= 10 − 4 star Θ
LΘ
= 10 −0.4 (8.2−4.7 )
1
=
25
LΘ
Thus, we get, Lstar = .
25

3. a) Since LA > LB, the surface area of A is greater than the surface area of B. So, A
is bigger.

b) In Fig. 7.12, we find that the temperature of a typical giant star can be taken as
5000 K. So, we can write for a typical giant, T = 5000 K. Further, the
luminosity of a typical giant can be written as, L = 100 LΘ
Thus, using Eq. (7.5), we get:
2 4
L  R   T 
=    
LΘ  RΘ   TΘ 
or,
2 4
 R  L  TΘ 
  =   = 100 × (6 / 5)
4
R
 Θ  LΘ  T 
or,
R = 14.4 RΘ .

4. a) Lower part of the main sequence.

b) From Eq. (7.3), we have

bol bol  L 
M star − M Θ = − 2.5 log  star  = − 2.5 log 400 = ─ 6.5
 LΘ 
Thus,
bol
M star = ─ 6.5 + 4.7 = ─ 1.8

If we look at the H.R. diagram to identify the class of the star having this
value of absolute magnitude, we find that its class should be B7,
approximately.

Terminal Questions

1. (a) O (b) G (c) A (d) M (e) B

2. We have from Eq. (7.3):

bol bol  L 
M star − MΘ = −2.5 log  star 
 LΘ 
or,
bol
M star = 4.7 − 2.5 log (100) = − 0.3
63
The Solar System
and Stars
Further, we have:

bol bol  d 
m star − M star = 5 log  
 10 
9.7 − (− 0.3) = 5 log (d/ 10 )
or,
d = 1000 pc .

3. From Eq. (7.5), we have:

L1 L2 = R12T14 R22T24
or,
10 4 = R12T14 RΘ2 T14 ;
or,
R1 = RΘ 10 4 = 100 RΘ

4. [Hint: Use Fig. 7.12 to determine the absolute magnitudes.]

The answers are: a) 1; b) 3; c) 3; d) 5.

64
 Stellar Structure
UNIT 8 STELLAR STRUCTURE
Structure
8.1 Introduction
Objectives
8.2 Hydrostatic Equilibrium of a Star
8.3 Some Insight into a Star: Virial Theorem
8.4 Sources of Stellar Energy
8.5 Modes of Energy Transport
8.6 Simple Stellar Model
Polytropic Stellar Model
8.7 Summary
8.8 Terminal Questions
8.9 Solutions and Answers

8.1 INTRODUCTION

In Unit 7, you have learnt about the classification of stars on the basis of their spectra.
You know that the H-R diagram is obtained on the basis of the luminosity, and
effective temperature of stars and enables us to classify them in the most
comprehensive manner. A careful look at this diagram reveals that there are gaps
between families of stars, e.g., between the main sequence and giants. You may
wonder why such gaps should exist when we have such a large number of observable
stars! Further, you may like to know: What causes some ordinary stars to become
giants and others to become dwarfs? These and similar other questions cannot be
answered on the basis of observations alone. We require the knowledge of the
physical conditions in the interior of the stars. In other words, we need to know: How
the temperature, pressure and density of a star vary in its interior? In the present unit,
you will study the physical principles which form the basis for understanding the
internal structure of stars.

You know that the Sun is emitting radiation at a constant rate and its diameter shows
no significant variation with time. This implies that the Sun as well as other stars are
in mechanical and thermal equilibria. In Sec. 8.2, you will study about hydrostatic
equilibrium and its consequences for the variation of density and pressure inside a
star. In Sec. 8.3, you will learn how to estimate the internal temperature of a star on
the basis of the virial theorem: statement of relation between the kinetic and potential
energies of a system in equilibrium. In addition to the considerations of hydrostatic
and thermal equilibria, the mechanism of energy generation and transport play an
important role in deciding stellar structures. In Sec. 8.4, you will learn why only the
energy generated due to nuclear reaction needs to be considered as the source of
stellar energy. You will also learn some important mechanisms of nuclear energy
generation in stars. In Sec. 8.5, various modes of transportation of energy from the
interior to the surface of stars have been discussed. You will discover the conditions
for the formations of convective and radiative zones in the stellar interior.
Finally, in Sec. 8.6, we discuss the computation of a simple stellar model. We also
discuss how the results of this model compare with the observations.

Objectives
After studying this unit, you should be able to:

• list the basic assumptions for the theoretical study of stellar structure;

65
The Solar System • show that the values of interior pressure and temperature of a star are higher than
and Stars the values at the surface by several order of magnitudes;
• explain that the nuclear energy generation is the only important energy generation
process in stars;
• predict when a radiative or a convective zone will be formed in the stellar interior;
• compute polytropic stellar model and compare the theoretical results with
observations; and
• solve numerical problems based on these concepts.

8.2 HYDROSTATIC EQUILIBRIUM OF A STAR

You know that stars, including the Sun, are made of hot gas. We cannot probe the
interior of stars to determine their physical parameters and their variation with time
and distance because of their high temperatures and enormous distances from the
Earth. The question, therefore, is: How do we determine the internal structure of a
star? Astrophysicists construct theoretical models of stars and compare their
predictions with observations. To keep the theoretical analysis simple, the following
assumptions are made:

i) The star is spherically symmetric: You know that stars have rotational motion
which alters their spherical shape. Since the rotational motion is slow in most
cases, it does not have appreciable effect on the shape of the stars. Spherical
symmetry is, therefore, a valid assumption.

ii) The star is in dynamic equilibrium: Dynamic equilibrium means that the energy
radiated by a star is equal to the energy supplied from its core. This assumption
seems valid because luminosities of stars have been observed to be constant over
a considerable period of time.

iii) The star is in thermally steady state: This implies that the temperature at each
point within a star is constant over a considerable period of time. Note that this
assumption does not mean that the entire interior of a star is at the same
temperature.

Under these assumptions, the theoretical understanding of stellar structure is based on

four equations based on certain fundamental principles of physics. First of all, let us
consider the principle of hydrostatic equilibrium.

You know that the observable stellar parameters such as luminosity change very
slowly. We can, therefore, safely assume that a star is in hydrostatic equilibrium, that
is, it is neither expanding nor contracting (at least not very rapidly). This equilibrium
is maintained by a balance between the force of gravity acting inwards and that due to
the gradient of pressure of the gas acting outwards. To find out the consequences of
the hydrostatic equilibrium, let us consider an element of volume dV at a point A
inside a star at a distance r from its centre (Fig. 8.1). If ρ(r) is the density of matter
inside dV, that is, at a distance r from the centre, the mass enclosed in the volume
element dV is ρ(r) dV. Further, if M(r) is the mass inside the sphere of radius r, the
gravitational force acting on the mass inside dV is given by:

GM (r )
ρ(r )dV (8.1)
r2

66
 Stellar Structure

dV (=dA.dr)

dP.dA

r dr

Fig.8.1: An element of volume dV at a distance r inside a star in hydrostatic equilibrium

Now, in view of the spherical symmetry of the star, the pressure, density and
temperature may be taken as identical at all points over the spherical surface of radius
r. Therefore, the net hydrostatic force acting on the volume element dV and pushing it
outward can be written as:

dP . dA, (8.2)

where dA is the area of the volume element dV perpendicular to r and dP is the

pressure difference between two sides of the volume element along the radius. For
hydrostatic equilibrium, the gravitational force must be equal and opposite to the
hydrostatic force. Thus, on the basis of Eqs. (8.1) and (8.2), we can write:
GM (r )
ρ(r )dV = −dP.dA
r2
dP
=− .dV
dr

This yields the equation of hydrostatic equilibrium.

Equation of Hydrostatic Equilibrium

dP GM (r )
=− ρ(r ) (8.3)
dr r2

We can also write Eq. (8.3) as

dP
= − ρ(r ) g (r ) (8.4)
dr
GM (r )
where g(r) is the acceleration due to gravity given by g(r) = . From Eq. (8.3),
r2
it is obvious that it is the pressure gradient that supports the star and not the pressure.
If we denote the mass of the whole star by M and its radius by R, the mean density,
< ρ > of the whole star can be expressed as:
M
<ρ>= (8.5)
4 3
 πR 
3 

67
The Solar System Let us now consider a spherical shell of the star between radii r and r + dr. The
and Stars 2
volume of the matter enclosed in this shell is 4πr dr. Since ρ(r) is the density of
stellar matter at distance r, the mass of this spherical shell is:
dM (r ) = 4π r 2 ρ(r ) dr

Thus, we can express the total mass inside the sphere of radius r as:
r
M (r ) = ∫ ρ (r ) 4 π r 2 dr
0

Differentiating both sides of the above equation with respect to r, we get the mass
continuity equation for the star:

Mass continuity equation

dM (r )
= 4πr 2 ρ(r ) (8.6)
dr

Eqs. (8.3) and (8.6) constitute two basic equations of stellar structure. Further, the
state of hydrostatic equilibrium in a star enables us to obtain a relation between its
gravitational potential energy and the kinetic energy of its constituent particles. This
relation is known as the virial theorem. You will learn about it now.

8.3 SOME INSIGHT INTO A STAR: VIRIAL THEOREM

You may be aware that the virial theorem is applicable for a system of perfect gas
particles. However, this theorem can also be applied to a star because it (star) can be
considered as a system of free particles. To obtain the relation between the potential
and kinetic energies of a star, let us consider the equation of hydrostatic equilibrium
(Eq. (8.3)). On multiplying Eq. (8.3) by 4πr3 and integrating over the radius of the
star, we get:

R dP R GM (r )
∫0 dr
4πr 3 dr = − ∫0 ρ(r ) r 2
4πr 3 dr (8.7)

On integrating by parts, the left hand side of Eq. (8.7) gives:

Actually, the pressure on R R R

dP
∫0 ∫0 3P4πr
the surface of a star is not 3 3 2
4πr dr = P 4πr − dr
zero. Its value is, however, dr
much smaller than the 0
pressure in the interior.
Therefore, it can be Since P = 0 at r = R, the first integral on the right hand side vanishes and the above
assumed to be zero. This is equation reduces to:
also the case with density.
R dP R
∫0 dr
4πr 3 dr = −3 ∫0 P 4πr 2 dr

Thus, Eq. (8.7) can be written as:

R R GM (r )
−3 ∫0 P 4πr 2 dr = − ∫0 ρ(r ) r
4πr 2 dr (8.8)

Now, assuming that the star comprises of monoatomic gas, its total thermal (or
68 internal) energy U can be written as:
 R Stellar Structure
2U = 3 ∫0 P 4πr 2 dr (8.9)

SAQ 1 Spend
3 min.
Derive Eq. (8.9).

Substituting Eq. (8.9) in Eq. (8.8), we obtain:

R GM (r )
2U = ∫0 ρ(r ) r
4π r 2 dr (8.10)

The right hand side of Eq. (8.10) can be expressed in terms of the gravitational
potential energy. The gravitational potential due to the mass M(r) inside the sphere of
GM (r )
radius r is − . Therefore, total potential energy due to all mass elements dM (=
r
4πr2ρ(r) dr) of the star is:

M GM (r ) R GM (r )
Ω=− ∫0 r
dM = − ∫0 r
4πr 2 ρ(r )dr (8.11)

Thus, on the basis of Eqs. (8.10) and (8.11) we get the virial theorem.

Virial theorem

2U + Ω = 0 (8.12)

You can see from Eq. (8.11) that the gravitational potential energy Ω of the whole star
can be determined only if we know the variation of density ρ(r) inside. Assuming that
ρ(r) ≈ < ρ > , the mean density of stellar matter, we may write:

4
M (r ) ≈ πr 3 < ρ > (8.13)
3

Substituting < ρ > for ρ(r ) and M(r) from Eq. (8.13) in Eq. (8.11) and integrating, we
get:

3 GM 2
Ω≈− (8.14)
5 R

Now, to find an expression for the internal energy of a star, we make use of the
equation of state of a gaseous system:
4 3
M (r ) = πr < ρ >
ρ 3
P= k BT (8.15)
µm
4 2
dM = π . 3r < ρ > dr
−23 −1
where kB (= 1.38 × 10 J K ) is the Boltzman constant, T is the temperature and 3
µm is the mean mass of a gas particle. For pure hydrogen gas, µ = 1. Substituting
dM
Eq. (8.15) in Eq. (8.9), we get: = 4 π r dr
2
<ρ>
3 R 3 M k BT
U= ∫
2 0
P 4πr 2 dr = ∫
2 0 m
dM
69
The Solar System since 4πr 2 dr = dV and ρdV = dM . If we define the mean temperature of the star,
and Stars
1 M
<T >=
M 0 ∫ TdM , then the above expression can be written as:

3 kB
U= M <T > (8.16)
2 µm

Substituting the expressions for potential energy (Eq. (8.14)) and internal energy
(Eq. (8.16)) in the virial theorem (Eq. (8.12)), we get:

1 µmG M
< T >=
5 kB R
Thus
< T > ∝ M 2 / 3 < ρ > 1/ 3 (8.17)

where R has been expressed in terms of the mean density.

Spend SAQ 2
10 min.
a) Verify the results contained in Eqs. (8.14) and (8.17).

b) Assume that the Sun is made of pure hydrogen (µ = 1). Show that the mean
temperature of the Sun is < T > ≅ 4 × 106K.

On solving SAQ 2, you must have appreciated that the internal temperature of the Sun
can be estimated without making any detailed calculations. Further, it is obvious from
Eq. (8.17) that if two stars have the same mass, the denser one will be hotter. For a
sun-like star, the effective surface temperature is Te ≈ 5780 K. And from the solution
of SAQ 2, we find that the mean solar temperature, < T > ≈ 4 × 106 K. This means
that the internal temperature must be much higher. You may ask: What causes such
high internal temperature in stars? To answer this, we must investigate the sources
of energy generation in stars. The stars can have possibly three kinds of energy
sources: gravitational, chemical and nuclear. You will study about them now.

8.4 SOURCES OF STELLAR ENERGY

Let us first consider the gravitational energy as the source of energy for stars. From
Eq. (8.14), note that when the whole matter of the star is scattered at infinity, its
gravitational potential energy is zero. When the stellar matter is assembled to make a
3 GM 2
star of radius R, its potential energy becomes − . This means that during the
5 R
gravitational contraction of a star, that is, during the formation of a star, an energy
3 GM 2
equal to is released.
5 R

It can be shown (TQ 2) that the Sun’s energy would last only for about 107 yrs if
gravitational energy was its only source. But, the results obtained on the basis of
radioactive dating of different types of meteorites, deep terrestrial oceanic sediments
and lunar rocks suggest an age of ≈ 5 × 109 years for the Sun. Thus, the gravitational
potential energy cannot be the source for solar luminosity. We must look for some
other source of energy for stars like the Sun.

70 We are now left with two other possible processes, namely, chemical and nuclear.
The possibility of a chemical process as the source of energy in stellar interior is also Stellar Structure
ruled out on the basis of results obtained in the following example.

Example 1

Assume that the Sun consists of hydrogen and oxygen and the proportion of these
elements is such that the entire solar material could be burned and transformed to
water vapour. Show that the total energy available from this process would last only
4
for ~ 10 yr given that 10 eV is liberated in the formation of each water molecule.

Solution
−27 −27
The molecular weight of water (H2O) is 18 u = 18 × 1.6 × 10 kg = 28.8 × 10 kg.
30
The mass of the Sun is 2 × 10 kg. Therefore, the total number of water molecules
present in the Sun is given by:

2 × 10 30 kg
≈ 6 × 10 55.
− 27
28.8 × 10 kg

Thus, total energy liberated due to the formation of water vapours in the Sun is:

6 × 1055 × 10 eV × 1.6 × 10−19 J (eV)−1 ≈ 9.6 × 1037 J.

Since the Sun radiates energy at the rate ~ 4 × 1026 Js−1, the duration over which the
Sun would radiate all its energy generated due to formation of water molecules is:

9.6 × 10 37 J
≈ ≈ 2.4 × 1011 s ~ 10 4 yr .
26 −1
4 × 10 Js

In view of the fact that the Sun has an estimated age of ~ 5 × 109 yr, the result of the
above example clearly shows that chemical process cannot be responsible for
generation of energy in stars.

Let us now look at the possibility of nuclear processes for generating energy in a sun-
like star. You may recall from your school physics that nuclear reactions are of two
types: fission reactions and fusion reactions. In nuclear fission, large unstable nuclei
like 238U break into smaller nuclei and energy is released. Since the abundance of
such nuclei is negligible in stars, such a process is also ruled out as a source of stellar
energy. We are, therefore, left with only fusion process to be considered as possible
energy source.

In nuclear fusion process, two lighter nuclei combine and form a new nucleus and
energy is released. The amount of energy released depends on the binding energy per
nucleon of the elements involved in the reaction. You may recall that the binding
energy of a nucleus is the energy required to separate its constituent nucleons by a
large distance. Refer to Fig. 8.2 which depicts binding energy curve. Note that as the
mass number increases from zero, the value of binding energy per nucleon increases.
This implies that if two or more, lighter nuclei (such as hydrogen) are fused together
to create a relatively heavier nuclei (such as 3He or 4He), we will have surplus of
energy. This is precisely what happens in fusion reactions: enormous amount of
energy would be released due to fusion of hydrogen nuclei and consequent production
of helium. Further, Fig. 8.2 also shows that the value of binding energy per nucleon
saturates at around mass number 50 and shows a very slow decrease beyond. The
nature of the binding energy curve, therefore, indicates that small-mass as well as
large-mass nuclei are less tightly bound than medium mass (such as Fe) nuclei. 71
The Solar System Before proceeding further, you may like to convince yourself whether or not the
and Stars energy generated due to nuclear fusion can account for the observed luminosity of the
Sun. To do so, solve the following SAQ.

Fig.8.2: Variation of the binding energy per nucleon with mass number

Spend SAQ 3
5 min.
Assume that originally the Sun comprised only of hydrogen and that the inner 10
percent of the Sun’s mass could be converted into helium. For how long would the
Sun be able to radiate at the rate given by L = 4×1026 Js−1. The mass of the Sun is
MΘ = 2×1030 kg.

On solving SAQ 3 you find that the Sun would indeed be able to radiate energy at the
present rate for another 5 billion years by fusing hydrogen into helium.
To know the amount of energy
Now the next logical question is: How is the fusion of hydrogen nuclei into helium
released due to fusion of four nuclei made possible? In other words, you may like to know what the pre-conditions
hydrogen nuclei and formation for nuclear fusion to take place are and how these conditions are obtained in the stellar
of a helium nucleus, note that interiors. Nuclear fusion involves fusion of two positively charged nuclei against
the total mass of four hydrogen
atoms is 4.031280 u while that
Coulomb repulsion.
of a helium atom is 4.002603 u.
The mass defect is 0.02867u The possibility of such a process can be understood on the basis of the potential
and in accordance with the energy curve that an atomic nucleus would experience when it approaches another
2
Einstein equation, E = mc , the
atomic nucleus (Fig. 8.3). Note that the potential energy curve consists of two distinct
energy released due to fusion
of four hydrogen nuclei into a regions. Region I corresponds to the situation when separation between the two nuclei
helium nucleus is equal to is such that their potential energy is due to Coulomb repulsion. Region II represents
2
(0.02867 u) × c where c is the the situation when two nuclei are very close to each other and the curve is in the form
velocity of light. of a potential well. The potential well illustrates the strong nuclear forces that bind the
nuclei. From Fig. 8.3, it is obvious that the two nuclei can fuse only when the
approaching nuclei are able to overcome the repulsive Coulomb barrier. This means
that the approaching nuclei have sufficient energy to overcome the barrier.

The question is: At what temperature, the approaching nuclei will have sufficient
energy? Using classical dynamics, we find that the temperature required to provide
sufficient energy to two nuclei so that they overcome Coulomb’s barrier is much
higher than the core temperature Tc (~ 1.5 × 107 K) of the Sun. Thus, classical physics
72
fails to explain the possibility of fusion reactions in stars. You must convince yourself Stellar Structure
about the inadequacy of classical physics to explain nuclear fusion by solving the
following SAQ.

Coulomb barrier

← Ekin ∼ kT
Energy

H
Distance

He Nuclear potential well

(binding energy)

C
O
Si
Fe

Region II Region I

Fig.8.3: Schematic representation of the potential energy barrier experienced by an atomic nucleus
approaching another atomic nucleus

SAQ 4 Spend
5 min.
Suppose that two nuclei have charges Z1e and Z2e and in order to interact, they must
−13
be separated by a distance ~ 10 m. Calculate their mutual potential energy. If their
relative kinetic energy is 3 kBT, calculate the temperature required by two hydrogen
nuclei to overcome this potential barrier.

The temperature in the core of the Sun is ~ 1.5 × 107K. At this temperature, only a
few proton-proton fusion can take place. There is, however, a finite probability that
particles with insufficient energy can tunnel thorough the potential barrier and react.
Now, let us look at some thermonuclear fusion reactions through which lighter nuclei
fuse and release energy.

In stars like the Sun, one of the prominent thermonuclear reactions is the so-called
proton-proton chain or p − p chain. This reaction proceeds sequentially through
three steps as given below:

H 1 + H 1 → H 2 + e + v, (1.19 MeV)
H 1 + H 2 → He 3 + γ, (5.49 MeV)
73
The Solar System He 3 + He 3 → He 4 + H 1 + H 1 , (12.85 MeV)
and Stars

Note that in these reactions, four hydrogen nuclei combine to form a helium nucleus.
The amount of energy liberated at each step is given within brackets. The p − p chain
produces most of the energy in Sun and other such stars.

Further, a relatively smaller amount of energy in the Sun is also generated by the
thermonuclear reaction known as carbon-nitrogen-oxygen (CNO) cycle as given
below:

H 1 + C12 → N 13 + γ, (1.45 MeV)

N 13 → C13 + e + v, (2.22 MeV)
H 1 + C13 → N 14 + γ, (7.54 MeV)
H 1 + N14 → O15 + γ, (7.35 MeV)
O15 → N 15 + e + v, (2.71 MeV)
H 1 + N15 → C12 + He 4 , (4.96 MeV).

You may note that in the CNO cycle, the carbon destroyed in the first step gets
regenerated in the last step. Further, similar to the p − p chain, CNO cycle also
produces a helium nucleus from four hydrogen nuclei. A CNO cycle, however, cannot
begin unless carbon is present.

The thermonuclear energy generated in the stars depends on the abundance of

fusionable matter and the interior temperature of the stars. This provides us a basis to
link the luminosity of stars with its mass. Let ε denote the rate of energy generated per
unit mass in thermonuclear reactions. Then, the luminosity dL caused by an element
of mass dM can be written as:

dL = ε dM (8.18)
2
Since dM = 4πr ρ(r) dr, Eq. (8.18) can be expressed as:

dL
= ε 4πr 2 ρ( r ) (8.19)
dr

Eq. (8.19) is one of the basic equations of stellar structure.

The energy generated at the core of a star must flow towards its surface because the
temperature of the core is very high compared to the star’s surface. Further, since the
star’s surface continuously radiates energy, it will cool off unless the radiated energy
is replaced. Energy transport in the star has important consequences for its structure
because the transport process determines the temperature (and pressure) of different
layers of the star’s interior. Would you not like to know what the different
mechanisms of energy transport in a star are? This is the subject matter of the next
section.

8.5 MODES OF ENERGY TRANSPORT

There are three basic energy transport processes: conduction, radiation and
convection. Fig. 8.4 shows a schematic diagram for these three processes.

74
 Stellar Structure
Conduction
Convection

Radiation

Fig.8.4: Three modes of energy transport: conduction, radiation and convection

Conduction is the most familiar form of heat flow and this process works through
vibrating atoms. The energy of the vibrating atoms is transferred to the nearby cooler
atoms by collisions and energy transport through conduction works better in solids
and not in gases. You will learn later in this Unit that conduction is responsible for
energy transfer in stars like white dwarfs whose interior is in a crystallised form
having density ~ 106g cm−3. For ordinary gaseous stars, conduction process is not
important.

Radiation is the next most familiar mode of energy transport and is responsible for
energy transport in some layers of the interiors of almost all the stars. The process of
energy generation in the central region of a star produces very high energy photons
which are γ-rays. As these photons travel outwards, they collide with matter. At each
collision, γ-ray photon loses energy and when it reaches the surface, its frequency lies
in the visible range. The progress of photons is extremely slow as they travel outward
and this, in fact, regulates the solar luminosity at the level of ~ 1026 Js−1. The
absorption of the energy of γ-rays by the stellar gas, is characterized by the
absorption coefficient, also known as the opacity, kλ of the gas. The subscript λ
indicates that the absorption depends on the wavelength. The important sources of
opacity at high temperatures inside a star are:

i) Electron scattering: The scattering of photons by free electrons.

ii) Photoionisation: The energy of photon is used for successive ionisation of
atoms/ions.

We can obtain an expression for the opacity on the basis of qualitative arguments.

Consider a slab of the stellar gas of thickness dx. Let Fλ denote the flux of radiation
that strikes at one end. If the mass density of the gas is ρ, the amount of radiation
absorbed by the slab is proportional to i) the density of matter in the slab ii) the
incident flux and iii) the thickness of the slab. Thus, the amount of flux absorbed is:

dFλ ∝ ρFλ dx

= −k λ ρFλ dx, (8.20)

where the minus sign indicates absorption. Photons generated inside the Sun do not
reach the solar surface directly; they are scattered by the electrons and nuclei. This
scattering is isotropic and thus their forward and backward scattering is equally likely.
The travel of photons inside a star is, therefore, like that of a drunken person. It is also
called the random walk. It takes thousands of years for these photons to reach the
star’s surface. To have an idea about the time taken by a photon to reach star’s surface
from the core, go through the following example carefully.
75
The Solar System
and Stars
Example 2

Suppose that the energy transport due to radiation process is analogous to the
random walk. Compute the time taken by a photon, generated in the core of the Sun,
to reach the solar surface. Given that for the Sun, the mean free path is l ~ 0.5 cm for
−3 6
photon at an average density and temperature of 1.4 g cm and 4.5 × 10 K,
respectively.

Solution

You may recall from Thermodynamics and Statistical Mechanics (PHE-06) course
that, according to the theory of random walk, we can write the mean square distance,
< D 2 > moved in N steps as:

< D 2 > = Nl 2

< D2 >
This gives N = , the number of steps required to travel a distance < D 2 > in
2
l
< D2 >
steps of size l in one dimension. For three dimensional space, N ~ 3 . Now,
l2
for the photon at the core which has to reach the surface, we have < D 2 >= R 2 .
Hence, the time taken for the photon to reach the solar surface is given by

3R 2
t=
c.l

since, in each step, the time taken is l/c where c is the velocity of light. Substituting
the values of R, c and l, we get

r t ≅ 30,000 yr.

Now, you may like to know: How does the radiative transport of energy give rise
r + dr to temperature gradient in stars? To find out, let us consider a thin spherical shell
around a spherical surface of radius r as shown in Fig. 8.5. Let the temperature of the
sphere be T. Since the spherical region within radius r acts as a source of black body
radiation, the radiative flux incident on the inner side of the shell can be expressed by
Fig.8.5: Spherical shell of (Stefan Law):
thickness dr around
a surface of radius r
F(r) = σT 4, (8.21)

where σ is the Stefan’s constant. Similarly, the radiative flux emerging outward from
the shell surface at r + dr is:

F (r + dr) = σ (T + dT )4 (8.22)

where T + dT is the temperature at the surface of radius r + dr. It is important to

mention here that dT is negative because the surface at r + dr is cooler than the inner
surface of the shell located at r. Since dT is very small compared to T, we can expand
(T + dT)4 using binomial expansion. Doing so, we get:

F (r + dr) = σ T 4 + 4σ T 3 dT
76
Therefore, the flux absorbed by the shell can be written as: Stellar Structure

3
dF = F (r + dr) − F(r) = 4σT dT (8.23)

Combining Eqs. (8.20) and (8.23), we get:

4σT 3dT = − k(r) ρ(r) F(r) dr (8.24)

2
Since luminosity, L(r) = 4πr F(r), we can rewrite Eq. (8.24) as:

dT k (r )ρ(r )
=− F (r )
dr 4σT 3

k (r )ρ(r )  L(r ) 
=−   (8.25)
acT 3  4πr 2 

ac
since σ = , where a is a constant and c is the velocity of light. Eq. (8.25) gives the
4
temperature gradient within a star due to radiative transport of energy. This expression
3
needs to be multiplied by an extra factor of on the right hand side so that it
4
becomes consistent with the one obtained by incorporating the details of such a
process. We, therefore, write the temperature gradient as:

dT 3 k (r ) ρ(r )  L(r ) 
=−   (8.26)
dr 4 acT 3  4πr 2 

Eq. (8.26) is yet another basic equation of stellar structure.

Now, let us consider the third mode of energy transport, namely, convection which
plays an important role in stars. Convection refers to the process in which heat energy
is transported by mass motion, i.e., by transport of the hot/cool matter itself. You are
familiar with convection currents or bubbles moving up and down when water in a
beaker is gradually heated from below. Such a motion also takes place in certain
regions in stars with hot fluid masses rising outward releasing their heat energy and
the cooler matter sinking downward to receive more energy. Fig. 8.6 depicts the
photospheric granulation which is a strong evidence supporting convective transport
of energy at the base of Sun’s photosphere. Convection causes mixing of the stellar

Convective
zone

Radiative
zone

Fig.8.6: Photospheric granulation of the Sun which is caused due to convective transport of energy
in which the hot matter comes out to the solar surface from layers below the surface

constituents in certain regions inside a star and produces homogeneity of chemical

composition by transferring heavy elements from interior to the surface. You may ask: 77
The Solar System Under what condition(s) does convection become the dominant mode of energy
and Stars transport? The process is dominant when the temperature gradient becomes too
steep. Steep temperature gradients are generally created in regions with high opacity
which restrains the flow of energy through radiative process.

To determine the temperature gradient in a convective region of a star, we consider a

situation where hot bubbles of gas rise up and expand adiabatically. After rising
through a characteristic distance, the bubbles lose extra heat and get mixed up with the
surroundings. For such a process, the bubble’s adiabatic temperature gradient is given
by:

 dT   γ − 1  T dP
 dr  =  γ  P dr (8.27)
  ad  

γ
Eq. (8.27) can be obtained by using the adiabatic relation P = K ρ and the equation of
k
state of the gas, P = B ρT . Also, from the equation of state, we have P = NkBT,
µm
where N is the number of particles per unit volume. In astrophysics, it is usual to take
the mass of a particle as µm, where µ is called the mean molecular weight and m is the
dP
mass of a proton. Further, for we use Eq. (8.3) in Eq. (8.27). With these
dr
substitutions, it is possible to write Eq. (8.27) as:

 dT   γ − 1  µm GM (r )
 dr  = −   (8.28)
  ad  γ  kB r2

Eq. (8.28) is one of the basic equations of stellar structure. You should note that only
if the actual temperature gradient in a star is steeper than the adiabatic gradient given
by Eq. (8.27), convective transport of energy can take place. We call the actual
temperature gradient in such a case as superadiabatic. Therefore, for convection to
take place, we must have:

 dT   dT 
  >  (8.29)
 dr  actual  dr  ad

In fact, it can be shown that convection dominates the radiative transport of energy in
 dT 
a region if   is slightly superadiabatic. In any case, the actual mode of
 dr  actual
energy transport in a region inside a star depends on the temperature gradient existing
there.

Now, you should pause for a moment and think what you have learnt so far. You have
learnt to derive certain equations on the basis of the principles of physics connecting
various parameters of a star. These equations are known as the basic equations of
stellar structure. You may ask: Why did we do all this? How do equations of stellar
structure help enhance our understanding of stars? These equations are used to
develop theoretical models of stars. If the predictions of these models are in
agreement with observations, we can conclude that the assumptions made about the
parameters of stellar interior are valid. In case of disagreement between theoretical
prediction and observations, the models are ‘fine-tuned’ by modifying the initial
assumptions. You will indeed appreciate that this is the only way to investigate the
interior of stars because we simply cannot look into those interiors. In the next
section, you will learn to develop a stellar model on the basis of equations of stellar
78 structure.
 Stellar Structure
8.6 SIMPLE STELLAR MODEL
Developing a stellar model essentially involves solving the equations of stellar
structure for a star. As such, it is a very complex task because large numbers of
equations with several unknowns need to be solved. This does not mean that we
cannot get a physical picture of a star. We find that, with some valid approximations,
simple stellar models are easier to calculate. Such models help understand the basis of
some of the empirical laws, such as, mass luminosity relation. Before discussing any
stellar model, let us first list the basic equations of stellar structure.

Basic Equations of Stellar Structure

In the previous sections, we used the following basic physical principles to obtain the
equations of stellar structures:

• Hydrostatic equilibrium,
• Equation of state for stellar matter,
• Mechanism of stellar energy generation, and
• Modes of energy transport in stellar interior.

These basic equations are used to compute theoretical stellar models. This is
equivalent to “constructing a theoretical star”! Once different models are computed,
their location in H-R diagram is found out. It is so because the H-R diagram sets a
detailed standard to be met by any theory of stellar structure and evolution. A
theoretical model is wrong if the physical characteristics of the computed “star” are
such that its location falls in the gap or empty regions of the H-R diagram. The
physical parameters of a star at any point in its interior are temperature, T(r), pressure
P(r), density ρ(r), and luminosity L(r). The basic equations of stellar structure are:

dP GM (r )
Hydrostatic equilibrium: =− ρ(r ) (8.30)
dr r2

dM
Mass continuity : = 4πr 2ρ(r ) (8.31)
dr

dT  3  kρ L(r )
Energy transport: = −  3 2
(8.32)
(radiative)
dr  4ac  T 4πr

dT  1  µm GM (r )
Energy transport: = −1 −  (8.33)
(convective) dr  γ  kB r2

dL
Energy generation: = 4πr 2ρ(r )ε(r ) (8.34)
dr
(Thermal equilibrium)

Equation of state: P (r ) = Rρ(r )T (r ) (8.35)

In the above equations, ε(r) is the thermonuclear energy production rate per unit mass.
You have learnt about it earlier in relation with luminosity. Further, the opacity k
occurring in these equations depends on temperature and density of the gas. In fact,
the exact form of the opacity relation depends on the process responsible for it.
Computation of stellar opacity is, however, a complex process and is usually
approximated by Kramer’s opacity relation given as
79
The Solar System ρ
and Stars k = const.Z (1 + X ) (8.36)
T 3.5

where X denotes the amount of hydrogen in a gram of stellar matter (it is also called
the abundance of hydrogen), Z denotes the abundance of heavier elements. (In
astrophysics, elements heavier than helium are called heavier elements.) The opacity
relation given by Eq. (8.36) is valid for stars on the main sequence.

To obtain the values of physical parameters by integrating the stellar structure

equations, we invoke the following boundary conditions:

M(r) = 0 and L(r) = 0 at r = 0 (the centre of star), (8.37)

and

M(r) = M; L(r) = L and T(r) = Teff at r = R (surface of a star)

(8.38)

With this background information, we are now in a position to discuss a stellar model.

8.6.1 Polytropic Stellar Model

In such a model, there is no need to know the actual source of energy generation in the
star. Further, we assume that any change in the equilibrium structure of a star takes
place in such a way that the specific heat remains constant, i.e.,

dQ
= C = constant (8.39)
dT

where C denotes the heat capacity when neither pressure (P) nor volume (V) is
constant. We call such a change as a polytropic change. An adiabatic or an
isothermal change, therefore, represents a polytropic change of zero and infinite
heat capacities, respectively. Instead of the adiabatic relation dQ = 0, we now have
dQ = CdT. In such a situation, the first law of thermodynamics

dQ = CV dT + PdV

takes the form

CdT = CV dT + PdV

Now, using the equation of state for a perfect gas, PV = RT and the fact that
R = CP − CV, we can write the above expression as:

(C − CV ) dT = (C P − CV ) dV (8.40)
T V

where CP and CV are the specific heats at constant pressure and constant volume,
respectively. Let us now define an exponent γ′ similar to the adiabatic exponent γ:

CP − C
γ′ = (8.41)
CV − C

Thus, we can write from Eq. (8.40) that

80 TV γ′−1 = const, PV γ′ = const . (8.42)

SAQ 5 Spend Stellar Structure
5 min.
Derive Eq. (8.42).

It is usual to express the physical variables, e.g., density, pressure and temperature in
terms of the polytropic index n defined as:

1
n= (8.43)
γ′ − 1

Since PV γ′−1 = const. , we can write:

1
1+
P = Kρ n (8.44)

where K is a constant. Further, the density is expressed in terms of a non-dimensional

parameter θ defined in terms of the central density ρc as

ρ = ρc θ n. (8.45)

We, therefore, get the following expressions for pressure and temperature:

P = Pc θ n+1 (8.46a)

T = Tc θ (8.46b)

n +1
µm
where Pc = Kρ c n and Tc =   Kρ c . We shall see below that Pc and Tc are the
1/ n

 kB 
central pressure and temperature.

SAQ 6 Spend
5 min.
Derive Eq. (8.46a) and (8.46b).

With this formal introduction to polytropic changes, let us consider the following
stellar structure equations:

dP GM ( r )
Hydrostatic Equilibrium: =− ρ( r ) (8.47)
dr r2
dM
Mass continuity: = 4πr 2 ρ(r ) (8.48)
dr

Substituting Eq. (8.48) in Eq. (8.47) and rearranging terms we can write:

1 d  r 2 dP 
= − 4π G ρ(r ) (8.49)
r 2 dr  ρ dr 

Substituting for P and ρ from Eqs. (8.44) and (8.45) in Eq. (8.49), we get:

 1
− 1
 (n + 1) K ρ n  1 d  r 2 dP  = − θ n (8.50)
 4πG c  r 2 dr  dr 
 
81
The Solar System To write Eq. (8.50) in a simpler form, let us introduce a dimensionless variable ξ as
and Stars follows:

r = αξ; (8.51)

1
 1 2
 (n + 1) K n − 1 
where α = ρ . Substituting Eq. (8.51) in Eq. (8.50) and rearranging
 4πG c 
 
terms we get:

Lane-Emden equation

1 d  2 dθ 
ξ  = − θ n (8.52)
2 dξ  ξ
ξ  d 

Eq. (8.52) is known as Lane-Emden equation. Solution of this equation, for a given n,
provides the density and pressure profile inside a star. The boundary conditions
under which this equation must be solved are:

dθ
θ = 1 and =0 at ξ = 0. (8.53)
dξ

θ is also known as Lane-Emden’s function.

Analytical solutions of Eq. (8.52) with the specified boundary conditions are possible
only for n = 0, 1 and 5. The analytical expressions for the Lane-Emden functions for
these values of n are:

ξ2
n = 0; θ 0 = 1 −
6
sin (ξ)
n = 1; θ1 = (8.54)
ξ
1
−
 ξ 2  2

n = 5; θ 5 = 1 +
 3 


Fig.8.7 shows the density profile inside a polytrope for n = 1.5 and 3.

Spend SAQ 7
10 min.
Verify that the Lane-Emden equation (Eq. (8.52)) is satisfied by the solutions given by
Eq. (8.54).

The general solution of the Lane-Emden equation is in the form of a series for θn as
given below:

ξ2 n 4
θn = 1 − + ξ − ... (8.55)
6 120

82
 Stellar Structure
1
Polytropes

0.8

0.6 n = 1.5
ρc
ρ /ρ

0.4
n=3

0.2

0
0 0.2 0.4 0.6 0.8 1
r/R
Fig.8.7: Density profile of a polytropic star
Current stellar models calculated for the Sun that fit the observations indicate that a
large fraction of hydrogen at the centre of the Sun has been converted to helium
(estimated 40 percent hydrogen and 60 percent helium). Energy transport is still
radiative in the solar interior upto ~ 0.73 RΘ and beyond this distance, the temperature
gradient reaches a value which sets up convection. Fig. 8.8 illustrates the internal
structure of the present Sun.
Photosphere

Chromosphere

2000 km
Convection
Zone
330 km
Radiative
Zone
1.6 × 107 K
160 × 103 kg m−3
≈ 1/4R 0.80R Corona
1R

Core

8 × 104

20 × 103 5 × 105
106
Temperature (K)

4 × 10−4
6 × 103 K

8 × 10−5

Density (kg m−3)

Fig.8.8: The internal structure of the Sun as per a stellar model which shows the main regions of
the Sun and values of its important physical parameters 83
The Solar System Computation of actual stellar models involves complex mathematics and needs
and Stars extensive computer resources. It is primarily due to the complexities of the equations
of stellar structure.

Now, let us summarise what you have learnt in this Unit.

8.7 SUMMARY
• To understand the internal structure of stars, theoretical models of stars are
developed and the predictions of these models are compared with observations.

• The principle of hydrostatic equilibrium is one of the fundamental principles of

physics invoked for investigating stellar structure.

• The equation of hydrostatic equilibrium is given as:

dP GM (r )
=− ρ( r )
dr r2

• The mass continuity equation is given as:

dM (r )
= 4πr 2 ρ( r ).
dr

• The relation between total thermal energy and the gravitational potential energy of
a system of perfect gas particles such as a star is given by the so called virial
theorem:

2U + Ω = 0.

• Of the three possible sources namely gravitational, chemical and nuclear of

energy generation in stars, only nuclear processes can give rise to such high
internal temperatures. Energy is generated in stars due to fusion reactions,
particularly due to fusion of hydrogen nuclei and consequent formation of helium.

• One of the basic equations of stellar structure, as given below, relates the
luminosity (an observable parameter) of a star with its density:

dL
= ε 4πr 2 ρ( r )
dr

where ε is the rate of energy generated per unit mass in thermonuclear reactions.

• Energy generated at the core of a star is transported to its surface through one or
more than one of the three basic energy transport processes namely conduction,
radiation and convection.

• The temperature gradient in a star is produced due to radiative transport of

energy and is given as:

dT 3 k (r ) ρ(r )  L(r ) 
=−  
dr 4 acT 3  4πr 2 

• Equations of stellar structure are used to develop theoretical models of stars.

If the predictions of a model are in agreement with observations, we conclude that
84 the assumptions made about the parameters of stellar interior are valid.
• Developing a stellar model basically involves solving equations of stellar Stellar Structure
structure. This is quite a complex process because a large number of equations
with several unknowns are to be solved.

• In polytropic stellar model – a relatively simple stellar model – we do not need to

know the source of energy generation. In this model, it is assumed that any change
in the equilibrium structure of a star do not alter its heat capacity:

dQ
= C = constant
dT

• The Lane-Emden equation is given as:

1 d  2 dθ 
ξ  = − θ n
2 dξ  dξ 
ξ 

Solution of this equation provides the density and pressure profile inside a star.

8.8 TERMINAL QUESTIONS Spend 25 min.

1. Show that when four protons combine to form helium, the energy released is
~ 26.7 MeV.

2. At present, the Sun radiates energy at the rate of ~ 4 × 1026W. Assuming that the
gravitational contraction is the only source of the Sun’s radiant energy, how long,
since its creation, it would have radiated energy at the present rate? Take
M Θ = 2 × 10 30 kg and RΘ = 7 × 10 8 m .

3. Show that Eq. (8.55) is consistent with Eq. (8.54).

8.9 SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs)

1. The left hand side of Eq. (8.8) is:

R R 3 R
3∫0 P 4πr 2 dr = 3 ∫0 PdV = 2
2 0 ∫ PdV = 2U

3
since the internal or thermal energy can be expressed as
2∫PdV .

2.a) From Eq. (8.11), we can write the total potential energy as:

R GM ( r )
−Ω= ∫0 r
4πr 2 ρ(r ) dr

Assuming that ρ(r) ≈ < ρ > , the mean density of stellar matter, we can write:

R GM ( r ) RG 4π 3
−Ω= ∫0 r
4πr 2 < ρ > dr = ∫ 0 r 3
r < ρ > 4πr 2 < ρ > dr

85
The Solar System R 4 π 4π 4 R M2
and Stars = ∫0 3G
3 3 0
∫
r < ρ > 2 dr = 3 G.r 4
R6
dr

(using Eq. (8.13))

2 R
3GM
∫0 r
4
= dr
R6
or,
3 GM 2
Ω=−
5 R

This is Eq. (8.14). Further, from the virial theorem (Eq. (8.12)), we can write:

Ω
U =−
2

3 GM 2
= (using Eq. (8.14))
10 R

Also, from Eq. (8.16), we have:

3 kB
U= M <T >
2 µm

Comparing the above expressions for U, we can write:

3 GM 2 3 kB
= M <T >
10 R 2 µm
or,
1 GM µ m
< T >=
5 R kB

Further,

4
M= πR 3ρ
3

or,
1
 M 3
R ∝  
 ρ 

Substituting this value of R, we get

2 1
<T > ∝M 3 <ρ> 3

which is Eq. (8.17).

b) Eq. (8.17) can be written as:

1 GM µ m
<T > =
5 R kB
86
 Substituting the values of G, M, R and kB, we get: Stellar Structure

1 (6.7 × 10 −8 cm 3 g −1s − 2 ) × (2 × 10 33 g) × (1.6 × 10 −24 g)

<T >= .
5 (1.4 × 10 −16 erg K −1 ) × (7 × 1010 cm)

6
≈ 4 × 10 K.

3. Mass of four hydrogen atoms = 4.031280 u

Mass of a helium atom = 4.002603 u

So, the mass defect = 0.02867 u

Thus, the energy released per gram when four protons combine to form helium
is

(0.02867 u) × (9 × 10 20 cm 2 s − 2 )
=
4.0328

≈ 6 × 1018 erg.

Thus, we can write:

(2 × 10 33 g) × (6 × 1018 erg g −1 )
lifetime of the Sun =
( 4 × 10 33 erg s −1 )

= 3 × 1017 s

≈ 1010 yr.

4. We can write the potential energy of two nuclei Z1e and Z1e separated by a
distance r as:

Z1 Z 2 e 2
P.E. =
r

e2
= if Z1 = Z2 = 1
r

Since the potential barrier will be overcome by the relative kinetic energy of
the two nuclei, we can write:

e2
3k B T =
r
or,
e2
T=
3k B r

4.8 × 4.8 × 10 −20

=
3 × 1.4 × 10 −16 × 10 −11

≈ 5 × 107 K.
87
The Solar System 5. From Eq. (8.40), we have:
and Stars

(C − CV ) dT = (C P − CV ) dV
T V

If we define an exponent γ ′ as

Cp −C
γ′ =
Cv − C

we can write:

CP − C
γ′ − 1 = −1
CV − C

C P − C − CV + C
=
CV − C

C P − CV
=
CV − C

Further, Eq. (8.40) can be written as:

dT dV
(CV − C) + (C P − CV ) =0
T V
or,
dT dV
+ ( γ ′ − 1) =0
T V

′
TV γ −1 = Const.

Further, using the equation of state of a perfect gas:

PV = RT

we can write:

dP dV dT
+ = .
P V T

dT
Substituting for , we get:
T

dP dV
+ γ′ =0
P V
or,
′
PV γ = Const.

6. From Eq. (8.44), we have:

n +1
P = Kρ n

88
 n +1 Stellar Structure
= K ρc θ( n
) n (using Eq. (8.45))

n +1
n
= Kρ c θ n +1

= Pc θ n +1

n +1
n
where Pc = Kρ c

Further, we can write pressure as:

kB
P= ρT
µm
or,
µm P
T=
kB ρ

n +1
µm ρ n
= K (substituting Eq. (8.44))
kB ρ

µm µm
= Kρ1 / n = K (ρ c θ n ) 1 / n (substituting Eq. (8.45))
kB kB

µm
= Kρ1c/ n θ
kB

µm
= Tc θ where Tc = Kρ1c/ n
kB

7. Lane-Emden equation (Eq. (8.52)) is:

1 d  2 dθ 
 ξ  = − θ n
2 dξ dξ 
ξ 

For n = 0, this equation reduces to:

1
d  2 dθ 
 ξ  = −1
ξ 2 dξ  dξ 

ξ2
And, for n = 0, we have θ = 1 −
6

Substituting this value of θ in the left hand side of the above equation, we get:

1
d  2  ξ  1 d  ξ 3  1  3ξ 2 
 ξ  −   = − = − = −1.
ξ 2 dξ
  3   ξ 2 dξ  3  ξ 2  3 

89
The Solar System sin ξ
and Stars Further, for n = 1, we have θ = from Eq. (8.54). So, we get:
ξ

dθ cos ξ sin ξ
= −
dξ ξ ξ2

dθ
Again, substituting this value of in the left hand side of Lane-Emden
dξ
equation, we get:

1 d
2
(ξ . cos ξ − sin ξ )
ξ dξ

1 sin ξ
=
2
[− ξ sin ξ + cos ξ − cos ξ] = − = −θ
ξ ξ

as required.
−1 / 2
 ξ 2 
For n = 5, we have, θ = 1 +
 3 

or,
−3 / 2
dθ 1 ξ 2  2ξ
= − 1 + .
dξ 2  3  3

dθ
Substituting the value of in the left hand side of Lane-Emden equation, we
dξ
get:

  2
−3 / 2 
1 d  ξ3 1 + ξ  
−   
ξ 2 dξ  3  3  
 

  2
−3 / 2
 2
−5 / 2 
1  3ξ 2 1 + ξ  ξ 3
3  ξ  2ξ 
= − + . 1+ .
ξ 2  3  3  3 2  3  3
  

−3 / 2 −5 / 2 −5 / 2
 ξ 2  ξ2  2   ξ 2   ξ2 ξ2 
= − 1 + + 1 + ξ  = − 1 + 1 + − 
 3  3  3   3   3 3 
  

−5 / 2
 ξ 2 
= − 1 +
 3 


= − θ5 , as required.

Terminal Questions

1. When four protons combine to form a helium atom, the mass defect is 0.02867 u.
Thus, energy released in this process:
90
 E = mc 2 Stellar Structure

= (0.02867 ×1.6 ×10 − 27 kg) × (9 ×1016 m 2 s − 2 )

= 4.128 ×10 −12 J

4.128 ×10 −12

= eV
1.6 ×10 −19

= 26 MeV.

2. Assuming the Sun to be a sphere, its gravitational potential energy can be written
as:

3 GM 2
Ω=
5 R

So, the time τ for which the Sun will radiate with its luminosity, L can be written
as:

3 GM 2 1
τ= .
5 R L

3 GM 2 1 1
= . . yr .
5 R L 3 × 10 7

3 (6.7 × 10 −11 m 3 kg −1 s −1 ) × (4 × 10 60 kg 2 ) 1
= × yr .
5 (7 × 10 8 m) × (4 × 10 26 kg m 2 s −1 ) (3 × 10 7 )

1
= 10 8 × yr
5

≈ 2 × 10 7 yr .

3. Eq. (8.55) is:

ξ2 n 4
θn = 1 − + ξ − ...
6 120

For n = 0, this equation reduces to:

ξ2
θ0 = 1 − → a constant
6

for n = 1,

1 1 ξ3 ξ5 
θ1 = sin ξ =  ξ − + − ... 
ξ ξ  3! 5! 


91
The Solar System
1  ξ3 ξ5  ξ2 ξ4
and Stars = ξ− + − ... = 1 − +
ξ  6 120 
 6 120

ξ2 n 4
=1− + ξ − ...
6 120

for n = 5,

−1 / 2
 ξ 2  1 ξ2 − 1 − 3 1 ξ4
θ 5 = 1 + =1− . + . . .
 3  2 3 2 2 2 9


ξ2 ξ4 ξ2 5 4
=1− + =1− + ξ − ...
6 24 6 120

ξ2 n 4
=1− + ξ − ...
6 120

Thus, we find that Eq. (8.55) is consistent with Eq. (8.54).

92
 Star Formation
UNIT 9 STAR FORMATION
Structure
9.1 Introduction
Objectives
9.2 Basic Composition of Interstellar Medium
Interstellar Gas
Interstellar Dust
9.3 Formation of Protostar
Jeans Criterion
Fragmentation of Collapsing Clouds
9.4 From Protostar to Pre-Main Sequence
Hayashi Line
9.5 Summary
9.6 Terminal Questions
9.7 Solutions and Answers

9.1 INTRODUCTION
In Unit 8, you have learnt the basic physical principles governing the processes in the
interior of stars. The existence of stars was taken for granted and their coming into
being was not discussed. It is, however, logical to ask: Where do the stars come from?
How are the stars formed? In general terms, you did study about the formation of the
solar system including the Sun (the only star in our solar system) in Unit 6. But the
focus of that unit was to understand the formation and characteristics of the planets. In
the present unit, you will learn about the formation of stars.
You must have observed stars in the night sky. On a clear dark night away from city
lights, you can observe that the space surrounding the bright stars is fairly luminous.
This is due to the scattering of light from the star by the gas and dust in its
surroundings. Further, astronomical observations suggest that the more luminous and
massive stars are younger and have been formed recently. Since these stars are It is usual to write our
completely surrounded by gas and dust, it is believed that the stars are formed from it. galaxy with the upper case
The gas and dust in the interstellar space is called the Inter Stellar Medium (ISM) and G, that is, as Galaxy.
it is the major constituent of our galaxy − the Milky Way Galaxy. You will learn the
basic composition of ISM and methods of its detection in Sec. 9.2. Under suitable
conditions, a cloud of gas and dust condenses or collapses due to its own gravity and
forms protostars. In Sec. 9.3, you will learn the Jeans criterion which is a rough
measure of the mass and size of the interstellar cloud which may collapse and give
birth to a star. You will discover that the interstellar cloud must fragment repeatedly to
form stars. In Sec. 9.4, you will learn how a protostar evolves into a full fledged star
and becomes a member of the main sequence (in the H-R diagram discussed in
Unit 7).
The various stages from the initial collapse of a cloud of ISM upto the pre-main
sequence are collectively considered as birth of a star, that is, the process of its
formation. You may ask: What happens afterwards? This issue is addressed in the
next Unit on Nucleosynthesis and Stellar Evolution where we discuss the life of stars.
Objectives
After studying this unit, you should be able to:

• describe the basic composition of ISM;

• derive Jeans criterion for the stability of a gas cloud;
• explain the necessity of repeated fragmentation of a collapsing cloud; and 5
• discuss the evolution of an interstellar cloud into a pre-main sequence star.
From Stars to Our
Galaxy 9.2 BASIC COMPOSITION OF INTERSTELLAR
MEDIUM

The interstellar medium (ISM) makes up only 10 to 15 percent of the visible mass of
the Milky Way Galaxy. It comprises matter in the form of gas and dust (very tiny
solid particles). About 99 percent of ISM is gas and the rest is dust. You may like to
know: Which elements are present in the ISM? An important clue for investigating
the composition of the ISM is the fact that the birth and death of stars is a cyclic
process. This is so because a star is born out of ISM, and during its life, much of the
material of the star is returned back to the ISM by the process of stellar wind (in case
of the Sun, it is solar wind; see Unit 5) and other explosive events such as Nova and
Supernova. The material thrown back into the ISM may form the constituents of the
next generation of stars and so on. To know the basic composition of ISM,
astronomers use photographs and spectra. In the following discussion, we shall confine
ourselves to the composition of the ISM of the Milky Way Galaxy in the
neighbourhood of the Sun. Further, for simplicity, we will first discuss the
composition of interstellar gas and then come to interstellar dust.

9.2.1 Interstellar Gas

Hydrogen and helium are the two major constituents of interstellar gas; hydrogen
constitutes about 70 percent and the rest is helium. Indeed, other elements are also
present but in very small quantities. The analysis of the radiation received from ISM
has enabled astronomers to classify the gaseous matter filling the interstellar space
into the following four types:
i) H II region,
ii) H I region,
iii) Inter-cloud medium, and
iv) Molecular cloud.

We now briefly describe these regions and their possible roles in the formation of
stars.

i) H II region: As the name suggests, these regions of ISM mainly consist of

(singly) ionised hydrogen. In addition, these regions contain ions of other
elements such as oxygen and nitrogen and free electrons. Since the ionisation
energy of hydrogen atom is very high, such regions can exist only in the
vicinity of very hot stars. H II region can be viewed even with naked eye in the
constellation of Orion. If you look carefully at Orion’s sword, you will see that
one of the objects is a hazy cloud of gas (Fig. 9.1). This bright cloud is called
Orion nebula. The spectrum of this nebula shows lines of hydrogen and some
other elements in its bright line spectrum. Such bright nebulae are also called
emission nebulae.

It has been discovered that emission nebulae do not shine by their own light. They
absorb high energy ultraviolet photons from hot stars. These photons ionise the
gas in the nebulae and subsequently, low energy photons are radiated. Since the
spectrum of the nebula consists of many emission lines of hydrogen, it indicates
that the light must have been emitted by a low-density gas. Further, the emission
lines of hydrogen are very strong and the red, blue and violet Balmer lines blend
together resulting in the characteristic pink-red colour of the nebula (Fig. 9.1).

6
 Star Formation

Fig.9.1: Orion nebula − a typical bright and diffuse nebula

At this stage, you may ask: Why are the H II regions always found in the
emission nebulae? Note that only those photons which have wavelengths shorter
than 91.2 nm have sufficient energy to ionise hydrogen. Such high energy photons
can be produced in sufficiently large numbers by hot (~ 25,000 K) stars only. And
stars (such as hot O or B star) having temperature of this order are located in or
near the emission nebulae. Further, H II regions have very low density
(~ 109 particles m−3). They provide observable evidence supporting the existence
of matter in interstellar space.

ii) H I region: Although it has been generally believed by astronomers that hydrogen
atoms populate interstellar space, they could not detect H I gas till 1951. The
reasons are obvious: it is not possible for the neutral hydrogen in ISM to produce
emission line as it is in the ground state. However, with the development of radio
telescopes, it is now possible to detect H I region. The detection of H I in the
ISM is based on the detection of a unique radiation of wavelength
21 cm.

SAQ 1 Spend
5 min.
Calculate the energy of electromagnetic radiation having wavelength 21-cm.

On solving SAQ 1, you must have found that the value of the energy of radiation
corresponding to wavelength 21-cm is very small. You may ask: What kind of
transition produces such a low energy photon? In a hydrogen atom, an electron
revolves around the proton. Since the proton and the electron possess spin, there
are two possible ways for their spins to align with respect to each other. The two
spins may be parallel (aligned) or anti-parallel (anti-aligned) (Fig. 9.2). It is known
that the parallel spin state of hydrogen atom has slightly more energy than the
anti-parallel spin state. Therefore, if there is a flip from the parallel to the anti- 7
parallel state, there is a loss in energy of the hydrogen atom and it results in the
From Stars to Our emission of a photon. The frequency and the wavelength of such emissions are
Galaxy
1420 MHz and 21 cm, respectively.

21-cm
radiation

Proton Electron

Parallel spins Anti-parallel spins

Fig.9.2: Parallel and anti-parallel spin alignments of the electron and proton in a
hydrogen atom

Further, the question is: Can we obtain 21-cm radiation in laboratory

conditions? It is not possible because the best vacuum that can be produced in a
laboratory has a much higher density of atoms than that found in the ISM. So, in
laboratory conditions, practically all the hydrogen atoms get de-excited due to
higher collision rate and we cannot obtain 21-cm line. Thus, the only place where
favourable conditions for emission of such radiations can exist is outer space. The
21-cm line was first detected in the year 1951 using a radio telescope even though
it was predicted as early as in the 1940s. Since its detection, it has become a
common tool in astronomy to map the location and densities of H I regions. This
helps in determining the structure of galaxies including our own. Investigations of
diffuse interstellar H I clouds suggest that their temperatures are in the range of 30
– 80 K, masses in the range of 1 – 100 solar masses and the number densities in
the range of 108 – 109 cm−3.

iii) Inter-cloud Medium: Having read about clouds of ionised and neutral hydrogen
in ISM, you would like to know: Is the space between the interstellar clouds
empty? It is not so; the inter-cloud space consists of

a) neutral hydrogen gas with a density of 105 atoms m−3, and

b) hot (~ 8000 K) ionised gas with very low density (~ 104 ions m−3).

You would further like to know: Is there any interaction between the H I
clouds and the inter-cloud medium? The H I clouds are very cool and have high
densities whereas inter-cloud medium has very low density and high temperature.
The pressure of a region, being a function of its density and temperature, in the
H I clouds and in the inter-cloud medium is about the same and they are in
equilibrium.

iv) Molecular Cloud: Analysis of optical spectra of interstellar medium reveals that
matter exists in molecular form in ISM. Since hydrogen is the most abundant
matter in ISM, it mainly consists of the hydrogen molecules (H2). However,
molecules of hydrogen do not emit photons of radio wavelength and vast clouds
of hydrogen molecules remain undetected by radio spectroscopy. Other
molecules, such as CO (carbon monoxide), capable of emitting in radio
Hydrogen molecules have wavelength, can indeed be detected. In fact, nearly 100 such molecules have been
been detected in the ISM by
detected. But the basic question is: How do these molecules form in ISM? It is
infrared spectroscopy.
believed that the atoms come in the vicinity of each other and bond to form
molecules on the surfaces of the dust grains (about which you will learn later in
this Section). These molecules are very weakly bonded and can be easily broken
by high energy photons. Thus, they can exist only deep inside dense clouds. Also,
8 efficient release of energy by molecules makes these dense clouds very cool.
These dense, cold clouds are called molecular clouds.
 In our Galaxy, the largest of these cool, dense molecular clouds are called giant Star Formation
molecular clouds (GMC). They are 15 to 60 pc across and may contain 100 to
106 solar masses! The internal temperature of GMC is very, very low (~ 10 K).
The question is: Do the giant molecular clouds have any role in the formation
of stars? You know that young stars are surrounded by H II regions. And H II
regions are invariably found near giant molecular clouds. This proximity indicates
that the GMC plays an important role in the formation of stars. Thousands of
GMC exist in the spiral arm of our Galaxy. The association of O and B main
sequence stars with GMC suggests that star formation takes place in these regions.
We will talk more about it later in this Section.

9.2.2 Interstellar Dust

Interstellar medium also contains dust grains, which constitute about one percent of
the interstellar mass. Although the temperature of interstellar dust region is very low
(~ 100 K), it can be detected by infrared telescopes. A typical dust grain is made of
thin, highly flattened flakes of silicates and graphites coated with ice. Fig. 9.3 shows a
typical dust grain, which is of the size of the wavelength of blue light.

Mantle (ice)

Core
(Silicates,
Graphite)

0.1 µm

Fig.9.3: A typical dust grain

Now, you may ask: What is the evidence supporting the existence of dust in ISM?
The two observable effects due to dust are extinction and reddening. Refer to
Fig. 9.4. Note that, besides the brighter gas and dust regions surrounding the stars,
there are darker regions as well. You may think that these regions are devoid of stars.
It is not so. In fact, these darker regions are so dense that they completely stop the
light emitted by stars behind them and therefore no light is able to pass through. This
phenomenon is known as interstellar extinction. The extent to which light is
scattered or absorbed in a dust cloud depends on the number density of particles and
on its thickness.

9
From Stars to Our
Galaxy

Dark region

Fig.9.4: Dark interstellar space caused due to extinction

To obtain an expression for the apparent magnitude of a star, located behind the dense
cloud of dust, recall (from Unit 1) that the relation between the apparent magnitude
(mλ), the absolute magnitude (Mλ) and the distance d in parsec of a star at wavelength
is written as:

mλ = M λ + 5 log10 d − 5 (9.1)

since the absorption and scattering of light is dependent on the wavelength. For stars
suffering extinction, we can write Eq. (9.1) as:

mλ = M λ + 5 log10 d − 5 + aλ (9.2)

where aλ is the magnitude of light scattered or absorbed along the line of sight.
Eq. (9.2) indicates that absorption increases the magnitude of a star. A star which
would be visible to the naked eye, for instance, may be invisible due to the large
extinction i.e., sufficiently large aλ. Further, the extinction may be expressed in terms
of the optical depth as:

aλ = 1.086 τλ (9.3)

Spend SAQ 2
10 min.
The relation between optical depth, τλ and intensity Iλ for a star is given by
I λ = I λ 0e − τ λ .

Show that aλ = 1.086 τλ.

(Hint: Remember that apparent magnitude may be written as m = K− 2.5 log I, where
K is a constant.)

To appreciate the fact that extinction depends on the density of dust grains, we can
express the optical depth τλ in terms of the number density of particles, n and
scattering cross section σλ as:
s

10
τλ = ∫0 n ( s) σ λ ds. (9.4)
Assuming that the scattering cross-section σλ is constant along the line of sight, we Star Formation
obtain from Eq. (9.4):

τλ = σλ Nd (9.5)

where Nd is the column density of particles, i.e., the number of particles in a cylinder
of unit cross section stretching from the star to the observer. Eq. (9.5) shows that
extinction depends on the amount of interstellar dust present in the path of light from
the star.

BD + 56524

HD 48099
aλ / a ν

Herschel 36

0 2 4 6 8
l/λ (µm-1)

aλ 1
Fig.9.5: Schematic diagram showing the variation of with
aν λ

aλ 1
Fig. 9.5 shows the variation of with . Here, aν is the amount of extinction in
aν λ
o
the visual band of wavelengths centered at 5500 A . In Fig. 9.5 we observe a peak in
the ultraviolet region which indicates that radiations of corresponding wavelengths are
strongly absorbed by ISM. Such a peak, therefore, provides a basis for determining
the composition of ISM. It is now known that graphite interacts strongly with
o
electromagnetic radiation of wavelength around 2175 A . Therefore, the occurrence of
o
peak at λ = 2175 A in Fig. 9.5 suggests the presence of graphite as one of the
constituents of ISM. Further, the presence of absorption bands at 9.7 µm and 18 µm in
the observed spectrum (not shown in Fig. 9.5) indicate the presence of silicate grains
in the ISM.

Yet another manifestation of interstellar dust is in the form of interstellar reddening.

You know that an O star should be blue in colour. But, it has been observed that some
stars with the spectrum of an O star look much redder. This is caused due to scattering
of light from stars by interstellar dust.

11
From Stars to Our Blue
Galaxy light

Red
light

Star
Telescope

Blue
Interstellar
light
dust

Fig.9.6: Interstellar reddening caused due to the scattering of blue light coming from star

Refer to Fig. 9.6. The light coming from a star behind the dust cloud is scattered.
Since the typical size of dust grains is of the order of the wavelength of blue light, the
blue light from star is scattered more than the red light. As a result, some of the blue
light from the star is lost and it appears redder.

You might wonder as to why we discussed ISM in such detail. It is because

astronomers suspect that stars are formed from ISM. The most important observation
supporting this hypothesis is the association between young stars and clouds of gas:
wherever we find the youngest group of stars, we also find large clouds of gas
illuminated by the hottest new stars. In the following section, you will learn about
the processes and principles involved in the formation of stars from ISM.

9.3 FORMATION OF PROTOSTAR

In the previous Section, you learnt that the giant molecular clouds (GMCs) are the
likely sites where star formation can take place. However, GMCs are very unlike
stars: their typical diameter is 50 pc, typical mass exceeds 105 solar masses and their
density is 1020 times less than a typical star. Thus, the question is: How do stars
form from such molecular clouds? The simplest answer to this question is that the
stars form due to gravity: a cloud of gas contracts due to self-gravity. This contraction
increases density and temperature of the core, initiating generation of nuclear energy.
You may ask : How do we reconcile the typical mass of a star with the mass of a
GMC (~ 105 solar mass)? This problem was addressed by Jeans who proposed a
criterion for the mass of a contracting cloud which may evolve into a star.

9.3.1 Jeans Criterion

Jeans proposed that there are two competing processes in the gravitational collapse of
a molecular cloud. On the one hand, the gravitational contraction increases the
internal pressure of the cloud which tends to expand the cloud. On the other hand,
gravity acts on the cloud and tends to further contract it. Which of these two processes
will dominate is determined by the mass of the cloud. If the internal pressure is more
than the gravitational force, the cloud will break up. A clump of cloud must have a
minimum mass to continue collapsing and give birth to a star. This minimum mass is
called the Jeans mass. It is a function of density and temperature.

To obtain an expression for the Jeans mass, we make the following simplifying
assumptions:

i) the cloud is uniform and non-rotating;

ii) the cloud is non-magnetic; and
12 iii) the gas and dust is confined to a certain region of space by the gravitational force
and is in hydrostatic equilibrium.
For such a system, we may write the relation between kinetic and potential energies as Star Formation
(see virial theorem, Unit 8):

2U + Ω = 0, (9.6)

where U is the internal kinetic energy and Ω is the gravitational potential energy of
the cloud. If M and R are the mass and radius of the cloud, respectively, the potential
energy of the system can be written as (Eq. (8.14), Unit 8):

3 GM 2
Ω=− . (9.7)
5 R

If the number of particles in the cloud is N and its temperature is T, the internal kinetic
energy of the cloud can be written as:

3
U= Nk BT , (9.8)
2

M
where kB is the Boltzmann constant. Further, the number of particles N = ,
µm
where µ m is the mean molecular weight and m is the mass of a hydrogen atom. If the
total internal kinetic energy is less than the gravitational potential energy, the cloud
will collapse. This condition reduces Eq. (9.6) into

2U < Ω

Substituting for Ω and U from Eqs. (9.7) and (9.8), respectively, we get:

3k BTM 3 GM 2
< .
µm 5 R

or,

5k BTR
M> (9.9)
µ mG

1
 3M  3
On substituting R =   in Eq. (9.9), we find that the minimum mass that will
 4πρ 
initiate a collapse is given by:

3 1
 5k T  2  3  2
M ≈ MJ =  B    (9.10)
 µ m G   4πρ 

Here MJ is called the Jeans mass and Eq. (9.9) is known as Jeans criterion. The
Jeans mass is the minimum mass needed for a cloud to balance its internal pressure
with self-gravity; clouds with greater mass will collapse.

Jeans criterion can also be expressed in terms of the Jeans length, RJ given by:

1
 15k B T  2
R J ≈   (9.11) 13
 4π G µ m ρ 
From Stars to Our 4π 3
Galaxy which can be obtained by putting M = R ρ in Eq. (9.9). For pure hydrogen,
3
µ = 1.

Spend SAQ 3
5 min.
A collapsing cloud is made of neutral hydrogen (H I) only. If the temperature of the
5 −3
cloud is 50 K and its number density is 10 m , calculate its Jeans mass.

In a situation where Jeans criterion is satisfied, the cloud must collapse

gravitationally. When the collapsing could attain high density, it becomes
gravitationally unstable and breaks up into smaller pieces. Thus, the general picture is
that stars form in groups due to fragmentation. You will learn more about it later in
this section.

At this stage, you may ask: How long does it take for a cloud to collapse? It is a
good idea to estimate the minimum time which will be taken if we assume that the
cloud collapses only under the influence of self-gravity and there is no other process
taking place to slow down the collapse. This assumption is called free-fall collapse,
and it implies that the pressure gradient in the interior of the clump is negligibly small.

To obtain a rough estimate of the free-fall time, you may recall that a particle on the
surface of a star of mass M and radius R experiences an acceleration g given by:

GM
g= . (9.12)
R2

If the time taken by this particle to fall through a distance R is tff we can write:

1
R= g t ff 2
2
or
1
 2R 2
t ff =   (9.13)
 g 

Substituting Eq. (9.12) in Eq. (9.13), we get:

1
 2R 3 2
tff =  
 GM 
 
1
  2
 
2R 3
= 
 4π 3 
G R ρ 
 3 
1
 3 2
=   (9.14)
 2πGρ 
Eq. (9.14) shows that the free-fall time, tff is a function of the cloud’s initial density
and it does not depend on the initial radius or the initial mass of the cloud. If some
part of a collapsing cloud, say in the central region, is denser than its surroundings, the
collapse of such a region is likely to be faster than that of the surrounding region.
14
SAQ 4 Spend Star Formation
5 min.
Calculate the free-fall time for a molecular cloud whose initial density is
10−17 g cm−3.

Now you know that the larger gas clouds collapse if their masses exceed the Jeans
mass. In a typical situation of an H I cloud with T = 100 K, ρ = 10−24 g cm−3 and
µ = 1, we find that for gravitational collapse, the mass of the cloud must be greater
than 105 times the solar mass. So, you may be led to believe that the stars could be
formed with masses of this order. However, observations suggest that the stars are
formed in groups and the masses of the stars are in the range of 0.1 − 120 MΘ. Thus,
the range of stellar mass is much smaller than 105 MΘ. This has led astronomers to
propose the fragmentation of interstellar clouds during collapse. You will now learn
about it.

9.3.2 Fragmentation of Collapsing Clouds

As mentioned earlier, Jeans criterion provides the theoretical justification for
fragmentation of a collapsing cloud. Refer to Fig. 9.7 which shows the fragmentation
of interstellar cloud of mass M and Jeans mass MJ such that M > MJ. Since M is
greater than MJ, the interstellar cloud collapses. This results into the increase of the
density and temperature of the cloud which may, in turn, change the Jeans mass for
the cloud. Let the changed Jeans mass be M ′J (≠ MJ ).

Interstellar
cloud
M > MJ

Collapse

M'J ≠ MJ Stage I

Stage Stage
M'J > MJ M > MJ > M'J
II III

M, M'J M'J < M1 M'J < M2

M, M'J M < M'J
M > M'J Stable M'J < M3

Stage
IV
Fragmentation
Stage
V
M'J < M1 M'J < M2

M'J < M3

Fig.9.7: Fragmentation of interstellar cloud

15
From Stars to Our Further, from Eq. (9.10) it is evident that, if the cooling of this cloud is not efficient,
Galaxy
the Jeans mass will increase (stage II in Fig. 9.7). If the mass of the cloud is less than
the new (increased) value of the Jeans mass, the collapse will stop and if M > M ′J ,
the cloud will collapse further. On the other hand, if the cooling of the collapsing
cloud is efficient, the temperature of the cloud will fall and the Jeans mass would
decrease (stage III in Fig. 9.7). In such a situation, it is possible that the mass in
certain regions of the cloud is more than the reduced Jeans mass (stage IV in Fig.
9.7). This may trigger further collapse of such regions resulting into fragmentation of
the cloud (stage V in Fig.9.7). Fragmentation of clouds continues until clouds of still
smaller masses, which are gravitationally stable, are created. These fragments of
molecular clouds are the birth places of the stars in the mass range of 0.1 − 120 MΘ.

You must note that for the scenario discussed above to be true, the Jeans mass should
not be a constant during collapse. It must decrease with increase in density in a local
region during collapse. This would lead to further gravitational instabilities which
may lead to separate individual collapsing regions.

Further, while discussing the collapse of interstellar clouds, we assumed that the
collapse process is isothermal. This can be considered to be a valid assumption as far
as the initial stages of collapse are concerned. As the collapse begins, the cloud is
likely to be optically thin due to its low density. Therefore, the gravitational energy
released during collapse is radiated away completely, keeping the temperature of the
cloud unchanged.

You may ask: What happens if the collapse is adiabatic instead of isothermal? In
that case, the energy released during the collapse is used up in raising the internal
energy of the cloud and its temperature increases. The increase in temperature thus
affects the Jeans mass. Let us now obtain an expression for the Jeans mass for
adiabatic collapse.

You may recall from the course entitled Thermodynamics and Statistical Mechanics
(PHE-06) that, for an adiabatic process, the relation between temperature T and
density ρ of a system is given by:

T = K1 ργ−1 (9.15)

where K1 is a constant and γ is the ratio of heat capacities. Using this relation in the
expression for the Jeans mass (Eq. (9.10)) we can write:

3 1
 5k T  2  3  2
MJ =  B   
 µmG   4πρ 

3 1
 5k K  2  3  2 ( 3 γ − 4 ) / 2
= B 1    ρ
 µmG   4π 

Thus
M J ∝ ρ ( 3 γ − 4) / 2 (9.16)

For the cloud comprising only atomic hydrogen, we have γ = 5/3. Therefore, for
adiabatic collapse of the cloud, we get:

M J ∝ ρ1/ 2 (9.17)
16
Eq. (9.17) shows that Jeans mass increases with increase in density. However, the Star Formation
relation between the Jeans mass and density for the isothermal collapse is given as
MJ ∝ ρ−1/2. Comparison of these two results indicates that in a switchover from
isothermal to adiabatic collapse, the Jeans mass reaches a minimum for the fragments
of molecular clouds.

The gravitational collapse is a complex process. It is difficult to ascertain exactly

when a switchover from isothermal to adiabatic collapse takes place. Moreover, the
transition is neither instantaneous nor complete. The minimum Jeans mass has been
estimated to be ~ 0.5 MΘ. This is the right order of magnitude for the mass of the
fragments of clouds which become stars.

The fragmentation of collapsing clouds leads to the formation of protostars − pre-

stellar objects which are hot enough to radiate infrared radiation but not hot
enough to generate energy by nuclear fusion. Due to gravitational collapse, the
fragmented clumps of cloud contract and at the core of each of them, high-density
region develops which is surrounded by a low-density envelope. Matter continues to
flow inward from the outer parts of the clumps. The protostar begins to take shape
deep inside the enveloping cloud of cold, dusty gas. These clouds are called cocoons
because they hide the forming protostar from our view. So, these are the initial stages
of the formation of stars. Let us now discuss this further.

9.4 FROM PROTOSTAR TO PRE-MAIN SEQUENCE

We have seen above that initially a cloud collapses, perhaps isothermally, and with the
passage of time the cloud collapse tends to become adiabatic. As a result, increased
pressure inside tends to slow the time scale of collapse particularly near the centre of
the cloud. Further, fragmentation of collapsing cloud leads to a stage when a
protostar is formed and its interior attains hydrostatic equilibrium. The interior of the
protostar at this stage becomes almost convective, that is, the temperature gradient
inside becomes larger than the adiabatic gradient. The stellar object so formed is
called a pre-main sequence star. To understand the evolution of a pre-main sequence
star, its position on the HR diagram and changes in its interior during evolution, you
must study the concept of a Hayashi line.

9.4.1 Hayashi Line

In the HR diagram, Hayashi line or track runs almost vertically in the temperature
range of 3000 to 5000 K as shown in Fig. 9.8. This line is important in the discussion
of pre-main sequence evolution of stars because of the following features:

i) When a protostar is formed, the interior of the cloud attains hydrostatic

equilibrium. At this stage, it is fully convective. The position of such an object
must fall on this line.

ii) For a given mass and chemical composition, this line represents a boundary in the
HR diagram. It separates the HR diagram into allowed and forbidden regions. The
forbidden region occurs to the right of this line and the stars in this region cannot
attain hydrostatic equilibrium. For stars falling to the left of this line, energy
transport due to convection/radiation or both is possible.

17
From Stars to Our
Galaxy

Luminosity (Log scale)

Hayashi line

Main
Sequence

3.8 3.7 3.6 3.5

Log Teff

Fig.9.8: The Hayashi line

The internal temperature of pre-main sequence stars is quite low and cannot ignite
thermonuclear reactions. To compensate for the loss of energy through radiation, a
protostar must contract and this leads to increase in its thermal energy. Because of
increase in temperature and pressure in the interior of the protostar, its collapse slows
down. In other words, we expect longer evolution time in the pre-main sequence
stage for smaller mass protostars. For more massive stars, the pre-main sequence
evolution is faster and their pre-main sequence lifetime is shorter.

Now, let us sum up what you have learnt in this unit.

9.5 SUMMARY
• The space between the stars, called the interstellar medium (ISM), is not empty.
It contains gas and dust, and the gas is mostly hydrogen.
• The interstellar gas is not uniformly distributed. At places, it is highly
concentrated. These regions are called gas clouds. It is in these gas clouds that
new stars are formed.
• The gaseous matter in the interstellar space is classified into four types, namely,
HI region, H II region, inter-cloud medium and molecular clouds.
• The interstellar dust, which constitutes about one percent of the interstellar mass,
gives rise to extinction and reddening. Extinction depends on the density of dust
grains and reddening is caused due to scattering of light from stars by interstellar
dust.
• Jeans proposed that a molecular cloud must have certain minimum mass, called
Jeans mass, for its collapse due to self-gravity. The expression for the Jeans
mass is:

3 1
 5k T 2  3 2
MJ =  B   
 µ mG   4πρ 
   

• In terms of the size, the Jeans criterion implies that if a gas cloud becomes larger
than a certain size, called Jeans length, it collapses under the gravitational force.
18 The expression for the Jeans length is:
 1 Star Formation
 15k B T  2
R J =  
 4πGµ mρ 

• The expression for the free-fall time taken by a cloud to collapse under the
assumption of free-fall collapse is:

1
 3 2
t ff =  
 2πGρ 

• Repeated fragmentation of collapsing clouds leads to the formation of protostars.

A protostar contracts due to gravitational force and its temperature and density
increases to such an extent that the nuclear reaction at its core becomes feasible.
A star is then said to be born. Such stars are called pre-main sequence stars.

9.6 TERMINAL QUESTIONS Spend 25 min.

1. What is the origin of the 21 cm line of hydrogen? Why can we not obtain this line
in a terrestrial laboratory? Explain the importance of this line in astronomy.

2. What is interstellar reddening? What does it tell us about the composition of

interstellar matter?

3. Derive Eq. (9.11). For the data given in SAQ 3, calculate the Jeans length.

9.7 SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs)

1. You know that the energy of electromagnetic radiation of frequency ν and

wavelength λ is given by:

E = hν

c
=h
λ

where h is the Planck’s constant and c is the velocity of light. Substituting the
values of h, c and λwe can write.

(6.63 × 10 −34 Js) (3 × 10 8 ms −1 )

E=
(0.21m)

6.63 × 3
= ×10 − 24 J
21

1
≈ 10 −24 × eV
1.6 × 10 −19

≈ 6.25 ×10 −6 eV

19
From Stars to Our 2. As per the problem,
Galaxy

I λ = I λ 0 e − τλ

Taking logarithm on both sides and multiplying by 2.5, we get :

2.5 log I λ = 2.5 log I λ 0 − 2.5 τ λ log 10 e

If m0 is the original magnitude and m is the increased magnitude because of

extinction, we have:

K − m = K − m0 − 2.5 × 0.4343τ λ
or,

m − m0 = 1.086τ λ

because we can write the apparent magnitude as:

m = K − 2.5 log I .

Further, the magnitude of light scattered or absorbed along the line of sight can
be written as:

a λ = m − m0 = 1.086 τ λ

3. From Eq. (9.10), we have the expression for Jeans mass:

3 1
 5k T  2  3  2
M J =  B   
 µ mG   4 π ρ 

For pure hydrogen, µ = 1 . So, we have:

3

MJ = 
( )
5 × 1.38 × 10 − 23 JK −1 × (50 K ) 2
−1 − 2 
( − 27
) ( −11
 1.67 × 10 kg × 6.67 × 10 m kg s 
3
)
1
 3 2
× −3 
5
( − 27
 4 π × 10 ×1.67 ×10 kgm  )
3 1

 5 × 1.38 × 5  10  300  2
2
= 10 24   × 10   kg
 1.67 × 6.67   4 π ×1.67 

10 34 × 5.45 × 3.75 × 10 3
= MΘ
2 × 10 33

≈ 10 5 M Θ

4. From Eq. (9.14), we have the expression for the free-fall time for a collapsing
cloud as:

1
 3 2
t ff =  
 2πGρ 
20  
 Substituting the values of G and ρ, we get: Star Formation

1
 2
 
 3 
t ff =  
 2π ×  6.67 ×10 −11 m 3 kg −1s − 2  × 10 −14 kgm −3  
     
 
1
  2
= 
30
×1012 s
 2π × 6.67 
 

1012
≈ yr ≈ 3 × 10 4 yr
7
3 × 10

Terminal Questions

1. See text.

2. See text.

3. The expression for Jeans mass is given by (Eq. (9.10)):

3 1
 5k T  2  3  2
MJ = B   
 µ m G   4π ρ 

Now, for a spherical cloud of mass M and radius R, we have:

4 
M =  π R3  × ρ
 3 

So, we can write:

4 
M J =  π R J3  × ρ
3 

Substituting the above expression for MJ in Eq. (9.10), we can write:

3 1
4 3  5k B T  2  3 2
 π RJ  × ρ =    4πρ 
3   µ mG   
or,
1
 15k B T 2
R J =  
 4πG µ m ρ 

As per the data given in SAQ 3, we have T = 50K and the number density
105 m−3. Further, for pure hydrogen, µ = 1 . Substituting these values in the
expression for RJ, we get:

21
From Stars to Our
Galaxy 1
 15 × (1.38 ×10 − 23 JK −1 ) × (50 K) 2
R J =  −11 3 −1 − 2 5 − 27 −3 − 27

−3 
 4π × (6.67 × 10 m kg s ) × (10 × 1.67 × 10 kg m ) × (1.67 × 10 kg m ) 

≈ 1019 m ~ 300 pc

22
 Nucleosynthesis and
UNIT 10 NUCLEOSYNTHESIS AND Stellar Evolution

STELLAR EVOLUTION
Structure
10.1 Introduction
Objectives
10.2 Cosmic Abundances
10.3 Stellar Nucleosynthesis
10.4 Evolution of Stars
Evolution on the Main Sequence
Evolution beyond the Main Sequence
10.5 Supernovae
10.6 Summary
10.7 Terminal Questions
10.8 Solutions and Answers

10.1 INTRODUCTION
In Unit 9, you have learnt that the gravitational collapse of interstellar cloud leads to
the formation of stars. You may recall that a stable star is formed when an equilibrium
between gravitational force in the collapsing cloud and the radiation pressure due to
nuclear energy generated in its core is attained. In this regard, some logical questions
which might come to your mind are: Since nuclear fuel burning at the core of a star
like the Sun is finite, what happens when the fuel is exhausted? Does the same type of
nuclear energy generation process take place in all stars? What happens to the material
produced in the nuclear reactions in the stars? If all the new born stars find place on
the main-sequence of the H-R diagram, how do we explain the existence of red giants,
supergiants and white dwarfs? To answer these questions, you need to learn about the
life of a star (that is, stellar evolution) after it has been formed and found its place on
the main-sequence. This is the subject matter of the present Unit.

An interesting approach to understand stellar evolution is to ask ourselves: How did

the ninety-three natural (chemical) elements, which are the building blocks of all
living and non-living matter around us, come into existence? All the elements,
except hydrogen and most of helium, were made inside stars through the process of
nucleosynthesis. The stars make elements during their life-time. Interestingly, all the
stars produce the same elements which we find on the Earth. Further, we find the
same elements everywhere in the universe and more or less in the same proportion.
The relative proportion of these elements in the Universe is called cosmic
abundances about which you will learn in Sec. 10.2. In Sec. 10.3, you will learn
about various nucleosynthesis processes taking place inside the stars at different
stages of their life. You will discover that the conditions for different processes are
different. This provides a basis to track the evolution of stars along the main-sequence
and afterwards. The low-mass and high mass stars evolve differently leading to
different end products such as white dwarfs, neutron stars and black holes. You will
learn about this in Sec. 10.4. In Sec. 10.5, you will learn that, when the entire nuclear
fuel at the core of a star is exhausted, rapid gravitational collapse takes place which
may result in a violent explosion, called supernova, of the stellar envelope.

Objectives
After studying this unit, you should be able to:
• list the various methods of determining cosmic abundances;
• list and explain various nucleosynthesis processes in the context of stellar
evolution; 23
From Stars to Our • describe where and how chemical elements are formed;
Galaxy
• explain the significance of stellar mass in respect of evolution of stars;
• estimate the lifetime of stars on the main sequence;
• describe the conditions for the formation of red giants, supergiants and white
dwarfs; and
• discuss conditions for the supernova explosions and the fate of left over stellar
cores.

10.2 COSMIC ABUNDANCES

You know that the atom is In school chemistry, you must have learnt about elements, atoms and molecules.
the smallest unit of an
element and atoms of each Elements are the simplest substances of ordinary matter. Scientists recognise
element are unique. Atoms 93 elements as natural. Besides these, there are 20 elements which can be
may be said to be the basic manufactured in the laboratory by bombarding nuclei with α-particles or other
building blocks of matter. high energy particles. These 20 artificial elements are not found in nature because
Molecules are formed when
atoms combine due to they are unstable and live for extremely short times.
electrical forces between
them. More complex Almost all the natural elements are found on the Earth. A few elements which are not
substances are formed when found on the Earth are found in other bodies of the solar system. You may be
there is chemical reaction surprised to know that the same 93 elements are also found elsewhere in the universe
between elements.
and their proportion is more or less the same as their proportion in the solar system.
The chemical composition of the universe refers to the presence of different types of
elements and their proportions in the universe. Since the chemical composition of the
universe and that of the solar system are similar, it is one and the same whether we
talk of the chemical composition of the universe, or the chemical composition of the
solar system. Further, the relative proportions of elements in the universe are called
cosmic abundances.

Determination of Cosmic Abundances

Now, the question is: How do we determine abundances in the solar system, that
is, how do we determine cosmic abundances? Some of the important methods to
obtain information about abundances are as follows:

i) In the solar system, the immediate source for obtaining such information is
obviously the Earth. Samples from many locations on the Earth are analysed in
the laboratory. Care is taken that these locations are as diverse as possible.
ii) The next obvious source is the Sun. You may recall from Unit 5 that the dark lines
in the solar spectrum, called Fraunhofer lines, are actually absorption lines due
to elements present in a slightly cooler layer above the photosphere. Each line in
the spectrum is checked against the sample spectra of elements and the elements
are identified. The intensity of a particular line gives the abundance of the
corresponding element. You may also recall from Unit 5 that the higher layers of
the solar atmosphere, the chromosphere and the corona, are at relatively higher
temperatures than the solar surface. The spectra of these layers show emission
lines due to elements present in these layers. Analysis of these lines also helps in
determining solar system abundances.
iii) The Sun also emits streams of particles in the form of solar wind. Occasionally,
the Sun emits very high energy particles, called the solar cosmic rays. The
compositions of the solar wind and the solar cosmic rays are directly determined
by instruments on-board many spacecrafts orbiting the Earth.
iv) The spectrum of other objects in the solar system such as the moon and planets
are other sources of information about abundances in the solar system. Samples of
dust brought from the moon and chemical analysis of the Martian surface has
added significantly to this information.
24
v) You must be aware that small rocky pieces wandering in the solar system, called Nucleosynthesis and
meteors, occasionally enter the atmosphere of the Earth. If meteors are not burnt Stellar Evolution
completely by the heat generated due to atmospheric friction, they reach the Earth.
These pieces are called meteorites. Analysis of their composition provides
valuable information about abundances. The spectra of comets are yet another
source of information about the solar system abundances.
vi) Outside the solar system, spectra of other stars and interstellar clouds are
important sources of information about the cosmic abundances.
Cosmic abundances of various elements have been determined using a variety of
methods including those discussed above. Refer to Fig. 10.1 which depicts the
variation of cosmic abundances with mass number of elements.

10
Abundance (Log scale)

-2
0 50 100 150 200
Mass number
Fig.10.1: Abundances of various elements in the universe as a function of mass number

The same data is shown in greater detail in Fig. 10.2. The abundances have been
expressed in terms of a unit in which the abundance of silicon (Si) is exactly 106. This
is because the abundance of Si is very close to this number.

10
Abundance (Log scale)

-2
0 20 40 60 80
Mass number

Fig.10.2: A detailed version of Fig. 10.1 for elements up to mass number 80

The salient features of Figs. 10.1 and 10.2 are as follows:

1. Hydrogen (H1) and helium (He4) are the most abundant elements in the universe.
About 90% of the particles in the universe are hydrogen atoms. Helium is the next
most abundant element, accounting for about 10% of all the particles.

2. Heavier elements constitute less than 1% of the total matter in the universe. 25
From Stars to Our 3. If we leave out H1 and He4, we observe that abundances generally increase with
Galaxy
mass number up to the mass number around 60. This is in the neighbourhood of
56
iron (Fe ). Around this mass number, there is a broad peak.
In the language of astronomy,
any element heavier than He4
is called a heavy element. 4. Beyond the mass number 60, the abundances decrease. At first, the decrease is
faster and then it gradually tapers off.

5. There are peaks of abundances corresponding to elements with mass numbers 12,
16, …. and so on (multiples of 4). Moreover, elements with mass numbers, 14, 18,
… and so on (multiples of 2) are more abundant as compared to those with odd
mass numbers.

On the basis of these features of cosmic abundance data, we can conclude that:

a) The origins of hydrogen and helium are perhaps different from the origin of
heavier elements in the universe.

b) Peaks of abundances at mass numbers that are multiples of four could involve a
particle such as the α-particle or the helium nucleus, which has mass equal to 4
atomic mass unit (amu).

Spend SAQ 1
5 min.
On the basis of relative number of atoms of hydrogen and helium in the universe,
calculate the fractional mass of the matter in the universe contributed by hydrogen and
helium.

Having learnt about the cosmic abundances, a logical question that may come to your
mind is: Where and how are these elements formed? Astronomical studies tell us
that all the elements, except hydrogen and helium, have been synthesised in the stars
during their evolution. This is also reinforced by the observation that the older stars in
our Galaxy contain much less heavier elements than the younger stars. Thus, we can
visualise the following roadmap for creation of elements and how the process is
related with the evolution of stars:

a) Elements are formed inside the stars. Since the birth and death of stars is a
continuous process, the formation of elements is also an on-going process.

b) The oldest stars in the Galaxy, called Population II stars, were formed from the
original matter of the Galaxy which was mostly hydrogen. These stars had to
manufacture their own heavy elements. Therefore, they are relatively poor in
heavier elements.

c) At the end of their life, some of these stars explode and return the heavier
elements formed by them to the interstellar medium.

d) From this enriched interstellar material, new stars are formed. These relatively
younger stars, also called Population I stars, are rich in heavier elements. In
addition, they also manufacture elements in their cores which constitute the raw
material for the subsequent generations of stars.

The hypothesis that elements are made inside the stars gets support from the detection
of elements like technetium in the spectra of some stars. This element is not found in
the solar system. Where could this element have been formed except in the stars in
which it is observed?

26
Now, the question is: What is the origin of the major constituents, namely Nucleosynthesis and
hydrogen and helium, of interstellar medium? An acceptable theory in astronomy Stellar Evolution
tells us that hydrogen and helium were formed in a different process (see Unit 15).
You could question this theory since you have learnt earlier that He4 is formed from
H1 in the core of the Sun and the other main-sequence stars. The fact is that if we take
account of all the helium that could have been formed in the stars in all the galaxies,
it falls much short of the total helium estimated to be present in the universe (about
30% by mass). To appreciate this statement, solve the following SAQ.

SAQ 2 Spend
10 min.
The atomic weights of hydrogen and helium are 1.0079 and 4.0026, respectively. In
the fusion reaction converting hydrogen into helium, one gram of hydrogen produces
about one gram of helium and approximately 6 × 1018 ergs energy is released. Given
33 −1 9
that the luminosity of the Sun is 4 × 10 erg s and its estimated age is 5 × 10 years,
show that only about 5% of its mass has been converted into helium. Take the solar
mass as 2 × 1030 kg.

Having solved SAQ 2, you might conclude that only a small fraction of the total
helium present in the universe has been manufactured in the stars. It is, therefore,
reasonable to believe that light elements like hydrogen, helium, deuterium (D2), He3,
and Li7 did not form inside the stars. You may ask: Do we have any clue about the
origin of these elements? According to one theory about the origin and evolution of
the universe, these elements were formed in the first minute after the birth of the
universe. At that time, the universe was hot and dense and the conditions were
suitable for the formation of light elements. (This issue is discussed in detail in unit 15
of this course.) This theory is supported by the fact that the abundances of light
elements predicted by it in the early universe agree very well with the observed
abundances. The coincidence is considered a very strong evidence supporting the
idea that the early universe was very hot and dense and that it was born in a
violent event called the Big-Bang.

A clue to support the hypothesis that heavier elements are manufactured inside the
stars was provided by Bethe (in US) and Weizsacker (in Germany) in 1938.They
showed the possibility of converting hydrogen into helium through nuclear reactions
which would take place at high temperatures and high densities. Such conditions are
readily available in the interior of stars such as the Sun, which also has plenty of
hydrogen. The work of Bethe and Weizsacker gave birth to the field of nuclear
astrophysics and subsequently, scientists were able to show that other heavier
elements could also have been formed in stars through the process of nucleosynthesis
involving a variety of nuclear reactions. Would you not like to know about
nucleosynthesis? We discuss this in the next section.

10.3 STELLAR NUCLEOSYNTHESIS

You may recall from your school physics that nuclear fusion is the process of coming
together of two or more light nuclei to form a new nucleus. Since a fusion reaction
involves coming together of charged particles, it requires that they have sufficient
energy to overcome the Coulomb repulsion. Such energies are readily available in the
form of thermal energy in stellar interiors. The nuclear reactions taking place in such
environments are called thermonuclear reactions. The process of creation of new
elements in such reactions is called nucleosynthesis.

A variety of nuclear reactions can take place depending upon the elements present,
temperature and density in the interior of stars. The information about these reactions
helps us understand stellar evolution better because on the basis of this information,
27
From Stars to Our we can determine the age of the stars as well as their future. Let us now discuss the
Galaxy
major processes by which elements are synthesised in the stellar core.
The temperature inside the stars
is very high. Therefore, all the 1) Hydrogen Burning
atoms are ionised. The nuclear
reactions take place between The first stage of nucleosynthesis is the fusion of hydrogen nuclei and consequent
nuclei and the products are also
nuclei. In this context, therefore,
formation of helium. You have already learnt in Unit 5 about the chain of
whenever we talk of atoms or reactions, called the pp-chain, which causes fusion of four hydrogen nuclei
elements, we really mean nuclei. (protons) to form helium. This is the process by which the Sun and other similar
stars generate their energy:

H1 + H1 → H 2 + e + + ν e

H 2 + H1 → He 3 + γ

He 3 + He 3 → He 4 + 2 H1

Another chain reaction has helium as its end product and it produces energy in the
main sequence stars. It starts with carbon and is called the carbon-nitrogen cycle
(CN-cycle). Obviously, for this reaction, it is necessary that the stars have some
carbon to begin with. Such stars are called the second generation stars. The stars
which start life with only hydrogen and helium are called first generation stars.

The temperature required for CN-cycle is higher than the temperature required for
the pp-chain. The nuclear chain reactions involved in the CN-cycle are given
below:
C12 + H1 → N13 + γ
N13 → C13 + e+ + νe
C13 + H1 → N14 + γ
N14 + H1 → O15 + γ
O15 → N15 + e+ + νe
N15 + H1 → C12 + He4
You may note that the end result of the CN-cycle is to combine four hydrogen
nuclei to form helium; carbon merely acts as a catalyst for the reaction and is not
consumed in the process. You may ask: If we know that CN-cycle is active in a
star, what information can we obtain about that star? Recall from Unit 7 that,
as we go up in the H-R diagram, the luminosity increases. Since luminosity of a
star is proportional to some power (generally 3.5) of its mass, the mass also
increases upwards. Further, the internal temperature of a star is generally
proportional to its mass. Therefore, the stars in which energy is generated by the
CN-cycle are generally found in the upper region of the main sequence. They
have high internal temperature and were formed from material enriched in heavy
elements.

Spend SAQ 3
2 min.
On the main sequence in the H-R diagram, where would you find stars which have
internal temperatures lower than that of the Sun?

2) Helium Burning

When all the hydrogen in a stellar core has been used up, pp-reactions and
28 CN-cycle are no longer possible. As a result, the energy generation stops and the
 pressure in the core decreases. It is no longer able to match the inward Nucleosynthesis and
gravitational pull of the particles and the core contracts. In some stars, the Stellar Evolution
contraction continues till the temperature has risen to about 2 × 108 K. When the
core temperature of a star attains this value, it is possible for helium nuclei to fuse
together and produce carbon.

If you look at the periodic table carefully, you will find that at mass number eight,
there is a gap. This means that the element at this position is unstable. The
question is: How do reactions between helium nuclei overcome this gap? The
4 8 8
two He nuclei (or α-particles) combine to form Be . Since Be is unstable, it
disintegrates into two α-particles in a very short time. In the presence of to and fro
4 8
reactions of this kind, the stellar core becomes a sea of He nuclei with a few Be
8 12
nuclei floating. These floating Be nuclei combine with α-particles to form C
8 12
nuclei. Despite very low population of Be nuclei, C is formed because the
8
reaction between α-particle and Be has a very high probability of occurrence. So,
carbon is formed when three α-particles combine according to the following
reactions:
He4 + He4 → Be8 + γ
Be8 + He4 → C12 + γ
The above nuclear reaction is also called the triple-αα reaction. Once carbon is
formed in the stellar core, formation of heavier nuclei becomes possible. You may
recall that carbon and some other heavier elements are absolutely essential for the
origin of life. Thus, it can be argued that we are here and discussing
nucleosynthesis today because triple-α α reactions took place in some stellar cores
in the distant past!

SAQ 4 Spend
5 min.
a) Why are three α-particles needed to initiate helium reactions?

b) Explain the importance of triple-α reaction in the formation of heavy nuclei. In

what way is triple-α reaction related to the origin of life on the Earth?

3) Burning of Carbon and Heavier Nuclei

Once carbon is formed, it can combine with an α-particle and form oxygen (O16).
16 20
Afterwards, O can react with an α-particle to form Ne . When all the helium
has been converted into carbon in the core of the star, helium reactions stop and
energy generation is terminated once again. This leads to further gravitational
contraction of the core and temperature of the interior of the star increases. At the
enhanced temperature, it becomes possible for two carbon nuclei to combine and
form Mg24.

By now, you must have noted that every time a particular type of nuclear fuel is
used up completely, core contraction due to gravitation takes place. As a
consequence, there is a rise in the temperature of the core. When the temperature
has risen to the required level, yet another nuclear reaction becomes possible. This
cycle continues till all the nuclei in the core have become iron nuclei.

You may ask: Why does the series end at iron? To answer this question, refer to
Fig. 10.3 which depicts the binding energy curve for nuclei. Note that the average
binding energy per nucleon increases with mass number till we reach the mass
number of iron. This means that with increasing mass number, the nuclei are more
tightly bound and are more stable. It also means that iron is the most stable
element. It cannot combine with other nuclei to produce still heavier nuclei. The
binding energy curve also gives us a clue for understanding why energy can be
29
From Stars to Our derived by fusing light elements as well as by splitting (fission) the nuclei of
Galaxy
heavy elements.
9

Binding energy per nucleon (MeV)

8

5
0 50 100 150 200 250

Mass number

Fig.10.3: Average binding energy per nucleon as a function of mass number

It is, therefore, understandable why elements heavier than iron cannot form as a
result of thermonuclear fusion reaction. However, the fact of the matter is that
elements heavier than iron do exist in nature. The question, therefore, is: How did
they form? It is believed that such heavier elements formed in some special types
of nuclear reactions called s- and r-processes.

4) s- and r- processes

The elements heavier than iron are probably synthesized by the absorption of one
neutron at a time. For example, let us consider the nucleus of atomic number Z
and mass number A. It is written as (Z, A). When it absorbs one neutron, its mass
number increases by one and it becomes (Z, A+1). If the new nucleus absorbs yet
another neutron, it becomes (Z, A+2). If this nucleus emits a β-particle before it
can absorb another neutron, it becomes (Z+1, A+2). The last one, nucleus (Z+1,
A+2), may absorb another neutron to become (Z+1, A+3). In this way, all the
elements up to Bi209 are formed. This process of absorption of one neutron at a
time is a slow process and is named as s-process. The neutrons required for this
process are produced as a by product of reactions of the following types:

He4 + C13 → O16 + n

O16 + O16 → S31 + n

Well, you can further ask: Why does the s- process stop at Bi209? It is because the
elements heavier than Bi209 are unstable and emit β-particles before they can absorb a
neutron. However, if neutrons become available in large numbers, then these nuclei
can absorb neutrons rapidly. Due to this rapid process, or the r-process, elements
right up to uranium are synthesized in stars.

So far, you have studied about cosmic abundances and stellar nucleosynthesis. You
now know that all the heavy elements are formed in stars due to thermonuclear
reactions. The relative proportion of heavier elements in stars tells us whether it is a
first or a second generation star. The various types of nuclear reactions are associated
with different stages in the life of a star and also with the location of the star on the
30 HR-diagram. You will now learn about the evolution of stars.
 Nucleosynthesis and
10.4 EVOLUTION OF STARS Stellar Evolution

You may recall from Unit 9 that gravitational collapse of a gas cloud gives birth to
stars. A stable star is formed and finds its place on the main sequence only when it
attains hydrostatic equilibrium, that is, when the pressure inside the star balances the
gravitational force acting inwards. In the following, we shall discuss the life of stars
on the main sequence and afterwards.

10.4.1 Evolution on the Main Sequence

For most of their lives, stars live on the main sequence on the H-R diagram. We
would, therefore, like to know: i) How long does a star live on the main sequence?
and ii) What happens in the core of a star while it is on the main sequence and
when it departs from the main sequence?

When a star arrives on the main sequence, it is said to have been born. That is why the
main sequence is called zero-age main sequence (ZAMS). While the star is on the
main sequence, it burns hydrogen either through pp-chain or through CN-cycle. There
is little change in its luminosity. The best example of the main sequence star is the
Sun. It is known that the luminosity of the Sun has not changed much during its
lifetime of 5 billion years. You may ask: How do we estimate the life of a star on
the main sequence? Fortunately, for the Sun, we have enough data to make an
intelligent guess. It is estimated that the core of the Sun has about 10% of its total
mass and from SAQ 2, we know that so far it has burnt only about half of this
(hydrogen) mass to make helium. Therefore, it is estimated that the Sun would stay on
the main sequence for another 5 billion years.

Let us now ask ourselves a more general question: How long does a star live on the
main sequence? To address this question, we need to look critically at the mass-
luminosity relation of stars. We know that the ultimate source of energy for a star is its
mass. When it is burning hydrogen to form helium, it is actually converting its mass
into energy. The relation between luminosity (L) and mass (M) of a main sequence
star is given by:

L ∝ M 3.5 (10.1)

Since luminosity is the energy radiated by a star per second and M is its total mass
which can be converted into energy, the time for which it will stay on the main
sequence can be written as:

τ ≈ M / L. (10.2)

From Eqs. (10.1) and (10.2), we can write:

τ ∝ M −2.5 (10.3)

The constant of proportionality is determined by the expected lifetime of the Sun.

Eq. (10.3) shows that the more massive a star is, the shorter is its life on the main
sequence. To get a feel of the numerical values of the age of stars on the main
sequence, you should solve the following SAQ.

SAQ 5 Spend
5 min.
The estimated lifetime of the Sun on the main sequence is ~ 1010 years. Calculate the
main sequence lifetime of a star of mass (i) 10 MΘ and (ii) 0.5 MΘ.
31
From Stars to Our
Galaxy
10 4
10 10

Luminosity

Luminosity (solar unit)

Lifetime (yr)
8 2
10 10

Lifetime

6 0
10 10

4 -2
10 10
0 5 10 15 20
Mass (solar unit)

Fig.10.4: Lifetime and luminosity of stars on the main sequence as function of mass

Refer to Fig. 10.4 which shows the lifetime of stars on the main sequence as a
Helium in function of the stellar mass. You may note that a star of mass 10 MΘ will stay on the
the core Expanding
main sequence for about 3 × 107 years and a star of mass 0.5 MΘ will stay on the main
outer layers
sequence for about 6 × 1010 years. You may ask: How come a more massive star
which has more nuclear fuel has shorter lifetime on the main sequence? Recall
that massive stars must burn their fuel at a faster rate to emit more energy from their
surfaces per second. Thus, they run out of fuel sooner. By the same token, lower mass
stars burn fuel at a slower rate. In this context, it is interesting to note that the
estimated age of the universe is ~ 1 to 2 × 1010 years, a time period much shorter than
the estimated lifetime of low mass stars on the main sequence!

Having got a fairly good idea about lifetimes of stars of different masses on the main
sequence, you may like to know: How does the core of a main sequence star
Hydrogen in
the shell fusing evolve?
to helium
While a star is on the main sequence, its core becomes progressively richer in helium.
Fig.10.5: When hydrogen in When the core consists only of helium, no nuclear reaction takes place because the
the core of a star is
exhausted, the core temperature is not high enough for the next stage of nuclear reactions. As a result, the
contains pure pressure in the core decreases and the core must contract under its own weight. The
helium and gravitational energy released due to contraction raises the temperature of hydrogen in
hydrogen burns in
a thin shell around the core so much that it starts burning (Fig. 10.5). The helium
a thin shell around
the core made in the shell adds to the mass of the core whose contraction is accelerated. The
energy produced in the shell and the gravitational energy due to the contracting core
push out the envelope of the star due to radiation pressure. As a result, the star
expands in size. Its surface becomes cooler but its luminosity increases enormously
because of increased surface area. It becomes a giant star.

Spend SAQ 6
3 min.
It is estimated that after its life on the main sequence, the Sun will swell to 200 times
its present radius. If, at that time, its surface temperature is half of its present
temperature, calculate the luminosity of the Sun in terms of its present luminosity.

Let us now look at the events that take place in the life of a star when it leaves the
main sequence.
32
10.4.2 Evolution beyond the Main Sequence Nucleosynthesis and
Stellar Evolution
When a star leaves the main sequence, it becomes a giant star because of its increased
size. Since the surface temperature of these stars is low, they appear red and are also
known as red giant stars. The more massive (> 10 MΘ) stars become so huge because
of expansion that they are called supergiant stars. The time taken by a star to travel
from the main sequence to the giant or supergiant branch is much shorter than its stay
on the main sequence. The Sun will also become a giant star after about 5 billion
years.

Meanwhile, the cores of these stars continue to contract and their temperatures rise
further. When the temperature of the core is about 2 × 108 K, triple α - reactions
become possible. The energy released in these reactions builds up the pressure in the
core and further contraction of the core is halted. Further details of evolution depend
on the mass of the star. Let us discuss some illustrative cases now.

Mass of the star is ~ 1 MΘ

If the mass of the star is approximately 1 MΘ, the helium burning is rather abrupt and
a large amount of energy is suddenly released. This phenomenon is called a helium
flash. Several circumstances, some not yet completely understood, combine at this
stage to force the star to throw away its outer envelope. The ejected matter surrounds Fig.10.6: A planetary nebula
the star. This object is called a planetary nebula (Fig. 10.6). In a small telescope, it
appears like a planet, hence the name planetary nebula.

Within a short time, the gas surrounding the star vanishes due to interaction with the
interstellar medium. Simultaneously, the core again begins to contract and becomes
very dense and the matter in the core becomes a degenerate gas.

Degenerate gas is a particular configuration of a gas composed of fermions (electrons,

neutrons, etc.) whose behaviour is regulated by a set of quantum mechanical laws, in
particular, Pauli’s exclusion principle (explained in the physics electives PHE-11
entitled Modern Physics). This configuration of a gas is usually reached at high
densities. Recall that according to Pauli’s exclusion principle, no more than two
fermions (of opposite spin) can occupy the same quantum state of a system. As
the density of electrons in the core of a star increases in a fixed volume, these particles
progressively fill the lower energy states. The additional electrons are forced to
occupy states of higher and higher energy.

This process of gradually filling in the higher-energy states increases the pressure of
the electron gas. The pressure of the degenerate gas depends only on the density of the
gas and is independent of its temperature. Since degeneracy occurs when the density
is high, the pressure of the degenerate gas is high. The equation of state of such a gas
is independent of temperature unlike the normal gas.

In stars of mass ~ 1 MΘ , only the electron component of the matter becomes

degenerate. The pressure of degenerate electrons is sufficient to halt the contraction of
the star. A state of equilibrium is established. The star is then called a white dwarf
star. The maximum mass of a white dwarf star is ~ 1.4 MΘ. This limit on the mass of a
white dwarf star was predicted theoretically by S. Chandrasekhar, an Indian
astrophysicist. It is, therefore, called the Chandrasekhar limit.

In the H-R diagram, the white dwarf stars occupy the left bottom corner, much below
the main sequence as shown in Fig. 10.7. The numbers above the lines in Fig. 10.7
indicate the mass of white dwarf stars in terms of solar mass.

33
From Stars to Our
Galaxy Main sequence Sun
0

log (L/L Θ)
-1.5
0.89 0.51 0.22

-3.0
White dwarfs

-4.5

4.5 4.3 4.1 3.9 3.7 3.5

log T
Fig.10.7: Location of white dwarf stars on the H-R diagram

There is no source of nuclear energy inside a white dwarf star. It is a dead star. It just
utilises the thermal energy of its particles to radiate from its surface. It can live like
this for several billion years and, afterwards, it becomes a cold object. The path of
evolution of such stars on the H-R diagram is shown in Fig. 10.8a. Note that, after the
star leaves the main sequence, it swells and heads towards the giant region.
Subsequently, depending upon its mass, it may become a white dwarf star.

105

Formation of planetary Giant region 10000

nebula
103
1000 10 solar radii
Luminosity (solar units)

Luminosity (solar unit)

Horizontal branch Red

10
Main 100 giant
sequence

10-1
White 10
dwarfs 1 solar
radius

10-3 1
Main sequence

10-5 0.1
80000 40000 20000 6000 2000 4.1 4 3.9 3.8 3.7 3.6
Surface temperature of star (K) Surface Temperature (K; Log scale)

(a) (b)

Fig. 10.8: a) Evolutionary track of a general mass star; and b) evolutionary track of a star of a few
solar mass

Mass of the star is ~ 5 MΘ

After becoming a giant star, a star having mass approximately equal to 5 MΘ traces its
path back on the H-R diagram towards the main sequence along a horizontal line,
called the horizontal branch (see Fig. 10.8b). This happens because of the
commencement of the helium fusion (also called triple alpha process). As helium is
34
exhausted and the next set of nuclear reactions involving carbon starts, the star
changes its evolutionary course backwards towards the giant branch. It wriggles Nucleosynthesis and
several times between the giant and the horizontal branch. During this time, it also Stellar Evolution
sheds a lot of mass. Eventually, it also becomes white dwarf after passing through the
planetary nebula phase.

Mass of the star is > 10 MΘ

If the initial mass of the star is several times the solar mass, the helium reactions start
gradually. When helium in the core is exhausted, the core contracts once again. The
helium reactions are now ignited in a thin shell around the core. The temperature of
the contracting core rises and becomes high enough for carbon-carbon reactions to
occur. The cycle of core contraction and burning of the next heavy element continues
till the core consists of iron. No more fusion reactions are possible now.

SAQ 7 Spend
3 min.
Why do fusion reactions stop at iron?

Let us pause for a moment and think of what we have learnt about the evolution of
stars. We have traced the evolution of stars and simultaneous formation of elements.
Depending upon its mass, a star can take different evolutionary courses. Well, you can In the nuclear reaction taking
place in the core of stars, a
further ask: What happens after all the nuclear reactions have stopped and the large number of neutrinos are
stellar core consists of iron only? We have seen from the binding energy curve (Fig. also produced. You may be
10.3) that iron has the highest binding energy per nucleon. It cannot, therefore, burn to aware that the neutrino has no
give off energy. Thus, the core is forced to contract. The gravitational energy heats charge and very small mass, if
at all. These particles react
the core resulting in the disintegration of iron nuclei into nuclei of helium. The break with matter extremely weakly.
up of iron is an endothermic process, that is, the reaction absorbs energy rather than So, their mean free path is very
giving it out. To feed this process, the only source of energy is the gravitational large and these were
collapse. As a result, the collapse of the core is accelerated. Faster release of considered to be the carrier of
energy from the core to the
gravitational energy due to collapse gives rise to further break up of iron nuclei into envelope. It has, however,
helium nuclei. This, in turn, accelerates the collapse. As the density of the core been found now that at the
continues to increase, even the helium nuclei cannot remain intact as nuclei. They extremely high densities
break up into protons and neutrons. developed in the core, the
mean free path of neutrinos is
shorter than the size of the
Coming back to the evolution of the collapsing star, let us first consider its envelope. core. So, they cannot transfer
Due to transfer of energy from the core to the envelope, very high temperatures are energy to the envelope. A
produced in the envelope and nuclear reactions in various layers of the envelope are different process is now
proposed for the transfer of
ignited. The energy released due to these reactions heats the envelope so much that it energy from the core. When
becomes prone to explosive disintegration in a matter of seconds. In this short time the density of matter
before explosion, nuclear reactions produce heavy nuclei all the way up to the iron approaches the nuclear density
15 −3
group. In addition, the large numbers of neutrons released in the nuclear reactions (~ 10 gcm ), the short range
nuclear forces come into play.
participate in the r-process and heavy nuclei beyond Bi209 are produced. Finally, the The particles repel one
envelope explodes. The explosion is called a supernova. The elements built inside another strongly and rebound.
the star over hundreds of millions of years of its life are thrown into the interstellar A bounce is said to travel from
medium. The new stars born from this enriched medium contain a small proportion of the core and into the envelope.
It is the bounce that is thought
heavy elements and these stars are called the second generation stars. It has been to transfer energy to the
suggested that the interstellar material from which the solar system was formed envelope.
contained heavy elements released in a supernova that took place nearby.

Well, the envelope explodes but what happens to the core? The core keeps
collapsing. According to our present knowledge, if the initial mass of the star is up to
about 12 MΘ, then the core is left with a mass of about 2 MΘ to 3 MΘ. The matter in
such stars is mostly neutrons. Like electrons, neutrons also obey Pauli’s exclusion
principle. At the extremely high density which exists in the core, neutrons become
degenerate. The pressure exerted by the degenerate neutrons is sufficiently high to halt
the collapse of the core. The core stabilises in the form of a neutron star.
35
From Stars to Our If the initial mass of the exploding star is close to 15 MΘ or more, the core that is left
Galaxy
behind has a mass more than 3 MΘ. A core of this mass cannot attain equilibrium of
any kind. It keeps contracting. Eventually, its gravitational field becomes so strong
that even light cannot escape it. It becomes a black hole.

10.5 SUPERNOVAE
Supernovae are extremely violent and bright stellar explosions. The energy released in
one such explosion is equivalent to the conversion of 1 MΘ into energy, or about
47
10 J! The luminosity of a supernova is of the order of the luminosity of a whole
galaxy. In all supernova explosions, the brightness reaches a peak within a few days
of the explosion. During the short time of its maximum brightness, it shines as a very
bright object. Thereafter, its brightness decreases, first rapidly and then slowly. The
variation of the brightness of an object with time is called the light curve of the
object.

Light curves of supernovae have been studied extensively because of their importance
in many areas of astronomy. Based on the type of light curves, supernovae have been
classified as type I and type II. Fig. 10.9 shows the light curves of type I and type II
supernovae. You may note that the light curve of type I supernova reaches higher
brightness but show a rapid decline. On the other hand, the light curve of type II
supernova shows a lower maximum brightness but slower decline. The difference in
the light curves of the two types of supernovae indicates differences in the stars which
explode.
Brightness

Type I

Type II

0
Time (days)

Fig.10.9: Schematic light curves of supernovae of type I and type II

You may argue: Since supernovae are so bright, it must be easy to observe them.
Surprisingly, it is not so; observations of only four supernovae in our Galaxy have
been recorded in the last two thousand years! The most famous supernova occurred in
the year 1054 A.D. and was observed by the Chinese astronomers in the constellation
Taurus.

Type I supernovae are further subdivided into two classes: Ia and Ib. The light curve
for a supernova of type Ia is shown in Fig.10.10. Again, note that the brightness of
the supernova is maximum at the time of explosion and it drops drastically afterwards.
The light curves of supernovae of type Ia are almost identical. These supernovae are
believed to be caused due to the explosion of white dwarf stars. You may ask: How
can a white dwarf give rise to a supernova explosion because its mass cannot be
greater than 1.4 MΘ? This would be possible if we imagine a white dwarf star in a
binary system with a main sequence or a giant star as its companion. Because of its
36 strong gravitational field, the white dwarf star sucks matter from the companion. As
its mass increases beyond 1.4 MΘ, it explodes as a supernova. Since all stars have the Nucleosynthesis and
same mass at the time of explosion, it is possible that supernovae Ia reach the same Stellar Evolution
brightness or absolute magnitude. Further, the observed apparent magnitude of a
supernova can be used to determine its distance using its distance modulus (defined
as m − M) because all type Ia supernovae have the same absolute magnitude at the
time of the maximum brightness. In recent years, type Ia supernovae have been used
successfully to find distances of distant galaxies. This has been very useful in
understanding the nature and the ultimate fate of the universe.
Brightness

Supernova explosion

0 Time (days)

Fig.10.10: Schematic diagram of the light curve of type Ia supernova

Type Ib supernovae are thought to be due to carbon burning in the degenerate core of
a star. Carbon burning in this situation is abrupt and very rapid. It is called a carbon
flash. It generates so much energy that the star explodes.

The stars which explode to become supernovae of type II have usually masses higher
than 10 MΘ. Their light curves are all distinct. These supernovae leave behind neutron
stars or black holes, which are detected several years after the explosion.

You may ask: How do we know all that we have said above about supernovae? At
the time of explosion, a star throws a cloud of gas which travels into the interstellar
medium with speeds of the order of 10,000 kms−1. The cloud expands and merges
gradually into the interplanetary medium. It seeds the interplanetary medium with
heavy elements. The explosion also generates ripples in the surrounding medium
which create conditions favourable for the formation of new stars. An example of the
what remains behind a supernova explosion is the Crab Nebula. It is the remnant of
the supernova of 1054 A.D. A neutron star is located at the centre of the nebula which
was formed as a result of the explosion.

Now, let us summarise what you have learnt in this unit.

10.6 SUMMARY
• Chemical composition of the universe refers to the presence of different types of
elements and their proportions in the universe. The chemical composition of the
universe is the same as that of the solar system.
• Relative proportions of elements in the universe are called cosmic abundances.
Observed abundances of elements show that (i) hydrogen and helium are the most
abundant elements in the universe, (ii) there is a broad peak of abundances near
37
From Stars to Our A = 56 (iron group of elements), and (iii) there are peaks at mass numbers which
Galaxy
are multiples of 4.
• All elements except hydrogen and helium are formed inside the stars due to
thermonuclear reaction and creation of new elements in such reactions is called
nucleosynthesis.
• Depending upon the elements present, density and temperature in the interior of
stars, nucleosynthesis takes place through one of the three major processes,
namely hydrogen burning, helium burning and burning of carbon and
heavier nuclei. As long as a star stays on the main sequence, it burns hydrogen.

• The life span of a star on the main sequence depends on its mass: τ ∝ M −2.5 .
Also, the evolution of a star away from the main sequence depends on its mass.
Low mass stars become white dwarfs at the end of their lives.
• Massive stars (> 10 MΘ) explode as supernovae at the end of their lives leaving
behind neutron stars or black holes.

10.7 TERMINAL QUESTIONS Spend 30 min.

1. List the major processes of formation of elements inside stars. Why can elements
beyond iron not be formed by fusion?

2. Suppose that a supernova explosion takes place at the distance of Proxima

12
Centauri (~ 3 pc). If its luminosity equals the luminosity of our galaxy (~10 LΘ),
show that it would appear about as bright as the Sun. Take the absolute magnitude
of the Sun as 5 and its apparent magnitude as − 27.
3. For stars having more mass than 10 MΘ, the luminosity is directly proportional to
their masses. Show that their lifetime on the main sequence is independent of their
masses.
4. Aldebaran (α-Tauri, Rohini in India) is a red giant star of spectral class K5. Its
radius is 22 times the radius of the Sun. Its surface temperature is 3800 K.
Calculate its luminosity (in units of solar luminosity) and its absolute magnitude.
The absolute magnitude of the Sun is 5.

10.8 SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs)

1. 90% hydrogen and 10% helium by number implies that the universe comprises 90
units of mass of hydrogen and 40 units of mass of helium because the mass of a
helium atom would be 4 units if mass of a hydrogen atom is taken as 1 unit.
So, out of 130 units of mass, 90 units is hydrogen and 40 units is helium.

Thus,
90
Percentage of hydrogen = × 100 ~ 70
130

40
Percentage of helium = × 100 ~ 30
130

That is, approximately 70 percent of the mass of the matter in the universe is
contributed by hydrogen and remaining by helium.

38
2. Total energy radiated by the Sun during its lifetime Nucleosynthesis and
Stellar Evolution
= Luminosity of the Sun × Age (in s) of the Sun
33 −1 9 7 7
= (4×10 erg s ) × (5×10 ×3×10 s) [Q 1 yr = 3×10 s]
50
= 6×10 erg

Now, this energy was generated in the nuclear reactions within the Sun converting
hydrogen into helium. So, to produce this much energy, mass of hydrogen
6 × 10 50 erg
consumed = = 10 32 g
18 −1
6 × 10 erg g

Therefore, percentage of the total mass of the Sun which has been converted into
10 32 g
helium till date = ×100
2 × 10 33 g
= 5%

3. The internal temperature of a star is roughly proportional to its mass. Since

luminosity of a star is proportional ~ M 3.5, the stars with lower internal
temperature than the Sun have lower luminosities than the Sun. So, these stars
will be found in the H-R diagram to the right of the Sun and lower than the Sun.

4.a) Since the nucleus with mass number 8, which nucleosynthesis of two α-particles
will produce, is unstable, three α-particles are needed to produce stable nuclei
beyond mass number 8.

b) If triple - α reaction could not take place, carbon and heavier nuclei would not
have been synthesized. Further, since carbon is essential for the origin of life,
without triple - α reaction, life on the Earth would not have been possible.

5. The lifetime of a star on the main sequence is given by (Eq. (10.3)):

τ ∝ M −2.5
Thus, for the Sun, we have:
τ Θ ∝ M Θ −2.5

On the basis of above expressions for τ , we can write:

−2.5
 M 
τ = τΘ  
 M 
 Θ 

i) For a star whose mass is 10MΘ, we can write its lifetime on the main sequence
as:

−2.5
 10 M 
τ =10 10
yr  
 M 
 Θ 

=1010 ×10 −2.5 yr

= 10 7.5 yr = 10 7 10 yr
≅ 3 × 107 yr 39
From Stars to Our ii) For a star whose mass is 0.5MΘ, we can write:
Galaxy

M 1
=
MΘ 2
Thus,
− 2.5
1
10 
τ =10  yr
2
 
10
≅ 6 × 10 yr

6. We know that the luminosity of a star is given as:

L = 4 π R2 σ T 4

So, we can write for the Sun:

LΘ = 4π RΘ 2 σ T 4

After the Sun has expanded, we have, as per the problem:

R = 200 RΘ
and
TΘ
T=
2

Thus, the luminosity of the expanded Sun is:

4
T 
L = 4 π (200 RΘ ) σ  Θ
2 
 2 
 
4 π × 4 × 104 RΘ 2 σ TΘ 4
=
16

L 4 π × 4 × 10 4 RΘ 2 σ TΘ 4
=
LΘ 16 × 4 π RΘ 2 σ TΘ 4
L = 2.5 × 103 LΘ

7. Fusion reaction stop at iron because iron is the most stable atom/nucleus. It cannot
combine with any other nucleus to produce heavier nuclei.

Terminal Questions

1. See text.

2. As per the problem, luminosity of the supernova, L = 1012 LΘ .

If the absolute magnitude of supernova is M and that of the Sun is MΘ, we can
write:

L
= 10 − 0.4 ( M − M Θ )
LΘ
40 1012 = 10 −0.4 ( M − M Θ )
 12 = − 0.4 (M − MΘ) Nucleosynthesis and
Stellar Evolution
M = − 30 + 5
= − 25

If m is the apparent magnitude of the supernova, we can write:

m = M + 5 log r − 5
= − 30 + 5 × (log (3)) (r = 3pc)
= −30 + 5 × 0.48
≈ − 27.5

The value (−27.5) of apparent magnitude of the supernova is approximately same

as that of the Sun. So, the supernova will be as bright as the Sun.

3. As per the problem,

L∝M
Now, the lifetime of a star can be approximated as:

M
τ~
L

M
or τ∝
M
= Constant.

So, the life time of such stars is independent of their masses.

4. You have seen in SAQ 2 of Unit 5 that, using Stephan - Boltzmann law, we can
express the luminosity of a star as

L = 4π R 2 σ T 4

where R is the radius of the star and T is its temperature.

We can, therefore, write the luminosity of a star like Aldebaran as:

L Ald = 4π R 2 σ T 4

= 4 π × (22 RΘ ) 2 σ (3800 K) 4

And, the luminosity of the Sun can be written as:

LΘ = 4π × RΘ 2 σ (6000 K ) 4

Thus, we have:

L Ald 484× (3.8) 4

=
LΘ 64
= 95

41
From Stars to Our Further, the ratio of luminosity of Aldebaran and the Sun can be expressed in
Galaxy
terms of their absolute magnitude MAld and MΘ, respectively, as:

L Ald
=10 − 0.4( M Ald − M Θ )
LΘ

Thus, we get:

95 =10 −0.4( M Ald − 5)

because MΘ is 5. Taking logarithm on both sides, we get:

log 95 = − 0.4 (MAld − 5)

1.98 = − 0.4 (MAld − 5)

or, M = − 4.95 + 5
= 0.05

42
 Compact Stars
UNIT 11 COMPACT STARS
Structure

11.1 Introduction
Objectives
11.2 Basic Familiarity with Compact Stars
Equation of State and Degenerate Gas of Fermions
11.3 Theory of White Dwarf
Chandrasekhar Limit
11.4 Neutron Star
Gravitational Red-shift of Neutron Star
Detection of Neutron Stars: Pulsars
11.5 Black Hole
11.6 Summary
11.7 Terminal Questions
11.8 Solutions and Answers

11.1 INTRODUCTION

In Unit 10, you have learnt about the evolution of stars, that is, how they live their
lives after being formed and take their respective positions on the H-R diagram. You
know that the evolution of a star is governed by two competing forces, namely, the
gravitational contraction and the radiation pressure due to energy generating
thermonuclear fusion reactions. In spite of the huge mass of a star, the amount of its
nuclear fuel is finite and it has to end ultimately. When it happens, a star can no longer
be prevented from contraction due to gravity. The density of the star increases
manifold and it turns into a compact object. In the present Unit, you will study about
such compact objects, also called compact stars.

There are many interesting questions pertaining to compact stars, such as: Does the
gravitational collapse of a star take place uninterrupted? If not, what is the mechanism
which balances the force of gravity? Do all stars end their life by becoming compact
stars of similar type? If not, what determine(s) the nature of the compact stars? You
will discover the answers to these and other related questions as you study this Unit.

A compact star can become a white dwarf, a neutron star, or a black hole depending
upon its initial mass. Further, the gravitational collapse of compact stars like white
dwarf and neutron stars is halted by the degeneracy pressure of fermions − a quantum
mechanical phenomenon. You will learn about the degeneracy pressure in Sec. 11.2.
The theoretical analysis of the relation between the nature of a compact star and its
mass was done by S. Chandrasekhar. This led him to predict a limiting mass for white
dwarf stars. You will learn about the theory of white dwarfs in Sec. 11.3. In Sec. 11.4,
you will study various characteristics of neutron stars and also understand why it is
difficult to detect them optically. All the theoretical predictions about neutron stars
could only be put to test when they were observed in the form of pulsars. In Sec. 11.5,
you will learn about one of the most interesting objects, called the black hole, which
physics has ever predicted. You will discover that the black hole signify the ultimate
victory of gravity in the evolution of stars.

43
From Stars to Our Objectives
Galaxy
After studying this unit, you should be able to:
• understand the role of mass in deciding whether a compact star becomes a white
dwarf, a neutron star or a black hole;
• explain the concept of degeneracy pressure of fermions and its role in compact
stars;
• discuss the concept of Chandrasekhar limit and obtain an expression for it;
• understand the formation of neutron stars and its internal structure;
• derive an expression for the gravitational red shift of the neutron stars;
• describe the detection of neutron stars in the form of pulsars and discuss its
properties;
• explain the concept of Schwarzschild radius for black holes; and
• describe the geometry of space-time around the non-rotating black holes.

11.2 BASIC FAMILIARITY WITH COMPACT STARS

You may recall from Unit 10 that the gravitational force is balanced by the outward
force due to pressure gradient inside a star. The pressure inside is generated due to
thermonuclear reactions. With the passage of time, a star exhausts all its nuclear fuel
by ‘burning’, that is, due to nuclear reactions. The star cannot support itself against
gravity and begins to collapse. Eventually, the density of the star increases
tremendously and the star turns into a very compact object. At this stage, you would
like to know:

a) In the absence of radiation pressure due to nuclear reactions, what enables

compact stars to withstand gravitational squeeze?

b) Are all the stars similar to one another after their collapse?

To address the first issue, recall from the course entitled Physics of Solids (PHE-13)
that when atoms are very close to one another, their quantum energy states overlap
and the electrons in those orbitals behave as if they are free from their parent atoms. A
similar situation is obtained in compact stars where, due to very high pressure, all the
electrons are separated from their parent atoms. This phenomenon is called pressure
ionisation. A compact star is, therefore, a collection of nuclei and free electrons. You
have studied in Unit 10 how the pressure of the degenerate gas of free electrons
restrains the seemingly unstoppable gravitational contraction. This is because a
quantum state cannot accommodate more than one fermion (e.g., electron, proton,
neutron, etc.). This implies that, when the density of electrons is high, they are forced
to occupy quantum states with higher energies because lower states are full. In such a
situation, the pressure of the gas depends only on the density and is independent of the
temperature. You have learnt that gas of free electrons in such a state is called
degenerate electron gas and the pressure exerted by it is called degeneracy
pressure.

So, the gravitational collapse of compact stars is balanced by the degeneracy pressure
of electrons. It is, however, important to note that Pauli’s exclusion principle holds for
all fermions. You will learn later in this Unit that degeneracy pressure due to neutrons
plays an important role in stabilizing some compact stars.

The answer to question (b) raised above is: No. The nature of the remains of a star
after death depends on its mass. You may recall from Unit 10 that the mass of a star
plays a crucial role in its evolution and determines its luminosity. Similarly,
depending upon its mass, a dying star can turn into any one of the three kinds of
compact stars, namely a white dwarf, neutron star or black hole. You will study about
44 these compact stars later in the Unit. Here, it should suffice to say that it was genius of
the Indian astrophysicist, S. Chandrasekhar, who first showed that a degenerate star Compact Stars
cannot have mass larger than a certain maximum mass. He suggested, on the basis of
theoretical calculations, that the degeneracy pressure of electrons will be able to stop
further collapse of a star and its mass is less than a certain mass called Chandrasekhar
limit. The resulting star is called a white dwarf. If the mass of a collapsing star is
more than the Chandrasekhar limit, but less than 3 MΘ then the degeneracy pressure
of neutrons can halt the collapse. These stars are known as neutron stars. Further, if
the mass of the collapsing star is even higher, there is no way that the collapse can be
halted and the collapsing star becomes a black hole.

Thus, compact stars are simply the end products of ordinary stars and are
characterised by smaller sizes and higher densities. To compare and contrast their
10
sizes with that of the Sun, recall that the radius of the Sun is 7 × 10 cm and its mass
33
is ~ 2 × 10 g. A white dwarf of the same mass as that of the Sun would have a radius
about 100 times smaller, that is, around 109 cm. A neutron star of similar mass may
have radius of about 10 km only. And a mass equal to that of the Sun is too small for a
black hole! Generally, it is believed that the minimum mass of a black hole is three
times the mass of the Sun (3MΘ). You may ask: What is the radius of a black hole?
Well, that is a somewhat difficult concept. We will talk about it in Sec. 11.5 of this
unit.

1016

1014
Neutron
stars
1012

1010
Density (g cm-3)

108
White
106 dwarfs

104

102 Earth
Sun
100
Main sequence
10 -2 stars

10-4
101 106 1011 1016

Radius (cm)

Fig.11.1: Typical range of densities of celestial objects as a function of their typical radii

To get an idea about the average densities of compact stars vis-à-vis their radii, refer
to Fig. 11.1. Note that the main sequence stars, such as the Sun, have
density ~ 1 g cm−3. On the other hand, the density of a white dwarf could be ~ 107 to You know from the course
entitled Thermodynamics and
108 g cm−3 and that of a neutron star ~ 1015g cm−3. A black hole, however, has no Statistical Mechanics (PHE-06)
defined average density, as you will learn later in this Unit. Also, the radius of a black that the behaviour of a gaseous
hole is a theoretical construct which means that if you come within this distance from system can be understood on the
the centre (where all the mass is concentrated), you cannot escape from the basis of its equation of state.
tremendous attraction of its gravitational force. Why only you? Even photons, the
fastest particles, cannot escape. 45
From Stars to Our 11.2.1 Equation of State and Degenerate Gas of Fermions
Galaxy
In the preceding paragraphs, you have learnt that the densities of white dwarfs and
neutron stars are very high compared to the densities of objects we come across in
everyday life, or even the densities of ordinary stars like the Sun. Therefore, to know
more about the compact stars, you should have some idea about how matter behaves
when the density is extraordinarily high. You may ask: Why should matter behave
differently? It is because the equation of state of matter at high densities is different
from the equation of state at ordinary densities. When the molecules of a gaseous
system are few and far between, as in our room, or in the atmosphere, collision is the
only way the molecules can interact with one another. Such a gas of molecules is
called an ideal gas and its equation of state is given by:

PV = NRT, (11.1)

where P is the pressure, V is the volume, N is the total number of particles, R is the
gas constant, and T is the temperature. It is evident from Eq. (11.1) that an equation of
state relates different thermodynamic quantities (such as pressure, volume and
temperature) of the gas. Such a relation is very useful in investigating the behaviour of
gaseous systems.

The velocity distribution of When the density of the gaseous system becomes as high as in compact stars, the
the particles of an ideal gas is distance between two atoms/molecules becomes comparable to the size of the
given by the well known atoms/molecules or even the size of the nucleons (i.e., protons, neutrons)! At such
Maxwell’s velocity densities, other forces such as the Coulomb force and nuclear forces begin to
distribution law which you
know already from the influence the dynamical behaviour of the atoms/molecules. As a result, the equation of
kinetic theory of gases. state of an ideal gas cannot describe the behaviour of high density gases.

Now, to get an idea about the equation of state which can describe matter at high
densities such as that in compact stars, we note that compact stars consist mainly of
degenerate gas of electrons and neutrons. Further, you may recall from the course
Thermodynamics and Statistical Mechanics (PHE-06) that we need to use statistics to
describe the collective behaviour of a large number of particles (atoms or molecules)
of a system. Ordinary gases obey the so-called Maxwell-Boltzmann (M-B) statistics.
But, the fermions, which make up the compact stars, obey the so-called Fermi-Dirac
(F-D) statistics. So, in the context of the equation of state, our problem reduces to
know how F-D statistics influences the physical parameters such as pressure inside a
compact star.

 dN 
You may further recall from this course that the number density  = of
 d 3 p d 3x 
 
particles in phase space can be written as:

dN g
= f (11.2)
d 3 p d 3 x h3
When the particle density is
low and temperature is high, where p and x are the momentum and position variables of the particle, respectively,
F-D as well as B-E (Bose-
Einstein) distributions become h is the Planck’s constant and f is the distribution function. In Eq. (11.2), d 3 p d 3 x is
identical to:
the volume element in the phase space, with d 3p containing the momentum elements
f (E) = e−(E−µ)/kT. and d 3x containing the position elements; g is called the statistical weight, i.e.,
which is known as the number of states that a single particle can have for a given value of the momentum p.
Maxwell-Boltzmann If S denotes the spin of particle, then g = 2S + 1. The distribution function f denotes
distribution. the average occupation number of a cell in the phase space.

46
You may also recall that, for an ideal gas in equilibrium at temperature T, the F-D Compact Stars
statistics gives the distribution function f as:

1
f (E ) = (11.3)
exp [( E − µ) / k BT ] + 1

where kB is the Boltzmann constant, µ is the chemical potential and E is the energy.
The energy distribution function given by Eq. (11.3) is valid for fermions such as
electron, proton and neutron which have half-integral spins (i.e., 1/2, 3/2 ..). From
Eq. (11.3), it is obvious that for T → 0, f (E) becomes a step function (Fig. 11.2), that
is,

f (E) = 1 when E < µ (11.4a)

and

f (E) = 0 when E > µ. (11.4b)

1.0 T=0
Fraction of states occupied

0.5

T>0

0.0
0.0 1.0

Energy (E/FF)

Fig.11.2: Fermi distribution function at T = 0 and at temperature, T > 0

When the distribution of fermions is similar to the T = 0 case shown in Fig. 11.2, the
gas of fermions is called completely degenerate. Physically, this means that all energy
states below a certain energy are fully occupied and the occupancy above this energy
is zero. The energy up to which all states are occupied is called the Fermi energy, EF
and for fermions at T = 0, EF = µ.

SAQ 1 Spend
5 min.
Explain how Eqs. (11.4a) and (11.4b) are satisfied.

So, now you know the nature of f for a degenerate gas. The question is: How can we
use Eq. (11.2) to arrive at the equation of state for compact stars? Suppose you
want to know how many particles are present in a unit volume having all permissible
values of momentum. You have to simply integrate the number density in phase space
(left hand side of Eq. (11.2)) over all possible momenta, i.e.,

dN
∫ d 3 pd 3 x d
3
n= p (11.5)
47
From Stars to Our Now, if you want to know the energy density ε of particles, you have to integrate over
Galaxy the energies of all the particles:

dN
ε=∫E 3 3
d 3 p, (11.6a)
d pd x

where E is the energy of the particle given by:

1
(
E= m c + 2 4
p 2 2
) (11.6b)

m being the mass of each particle and c being the velocity of light. Similarly, the
pressure of the gas defined as the rate of momentum exchanged across an ideal
surface of unit area, can be written as:

1 dN
3 ∫ d 3 p d 3x
3
P= pv d p (11.7)

Actually, Eq. (11.7) shows that pressure can be expressed as the momentum flux of a
gas. Since we are discussing only the isotropic pressure here, the said flux in any one
direction (out of three) is one-third of the net momentum flux. That is why a factor of
1/3 has been put before the integral. Assuming that the motion of fermions in compact
stars is non-relativistic, it can be shown (we have avoided giving the mathematical
details) that Eq. (11.7) reduces to:
5

P = K ρe3 (11.8)

The planetary nebula, so where K is a constant and ρe is the density of electrons. Eq. (11.8) is the equation of
called because of its planet-
like appearance, is visually state of a degenerate gas of electrons at high density. Note that, unlike the ideal gas,
one of the most attractive the pressure of a degenerate gas is independent of temperature. This implies that even
astronomical objects. when temperature of a degenerate gas of fermions is very very low, it can exert
tremendous pressure.
Let us now discuss white dwarf stars in which only the electrons are degenerate.

11.3 THEORY OF WHITE DWARF

After a star has exhausted its nuclear fuel, it begins to collapse due to gravity. If the
mass of the star is less than about 8 MΘ, the gravitational contraction is accompanied
by expulsion of matter from its outer envelope. The discarded matter forms a ring-like
structure around the collapsing star and is called planetary nebula (see Fig. 11.3).

48 Fig. 11.3: A planetary nebula

The core of the star continues to contract till its density attains a value in the range Compact Stars
5 8 −3
10 − 10 g cm . At this density, a new equilibrium sets in and these stars are called
white dwarfs. The evolution of a medium mass star into a white dwarf is shown
schematically in Fig. 11.4.

105

Giant
Formation of planetary nebula region

103
Luminosity (solar units)

10
Main
sequence

10-1
White
dwarfs

10-3

10-5
80000 40000 20000 6000 2000
Surface temperature of star (K)

Fig.11.4: Evoluation of medium mass star leading to the formation of white dwarf

One of the interesting features of white dwarfs is that they have mass almost equal to
the mass of the Sun, but their radii are only about five to ten thousand kilometres. The
characteristic features of white dwarfs were explained theoretically by
S. Chandrasekhar.

11.3.1 Chandrasekhar Limit

S Chandrasekhar investigated the effect of the gravitational field on a degenerate gas

of electrons to estimate the mass-radius relation of white dwarfs. One of the
characteristics of a degenerate gas is that its equation of state does not involve
temperature. When particles are non-relativistic, the pressure exerted by such a gas is
given by Eq. (11.8). Now, inside a star in hydrostatic equilibrium, the pressure must
balance the gravitational pull towards the centre. For a star in hydrostatic equilibrium,
(recall from Unit 8 that) the pressure gradient can be expressed as:

1 dPe GM
= (11.9)
ρ e dr R2

where M and R are the mass and the radius of the star respectively. If we assume that
the density of the star is uniform, we can write:

4 3
M = πR ρ e (11.10)
3
49
From Stars to Our Substituting for ρ e from Eq. (11.10) in Eq. (11.9) and integrating from the centre to
Galaxy
the surface, we get:

M2
Pe ∝ (11.11)
R4

But, from Eq. (11.8) and Eq. (11.10), we find that:

5
3
M
Pe ∝ (11.12)
R5

Comparing Eqs. (11.11) and (11.12), we obtain:

1
R ∝ (11.13a)
M 1/ 3

We can also write Eq. (11.13a) as

ρ ∝ M2 (11.13b)

Eq. (11.13) shows that, as the mass of the white dwarf increases, its radius decreases.
The mass-radius relation is plotted in Fig. 11.5. The shrinking of radius is
understandable because increase in mass would mean increase in the force of
S. Chandrasekhar was gravitational contraction.
awarded Nobel Prize in
1983 for his extensive
theoretical work on white
dwarfs. 0.02
Radius (solar unit)

Chandrasekhar limit

0.01

0
0 0.5 1.0 1.5 2.0 2.5
Mass (solar unit)

Fig.11.5: Mass-radius plot for white dwarfs

A logical question at this point is: What will happen if we go on adding mass to a
white dwarf? Eq. (11.13a) indicates that if enough mass is added to a white dwarf, its
radius would ultimately shrink to zero! What is the value of this mass?
Chandrasekhar showed that if mass of the white dwarf is about 1.4 times the solar
mass, its radius will shrink to zero. This is called the Chandrasekhar limit. If the
mass of the white dwarf is more than this limiting mass, its gravitational contraction
cannot be balanced by the degeneracy pressure of the electrons. Thus, a star of mass
greater than 1.4 solar mass cannot become a stable white dwarf unless it ejects mass in
some way. No white dwarf has been discovered which has a mass higher than
Chandrasekhar limit.

In the absence of energy generating nuclear reactions, there is actually no source of

energy left in a white dwarf. Therefore, these stars would go on shining by radiating
50
their thermal energy and in the process, their temperature would decrease. Ultimately, Compact Stars
the entire thermal energy will be lost. The star becomes a cold object.

Would you not like to know how long a white dwarf star takes to cool down? The
internal temperature (say, Twd), of a white dwarf is almost constant. Its total thermal
energy can, therefore, be written as:

3
U= N e k B Twd
2
3 M
= k B Twd (11.14)
2 µ mp

where M is the mass of the white dwarf. Eq. (11.14) has been written by taking the
energy of an electron at temperature Twd as (3/2)kBTwd, (1/2)kBTwd for each degree of
freedom. In Eq. (11.14), µm p is the mean molecular weight of nuclei inside the star.
For stars which consist entirely of heavy elements, µ = 2 . If we take M = 1MΘ, and
Twd = 107K, the total thermal energy of a white dwarf is ~ 1048 ergs! This is a very
significant amount of energy. The question is: How long will this amount of thermal
energy enable a white dwarf to shine? To know this, solve the following SAQ.

SAQ 2 Spend
5 min.
Suppose the luminosity of a white dwarf star of mass 1MΘ is 10−3LΘ. If the luminosity
of the Sun, LΘ is 4 × 1026 Js−1, calculate the time for which the white dwarf will keep
shining with its present luminosity.

Having solved SAQ 2, you know that the thermal energy of a white dwarf can keep it
shining for billions of years! Sirius B, the companion of the ‘Dog Star’ Sirius A, is the
earliest white dwarf star to be observed. Fig. 11.6 shows a white dwarf star.

Fig.11.6: A white dwarf star shown by arrow

Fig. 11.7 shows the distribution of the observed white dwarfs on the H-R diagram.
Different lines such as 0.89, 0.51 and 0.22 in the Figure indicate the masses of stars in
terms of the solar mass. White dwarfs are dimmer compared to the main sequence
stars of the same surface temperature because of their smaller size. The difference in
absolute magnitudes of white dwarfs compared to main sequence stars is in the range
5 – 10. 51
From Stars to Our
Galaxy
Main sequence Sun
0

log (L/L Θ)
-1.5
0.89 0.51 0.22

-3.0
White dwarfs

-4.5

4.5 4.3 4.1 3.9 3.7 3.5

log T
Fig. 11.7: Location of white dwarfs in the H-R diagram

As we have mentioned earlier, white dwarfs are compact stars resulting from the death
of medium mass stars like the Sun. You may ask: What are the remnant of stars
which are much more massive than the Sun? Such stars end their lives as neutron
stars. You will learn about it now.

11.4 NEUTRON STAR

To understand the formation of neutron stars, you may recall from Unit 10 that when
the nuclear reactions stop, the core of a massive star (~ 10 MΘ to 100 MΘ) collapses
and a supernova explosion takes place. The supernova explosion blows away the outer
shell of the star. The question is: What is the remnant of a supernova explosion? In
1934, Walter Baade and Fritz Zwicky suggested that the core of a supernova
explosion could be a small, high density neutron star.

When the core of a star begins to collapse in the absence of nuclear reactions, its
density increases and attains a value comparable to the density of white dwarfs. The
degeneracy pressure due to electrons can balance the gravitational contraction only
when the mass of the object is smaller than the Chandrasekhar limit. If the mass is
more than this limiting value, the gravitational collapse continues and the density of
the star increases further. When the density is in the range ~ 1014 − 1015 g cm−3, the
following two things happen:

a) protons and electrons combine to produce neutrons and neutrinos according to the
following inverse β-decay reaction:

e+p→n+v

b) atomic nuclei begin to disintegrate due to pressure ionisation.

Since neutrons are fermions, their degeneracy pressure halts the gravitational
contraction and a stable neutron star is formed. It has been calculated that the neutron
stars have radii of about 10 km and densities of their core is about 1014 to 1015g cm−3.
Note that neutron stars are much denser than the white dwarfs.

You may ask: Is there any limiting mass for neutron stars similar to the
Chandrasekhar limit for white dwarfs? Yes; but the precise value of the limiting
mass for the stars is difficult to determine because the behaviour of matter at such
high densities is not well understood yet. The estimated limiting mass is in the range
52 2 – 3 MΘ.
 Compact Stars
11.4.1 Gravitational Red-shift of Neutron Stars

You have just learnt that the neutron stars are very compact. Their gravitational pull
must be very strong indeed. How strong is its gravitational pull? To get an idea, we
may estimate the value of gravitational acceleration, g, on a neutron star, say, of mass
Mn ~ 1.5MΘ and radius Rn ~ 10 km. The expression for gravitational acceleration on a
neutron star can be written as:

GM n
gn =
Rn2

Substituting the values of Mn and Rn, we get, gn ~ 2 × 1014 cm s−2. The value of gn is
11
~ 2 × 10 times higher than the gravitational acceleration on the Earth! That is, the Very often, we do thought
gravitational force on a neutron star is 1011 times stronger than that due to the Earth. experiments: the experiments
which cannot be done for
Well, to get a feel for the perceptible effect of such a strong gravitational field, let us practical reasons, but they
could be done, in principle.
do a thought experiment. Suppose you are standing on the surface of the star which is
o
emitting a yellow light at λ = 5800 A . The light reaches your eyes approximately one
metre above the ground. The interesting question is: Will you see the light as yellow?
If you think you will, you are wrong! Let us find out the reason.

A photon leaving the surface of the neutron star has energy hν and equivalent mass
hν/c2. The neutron star will attract this photon and, as a result, the photon has to work
‘hard’ against the star’s gravity to reach your eyes (just as a stone thrown upward
from the earth slows down as it goes higher). The work done by the photon to reach
hν
your eye is, W = g n H , where H ~ 100 cm, is the height of your eyes from your
c2
feet. Further, ν can be calculated using the relation λ = c / ν, i.e.,
ν ~ 5.172413 × 1014 Hz. Now, to arrive at your eyes, the photon works against gravity
and hence its energy, that is, frequency will decrease and wavelength will increase. To
calculate the change in wavelength or frequency, we can use the energy conservation
principle. Let ν′ be the frequency of the light (photons) that you observe. Thus, from
the energy conservation principle, we can write:

hν ′ = hν − W (11.15)

hν
= hν − gnH
c2

 g H 
= hν 1 − n 
 c2 

In terms of the wavelength of photons, Eq. (11.15) can be written as:

dλ gnH
~ (11.16)
λ c2

where dλ = λ′ − λ and λ′ = c / ν′. Substituting the values of gn, c and H in Eq.

(11.16), we get:

dλ
~ 2.2 × 10 −5 (11.17)
λ 53
From Stars to Our o
Galaxy Thus, yellow light of wavelength λ = 5800 A leaving your feet (the surface of the
neutron star) will arrive at your eyes as light of
 g H  o
wavelength λ ′  = λ + λ n  ~ 5800.13 A . Such differences in the wavelengths can
 c 2 

be measured easily these days.

The increase in the wavelength of a photon when it comes out of a strong gravitational
field is called gravitational red-shift. The significance of this phenomenon lies in
the fact that scientists must make necessary corrections in the observed frequency of
photon presumably coming from objects such as neutron stars in order to understand
physical processes on their surface.

It is difficult to observe neutron stars optically because, being very small in size, these
are very faint objects. However, with radio telescopes, neutron stars have been
detected in the form of pulsars: the stars which emit regular pulses of radiation, very
often several times a second.

11.4.2 Detection of Neutron Stars: Pulsars

In 1967, a remarkable discovery was made in the history of astronomy by a student

named Jocelyn Bell who was doing her doctoral work under Prof. Anthony Hewish in
Cambridge, England. She observed certain periodic pulses of radio waves coming
from a certain direction in the sky which were repeated precisely every 1.337s
(Fig. 11.8). Very soon, she discovered a few more such objects with different
periodicities. These objects are known as pulsars. A year later, it was already clear
that these pulses must have been emitted by rotating neutron stars. Periodicities of
the pulsars have been found to be between 10−3s and 4s.
Number of photons

Time
(a) (b)

Fig. 11.8: a) Detection of pulses of radio waves, an evidence of the existence of neutron stars; and b)
pulsar at the centre of a nebula (Image credit: Chandra X-ray Observatory)

You may ask: How can we be sure that a strong pulse, whose periodicity is about
a second, is emitted by a neutron star and not by a white dwarf? When a star
rotates, each of its layers experiences centrifugal force directed outward as well as the
pointing gravitational force directed inward. If Ω be the angular velocity and R be the
radius of the rotating star, the two forces balance in a Keplerian orbit and we can
write:

GM
Ω2 R = (11.18)
R2

54
 M Compact Stars
For a white dwarf, we can write 3
~ ρ ~ 10 7 gcm −3 . So the typical value of Ω is:
R
Ω ~ 2.58 s−1. (11.19)
Therefore, the time period can be written as:

2π
T=
Ω

~ 2.433s. (11.20)

From Eq. (11.18), it is clear that for objects with lower densities, the corresponding Ω
will also be lower, and hence the time period would be higher! Therefore, white
dwarfs and other bigger celestial objects are too big, their densities are too low and
they cannot emit pulses with periods as short as those observed in pulsars. In addition,
it has been observed that there are pulsars having periods of only a few milliseconds!
It is, therefore, almost impossible for white dwarfs to show such periodicities. So, the
belief that pulsars must be neutron stars became even stronger. Now, before
proceeding further, you should solve an SAQ.

SAQ 3 Spend
o 5 min.
Calculate the gravitational red shift for the yellow light (λ = 5800 A ) on the surface of
Sirius B when the photon travels a distance of 1m. Take the mass of Sirius B as
M SiB = 1M Θ , and its radius as R SiB = 16000 km.

Emission mechanism and observed profiles

Regarding pulsars, a logical question could be: What causes emission of pulses from
a neutron star? It is suggested that a neutron star is like a gigantic light house. In a
light house, a powerful light-source rotates and ships see it periodically. In exactly the
same way, emission of radiation takes place continuously, but due to rotation of the
neutron star the emission is detected only periodically by an observer on the Earth
(Fig. 11.9).

Magnetic axis
Radiation

Neutron star

Rotation axis

Radiation

Fig. 11.9: Light-house model of pulsars

55
From Stars to Our You may argue: If emission takes place from the surface of a neutron star, it
Galaxy should have been detected all the time and not periodically! Well, to understand
the periodic emission, you need to know that it is linked to the rapid rotation and
strong magnetic fields of neutron stars. Suppose a molecular cloud of radius, Rmc ~ 1
pc having a typical magnetic field, Bmc ~ 10−6G collapses into a neutron star of size,
Rn ~ 10 km. From Unit 8, you know that if the matter is highly conducting, the
magnetic flux linked to it is conserved:
2
B r ~ constant, (11.21)

Thus, we can write:

Bn Rn2 = Bmc Rmc

2
(11.22)

Substituting the typical values of Rn, Bmc and Rmc, we get:

Bn ~ 9.5 × 1018G.

Thus, we find that the value of magnetic field associated with a neutron star is very
high compared to the Earth’s magnetic field which is only a fraction of a Gauss! The
above estimate has been arrived at under the assumption that there is no loss of
magnetic field and thus the value indicates the upper limit. In reality, after some
dissipation in the process of the formation of a neutron star, the field is smaller, close
to 1011 to 1014 Gauss or even less. Generally, this field is bipolar and the magnetic axis
is not aligned with the spin axis of the star (Fig. 11.9).

As the neutron star spins at great speeds, its magnetic field induces a very strong
electric field. As a result, electrons present in the atmosphere of the neutron star are
accelerated and attain very high energies. These electrons then gyrate round the
magnetic lines of force and emit radiation called synchrotron radiation which is
directed along the lines of force. Every time the neutron star rotates, each of the two
radiating magnetic poles may point towards us (Fig. 11.9) and we may see two pulses
of radiation per rotation of the star. In many objects, you can see these two peaks
very distinctly (Fig. 11.8). Thus, the lighthouse model tells us that pulsars are not
mysterious objects at all; they are rotating neutron stars.

You may further argue: If the neutron star is emitting radiation continuously, it
would lose energy; in the absence of nuclear energy, what energy is it losing − the
gravitational energy, the rotational energy or the magnetic energy? If it loses
gravitational energy, the star would collapse further and the spin period may go down.
If it loses magnetic energy, the intensity of radiation will go down with time.
However, if it loses rotational energy, the pulsar would slow down and its time period
would increase. Observations support the argument that the most dominant component
of the energy loss is the loss of rotational energy.

So far, you have learnt that stars having mass up to 1.4 MΘ stabilise as white dwarfs
and those having mass up to 3 MΘ stabilise as neutron stars. Now, suppose that a star
has mass greater than 3 MΘ and all thermonuclear reactions have ceased in it. The
question is: What is the fate of such a star? What force will oppose gravity and
prevent the complete gravitational collapse of such stars? Obviously, the degeneracy
pressures due to electrons and neutrons are insufficient to halt the collapse because the
stellar mass is greater than 3 MΘ. In fact, there is no force which can halt the complete
gravitational crunch of such a star. This theoretical collapse of a star into a singular
point of zero volume and infinite density is called a black hole. You will learn about
black holes now. But, before that, how about solving an SAQ?

56
SAQ 4 Spend Compact Stars
5 min.
6
Suppose the Sun shrinks to the size of a neutron star of radius 10 cm. Calculate the
magnetic field strength at the surface of the neutron star. Take the radius of the Sun to
be 1011 cm and the magnetic field at its surface equal to 1 gauss.

11.5 BLACK HOLE

When the mass of a star is more than the limiting mass (~3 MΘ) for neutron stars and
all nuclear reactions in its core have stopped due to the lack of nuclear fuel, there is no
force which can stop its continuous gravitational collapse. Due to such an unhindered
collapse, the object collapses to zero radius and infinite density! It is very difficult to
visualise such an object physically. However, such an object has an interesting
property. Its gravity is so strong that not even light can escape from it. Therefore, such
a body is called a black hole.

The physics of black hole involves very sophisticated mathematics which is beyond
the scope of this course. Nevertheless, we can apply simple principles of physics to
conclude that black holes do exist. Let us ask ourselves: For a given mass, what
should be the size of a body so that even light (that is, photons - the fastest
moving thing) cannot escape from it? The answer to this question has great bearing
on our understanding of black holes. The first theoretical attempt to address this
question was made by Schwarzschild who used the principles of General Theory of
Relativity given by Einstein. The main conclusions of Schwarzschild can also be
derived using Newtonian mechanics and the concept of escape velocity. You may
recall from school physics that, for a body of mass M and radius R, the escape velocity
is given by:

2GM
v esc = (11.23)
R

Since we are interested in a black hole and trapping of light by it, we put vesc = c, the
velocity of light. Then, Eq. (11.23) reduces to:

2GM
Rg = (11.24)
c2

Rg in Eq. (11.24) is called the Schwarzschild radius. Eq. (11.24) signifies that the
size of an object of mass M must shrink to Schwarzschild radius to become a black
hole. For example, if the object has mass equal to 1MΘ, its radius must be 3 km if it
has to behave like a black hole. In other words, the Sun must shrink to a radius of 3
km to be able to trap light and become a black hole!

SAQ 5 Spend
3 min.
Calculate the Schwarzschild radius for the Earth.

You may ask: If black holes are point like objects, what is the physical
significance of Schwarzschild radius? Schwarzschild radius essentially means that
any object (including a photon) which comes within Rg (= 2GMbh /c2) of a black hole
is trapped. This limiting value Rg is also called the event horizon, since no event that
takes place within Rg from the centre of the black hole can be viewed by any observer
at R > Rg. Why is it so? This is because the gravity of the point like object is so strong
that even radiation (i.e. photons) cannot escape from the region inside the
Schwarzschild radius. 57
From Stars to Our Although the concept of Schwarzschild radius is helpful in visualising a black hole,
Galaxy you have to be careful in not stretching the meaning of escape velocity too far. This is
because of Einstein’s General Theory of Relativity (GTR) which treats gravity
entirely differently from Newtonian gravity. According to GTR, we cannot talk of
escape velocity at all. Einstein proposed that near a gravitating object, the photons
move in a curved trajectory as seen by an observer at a large distance. The black hole
being compact and massive, its gravitational force is very strong which bends the
photon path so much that a photon trying to escape would immediately return back.

Bending of light close to a black hole

As you know from the well known Einstein mass-energy equation (E = mc2), every
form of energy E has an equivalent mass m. So, the photons with energy E = hν will
also have a mass, mphoton = hν/c2. This is not the true mass since photons are really
massless. Nevertheless, this mass can be attracted by any other massive body such as
the Sun. This will bend the path of a photon. In fact, the bending of light was observed
by the famous British astronomer Arthur Eddington in the year 1919. He observed a
star during the total eclipse whose location in the sky was near the edge of the Sun and
found that the apparent location of the star has been changed by a small angle. This
convinced him that the light from that star must have been bent by the Sun. Thus,
strong bending of light can be taken as an evidence that black holes exist.

Types and location of black holes

The exact process of the formation of a black hole is not known but there are several
possibilities. Unlike the other types of compact stars, black holes do not have any
narrow range of masses. There seem to be two populations of black holes: one
population has a mass typically 6 to 14 times the mass of the Sun. It is formed due to
gravitational collapse after a supernova explosion. These so-called stellar mass black
holes could be detected all around the galaxy. Black holes of the second population
are very massive, with masses ranging from a few times 106MΘ to a few times 109MΘ.
These seem to be located at the centres of galaxies.

Detection of a black hole

Since radiation cannot escape from a black hole, it cannot be observed / detected
directly. However, it can be detected indirectly through its gravitational field. Any
matter close-by is strongly attracted by it. Suppose that the black hole is a member of
a binary. Then, it sucks matter from its companion. The matter flowing into the black
hole gets heated to a very high temperature and emits X-rays. Thus, to search for
black holes, we should look for X-ray binaries.

Now, let us summarise what you have learnt in this unit.

11.6 SUMMARY
• When the nuclear fuel of a star is exhausted completely, it contracts due to self-
gravity and becomes a compact star having density much higher than a main
sequence star.

• In the absence of radiation pressure due to nuclear reactions, the gravitational

force in the compact stars is counter-balanced by the degeneracy pressure of
fermions like electrons and neutrons.

• The nature of the remnants of a star after death depends on its mass; the dying
star turns into any one of the three kinds of compact stars, namely, white dwarf,
58 neutron star or black hole.
 Compact Stars
• If the mass of a dying star is less than the Chandrasekhar limit (~ 1.4MΘ), it
turns into a white dwarf in which the gravitational force is balanced by the
degeneracy pressure of electron gas.

• If the mass of the dying star is more than the Chandrasekhar limit but less than
another limiting value (~ 3MΘ), it turns into a neutron star in which the
gravitational force is balanced by the degeneracy pressure of neutrons. If the mass
of the collapsing star is even higher, the gravitational collapse cannot be halted
and the collapsing star becomes a black hole.

• The degeneracy pressure of a degenerate gas of electrons is given by:

5
3
P = K ρe

• Chandrasekhar showed that, for white dwarfs, the mass-radius relation is:

1
R∝ 1
3
M

and the density-mass relation for such stars is:

ρ∝ M 2

• Pulsars are stellar objects which emit periodic radio frequency pulses. Pulsars are
rotating neutron stars, because white dwarfs cannot emit pulses of this
periodicity.

• Emission of pulses from a rotating neutron star is explained on the basis of light-
house model.

• Black holes are objects of zero radius and infinite density. Such an object is
difficult to visualise physically.

• The Schwarzschild radius is given as:

2GM
Rg =
c2

and it signifies that the size of an object of mass M must shrink to Rg to become
a black hole.

• The Schwarzschild radius is also called the event horizon because an event that
takes place within Rg from the centre of the black hole cannot be observed.

• The physics of black hole cannot be understood on the basis of Newtonian

mechanics; we need to invoke Einstein’s General Theory of Relativity (GTR)
for this purpose.

11.7 TERMINAL QUESTIONS Spend 30 min.

1. For a completely degenerate electron gas, the number density of particles with
momenta in the range p and p + dp is given by: 59
From Stars to Our
Galaxy 8π
n( p )dp = p 2 dp p≤ p
3 F
h

=0 p> pF

where p F is the Fermi momentum (corresponding to Fermi energy). Show that

1
p F increases with n as n 3 . Also show that the average momentum < p > varies
1
as n 3 .

2. The masses and radii of a typical neutron star (NS), a typical white dwarf (WD)
and a typical main sequence star (MS) are given below:

Mass Radius

NS 1M Θ 10 km
WD 1M Θ 104 km
MS 1M Θ 106 km

Calculate the rotational time periods in all these cases and show that only neutron
stars satisfy the pulse time periods observed for pulsars.
o
3. Assume that the peak wavelength of X-rays coming from an X-ray binary is 1 A .
Calculate the temperature that the falling matter onto a black hole must attain to
emit this radiation.

11.8 SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs)

1. At T = 0, the exponent ( E − µ) / k B T in the denominator of Eq. (11.3) goes to

infinity. Thus, for E > µ , the denominator of f (E) becomes infinite and we have
f (E) = 0.

For E < µ , the exponent ( E − µ) / k B T in the denominator of Eq. (11.3) becomes

− ∞ at T = 0. Thus, the denominator of f (E) becomes 1 and f (E) = 1.

2. On the basis of Eq. (11.14), we find that the thermal energy of a star of mass
1M Θ is 10 48 erg. As per the problem, luminosity of the white dwarf star is:

L = 10 −3 LΘ

= 10 −3 × ( 4 × 10 26 Js −1 )

= 4 × 10 30 erg . s −1

Thus, the time for which the white dwarf will keep shining can be written as:

10 48 erg
t =
4 ×10 30 erg . s −1
60
 1018 1  7 
Compact Stars
= yr Q 1 yr = 3 ×10 s 
4 3 × 10 7  

≅ 1010yr.

3. On the basis of Eq. (11.16), we can write the expression for the change in
wavelength on the surface of white dwarf (Sirius B) as

g wd H
dλ = λ.
c2
λH GM wd
= ×
2
c R2
o
(5800 A ) × (100 cm) (6.67 ×10 −8 cm 3 g −1s − 2 ) × (2 ×10 33 g)
= ×
(9 ×10 20 cm 2 s − 2 ) (1.6 ×10 9 cm) 2

5.8 × 6.67 × 2 o
= ×10 −8 A
9 × 1 .6 × 1 .6

o
≅ 3.3 ×10 −8 A (Negligible)

4. From Eq. (11.21), we can write the magnetic field linked to the Sun and to the
neutron star (to which the Sun converts) as:

BΘ R 2 Θ = B NS R 2 NS = Constant

where BNS and RNS , respectively, are the magnetic field and radius of the neutron
star. Thus,

R 2Θ
B NS = BΘ
R 2 NS

(1011 cm) 2
= 1 gauss ×
(10 6 cm) 2

= 1010 gauss

5. From Eq. (11.24), we can write the expression for Schwarzschild radius for the
Earth as:

 R  = 2GM E
 g
 E c2

2 × (6.67 × 10 −8 cm 3 g −1s −2 ) × (6 × 10 27 g)
=
9 × 10 20 cm 2 s − 2

= 0.9 cm

61
From Stars to Our Terminal Questions
Galaxy
1. As per the problem,

8π
n ( p) dp = p 2 dp
3
h
pF
8π
∫p
2
n= dp
3
h 0

8π p3 F
=
h3 3

Thus, we can write:

1
pF ∝n 3

Now, the expression for average momentum can be written as:

pF
1
<p> =
n ∫ p n ( p) dp
0

pF
8π1
∫p
3
= × dp
h3 n
0

4
8π 1 p
= 3× × F
h n 4
4
8π 1 n 3
= 3× ×
h n 4

Thus, we can write:

1
<p> ∝ n 3

2. From Eq. (11.18), we can write the expression for the angular velocity of a
rotating star as:

GM
Ω=
R3

So, the time period is:

2π
T=
Ω

R3
= 2π
GM
62
Thus, the time period for the neutron star is: Compact Stars

1018 cm 3
T NS = 2π
(6.67 ×10 −8 cm 3 g −1s − 2 ) × (2 ×10 33 g)

10
= 2π × 10 − 4 s
6.67 × 2

= 5.4 ×10 − 4 s

Time period for white dwarf is:

10 27 cm 3
TWD = 2π
(6.67 ×10 − 8 cm 3 g −1s − 2 ) × (2 × 10 33 g)

≈ 17 s

And the time period for a main sequence star is:

10 33 cm 3
TMS = 2π
(6.67 ×10 −8 cm 3 g −1s 2 ) × (2 × 10 33 g)

≈ 17 ×10 3 s

Thus, we find that the time period of the pulses emitted by a neutron star is of the
same order as of the pulses from pulsars.

3. We know that,

λ m T = 0.3 cm K

As per the problem:

o
λ m = 1 A = 10 −8 cm
Thus,

0 .3
T= K
10 −8
= 3 ×10 7 K

63
From Stars to Our
Galaxy
UNIT 12 THE MILKY WAY
Structure
12.1 Introduction
Objectives
12.2 Basic Structure and Properties of the Milky Way
12.3 Nature of Rotation of the Milky Way
Differential Rotation of the Galaxy and Oort Constant
Rotation Curve of the Galaxy and the Dark Matter
Nature of the Spiral Arms
12.4 Stars and Star Clusters of the Milky Way
12.5 Properties of and Around the Galactic Nucleus
12.6 Summary
12.7 Terminal Questions
12.8 Solutions and Answers

12.1 INTRODUCTION
In Unit 11, we discussed the death of stars and its consequences and thereby
completed our discussion about stars. You now know how stars are formed, how they
live their lives and how and why they die. If you wish to know more about the
Universe, you may ask questions like: What is the structure of the Universe? Is it a
single entity comprising unrelated and independent stars or does it consist of
substructures in which stars are arranged according to a scheme? If stars constitute
communities, what is the nature of such communities and what mechanism brings
them into existence? It is now believed that stars arrange themselves into billions
of self-contained systems called galaxies. Our star, the Sun, is one of about 200
billion stars in the galaxy called the Milky Way – our home galaxy. In the present
Unit, you will learn about the Milky Way galaxy also called the Galaxy.

On a clear night sky, you can see a broad white patch running across the sky. It is seen
during winter months in the northern hemisphere. The patch is actually made up of
hundreds of billions of stars, so close to each other that they cannot be seen
individually. This is a part of our home galaxy, the Milky Way. Just as Copernicus
discovered that the Earth and the planets revolved around the Sun, astronomers in the
early twentieth century discovered that the Milky Way system is indeed a galaxy in
which our solar system is situated.

In Sec. 12.2, you will learn the basic structure and properties of the Milky Way. It is
found that the Sun revolves once in 200 million years around the centre of this galaxy
in a nearly circular orbit. In addition, the Galaxy itself is rotating. You will learn the
consequences of the rotation of the Galaxy in Sec. 12.3. You will also discover that
the Galaxy is a spiral galaxy with several spiral arms. In Sec. 12.4, you will learn
about the various constituents of the galaxy, namely the stars, globular clusters,
compact stars and the interstellar medium. At the present time, we know a lot about
our Galaxy, its shape and its size. Much of the contemporary focus has been to
understand the central region of the galaxy and the activities surrounding it. Today, it
is believed that the central region contains a massive black hole of mass M ~ 2.6 ×
106MΘ. In Sec. 12.5, you will learn about the nature and characteristics of the central
region of the Milky Way.

64
Objectives The Milky Way

After studying this unit, you should be able to:

• describe the shape and size of the Milky Way galaxy;

• explain the consequence of rotation of the Galaxy;
• describe the major building blocks of the Galaxy including the stars and star
clusters;
• explain the nature and persistence of the spiral arms of the Galaxy; and
• describe the activities taking place near the centre of the Galaxy particularly the
nature of the central compact object.

12.2 BASIC STRUCTURE AND PROPERTIES OF THE

MILKY WAY
The Milky Way is a highly flattened, disk shaped galaxy comprising about 200 billion
(2 × 1011) stars and other objects like molecular clouds, globular clusters etc. It is so
huge that, to travel from one edge of the Galaxy to the other, light takes about
100,000 years! Its radius is about 15,000 pc. The solar system is located roughly at a
distance of about 8.5 kpc from the centre of the Galaxy. The total mass of Milky Way
has been estimated to be about 2 × 1011MΘ. Refer to Fig. 12.1 which shows a
schematic diagram of the Milky Way Galaxy. You may note that, broadly speaking,
the Galaxy can be divided into three distinct parts: a central bulge (B), the flattened
galactic disk (A) and a halo (H) which surrounds the Galaxy.

H
B
A

Sun
8500 pc

30,000 pc

(a) (b)

(c)

Fig. 12.1: a) A schematic diagram of the Milky Way Galaxy viewed edge on, b) view of the Galaxy
from the top; and c) part of the Milky Way visible in the sky

The Milky Way belongs to the Local Group, a group of 3 big and 30 odd small
galaxies. After the nearby Andromeda galaxy (M31) shown in Fig. 12.2, ours is the
65
From Stars to Our largest galaxy in the group. It has a few dwarf galaxies as satellites or companions.
Galaxy Prominent among them are Large and Small Magellanic clouds.

(a)

(b) (c)

Fig. 12.2: a) The Andromeda galaxy; b) the Large Magellanic Cloud; and c) the Small Magellanic
Cloud

Let us now discuss the nature of the components of Milky Way.

The Central Bulge

The central bulge (Fig. 12.1a) is a more or less spherical cloud of stars. Being located
in the disk region of the Galaxy, we cannot see this region in optical wavelengths. It is
so because the disk region consists of gas and dust which absorbs optical wavelengths
and obstructs our view. The total mass of the bulge is estimated to be about 1010MΘ.
Apart from stars, this region consists of gas in the form of molecular clouds and
ionised hydrogen. The motion of the stars and the gas near the centre of the bulge
suggests that there could be a massive black hole at the centre.

The Disk Component

The flattened disk component has a radius of about 15,000 pc. But its thickness is very
small. Most of the stars are located along the central plane of the disk and as we move
66
away from this plane, the density of stars decreases.
The most significant feature of the disk component is the existence of spiral arms. The Milky Way
Condensation of stars has been observed along the spiral arms. These arms have very
young stars called Population I stars, star-forming nebulae, and star clusters. The arms
are named after the constellations in the direction of which a large portion of the arm
is situated. Our solar system is located on a Local or Orion arm.

The Halo Component

The bulge and the disk components are surrounded by another, not so well defined,
and not so well understood spherical component called the halo component. This is
mainly made up of gas and older population of stars. These stars exist in very dense
5 6
clusters; each cluster having 10 to 10 stars. These are called globular clusters. Stars
in these clusters are so densely packed that they cannot be resolved, and clusters
appear like a circular patch of light (Fig. 12.3).

Fig. 12.3: A globular cluster (NGC 1916) containing millions of stars

The nature of galactic rotation (about which you will learn later in the Unit) suggests
that there is a large amount of matter which is governing the motion of the stars in the
disk. This matter is not visible in any wavelengths and scientists call it the dark
matter. It is believed that the halo contains at least an equal amount of matter as the
disk itself, if not more, in the form of dark matter. What could be the nature of this
matter? We do not know yet.

12.3 NATURE OF THE ROTATION OF THE MILKY WAY

In the early nineteenth century, Jan Oort and Bertil Lindblad studied the motion of a
large number of stars located near the Sun. Their study indicated that stars in the
Galaxy constitute a gravitational system. Like many other gravitating systems, our
galaxy also rotates, though very slowly. The rotational velocity enhances the stability
of the Galaxy since the outward centrifugal force can counterbalance the inward pull
due to gravity.

The assumption that the Galaxy is a gravitating system is at the core of the
traditional hypothesis about how the Galaxy came into being. Since the centrifugal
force is only along the equatorial plane, there is no obstacle for matter (other than the
pressure) to fall along the vertical direction. As a result, the initially spherical
distribution of matter has become, after over 10 billion years, the highly flattened
galaxy of today.

67
From Stars to Our Refer to Fig. 12.4 which shows the initially spherical distribution of matter (Fig.
Galaxy 12.4a) gradually evolving into the present day flattened Galaxy (Fig. 12.4d).

(a) (b) (c) d)

Fig. 12.4: Evolution of the Galaxy (from a to d) starting from a spherical cloud of gas and settling
into the flattened disk shape of today

The rotational velocity of stars in the Galaxy is very slow compared to what we are
familiar with in our solar system. For instance, at a distance of roughly 8 kpc away
from the centre, the Sun takes about 240 million years to rotate once around the
galactic centre. The question is: How do we know about this rotation? How do we
measure it? What are the consequences of this rotation on the structure and
further evolution of the Galaxy? Let us now learn about these aspects of the Galaxy.

12.3.1 Differential Rotation of the Galaxy and Oort Constants

To appreciate the rotation of stars and other objects in the Galaxy, you need to
understand the concept of differential rotation. Let us consider the motion of a star in
the Galaxy. Suppose, for the sake of argument, that the whole mass of the galaxy is
concentrated at its centre and the stars move like planets round the Sun on orbits
called the Keplerian orbits. The angular velocity of a star at a distance r from the
centre can, therefore, be written as (Eq. (5.1), Unit 5):

1/ 2
 GM Gal 
ωKep (r ) =   . (12.1)
 r3 

where MGal is the mass of the Galaxy. We see from Eq. (12.1) that the angular
velocity is not a constant. Further, using Eq. (12.1) and the relation v = ωr, we can
write the rotational velocity of the star as:

vφ (r) ∝ r−1/2. (12.2)

Now, if stars in the Galaxy are embedded as particles in a rigid body, then the angular
velocity, ω of stars would have been constant, independent of its distance from the
centre and the rotational velocity vφ (r) of a star in such a system is given by:

vφ (r) ∝ r. (12.3)

Comparison of Eqs. (12.2) and (12.3) clearly shows that the nature of the dependence
of rotational velocity on the distance from the centre is different in the two cases – the
Keplerian motion and rigid body rotation.

When different components of a system rotate independently, the rotation is known as

differential rotation. On the basis of observations, we can say that the stars indeed
have differential rotation. We do see the signature of this differential rotation in
astronomical observations.

68
 The Milky Way
S V0

l
D

Q
90o + α
V
R0 α

Fig.12.5: Geometry of the differential galactic rotation for stars closer to the Sun

Let us now obtain expression for the velocity of an object in the Galaxy. To do so,
refer to Fig. 12.5, which depicts the velocity vectors of the Sun (S) and a star (Q) with
respect to the galactic centre (C). Let us assume that the Sun and the star are at a
distance of R0 and R , respectively from C and let D be the distance between the Sun
and the star. Let V0 be the Sun’s rotational speed and V be the star’s rotational speed;
both assumed here to be on circular orbits for simplicity. Let us also assume that l is
the angle between the direction of the galactic centre and the direction of the star from
the Sun.

This is the so-called galactic longitude of the star. (Galactic longitude of the centre is l
= 0 by this definition). The Sun-star direction makes an angle α with the velocity
vector of the star. If ω0 and ω be the angular velocities of the Sun and the star,
respectively, we can write:

ω 0 = V0 / R0 (12.4a)
and
ω =V / R (12.4b)

We are interested in finding the radial speed of the star with respect to the Sun. The
radial velocity is the velocity along the line joining the Sun and the star. We can,
therefore, write the star’s radial velocity with respect to the Sun, Vr, as:

Vr = V cosα − V0 sin l . (12.5)

where the first term on the right hand side is the component of the star’s space
velocity along the Sun-star direction and the second term is the component of the
Sun’s own velocity along the Sun-star direction. Thus, radial velocity is essentially the
difference between the projected velocities along the Sun-star direction. Substituting
Eq. (12.4) in Eq. (12.5), we get:

Vr = ω R cos α − ω0R0 sin l (12.6)

In triangle QSC in Fig. 12.5, the angle SQC is 90 + α.

69
From Stars to Our Thus, using the law of sines, we can write:
Galaxy
sin (90 + α) sin l
=
R0 R
or,
R0
cos α = sin l
R

Substituting this value of cos α in Eq. (12.6), we get:

Vr = (ω − ω0) R0 sin l (12.7)

Similarly, the tangential component of the velocity of star with respect to the Sun can
be written as:

Vt = V sin α − V0 cos l = (ω − ω0) R0 cos l − ωD. (12.8)

In the solar neighbourhood, D << R0, and we can approximate the angular velocity of
the star, ω in terms of ω0 by using the Taylor series expansion. Thus, we can write:

 dω 
R0 - R = D cosl ω = ω 0 +   ( R − R0 ) (12.9)
S  dR  R0
D
l
If we define a constant A as,
Q

 
 dω   
R
A =− 0   = 1  V0 −  dV   (12.10)
2  dR  2  R0  dR  
R0 R   R0 R0
 

then, the radial component of the velocity of a nearby star can be written as:

Vr = − 2A (R − R0) sin l = AD sin (2 l ) (12.11)

where, we have made use of the fact that for D << R0, (R0 − R) ~ D cos l (Fig. 12.6).
C
Further, using the same procedure and approximations, you can show that the
Fig.12.6 tangential velocity of the star is given by:

Vt = D [A cos (2 l ) + B] (12.12)

where B (= A − ω0 ) is another constant given as:

R 0  dω 
B =−   − ω0 (12.13)
2  dR  R
0

Since V0 = ω0 R0, Eq. (12.13) can be written as:

 
1  V0  dV  
B=−  +  (12.14)
2  R0  dR  
 R0 
 

The constants A and B are called Oort constants. Fig. 12.7 depicts the observed
70 variation of a star’s radial velocity as a function of galactic longitude. Note that the
variation of radial velocity is periodic in longitude with a period of 180o and this The Milky Way
observation is consistent with Eq. (12.11). The value of R0 is 8.5 kpc. The estimated
values of Oort constants are:

A = 15 kms−1 kpc−1

and
−1 −1
B = −10 kms kpc

+40

+20
Radial speed (kms-1)

-20

-40
0 90 180 270 360
Galactic longitude (degrees)
Fig. 12.7: Variation of the radial velocity of stars as seen from the Earth

SAQ 1 Spend
10 min.
a) Derive Eq. (12.12).

b) Show that
 
1  V0  dV  
A=  − 
2  R0  dR  
  R0 
 
and
 
1  V0  dV  
B= −  + 
2  R0  dR  
  R0 
 

On the basis of the above discussion, you know that the galactic disk in our
neighbourhood is indeed differentially rotating. You may ask: Do the stars rotate as
if the entire mass is concentrated at the galactic centre? To answer this question,
we must know the rotation velocity of galactic objects as a function of their distances
from the centre of the Galaxy. This is known as the rotation curve of a galaxy which
contains useful information about the mass distribution in a galaxy. Let us learn about
it now.

12.3.2 Rotation Curve of the Galaxy and the Dark Matter

The rotation curve of a galaxy is obtained by plotting the rotational velocities of

galactic components against their distances from the galactic centre. Refer to Fig. 12.8
which depicts the rotation curve of our galaxy. Note that it is not a smooth curve;
rather, it has various ups and downs. This nature of the curve is not understandable in 71
From Stars to Our terms of Keplerian motion of stars in the vicinity of the Sun which assumes that the
Galaxy galactic mass is concentrated at the centre and the rotation velocity must be
proportional to R −1/2 (Eq. (12.2)).

300

Rotational velocity (kms-1)

Keplerian motion

250

Rotation curve

200 Sun

150
0 4 8 12 16
Radius (kpc)

Fig. 12.8: Rotation curve of the Milky Way

Rotation curves, such as the one shown in Fig. 12.8, indicate that the matter is spread
out all across the galaxy. This inference is, however, contrary to the concentration of
stellar mass around the nucleus of the galaxy. You may ask: What is the way out of
this paradoxical situation? If the rotation curve flattens out instead of falling
monotonically, we can infer that much of the matter in the disk cannot be in the form
of stars. How do we infer this? Let us try to understand it using the fact that the
centrifugal force Vφ2 R must balance gravity and we can write:

Vφ2 GM ( R )
= (12.15)
R R2

where, M(R) is the mass of the Galaxy up to radius R and Vφ is the rotational velocity
of a stellar object at a distance R from the galactic centre. If Vφ = V0, a constant, we
have from Eq. (12.15):

V 2R
M ( R) = 0 . (12.16)
G

Eq. (12.16) indicates that the mass within radius R must increase linearly with R, and
not be concentrated entirely at the centre. But we do not observe this spread out mass.
This means that sufficient amount of matter in the Galaxy exists in a form which is
not detectable. What is this matter made up of ? No one knows. In the field of
Astronomy and Astrophysics, this is known as the missing mass problem. The
matter perhaps exists in a form that does not emit radiation and hence is also called
dark matter.

12.3.3 Nature of the Spiral Arms

In Fig. 12.1, we showed a sketch of our galaxy as it would be seen by someone

situated very high above and outside our galaxy. Note that it consists of spiral arms.
The prominent arms are (a) Norma Arm, (b) Scutum-Crux Arm, (c) Sagittarius Arm,
(d) Orion Arm, (e) Perseus Arm and (f) Cygnus Arm. The Sun’s location is on Orion
72 Arm (see Fig. 12.9). The spiral arms do not form a single entity that was originally
present. These are the result of dynamical interaction of the Galaxy with other The Milky Way
galaxies and the matter present in the inter-galactic space.

Perseus arm 1 kpc

Orion-Cygnus arm

Sagittarius arm

Sun

To centre

Fig. 12.9: Schematic diagram depicting some of the spiral arm structures of the Galaxy in which
dots denote stars

SAQ 2 Spend
7 min.
How many times would the Sun have revolved around the centre of the Galaxy if it is
rotating with a velocity of 250 kms−1 at a distance of 8.5 kpc from the galactic centre?
Assume the age of the Sun to be 4.6 × 109 years. This age is different from the age of
the Galaxy, since the Sun is relatively younger.

To have a qualitative understanding of the spiral arms of the Galaxy, we need to ask
ourselves: How do we determine that the disk of the Galaxy comprises spiral
arms? In view of the differential rotation of the galactic disk, how do the spiral
structures persist for so long?

The first evidence of the spiral nature of the Galaxy came from the observations that
the distribution of O and B stars is not uniform in the Galaxy. To appreciate the
significance of this observation, you should recall from Unit 10 that O and B stars are
very massive, bright and short-lived. These massive stars, after their death, return
some of the stellar materials back to the interstellar medium. It is observed that these
stars occur around the locations where star formation can take place; that is, the
interstellar medium containing gas and dust. Therefore, it was suggested that the spiral
arms of the Galaxy contain gas and dust in the form of molecular clouds, and young
stars like O and B stars. But the problem was how to detect gas and dust at such
distances? This could be made possible by radio astronomy, particularly after the
discovery of 21 cm radiation. This enables astronomers to map the Galaxy and probe
its spiral nature. Astronomers use the intensity of the neutral hydrogen and carbon
monoxide emission lines to trace the spiral arms. Further, the young stars emit
ultraviolet light and ionise the gas which surrounds them. These ionised gaseous
regions are known as the H II (H II stands for ionised hydrogen) regions. They are
very luminous. Simultaneously, due to the recombination process taking place in these
H II regions, neutral hydrogen (H I) regions are formed which are detected by their
21-cm radio emission.
73
From Stars to Our Next logical question could be: How do these spiral structures persist because the
Galaxy differential rotation of the galaxy should have destroyed them? To answer this
question, C.C. Lin and Frank Shu proposed the so called density wave model.
According to this model, spiral arms are not a simple fixed array of stars; rather, spiral
arms are the areas where the density of gas is greater than in other places. As such, the
arms and the space between the arms contain roughly the same number of stars per
unit volume. However, the arms contain larger number of brighter (O and B) stars.
According to Lin and Shu, the high density waves move through the galactic disk
which gives rise to the formation of stars. As interstellar clouds approach the density
wave, it is compressed, the collapse of interstellar gas is triggered and new stars are
formed. Thus, density waves are capable of generating all the constituents of spiral
arms. You must note that according to the density wave model, a spiral arm is not a
static collection of slow moving gas and stars; rather it is a dynamic entity which
always contains the same type of objects.

This brings us to the question: How does a density wave come into existence? The
density wave model does not provide a satisfactory answer to this question.
Astronomers believe that the death of massive stars causing supernovae explosions
may produce density waves. It has been found by computer simulation (experiments
on the computer) that the supernovae explosions combined with the Keplarian motion
can lead to the formation of spiral arms.

So far, you have learnt how the Milky Way looks like from the edge as well as from
above and what are its large scale basic components. Now we discuss the nature of the
building blocks, namely, the stars and the star-clusters of these components.

12.4 STARS AND STAR CLUSTERS OF THE MILKY WAY

You have learnt in the previous section that the components of the Galaxy consist of
stars and star clusters. The question is: Are the stars in all the regions of the Galaxy
similar? A detailed knowledge about the types of stars in each region can provide
valuable information about various aspects of the Galaxy itself.

Types of Stars in our Galaxy

Stars can be classified into the so-called population (I and II) depending on the
abundance of the metals (any element heavier than hydrogen and helium is called a
metal in astronomy) in them. To quantify, let us represent the fraction of hydrogen per
unit mass by X, of Helium by Y and of metals (all metals combined) by Z. The stars
which are very much metal deficient are called the population II stars. For such stars,
Z < 0.001. Population II stars are observed far from the galactic plane and are
primarily located in the halo. On the other hand, the metallicity of the stars increases
(Z → 0.01) as we approach the disk closer to the plane. These stars constitute the disk
population. Stars in the plane of the disk and in spiral arms are relatively young,
bright and blue in colour. They are called population I stars and they are metal rich.

You may ask: Why do the different regions of the Galaxy have different
metallicity? Actually, when the galaxy was originally formed, there was very little
metal in it, since no metal had been produced by that time. Initially, the stars formed
from material which had no metal (Z ~ 0). But as the density of the initial cloud, from
which the Galaxy was formed, increased towards the plane of the disk, massive stars
were forming closer to this plane and they evolved more rapidly than the halo stars.
They underwent supernova explosions and spewed out metals formed in them (refer to
Unit 10) in the interstellar gas. The interstellar gas was thus enriched with metals as it
approached the galactic plane. The second generation stars formed from this gas are
naturally more metal rich. The population I stars, therefore have higher metallicity
74 (Z ≥ 0.01).
Further, the globular clusters (consisting of roughly 104 to 105 population II stars The Milky Way
each) which formed a spherical halo around the Galaxy are found throughout the
Milky Way and other galaxies. Like the stars, the globular clusters also display a high
degree of gradient in their metal content depending upon their location in the Galaxy.
They are found in almost every component of the Galaxy: from galactic centre to very
far away in the halo. There are a few hundreds of these objects in our galaxy. Fig. 12.3
shows one such globular cluster. The typical size of a globular cluster is about 5 pc. It
has been found that the majority of stars in the globular clusters are the main sequence
stars while other stars in them belong to the red-giant branch, sub-giant branch and the
horizontal branch.

Since our galactic centre is about hundred times closer to us than the nearest well-
known galaxy Andromeda, we may assume that the astronomers would be most
excited to know the nature of the galactic centre. However, as we said earlier, we
reside on the plane of the Galaxy, and millions of stars and dust obscure our direct
view of the centre in optical, ultraviolet and soft X-rays. In infrared and radio waves,
the observation becomes easier. Given that the galactic centres are usually bright and
contain many stars, you may naturally be curious to know as to what may be going on
in and around the galactic centre of the Galaxy. In the next section, we shall describe
the motion of matter and gas within about 100pc of the galactic centre including the
properties of the possible black hole at the centre of the Galaxy and how it can be
detected.

12.5 PROPERTIES OF AND AROUND THE GALACTIC

NUCLEUS
Astronomers believe that at the very centre of the Galaxy, there exists a black hole of
mass 2.6 × 106MΘ. Modern radio telescopes using very large baseline arrays have
been able to resolve features of the galactic centre in great detail. It has been estimated
that about 10% of the entire mass of the galactic interstellar matter (that is, about
108MΘ) resides within a 100pc of the centre of the Galaxy. The number density of
−3
hydrogen is about 102 cm which is about a hundred times greater than the average
density in the entire Galaxy. This interstellar gas is mostly concentrated in giant
molecular clouds (GMCs), of few parsec size, which are very hot (40-200K)
compared to the average gas in the galaxy.

Since the observed X-rays and γ-rays have energies ≥ keV, the gas must be very hot
(~ 107K or more). At least a dozen, very compact and energetic sources have been
detected by satellites such as EINSTEIN and ROSAT (sensitive to X-rays/γ-rays)
within the central region. Many of these compact objects have been identified to be
black holes and neutron stars. At the centre, Sgr A* itself is not very strong in X-rays,
and it is believed that this is due to the fact that not much matter is falling on it.

As we go closer to the centre, the activity increases and there is evidence of very high
rates of star formation. But since the disk does not contain enough mass supply to
continuously carry on star formation, the matter has to come from outside the region.
It is believed that interstellar gas accumulates at a rate of about 10−2MΘ for some time
(say 107 years) and the accumulated mass then collapses and a fresh star formation
event is triggered due to density waves. This is probably what is happening from time
to time close to the centre of our galaxy.

One can easily compute the mass of the central compact object by using the well
known virial theorem:

2U + Ω = 0, (12.17)

where U is the total kinetic energy and Ω is the total potential energy. That is, 75
From Stars to Our 2
M < v > + Ω = 0, (12.18)
Galaxy
where < v2 > is the mean square velocity and

R M ( R)
Ω = −G ∫ dM ( R ) (12.19)
0 R

Using Eqs. (12.18) and (12.19), we can determine the enclosed mass at a given radius
if we know the stellar velocities data.

107
Enclosed mass (solar unit)

106

105
0.001 0.01 0.1 1.0 10.0
Radius (pc)

Fig. 12.10: Graph showing variation of the enclosed mass as a function of the radial distance close
to the galactic centre

In Fig. 12.10, we show the enclosed mass as a function of distance from the centre of
the Galaxy. It is clear that even though there are no measurements below R < 0.015pc,
the enclosed mass seems to have converged to a fixed number. The mass can be read
out from the plot itself: it is 2.6 ×106MΘ. The only acceptable solution seems to be that
a massive black hole of this mass resides at the centre. For a spiral galaxy this is not
very small, but probably in the lower end. Generally, in the centres of spiral galaxies,
black holes of mass M ~ 107MΘ are quite common.

Spend SAQ 3
5 min.
What is the mass density of the region around the galactic centre in units of (MΘ pc−3)
below r = 0.015 pc?

Now, let us summarise what you have learnt in this unit.

12.6 SUMMARY
• The Universe comprises billions of self contained systems called galaxies. The
Sun – only star in our solar system – is one of about 200 billion stars in the galaxy
called the Milky Way, our home galaxy.
• The Milky Way is a highly flattened, disk shaped galaxy. Its radius is 15,000 pc
and its total mass is estimated to be about 2×1011MΘ.
76
• Broadly, the Milky Way can be divided into three distinct portions: a central The Milky Way
bulge, the flattened galactic disk, and a halo.
• The central bulge is a spherical cloud of stars and most of the mass (~1010MΘ) of
the Galaxy is contained in it.
• The disk component consists of spiral arms; most of the stars are located along
the central plane of the disk.
• The halo is made up of gas and older population stars; these stars exist in very
dense clusters called globular clusters. The halo contains an equal amount of
matter as the disk itself in the form of dark matter.
• Like many gravitational systems, our galaxy also rotates. The rotational velocity
of stars in the Galaxy is very slow compared to what we are familiar with in our
solar system. The stars in the Galaxy undergo differential rotation.
• Rotation curve of a galaxy contains useful information about the distribution of
mass in it and it helps us understand whether or not most of the mass of the galaxy
is concentrated at the galactic centre.
• The spiral arms of the Galaxy do not form a single entity that was originally
present; rather, they result from the dynamical interaction of the Galaxy with
other galaxies and the matter present in the inter-galactic space.
• The persistence of spiral arms, despite differential rotation, is explained on the
basis of density wave model. According to this model, spiral arms are the areas
where density of gas is greater than other places. The arms and the space between
them contain roughly the same number of stars per unit volume. However, the
arms contain larger number of brighter (O and B) stars.
• Investigations indicate that a massive black hole, having mass 2.6×106MΘ,
resides at the centre of the Milky Way.

12.7 TERMINAL QUESTIONS Spend 30 min.

1. Suppose the contribution of the stellar disk to the mass of our galaxy is negligible
and the mass of the entire Galaxy is concentrated at the centre located at a
distance of 8.5kpc. What would be the mass you need to put at the galactic centre
in order that the estimated Oort constants agree roughly with those observed?
2. Describe the spiral arms of the Galaxy. Explain how they can persist for a long
time.
3. Distinguish between stars of population I and II. Where are they found in the
Galaxy?
4. It is believed that, in its early phase, the universe was very hot. The average
energy of particles at that time was ~ 15 GeV. What was the temperature in
degree Kelvin?

12.8 SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs)

1. a) From Eq. (12.8), we have the expression for the tangential component of the
velocity of star as:

Vt = (ω − ω0 ) R0 cos l − ωD
77
From Stars to Our
Galaxy And, from Eq. (12.9) we have

 dω 
ω − ω0 =   (R − R )
0
 dR 
  R0

 dω 
= − D cos l  
 dR  R0

because ( R0 − R) ~ D cos l . Thus, we can write:

R0  dω 
Vt = − 2 D cos 2l   − ω0 D
2  dR  R0

because we can approximate, ω ≈ ω0 . Thus, we get

 R0  dω  
Vt = D  − 2 cos 2l   − ω0 
 2  dR  R 
0

 − R0  dω  
= D (1 + cos 2 l )   − ω0 
 2  dR  R0 

 R  dω  R  dω  
= D − 0   − cos 2 l 0   − ω0 
 2  dR  R 2  dR  R 
0 0

 R  dω  R  dω  
= D − 0   cos 2 l − 0   − ω0 
 2  dR  R 2  dR  R 
0 0

= D [ A cos 2 l + B ]

b) We can write:

 dω   d V 
  =  
 dR  R0  dR  R  R0

1  dV  V0
=   −
R0  dR  R R0 2
0

where ω and V, respectively, are the angular velocity and radial

velocity of a star. Thus, we can write:

1  V0  dV  
A=  −  
2  R0  dR  R 
 0 

and

1  V0  dV  
B=−  +  
2  R0
  dR  R0 
78
2. We can write the time period of the Sun’s rotation as: The Milky Way

2 π R0
T=
V0
9
Since the age of the Sun is 4.6×10 yrs, the number of times (n) it would have
revolved around the galactic centre can be written as:

4.6 × 10 9 × 3 × 10 7 7
n = s (Q 1 yr = 3 × 10 )
T

(4.6 × 3 × 1016 s) × (250 × 10 3 ms −1 )

=
2 π × (8.5 × 103 × 3.1 × 1016 m)

4.6 × 3 × 250
=
2 π × 8 .5 × 3 .1
≈ 21

3. Assuming that the mass of the galactic centre is ~ 2.6 ×10 6 M Θ , we can write:
2.6 × 10 6 M Θ
Mass density =
4π
× (0.015 pc) 3
3

if we take the mass enclosed within r ≈ 0.015 pc to be 2.6 × 10 6 M Θ . So, we

have:

2.6 ×10 6 M Θ
mass density =
4π
× (0.15 pc) 3 ×10 − 3
3

3 × 2 .6
= ×10 9 M Θ pc − 3
3
4π × (0.15)

≈ 1.8 ×1011 M Θ pc − 3

Terminal Questions

1. From Eqs. (12.10) and (12.14), we have:

V0
= A− B
R0

= (15+10) kms−1. kpc−1

= 25 kms−1.kpc−1

substituting the values of Oort’s constants A and B. Thus,

V0 = 25 kms − 1 . kpc − 1× (8.5 kpc)

−1
= 212.5 kms 79
From Stars to Our So, for the given values of Oort’s constant A and B, the value of the star’s (the
Galaxy −1
Sun) rotational speed (V0) is 212.5 kms . Now, to determine the mass, MG at the
galactic centre whose gravitational force needs to be counter balanced by the
centrifugal force experienced by the Sun rotating with speed V0, we can write:

G MG V 2
= 0
R0 2 R0

V0 2 R0
MG =
G

(212.5 ms −1 ) 2 × 10 6 × (8.5 × 10 3 × 3.1 × 1016 m)

=
6.7 × 10 −11 m 3 kg −1s − 2

(Q 1 pc = 3.1×1016 m)

( 212.5) 2 × 10 6 × 8.5 × 10 3 × 3.1 × 1016

= MΘ
6.7 × 10 −11 × 2 × 10 30

( 2.125) 2 × 10 4 × 10 6 × 8.5 × 10 3 × 3.1 × 1016

= MΘ
6.7 × 10 −11 × 2 × 10 30

(Q M Θ = 2 ×10 30 kg)

(2.125) 2 × 8.5 × 3.1

= 1010 . MΘ
6 .7 × 2
≈ 1011 M Θ

2. See text.
3. See text.
4. You know from the kinetic theory of gases that, the average energy of the
particles at temperature T is given by:

3
E= k BT
2
As per the problem, average energy

E = 15 GeV

=15 ×10 9 ×1.6 × 1019 J

Thus, we have:
3
15 × 109 × 1.6 × 10 −19 J = k BT
2
or,
(1.5 × 1.6 × 10 −9 × 2 J)
T=
3 × (1.38 × 10 − 23 JK −1 )

≈ 1014 K
80
 Galaxies
UNIT 13 GALAXIES
Structure
13.1 Introduction
Objectives
13.2 Galaxy Morphology
Hubble’s Classification of Galaxies
13.3 Elliptical Galaxies
The Intrinsic Shapes of Ellipticals
de Vaucouleurs Law
Stars and Gas
13.4 Spiral and Lenticular Galaxies
Bulges
Disks
Galactic Halo
The Milky Way Galaxy
13.5 Gas and Dust in the Galaxy
13.6 Spiral Arms
13.7 Active Galaxies
13.8 Summary
13.9 Terminal Questions
13.10 Solutions and Answers

13.1 INTRODUCTION
On a clear, dark night, you can see the diffuse faint, narrow band of the Milky Way
stretching across the sky. It becomes broadened towards the constellation Sagittarius,
and is seen to be covered here and there with dark areas. The Italian astronomer and
physicist Galileo Galilei was the first to observe the Milky Way through a small
telescope. He saw in the faint band an array of stars and clusters of stars, interspersed
with dark patches.

Following Galileo’s pioneering observations in the 17th century, the Milky Way has
been studied extensively with a variety of telescopes of increasing sensitivity and
sophistication. We now know that the Milky Way is composed of more than a
hundred billion stars, spread in a large, thin disk with a bloated centre, which is
known as the bulge. The disk has a diameter of about 30,000 pc, but it is only about
1000 pc thick. There are large spiral features, known as spiral arms in the disk. An
object with a structure like that of the Milky Way is called a spiral galaxy. The Sun is
an inconspicuous star, situated in the disk of the Milky Way, at a distance of about
8,500 pc from the centre.

Our Galaxy is not the only one in existence. Within the observing limits of even a
moderately large telescope there are about 10 billion galaxies, covering a wide range
of sizes and shapes. While about half the galaxies have shapes like the Milky Way, a
large fraction of the rest have the appearance of ellipses. Many galaxies have irregular
shapes, while some are mere dwarfs compared to the larger systems. The spiral galaxy
NGC 4622, shown in Fig. 13.3, is similar to the Milky Way. If we could move out of
our Galaxy, and observe it from a great distance along a line of sight which is
perpendicular to the disk, it would show a more or less similar appearance.

Some galaxies occur as single objects, while others occur in groups of a small number
of galaxies or in large clusters containing thousands of galaxies. It is not unusual to
find two galaxies in collision with each other, or interacting with each other from a
5
Galaxies and the distance. The centres of a small fraction of galaxies contain what is known as an
Universe active galactic nucleus. This is a tiny object compared to the whole galaxy, but emits
energy which can exceed by far the entire energy output of the rest of the galaxy.

Using very large telescopes on the Earth, or the Hubble Space Telescope (HST) which
is in orbit around the Earth, it is possible to obtain images of galaxies which are
located extremely far away. Light from these galaxies takes a long time to reach us.
Light from the galaxies that we detect now, began its journey towards the Earth so
long ago that the galaxies, and the Universe itself, were significantly younger at that
time. The galaxies at these early epochs are found to be significantly smaller and less
well-formed than the galaxies closer to us, which we observe at a much later time in
the history of the Universe. From these observations of distant galaxies, some idea is
now emerging about the formation of galaxies, and their subsequent evolution to their
present state. We shall consider some of these matters in some detail in the following
sections.

Objectives
After studying this unit, you should be able to:
• explain Hubble’s classification of galaxies;
• describe the properties of elliptical galaxies;
• state de Vaucouleurs law;
• describe the properties of lenticular and spiral galaxies; and
• distinguish between normal and active galaxies.

13.2 GALAXY MORPHOLOGY

A galaxy is a system of a very large number of stars, which are bound together by
their mutual gravitational attraction. A typical galaxy, like our own Milky Way,
contains ~ 1011 stars, spread over a region of size ~ 30 kpc, and has luminosity
~ 1011 LΘ, where LΘ is the luminosity of the Sun (Fig. 13.1).

6 Fig.13.1: The Milky Way as seen in the sky

 Galaxies

Fig. 13.2: An artist’s sketch of the Milky Way Galaxy

In spite of their very great energy output, such galaxies appear to be very faint when
observed from the Earth, because of their vast distances from us. A few galaxies are
visible to the naked eye as faint patches of light on dark nights, and many more are
visible when a small telescope is used.

Faint, diffuse objects observed in the night sky are called nebulae. The nature of these
nebulae was the subject of intense debate in the 1920s. While some of the nebulae are
clearly objects in our own Galaxy, like remains of supernovae, it was not clear
whether some of the objects were inside our Galaxy, or at great distances outside it.

The matter was finally settled by Edwin Hubble who determined that the distance to
the Andromeda nebula is about 700 kpc by observing variable stars in it. The modern
value of this distance places the nebula far outside the confines of our own galaxy.
Hubble’s observation established that the nebula was an independent spiral galaxy; he
went on to study and classify many other galaxies.

×106 pc from us
Fig.13.3: The spiral galaxy NGC 4622. The galaxy is located at a distance of about 70×
7
Galaxies and the 13.2.1 Hubble’s Classification of Galaxies
Universe
The observed images of galaxies show that they come in a wide variety of brightness,
shape, size and structure. Each observed galaxy is different in detail from other
galaxies, and when a large number of galaxies are examined, it becomes obvious that
there are some basic types into which galaxies can be classified. The first detailed
classification system was introduced by Edwin Hubble in 1936. This pioneering work
has been followed by other more sophisticated classification systems which take into
account the observed properties of galaxies in greater detail, but Hubble’s scheme is
the most widely used even at the present time, because of its simplicity and the insight
that it provides from observational details which can be easily obtained even with a
modest sized telescope.

Hubble’s scheme can be illustrated by his tuning fork diagram shown in Fig. 13.4.
At the left of the diagram, along the base of the tuning fork, are the elliptical galaxies,
which have simple elliptical shapes and appear to be smooth and without any
additional structures. Starting with the almost spherical galaxies of type E0, as one
moves towards the right, the images of galaxies become increasingly elliptical. The
sequence of elliptical galaxies terminates at the point where the two arms of the tuning
fork begin. Along the upper arm are the so called normal galaxies, which come in two
types: lenticular or S0 galaxies and spiral galaxies. As we move to the right along the
upper arm, from spiral type Sa to Sb to Sc, the bulges in the spiral galaxies become
less prominent, and the spiral arms appear to be more open. We will describe the
various galaxy types in greater detail in the following sections.

Normal spiral galaxies

Elliptical galaxies
Sa Sb Sc

E0 E4 E7 S0
Lenticular
galaxy

SBa SBb SBc

Barred spiral galaxies

Fig.13.4: Hubble’s tuning fork diagram

The lower arm of the fork again has lenticular and spiral galaxies, but with a linear
central feature called a bar. These barred galaxies constitute about half of all
lenticular and spiral galaxies. The distinction between normal and barred galaxies is
not absolute, in the sense that most galaxies have some faint bar like features, but a
galaxy is called barred only when the bar is very prominent. Every galaxy type along
the upper arm of the fork has a barred counterpart along the lower arm.

The classification of galaxies as suggested by Hubble is based on very luminous

galaxies, with absolute magnitude MB ≤ − 20, which he called giant galaxies.
However, when galaxies in our neighbourhood are observed, it is found that the most
numerous galaxies are significantly less luminous and more compact than the giants.
These dwarf galaxies are designated as dE.
8
Many galaxies have highly irregular shapes, and prominent features like jets, tails and Galaxies
rings. It is believed that these features are often produced because of interactions
between galaxies, which can lead to large scale disturbances in the distribution of stars
and gas in the galaxies. Many examples of galaxies in on-going close interaction can
be seen. Large galaxies can also swallow significantly smaller companions, and this
process of cannibalisation can lead to significant changes in the structure of the large
galaxy. It is difficult to classify galaxies with highly irregular structures because of
their complexity.

SAQ 1 Spend
5 min.
A galaxy of absolute magnitude M = − 20 is at a distance of 700 kpc. Would it be
visible to the naked eye?

13.3 ELLIPTICAL GALAXIES

When elliptical galaxies were first photographed, they were observed to have elliptical
shapes and a smooth distribution of light (see Fig. 13.5). They were lacking in features
which are very prominent in spiral galaxies, like spiral arms and dark patches and
lanes of dust. Later observations with highly sensitive detectors have shown that
elliptical galaxies often do have faint features produced by dust and other factors.
These faint features turn out to be important indicators of the origin and evolution of
elliptical galaxies to their present form.

Fig.13.5: The elliptical galaxy M87

Elliptical galaxies have an enormous range of optical luminosities. The so called giant
ellipticals have luminosities L ≥ L*, where L* ≈ 2 × 1010LΘ is a characteristic galaxy
luminosity. The number density of galaxies declines sharply for luminosity L > L*. A
galaxy with luminosity L ≈ L* has an absolute magnitude of M ≈ − 20. The most
luminous elliptical galaxies can have L ≥ 100L*.

Elliptical galaxies with L ≤ 3 × 109 LΘ, i.e, M ≥ − 18, are called dwarf ellipticals. 9
Galaxies and the 13.3.1 The Intrinsic Shapes of Ellipticals
Universe
When the distribution of light in an elliptical galaxy is studied, it is found that the
isophotes, or curves of equal light intensity, are elliptical in shape; this in fact gives
this type of galaxy its name. In the simplest elliptical galaxies, all the ellipses have the
same centre, their major axes are oriented in almost the same direction, and the
ellipticities are nearly constant.

It was believed at one time that elliptical galaxies acquired their shape due to rotation.
The rotation would cause the galaxy to bulge in directions normal to the axis of
rotation, because of the centrifugal force. However, observations of luminous
elliptical galaxies show that they do not rotate fast enough for the observed flattening
to be due to the rotation. It is now known that the shape comes about because of the
way the stars in the galaxy move.

13.3.2 de Vaucouleurs Law

Most galaxies are so far away that the stars in them cannot be seen individually, and
we can only observe the integrated light from stars in different regions of the galaxy.
The appearance of an external galaxy is therefore like that of a diffuse object. It is
therefore appropriate that we measure the surface brightness of the galaxy, which is
the amount of light received per unit angular area of the galaxy, say one square arc
second.

The surface brightness of light along the isophotes of an elliptical galaxy decreases as
we move away from the centre. It was discovered by G. de Vaucouleurs that the
surface brightness is a very simple function of the length of the semi-major axis of the
isophote. If r is this length, then the surface brightness I(r) is given by

1/ 4
I (r ) = I (0)10 −3.33( r / re ) (13.1)

where re is called the effective radius and I(0) is the surface brightness at r = 0. The
total light emitted by the elliptical galaxy is given by
∞
∫0 I (r )2πr dr = 3.37 × 10
−3
LE = πre2 I (0) (13.2)

Half the total light of the galaxy is emitted from inside re:

re 1
∫0 I (r )2πr dr = 2 LE (13.3)

de Vaucouleurs’ law takes on a particularly simple form if the intensity is expressed in

magnitudes.

It can then be written as:

 8.325 1/ 4 
µ ( r ) = µ ( 0) +  r  (13.4)
 r 1/4 
 e 

Here µ(r) is the surface brightness of the galaxy expressed in magnitudes per square
arc second, and µ(0) is the corresponding magnitude at the centre of the galaxy. It
follows from this equation that a plot of the surface brightness against r1/ 4 should be a
straight line. The surface brightness becomes fainter by 8.325 magnitudes in going
10 from r = 0 to r = re.
The light distribution in most elliptical galaxies does follow de Vaucouleurs law fairly Galaxies
closely. NGC 661 is an excellent example of this; we have shown, in Fig. 13.6, a plot
of the surface brightness in magnitude against r1/4. The plot is seen to be a straight
line, except in the central bright region where there is significant flattening of the
curve. Much of the deviation seen here is due to the effect of the Earth’s atmosphere.

− 10

−9 NGC 661
Surface brightness in magnitudes

−8

−7

−6

−5

−4

−3
1 2 3
1/4
(Distance from centre)

Fig.13.6: The surface brightness distribution of the galaxy NGC661 (centred around 5500 Å
1/4
wavelength). On the x axis is shown r , where r is the major axis distance of the elliptical
isophotes from the centre. On the y axis is shown the surface brightness in magnitude,
with the origin shifted for convenience

13.3.3 Stars and Gas

A study of the colour and spectrum of elliptical galaxies shows that they lack blue
stars. These stars are highly luminous, and are much more massive than the Sun. The
total lifetime τN of a star, during which the usable nuclear fuel in it is exhausted,
depends on its mass,

−3.5
M 
 10 
τ N ~ 10   yr (13.5)
 MΘ 

The blue stars being more massive than the Sun have a lifetime significantly less than
1010yr, with the most massive stars having a lifetime as short as ~ 107 yr. The lack of
such stars means that there has been no star forming activity in ellipticals in relatively
recent times. This points to the absence of substantial amounts of cool gas mixed with
dust from which stars can be formed.

Giant ellipticals contain ≤ 108 − 109 MΘ of cool gas. Observations from X-ray
satellites have, however, shown that ellipticals can be highly luminous at X-ray
wavelengths, which indicates the presence of significant amounts of hot gas, at
temperatures of a few times 107 K. In very bright ellipticals, the mass of the gas can
be as high as ~ 1011 MΘ, which constitutes ~ 10 − 20% of the visible mass of the
galaxy.

SAQ 2 Spend
5 min.
Explain in your own words why we expect the gas in elliptical galaxies to be hot.

11
Galaxies and the
Universe
13.4 SPIRAL AND LENTICULAR GALAXIES
Spiral galaxies are identified by a disk-like structure in which are present the spiral
arms. An image of the famous spiral galaxy M31, which is also known as the
Andromeda galaxy, is shown in Fig. 13.7.

A characteristic of the disk is that it is much extended but rather thin. When a spiral
galaxy is viewed face-on, i.e., when the normal to the disk is along the line of sight,
the disk appears to be circular, as in the case of NGC 4622 (see Fig.13.3). When there
is a non-zero angle between the normal and the line of sight, the disk then appears to
be elliptical, as in the case of the Andromeda galaxy. When the disk is viewed edge-
on, the disk appears to be rather thin, as in Fig. 13.8.

Apart from the disk, spiral galaxies contain a central bulge, which is quite obvious in
all the spiral galaxies. They also have a very large but faint halo, whose existence
becomes apparent from a detailed study of the distribution and motion of stars.

Lenticular galaxies, like the spirals, have a bulge and a disk, but the disk does not
contain spiral arms. The bulge and the disk here are of approximately equal
prominence. The bulge has properties very similar to elliptical galaxies, except that it
contains more gas and dust.

Fig.13.7: The Andromeda galaxy, which is spiral galaxy of type Sb, at a distance of 2.2 million light
years from us. Two dwarf galaxies, which are satellites of the Andromeda galaxy are seen
in the image

13.4.1 Bulges
The bulge of spiral galaxies is a dense system of stars in the inner region of a galaxy,
more or less spherical in shape. The bulge of our own galaxy, the Milky Way, can be
seen, towards the constellation Sagittarius, and a bulge is clearly visible in the spiral
galaxies M31 and NGC891 (see Figs. 13.7 and 13.8). The properties of the bulges are
rather similar to the properties of elliptical galaxies.

In addition to the systematic motion, the stars in the bulges also have significant
random motion, which supports them against gravity. The number density of stars in
12 the bulge is about 104 higher than the number density in the neighbourhood of the
Sun. Bulges have very little gas in them, except near the centre, and so there is not Galaxies
much ongoing star formation. This means that the bulge does not have the short lived,
high mass blue stars which are present in the disks, because of the continuous star
formation taking place there.

The scale size of the bulges is typically a few kiloparsec (kpc), while the radius of the
disk of a spiral galaxy like ours is about 15 kpc. The prominence of the bulge relative
to the disk decreases along the Hubble sequence (see Fig. 13.4), with the bulges being
most conspicuous in spiral galaxies of type Sa and being almost absent in the galaxies
at the end of the Hubble sequence and beyond.

Fig.13.8: The spiral galaxy NGC 891, which is seen edge on. The thin disk and the bulge are clearly
seen. Along the plane of the disk is seen a dark band, which is a layer of dust present in
the galaxy

13.4.2 Disks
The disks in the more luminous (MB ≤ − 20) spiral galaxies extend to ~ 15 kpc or
more, while their thickness is only a few hundred parsec, which makes them rather
thin. The disks are generally taken to be circular in shape (even though they may
appear to be elliptical, due to the projection effect mentioned above). The surface
brightness of the disk follows a simple exponential law. At a distance r from the
centre, measured along the mid-plane of the disk, it is given by

r
−
I d ( r ) = I d (0) e rd (13.6)

where rd is the disk scale length. This is a few kpc for the typical spiral galaxy.
If we move normal to the disk keeping r constant, the surface brightness decreases,
and at a distance z from the mid-plane, it is given by

Id (r, z) = Id (r) exp −z/h (13.7)

where Id (r) is as in Eq. (13.6). The scale length h is a few hundred parsec. A galaxy,
whether spiral, elliptical or of any other type, does not have a sharp boundary. The
surface brightness, and therefore the number density of stars which produce the light,
decreases as one moves away from the centre, and gradually reduces to zero. In such a
13
Galaxies and the circumstance, it is best to characterize disk size by scale lengths like rd and h, and the
Universe sizes of bulges and ellipticals with the effective radii.

The total light emitted by the disk is obtained by integrating Eq. (13.6) over the
surface of the disk:
∞
∫0 2πr I d (r ) dr = 2πrd I d (0)
2
Ld = (13.8)

If we assume that the distribution of light in the bulge is of the de Vaucouleurs type,
then the total light Lb emitted by it is given by Eq. (13.2). The ratio of the emission
from the disk to the emission from the bulge is then

2
Ld  I ( 0)   rd 
= 5.94 × 10 2  d 


r

 (13.9)
Lb  I b ( 0 )   e 

For a galaxy without a significant disk component, like an elliptical, this ratio is zero.
For lenticular galaxies it is ~ 1, and it increases along the Hubble sequence towards
the late type spirals.

As star formation is an ongoing process in the disk, massive blue stars can be
observed in it. Since such stars have a short lifetime, they cannot be found in
environments where there is no star formation taking place like bulges of spiral
galaxies or elliptical galaxies. The presence of massive stars makes the disk bluer on
the whole than the bulge. Stars in the disk are in differential rotation around the centre
of the galaxy. You have already read about the differential rotation of stars in our own
galaxy, the Milky Way in Block 3.

About half of the disks of spiral galaxies, as well as lenticulars, have a linear structure
which is known as a bar. We have seen in Section 13.2.1 that the lower arm of
Hubble’s tuning fork diagram contains these objects.

Spend SAQ 3
5 min.
Explain in your own words why older galaxies should be redder.

13.4.3 Galactic Halo

A halo of stars and globular clusters is found to be surrounding the bulge and disk of
our Galaxy. The halo extends to substantial distances beyond the disk, and has the
shape of a flattened spheroid (see Fig. 13.9). Such halos are believed to exist in all
galaxies to a greater or lesser extent.

Globular clusters are gravitationally bound systems of 105 − 106 stars. These clusters
are found in the disk plane and close to it, as well as far from the plane. There are
about 150 globular clusters associated within our Galaxy, and these are found to be
distributed approximately in a sphere around the centre of the galaxy. A characteristic
of the globular clusters is that they are very old stars. The clusters far from the plane
are very metal poor, the abundance of the heavy elements in them being as small as
1/300 of the solar value. These clusters are estimated to be at least 11 − 12 Gyr old,
which makes them the oldest structures in our Galaxy.

We expect that other galaxies too would have halos like our own galaxy. One way of
tracing a halo in external galaxies is through the system of globular clusters, which
can be observed to great distances. Globular clusters have been observed around many
14 galaxies, and elliptical galaxies are found to be particularly rich in these objects.
SAQ 4 Spend Galaxies
5 min.
Explain why metal poor stars are very old.

13.4.4 The Milky Way Galaxy

Our Galaxy, called the Milky Way, is a spiral galaxy of type Sbc, which is
intermediate to Hubble types Sb and Sc. The bulge of the galaxy is clearly visible to
the naked eye in the direction of the constellation Sagittarius. The disk of stars is
visible as a diffuse band going across the night sky. A sketch of the structure of the
galaxy is shown in Fig. 13.9.

Fig.13.9: A sketch of our Galaxy, showing the bulge, disk and the halo

The visible bulge is about a few kiloparsec in radius, while the extent of the radius of
the disk is ~ 15 kpc. The Sun is located about 8 kpc or 28000 lys from the galactic
centre, some distance away from the mid-plane of the disk. The luminosity of the disk,
i.e., the total amount of energy per second emitted by all the stars in the disk,
is ~ 2 × 1011LΘ, where LΘ = 4 × 1033 erg s−1 is the luminosity of the Sun. The total
mass of stars in the disk is ~ 6 × 1011 MΘ, where MΘ = 2 × 1033 g is the mass of the
Sun.

13.5 GAS AND DUST IN THE GALAXY

It was discovered through observations in the early decades of the 20th century that
inter-stellar space, i.e., the space between the stars in our Galaxy contains matter in
the form of gas and dust. The density of the gas is very low and it is difficult to detect
it. The dust usually occurs well mixed with the gas, and constitutes only about one
percent of the total material. The dust nevertheless is able to significantly obscure the
light coming to us from distant stars, and therefore can be easily detected. The gas and
dust together are called the inter-stellar medium (ISM) of the Galaxy.

Inter-stellar dust often occurs in the form of clouds, as is evident from the many dark
nebulae, like the Horsehead nebula (Fig. 13.10a), which are observed in the Galaxy.
The Eagle nebula and the details of the great clouds of dust observed by the Hubble
Space telescope are shown in Fig. 13.10b and c. The dust is mostly confined to the
disk of the Galaxy, and can be clearly seen in the Milky Way, even with the naked
eye, as great dark patches covering the galaxy here and there. It is specially so towards
the Galactic centre.
15
Galaxies and the
Universe

(a)

(c)

(b)

Fig.13.10: a) The Horsehead nebula; b) great columns of dust and gas in the Eagle nebula with c)
its details as observed by the Hubble Space Telescope

The distribution of dust in the disk is particularly evident when a spiral galaxy is
viewed close to the edge of the disk, as in the case of the galaxy NGC891, which is
shown in Fig. 13.8. Dust affects light passing through it by scattering as well as by
absorption. Both absorption and scattering by dust remove a fraction of light coming
to an observer from a star, and the effect of the two together is termed as extinction.
The fraction of light lost due to extinction depends upon the wavelength of light, the
size of the dust grains, and the quantity of dust in the path of light.

For the dust found in the ISM of our Galaxy, the extinction is approximately
proportional to the reciprocal of the wavelength. This means that blue light suffers the
extinction most and near-infrared light the least. Because dust is concentrated in the
Galactic plane, a star in the plane observed by us is significantly dimmed and
reddened.

Gas and dust have been observed in other spiral galaxies as well. It was believed for
long that elliptical galaxies are free of gas and dust. But X-ray observations have
shown that these galaxies contain hot gas. Careful observation in recent years has
shown that a surprising number of elliptical galaxies also contain small quantities of
dust. This can be distributed in the form of a disk in the central region of the galaxy
16 and sometimes larger disks are also present. Some elliptical galaxies contain a large
quantity of dust, which can be distributed over a region larger than the visible extent Galaxies
of the galaxy. A good example of a giant elliptical galaxy containing dust very
prominently is the radio galaxy Centaurus A shown in Fig. 13.11.

Fig.13.11: The elliptical galaxy Centaurus A which is a highly luminous radio source. A prominent
dust lane is seen to be running across the face of the galaxy

SAQ 5 Spend
5 min.
Explain why stars towards the centre of our galaxy appear fainter and redder.

13.6 SPIRAL ARMS

The disks of spiral galaxies contain spiral arms, which make these galaxies very
photogenic. The typical spiral galaxy has two arms, but many galaxies have three or
four arms. Spiral arms are found to contain many massive, blue stars. Since such stars
can only live for ≤ 10 Myr, the arms must be sites of continuing star formation. The
young blue stars are hot, and they emit radiation which can ionize any gas which may
be present around them. The gas produces emission lines which can be detected. Such
regions ionized by stars are known as HII regions. HII regions are abundant in spiral
arms, indicating the presence of large quantities of gas, as well as populations of
young massive stars. The arms also contain significant quantities of dust.

The disk in a galaxy rotates differentially, that is, stars which are closer to the centre
generally rotate with higher angular speeds than those which are further away. This
differential motion should lead to tighter winding of spiral arms. It can be shown that
for our Galaxy, the arms should have tightened significantly in less than 109 yr. Since
the Galaxy was formed more than 10 billion years ago, the arms should have been
much more wound up than what is observed. The fact that we do not observe such
winding up in our Galaxy and other similar galaxies is explained by the density wave
theory of spiral arms. According to this theory, the stars and gas present in the arms
are not fixed there. As stars and gas move along their orbits in the disk, at some point
they pass through the arms and slow down. This leads to crowding of stars and gas in
the arms, and therefore to star formation. The arms can therefore be looked upon as
density waves.

17
Galaxies and the
Universe

Fig.13.12: Spiral arms in galaxies

13.7 ACTIVE GALAXIES

A very small fraction of all galaxies are found to be active, in that they emit very
significant quantities of energy which is not produced by the stars. The energy is
produced in a very compact region at the centre of the galaxy, and in many active
galaxies exceeds by far the energy produced by the stars. The compact region from
which the energy is produced is known as the active galactic nucleus (AGN).

A “normal” galaxy like the Milky Way contains ≥ 1011 stars, each of which emits
1033 erg s−1. The total emission from all the stars is therefore ~ 1044 erg s−1. This
energy is produced from a region which extends over a few tens of kiloparsec. An
AGN, on the other hand, produces energy at the rate of 1044 − 1047 erg s−1, from a
region which is a fraction of parsec in size. The radiation produced by stars is thermal
in character, i.e., the spectrum of the radiation is similar to the spectrum produced by a
hot gas. The spectrum of radiation produced by an AGN is quite different, and in
simple cases has a power-law form. This means that the intensity varies as some
power of the frequency:

Iν ∝ νγ (13.10)

where γ is a constant.

The spectrum of active galaxies is very broad, extending from the radio region to the
infra-red, optical, ultraviolet, X-ray and gamma-ray regions. The shape of the
spectrum indicates that the physical processes that produce the radiation are different
from the processes which produce the spectrum of a hot gas. These processes are
called non-thermal. The spectrum from stars has a number of absorption lines in it,
which are produced by the absorption of thermal radiation by atoms of specific
chemical elements in the cooler region of the stars. While the spectra of AGN do have
absorption lines, the spectra also exhibit very prominent emission lines, which are
produced by non-thermal processes.

There are different kinds of active galaxies. Seyfert galaxies are generally spiral
galaxies with an AGN which is moderately luminous. Radio galaxies are always
18
elliptical, and are associated with highly luminous radio emission. BL-Lacs are active Galaxies
galaxies which are ellipticals. Their characteristic is that, unlike as in the other AGN,
emission lines are either completely absent from their spectra, or are weak. The most
luminous AGN are the quasars. The luminosity of the AGN here is so high that it
outshines the galaxy. The appearance of the object is therefore that of a point source,
like a star. Observations by very large telescopes are needed to discern the galaxy
associated with a quasar. You will study more about active galaxies in Unit 14.

The source of the immense quantity of energy produced by an AGN is believed to be

a black hole, resident at the centre of the galaxy. The black hole is thought to be super
6 9
massive, i.e., it has mass in the range of ~ 10 − 10 MΘ. Matter from the region
surrounding the black hole falls onto it, and gravitational energy is released in the
process. This energy is taken up by electrons, which in turn emit the energy through
different processes, producing the observed spectrum. The different kinds of AGN
observed can be explained in terms of variations of this general theme. While the
above picture is more or less accepted, it is not completely proven yet.

It has been established that AGN were far more common early in the life of the
Universe than at present. Many galaxies which are normal now must therefore have
been active in the past. They must therefore contain super-massive black holes which
are dormant. Recent observations have shown that a number of normal galaxies may
indeed host such black holes, and it is even thought possible that all galaxies have
super-massive black holes at their centres. The role of black holes in the formation of
galaxies, and whether the black hole or the galaxy came first, is not understood yet
and is an area of active investigation.

SAQ 6 Spend
5 min.
Explain the difference between thermal and non-thermal radiation.

In this unit you have learnt about galaxies. We now summarise its contents.

13.8 SUMMARY
• A galaxy is a system of stars, gas and dust, held together by the mutual
gravitational pull of these components. There are billions of galaxies in the
universe.
• Hubble classified these galaxies as elliptical, lenticular and spirals, both normal
and barred spirals.

• The surface brightness of ellipticals varies according to the de Vaucouleur’s law:

 8.325 1/ 4 
µ ( r ) = µ ( 0) +  r 
 r 1/4 
 e 

• Spiral galaxies are characterised by spiral arms and bulges. The spiral arms
contain clouds gas and dust and new stars are continuously being formed there.
Most galaxies have halos around them which contain extremely old stars.

• In the nuclei of many galaxies unusual phenomena take place involving release of
huge amounts of energy. These nuclei are called active galactic nuclei. It is
believed that this activity is caused by massive black holes sitting in the nuclei.
The Milky Way galaxy also has a 106 MΘ black hole at its centre.

19
Galaxies and the
Universe
13.9 TERMINAL QUESTIONS Spend 30 min.

1. Explain Hubble’s scheme of galaxy classification. Why has this scheme proved
enduring? What class has been assigned to the Milky Way Galaxy?

2. State de Vaucouleurs law and define the effective radius of an elliptical galaxy. Is
this law obeyed by elliptical galaxies?

3. Explain how the surface brightness in the disc of a spiral galaxy varies with the
distance from the centre of the galaxy. Show that the total light emitted by the disc
is 2πrd2 I d (0) , where rd is the scale length of the disc and Id (0) is the surface
brightness at r = 0. What is the meaning of rd?

4. Define an active galaxy. What is the source of its activity?

13.10 SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs)

1. M = m − 5 log r + 5 , where r is in parsec, here 7 × 105 pc.

− 20 = m − 5 (5.8451) + 5 ⇒ m = − 20 + 29.2255 − 5 ≈ 4.2

Since human eye can see up to magnitude 6, this object will be visible to the
naked eye.
2. See Text.
3. In older galaxies, there are fewer younger (blue) stars but a large number of red
stars as you have studied in Unit 10.
4. See Text.
5. See Text.
6. See Text.

Terminal Questions
1. See Text.
2. See Text.
3. For the first part, see Text. For the second part,

∞
∫0 2πr I d (0) e
− r / rd
Ld = dr put r/rd = x

∞
= 2π I d (0) rd2 ∫0 e − x x dx

= 2π I d (0) rd2 , since the integral = 1

For the third part,

I d (r ) = I d (0) e − r / rd

At r = rd, Id becomes 1/e of Id (0), i.e., intensity reduces to 1/e of its initial value.
4. See Text.
20
 Active Galaxies
UNIT 14 ACTIVE GALAXIES
Structure
14.1 Introduction
Objectives
14.2 ‘Activities’ of Active Galaxies
How ‘Active’ are the Active Galaxies?
Classification of the Active Galaxies
Some Emission Mechanisms Related to the Study of Active Galaxies
14.3 Behaviour of Active Galaxies
Quasars and Radio Galaxies
Seyferts
BL Lac Objects and Optically Violent Variables
14.4 The Nature of the Central Engine
Unified Model of the Various Active Galaxies
14.5 Summary
14.6 Terminal Questions
14.7 Solutions and Answers

14.1 INTRODUCTION
In Unit 13, you have become familiar with various types of galaxies. However, there
are galaxies which do not fit into the scheme we described in the last unit. They
require more detailed understanding. In this Unit, we shall study the nature of these
galaxies. These galaxies are ‘active’, namely, there are signs of activity in them. For
instance, their intensities may change significantly in a matter of hours to days. The
activities may be manifested in the entire bandwidth of the electromagnetic waves,
ranging from radio waves to γ-rays.

One important point should be clear to you by now and that is: nothing in this
Universe is truly inactive. So why should we learn about ‘active’ galaxies in a
separate Unit, when all the galaxies are ‘active’? Truly, every galaxy is evolving and
therefore changing with time. Only the degree of activity varies. Roughly speaking,
we call those galaxies to be active which show properties such as very strong intensity
variation in a relatively short period, and show ejection of matter at very high speeds,
etc. Since they are compact, the nuclei of these galaxies show more ‘activity’. So,
naturally, we are interested in studying these compact regions, which are known as
Active Galactic Nuclei or simply AGNs. Several classes of objects may be called
‘AGNs’: These are: Radio Galaxies, Quasars, BL Lac Objects, Optically Violent
Variables, Seyfert Galaxies, Star-Burst Galaxies, etc. In this unit, we will try to
understand how some of these objects behave and why they behave that way.

Objectives
After studying this unit, you should be able to:

• describe the nature of activities near the active galactic nuclei;

• explain how these objects are classified;
• explain the importance of synchrotron radiation for the study of AGN; and
• describe the structure of matter very close to the active galactic nuclei.

21
Galaxies and the
Universe 14.2 ‘ACTIVITIES’ OF ACTIVE GALAXIES
You have learnt in the last Unit that there is something unusual going on in the nuclei
of many galaxies. The ‘unusual’ may be the tremendous amount of energy released, or
may be the emission of extremely high energy particles in the form of narrow jets. It is
obvious that all galaxies are not equally active. Therefore, there is a need to discuss
further how active an active galaxy really is.

14.2.1 How ‘Active’ are the Active Galaxies?

Active galaxies exhibit evidences of activity in a variety of ways. Their luminosities
may be around 1044 erg s−1 (compare this with the luminosity of the Sun, which is
33 −1 48−49 −1
4 × 10 erg s ) and can go up to 10 erg s . This tremendous energy release is
possible only if these objects swallow a few stars per year and convert them into
energy. Sometimes the luminosity may vary by tens of percents in a matter of hours!
This gives you an idea of the compactness of the size of the nuclei of these galaxies.
Recall that light travels at a speed of 3 × 1010cm/s and at this rate in, say, three hours
4 14
time (~ 10 s), it will travel 3 × 10 cm. This is roughly the size of the region from
where most of the energy is released. You already know from Unit 11 that a black
hole of mass M has a size of 3 × 105 (M/MΘ) cm. Thus, if the activity of the Active
Galaxies is governed by a black hole at the centre, it must harbour a black hole of
mass of around 109 MΘ or so.

Another signature of the activity is that these galaxies emit non-thermal emissions in
all the wavelengths, i.e., radio to γ-rays. Compared to a normal galaxy, where the
nuclear brightness never masks the spiral arms or the rest of the galaxy, in active
galaxies, the nucleus is so bright that the rest of the galactic structure is not seen in
most cases. Active galaxies also produce (generally) jet-like outflows of very high
energy particles on both sides, normal to the galactic plane. These jets are highly
collimated and are continuously ejected. Normal galaxies do not produce these jets.
Apart from these, sometimes the emission lines are broad and the state of polarization
of the emitted light changes with time in a matter of minutes!

Since a tremendous amount of energy is released continuously (rather than a one-time

event as in a super-novae), it is natural to assume that the main reason of this is the
gravitational energy released by the accretion of matter. What is the source of this
matter? What is the nature of the accretion disks? How would the radiation be emitted
and at what frequency? Why do the jets form? How are the jets collimated? All these
important questions still puzzle astrophysicists today and only some of these questions
have been answered satisfactorily.

14.2.2 Classification of the Active Galaxies

Active galaxies are classified according to some, often very vague, criteria. Broadly
speaking they may be classified in the following way:

a) Radio Galaxies: In these galaxies, the emission in radio waves is very strong,
larger than 1040 ergs/s. They could be further classified into extended radio
galaxies and compact radio galaxies, depending on the size of the emission region.

b) Quasars: These objects were called Quasi-Stellar-Objects or Quasars because

they ‘looked’ like stars. Basically, in these objects, the nucleus emits most of the
radiation. The behaviour in radio is often similar to that of the radio galaxies. In
some, radio emission is not prominent and the quasars are called radio quiet
quasars.

22
c) Seyfert Galaxies: These are usually spiral galaxies with unusual nuclear Active Galaxies
brightness. Some, say, 5 to 10% of the Seyfert galaxies may be ellipticals as well.
They have very broad emission lines in their spectra which distinguish them from
other classes.

d) BL Lacertae Objects: These are commonly known as BL Lac objects. They are
identified by very rapid variability in radio, infrared and optical emissions. The
light is strongly polarized and the polarization varies rapidly. They have no
prominent emission lines and the spectrum is dominated by a continuum.

e) Optically Violent Variables: These are similar to the BL Lac Objects, but they
have weak, broad emission lines.

f) Star-Burst Galaxies: In these galaxies the star formation rate is much higher
compared to the rates in normal galaxies. The radiation is emitted in the infra-red
region. Star formation may be triggered by merger or collision of galaxies.

(a) (b)

(c) (d)

Fig.14.1: a) Radio galaxy 3C433 (image courtesy of NRAO/AUI sourced from www.nrao.edu/.../3c433_
3.5cm_opt_6in_med.jpg; b) Quasar 3C273; c) central region of the Seyfert galaxy NGC 1068; d) Starburst galaxy
M82 23
Galaxies and the Let us now learn about some of these objects which show major activities.
Universe

14.2.3 Some Emission Mechanisms Related to the Study of Active

Galaxies

One of the most important processes that you need to know while studying active
Galaxies is the synchrotron radiation. You may also know this radiation as cyclotron
radiation, a terminology used for radiation emitted by low energy electrons in the
presence of a magnetic field. To emit this radiation two ingredients are required,
namely, large number of high energy electrons and a strong magnetic field. Since the
magnetic field is present everywhere in AGN, synchrotron radiation is a very common
phenomenon.

A relativistic electron of rest mass m0, charge e, velocity v (and Lorentz factor
2 2 −1/2
γ = (1 − v /c ) ) in a magnetic field B feels the force,

d e
( γm0 v ) = ( v × B) (14.1)
dt c

The electron moves in a helical path in the magnetic field (Fig. 14.2) with an angular
frequency

eB  B   10 
ωg = = 1.8 × 1011     rad s −1 (14.2)
γm0 c  10 5 G   γ 

Magnetic field

Electrons in a
helical path

Fig.14.2: Motion of an electron in a magnetic field and the emission of synchrotron radiation
(image credit Gemini Observatory)

The emission is in the shape of a broad spectrum with a peak at critical frequency
ωc ~ γ2 ωg.

Because of this continuous radiation, the electron loses energy and starts cooling with
time. The cooling time scale is clearly the time in which the whole energy is radiated
away:
2
Tcool ~ γm0c /P s. (14.3)

where P is the energy emitted by an electron in one second.

24
To obtain the net emission from the entire gas you need to consider the velocity Active Galaxies
distribution of the electrons, which may or may not be thermal. In AGN the
distribution is non-thermal. For the non-thermal electrons, the emission spectrum
looks like a power-law, i.e., the intensity of radiation is expressed as:

F(ν) = F0 ν−α (14.4)

Flat
where α is a constant known as the spectral index. When α ≤ 0.4, we call the
spectrum a flat spectrum, and when α is larger, we call the spectrum a steep
spectrum (Fig. 14.3).
Steep

F (ν)
Synchrotron radiation is largely linearly polarized. Crudely speaking, an observer
viewing the gyrating electron sideways will see an oscillating electron which emits a
linearly polarized radiation. From above and below, oppositely directed circular
polarization will be seen. These circular polarizations will cancel each other when
overall integration is made but the linear polarization will survive and it can be
detected easily. Another form of radiation which is very important is the emission line
component. Emission lines come from atomic transitions from one bound state to ν
another (Fig. 14.4). Fig.14.3: Schematic diagram
of flat and steep
etc… spectra

Level 4

Level 3

Level 2
Energy

Level 1

Fig.14.4: Schematic diagram of an atomic transition from an excited state to a lower state and
photon emission

Similarly, when a photon of the ‘right’ frequency ν = (Ef − Ei)/h is incident on an

atom, the photon may be absorbed. This absorption line will be very thin if the
emitting atom is at rest, but if the atoms are moving around, the line may become
broad due to Doppler effect: The line width dν will then correspond to the velocity v
of the emitting matter: dν/ν ~ v/c, where ν is the frequency of the radiation emitted in
the rest frame and c is the velocity of light. From the broadening of the lines from
Seyfert galaxies, sometimes the velocity inferred is anywhere close to
7 × 106 − 2 × 107 m/s.

In the next few sections, we shall discuss some of the important classes of the Active
Galaxies in more detail. After that, you will learn about the detailed nature of the
activities close to the centre, i.e., the nature of the ‘central engines’ which drive these
powerful sources in space. But before that, you may like to try an SAQ!

SAQ 1 Spend
5 min.
Explain why synchrotron radiation is largely linearly polarized.
25
Galaxies and the
Universe 14.3 BEHAVIOUR OF ACTIVE GALAXIES
In this section, you will learn about the properties and behaviour of quasars, radio
galaxies, Seyferts, BL Lac objects and optically violent variables.

14.3.1 Quasars and Radio Galaxies

A quasar (contraction of QUASi-stellAR radio sources) is an astronomical source of

electromagnetic energy, including light that dwarfs the energy output of the brightest
stars. A quasar may readily release energy in levels equal to the output of dozens of
average galaxies combined. The best explanation for quasars is that they are powered
by supermassive black holes.Quasars can be observed in many parts of the
electromagnetic spectrum including radio, infrared, optical, ultraviolet, X-ray and
even gamma rays while most quasars are found to emit in the infrared.

The spectra of quasars exhibit characteristic strong emission lines rising above a broad
continuum, which are red-shifted due to the high recession velocity of the quasar
relative to us. For example, the Lyman α line formed by transitions between the n = 1
and n = 2 energy levels in neutral hydrogen normally produces spectral lines with a
wavelength of 121.6 nm or 1216 Å which is in the ultraviolet part of the spectrum. In
quasar spectra, these are seen at wavelengths in the visible part of the spectrum (see
Fig. 14.5).

Fig.14.5: Spectrum of a quasar showing relative intensity on the y axis as a function of wavelength
(from www.eso.org)

The nuclei of quasars emit UV radiation so strongly that the rest of the galaxy is
masked. The radiation that is observed from radio wave to X-ray is totally non-
thermal. Synchrotron radiation is the most likely source of this emission. Some
26 quasars could be intense in radio emission while other quasars may have no
significant radio emission. The power emitted in radio waves is usually smaller than Active Galaxies
that emitted in other wavelengths. In the infra-red region the emitted power is
significant.

Fig. 14.6: Spectrum of a quasar in the X-ray region showing relative intensity on the y axis
(Source: www.estec.esa.nl)

Quasars are found to vary rapidly in a matter of few minutes to weeks, months and
even years especially in the optical and radio regions. Since nothing can travel faster
than light, the shortest time-scale of appreciable variation gives an idea of the upper
limit of the size of the object as well. For instance, if a quasar intensity doubles in a
day (tvar ~ 105s), the size of the region emitting most of the radiation must be smaller
than the light-crossing time, i.e., Lsize ~ ctvar ≈ 3 × 1015cm. So the central region must
be even smaller. This indirectly gives an estimate of the mass of the black hole:

Lsize
M= 1010 M Θ ,
3 × 1015

where Lsize is in units of cm.

Emission lines from the quasars are very broad and show characteristic red shifts
because these objects are located at very large distances. The red shift indirectly tells
you how far the object is situated.

The width of the emission lines are determined by large scale motions of the emitting
matter. The emitted frequency ν0 is related to the observed frequency ν by,

ν = ν0 (1 + v/c).

If the emitting atoms follow the Maxwell’s velocity distribution at temperature T, then
we can obtain a simple expression for the intensity distribution I(ν) dν.

27
Galaxies and the We can write
Universe

 M c 2 (ν − ν )2 
p 0 
I d (ν) = I 0 exp − (14.5)
 2 k BT 2 
 ν0 

Here I0 is the maximum intensity which occurs at ν = ν0, and Mp is the mass of the
atom of the gas. As mentioned earlier, the width directly gives the information about
the velocity, and therefore the temperature of the emitter. If we define half-width of
the line as the width at half of maximum intensity, then we can show that

2k B T
half width = 2(ν − ν 0 ) = 2ν 0 ln 2 (14.6)
M pc2

Spend SAQ 2
10 min. o
What is the half width dν of a line of wavelength λ = 6300 A when the temperature of
the gas is 105K? Assume O atoms to be the emitters.

The quasars which are located far away, i.e., quasars with large red shifts also exhibit
a strong degree of absorption features. The red shifts of the absorption lines are almost
always found to be less than or equal to the red shift of the emission line itself. This
indicates that the absorbing gas is located in between the quasar, or is a part of the
quasar itself.

Radio Galaxies

Those Active Galaxies which emit very large amount of energy in radio waves, in
excess of, say, 1040 erg/s, are known as radio galaxies. Some of these could be
compact radio galaxies, i.e., the emission is concentrated near the nuclear region. The
others could be extended radio galaxies, i.e., the emission is seen in the form of two
very large blobs which are well separated and usually co-aligned with the central,
optically bright nucleus. Fig. 14.7 shows the nature of the radio emitting regions in a
radio galaxy.

Fig.14.7: Radio emitting regions in a radio galaxy: jets from the radio galaxy 3C296. The infrared
28 regions are shown in red
 43 44
A typical radio lobe in a bright radio galaxy has a luminosity of about 10 or 10 Active Galaxies
ergs/s. But as we discussed in the last section, this is due to synchrotron radiation and
it cools rapidly. In order that the blobs are produced and propagated for millions of
years, they must be supplied with energy constantly. Indeed, higher resolution radio
images point to an important finding: These radio emitting blobs are supplied with
energy by very narrow jets of rapidly moving matter.

If one observes these jets at higher and higher resolution, the matter will be found to
propagate in the same direction even close to the centre. This indicates that the prime
cause of the jet production lies in the activity at the centre. We shall discuss this in
Section 14.5.

14.3.2 Seyferts
Seyfert galaxies are named after Carl K. Seyfert who, in 1943, described their central
regions as having peculiar spectra with notable emission lines. Seyfert galaxies
appear to be normal spiral galaxies, but their cores fluctuate in brightness. It is
believed that these fluctuations are caused by powerful eruptions in the core of the
galaxy. In some Seyferts the nucleus outshines all the stars in the host galaxy.

These active galaxies are distinguished from the others because of the presence of
very ‘broad’ and ‘narrow’ (but broader compared to those from the normal galaxies)
emission lines. These are known as the Type-I and Type-II Seyferts, respectively.
Often in a given galaxy, features of both the types are seen. Fig. 14.8 shows the
emission sectrum of a Type-I Seyfert galaxy.
Relative flux

Wavelength

Fig.14.8: Emission spectrum of the Type I Seyfert NGC 5548

In the presence of rapidly moving emitters, a line emitted at a fixed frequency ν in the
rest frame of the atom, will be observed at a different frequency ν′ and as we have
discussed before, the shift in frequency is related to the velocity of the atom by

dν ν′ − ν
= ~ v / c. (14.7)
ν ν

Since the sign of v could be both negative and positive, the shift occurs on both the
sides of the emitted frequency.

Close to the centre of the Seyferts, there are emitting clouds randomly rotating around
the nucleus and are responsible for the broad emission lines. These lines show 29
Galaxies and the variability in a matter of weeks to months. From this, one can estimate the distance
Universe
from the centre where the emitting clouds must be located. It turns out that they could
be as close as 1016cm. The corresponding velocities could be as high as
6 7 −1
7×10 − 2×10 m s .

Spend SAQ 3
5 min.
Estimate the distance that an emission cloud must have from the centre of a Seyfert
galaxy (having a central compact object of mass 107MΘ) in order to produce a velocity
of 106 m s−1. Use Kepler’s law.

Lines emitted from the clouds which are located very far away are narrow and they
correspond to velocities of the order of only 200 − 500 km/s. Assuming that the
rotation velocities of these clouds are Keplerian, a factor of twenty lower in velocity
corresponds to a factor of four hundred larger in distance (for a given central object).
Thus, narrow line regions (NLRs) are located very far away.

14.3.3 BL Lac Objects and Optically Violent Variables

These types of active galaxies are characterized by a high degree of variability in

radio, infra-red and optical wavelengths. For instance there are cases in which the
intensity varies by a factor of 10 or more in a matter of weeks. From one night to
another, variation up to 10-20% is very common. These objects have no emission
lines and the polarization is large and rapidly varying. The emitted radiation is most
concentrated at the centre and there is a weak sign, if any, of jets or radio lobes. In
optically violent variables, there is a weak broad emission line component. Together
these are known as Blazars. Fig. 14.9 shows a typical spectrum from a BL Lac object.

Fig.14.9: Spectrum of the BL Lac object Markarian 421 (Source: www.astr.ua.edu/ keel/agn)

30
 Active Galaxies
14.4 THE NATURE OF THE CENTRAL ENGINE
So far you have learnt about various classes of active galaxies. But what is really
happening close to the centres of these galaxies? We have already mentioned that
from the nature of rapid variability, one can infer that the masses of the central objects
must be very high and must be concentrated in a compact region. Also, you have
47 −1
learnt that energy is released at a tremendous rate. For instance, a rate of 10 erg s
26 −1
corresponds to a destruction of matter of about 10 g s (using the famous formula
2 7
of Einstein E = mc ). So in one year, i.e., in 3.15 × 10 s, these galactic centres
33
completely destroy about 3 × 10 g of matter per year (which is equivalent to 1.5
times the mass of the Sun!).

One of the simplest solutions to this energy budget is that the matter can be accreted
on to the central compact object and the gravitational energy could be released. This
energy is eventually converted into kinetic energy and then to thermal energy and
radiation:

GMm p 1
~ m p v 2 ~ k B T ~ hν (14.8)
r 2

where, M is the mass of the central compact object, mp is the mass of the particle, T is
the temperature of the gas, kB and h are the Boltzmann constant and Planck’s constant,
respectively, v is the velocity of matter and ν is the emission frequency. As matter
falls closer and closer to the centre, the velocity is increased. The gas becomes hotter
and radiation is emitted at a much higher frequency.

Now, what could the nature of the compact object be? In Unit 11, you have seen that
only stable compact objects which are also massive are necessarily black holes. So,
for the sake of argument, assume that the main cause of the activity is due to matter
falling on to a super massive black hole. Does this picture explain everything?
Roughly speaking, the answer is ‘yes’!

14.4.1 Unified Model of the Various Active Galaxies

So far, you have learnt how matter is accreted on to a black hole and how the radiation
is emitted. What we have described so far is what goes on, say, within the inner 1000
rg where rg = 2GMBH /c2 is the Schwarzschild radius of a black hole. When the black
hole is super-massive, as in the case of our present discussion, the region farther out
emits optical and infra-red radiation since there is supposed to be a large dusty torus.
There are smaller clouds of gas, some of which are closer than this torus and move
very rapidly. They are responsible for broad line emissions (BLR) since high
velocity causes larger broadening of lines.

Similarly some of the clouds move around very farther away and move slower. These
clouds produce narrower emission lines. Thus, depending on the angle at which the
observation is made, the dusty torus may block the BLR only selectively. The
possibility that different classes of active galaxies may be actually due to such
observational effects has been proved in the past most conclusively. This is because
after prolonged observation, objects of a given class have been found to change their
class altogether.

In Fig. 14.10 we draw the unified model of AGN that is widely adopted in the
literature, i.e., all AGN are the same. The differences in lines, jets and spectra may be
due to different viewing angles. As the plane of accretion can be randomly oriented in
the sky, different lines of sight will result in different kinds of observations. So, the
same object may be seen differently and classified differently. It all depends on our
line of sight! 31
Galaxies and the
Universe
HBLR
Seyfert 2

Narrow line clouds

Jet
Dusty
torus Accretion
Broad line disk
clouds
Cool Warm Hot Hot Warm Cool
dust dust dust dust dust dust

non-HBLR
Seyfert 2
Black hole

Seyfert 1
Fig.14.10: The unified scheme for understanding AGN

In this unit you have learnt about the active galactic nuclei. Let us now summarise the
contents of this unit.

14.5 SUMMARY

• Active galactic nuclei (AGN) are broadly classified as Radio Galaxies, Quasars,
BL Lac Objects, Optically Violent Variables, Seyfert Galaxies, Star-Burst
Galaxies, etc.
• The spectra of AGN reveal the nature and extent of activities in them, which
serves as the basis of their classification.
• The emission spectrum of AGN obeys the power law in the radio region:
F(ν) = F0 v−α
where α is a constant known as the spectral index. When α ≤ 0.4, the spectrum is
called a flat spectrum, and when α is larger, it is called a steep spectrum.
• One of the most important processes occurring in AGN is the synchrotron
radiation.
• According to the current understanding, all the AGN may actually be the same
object seen at different angles.
32
 Active Galaxies

14.6 TERMINAL QUESTIONS Spend 30 min.

1. Electrons having Lorentz factor γ = 100 are gyrating in a magnetic field of 1010G.
At what frequency would its synchrotron emission peak?

2. Derive the expression for the half width (Eq. 14.6) caused by Doppler effect.

3. Explain the origin of broad-line and narrow-line regions in a Seyfert galaxy.

4. Describe briefly the nature of the central engine in an AGN.

14.7 SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs)

1. See text.

2k B T
2. Half width = 2ν 0 ln 2
M pc2

2 2 × 1.38 × 10 −16 × 10 5
= ln 2 (in cgs units)
6300 × 10 −8 16 × 1.6 × 10 − 24

2 × 10 6 × 10 6 2 × 1.38
= × 0.693 × 10
63 16 × 1.6

200 2 × 1.38
= 1010 × 0.693 × 10
63 16 × 1.6

= 2.7 × 1010 Hz.

mv 2 GMm GM GM
3. = ⇒ v2 = ⇒R=
R R2 R v2

6.7 × 10 −8 × 10 7 × 2 × 10 33
= (in cgs units)
1016

= 1.34 × 1017 cm = 0.043 pc

Terminal Questions

B 10
1. ω g = 1.8 × 1011 rad s −1
5 γ
10

1010 10
ω c = γ 2 ω g = 10 4 ω g = 10 4 × 1.8 × 1011
10 5 100

= 1.8 × 1019 rad s - 1

33
Galaxies and the
Universe
 M c 2 (ν − ν )2 
p 0 
2. I d (ν) = I 0 exp −
 2 k BT 2 
 ν0 

I max = I 0 at ν = ν0

At half the maximum intensity

I
I= 0,
2

Therefore,

 M c 2 (ν − ν )2  1
p 0 
exp  − =
 2 k BT 2  2
 ν 0 

 M c 2 (ν − ν )2 
p 0 
⇒ exp  =2
 2 k BT 2 
 ν 0 

M p c 2 (ν − ν 0 ) 2
⇒ = ln 2
2k B T ν 02

2k B T
Thus, Half width = 2(ν − ν 0 ) = 2ν 0 ln 2
M pc2

3. See Text.

4. See Text.

34
 Large Scale Structure and
UNIT 15 LARGE SCALE STRUCTURE AND the Expanding Universe

THE EXPANDING UNIVERSE

Structure
15.1 Introduction
Objectives
15.2 Cosmic Distance Ladder
An Example from Terrestrial Physics
Distance Measurement using Cepheid Variables
15.3 Hubble’s Law
Distance-Velocity Relation
15.4 Clusters of Galaxies
The Virial Theorem and Dark Matter
15.5 Friedmann Equation and its Solutions
15.6 Early Universe and Nucleosynthesis
Cosmic Background Radiation
Evolving vs. Steady State Universe
15.7 Summary
15.8 Terminal Questions
15.9 Solutions and Answers

15.1 INTRODUCTION
In previous units you have learnt about stars and galaxies and their properties. You
have learnt that there are billions of galaxies in the universe. It is now time to estimate
the size of the universe and discuss the current ideas about its major components and
its origin. The science which deals with the origin of the universe is called cosmology.

In this unit, we discuss some aspects of cosmology. First we learn how distances of
distant objects can be estimated so that we can get an idea of the size of the universe.
Estimation of distances is a very tricky problem because there is no way of verifying
these distances directly. So, we look for internal consistency in various methods,
which means that the distance estimates given by them agree with one another.

In this unit we discuss the distribution of matter on very large scales also called the
large scale structure of the universe. We also look at the kind of matter that may be
forming the bulk of the universe. This matter is not visible and is called the ‘dark
matter’. It shows itself up only through its gravitational effect.

Objectives

After studying this unit, you should be able to:

• explain how astronomical distances are measured;
• state Hubble’s law and explain how Hubble’s constant indicates the age of the
universe;
• explain the need to postulate the existence of dark matter in the universe;
• derive Friedmann equation and solve it in simple cases; and
• explain why a hot and dense phase in the early universe is needed to explain the
existence of cosmic background radiation and to synthesise light elements.

35
Galaxies and the
Universe 15.2 COSMIC DISTANCE LADDER
An important step in this direction is to estimate the physical size of distant objects
and their distances from us. You know that to specify the position of an object in
three-dimensional space, we need 3 coordinates. Since we are observing from the
Earth (or its vicinity from satellites), it is best to use spherical polar coordinate system
with us as the origin. Two of the coordinates, namely, θ and φ (which are related to
the declination and right ascension, respectively) are easily fixed by pointing a
telescope in the direction of the object. Fixing the distance to an astronomical object is
relatively trickier.

In this section, we give an introduction to the basic principles that are used to measure
distances. The basic principle involved is to use the properties of the nearby objects
and deduce distances of similar objects farther off using these properties. Then we use
the latter objects to deduce the distances to objects still farther, and so on. This series
of steps which takes us from one step (in terms of distances) to the next step (in terms
of distances even farther) is termed the Cosmic Distance Ladder.

All this is best understood through an example of estimating distances on the Earth.

15.2.1 An Example from Terrestrial Physics

Consider the following situation. You are in a house and there are small plants around
the lawn in your house. Outside the house also there are similar plants and in addition
there are also some trees. We do not know the height of these trees. Very far away
there is the sea and on the beach also there are a number of such trees. The problem is
to estimate the distance of the sea from your house. To begin with we do not know
about the height of the trees.

The obvious thing would be to walk down to the sea (keeping in mind that all the
steps should approximately be of the same size) and count the number of steps. The
length of the step multiplied by the number of steps gives us the distance to the beach.
But you cannot use this method if you are not allowed to, or are unable to come out of
the house and go to the beach. The situation is more like this in the context of distance
measurement in astronomy. We can make direct measurement only on the Earth’s
surface. Since it will take about 1,00,000 years, even to reach the other end of the
galaxy by moving at the speed of light, we need to devise a better strategy.

Let us come back to our problem of measuring the distance to the sea beach. We will
put an extra condition that we are not allowed to come out of the gates of the house.
With this constraint we can proceed in the following way.

We first measure the heights h of the plants inside the garden of the house. Next we
measure the angle θ subtended by similar plants outside which are in the vicinity of
h
the trees. The distance of these plants from us equals d = . Since the trees are in the
vicinity of these plants we know that their distance from usθ is also d. Further, we can
measure the angle θ1 subtended by the heights of these trees. The height h1 of these
trees is now given by h1 = d θ1. In this way we have achieved the first step in the
ladder of finding distances. We next use this information to estimate the distance to
the trees near the seaside. Since the trees near the beach are similar to the ones nearby,
we assume that their height is also h1. By measuring the angle subtended by these
trees we can compute the distance of these trees from us. Because of the physical
association of these trees with the sea side we know the distance of the sea from your
house. This is the second step of the ladder. Like this we can add more steps and go
on to estimate the distances of far off objects.
36
15.2.2 Distance Measurement using Cepheid Variables Large Scale Structure and
the Expanding Universe

We will now give an example of using this principle to measure distances in

astronomy. In Unit 1 you have learnt the technique of measuring distances by parallax
method. In this section we will discuss how this method can be used to calibrate
another technique for distance measurements using Cepheid variable stars, which in
turn are used to measure even further distances. The brightness of Cepheid variable
stars is a periodic function of time (Fig. 15.1b).

3.0

Visual Magnitude
3.5

4.0

4.5

5.0
0 5 10
Days
(a) (b)

Fig.15.1: a) Cepheid variable stars in NGC 300; b) the brightness of Cepheid stars as a function of time

We have a large number of such stars in our neighbourhood. Their distances can be
measured by the parallax method. From these distances and from their observed
apparent magnitudes, their absolute magnitudes and luminosities can be calculated. It
turns out that their absolute magnitudes are directly proportional to their periods
(Fig. 15.2). This is called the period-luminosity relation for the Cepheids.
−8

−6
Absolute magnitude

−4

−2

0
0 0.5 1.0 1.5 2.0 2.5

Log period (days)

Fig.15.2: Period-luminosity relation for Cepheids
37
Galaxies and the This relation is firmly established for the local sample. Now we assume that the
Universe
distant sample of these stars also obeys this relation. So, the observed period of a
member of the distant sample is used to find its absolute magnitude. The apparent
magnitude can be observed directly. Using the relation between the absolute
magnitude and the apparent magnitude:

M = m + 5 − 5 log r, (15.1)

we can find the distance of this star. In this way, the Cepheid variable stars have been
used to find distances of nearby galaxies. Used in this manner, Cepheids are called
standard candles.

We can now use Cepheids to define some other objects, such as supernovae, to act as
standard candles to estimate even larger distances.

What is the result of these investigations? We find that distant galaxies are rushing
away from us with velocities which are proportional to their distances. This is called
Hubble’s law.

Spend SAQ 1
5 min.
Explain the concept of a distance ladder.

15.3 HUBBLE’S LAW

Hubble’s law is probably the single most important step in our attempt to understand
the Universe. This law was discovered by Edwin Hubble and it relates the distances of
galaxies with the velocities with which they are receding away from us (Fig. 15.3).

300,000

250,000

200,000
Velocity (km s− )
−1

150,000

100,000

50,000

0 500 1000 1500 2000 2500 3000 3500 4000

Distance (Mpc)
Fig.15.3: Hubble’s law

The velocity of any object can be split into a component that is along the line-of-sight
and another component that is transverse (perpendicular to) the line-of-sight. The line-
of-sight component of the velocity can be determined very accurately by Doppler shift
of the light that we received from the object. Hence this gives the velocity with which
38 the object is coming towards us or receding away from us. Let us look at this process.
15.3.1 Distance-Velocity Relation Large Scale Structure and
the Expanding Universe

Using the methods similar to those mentioned above, Hubble estimated the distances
and velocities of a set of galaxies and plotted them. He found that the galaxies in
general seem to be receding away from us. This is popularly known as the expanding
universe. Further, the velocities with which they are receding away, are directly
proportional to their distances from us (Fig. 15.3). This prompted him to propose a
law, now known by his name.

Hubble’s law

v = Hr, (15.2)

where v is the line-of-sight velocity of an object, r its distance from us and the
proportionality constant H is called the Hubble constant.

The importance of this relation is that, once we know the velocity of a galaxy (by red
shift measurement), we can calculate the distance at which it is located. It is important
to point out here that Hubble’s law holds even if we were on some other galaxy. Our
location in the universe does not have any special importance.

Notice that in the above relation H is the slope of the curve shown in Fig. 15.3. Notice
also that 1/H has the dimensions of time. In a very simple picture, we can imagine that
all the galaxies which are today moving away from one another were at some time in
the past together at one point. Some event occurred at that time which triggered the
expansion of the universe. This event is usually called the Big Bang. The quantity 1/H
measures the time since that even, or the age of the universe.

SAQ 2 Spend
5 min.
In astronomy, the velocity is measured in km s−1. The distance of galaxies is measured
in Mpc or million parsec. Find the dimensions of H.

Unfortunately the measured value of H has lots of errors. But we know today that it is
roughly 70 km s−1/Mpc−1. Estimate the age of the universe.

15.4 CLUSTERS OF GALAXIES

Galaxies mostly exist as members of large groups called galaxy clusters
(see Fig. 15.4). A cluster of galaxies contains about a thousand galaxies. A galaxy
cluster consists of a variety of galaxies. It is an observed fact that the gross features of
these clusters are very similar. This fact alone has a far reaching implication. We
expect that roughly the same kind of physical processes are responsible for their
evolution. Further, it is natural to assume that they were created at different times and
began to evolve. This is because, in order to initiate the process of creation and
subsequent evolution of these clusters one needs a certain combination of
astrophysical conditions.

These conditions need not be the same at all points at a given time. Let us illustrate
this situation in the following manner:

Consider a set of systems, A, B, C, … Let the evolution in each of these systems be

governed by the same physical processes. If they started to evolve at different times,
we would expect that at any time, in particular today, they should be in different
stages of evolution. Hence, we would expect they should not show great similarity. 39
Galaxies and the
Universe

Fig. 15.4: Cluster of galaxies

Alternatively, if we find that they look similar today, we may naively think that they
must have got created at the same time, so that they have had the same time for
evolution. For the systems under consideration, namely, the clusters of galaxies, we
know that they look similar but at the same time we have reasons to believe that they
started their evolution at different times.

Our simple-minded reasoning leads to the suggestion that they may have been created
at different times but they have reached some kind of steady state today. Such systems
have a simple, but important relationship for their kinematic parameters such as their
mass and velocity. This relationship is called the virial theorem which we now
describe.

15.4.1 The Virial Theorem and Dark Matter

The virial theorem says that if a system is bounded and is in equilibrium, then its
moment of inertia does not change with time. Such a system will have the following
relation between its total kinetic and potential energies:

2T + V = 0, (15.3)

where T is the total kinetic and V is the total potential energy of the system.

Suppose we have a spherical system consisting of N particles that are interacting

gravitationally. Let the position of the ith particle be ri and its velocity vi. The total
kinetic energy is then

1
∑i=1 vi2 .
N
T= m (15.4)
2

The total potential energy is obtained by summing over the potential energy of all the
pairs. We have

Gmi m j
V= ∑i> j ri − r j (15.5)

40
For a spherically symmetric distribution, V will come out to be proportional to Large Scale Structure and
−1 the Expanding Universe
M (R) R , where R is the radius of the system and M the total mass enclosed in a
sphere of radius R. The total kinetic energy can be estimated in the following manner.
If the velocities are in random directions, some of the particles of the system will
contribute a blue shift and some red shift. This will result in broadening of the spectral
lines. Hence, from the width of the spectral lines we can estimate the root mean
square (rms) velocity. This gives the total kinetic energy. If the system is in
equilibrium, we should have

T = − V/2 (15.6)
or

2
v rms ∝ R −1 . (15.7)

When one tries to estimate the mass of the system in the above manner, one finds that
there is much more mass in the system as compared to that suggested by the luminous
mass alone. Hence, it is postulated that there should be a significant fraction of mass
in the form of dark matter. What form this dark matter takes, we do not yet know.
One possibility is that it consists of cold, burnt out stars which emit very little
radiation. Another possibility is that it is in the form of particles which interact very
weakly with normal matter. It is a very active area of research at present.

15.5 FRIEDMANN EQUATION AND ITS SOLUTIONS

Cosmology is the study of the overall features of the universe. In this framework, one
does not bother about local features of the universe like planets or stars. In cosmology,
we set up equations and solve them to obtain the very large scale features of the
universe.

To set up the basic equations governing the evolution of the universe as a whole,
Newtonian theory of gravity is inadequate and is, rigorously speaking, inapplicable.
The correct theory to use is Einstein’s General Theory of Relativity. This, however, is
outside the scope of this course. All is, however, not lost. It so happens that if we go
ahead and apply Newton’s laws (which strictly speaking is not the correct thing to do)
the final equations which we get are the same equations that result from the correct
theory, namely, the General Theory of Relativity.

The aim of this Section is to study the evolution of the Universe using these equations.
Hence, we will not be disturbed by the fact that the derivation of these is not rigorous.
We will happily go ahead and use these equations since we are aware that the final
equations are the correct ones.

We know that over a variety of scales the universe is not homogeneous. We have
planets, stars, galaxies and clusters of galaxies, which indicate that the universe is far
from being homogeneous. However, if we consider the universe as a whole, then at
large enough scales the universe seems to be homogeneous and isotropic. This last
point needs some explanation.

Consider a well maintained lawn. From a distance, the lawn appears uniformly green.
But as we start analysing the lawn on smaller scales, it begins to lose its homogeneity.
Consider a grass hopper sitting in the lawn. It observes the lawn at a scale which is
about the blade of grass. Clearly it will notice that the lawn is not at all homogeneous.
This means that the lawn is homogeneously green over distance scales which are
much bigger than the size of a blade of grass. In a similar way, we say that over scales
of sizes, much bigger than the size of galaxies, the universe is homogeneous.
41
Galaxies and the The equations that we now derive are the equations that govern the overall evolution
Universe
of a homogeneous universe. Consider two points A and O. With O as centre, and AO
as radius, draw a sphere. The force with which a test particle at A is gravitationally
pulled towards O can be calculated by just using the mass enclosed by the sphere. As
we know, the mass outside the sphere will not exert any net force on A.

Let the density of matter be ρ. Notice that density does not depend upon space. This is
because of our condition of homogeneity. However, there is no such restriction on
time dependence. So we will include time dependence for generality. The mass of the
sphere of radius R = OA is

4π
M= ρ(t ) R 3 (t ), (15.8)
3

& be the
The gravitational potential at A due to this sphere is − GM/R. Further, let v = R
velocity with which A is moving with respect to O. Depending on the magnitudes of
the kinetic energy and the potential energy, the particle at point A may keep moving
away from O or may turn back and fall towards O. This is the well known condition
for escape velocity.

The total energy per unit mass of a test particle at A is

GM v2
E=− + (15.9)
R 2

or

2GM
R& 2 − 2 E = (15.10)
R

If the energy E is positive, then the distance between A and O will keep increasing and
if E is negative, A will attain a maximum distance from O and then begin to fall
towards O.

This condition on E can alternatively be expressed by replacing −2E by k .2 E ; k = +1

corresponds to the case E < 0.

An alternative way of writing the last condition is then

2GM
R& 2 + k .2 E = (15.11)
R

Dividing both sides by R 2 and expressing the mass M in terms of the density ρ as

4 3
M= πR ρ
3

we get,

R& 2 k .2 E 8πG
+ = ρ (15.12)
2 2 3
R R

Defining a new variable a = R 2 E , we can now write the Friedmann equation

42 governing the evolution of the distance between two particles.
 Large Scale Structure and
the Expanding Universe
Friedmann equation

a& 2 k 8πGρ
+ = (15.13)
a2 a2 3

We get the same equation from the general theory of relativity. The parameter k then
signifies the curvature of space. In the general theory of relativity, the effect of the
gravitational field is to make the space curved. The curvature of space is denoted by k,
which can take values +1, 0 or −1, depending on the overall density of the universe.
The quantity a is called the scale factor and the nature of a as function of t indicates
the nature of the expansion of the universe.

Fig. 15.5 shows the behaviour of a as a function of t for the three values of k, i.e., +1,
0 and −1. These curves are the solutions of the Friedmann equation. We see that
when k = −1 (which according to general theory of relativity implies that the overall
density of the universe is less than a certain critical density), or k = 0 (the overall
density of the universe equals the critical density), the universe keeps expanding.
When k = +1 (the overall density of the universe is greater than the critical density),
the universe expands up to a point and then starts contracting. Present observations
indicate that the universe will keep expanding, and its expansion will not be followed
by contraction.

In the early phase of the universe, the curvature must have been small, so it is
sufficient to consider the case of k = 0. The solution of the Friedmann equations, of
course, depends on the nature of energy density. If ρ ∝ a−n, the equation can be solved
for k = 0 and the result comes out to be

a ∝ ( n / 2) 2 / n t 2 / n (15.14)

k<0
k =0
a(t)

k>0

t
Fig.15.5: The variation of a with time
43
Galaxies and the SAQ 3
Universe Spend
5 min.
−n
Verify that Eq. (15.14) is a solution of Eq. (15.13) when ρ ∝ a .

Time-temperature Relationship
We know from the thermodynamics of radiation that it has pressure and energy
density which we denote by prad and ρrad, respectively. They are related to each other
through the relation,

p rad = ρ rad c 2 3 (15.15)

where c is the speed of light. The distance between points increases as R which, in
3
turn, is proportional to a. Hence, the volume increases as a and the energy contained
3
in it as ρ a . Now, using the first law of thermodynamics, dU + pdV = 0, we have

d [ρ c 2 a 3 ] + pd [a 3 ] = 0. (15.16)

Applying it to radiation, we get

ρ rad ∝ a −4 (15.17)

Using this relation in the expression for ρ derived in the last section (with n = 4) we
get,

a∝ t (15.18)

At the same time we know that the temperature of radiation is related to its energy-
density by

ρ rad ∝ T 4 (15.19)

for isotropic and homogeneous radiation field.

From the last three equations, we get the relationship of temperature with time as

T ∝ 1/ t (15.20)

We see that t = 0 is both interesting as well as disturbing. This is because at that point
of time, the temperature shoots to infinity and so does the energy density. It is this
epoch which is termed as the Big Bang.

15.6 EARLY UNIVERSE AND NUCLEOSYNTHESIS

We saw that the temperature of radiation varies as inverse of the square root of time.
The immediate consequence of this is that the radiation temperature in early times
should have been very high. If matter and radiation were in thermal equilibrium, the
above statements imply that the temperature of matter was also high in early times.

We know from thermodynamics that temperature is a measure of the mean kinetic

energy which in turn implies high energy collisions between particles. The earlier the
epoch, the higher the energy with which these particles collide with each other. Today
we know about the physical phenomena at high energies from experimental
investigations using high energy particle accelerators and colliders. We should expect
44
that the same phenomenon must have taken place in the early universe. In fact, for this Large Scale Structure and
the Expanding Universe
reason, early universe is often called the poor man’s laboratory.

An important class of reactions at high energy is those which lead to the synthesis of
nuclei of elements. Very early on, there were no complex nuclei. The only ones were
the hydrogen nuclei, i.e., protons. At those energies, even if the protons and neutrons
combined to form higher atomic number nuclei (e.g., the Helium nucleus), the kinetic
energies of the particles were so high that the collisions would have immediately
disintegrated them.

As the universe cools, a certain temperature is reached when the energies are low
enough that this backward reaction (namely disintegration) begins to get suppressed.
Hence, stable helium nucleus begins to get formed. As the temperature lowers further,
we expect that higher atomic number nuclei will begin to get synthesized. So the
question is, “Can we proceed in this way and synthesize all the naturally occurring
nuclei”? The answer unfortunately is no!

This line of reasoning works for only the elements with first few atomic numbers. As
the temperature decreases, we can form Lithium and some Boron. The problem comes
up when we need to form Beryllium. In the process of the formation of the stable
Beryllium nucleus, one passes through an intermediate stage where, spontaneous
disintegration is faster than the fusion. So even before there can be fusion the nucleus
which is supposed to participate in the fusion, disintegrates. Hence one cannot form
Beryllium by this procedure. Only after the formation of Beryllium, can the nuclei of
higher atomic numbers be formed. Hence this is called the “Beryllium Bottleneck”.
The universe has to wait for a very long time, namely, till stars form, in order to
synthesize elements of atomic number 5 and higher. (Refer to Unit 10 for details of
nucleo-synthesis inside the stars).

15.6.1 Cosmic Background Radiation

Do we have any signature of the early hot phase of the universe? The answer is yes. In
1965, two scientists at the Bell Telephone Laboratories in America discovered
accidentally a radiation at a very low temperature of only 3 K which seemed to come
from all directions. It was highly isotropic. It was suggested that the radiation fills the
whole universe. Since the wavelength of the peak radiation, ~ 1 mm, falls in the
microwave region, it was called the cosmic microwave background radiation
(CMBR). This is the relic of the era when the universe was very hot and dense. It is
argued that the radiation was once very hot and has been cooled to its present
temperature due to the expansion of the universe over billions of years (recall from
4
Eqs. (15.20) and (15.19) that T ∝ 1 t and ρrad ∝ T ). Put in another way, the same
energy fills an every increasing volume, so its energy density decreases and so does its
temperature.

The discovery of the CMBR has great significance for cosmology. (The discoverers of
the radiation were awarded the Nobel Prize for their work.) It shows that the universe
was once very hot and dense. This lends great support to theories of the universe
which maintain that the universe is changing with time, that is, it is evolving. Its
existence is a very powerful argument against theories which propose that the
universe is steady, that is, it is unchanging. In the theories of the latter type, it is
extremely hard to produce such a radiation.

CMBR is a topic of intense research today.

15.6.2 Evolving vs. Steady State Universe

You have just seen that CMBR points to a phase of the universe when it was very hot
and dense. This phase is generally known as the ‘hot Big Bang’. The idea is that some 45
Galaxies and the violent event took place at that time which sent the universe expanding. You have also
Universe seen above that if the early universe had not been hot and dense, it would not have
been possible to synthesise light elements, such as H2, H e4 and Lithium. In fact, the
prediction of the precise observed abundances of these light elements is a very
powerful argument in favour of the universe that changes with time: an evolutionary
universe.

Yet there is a set of scientists who believe in a steady state universe, a universe
which has no beginning and no end, which appears the same at all points in space and
at all times. Historically, the steady state theory emerged in the1940s and 1950s, when
observational techniques were not much developed, and so the Hubble constant H
could not be measured accurately. Recall that 1/H gives a rough time scale of the age
of the universe. So, the age of the universe inferred from the value of H at that time
turned out to be less than the age of some fossils on the Earth. This was quite
embarrassing. To overcome this age problem, the scientists proposed the steady state
universe. However, CMBR and synthesis of light elements in the early universe are
very powerful arguments against this theory and in favour of the hot Big Bang theory.

With this we come to an end of this unit in which you have studied about the large
scale structure of the universe. We now present its summary.

15.7 SUMMARY
• Distances of far-off objects inform us about the large scale structure of the
universe. To find distances of other galaxies we employ the concept of the
distance ladder. The first step of the ladder is the Cepheid variable stars found in
nearly galaxies. Subsequent steps include objects such as supernovae which can
be detected even in galaxies which are very far off.

• The outcome of the exploration of the universe is that the galaxies are rushing
away from one another and the universe is expanding.
• The expansion of the universe is in accordance with Hubble’s law, v = Hr, so that
1/H gives a rough estimate of the age of the universe.
• The behaviour of the universe is governed by the Friedmann equation:

a& 2 k 8πGρ
+ =
a2 a2 3

• The need for a hot and dense early universe can be explained in connection with
the synthesis of light elements and the existence of the cosmic microwave
background radiation.
• The present understanding is that if the universe was once hot and dense, then it
must be an evolving universe, and not a steady state universe.

15.8 TERMINAL QUESTIONS Spend 30 min.

1. Explain how Cepheid variables have been used to measure astronomical

distances.

2. State Hubble’s law. How can this be used to get an estimate of the age of the
universe?

3. Explain why at one time, the steady state theory appeared necessary. What is its
46 status now?
 Large Scale Structure and
15.9 SOLUTIONS AND ANSWERS the Expanding Universe

Self Assessment Questions (SAQs)

1. See Section 15.2.

−1 −1
2. H = 70 km s Mpc

= 70 × 10 5 cm s −1 / 3 × 1018 × 10 6 cm

7 × 10 6
= s −1
24
3 × 10

1 3 × 10 24
∴ Age of the universe = = s
H 7 × 10 6

3 ×10 24
= yr
7 ×10 6 × 3 ×10 7

= 1.4 ×1010 yr = 14 billion years.

a& 2 8πG
3. Eq. (15.13) ⇒ = ρ
a2 3
a& 2 8πG
Since ρ = Ca − n ⇒ = C a −n [C is a constant]
2 3
a

n
a& 8πGC −n da 1−  8πGC 
∴ = a 2 ⇒ a& = = Aa 2 A = 
a 3 dt  3 

n−2 n
a 2
∴ ∫ a 2 da = A dt ⇒
n ∫∝t ⇒ a ∝ n ( 2) 2n . t 2n .
2

Hence, proved.

Terminal Questions

1. See Text.
2. See Text.
3. See Text.

47