(Erika Böhm-Vitense) Introduction To Stellar Astr
(Erika Böhm-Vitense) Introduction To Stellar Astr
(Erika Böhm-Vitense) Introduction To Stellar Astr
Volume 2
Stellar atmospheres
Introduction to Stellar Astrophysics
Volume 1
Basic stellar observations and data
ISBN 0 521 34402 6 (hardback)
ISBN 0 521 34869 2 (paperback)
Volume 2
Stellar atmospheres
ISBN 0 521 34403 4 (hardback)
ISBN 0 521 34870 6 (paperback)
Volume 3
Stellar structure and evolution
ISBN 0 521 34404 2 (hardback)
ISBN 0 521 34871 4 (paperback)
Introduction to
stellar astrophysics
Volume 2
Stellar atmospheres
Erika Bohm-Vitense
University of Washington
CAMBRIDGE
UNIVERSITY PRESS
PUBLISHED BY THE PRESS SYNDICATE OF THE UNIVERSITY OF CAMBRIDGE
The Pitt Building, Trumpington Street, Cambridge CB2 1RP, United Kingdom *
CAMBRIDGE UNIVERSITY PRESS
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40 West 20th Street, New York, NY 10011-4211, USA
10 Stamford Road, Oakleigh, Melbourne 3166, Australia
Typeset in Times
A catalogue record for this book is available from the British Library
Bohm-Vitense, E.
Introduction to stellar astrophysics
Includes index.
Contents: v. 1. Basic stellar observations and data -
v. 2. Stellar atmospheres.
1. Stars. 2. Astrophysics. I. Title.
QB801.B64 1989
523.8 88-20310
ISBN 0 521 34403 4 hardback
ISBN 0 521 34870 6 paperback
Preface page ix
2 Stellar spectra 15
2.1 The spectral sequence 15
Problems 231
Appendix LTE model stellar atmospheres 241
References 245
Index 247
Preface
ix
x Preface
where the integral is taken along the path s of the light beam. KX is the
absorption coefficient of the gas through which the beam passes. We first
have to determine the nature of this gas, and then we have to measure the
absorption coefficient of a similar gas in the laboratory to determine K} or
we have to calculate it from atomic theory. Without further preparation
at this point we can only say that the KX for stellar atmospheres have been
determined (we will discuss how later) and we find that the atmosphere of
the sun is roughly 100 km thick. For the hottest main-sequence stars it is
about 1000 km thick. This thickness has to be compared to the radius of
the sun, for instance, which is 700 000 km. The atmosphere is only a very
thin surface layer. If we compare the thickness of the skin of an apple with
the size of the apple, then this skin is relatively thicker than the atmosphere
compared to the size of the star. It is somewhat disappointing that this is
all we can see of the stars - such a very thin surface layer - yet we have
to get all our information about the stars from the study of the radiation
coming from this very thin surface layer. How much does this tell us about
the interior of the star, about those very deep layers which we cannot see?
This we will find out in Volume 3.
From the discussion above, it appears clear that we have to make a very
sophisticated study of the starlight in order to obtain all the information
we possibly can. This will involve quite a bit of theory. I will do my best
to make this theory as transparent as possible. In order to do this I have
to make some simplifications, but still retain the important ingredients. I
am sure that some knowledgeable astrophysicists will object to some of
the simplifications, but my experience has taught me that it is more
important for the beginner to understand the basic principles, even at the
expense of accuracy, than it is to see all the complications immediately.
Once the basic physics is understood, the complications can always be
added later.
I am also sure that several readers will think that some of the discussions
are too elementary. Again, my experience has shown that many students
have never heard about these elementary facts about atomic physics and
the formation of spectral lines. In fact, I have talked to many well-known
physicists who did not know that stellar spectra show absorption lines and
not emission lines, as do laboratory spectra. I therefore feel that these basic
facts need explanation, so that some readers are not lost from the very
beginning.
Before we start our theoretical discussions, we review briefly those
sections of Volume 1 which are important for the following discussion. A
few new considerations will be inserted. The reader who has studied Volume
1 is advised only to skim through the first two chapters of Volume 2.
Preface xi
I apologize for not giving all the references for quoted investigations.
Instead, I have chosen to give only references for other textbooks or reviews
in which these references can be found. In addition, I give references for
figures and tables which were reproduced from other books or papers. I
also give references of recent articles, for which the references cannot be
found in older textbooks.
In many sections this text relies heavily on the book Physik der
Sternatmosphdren by A. Unsold, my highly respected teacher to whom I
am very grateful.
I am very much indebted to Ms Sandi Larsen for typing this manuscript
and for her patience with many changes.
1
1
2 1 Stellar magnitudes and stellar colors
\
(b)
Fig. 1.1. The amount of energy received to heat the water in the bowl depends on
the angle at which the sunbeams enter the bowl. If the beam is inclined with
respect to the normal on the water surface s by an angle <5, the effective beam
width is reduced by a factor cos d = t/s.
1.2 Stellar colors 3
or, if we refer to a given wavelength X and a bandwidth of 1 cm
Ex = nfx, 1.1b
where fx is called the apparent flux.
The magnitudes referring to the energy received here on Earth (that is,
to the brightness as it appears to us) are called apparent magnitudes. They
are designated by a lower case m, or mx if they refer to a given wavelength.
These magnitudes are related to the energy received by
, 1.2
Obviously, this does not determine the magnitudes of the stars unless
we give the magnitude for one star. This star is Vega; it always has the
magnitude 0 by definition, regardless of the wavelength we are looking at.*
The only exception is the bolometric magnitude, which refers to the total
radiation of the stars, including the invisible part.
The apparent magnitudes refer to the energies received above the Earth's
atmosphere, i.e., after correction for the absorption occurring in the Earth's
atmosphere. In the next section we will see how we can make this correction.
We want to emphasize here that this definition of the zero point for all
magnitude scales is quite arbitrary and has nothing to do with the actual
energy distribution of Vega. It has its foundation only in history. When
the magnitude measurements were made, one could only observe first with
the eye, then later with photographic plates, and it was not known how
the energies in these wavelength bands were related. It therefore seemed
best to keep the same normalization.
In Fig. 1.2 we show the actual energy distribution of Vega as a function
of wavelength X, i.e., fx is plotted as a function of X. It is obvious that fx
is not independent of X, even though all magnitudes mx = 0 by definition.
It is very important to remember this if we use the magnitudes to infer
anything about the energy distribution in the stars.
emits much more energy at the shorter wavelengths than at the longer
wavelengths. If we observe a star which is fainter than Vega in the blue
region of the spectrum, but is just as bright as Vega in the visual region,
then this star has a visual magnitude equal to that of Vega, namely m v = 0,
but in the blue it has a larger magnitude than Vega because it is fainter.
This may well happen if we compare a star like the sun with Vega. In
Fig. 1.3 we compare the energy distribution of a solar type star with that
of Vega. We have assumed the solar type star to be at such a distance that
the observed fluxes in the visual spectral region are the same. In the blue
and in the ultraviolet the solar type star is then fainter than Vega. For the
solar type star the blue magnitude is larger than the visual magnitude. For
a given star the magnitudes in different wavelength bands can be quite
different. The difference in these magnitudes tells us something about the
energy distribution in the stars. The solar type star has relatively more
energy in the visual region than Vega; it is more red than Vega. If we take
the difference between the magnitude measured in the blue, mB, and in the
visual, m v , we find for the solar type stars
lYln Wl\r > U . 1 .^T
i i I i i
8.5
8.0 -
\
7.5 -
7.0 -
\
6.5
i i i i i i i i
3.5 3.6 3.7 3.8 3.9 4.0 4.1 4.2
log A
Fig. 1.2. The energy distribution in Vega, log j \ + const, is shown as a function of
wavelength Lfk is strongly variable with wavelength, even though all mA are zero
by definition.
1.2 Stellar colors 5
According to our discussions in Volume 1, such stars are generally
expected to be cooler than Vega. Stars with relatively more energy in the
blue than Vega have smaller magnitudes in the blue; they therefore have
B V < 0. These stars are expected to be hotter than Vega.
We can, of course, determine magnitudes for many wavelength bands
and correspondingly determine many colors. Different astronomers prefer
different color systems, depending on what they want to infer from these
colors. There are, therefore, many different color systems in use. The most
widely used system is still the UBV system, using ultraviolet, blue and
visual magnitudes. The bandwidths for these magnitudes are roughly
1000 A, which means they are broad bands. The sensitivity curves for this
system are reproduced in Fig. 1.4. Of course, we receive more light in such
broad bands than in narrow bands, which is an advantage if we want to
observe faint stars. On the other hand, much information is smoothed out
7.7
7.6 -
/ x\
7.5
x Solar type star
if
7.4
7.3
7.2
7.1
\\ \
x\
\ \
{2 7.0
o _/ Vega type^ A \\ \
\
6.9 star \
\
1 \\ \
i \
5
6.7 4
1
\\ \
\
6.6 I \\ \
6.5
I \ \
I \ \
6.4
L \
\
\
\
6.3 \
-
6.2
6.1
_ \
\
3600 A 4300 A 5000 A x\
6.0 , 1 1 1
Fig. 1.3. l o g / ; + const, for Vega is again shown as a function of wavelength. Also
shown is the relative energy distribution of a star like the sun* This star is assumed
to be at a distance such that its brightness in the visual is the same as that of
Vega. We see that it then has much less flux than Vega in the blue wavelength
region around 4300 A. Its mB must then be larger than zero. For this star B V ~ 0.6 > 0.
6 1 Stellar magnitudes and stellar colors
in such broad bands. For brighter stars narrower bands give us more
information. We will discuss other color systems later, such as Stromgren's
color system.
I I i
U B V
1.0
r\
0.8
\ \
0.6
\ \ \ -
0.4
\ -
0.2
1
3
/
A 4
j 5
A(103A)
6 7
Fig. 1.4. Sensitivity curves for the receiving instruments to measure UBV
magnitudes. (From Johnson, 1965.)
Zenith
Top of the
atmosphere
- Earth's surface
Fig. 1.5. The light beam from the star passes through the Earth's atmosphere
along the path 5, which is inclined with respect to the zenith direction by the angle
6. Along path s the intensity is reduced due to absorption. The absorption
increases if 5 becomes longer for larger zenith distances.
1.3 Correction for absorption in Earth's atmosphere 7
energy there is in the beam, i.e., the more photons there are, the greater
the chance that one will be absorbed. The change in energy d/A is therefore
proportional to Ik, where Ik is the intensity of the beam at wavelength L
The number of photons absorbed at the wavelength k also depends on the
special properties of the gas in the atmosphere, which is described by the
so-called absorption coefficient Kk. The change in intensity along the path
length ds is then given by
d/A = KkIk ds. 1.6
Since the intensity is decreasing, the d/A is, of course, negative. Dividing
by Ik and remembering that dl/I = d In /, where In denotes the logarithm
to the base e, we find
dQnIx)=-Kxds=-dTx. 1.7
Here we have defined the so-called optical depth xk by
rs0
dxk = Kkds and s 0o))=
T A (S = KKxds.
kds. 1.8
Jo
Equation (1.8) can be integrated on both sides and yields
^ T A s ( s ) . 1.10
/A(0)
Taking the exponential on both sides gives
The optical depth along the path of the light xks depends on the zenith
distance 9, as can be seen in Fig. 1.5. We see that cos 9 = t/s = dt/ds or
At
c9dt. 1.12
cos 9
It then follows that
xks=
Kk ds = sec 9 Kk dt = sec 9xkt, 1.13
Jo Jo
where xkt is the optical depth measured perpendicularly through the
atmosphere. We can then write equation (1.10) in the form
1.14
8 1 Stellar magnitudes and stellar colors
where TXX is now independent of 9. xXt is called the optical depth of the
atmosphere at wavelength X and is usually written as xx.
In order to derive from /A(s, 9) the intensity above the Earth's
atmosphere, /A(0), we have to know TA, which may depend strongly on
wavelength.
Just as we discussed for the sun in Volume 1, we can determine the
optical depth xx by measuring the intensity received from a given star by
measuring Ix(s) for different angles 9, when the star moves to different
places during the course of the night (due to the Earth's rotation). In
principle, two measurements are enough; they provide us with two
equations for the two unknowns /A(0) and TA, as we will see. Let
Ixl = / A (#i, s) be the intensity measured for the zenith distance dl of the
star and IX2 the intensity measured at zenith distance 92. We then find,
according to equation (1.10),
In Ixl In IX2 = iA(sec 91 sec 0 2 ), 1.15
and
In A
In MO)-
1 2
secO
Fig. 1.6. In order to determine /A(0) it is best to plot the measured values of
In lx(0) as a function of sec 9. The best-fitting straight line through the points
intersects the In / ; axis at the value In /A(0).
1.4 Absolute magnitudes of stars 9
that sec 9 = 0 does not exist; this is just a convenient way to read off the
value for In /A(0).)
In fact, Fig. 1.5 is a simplification of the geometry. The Earth's surface
is not plane parallel. In addition, the beam of light is bent due to refraction.
These effects give a relation xXsjxXt ^ sec 9. The actual ratio is called air
mass. For sec 9 < 2 the difference is in the third decimal place and is in
most cases negligible.
It is important to note that the above derivation only holds for each X
with a given KX, or for a broad band if KX happens to be wavelength
independent. It cannot be applied to a broad wavelength band with varying
KX. In the ultraviolet the variations of KX with A are especially strong.
People tend to forget the variations of KX when determining the extinction
correction for broad band colors like the UBV magnitudes. Using the same
method yields wrong values for xx.
How can we avoid this problem? We have to remember that ultimately
we want to determine the ratio of the corrected intensity of one star to the
corrected intensity of a standard star. If we correct both stars by the same
wrong factor, the wrong factor does not do any harm because it cancels
out. According to equation (1.14), the correction factor is esec6>T;. If we
measure both stars for the same zenith distance, 9, the errors will almost
cancel out. (They do not completely cancel if the stars have different energy
distributions, because then different errors in xx will occur.) In addition,
errors will be minimized if the stars are measured at small zenith distances.
In Table 1.1 we list the relation between U B and B V colors for
normal main sequence stars and for supergiants. In the next section we
will briefly review what these stars are. In the same table we also list the
UBV colors for black bodies of different temperatures. As can be seen, the
stars do not radiate like black bodies. One of the aims of Volume 2 is to
understand why the stars have a different energy distribution from that of
black bodies. As can be seen from Table 1.1, the stars have less energy in
the ultraviolet.
In Fig. 1.7 we have plotted the U B colors as a function of B V for
stars and for black bodies.
-16000 K
-1.0
8000 K
6000 K
4000 K
1.0
_L
-0.3 0.5 1.0
B-V
Fig. 1.7. The two-color diagram for the U B, B V colors for stars and for black bodies.
1.5 Luminosities of stars 11
determine the distances of the stars or we have to compare stars which
happen to be at the same distance, namely stars in star clusters. We saw
in Volume 1 how we can determine distances for nearby stars by means
of trigonometric parallaxes. If for these nearby stars we determine the
magnitudes they would have if they were at a distance of 10 pc (i.e., if we
determine m (10 pc) by comparison with the actual brightness of Vega
which is m = 0), then this m (10 pc) is called the absolute magnitude of
the star. Since Vega is closer than lOpc, its absolute magnitude is larger
than 0; it would be fainter than it actually is if it were at 10 pc. Its absolute
magnitude is M v = 0.5. Here we have used a capital M to indicate that we
are talking about absolute magnitudes. Absolute magnitudes are always
given by capital M; the index refers again to the wavelength band used to
determine the absolute magnitude. As was derived in Volume 1, the absolute
magnitude can be obtained from the apparent magnitude and the distance d
(in parsecs) of the star as follows:
Mv-mv=:5-51ogd. 1.17
The abbreviation log means logarithm to the base 10. Similar relations
hold, of course, for the magnitudes in other wavelength bands. It is obvious
that the colors determined from the absolute magnitudes are the same as
the ones determined from the apparent magnitudes.
B-V Tcff BC T ef f BC
-0.4 2.0
Fig. 1.8. The absolute magnitudes of nearby stars and some clusters are plotted as
a function of the B - V colors. Most stars fall along a sequence, called the main
sequence. Some stars are brighter than the main sequence stars; they are called
giants. A few stars are much fainter than the main sequence stars; they are called
white dwarfs. The open circles refer to probable binaries.
Stellar spectra
18
3.1 The black body 19
the hole. The light which we see is emitted by the black body itself. But,
of course, the room is not in thermodynamic equilibrium. If we look at
the radiation from different parts in the room we see different radiation
and different temperatures. The window is black, but it is not what we
would call a black body to be used as an ideal light source. In our ideal
light source the radiation in the box cannot depend on the properties of
the gas or the walls. For instance, if one wall were to emit more radiation,
say at 5000 A, than the other walls, then there would be a flow of photons
from this wall to the other walls, which would then be heated up. We could
also use this flow of photons to power a perpetuum mobile. Yet, the second
law of thermodynamics tells us that we cannot power a thermodynamic
machine unless we have temperature differences such that a heat flux goes
from the higher temperature region to the lower one. A black body in
complete thermodynamic equilibrium by definition does not have
temperature differences. If it has, it is not a black body. We have to insulate
it better and wait longer until all temperature differences have disappeared.
Then we find that the radiation coming out of the tiny hole always looks
the same, no matter which gas is in the box nor what the box is made of.
Obviously, the hole has to be tiny, otherwise the gas next to the hole will
cool off and we will no longer have the same temperature everywhere. If
we now measure the radiation leaving the hole we find, of course, that the
amount getting out increases in proportion to the size of the hole. In order
to get a measurement independent of size, we have to divide the amount
of energy received by the area of the hole, and then determine the amount
of radiation which we would receive if the hole had a cross-section of 1 cm2.
We also have to measure the amount of energy which we receive in each
unit of time. The amount of energy which leaves the hole per cm2 each
AT
/
/ /
Fig. 3.1. The box with a small hole will act as a black body, i.e., will absorb all
the light falling into the hole. If such a black body is well insulated and left to
itself for a long time, complete thermodynamic equilibrium is established. The
radiation of such a black body serves as an ideal light source with which stellar
radiation can be compared.
20 3 Temperatures estimates for stars
second into all directions and at all wavelengths is called TLF, where F is
called the flux. It is found that the flux increases with increasing temperature
of the black body. Stefan and Boltzmann measured that
A
3.1
where a is the Stefan-Boltzmann constant,
a = 5.67 x 10"5 erg cm"2 degree"4 s~\
provided we measure the temperature T in degrees Kelvin.
Equation (3.1) is called the Stefan-Boltzmann law.
If the temperature is high the hole no longer looks black to us because
of its own radiation. Nevertheless, it is still a black body because it still
absorbs all the light falling into the hole. The experiment verified that all
black bodies, no matter what they are made of, emit the same kind of
radiation if they have the same temperature. We can then once forever
measure the energy distribution of a black body with a given temperature.
If we do not measure the total radiation coming out of the hole, but
only the amount which comes out per second perpendicularly to the hole
and into a cone of opening dco (see Fig. 3.2) and in a wavelength band of
width dX, then the amount of energy is given by
Ex = Ix dX dcoAa. 3.2
where ACT is an area perpendicular to the beam of radiation.
Ix is called the intensity. For a black body the intensity distribution is
given by the well-known Planck formula:
w 2hc2 1
3.3
Fig. 3.2. The energy leaving the black body through a hole with an area da into a
cone with opening dco and a band width dk is given by Ek = lx dX do) do. For
A<T = 1 cm 2 , A/ = 1 cm and Aco = 1 this energy is equal to the intensity /A, which
for a black body is called Bx.
3.1 The black body 21
This is the amount of energy measured per wavelength interval AX = 1. If
the energy per unit frequency interval is measured, we find
. 2hv3 1
= Bv = black body intensity. 3.4
14.8
log BA
nF = nf 3.8
a = R/d is the angular radius of the star (see Fig. 3.5), measured in radians.
We therefore only have to measure the angular radius of the star in order
Fig. 3.4. The amount of energy leaving the star per second is given by
L = 4nd2 x nf. It is also equal to 4nR2 x nF.
3.3 Wien temperatures 23
to determine nF. We thus know nF if we know nf the amount of energy
received above the Earth's atmosphere, and the angular radius of the star.
For the sun we determined nFQ = 6.3 x 1010 erg cm" 2 s" 1 . The relation
nF = aTtu 3.9
defines the effective temperature of a star. It gives us the temperature which
a black body would need to have if it were to radiate the same amount of
energy per cm2 s as the star. For the sun we measure nf = S = 1.37 x 106
erg cm" 2 s" 1 . This is called the solar constant. With an angular radius of
a o = 959.63 arcsec, we find nFQ = 6.3 x 1010 erg cm" 2 s~ \ which leads to
Tcffo = 5800 K for the sun.
For a Lyrae we can also measure the flux nf received above the Earth's
atmosphere; the angular diameter for a Lyrae was measured by Hanbury-
Brown to be 3.2 x 10~3 arcsec. From these values and also other
measurements we determine that the effective temperature of a Lyrae is
9500 K.
In Table 3.1 we list angular radii and effective temperatures for some
representative stars, according to Schmidt-Kaler (1984).
Fig. 3.5. The ratio R/d is the angular radius a of the star if a is measured in
radians, sin a = a because a 1.
24 3 Temperatures estimates for stars
Table 3.1. Angular diameters and effective temperatures for some bright stars
' Intrinsic color of the star, i.e., corrected for interstellar reddening (see Chapter 4).
for a Lyrae which is 7600 K. This is much lower than the effective
temperature. We suspect that this discrepancy has something to do with
the abrupt decrease in energy output seen for a Lyrae for wavelengths
shorter than 3700 A, which looks as if something suddenly absorbs all the
energy which was supposed to come out at shorter wavelengths and which
might have given us a higher Wien temperature.
We will come back to this question later when we understand better
what happens to the radiation in the stellar atmospheres.
i.e., it increases proportionately to the opening of the cone, the size of the
area, and the width of the wavelength band dX. According to the definition
(4.1), the intensity Ik is then the energy which passes per second through
an area of 1 cm 2 into a solid angle Aco = 1 in a wavelength band AX 1.
Clearly, not all the energy passing into Aco = 1 goes perpendicularly through
dcr, since Aco = 1 describes a rather large cone. Rather, we should think
about the energy passing in a narrow cone divided by the size of the solid
angle of the cone in the limit of an infinitely small cone opening.
26
4.2 Radiative energy transport 27
Let us now consider the energy passing through a surface area da under
an angle 9 with respect to the normal of this surface area (see Fig. 4.2).
The effective beam width is reduced by a factor cos #, so now
Ex = Ix da cos 0 dcodX. 4.2
Fig. 4.2. The amount of energy going through da under an angle 0 is reduced by
a factor cos 0. The amount of energy going through da is the same as the amount
going through the surface area da\ where da' is the projection of da onto a plane
perpendicular to the beam of light, i.e., da' = da cos 0.
J a
/
/
/ \ *~i^
dsj_
/
/
/
/
/
/
/
/
Fig. 4.3. A beam of light passes through a box of cross-section a and length 5
corresponding to an optical depth xk = j j KX ds. The intensity IXo entering the box
on the left-hand side will be changed by absorption and emission along the path s
and will leave the box on the right-hand side with an intensity /A, which depends
on the optical depth xk of the box and on its emission properties.
28 4 Radiative transfer
When the light beam passes along the path element ds there is some
absorption and the energy Ex is reduced by the amount
d A = KXEX ds = KX ds Ix dco dX do. 4.3
At the same time, there is also some emission from the volume dV = da ds,
which is given by
d A = sx dco dX ds da, 4.4
where eA now is the amount of energy emitted each second per unit volume
into Aco = 1 per wavelength band AX = 1. Combining absorption and
emission, we find
d A = d/A da dX dco = KXIX dco dX da ds + A dco dX ds da. 4.5
Sx = ~ or SX = KXSX. 4.9
Emission lines
Hot gas
Absorption line
spectrum
Cold gas
Fig. 4.4. Kirchhoff and Bunsen observed that the spectrum of a hot gas shows
emission lines, while the spectrum of the light from a hot continuous source,
passing through cold gas, shows absorption lines. The wavelengths of the emission
and absorption lines are the same if the hot and the cold gas contain the same
chemical elements.
4.4 Absorption versus emission lines 31
giving the change of intensity over a path length ds in Fig. 4.3. We want
to integrate this equation in order to find / A (T A ) along the path 5. In order
to achieve this integration, we multiply both sides of the equation by e +T/
and obtain
4.18
This is now a form of equation (4.8), which can easily be integrated over
the interval xk = 0 to TA, giving
(a) TA 1, i.e., the gas volume has a small optical depth. This is also called
the optically thin case.
We can then expand the exponential in a Taylor series and obtain
e~T; ^ 1 TA forrAl, 4.23
and find
JA = S A ( 1 - 1 + TA) = T A S A , 4.24
or in LTE, i.e., if SA = BA,
Ik = xkBk with tA = KAS, if Kk is constant along s. 4.25
32 4 Radiative transfer
The intensity leaving the box on the other side will be large when KX is
large, and small when KX is small.
Suppose KX depends on the wavelength X, as shown in Fig. 4.5. The
frequencies at which we see the spectral lines are actually frequencies for
which the absorption coefficient KX is very large because they are resonance
frequencies in the atom. The wavelengths for very large KX in Fig. 4.5
therefore correspond to the wavelengths of spectral lines. From equation
(4.25) we see that the intensity will be large at those wavelengths where
the optical depth is large, i.e., at wavelengths for which KX is large, which
means in the spectral lines. For xx 1 we therefore expect to see emission
lines with large intensity at the wavelengths of large KX.
(b) The other limiting case is the optically very thick case, i.e., T A 1. In
this case, e~T; - 0 and we find from equation (4.22)
j A = sx, or in LTE, h = Bx. A26
In this case, the emitted intensity is independent of KX. We have the
situation shown in Fig. 4.6.
(a) (b)
Fig. 4.5. (a) The assumed wavelength dependence of /cA. (b) The intensity
distribution obtained for xx 1 and IXo = 0.
(a) (b)
SA or i
Fig. 4.6. (a) The assumed wavelength dependence of KX. (b) The intensity
distribution found for xk 1; the intensity is independent of KX.
4.4 Absorption versus emission lines 33
This is, of course, the case of the black body: TA oo means all the
incoming light is absorbed. The emitted intensity is only due to the emission
within the box and the intensity observed is given by the source function
which, in this case, is the Planck function.
The first case TA 1 is important for many astronomical observations.
Suppose we see a large interstellar gas cloud or nebula with a density of
a few particles per cm 3 up to about 10 000 particles per cm 3 (see Fig. 4.7).
Such a nebula usually shows an emission line spectrum, demonstrating
that it is optically thin (see Fig. 4.8), at least in the continuum and in the
line wings, not necessarily in the centers of the lines. If in the continuum
and in the wings T A 1, but in the line centers T A 1 we find in the line
centers /A = SA while in the wings we have /A = tASA SA.
The solar corona also corresponds to the case TA 1 and therefore shows
an emission line spectrum observable during solar eclipses (see Fig. 4.9).
This is due to the small optical depth (and not due to the high temperature,
at least not when observed beyond the sun's limb).
Clearly, stars do not show an emission line spectrum because they are
optically very thick, r ; -> oo. They do not, however, show a smooth intensity
distribution for all wavelengths, as does the black body, because the source
function Sk is not constant along the path s, as we assumed when we derived
equation (4.26).
We shall now discuss the case when JAo ^ 0.
Let us again distinguish two cases: the optically thin case with T A 1
and the optically thick case with TA 1.
In the optically thin case we can again replace eT; by 1 tA. Equation
(4.21) then reads
h = IJLOQ ~ *x) + S,TX = IXo+T;(S,-I;M), 4.27
which for IXo = 0 reduces again to the case discussed previously.
Hot nebula
Fig. 4.7. A large interstellar nebula of low density shows an emission line
spectrum, indicating that it is optically thin along the line of sight.
34 4 Radiative transfer
For lko ^ 0 the emerging intensity is given by the original intensity /A()
plus a term whose sign depends on whether /, o is larger or smaller than
Sx. If 7Ao > SA, then the last term in equation (4.27) is negative. We then
rewrite this equation in the form
/A = / A - ^ ( / A O - S J . 4.28
This means that from the original intensity there is something subtracted
which is proportional to the optical depths TA, SO there is more intensity
CO CNI CD
o CD
CD CD
o ts
h- r*.
O U u CJ
ID
Balmer
-3650 continuum
of hydrogen
-3727 [Oil]
3868 [NelllJ
3889 H
- 3 9 6 7 [NelllJ, H
-4100
_ 4340 Hy
- 4 3 6 3 [0111]
- 4 4 7 1 Hel
- 4 6 8 6 Hell
-4861 H/y
- 4 9 5 9 <N2)
- 5 0 0 7 (NJ L J
Fig. 4.8 shows slitless spectra of such emission line nebulae. An image of the
nebula is seen for each emission line. This is a positive photo. Bright images are white.
(From Unsold, 1977.)
4.4 Absorption versus emission lines 35
missing at those wavelengths for which KX is larger, i.e., in the spectral
lines. In this case we will see absorption lines on the background intensity
/; 0 (see Fig. 4.10). This case is obviously the second case of the
Kirchhoff-Bunsen experiment.
In the case that SA > IXo we find again for small optical depth
/ A = / AO + T A ( S A - / A O ) . 4.29
The last term is now positive. We will see emission lines on top of the
background intensity IXo (see Fig. 4.11).
For T ; 1 we see from equation (4.21) that in any case Ix = SA, no matter
what / ; o was. The original intensity is totally absorbed along the path s
with the very large optical depth.
The stars usually show an absorption line spectrum. They must therefore
correspond to the case when lko - the intensity coming from the deeper
layers - is larger than the source function SA for the top layers of the stars.
Let us assume LTE, which means that Sx = Bx. For this case the source
(a)
. ,..-.v..
]
m
i
>
mm
1 !
un \(tf 1 11 16 R sans
N . - I V I l H P Ft 11
HI
Hi-
KnXlV'l Me
Fig. 4.9. (a) The solar corona is optically thin and therefore shows an emission
line spectrum, (b) During solar eclipses the optically thin solar corona shows an
emission line spectrum. We clearly see the image of the corona in the 5303 A line
emitted by the Fe XIV ion, i.e., an Fe atom which has lost 13 electrons. The other
spectral lines seen originate in the solar chromosphere and transition region
between chromosphere and corona. The spectrum was obtained by Davidson and
Stratton, 1927. This is a negative photo. Bright lines are dark.
36 4 Radiative transfer
function increases with increasing temperature. We can now divide the star
into two regions: the optically thick deep layers and the optically thin
reversing layer at the top. JAo entering the reversing layer is then given by
the Planck function BXd for the deeper layers, and the source function SA
for the reversing layer is then given by the Planck function Bxt for the top
layers. The fact that the stars show an absorption line spectrum tells us
that the Planck function for the deeper layers is larger than the Planck
function for the top layers, i.e.,
Bu>BXt9 4.30
Thus the temperature in the deeper layers is higher than in the top layers.
Our discussion also shows that if we see layers of a star with temperature
increasing outwards, we would expect an emission line spectrum. This is
(a) (b)
Fig. 4.10. (a) The assumed wavelength dependence of KX. (b) The intensity
distribution is shown as a function of wavelength if KX(K) is given by (a) and IXo > Sx.
(a) (b)
0--
Fig. 4.11. (a) The assumed wavelength dependence of KX. (b) The wavelength
dependence of Ix if KX{X) is given by (a) and IXo < Sx.
4.4 Absorption versus emission lines 37
actually the case when we observe, for instance, the sun at ultraviolet
wavelengths. At the wavelengths longer than about 1700 A, we still see
deeper layers of the photosphere where the temperature is still decreasing
outwards, and we see the absorption line spectrum (see Fig. 4.12). For
k < 1600 A, the continuous absorption coefficient is so high that we only
get light from the chromospheric layers in which the source function is
increasing outwards because the temperature increases outwards. For these
wavelengths we therefore see an emission line spectrum.
We can now summarize our results.
There are, in principle, two possibilities for the formation of emission
line spectra:
1. An optically thin volume of gas in space with no background light,
i.e., an emission nebula, emits an emission line spectrum.
X
o
Lil *r^ *-
1
1 I i s B B
rial n m
in [fa
ii II
mwm
I
ji :, , - . 1
- / :
HKBW
:
,.
H m I I
* ^
F/gf. 4.72. The energy distribution for the solar spectrum is shown for the
wavelength region 1400 A < X < 2000 A. For the longer wavelength region the
absorption coefficient in the continuum is still low. We see deeper layers of the
photosphere in which the temperature is decreasing outwards. We therefore see an
absorption line spectrum. For k < 1600 A the continuous absorption coefficient is
much larger. For these wavelengths we see chromospheric layers in which the
temperature is increasing outwards. We see an emission line spectrum. (From
Zirin, 1966, p. 157.) This is a positive photo. Bright lines are white.
38 4 Radiative transfer
2. An optically thick volume of gas in which the source function is
increasing outwards. The solar spectrum in the ultraviolet is an
example.
An absorption line spectrum is formed in an optically thick gas in which
the source function decreases outwards, which generally means that the
temperature is decreasing outwards. (If the source function is not equal to
the Planck function, then it can also happen that the source function
decreases outwards while the temperature increases outwards. Such
exceptions have to be kept in mind.)
An absorption line spectrum can also form if an optically thin gas is
penetrated by a background radiation whose intensity Iko is larger than
the source function. A thin shell around a star might be an example, or
the interstellar medium between us and the star.
Radiative transfer in stellar
atmospheres
.= -KXIX{9)- 5.1
ds
With KX ds = dxXs we can introduce the optical depth TXS along s for the
outgoing beam of light, which will depend then on the direction 9 of the
beam. As in Chapter 1, we introduce the optical depth TX perpendicular
to the stellar surface going from the outside in. From Fig. 5.2 we see that
Fig. 5.1. A light beam passing through the stellar atmosphere from the inside to
the outside under an angle & with respect to the normal n to the surface
experiences absorption along the path s and also emission.
1
KA dt
Fig. 5.2 shows the relation between the optical depth xXs along the outgoing light
beam, leaving the surface under an angle 5, and the optical depth xx going from
the surface in, perpendicular to the surface.
5.2 Surface intensities 41
Introducing this into the transfer equation (5.1) gives
cos 9 - - = Ix(9) - SA, where /A = /A(rA, ), 5.3
di A
where all signs have changed, since we are now looking from the outside
in, i.e. opposite to the direction of light propagation.
Equation (5.3) is the radiative transfer equation for the plane parallel
case. Whenever we talk about changes in intensity due to absorption or
emission, this is the equation which describes what is happening. This
transfer equation enables us to derive an integral relation between the
intensity leaving the stellar surface and the depth dependence of the source
function. In order to determine / A (T A , 9) we need to integrate equation (5.3).
5.2 The surface intensities
In order to derive the intensity /A(0, 9) at the surface, we multiply
equation (5.3) by e~T^seCl9, similar to the derivation of equation (4.17),
and obtain
A\ ' p - T,. sec & __ j / Q\ g T ; . sec # o _ - T;. sec S c A
sec 9 drA
where Sk = S;(r?) and 10) = I;(r^ 9). This can be written as
T secg
d(/;e '- ) __ ^ c-TjSecg ^^
A
d(r ; sec#)
In order to derive the intensity /A(0, 9) at the surface, we now integrate
d(iA sec 9) from 0 to oc, and find
M C
[ A ( ) ] g A e - ^ * d ( T A s e c a ) 5.6
Jo
and
Ix(099)= S A (TA )e~ r ^ secd d(T A sec9). 5.7
Jo
where we have made use of the fact that, for TA -> oo, e " t ; s e c d -> 0.
Equation (5.7) can easily be interpreted if we remember that the emission
along ds* is determined by A dV = aA ds = SA/cA ds = SA/cA d(f sec 9) =
SA d(rA sec S), where d(tA sec 9) = dxXs, i.e., the optical depth along the path
s. From this emission at TA the fraction SA d(rA sec #)e~ r/ sec d arrives at the
surface; the rest has been absorbed on the way. /A(0, 9) results from the
summation of all the contributions of the volume elements along the path
of the light as expressed by equation (5.7).
Let us evaluate /A(0, S) for a simple case of S A (T A ). Assume
Sx{Tx) = ax + bxTx. 5.8
* Remember IA is the energy going through 1 cm 2 , therefore dcr= 1 cm 2 and
dV= lcm 2 ds = ds.
42 5 Radiative transfer in stellar atmospheres
We derive
T/sec d
rAe d(r A sec
we find
/A(0, 3) = X At cos1' with /!,. = aXii\ 5.11
(a) (b)
//f
/ 1/ \ K/ / \// \1
Fig. 5.3. Polar diagram of 1^), (a) for ak=l, bx=l, and (b) for A = \,bx = 2.
The larger ^A/aA, the more the radiation is concentrated in the forward direction.
Observer
Fig. 5.4. Light beams received from a star by a distant observer are essentially
parallel. The light coming from the center of the stellar disc leaves the surface
perpendicularly to the surface (3 = 0), while the light observed at the limb leaves
the surface essentially parallel to the surface (S = 90). The angular distribution of
IX(S) can be observed for the sun by means of the center-to-limb variation of the
intensity of the solar disc.
5.3 Fluxes 43
perpendicular to the surface, while at the limb the starlight which we receive
leaves the surface under an angle 3 = 90. Therefore, in the center we should
see an intensity Ix = ax + bx, while at the limb we should see only Ix = ax.
We get less light from the limb than from the center.
Unfortunately, the sun is the only star for which we can see the
center-to-limb variation of the intensity. In Fig. 5.5 we show a photograph
of the solar disc. The limb darkening is obvious.
Equation (5.12) tells us that in the center of the solar disc (S = 0, cos # = 1)
we see the source function Sx at a depth T A = 1, at the edge of the disc
(9 = %n9 cos $ = 0) we see Sx at the surface. Actually, since t/s = cos & in
the plane parallel approximation we see that xx = cos 9 if rAs = 1. Equation
(5.12) then means that we always see into the atmosphere down to a depth
where xks = 1, which corresponds to xx = cos 9 (see Fig. 5.6).
Fig. 5.5. This photograph of the solar disc clearly shows the limb darkening due
to the angular dependence of the intensity Ix($). (From Abell, 1982, p. 512.)
44 5 Radiative transfer in stellar atmospheres
amount of energy going through 1 cm 2 per second per AA = 1 into the solid
angle dco in the direction inclined by the angle 9 to the normal of the area
1 cm 2 is given by (see Fig. 4.2)
dEx = Ixcos9dco9 5.13
and
nFx = lk cos 9 dco, 5.14
Fig. 5.6. The radial optical depth equals cos if the optical depth along the line of sight is 1.
Fig. 5.7. The description of dw in polar coordinates, (a) Sphere viewed from the
side, (b) sphere viewed from the top.
5.4 Surface flux and effective temperature 45
distance (colatitude) 9 from the pole, corresponding to a change dcp in
longitude and d9 in colatitude, is given by
da = dco = rp d9 dcp; with p = r sin 9 = sin 9. 5.15
For r = 1 we find
d<j = dco = r2 sin 9 dS dcp = sin 9 d# dcp. 5.16
With this we find
longitude cp. We can therefore easily integrate over cp, which yields a factor
2ft, and obtain
7 A (r A ,9)cosSd(cos9). 5.19
nFk(0)\ == 2TT
2n\ (ax + bk cos #) cos 9 d(cos 3) = (ax + |fcA)7c. 5.21
Jo
Comparing this with our source function approximation (5.8), we see that
F A (0) = S 1 (T 1 = | ) . 5.22
This is the important Eddington-Barbier relation. It is very helpful for the
understanding of the formation of stellar spectra. The flux which comes
out of the stellar surface equals the source function at the optical depth
xk = f. This is accurate to the degree that (5.8) is a good approximation.
We now again assume LTE, i.e., that Sx{xx) = Bx(xx). We then find
F A (0) = B A (TA = ) . 5.23
For simplicity, let us now further assume that the absorption coefficient
Kk is independent of X, i.e., that it has the same value for all wavelengths
X. We call such a (hypothetical) atmosphere a grey atmosphere. For a grey
atmosphere we find with KX = K and xk = x that
FX(0) = BX(T(T = D). 5.24
This means the energy distribution of Fk is that of a black body
corresponding to the temperature at the optical depth T = f. Then
cos#d(cos#) = 5.28
i.e., in an isotropic radiation field the radiative flux is zero (Fig. 5.8). In
the integral of (5.18) we have cos 9 as a factor which changes sign for
9>^n. This means the contribution of the radiation from 9>\n is
subtracted from the contribution for 9 < \n.
The flux integral (5.19) measures the anisotropy of the radiation field.
Of course, the flux is in principle a vector with a direction, depending on
the 9 = 0 direction, as can be seen from Fig. 5.9(a) and (b).
In the spherically symmetric case of the stars, however, we will generally
expect the flux to go radially outward, and therefore it has only a radial
component. So we will only talk about the radial component of the flux
and call that F or Fk. In the spherically symmetric star the other components
are zero.
Thefluxobviously describes the net amount of energyflowingoutwards.
Let us now ask whether the observations of stellar spectra give us the
intensity distribution /A(0, 5) or the flux Fx(0) as a function of L
Since we cannot see the different parts of the star separately (except for
the sun), we can only observe the total amount of light coming from all
parts of the visible hemisphere of a star.
The amount of radiation leaving 1 cm2 each second is given by
/A cos S Aa>. For a given angle 3, we get radiation from an area 2npR d&
with p = R sin S (compare Fig. 5.10).
Horizontal layer
Fig. 5.8. In an isotropic radiation field, the net radiative flux is zero.
(a) (b) FA
Horizontal Horizontal
layer layer
Fig. 5.9. The flux is a vector as may be seen by comparing figures (a) and (b).
48 5 Radiative transfer in stellar atmospheres
The solid angle Aco, out of which we receive the measured amount of
energy per cm 2 at the Earth, is determined by the distance of the Earth
and is given by Aco = 1 cm 2 /d 2 (see Fig. 5.11). The total amount of energy
received by 1 cm2 per second at the Earth is then
rn/2
Ek = Aco JA(0, S) cos S x 2nR2 sin 9 d9
nil
= 2nR2 Aco JA(0, 9) cos 9 sin 9 d9. 5.29
(O,0)cos0A
observer
Arr- 1
R60
Fig. 5.10. Stellar observations give us the sum of the intensities leaving the stellar
surface under all angles of 3 between 0 and ^n. The dashed region indicates the area with
between 3 and S + d.
cm' =
Fig. 5.11. The solid angle Aco out of which we receive radiation from the star with
a receiver of area a is given by a/d2. This is the area that Aco would cut out of a
sphere with radius 1. For dR, Aco is the same for all points on the stellar surface.
5.6 Radiation density 49
2
the mean intensity coming from 1 cm of this disc as
F Cn/2
x
Ix = =2 /A(0, 9) cos 9 sin 9 d9 = FA(0). 5.31
AconR2 Jo
FA(0) is the mean intensity radiated from the stellar disc.
The spectra of stars give us the energy distribution of the flux Fx. While
the flux by definition describes the energy coming from 1 cm 2 and going
into all angles 9 and cp, we see from the star's hemisphere the radiation
for different 9 and cp coming from different areas of the stellar surface. The
net effect is the same as if we could see all directions from the same spot.
^ , 5.32
Fig. 5.12. The intensities IX(S, cp) enter the volume element AV through a surface
element da. At any given moment the fraction Ik da cos 9 da> x (s/c) is to be found
in the volume of gas under consideration.
50 5 Radiative transfer in stellar atmospheres
to the normal n on da. The amount is
d = /A(#, cp) cos 9 da dco. 5.33
In each second this energy is spread out over a distance c. At any moment
a fraction s/c is actually in the volume AK, where s is the path length of the
beam through AV. At any given time the amount of energy in A V is therefore
given by
s
da>-. 5.34
c
With da cos 9 s = d F w e find after integration over dV for the radiation
density
ir ,(p) dw, 5.35
CJco
or
uA = - J A . 5.36
The depth dependence of the source
function
where the relation between AXi and aXi is known to be (see (5.11))
AXi = aXii\ 6.2
If we can measure AXi for a given wavelength A, i.e., if we measure the
center-to-limb variation of 7A(0, S), then we can determine the depth
dependence of the source function. We can do this for all wavelengths X
and obtain SA(TA) for all L
As an example, we have chosen the wavelength X = 5010 A. Measured
data for the center-to-limb variation can be approximated by (see the table
in Chapter 6, problem 1, page 233).
jA(0, 9) = [flo(A) + ax{X) cos 9 + 2a2(l) cos 2 S]/ A (0,0). 6.3
From this expression we find, according to (6.1), that
SA(tA(5010 A)) = [flo(A) + a1(A)tA(5010 A) + a2{l)T2x{5010 A)]/ A (0, 0). 6.4
In Fig. 6.1 we have plotted the source function S A (T A (5010 A)) as a
function of the optical depth according to equation (6.4). If we now go
one step further and assume again that SA = BA(T (T A )), then we know at
each optical depth TA how large the Planck function Bx has to be. Since
Bk depends only on the temperature T and on the wavelength X we have
one equation at each optical depth to determine the temperature T, which
will give the required value for the Planck function, i.e.,
2hc2 1
s,(tA) = B,(r(O) = e ^ r T , 6.5
where, in our example, the wavelength k is 5010 A.
51
52 6 Depth dependence of source function
From equation (6.5) and the values of SA(TA) derived from the
center-to-limb variation, the temperature Tcan be determined as a function
of the optical depth for the wavelength of 5010 A.
The optical depth dependence of the temperature obtained in this way
for the wavelength of 5010 A is shown in Fig. 6.2. As we expected, the
temperature increases with increasing optical depths.
1.0
r, (5010 A)
Fig. 6.1. The dependence of the source function 5A (5010 A) on the optical depth
TA for the wavelength of 5010 A. From TA > 1 we receive little radiation. The values
for deeper layers therefore become less certain as indicated by the dashed line.
6.2 Wavelength dependence of absorption coefficient 53
t and a particular temperature, we have different optical depths, r ; , for
different wavelengths A. We know, however, that at a given geometrical
depth t we must have one value for the temperature. All points with the
same temperature To must therefore refer to the same geometrical depth.
We draw a horizontal line through Fig. 6.3 which connects all the points
with the temperature To = 6300 K. All these points must belong to the
same geometrical depth t. We now write the optical depths rA in the form
Tk = Kkt. 6.6
Here KX is an average over depth down to depth t. For all the points on
the horizontal line we know that the value of t is the same. We read off
at the abscissa the optical depths TAI, T ;2 , T ;3 , etc., which belong to this
depth t. Equation (6.6) now tells us that
/(6300 K) = ^ = 2~ = ^ , 6.7
7000
6500
6000
5500
5000
1.0
T, (5010 A)
Fig. 6.2. The dependence of the solar temperature T on the optical depth at
A = 5010 A is shown as derived from the observed center-to-limb variation of the
solar intensity at / = 5010 A. For tA > 1 the values become less certain, as
indicated by the dashed line.
54 6 Depth dependence of source function
or
T T
^ =^ and ^ = ^ , 6.8
T K T K
A2 X2 Ai X2
which means we can determine the ratio of all Kk. with respect to one Kko.
For a given temperature T(t) we can also plot the values T A (T) as a
function of X. Because t is the same for all X, the wavelength dependence
of TA shows directly the wavelength dependence of KX9 as was first
derived by Chalonge and Kourganoff in 1946.
If KX. does depend on the geometrical depth t, we will find an average
value of Kk. down to the depth t. For different temperatures, i.e., for different
depths, these averages will be somewhat different. The results are shown
in Fig. 6.4.
It turns out that the wavelength dependence of Kk agrees with the
absorption coefficient which was calculated for the negative hydrogen ion
6500
T T 6300 ^ A = 8660 A
6000 -
/
5500
Fig. 6.3. The dependences of the temperature T on the optical depth TAI, T^2, TA3,
where kx = 5010 A, X2 = 3737 A, X3 = 8660 A. The horizontal line connects points
of equal temperature To = 6300 K. These points must all belong to the same
geometrical depth t. The corresponding optical depths for the different
wavelengths can be read off at the abscissa.
6.3 Radiative equilibrium 55
H" (see Fig. 6.4). Rupert Wildt first suggested in 1938 that H" might be
most important for the continuous absorption coefficient in the solar
photosphere. It is still very difficult to measure the absorption coefficient
for H~ in the laboratory because there are so few H~ ions. The solar
photosphere needs a depth of roughly 100 km to reach an optical depth of
1. In the laboratory we can use higher densities, but the required path
lengths are still very large. Therefore, it was very important when Chalonge
and Kourganoff (1946) showed that the wavelength dependence of KA could
be measured for the sun and that it did agree with the one calculated for H~.
We pointed out above that the energy transport in the star, which
has to bring the energy from the interior to the surface, requires that the
temperature decreases from the inside out, because a thermal energy flow
goes only in the direction of decreasing temperature. This discussion
indicates that the energy flux through the star is related to the temperature
gradient. Here we will show that we can actually calculate the temperature
gradient if we know how much energy is transported outwards at any given
point and by which transport mechanism.
+0.5
1
nF
4-
nF
nF
Fig. 6.5. The same amount of energy must be transported through each horizontal
layer if the temperature is to remain constant in time.
58 6 Depth dependence of source function
cases are very rare in stars, so these concepts are hardly ever used. The
scheme in Fig. 6.6 demonstrates the situation.
Generally, in the stellar photospheres we find that the condition of
radiative equilibrium is a very good approximation. In the solar
photosphere we see granulation showing that there is also energy transport
by mass motion, but the convective flux is of the order of 1% of the total
flux. If there is a temperature gradient there is also some conductive flux,
but again, this is much smaller than both the radiative energy flux and
the convective flux. When we know how large the temperature gradient is
in the solar photosphere and in other stellar atmospheres, we will come
back to the discussion of this question. In the following chapters we will
only be concerned with radiative equilibrium.
For radiative equilibrium we now consider the energy flux F = FT to be
radiative flux only. Equation (6.9) does not, of course, imply that the same
photons keep flowing across the different horizontal planes. There is much
absorption and re-emission, but the surplus of radiation in the outward
direction, i.e., the net flux F(T), has to remain constant.
In order to see what this means for the source function, we have to start
from the transfer equation (5.3) which we integrate over the whole solid
angle and obtain
Radiative equilibrium
1 Conductive equilibrium
Convective equilibrium
F Radiative ' 'convective ' ~ 'conductive
div f radiatlve = 0 diV f convec tive = 0 div F conductlve = 0
=
\'convective ** (Radiative 0 (Radiative = 0
'conductive ~ ^/ 'conductive ~ ^) 'convective ~ ^)
'radiative ^ '"convective "" '"conductive 'convective ^ '"radiative ' 'conductive conductive ^ radiative ' 'convective
/ A (t A ,S))dco. 6.11
(a)
1 !
1 t
I
t
i
1 i i
t t
1 1
1 i
i i_
+
t t t t
T 1
* _'l II.-*
! 11
*t t t t
t t 1 t t
1 | |
I 1 1 11 ii i
j i
Fig. 6.8. (a) illustrates the situation in an atmosphere with radiative equilibrium.
The source function must increase with depth if F has to remain constant, (b)
Here we plot the emissions into half-spheres for # < jn and # > \n. The integrals
over half-spheres have to grow by nF for each step AT = f.
6.4 Theoretical temperature stratification 63
(5.3) by cos 9 and integrate over all solid angles, a>, to obtain
-L cos 9 / A (T A ) dco
-L 6.19
= iF,(Tj). 6.20
drA
We know that the temperature gradient isfinallydetermined by the total
energy transport through the atmosphere, i.e., by nF. For radiative
(b)
t I
1 I I
1 t
1 1 ! t 1 t
! 1 I 11 t 1
1
J i J 1 1
J J I
t !1 t I t ! t1
1 1 i J1 I J
Fig. 6.(b)
64 6 Depth dependence of source function
equilibrium, we therefore have to introduce the equation for radiative
equilibrium, namely dF/dt = 0, into our equations. Since F = J^ Fk dA, we
have to integrate the equations over wavelengths k. However, in
equation (6.20) KX occurs in the denominator. If we want to integrate over
k we have to specify the wavelength dependence of KX. We have not yet
discussed this, except for the empirical numerical determination for the
sun; therefore, we will make the simplest specification; namely, assume
that K is independent of wavelength, which means we now make the
approximation of a grey atmosphere. Even if the real stellar atmospheres
are not grey, we may hope that our results are still approximately correct
if we determine an appropriate average value for KX.
If we are dealing with a grey atmosphere we can integrate equation (6.20)
over k and obtain
AK C
= {F, whereX-
Jo Kxdk. 6.21
d J
This introduces a new unknown function K(T). In order to find an equation
for dF/dx we now differentiate with respect to T, which, after integrating
the transfer equation over k, yields
^^LJ(t)_s(t)=o,
dx2 4 di
because we are dealing with radiative equilibrium, for which dF/dx = 0.
We then find again that for a grey atmosphere S(x) = J(x). This is our first
important result. Integration of the equation with respect to x gives
co = 4-n
ff An
2
Sd ... x), 6.24
J (o
<o = 4-n ^
or, after division by An,
Kk = %Jk and K=^J, 6.25
when we integrate over / . This approximation for the K function is called
the Eddington approximation. It has received wide application and is
therefore very important.
Inserting the relation (6.25) into equation (6.21) we find
I I I I I I I I I I SX\ I
6000 -
5500 -
5000 -
4500
Excitation
energy
= 13.6eV
Fig. 7.1. Schematic energy level diagram for the hydrogen atom. Transitions
between two discrete bound levels lead to line absorption or emission.
68
7.1 Different absorption processes for hydrogen 69
The difference between the first excited level (n = 2) and the ionization
limit is, according to the Bohr model, 13.6/4 eV. The difference to the next
higher one (n = 3) is 13.6/9 ^ 1.5 eV. This yields for the excitation energy,
i.e., the energy difference to the ground level,
For the hydrogen atom, all orbitals of a given main quantum number
n but with different quantum numbers for angular momentum have the
same energy. This means we can realize the same energy by several different
sets of quantum numbers. The number of the different possibilities that
will lead to the same energy level is called the statistical weight for that
particular energy level. For a given n this turns out to be 2n2. The statistical
weight for each level n in the hydrogen atom is therefore 2n2. We can
realize a state with principal quantum number n in 2n2 different ways.
Since the Pauli principle does not permit two electrons with equal quantum
numbers in one quantum cell, i.e., for one n, we find that we can have 2n2
electrons at each level n. This, of course, also applies to atoms with larger
nuclear charges Z. This explains the shell structure of the atoms, which
leads to noble gas configurations if there are 2n2 electrons in the outer
shell. For atoms or ions with more than one electron, electron states with
the same n but with different other quantum numbers usually have different
energy values.
A transition of an electron from one bound energy level to another leads
to the emission or absorption of a photon, depending on whether the
final energy level has a lower or higher energy than the starting level. The
frequency of the emitted or absorbed photon is given by
7 2
hv = Xm-Xi> -
s s
if Xm i the energy of the higher energy level and Xi * that of the lower
energy level involved in the transition. Since this gives a definite value for
the frequency or the wavelength X = c/v, a transition between two bound
states can only lead to the emission or absorption of light of one particular
wavelength, i.e., to line emission or absorption. We will talk about this
later. We are presently concerned with the absorption between lines, i.e.,
an absorption mechanism that involves a continuum of frequencies. This
is only possible if at least at one end of the transition there is a continuum
of energy levels. We know that a free electron can have a continuous set
of velocity values. If the transition begins or ends at a free state of the
electron (i.e., an energy above the ionization limit) we can have a continuum
of transition energies, which means a continuum of absorbed and emitted
70 7 The continuous absorption coefficient
frequencies or wavelengths (see Fig. 7.2). We have basically two possibilities
for continuous absorption. One possibility is a transition from a bound
state at level n, to a free state with a velocity v. These transitions are called
bound-free transitions. The frequency of the absorbed photon is given by
We can also get a continuum of transitions if the electron goes from one
free state with velocity vx to another free state with velocity v2. The absorbed
frequency for these free-free transitions is given by
hv = \mv\ \mv\. 7.4
There are an infinite number of free-free transitions that will absorb the
same frequency for different starting energies \mv\. This holds for every
frequency. For the calculation of the free-free absorption coefficient of a
given atom or ion an infinite number of transition probabilities has to be
computed at each frequency. For each discrete energy level n there is only
one bound-free transition which leads to an absorption of frequency v in
the continuum. However, for each frequency the transition probability has
to be computed for each level n.
Each bound-free transition corresponds to an ionization process, since
the electron is free afterwards. The emission of a photon by a free-bound
transition corresponds to a recombination process.
For the hydrogen atom, the continuous absorption coefficient for one
hydrogen atom at level n was computed to be
1 64;r4 1 Z W
3
ch6n3
\mv\-\mv]
= X3-X2
Excitation
energy
Fig. 7.2. A continuum of absorption processes can occur, if the electron transfers
from a bound state into a free state, or if it goes from one free state into another
one. We call these transitions bound-free or free-free transitions.
7.1 Different absorption processes for hydrogen 71
where G is a correction factor, the so-called Gaunt factor, which does not
deviate from 1 by more than a few percent. Here m and e are electronic
mass and charge. Ze is the effective nuclear charge attracting the absorbing
electron.
In order to compute the total absorption at frequency v, we have to
multiply an by the number of atoms in the considered state with quantum
number n and then sum up over all states n that contribute at the given
frequency v. Finally, we have to add the free-free contributions.
Which states n contribute at a given frequency? From Fig. 7.2 it is
obvious that the minimum photon frequency necessary for a transition of
an electron into the continuum - or for an ionization process - from level
n is given by Xion-Xn- For the hydrogen atom xn = * ion (l - (1/n2)). For
continuous absorption from level n we therefore need a photon with an
energy large enough to overcome this energy difference, i.e.,
he
hv>Xion-Xn or X< . 7.6
vAion An)
For very long wavelengths only energy levels with very large %n can
contribute to the continuous absorption. For very large X mainly free-free
transitions will contribute. The contribution of level n will start at the
frequency vn = (xion - xn)/h or Xn = hc/(xion - Xn) and will continue for
shorter X. At these frequencies the KX increases discontinuously because the
number of absorbing atoms suddenly increases. The wavelength
dependence of the continuous K for the hydrogen atom is shown in Fig.
7.3. In Fig. 7.4 we show the qualitative run of the He + continuous
absorption coefficient. For this ion the ionization energy is larger by a
factor of Z 2 = 4 than that of the hydrogen atom. All excitation
energies are larger by the same factor, and all discontinuities occur at
wavelengths shorter by that same factor of 4. In both plots the quantum
mechanical correction factors - the so-called Gaunt factors - were omitted.
They change the K by a few percent only. In accurate calculations they
have to be included.
We see in Fig. 7.3 that in the visual spectral region it is mainly the
Paschen continuum, i.e., absorption from the level n = 3, which determines
the hydrogen absorption coefficient. In the ultraviolet it is the Balmer
continuum, i.e., absorption from the level n 2. The discontinuity at
X = 3647 A is determined by the ratio of the hydrogen atoms in the second
and third quantum levels.
Another interesting fact is seen in Figs. 7.3 and 7.4; namely, the
absorption coefficient per atom at the Lyman limit always has
approximately the value of the cross-section of the lowest orbital in the
72 7 The continuous absorption coefficient
atom, which for the hydrogen atom is 0.5 x 10~ 8 cm. The cross-section a
is therefore cr ~ 0.785 x 10~16 or log a = 0.895 17, which is nearly equal to
log AT; at the Lyman limit as seen in Fig. 7.3. For the He + ion the radius of
the orbital is smaller by the factor Z. The cross-section is then smaller by a
factor of 4, as is the case for K} at the Lyman limit for the He + ion.
log
0-0.2
-18
-19
-20
H
-21
-22
-23 -
-24 -
-25 -
-26 -
-27
0 = 1.0
-28
A? = 3
I ii i i
2.5 3.0 3.5 4.0
log;. [A]
Fig. 7.3. The hydrogen absorption coefficient KX per hydrogen atom is shown as a
function of wavelength for two temperatures 0 = 5040/T = 0.2 and for
0 = 5040/T = 1.0. Higher temperatures lead to higher values of KX in the visual
spectral region.
7.2 Boltzmann formula 73
Boltzmann showed that the probability of finding an atom or electron
in an excited state with energy %n decreases exponentially with increasing
Xn but increases with increasing temperature T, i.e.,
n 9n a~XnlkT 7 7
where %n is the energy difference between the level n and the ground level,
gn and gx are the statistical weights for level n and the ground state,
respectively. Equation (7.7) is called the Boltzmann formula; k is the
Boltzmann constant, k = 1.38 x 10~ 16 erg degree" 1 .
The Boltzmann formula in equation (7.7) is only a special form of the
general case that the number of particles with an energy E is always
proportional to e~E/kT (see, for instance, the Maxwell distribution for the
kinetic energies of particles).
-17,
log/c/He+
-18 H:
= 0.1
-22
-23
-24
5040
1.160 x 10 4 x 0.4343= - ^ ( e V ) . 7.8
Here m is the mass of the electron, Nf is the number of ions in their ground
state, and Nx is the number of atoms in the ground state. g\ is the statistical
weight of the ground state of the ion.
Similarly, we can write down the Saha equation for the next higher state
of ionization, if an additional electron is removed from the ion and an ion
with two positive charges is formed. In this case we have to write
Nt +ne _ gi + 2(2nmkf>2 iL/kT
7.3 Saha equation 75
+
where gf is the statistical weight of the ground state of the twice ionized
ion and Nf + is the number of twice ionized ions in the ground state. x^m
is the amount of energy necessary to remove the additional electron from
the ion with one positive charge. It does not include the energy necessary
to remove the first electron.
If we want the ratio of the total number of ions N+ to the total number
of atoms iV, then we have to calculate N/N1 and N+/Nf. We can do this
by using equation (7.7). Clearly, N is the sum of all the particles in the
different quantum states, Nn9 and similarly, N+ is the sum over all the ions
in their different quantum states, N*. With
0i
and
iV+=iVr+ N:=Nt+^l g^-^i", 7.11
n=2 Q\ n=2
we find
N+=^(gt+l9:^lkT) = ^u\ or ?- = "-, 7.12
01 V .-2 / 01 ^1 01
where
JV+ N
Multiplication of (7.9) by - x yields
N u h5
This is the well-known Saha equation. It is the equivalent of the mass
reaction equation used in chemistry to compute the concentrations of
different reacting particles.
For the next higher degree of ionization, we obtain
N+ ~ u+ h3 ' b
log N H(" = 3) = 0.95 - 12.10 = 0.95 - 12.1 x 0.84 - 0.95 - 10.16 = -9.21.
NH(n=l)
NH(n=l)
Since
__ ) NH(n =
JVH(n = 3) NH(n=l)NH(n =
we find
N(H~) 3xlO"8
As the atomic absorption coefficients per absorbing atom are generally
of the same order of magnitude, we expect the H" absorption to be roughly
100 times more important than the Paschen continuum.
On the other hand, if we compute the number of atoms in the second
quantum level, we find
7
1(
78 7 The continuous absorption coefficient
There are 10 times as many H~ ions as hydrogen atoms in the second
quantum level, but the number of hydrogen atoms in the second quantum
level is not completely negligible. At the Balmer limit, i.e., at ?3647 A, the
continuous K in the sun is expected to increase by about 10%. There is a
small discontinuity observed in the continuous *c, as seen in Fig. 7.5.
What is the wavelength dependence of the H~ absorption coefficient?
We have one bound-free continuum from the ground level of H ". There
are no excited levels in H~. The bound-free continuous absorption can
take place for all k shorter than that corresponding to hv = xion = 0.7 eV.
This corresponds to a wavelength of 17 000 A. For longer wavelengths, we
can only have free-free absorption, which goes roughly ~ v ~ 3 .
The bound-free K ( H ~ ) shows a different behavior from /c(H). It first
increases when going to shorter wavelengths, has a maximum around
8000 A, and then decreases towards the violet. The sum of the free-free
and the bound-free absorption shows a minimum around 16 000 to
-24.0
= 0.9
= 1.0
bound-free H
log KA
log KR -
-24.5
free-free
H
-2E
J.5 4.0 4.4
log/*
Fig. 7.5. The continuous absorption coefficient per heavy particle is shown for 0 = 0.9,
which means T=5600, and log Pe= 1.0. The contributions from the bound-free
and free-free continua of H" are dominant. The small contribution from the
Balmer continuum of hydrogen is visible at log X = 3.562. kR is the Rosseland
mean absorption coefficient; see section 8.4.
7.6 Metallic absorption in the sun 79
16 500 A, depending on the temperature and the relative importance of the
free-free and bound-free absorption.
We have seen earlier how we can test the frequency dependence of the
absorption coefficient in the sun by means of the center-to-limb variation
of the intensities in different frequencies (see Section 6.2). We can now see
that the measured and the theoretical wavelength dependences do indeed
agree quite well.
Continuum
\hv ^ h c Ay ^ 3.1 eV
= =4000 A
Fig. 7.6. Schematic energy level diagram for iron. All iron atoms with excitation
energies / > 4.8 eV can contribute to the bound-free transitions for k < 4000 A.
7.7 Scattering by atoms and ions 81
Even more important is the absorption by the metal atoms in the ground
level which for the iron atoms occurs for wavelengths shorter than 1570 A,
and for the Si atoms with an ionization energy of 8.15 eV at a wavelength
of 1520 A. We find a very pronounced discontinuity in the overall
absorption coefficient at these wavelengths which leads to very pronounced
discontinuities in the observed energy distributions of A, F and G stars,
as we shall soon see.
Ze+
Fig. 7.7. The electron bound to the nucleus of an atom will start oscillating in the
electric field E of the lightwave with intensity Imo. This oscillation will cause it to
radiate into all directions. The emission JM has the frequency a> of the forced
oscillation, which means the emitted light has the same frequency as the original
light beam. The light is scattered, but mainly in the directions perpendicular to the
direction of oscillation.
82 7 The continuous absorption coefficient
waves, which means that in the presence of light the electron bound to the
atom experiences an oscillating electric field which excites the bound
electron to oscillations. The amplitude of these oscillations is also
proportional to (co2 co2)'1. The electron in the atom has many
eigenfrequencies cor, namely, all the Lyman lines in the case of the hydrogen
atom. All the resonance frequencies contribute to the amplitude of
oscillation. The probability of each resonance frequency contributing is
given by the so-called oscillator strength / r , describing the probability for
that particular wavelength to be absorbed or emitted by the atom. This
oscillator strength fr is, of course, the same quantity which also determines
the relative strengths of the Lyman lines.
If an electron is oscillating, it is accelerated. Each accelerated charge
radiates; therefore, the oscillating electron also radiates. The radiated
intensity is proportional to the square of the acceleration, which is 5c if the
displacement is called x. We describe the oscillating displacement x by
x = A cos cot, where co = 2nv,
then
x= coA sin cot
which gives
x= -co2A cos cot. 7.18
The radiated intensity is then proportional to
x2 = x2co4. 7.19
If damping is neglected, the emission corresponding to each eigenfrequency
cor turns out to be
Emission = / w oc . 7.20
2 2 2
(co -co )
As we said earlier, for the neutral hydrogen atom the resonance frequencies
correspond to the Lyman lines which are in the far ultraviolet. For the
optical region we therefore find corco such that the scattering coefficient,
also called the Rayleigh scattering coefficient, crR, is
CO?
with the definition
where C 7 22
-7 = Z A ' = ( ~y ) ' *
2
cbz r coz 3 Vmc /
7.8 Thomson scattering by free electrons 83
and m and e are again electron mass and charge. Since for the hydrogen
atom Yafr= 1 f r a ^ transitions from the ground level, one finds that
cbr = co0 differs only little from co for Lyman a; it actually corresponds to
1026 A. The Lyman a line has a wavelength of 1215 A.
For the Rayleigh scattering coefficient <rR the final expression is
) = C ( ) oc , 7.23
ffei = T 7.25
3 \mczj
per electron, or
<rel = 0.66x 10" 2 4 cm 2 .
This is the Thomson scattering coefficient. The Rayleigh scattering coefficient
can then be written as
)
Quantum mechanical calculations give slightly lower values, especially if
the frequencies are close to the resonance frequencies. We should have
considered the damping due to the radiation.
84 7 The continuous absorption coefficient
-19
-H Lyman
limit
-20
-21
Si
-22
Mg's
81
-23
Mg3P
-2L
H~ bound -free
-26 In I I I
1000 10000 A 100000
Fig. 7.8. The wavelength dependence of the continuous absorption coefficient per
particle is shown for Teff = 5040 K ( 0 = 1.0) and log Pe = 0.5.
7.9 Absorption coefficients for A and B stars 85
determined empirically by Chalonge and Kourganoff (1946) for the solar
atmosphere.
Fig. 7.9. The continuous absorption coefficient per particle is shown as a function
of wavelength for a temperature of 28 300 K and log Pe = 3.5 as appropriate for a
main sequence B0 star. (From Unsold, 1977, p. 159.)
86 7 The continuous absorption coefficient
atoms in the level with n = 3, i.e. in the Paschen level, we find
.-?!. 8.3
3K
where k is the average absorption coefficient, averaged over wavelengths.
TA(3647 ) =I
rA (3647+) = I KA <
Fig. 8.1. For wavelengths with large KA we see high layers with low temperatures.
For wavelengths with small Kk we see deep layers with high T. f is the optical
depth corresponding to k.
87
88 8 Non-greyness of the absorption coefficient
Thus, at this wavelength of 3647+ A, we obtain radiation from a deeper
than average geometrical depth. At the deeper layer the source function
and the temperature are larger and so we get more radiation at this
wavelength with a small KX than we would have received in a grey
atmosphere with K = K.
On the other hand, at X = 3647" A, i.e., at the short wavelength side of
the Balmer discontinuity, where Kk is larger than /c, we receive radiation
from a higher level, where T and SA are smaller than at f = , and we
receive less radiation than for a grey atmosphere with K = K. Here f is the
optical depth corresponding to K.
We show the relation between the wavelength dependence of KX and Fx
in Fig. 8.2. The distribution of the flux Fk looks like a mirror image of the
wavelength distribution of KX. We find discontinuities in the continuum
energy distribution at those wavelengths where Kk has a discontinuity.
log*,
N on-grey energy
distribution
Grey case
log A
Fig. 8.2. Schematic diagram showing the influence of the non-greyness of the
hydrogen absorption coefficient on the observed energy distribution of the star.
8.2 Dependence on temperature and electron density 89
where ne is the number of free electrons per cm3, also called the electron
density, and f(T) is a slowly varying function of T. We find then that
K(3647 + ) KQT)NH(n = l)nJLT)
8.5
K(3647 (n = \)nJ{T) + K~(H)NH(n = 2)'
H (Balmer continuum)
0 = 0.8
Fig. 8.3. The continuous absorption coefficients K per gram of hydrogen and the
negative hydrogen ion H ~ for a temperature of 6300 K and an electron pressure of
log Pe = 0.5. For A < 3647 A, K is mainly determined by neutral hydrogen, while
for A > 3647 A, K is determined by H " .
90 8 Non-grey ness of the absorption coefficient
Generally, the discontinuity is a measure of the temperature and the
electron density, and can be used to determine one of these quantities.
For F stars the contribution of H~ to the absorption on the short
wavelength side of 2 = 3647 A is very small. In this case, equation (8.5)
reduces to
K(3647 H(n = \)nef(T) ^ NH(n=l)
8.6
ic(3647") K-(H)NH(n = 2) NH(n = 2) e'
For cool stars the Balmer discontinuity depends on both the temperature
and the electron density. If we can determine the temperature from another
observation we can determine the electron density from this discontinuity.
With increasing electron density the discontinuity becomes smaller because
the H~ absorption increases when more H~ ions are formed for higher
electron densities. In Fig. 8.4 we compare the energy distributions of an
7.5
6.5
7"eff = 8 0 0 0 K
6.0
3.5 3.6 3.7 3.8 3.9 4.0
log A [A]
-16000 K
0.5
1.0
-0.3 0.5 1.0
B-V
Fig. 8.5. The two-color diagram is shown for main sequence stars and for black
bodies. The solid arrows show qualitatively how the colors of grey stars (blaek
bodies) change due to the Balmer discontinuity. The dashed arrows indicate how
the line absorption changes the colors qualitatively. For hot stars the Balmer
discontinuity is more important than the line absorption. For cool stars the line
absorption becomes very important, while the Balmer discontinuity disappears.
94 8 Non-greyness of the absorption coefficient
This should then be equal to what is obtained in the grey case:
f F x d X = | B A ( T = 8.10
Jo Jo
As can be seen from Fig. 8.6, we can describe Bx(xk = ^) by a Taylor
expansion around the layer f = f. One then obtains
f = 2 l = 2 + Ai, 8.13
3KA 3
according to the definition of Af (see Fig. 8.6). From this equation we derive
3K, 3 A )
The condition (8.9) for Fk then requires
f 00 f0O p
B 4 (t 4 = f)dA= Ba(f = | ) d A +
Jo Jo Jo
f
BA(f = )dA. 8.15
Joo
For this to hold, we must require
^lWo or p ^ l c U - p ^ d i . 8.16
df 3 \ K A / Jo df KX J o dr
Ar, = i) T, = * , T = * - .
Af < 0
Fig. 8.6 demonstrates that BA(tA = f) = B,,(f = ) + - Af(TA = f), with Af < 0.
8.4 Influence of non-greyness on temperature stratification 95
or, after dividing by ic,
=
j 4*/l *X ~ ^A9 8.18
dtA
or, if we multiply first by cos 9 and then integrate over w and divide by n
we find with the Eddington approximation Kx = \Jk (which holds for an
isotropic radiation field) that
4.rx , s.iy
di, 3 dtA
for a deeper layer, where JA BA. Or, rewriting the second half of the
equation, we find
FA = 4 d ^ . 8.20
3dr A
This is a very important relation between Fx and the gradient of the source
function.
From the condition that
8.21
Jo
we now derive
3 f _ .. f ^ d B , .. [wdB, .. . . , dB
8.22
o '" Jo dxx Jo df dt
This yields
\ . 8.23
df
0 K KJ df
which is the same prescription as above. Using again
dBA dB AdT _, dB dB dT
2 and =
df dT df df dT df
96 8 Non-greyness of the absorption coefficient
we derive again
r oo 1
8.24
Jo dT
where G'k{T) is the Rosseland weighting function and a = hv/kT. From this
definition of G(a) we find
8.25
G ( o = , .a
ft c dB/dT 47i 4 (e -l) 2
This kind of mean absorption coefficient is called the Rosseland mean
absorption coefficient. It is a harmonic mean. Frequencies with small KX
are weighted most heavily, because the Rosseland mean k is a flux weighted
mean and the flux is largest in the regions of small KX, as we see from
equation (8.20) if we write it in the form
Fx = -. 8.26
KX dt 3
Because of this, the Rosseland mean k is always close to the smallest values
of KX. See Figs. 7.5 and 8.3, where we have indicated the Rosseland mean
values kR.
With the prescription (8.24) we can calculate mean values of k for all
desired temperatures and electron pressures.
For a given temperature and electron pressure we can calculate the
degree of ionization for all the atoms and ions present in stellar atmospheres.
With this we can add up all the numbers of particles to obtain the total
number of particles per cm3 from which the gas pressure can be obtained.
We can thus also determine k as a function of temperature and gas pressure.
In Fig. 8.7 we reproduce a plot showing icasa function of the gas pressure
for different values of the temperature. The values of & = 5040/T are given
on each curve. For low temperatures the continuous K is essentially due to
H" only. For a given temperature the average k increases proportionally
to the electron pressure. For very low pressures the electron scattering
becomes important and prevents a further decrease of k. At temperatures
above 6000 K hydrogen absorption becomes important for very low gas
pressures. At temperatures above 10 000 K hydrogen absorption is
important for all pressures considered here and gives a large increase in
k. At still higher temperatures hydrogen ionizes and k does not increase
any more, but decreases with increasing T. Finally, electron scattering takes
over.
8.4 Influence of non-greyness on temperature stratification 97
(I) (II)
Fig. 8.7. The Rosseland mean values KR for the continuous absorption coefficient
are shown as a function of the gas pressure. Each curve corresponds to a given
value of the temperature. The value of 0 = 5040/T is given for each curve. The
plot is double logarithmic. What is shown is actually the absorption coefficient per
gram of material, i.e., for a column of gas which contains one gram of material, or
per heavy particle, as shown on the outer left-hand scale. Solar element abundances
were assumed.
98 8 Non-greyness of the absorption coefficient
8.4.2 Changes in the temperature stratification T(z) due to the
non-greyness
By computing K in such a way as to make the integrated flux the
same for the grey and for the non-grey case, we hoped that the temperature
stratification
r 4 = f7i ff (f R + g(fR)), 8.27
with fR being the Rosseland mean value of TA, i.e., with
fR=
RRdt, 8.28
Jo
will still give a reasonable approximation. This is true qualitatively. Some
important modifications, however, have to be taken into account.
We can see what happens if we look at the condition for radiative
equilibrium
r oo r oo p oo
KxJxdX=\ /cASAd>l- = ^kBkdX forLTE. 8.29
Jo Jo Jo
If Kk varies strongly with 2, then these integrals are mainly determined by
the wavelengths with large Kk. What happens to Jk when Kk becomes very
large as compared to the grey case? Let us look at the surface first, where
/A(cos S) ~ Bk(xk = cos $). With the large KA, the Ik now comes from higher
layers and therefore is smaller. So Jk will also become smaller, but we must
require J^ KkJk dk = \Q KkBk dL If Jk becomes smaller, Bk must also
become smaller at the surface, which means the temperature must decrease
at the surface.
Therefore, generally the first effect of the non-greyness is a decrease in
surface temperature; this is also called surface cooling and is due to the larger
emission because of the larger KX (though under special circumstances this
may not be the case as for instance for strong lines in the red).
The second effect is the backw arming. We now have essentially many
strong lines in the spectrum. In the regions with large Kk the intensity Ik
and the flux Fk decrease, which means the total flux coming out of an
atmosphere with the same temperature stratification is decreased. This
means the same temperature stratification now leads to a smaller effective
temperature. If we want to obtain the same total flux, we have to get more
flux out of the wavelengths with the small Kk. If we have mainly many
lines the flux has to be obtained from the wavelengths between the lines.
This can only be done by increasing the temperature in the atmosphere.
Some people call this redistribution of energy: the energy absorbed at
some wavelengths comes out at other wavelengths. But remember, this can
8.4 Influence of non-greyness on temperature stratification 99
only be done by way of increasing the temperature in deeper layers. The
energy distribution in the continuum now resembles the energy distribution
of a somewhat hotter star, if the non-greyness is due mainly to line
absorption. The star cannot absorb the energy at one wavelength and
arbitrarily emit it at another wavelength, as we please.
A modified temperature distribution of the form
r* = i r ( f + 4(f)) 8.30
leads to a flux
j;
iff/ is the fraction of energy absorbed by the lines. If we require nF =
8.31
A (grey)
(grey)
fo-1 T0 7 0 +1 *f
F/^. 8.8. In wavelengths with large KX the ingoing average intensity 31 is increased
by AJ~ because the radiation originates at a deeper point with a larger Planck
function than in the grey case. The outgoing radiation originates at a higher layer
with a smaller Planck function than in the grey case. The average intensity Jl is
reduced by the amount AJ + as compared to the grey case. If the Planck function
is a linear function of depth tsJ\ = -AJk and for deep layers the sum J* + J~ remains
the same as in the grey case.
100 8 Non-greyness of the absorption coefficient
with a temperature increased by about 200 degrees. With an effective
temperature of 5800 K for the sun the energy distribution in the continuum
looks similar to that of a black body of 6000 K, as we saw earlier in
Volume 1.
What happens to the Jk in the deep layers?
For wavelengths with small KX the outgoing average intensity Jf is
increased because it now originates in deeper layers than in the grey case,
but the inward going intensity now originates at a higher layer with a
smaller Planck function. The sum of the ingoing and outgoing intensities
remains the same as long as we are far enough away from the surface such
that the Planck function is still a linear function of depth. See Fig. 8.8 for
the case of large KX.
The pressure stratification
(b)
mg
Fig. 9.1. (a) The net pressure force on a given volume of gas is the difference
between the pressure force pushing up from the bottom and the one pushing down
from the top. (b) The gravitational force equals the weight of the volume element
under consideration. In an atmosphere pressure forces and gravitational forces must balance.
101
102 9 Pressure stratification
we consider a volume element of vertical extent At. From the top the gas
pressure Pg(ti) presses the element down; from the bottom the gas pressure
Pg(ti + At) tries to push up the volume of gas considered. The net pressure
force on the volume element under consideration is then
Pg(t1) = Pg(tl) + d^At-Pg(ti) = d^At. 9.1
dt kcm kcm/p
kcm stands for k per cm 3, which has the dimension cm" 1
Here we need to pause and consider the units of the absorption coefficient
K. We introduced K when we considered the change of the intensity / along
a path of length As; we wrote A//As= KI. The dimension of K must
therefore be cm" 1 . It describes the fractional change of intensity per cm.
We note that the dimension and the value of K does not depend on whether
we use the intensity / v or Ik because K gives the fractional change.
Kcm = K [cm" 1 ]* obviously depends on the number of particles in 1 cm 3,
which changes with increasing density. It is often much better to use a
quantity which is less density dependent. We therefore often describe the
change of intensity which has occurred after the beam of light has passed
through a column of gas with a cross-section of 1 cm2 and which contains
1 gram of material. If no change in the degree of ionization occurs, then
the number of particles in 1 gram of material is always the same for a given
chemical composition. The K for this amount of material therefore depends
much less on temperature and pressure than does the Kcm. How long is the
column of gas which contains 1 gram of material? If the density is p then
the length s of the column has to be s = \/p. Clearly, K per gram = /cgr = Kcm s,
where s is the length of the column containing 1 gram of material. The
fractional change along this column is therefore Kcm sor/c per gram = K per
cm3 divided by p. We write this as
Kgr = Kcm/p. 9.8
Kgr has the dimension cm2 g"1, which is that of a cross-section. Many
astronomers, especially those that discuss stellar interiors, mean /cgr if they
just write k.
Now we can go back to equation (9.7). On the right-hand side we have
in the denominator the ratio K^/p, which can be replaced by /cgr and obtain
i _ JL 99
df kgr'
The gas pressure Pg(x) can be obtained by integrating this differential
equation.
9.2 Integration of the hydrostatic equilibrium equation
In a formal way we can integrate the hydrostatic equilibrium
equation by writing
* Kem = K per particle [cm 2 ] x N [cm" 3 ] = K per cm 3 , where N = number of particles per
104 9 Pressure stratification
1
" df, Pgo = Pg(ro). 9.10
/to *gr
Here /cgr is a complicated function of temperature and pressure (see Fig.
8.7). We would have to know the pressure in order to calculate the
right-hand side. In general, the integral can be evaluated by step wise
integration, as we will soon discuss, but let us first use some simple
approximations for kgT as a function of optical depth or of pressure. These
permit an analytical integration of the differential equation (9.9). The
simplest approximation is the assumption kgT = const. This is not as
unreasonable as it may seem at first sight, since the number N of particles
per gram of material is constant for a given chemical composition provided
the degree of ionization does not change. It is
N--L.
fimH
Of course, as we saw earlier, even for the same number of particles k may
vary greatly by changing degrees of excitation. So kgT = const, is not a good
approximation, but let us see what we obtain.
Integration of equation (9.9) yields
9.11
K
gr
P = f. 9.12
kgT
Knowing T(f) for a given Teff, we can assume a value for /cgr, insert this
into equation (9.12), and compute a value for the gas pressure. With this
first approximation to the gas pressure a value of kgr can now be calculated.
Inserting this into equation (9.12) gives a better approximation to the gas
pressure. With this we can calculate a better value for kgT and so on until
we achieve convergence, which means until kgT and Pg are consistent. Of
course, this does not mean that we have the correct gas pressure at the
depth f, since equation (9.12) is based on the erroneous assumption that
kgx is constant with depth. Nevertheless, this method is used in all modern
computer programs to find a starting value Pgo in equation (9.10). If we
were to follow the iteration procedure as described, it would take rather
long before convergence is achieved. There are various ways to speed up
the convergence, but the basic method remains the same.
Because the value Pgo obtained in this way is not the correct one, we
have to make sure that Pgo refers to a very high layer in the atmosphere
9.3 Dependence of gas pressure on gravitational acceleration 105
such that PgoPg(z) for any of the layers which are of interest to us.
Numerical integrations are usually started at an optical depth f = 10 ~ 4 or
f=10"5.
Once we have a starting value Pgo(To) for the integration we can continue
stepwise. Knowing T0(f0) and Pgo, we can, for this depth, calculate
/cgr 0(T0, Pgo). Assuming that /cgr remains constant over the next small
interval AT, we find for T1 = TQ-\- AT
W ) = ^O + ^ - ( T I - T O ) . 9.13
At fx we know T(f x) and now also Pgi = P^f x). We can therefore determine
/cgrtl = Jcgr(f x). In order to improve the value for Pg(Tx) we can now use the
average value of /cgr in the interval between f0 and fx and insert that into
equation (9.13) to obtain a better value for Pgi. There are more sophisticated
methods to get the best possible numerical value for Pgi. Knowing Pgi and
^gr,i w e c a n n o w proceed in the same way to obtain P^(f2) = Pg2 and then
7cgr 2 = Kgr(f2), where T2 = T1-\- At. This value can then be improved in the
same way Pgi was. We can then proceed to the next f and so on. In this
way we can obtain very good numerical solutions of equation (9.9). There
are many methods to give more accurate solutions than the one described,
but the principle is the same. However, these numerical solutions do have
a disadvantage in that it is difficult to estimate how the solutions are going
to differ for different input parameters, such as the gravity. This can be
seen much better from analytical solutions.
't.f
Fig. 9.2. In a given atmosphere temperature and pressure are given as a function
of optical or of geometrical depth. For each depth we can then calculate Jcgr(T, Pg)
and plot kgr also as a function of geometrical or optical depth.
T{PQ)
T(Pg)
Fig. 93. For a given value of zt in Fig. 9.2 we read off T^ij) and /^,(TI), and
plot ^(P^j). For this same value of Pgx we determine /cgr(ii) from Fig. 9.2 and plot
/cgr as a function of P9. This is the correct value of /cgr, which was obtained using
T{Pg). We follow this procedure for all values of f and Pg. We thus obtain the
curve Kgr(Pg) shown, which gives the correct values of /cgr in the atmosphere.
9.4 Electron pressure 107
Integration yields
_J_P + i=-JLf. 9.I6
n+ 1 Kgr0
Here we have used the boundary condition Pgo = 0.
As long as H ~ is responsible for the continuous absorption, n = 1 is a
good approximation. We then find
g
or Pg = -^L,yfgyfx.
g
9.17
K
gr,0
Thus, wefindthat if in the atmosphere /cgr is approximately proportional
to the gas pressure, the gas pressure for a given f increases proportionally
to yjg. Since for different atmospheres we always see approximately the
same f, the pressure in that layer varies approximately as y/g, and is not
proportional to the gravity itself, as one might think at first when looking
at the hydrostatic equation. The larger the pressure, the greater is /cgr.
Therefore, we see geometrically higher layers in stars with higher gravity
(see Fig. 9.4). The net effect is the square root law.
If a larger value of n in equation (9.16) gives a better approximation (for
instance, if n = 2 is better), then we find an even smaller dependence of the
pressure on the gravity, namely Pgocgi/3.
(a) (b)
9 = 90 9 > 9o
f=0 f = 0
Fig. 9.4. (a) For an atmosphere with a gravitational acceleration g = g0 we see the
layer f = f in which the pressure is equal to Pgo. (b) In another atmosphere for
which the gravitational acceleration is larger than g0, the pressures of all depths
are increased. If K is proportional to the gas pressure, K also increases. We again
see the layer where f = , which, however, now corresponds to a smaller
geometrical depth, i.e., a higher layer. The pressure at this higher layer is increased
only by a factor given by the square root of the gravitational acceleration as
compared to the pressure in f = in the lower gravity atmosphere.
108 9 Pressure stratification
With the gas pressure and the temperature known, we can now obtain the
electron pressure which enters into the Saha equation. We can generally say
P=NkT, 9.18
if
3
with N = sum of all particles per cm .
Pe = nekT, 9.19
3
with ne = number of electrons per cm . Thus ne must equal the number of
ions + 2 times the number of twice ionized ions +
ne = N+ + 2N2+ + 3N3+ + '- + iNi+. 9.20
In the simple case of pure hydrogen, we have
N = ne + JV(H) + N{H + ) = N(U) + 2ne since ne = JV(H + ), 9.21
and for a given Pg we have
N = Pg/kT. 9.22
The Saha equation tells us that
/ m or ?V2-flT). 9.23
JV(H) N(H)
Equations (9.20) and (9.21) yield
N(H) + 2JV(H + ) = PJkT. 9.24
Equations (9.23) and (9.24) are two equations for the two unknowns
Af(H + ) = ne and N(H), which can be solved for given T and Pg.
If all the hydrogen is ionized, we find, of course, Pg Pe = \Pg. If we
were to have a pure helium atmosphere for which at very high temperatures
all the helium is twice ionized, we would find
N = N(He2 + ) + ne and ne = 2N(He 2 + ), 9.25
which gives
p.jm^)j 926
2 +
Pe 2N(HQ ) 2
Table 9.1. Gas pressures and electron pressures in stars at optical depths f = and f = 0.1
. = 0.1
pressure and then calculate the gas pressure, but when we integrate the
hydrostatic equilibrium equation we first obtain the gas pressure and then
have to calculate the electron pressure.
In Table 9.1 we give gas pressures and electron pressures for stars of
different temperatures and different gravitational accelerations for the
optical depths f = | , which we see in the continuum. We also give the
pressures for an optical depth f = 0.1, a typical depth for spectral line
formation.
9.29
g ^ NkT. 9.30
where turb is now the reference velocity for the turbulent velocity
component perpendicular to the hypothetical wall. For a Maxwellian
velocity distribution we find
T2 _ is2 Q i->
Sturb 2Sturb- y.DZ.
df dr dt
9.6 Effects of radiation pressure 111
For the change in gas pressure Pg we thus find
9 ' Q 1C
eff
dt dt '
where
1 dP l d
< ?2 o
cos 9 dv da>.
hv
The total momentum transfer per cm 2 each second by photons with
frequency v hitting the wall (or being absorbed) is therefore given by
-cos29dvdco. 9.37
= 2n C
We now have to add the momentum transfer due to the recoil by the
photons emitted or reflected by the wall (or the atoms). The effect of these
photons can easily be included by extending the integration in equation
(9.37) over the radiation coming back from the wall, which means over
the whole solid angle da>. This leads to
2
AM = c^ cos 5 sin 3 dS dtp. 9.38
Jo Jo
Jo Jo c
dt dt dt
and
dPg dPt dPr QA.
= gp-l-i = gef{p9 9.44
dt dt dt
with
idP, ldPr
p dt p dt
and
p at c
If we are not dealing with a grey atmosphere we cannot evaluate the
integral in equation (9.41). We are left with the expression
1 dP if00
0r = - - T = - *v, g r ^vdv. 9.47
p dt c J o
At large depth, i.e., TV 1, we find
*v 3 dtv 3 Kv>cm dt
and we have
Xcm4dBv nA c 0 J D
9.49
9r c
P J o ^ c m 3 dt cp 3 Jo dt cp3dt
with
An dB
3d( - , F, 9.50
i?A = t ^ = l - ^ . 10.1
Fc Fc
Clearly, Rx depends on the wavelength within the line. The total area in
the line divided by Fc (see the hatched area in Fig. 10.3) is called the
equivalent width Wk. The division by Fc means we are measuring the flux
in units of the continuum flux, which means the continuum flux becomes
1. We then have the situation shown in Fig. 10.3: the equivalent width
describes the width a rectangular spectral line must have in order to have
the same total absorption as the actual line. From this definition we find
\ ' ^ c U , 10.2
line J line *c
where the integral has to be extended over the whole line. In the following,
Surface T= 0
1 TA = (in line)
Fig. 10.1. In the wavelength of the line, where KX KC, we see much higher layers
with T=TL for which TL < Tc and therefore Fx = BX(TL) <FC = BX(TC\ where Tc is
the temperature of the layer seen in the continuum.
114
10.1 Formation of optically thin lines 115
TC refers to the optical depth in the continuum, while TA is the optical depth
in the line; this means tA = i line + TC.
For weak lines, i.e., if K l i n e Kcontinuum, the layer TA = will be close to
the layer with xc = ; we can make a Taylor expansion of BX(T ) around
the layer for which TC = \ and find
_ 2\ __ D _ 2\ (&Bx\ = |-Arc). 10.3
AV A. 3/ A\ C of I J I Arc =
T
c/T c =2/3
It is
or 7" 7-
L
C ~ Lk
10.4
^1 and
Fig. 10.2. A schematic plot of the energy distribution around a spectral line as a
function of wavelength X. At the wavelength of the line the flux Fk is smaller than
the flux in the continuum Fc. Rx is the line depth measured in units of the continuum flux.
Fig. 10.3. A schematic plot of the energy distribution in the neighborhood of the
line. The equivalent width Wx gives the width which a rectangular spectral line
must have in order to have the same total absorption as the actual line.
116 10 Theory of line formation
Making use of this, we see that
2 KL
ATC = + - if KLKC. 10.6
3 KC
If KL KC, the line is called optically thin.
With equation (10.6) we find for the line depth
R _Fc
Fc
dB>
dic
10.7
or
2 KL d In Bk
10.8
3 KC Axc Tc = 2/3
= 1 - (1 - 1 4 ) = T, = 10.9
= K L.cH.fr
T c =2/3 AC c,cm
Heff can be looked at as the effective height of the atmosphere with respect
to line absorption, or in other words, the stellar atmosphere acts as a
laboratory absorption tube of length ffeff.
Equation (10.8) is a very important relation. It tells us that the line depth
is not only determined by the size of the line absorption coefficient, but
also by the value for the continuous absorption coefficient and, last but
not least, by the gradient of the source function, which can usually be
replaced by the gradient of the Planck function. / / there is no temperature
gradient with the temperature decreasing outwards, then there are no
absorption lines in the spectrum.
If we want to study the line profile, i.e., the energy distribution within
the line, then we have to study the wavelength dependence of /cL, as is
obvious from equation (10.8).
10.2 Line absorption coefficient 117
KL a x -7 . 10.12
Ua
Aco + (y/2)2
2
This so-called damping profile, or Lorentzian profile, has a total half width
2Aa>(!) = y (see Fig. 10.4). The larger the amount of radiation, the larger
and the broader the line.
From the classical damping theory of the oscillating and radiating
electron one finds
f. 10.14
me
Here we have introduced the factor /, which is called the oscillator
strength because this factor indicates how many classical oscillators the
line under consideration corresponds to.
In the quantum-mechanical approach, the natural width of the line is
determined by the finite width of the two energy levels. According to the
Heisenberg uncertainty principle, the uncertainty of an energy measurement
AE is given by the time Ar which is available for the measurement, namely
A At = h = h/2n, h = Planck's constant. 10.15
The maximum time available for the measurement of the energy E is given
by the time which the electron stays in that energy level. The larger the
transition probability, the shorter the lifetime, the stronger the radiation
and the larger A, which means the larger the natural line width. It turns
out then that y is proportional to the transition probability. This means
that the natural line width is of the order
Aa>/a> = AA/A oc transition probability. 10.16
Actually, it is the sum over all the transition probabilities for the level
under consideration. When calculating the line width we obviously have
to take into account the widths of both levels which are involved in the
line transition.
Q - y/2 + y/2
Fig. 10.4. A schematic plot of the classical damping profile for an oscillating
electron with eigenfrequency o>0 and a damping constant y.
10.3 Doppler profile 119
In most astrophysical cases the line width is, however, determined not
by the radiation damping, but by the thermal motion of the emitting and
absorbing atoms.
A particle moving with away from the observer (see Fig. 10.5) emits light
at a wavelength co0 + Aco with
J- or ^-J- 10.18
(Do C CD0 C
for motion away from the observer for motion towards the observer
Only the component of the motion in the direction of the line of sight
has to be considered for the Doppler effect. Therefore, we have to consider
the Maxwell distribution for one velocity component only.
The number of emission or absorption processes at u> 4- co/c is then
given by the number of atoms moving with a velocity component towards
us. This means that the emission or absorption profile must be given by
Emission at /\ = Ao - /\ 0 or M = w 0 + - w0
Fig. 10.5. Thermal motions of amount of the light emitting particles leads to
emission at frequencies OJ co0 a)0/c seen by us, the observers, to = w0 in the rest
frame of the particle.
120 10 Theory of line formation
the Maxwell distribution (10.17), or
with
and o = ^ = _ A c o , ^
c X CO C
or
2
e-(AA/AAD)
^ with AAD = A ^ . 10.20
e
me2
we can also write
oo
1AA =
oo
The integral over KX over the line must, of course, remain the same,
independently of the Doppler broadening. In a frequency independent
radiation field the amount of energy absorbed by the atoms does not change
when the atoms are in motion. The condition (10.21) then leads to the
value for KLa
L\
e X
e
me 2 TT
A/l *
Actually, the radiation for electrons in different energy states and for
different transitions is different in intensity, which is not obvious from
equation (10.21). As mentioned earlier, one therefore introduces a factor
/ describing the actual strength of the lines. This factor tells us to how
many classical oscillators the emission corresponds. Therefore, we call this
factor the oscillator strength of the line, or the transition.
For broadening by the Doppler effect only, we find a half width
2). 10.23
10.4 Voigt profile 121
For the sun with T~6000K we find for an iron atom with atomic
weight \i = 56 that
12 x 8.3 x 107 x 6 x 103
56
or
^o^likms"1 and -= ' *\ 0 ~ 4 x 10" 6 .
For X ~ 4000 A the AXD - 1.6 x 10" 10 - 0.016 A, i.e., the Doppler width
is much larger than the damping width of the line, which is about
1.1 x 10 ~ 4 A, as we saw above.
For the hydrogen atom at T = 6000 K, we find
1(2 x 8.3 x 107 x 6 x 103N
* VV 1
This is always a good reference point to estimate average thermal velocities.
The Doppler width of the Balmer lines is about 0.1 A. However, in most
of the stars the hydrogen lines are much wider because of Stark effect
broadening, which we shall discuss in Chapter 11.
Doppler profile
Fig. 10.6. A schematic plot of the different line profiles contributing to the Voigt
profile. Generally, the line profile consists of the Doppler core and the damping
wings. It is called a Voigt profile.
0 \ / 2 3 10 11 12
~- ^ r
<X= 1
V
^ - ^
0,1
V
1
.
.
N .
-
^
.
,
- ^ .
0,01
0,001
,
.
-
V\
* .
'=0
Fig. 10.7. Voigt profiles for different values of a are shown in a logarithmic plot.
(From Unsold, 1955, p. 265.)
10.4 Voigt profile 123
AA Aco y
r
Observer
Fig. 10.8. One half of a rotating star moves towards us. The other half moves away from us.
<j) = longitude
6 = colatitude
Fig. 10.9 shows the relation between the cartesian coordinates X, Y, Z and the
spherical polar coordinates, r, #, <f>.
10.6 Other distortions of line profiles 127
instrumental profile. It gives an additional line broadening, especially for
low dispersion spectra. For such low dispersion spectra the instrumental
broadening is frequently more important than any broadenings intrinsic
to the star. On such low dispersion spectra we cannot therefore measure
intrinsic line profiles. For high dispersion spectra we have to correct the
measured line profile for the instrumental broadening, which has to be
measured in the laboratory. Quite often we are unable to extract the true
line profiles, especially for spectra of faint stars, for which we can only get
low resolution spectra. We can then only work with the total absorption
in the lines. The total absorption in the lines remains unchanged due to
all these macroscopic broadening mechanisms. (This does not prevent us
Fig. 10.10. vy = const, for stripes on the star's disc parallel to the projected axis of rotation.
. . . . . . . . . I . . . I . . . .
I .. i . . . . . I . . . . I . . . . I
-AA
Fig. 10.11. A spectral line broadened by rotation only has an elliptical line profile
for a star with no limb darkening ( ). ( ) profile with strong limb darkening.
(From Gray, 1976, p. 398.)
128 10 Theory of line formation
from measuring slightly wrong values for lower dispersion spectra because
we tend to make wrong judgements for the position of the continuum on
such low dispersion spectra.)
9r0
we find that
**A C 9rsfrs^*'Xs'kT 10.28
9r0
If we measure Wk and grsfrs in the laboratory as well as #s, we can determine
Nro/gro and thereby the number of atoms or ions of the particular element
for which the line was observed. However, in order to make use of the
Boltzmann formula we must know the temperature. We have already
discussed several methods for the determination of atmospheric
temperatures. In this context it seems to be best to determine the
temperature directly from the study of lines which originate from different
levels with different excitation energies. If we measure the Wx for two such
lines, one with index 1 and another with index 2, we find for the equivalent
widths
Nr
W cr n f }2e~*llkT
X ^- *JT\ J 1 1
1
9r0
M
10.29
10.8 Optically thick lines 129
The ratio of the equivalent widths gives
t^Ai = 9rjl ^1 e-(Xi
(a) (b)
Fig. 10.12. Changes of the line profile with increasing KL/KC for (a) optically thin
and (b) optically thick lines.
130 10 Theory of line formation
8.30). (In cases of NLTE when Sx ^ Bk the SA(0) may tend towards zero,
for instance, in resonance lines, but these are exceptional cases, although
usually the strongest lines.)
In the wings the line is still optically thin. There the line depth still grows
proportionally to KL/KC9 SO the line becomes wider, and the profile becomes
more rectangular. Since the line flanks are very steep, the increase in total
absorption is very small with increasing values of KL/KC. The increase in
optical thickness thus gives an additional line broadening which might be
mistaken for an additional broadening due to turbulence, especially if the
line is broadened by rotation or by the instrumental profile such that the
line depth may not appear to be very deep (see, for instance, Volume 1,
Fig. 13.3, where we show an optically very thick line in a non-rotating star
and in a star with high rotational velocity). It appears similar to an optically
thin line with very high turbulent velocities. It is not always easy to
distinguish between these possibilities. The best way seems to be by Fourier
analysis, although extremely accurate measurements are necessary.
^ ^ [ RXU, 10.31
line *c Jline
where the integral has to be extended over the whole line. In the previous
sections we have discussed the wavelength dependence of Rk for optically
thin lines. We have also seen that for optically thick lines the limiting
central line depth is given by RXo with
10.32
l
Rk = . 10.34
For very large values of KJKC the X{k) is much larger than 1. It is then
also much larger than Rko such that l/X(k) is much smaller than l//?Ao
and we have /?A = RXo.
For small X(k), i.e., optically thin lines, l/X{k) l/RXo and Rk - X{k).
As we saw before for a Doppler profile
^ | s / e f , 10.36
me AAD
The integral
Rxdk=
line J line
Large y
Small y
Small
Fig. 10.13. Schematic curve of growth for different values of the Doppler width
AAD and damping constants y.
10.9 Curve of growth 133
SAO)
Fig. 10.14. For the same central line depth, the equivalent widths of optically
thick lines increase proportionately to the line widths.
Fc (0)
_v / c
(0)
Fig. 10.15. For saturated lines the equivalent width becomes proportional to the
central line depth.
\
$1
/ - I o g 2 a = ^r^O.5
,^-
^4.0
-logy/A W o
r
rl
: ! ^ < , i 2a = 0
. ' t=p-i||r-
' II -- aa
\ ' \ I i i
|
'
j
i
'
!
i
!
i i; i i ! i I
+2 +3 + 4 +5 +6 +7
Q
1 1
logC ^^-1.027
C\l A
S -1
1 VT i
-1 +1
Fig. 10.16. The so-called Universal curves of growth according to Unsold, 1955, depending only o n a = (y/2Aa)D).
Rc = R(X0) and Xo = X(X0).
10.9 Curve of growth 135
This is the form of the curve of growth which Unsold calls the 'Universal'
curve of growth, which we show in Fig. 10.16.
The hydrogen lines deserve special consideration, not only because they
are the strongest lines, but also because they are a very important tool in
analyzing stellar spectra. The reason is that they are broadened by the
so-called 'molecular' Stark effect, that is by the electric fields due to the
passing ions. When a radiating atom or ion finds itself in an electric field,
the energy levels are shifted to slightly different values of the excitation
energy xn o r Zexc The emitted lines therefore occur at slightly different
wavelengths. Even more important is that for the hydrogen atom the
different orbitals contributing to the energy level with a main quantum
number n, are normally degenerate, i.e., when there is no external force
field they all have the same excitation energy (this is, of course, indicated
by the statistical weight for each level). In' the presence of an external force
field this degeneracy is removed, so that in an external field the different
orbitals contributing to a given level with main quantum number n now
occur at slightly different values for the excitation energy. This in turn
means that, instead of one line being emitted in the force free case, there
are now several lines emitted or absorbed at slightly different wavelengths.
Fig. 11.1 shows the different components which are observed for the
different Balmer lines.
This shifting of the energy levels in the electric field of the neighboring
ions has mainly two effects. First, it shortens the lifetime of the undisturbed
level which, according to equation (10.15), results in a broadening of the
undisturbed energy level. Since this broadening is due to the electric field
of the ions passing nearby or due to 'collisions' with other ions, this
broadening is called impact broadening. Since this broadening is also due to
a shortening of the lifetime of the original level, its effect on the width of
the line is the same as that of radiation damping. It leads to a line profile
which is similar to that for radiation damping but with a larger y. Since
136
11 Hydrogen lines 137
the collision with the passing ion only takes a short time, the new, shifted
levels only have a very short lifetime and are also broadened by this effect.
In addition to this impact broadening of the unshifted lines, we have to
consider the effects of the new line components, for which the splitting
increases with increasing field strengths. Ions passing very close cause a
large splitting, while ions passing at larger distances only cause a small
splitting. The pattern of the absorbed or emitted line therefore depends on
the statistics of the electric fields which the absorbing or emitting atoms
experience at the moment of the absorption or emission of the photon. In
order to calculate the overall profile of the absorbed line, we have to look
at the statistics of the electricfieldsexperienced by the atoms at the moment
4
0,3 "a ~
az
I 1 11 I 1
-8 -6 - '2 0 +2 * <6 +8 Ak
0.2
0,7
J L
7V-72-70-8 -6 - 2 +6 *8>
0,75
A
070
005
I I i. jJLi
22 -20-78-76 7# -72 '70 S S -4 2 0*2
LL,
0.70 "6
0,05
I I II I 11,I,Ml I I I I1i1I
-32 -28 -21 -20 -76 72 '8 t 0 '* +8 +72*76 >20*2t <28 <32 AA
Fig. 11.1. The Stark effect splitting of the different Balmer lines (according to
Unsold, 1955, p. 320.)
138 11 Hydrogen lines
of absorption. This part of the line broadening is described by the so-called
statistical line broadening theory. Since it deals with the lines originating
from the shifted energy levels, they occur in the wings of the original line.
The larger the electrical fields or, in other words, the closer the encounters,
the further out in the wing this displaced line occurs. The line profile in
the wings of the hydrogen lines is therefore described by the statistical
theory of the broadening by the 'molecular' Stark effect. It is clear that
this broadening is not actually due to molecules, but rather to ions. This
name is only used to indicate that it is not due to macroscopic electric
fields, but rather to microscopic fields due to the charged particles in the
plasma.
If we want to calculate the distribution of the absorption coefficient for
the hydrogen lines, we have to calculate the probability for absorption at
a given wavelength difference A/I from the line center. If this A/I is still
within the impact broadened Lorentzian profile of the original line, then
we know the absorption coefficient from the Lorentzian line profile; but
if AA is larger and is in the wings where the statistical broadening theory
has to be applied, we have to determine the probability of finding an ion
at the correct distance d from the absorbing atom that creates a field
F = e/d2 leading to the shift AX under consideration. We see immediately
that the chance of finding a particle at the required distance d is larger if
there are more particles per cm 3 . The chance of finding at a given moment
an ion at exactly the required distance is proportional to the number of
particles N+ present per cm 3 . We therefore find that in the case of the
statistical line broadening in the hydrogen wings the absorption coefficient
at any given Al is proportional to the number of ions per cm 3 which equals
the number of free electrons per cm 3 , at least if all ions are ionized only once.
Why do we never mention the electric fields caused by the passing
electrons? We said that the broadening is due to the shifting of energy
levels in the atom due to the electric field experienced by the absorbing
atom. In the case of the passing electrons the velocities of the electrons are
too large. The collision times are too short. In other words, the phase shift
t] = j Aco dt experienced by a light wave is too small.
The actual profiles of the hydrogen line absorption coefficients have been
calculated in successively better approximations. The latest ones were
tabulated by Vidal, Cooper and Smith (1973). What is of interest to us here
is the fact that at any A/I in the wings of the hydrogen lines KX is proportional
to the number of free electrons per cm 3 , ne. For high density atmospheres
when ne is large, the hydrogen lines are much broader than for low density
atmospheres in supergiants (see Fig. 11.2).
11 Hydrogen lines 139
Another important fact is the increase in the splitting of the different
Stark effect components of the lines with increasing main quantum number
n (see Fig. 11.1). In fact, the splitting increases with n2. For a given electric
field the Balmer lines with a larger n of the upper level are broadened more
than Ha or H/J. On the other hand, we know that as the upper energy levels of
the hydrogen atom get closer and closer together, the energy difference
decreases as l/n3. For some level n = nu the line broadening becomes larger
than the difference in wavelength for successive Balmer lines. The lines
merge together and can no longer be seen separately (see Fig. 11.2). From
a simplified broadening theory for the Balmer lines - the original Holtsmark
theory - Inglis and Teller (1939) find that the highest quantum number nu
for which the energy levels are still separated is related to the electron
density ne by
log nc = 23.26 - 7 . 5 log nu. 11.1
This equation provides a very simple means of determining the electron
density ne. We only have to count the number of visible Balmer lines,
*
determine the upper quantum number of the last visible line nM, and calculate
ne from equation (11.1).
a
o
is
H
iH1 1
lii
1I
V
v |v v 1 v M V 1 X*
1
1
11
1) C Mo
11 85Io
Si
67Oph B5Ib
1IIn p 1II
TOri
HD B5Y
36936
1 Pi
1 i i t t i i M 1 I
Fig. 11.2. The spectra of a B5 main sequence star and a B5 supergiant are
compared. In the main sequence star the hydrogen lines are much wider than in
the supergiant. For the supergiant we can see more Balmer lines separately. (From
Morgan, Abt and Tapscott, 1978. This is a negative. White means no light.)
12
Spectrum analysis
3 KC dt c
For the hot stars the continuous absorption coefficient KC in the visual is
due to the Paschen continuum, which means that it is due to absorption
140
12.1 Balmer jump and hydrogen lines 141
from the third level of the hydrogen atom. This means KC is proportional
to NH(n = 3). The line absorption coefficient for the Balmer lines which
originate by absorption from the second quantum level is, of course,
proportional to the number of hydrogen atoms in the second quantum
level NH(n = 2). As we saw in the previous chapter, the line absorption
coefficient in the wings of the Balmer lines is also proportional to the
number of free electrons because of the Stark effect broadening of these
lines. Therefore, we find for the hydrogen lines in the B stars
A^2K 12.3
NH(n = 3)
Knowing the temperature from the Balmer discontinuity, the depth in the
Balmer line wings tells us the electron pressure and thereby the gas pressure.
Since the gas pressure must balance the weight of the overlying material,
the gas pressure also gives a rough estimate for the gravitational
acceleration in the stars. We must have approximately
Pg = mg, 12.4
where m is the mass of the overlying material.
Knowing that we are looking down into an optical depth TC = f and also
knowing that the continuous absorption is due to the hydrogen atoms in
the third quantum level, we can calculate the number of hydrogen atoms
in the third quantum level above TC = from
hNH(n = 3) = l 9 12.5
3KCQ
where Kca is the continuous absorption coefficient per hydrogen atom in
the level n = 3, and h is the effective height of the overlying atmosphere.
From the Boltzmann and Saha formula we can calculate the number of
H + ions which constitute the main fraction of the mass. Considering the
abundance of helium the total mass per H + ion is given by
m H ( l + 0 . 1 x 4 ) = 1 . 4 m H and the total mass m above the level TC = is
given by
m = h x 1.4 x Nu+ x mH. 12.6
An approximate value for the gravity g is then obtained from
g = Pg/m. 12.7
+
In determining the mass per H ion we have already made use of the
abundance of the elements which we only know after we have made a
complete analysis. We will discuss this shortly.
In Section 8.2 we discussed the dependence of the Balmer discontinuity
on pressure and temperature. We saw that for B stars the size of the
142 12 Spectrum analysis
discontinuity is determined by
*(3647 + ) K+(H)JVH( =
K(3647") K-(H)iVH(n = 2) ATH(n = 2) g2
which means the discontinuity depends only on the temperature, which
can therefore be determined from the Balmer jump alone. However, for O
stars and for supergiants the importance of the electron scattering
introduces some dependence on the electron density.
F2lo
F2Ib
F2EH2
78UMo F2Y
Fig. 12.1. For F stars the widths of the hydrogen lines are independent of the
luminosity of the stars. (From Morgan, Abt and Tapscott, 1978.)
Hot stars
Cool stars D = D(nel T)
Fig. 12.2 summarizes the dependences of the Balmer jump and the depths of the
hydrogen line wings on temperature and electron density for hot and for cool stars.
144 12 Spectrum analysis
Table 12.1. Central wavelengths and bandwidths of intermediate bandwidth photometry used
by Stromgren
A Halfwidth I Halfwidth
Band (A) (A) Band (A) (A)
u 3500 300 b 4670 180
v 4110 190 y 5470 230
^max Halfwidth
Filter (A) (A)
a Interference filter 5030 90
b Interference filter 4861 35
c Interference filter 4700 100
d Interference filter 4520 90
e Interference filter 4030 90
f Glass filter 3650 350
Intermediate bands
d c b a Narrow bands
/y
3647 4000 5000 6000
Fig. 12.3. The Stromgren bands describe the gradient of the energy distribution on
both sides of the Balmer jump. The comparison of the extrapolation from the long
wavelength side with the actual gradient over the discontinuity measures the Balmer jump.
This index makes use of the fact that the line intensities are stronger in
the blue and ultraviolet spectral region than in the yellow and red. It
compares the intensity gradient at the shorter wavelengths, where it is
reduced by the absorption lines, with the intensity gradient at longer
wavelengths, where it is influenced very little by the lines.
In the narrow band photometry a similar index m can be defined by
m = (e d) (d a),
which makes use of the same facts.
There are some problems with a simple interpretation of the m and mx
indices. As we saw in the previous chapter, the line strength on the flat
part of the curve of growth may be increased due to microturbulence. It
is not quite clear how much the m index might be influenced by this effect.
Stars with large microturbulence or with magnetic fields could possibly be
mistaken for stars with a very large abundance of heavy elements. All color
systems designed to measure heavy element abundances from the integral
effects of many strong lines suffer from these limitations.
12.3 Curve of growth analysis 147
Energy
Fig. 12.6. For given values of the total orbital momentum L and the spin S the
atomic energy level splits up into several fine structure levels according to the
different values of J which depends on the relative orientation of S with respect to
L. Lines due to transitions between the fine structure levels of two main levels are
called a multiplet. All the transitions belonging to one multiplet are shown. Many
of these transitions are 'forbidden', however, which means the transition
probabilities are extremely small.
12.3 Curve of growth analysis 149
nearly equal. The relative line strengths for the lines within each multiplet
are therefore completely determined by the transition probabilities or by
their relative oscillator strengths. The relative intensities are therefore
independent of temperature and pressure. This fact is very important for
the spectrum analysis by means of the curve of growth.
Let us go back to the problem of line identification. Suppose the measured
wavelength of a line at 3943.28 A, which you want to identify, almost
agrees with the wavelength of an iron line which belongs to the multiplet
with the number 72. In order to check the identification you have to go
to the front of the multiplet tables and look up the multiplet number 72
of iron. You find there are three lines in this multiplet which are stronger
than the line you have measured. If your identification of the line under
consideration with the iron line is correct, you should also see these three
stronger lines in your spectrum at ^3949.9, 3977.74 and 4003.71 A. They
must be stronger than the line you have measured. You have to find the
wavelengths of those three lines in your spectrum and see whether these
lines are there and whether they are indeed stronger than the line you have
measured. If they are, then the identification is possibly correct. If the three
lines are not there or are too weak, then the identification of the measured
line as this iron line is most probably wrong and you should check whether
the line next to it in the multiplet table - namely, the Sm II line at 3943.239
- might be a better possibility. Only if all the relative line strengths within
the multiplets in question match the expectations can you be relatively
certain that you have made the correct identification.
log
log WA/AAD2
log WJAAD
Fig. 12.7 shows a schematic curve of growth. Different values of AXD may lead to
very different values for the derived number Ns of absorbing atoms.
12.3 Curve of growth analysis 151
good oscillator strengths is not an easy task and we are still lacking good
values of oscillator strengths for many lines of complicated atoms. But
once we have reasonably good values for the oscillator strengths, we can
fit the lines of a given multiplet on the theoretical curve of growth shown
in Fig. 10.16. The abscissa of this curve of growth is the line absorption
coefficient per cm 3 in the line center times a factor x \/KC x d In B A /dt c and
then the normalizing factors l/A>lD and l/Rc9 where Rc is the largest possible
central depth of a line. If we consider only a narrow spectral region, then
all the last factors mentioned are nearly the same for the different lines.
For a small wavelength band we only have to consider the different values
for the line absorption coefficient when comparing different lines. The line
absorption coefficient in the line center is given by a constant K0 and the
number of absorbing atoms times the oscillator strength, i.e.,
KL = K0N(n = ni)fik, 12.9
where nt stands for the quantum numbers of the lower level involved in
the transition, and fik is the oscillator strength for the transition from the
level with quantum numbers i to the level with quantum numbers /c. If we
plot the abscissa in a logarithmic scale, then the difference in the abscissa
for different lines is given by
log N(n = ns) - log N(n = nt) + log/ s m - log/ 0 -
if the line transitions are from s to m and from / to j . The number of atoms
in the various states of excitation can be described by the Boltzmann
formula, namely
No Go
where gs is the statistical weight for the level with quantum numbers 5, xs
is the excitation energy for this level, No is the number of atoms in the
ground level, and g0 is the statistical weight of the ground level. For lines
of a given multiplet the xs a r e aU the same for different lines because the
fine structure splitting of the levels is very small. With this in mind we can
then say that the difference in the abscissa for lines of a given multiplet is
given by
A abscissa = log gs + log fsm - log gz - log flj9 12.11
or
A abscissa = log gjsm - log gjlj9 12.12
because No and g0 are also the same for one multiplet. In fact, they are
the same for all lines of a given atom or ion.
152 12 Spectrum analysis
It is always the product gf which we need and it is also this product
which is measured from the line intensities in the laboratory. Therefore,
this product is usually given in the oscillator strength tables.
For the lines in a given multiplet we then know the difference in the
abscissa. The equivalent widths have to be measured on the spectrum. To
begin with, we can plot the first line sm (going from quantum state s to
the state m) of the multiplet at an arbitrary position of the abscissa in the
plot with the curve of growth, because we do not know the exact value
for the abscissa. When we now plot the value for the second line Ij we
know that its abscissa must be decreased by the amount log gsfsm log gxfVp
as compared to the abscissa for thefirstline. We therefore know the relative
position of this line with respect to the first line. For the third and fourth
lines of the same multiplet we also know the relative values of the abscissa.
After we have plotted all the values for the different lines of the multiplet
we have a picture, as shown in Fig. 12.8 by the x . Clearly, these points
must fit on the theoretical relation between the equivalent widths and the
abscissa. We see that we have plotted all our points for the wrong absolute
value of the abscissa, which was to be expected, since we chose the abscissa of
the first line arbitrarily. Shifting the points over onto the theoretical curve
of growth without changing the relative values of the abscissa tells us where
these points really belong on the curve of growth.
1233 Determination o
We have still not talked about how to determine the ordinate which
is not solely determined by the equivalent width - which we can
log
1AORC
A log AAC
* logNsfsk +logC-logAA c
Fig. 12.9. If the assumed value for AAD was too small the points for the saturated
lines lie too high by the error in A log AAD, when the absolute value for the
abscissa is determined by the weak lines.
154 12 Spectrum analysis
Sth 12.17
\x = atomic weight.
Fig. 12.10. The equivalent width for the lines in multiplet n and m divided by
AADRC are plotted as a function of log gnifni and log gmjfmp respectively. Their
abscissae in the curve of growth differ by log Nm/gm log NJgn = @(xm ~ In)
which can be determined by the horizontal shift necessary to shift the points for
the two multiplets on top of each other.
12.3 Curve of growth analysis 157
For particles with large /i the thermal velocity is small, while particles
with a small molecular weight ft have large velocities. It is this difference
which, in principle, makes it possible to determine the kinetic temperature.
Suppose we have measured A/D for iron lines and for carbon lines. We
then know 2(Fe) and 2(C), where
^2(Fe) = 2 ^ + C 12.18
and
2
= ^ tfurb. 12.19
H(C)
From equations (12.18) and (12.19) we then find
J ' / pp ,2.20
W e ) (C)J
In practice, there are limits to the accuracy of this temperature
determination. First of all, this can only work if the microturbulent
velocities turb are not much larger than the thermal velocities. Otherwise,
the differences in AXD are too small to be measurable. The second problem
is that the atomic data, as well as the measured equivalent widths, are
often not very accurately known, which means that the best vertical shift
between observed and theoretical curve of growth is often not very
accurately determined, resulting in rather large uncertainties for AXD and
therefore for 2. However, in the future improved atomic data and higher
accuracy of the measurements will make this determination of a kinetic
temperature a nice tool to check on our assumption of LTE.
KAx^ocNH(n = 2
^ g = ^ e - ^ - ^ f e r n e = ^e^-^/fcrng, 12.22
KC NH(n = 3) g3 g3
if the continuous absorption is due to the neutral hydrogen atoms in the
third quantum level, as, for example, in early B stars. For increasing T the Rx
decreases. For increasing degree of ionization of hydrogen, both the line
absorption and the continuous absorption coefficients decrease and the ratio
12.3 Curve of growth analysis 159
only changes because the occupation number in the third quantum level,
doing the continuous absorption, increases relative to the one of the second
quantum level responsible for the line absorption. This is the reason for the
decrease of the Balmer line intensities with increasing temperature for the B
stars. The temperature dependence is very weak because Xz~li *s s o s m a U-
When comparing line strengths in supergiants with those of dwarfs, we
have to be aware of the influence of the gravity which determines the
pressure in the atmospheres. For instance, for G type stars the continuous
absorption is due to the H ~ ion. The Saha equation tells us that the number
of H~ ions relative to the number of neutral atoms increases with increasing
electron density ne. Supergiants therefore generally show stronger spectral
lines than dwarf stars (see Fig. 12.1), not because there are more heavy
elements in their atmospheres, but because the continuous absorption
coefficient is smaller in these atmospheres of low gravity and therefore
small electron density. Therefore, we have to be very cautious when looking
at stellar spectra. Weak lines do not necessarily mean a low abundance of
heavy elements and vice versa.
For the Balmer lines of hydrogen we find, for instance, for A and B stars
that the lines become stronger when the hydrogen abundance decreases if
the hydrogen is replaced by helium. The reason is the increase in gas
pressure coupled with an increase in the electron pressure in the
atmospheres with larger helium abundances. Because the line absorption
coefficient in the Balmer line wings is proportional to the electron density
and the continuous absorption coefficient is still due to the hydrogen, we
find an increase of the hydrogen line strength for stars with a smaller
hydrogen abundance, unless we go to extremely small hydrogen
abundances, when the continuous K becomes due to helium.
When making a computer analysis of stellar spectra, we have to be aware
that when trying different abundances of heavy elements in order to match
the observed spectrum we have to be sure to change the atmospheric model
each time we try a different element abundance because the change in the
continuous absorption coefficient due to the changing metal abundance is
as important as the change in the line absorption coefficient.
As an example, let us look at the dependence of an Fe + line strength
on the abundance of Fe for solar type stars. For such a line the line depth
is determined by the ratio
Fe + Fe +
i?A(Fe + )oc oc . 12.23
H" NHne
In solar type stars the electrons come from the ionization of the metals
like iron. We saw that in the sun iron is mainly ionized. The iron abundance
160 12 Spectrum analysis
is therefore given by
abundance (Fe) = e+/NH = ,4(Fe). 12.24
If the abundances of all the heavy elements vary together, as is roughly
the case, then we must also have
neocFe+ =A(Fe)NH. 12.25
Inserting this into equation (12.23), we find
NHne ne NH
The Fe abundance cancels out in a first approximation. Actually, NH
increases for decreasing metal abundances. Therefore, we do find a slight
decrease in the line strengths of the Fe + lines for decreasing metal
abundances, but the effect is very small. A rough glance at the spectrum
may be very misleading.
Population I Population II
we can see helium lines only in the chromospheric spectrum for which
analysis is much more difficult and uncertain because LTE cannot be
assumed. We have very incomplete knowledge about the helium abundance
in cool stars and must resort to indirect methods which will be discussed
in Volume 3. Unfortunately, the accurate knowledge of the helium
abundance in old stars, which are all cool stars, is very important for many
cosmological problems.
There is only a minor trace of heavy elements mixed in with the hydrogen
and helium - only about 10" 3 by number of atoms - but because these
elements are heavy their fraction by mass, Z, still amounts to about 2%,
i.e., Z ^ 0.02. This leaves for the hydrogen abundance by weight X
X = 0.67 0.1
for most of the 'normal' stars.
162 12 Spectrum analysis
Among the heavy elements the most abundant ones are C, N, O, Ne.
For each of these atoms or ions there are about 1000 hydrogen atoms or
ions. The elements next in abundances are Fe, Si, Al, and Mg, but they
are less abundant again by another factor of 10. Table 12.3 gives recent
abundance determinations for the more abundant elements for stars in our
neighborhood, the so-called Population I stars.
{ n(u^l)
oc/V, (/,</) ocNuA(uJ) ocNuB(uJ)
Fig. 13.1. For a given line we have to consider three kinds of radiative transitions,
absorption (left), spontaneous emission (center), and induced emission (right).
163
164 13 Non-local thermodynamic equilibrium
where A(u, I) is the transition probability for the spontaneous transitions
from u to /. It is also an atomic constant.
Generally, we do not know what the average intensity */v(/,M) is, but if we
are dealing with thermodynamic equilibrium we know that
where BV(T) is the Planck function for the temperature T. In the case of
thermodynamic equilibrium we must require that for each transition there
must be an equal number of radiative transitions going in both directions.
We call this detailed balancing. In thermodynamic equilibrium, detailed
balancing must be true because it can be shown that otherwise it would
be possible to build a perpetuum mobile of the second kind, which we know
is not possible. We also know that in thermodynamic equilibrium the ratio
of the occupation numbers is governed by the Boltzmann formula, i.e.,
N a
Nt gt '
If these upward and downward transitions were the only processes that
occur, the condition of detailed balancing would require that
n(u /) = n{l t / ) ,
which means
or
Av = 4n /c v J v dv, 13.11
I . Iv, 13.12
J/ llii n e
where we have assumed that the J v within the line can be replaced by an
average J v , and where the integrals have to be extended over the whole line.
For the apparent net number of absorption processes, we find similarly
n(l -> M) - n(u -> /) = NtB(l, u)Jv-NuB(u, l)Jv
Kvdv = 13.19
J line V une
J li N,B(l,u)
If we now make use of equation (13.9), we derive
) 13.20
^i_Ar^A_47t Jv
f Kv ' d v - 4 7 t J v K '
\ N,gJ hv]Une ^v
13.30
Remembering again that the energy emitted per photon is hv, we find that
the number of photons spontaneously emitted per cm3 s is
13.31
hvjn
Combining equations (13.30) and (13.31), we derive
t_
N,gJ
13.4 Excitation of energy levels 169
After simplifying we obtain
2
^3 1 1335
Note that in deriving this expression for the source function we have
not made use of any equilibrium condition. This expression holds quite
generally.
If the occupation numbers are given by the Boltzmann formula for the
kinetic temperature T, then
2
^ ^ 13.36
c _z_ L B (T ) 13 38
^ ~ ~^T (e/,v/fcreXc _ j) ~ ^vWexd- 1^0
For a given transition the source function is always given by the Planck
function for the excitation temperature for the two levels involved in the
transition. In non-LTE (NLTE) conditions we will generally find different
excitation temperatures for different pairs of energy levels.
The problem, of course, remains how to determine NJNt and Texc under
NLTE conditions.
Nu gu
which is the Boltzmann formula. Even if excitation and de-excitation are
due to radiative processes only, deviation from LTE occurs only if Jv
deviates from Bv.
13.5 Summary 173
In the general case we derive for radiative excitation and de-excitation
only that
**_, *3i, 13 .50
or
c (e*v/*lkin 1) Jv Jv
The left-hand side of this equation gives essentially the ratio of the actual
occupation numbers to the one obtained from the Boltzmann formula,
which is determined by the ratio of Bv/Jv. However, this is true only for
the two level atom, discussed here.
13.5 Summary
In summary, we may conclude that large deviation from LTE may
be expected for low density gas in which the radiation field deviates strongly
from the Planck function for the kinetic temperature. Such deviations occur
in hot stars due to the discontinuities in the continuous absorption
coefficients, especially in the ultraviolet, where most of the flux is emitted.
Large deviations are also expected in the low density chromospheres and
coronae of cool stars, where the radiation field corresponds to a diluted
Planck function for the effective temperature of the star, while the kinetic
temperature in the coronae may be several million degrees.
In the photospheres of cool stars, in which the densities are rather high
and for which the radiation field does not deviate much from the Planck
function, we may expect that the LTE approximation will be a good one,
and that our discussions of the photospheric structure will not suffer from
this approximation.
Actual calculations - mainly by Mihalas (e.g., 1978) and collaborators
- have shown that the LTE approximation is not too bad for all main
sequence stars and giants with temperatures less than about 25 000 K. For
main sequence O stars it does not seem to be a valid approximation. We
also have to check this assumption for supergiants.
14
The hydrogen convection zone
14.1 Introduction
In Volume 1 we saw that the surface of the sun is not smooth, but
that we see bright granules separated by darker intergranular lanes. The
structures appear to have dimensions of the order of 500 km diameter and
therefore can only be recognized under conditions of very good seeing
(1 arcsec corresponds to 700 km on the sun). There may be even smaller
structures which we cannot resolve because of the atmospheric seeing and
because the solar observation satellites so far do not have mirrors large
enough to resolve such small-scale structures. When we discussed these
structures on the solar surface, we pointed out that in the bright regions
the motions, measured by the Doppler shift, are mainly directed outwards,
while in the dark intergranulum the motions are mainly downwards. These
motions and temperature inhomogeneities seen in the granulation pattern
are due to the hydrogen convection zone just below the solar photosphere.
These motions in the hydrogen convection zone are believed to be the
source of the mechanical energy flux which heats the solar chromosphere
and corona. Similar convection zones in other stars are believed to be
responsible for the heating of the stellar chromospheres and coronae whose
spectra are observed in the ultraviolet and in the X-ray region by means
of satellites. Before we can discuss these outer layers of the stars, we have
to discuss briefly the reason for these convection zones and the velocities
expected to be generated by this convection, and how these and the
mechanical energy flux generated by these motions are expected to vary
for different stars with different Teff, gravity, and with different chemical
abundances.
smaller scale, we can witness convection on a hot summer day when the
asphalt of a highway is heated by the solar radiation and the hot air rises
above the asphalt, causing the light beams to be bent due to refraction in
the different temperature regions, showing us the flimmering image of the
road. We see the same phenomenon above a hot radiator. We always see
these convective motions if we have a large temperature gradient in a fluid
or gas. How large does the temperature gradient have to be in order to
cause convection? This was first studied quantitatively by Karl
Schwarzschild. We discuss here his results, which can best be understood
when looking at Fig. 14.1. Here we have plotted schematically the
temperature stratification in a stellar atmosphere as a function of the
geometrical depth t. Of course, we could also use the gas pressure Pg as
the abscissa. Suppose that at the point Px a small gas bubble rises
accidentally by a small amount. This small shift upwards brings it into a
layer with a slightly lower gas pressure. The bubble therefore expands and
in the process cools off adiabatically if it does not have any energy exchange
with the surroundings. Therefore, it experiences a temperature change
given by
fdT\ dPjdT^
14.1
\dtjbubble dt \dP,/ adiabatic
-rp (atmosphere)
Fig. 14.1. A gas bubble accidentally rises from Pl to the point P2. At the higher
layer it expands to the lower gas pressure of the new surroundings and cools off
nearly adiabatically. At P2 it still has a higher temperature than the surroundings if
the adiabatic temperature gradient dT/dPg is smaller than the corresponding
temperature gradient in the surrounding atmosphere.
176 14 Hydrogen convection zone
The question is now, will this gas bubble continue to rise? Or will it fall
back to its original position? If it continues to rise, then the atmosphere
will soon move everywhere, because further small disturbances are created
by the rising bubble; the atmosphere is then unstable against convection.
On the other hand, if the bubble experiences a force that pushes it back
to its starting point, then no motion can be kept up. The atmosphere is
stable against convection.
The only force accelerating a floating gas bubble is the buoyancy force. If
the bubble has a lower density than its surroundings, then the buoyancy
force will push it upwards and it will continue to rise. If the density of the
bubble is higher than in the surroundings, the bubble will fall back. Because
the bubble expands to pressure equilibrium with the surroundings it has a
lower density than the surrounding gas only if it has a higher temperature
than the surrounding gas. According to equation (14.1), the temperature of
the gas bubble at point P2 is given by
7"i is the temperature at point P1. At is t{P2) t(Px) (bu means bubble, ad
means adiabatic). The temperature of the surrounding gas is determined
by the average temperature stratification in the atmosphere, which, in the
absence of convection, is determined by radiative equilibrium, as discussed
in Chapter 5. At the point P2 the temperature of the surrounding gas is
therefore given by
^ ^ ^ A t 14.3
dthu dt dPgat
(at means surrounding atmosphere). Subtraction of equation (14.3) from
equation (14.2) shows that
<[_^r^_ . 14.6
vd log / y at
It has become customary to abbreviate the logarithmic gradient by
d log T
= v. 14.7
d log Pg
If we start out with an atmosphere in radiative equilibrium with V = Vrad,
then convection sets in if
V a d <V r a d . 14.8
We shall soon see that equation (14.6) always holds if equation (14.8)
holds. Equation (14.8) therefore is generally the condition for convective
instability in a stellar atmosphere, provided that the mean atomic weight
\x is constant in the layer under consideration. Without saying so, we have
used the assumption of constant \i when stating that the density is lower
only when the temperature of the bubble is higher than in the surroundings.
A changing /i, due to ionization, can easily be taken into account. A change
in \i due to different chemical abundances may be very difficult to take
into account, as we shall see in Volume 3.
For our purposes here, condition (14.8) is sufficiently accurate to
determine convective instability.
From Fig. 14.1, it is quite obvious that if condition (14.8) is satisfied
and the upwards displaced bubble keeps rising, then a downward displaced
bubble has a higher density than the surrounding gas because the falling
bubble has a lower temperature than the surroundings. Falling motions
therefore also continue.
L-*J.F,.**-T><*
F , T becauseB^r. 14.9
3Kcmdt 3dt 3 cm7t
3K dt
178 14 Hydrogen convection zone
Solving for d In T/dt, we obtain
dlnT nFr3kc
14.10
df 16<rT4 '
In order to calculate d In P/dt, we have to make use of the hydrostatic
equation
, . mi
it P,
Division of equation (14.10) by equation (14.11) yields
dh Ll;dlnf'_3rf > y
dt I dt \6aT*g
From this equation we see that there are two ways in which Vrad may
become very large:
1. If Fr becomes very large. This may happen close to the center of
the star if all the luminosity of the star is generated very close to
the center and has to get through a very small surface area 4nr2.
This leads to a central convection zone in hot stars.
2. If /cgr becomes very large, while Pg also remains large. This only
happens if icgr increases steeply with depth, as we shall see now.
The hydrostatic equation says
^-. 14.13
df kgr
If /cgr is constant with depth, then integration of this equation gives
Pg = ^rg; 14.14
which means
With T 4 ocf we see from equation (14.12) that Vrad becomes independent
of T 4 and Pg as well as of f, because T 4 is proportional to f for f f.
If we now approximate the depth dependence of/cgr (see equation 9.14) by
k%x = AP\ 14.15
then integration of the hydrostatic equation (14.13) yields
1
and ^ - 1 1 14 .16
b+l A dt
The radiative temperature gradient can now be written as (see equation
14.10)
djrLr = J ^ = i _ i _ 1 4 n
df 16ar4 4(f + )
14.3 Reasons for convective instability 179
Here we have used the radiative equilibrium temperature stratification for
a grey atmosphere, namely
crT* = fo"7j!ff(T + f) = fnFr(r + i). 14.18
Division of equation (14.17) by equation (14.16) yields
d In T Id In P,
Vrad
" ~dd f //~ ddf ~ 4 (, ^
)"
We now see that Vrad does not depend on A, which means it does not
depend on the absolute value of k, but only on the increase of k with depth
described here by the exponent b for the dependence of k on Pg. Vrad
becomes large only for large values of b. At the surface the k needs to be
small such that a large gas pressure is obtained. If k then increases towards
deeper layers the product kPg can become large because Pg remains large
even when k increases.
A look at Fig. 8.7 shows that k increases steeply for temperatures above
6000 K. At these temperatures the hydrogen just barely starts to ionize.
Because hydrogen is so abundant, even 0.1% ionization of hydrogen
increases the number of free electrons by a factor of 10. The increase in
the electron density leads to an increase in the H~ absorption coefficient.
At the same time the hydrogen absorption coefficient also increases because
of the increasing excitation of the third and second energy levels of the
hydrogen atom. For somewhat higher temperatures than 6000 K this latter
effect becomes more important than the increase in the H~ absorption.
( 1 + - L ^ ) + * = 0, 14.30
V \ C IP
or
Cp
d l n V= -dlnPg. 14.31
With this relation we can now easily determine Vad. For an ideal gas
pg = Rg Tp/fi and d In Pg = d In T + d In p or
dlnT dinp
d In Pg d In Pg
14.3 Reasons for convective instability 181
For an adiabatic change we know from equation (14.32) that
d In p 1 Ce
14.35
d l n P 9 y Cp
If we insert this into equation (14.34), we obtain
Vad=l-- =- . 14.36
v d In P J.
g dd y y
For a monatomic gas the internal energy per particle is generally /cT,
and for a mol it is Rg T, and therefore
--Cv = \Rg. 14.37
y = - = or y = 1 H- 14.40
y
cv cv cv
y approaches 1 if Cv becomes very large. Remembering that Cv is the
amount of heat energy needed to heat one mol of gas by one degree kelvin,
this means if a large amount of energy is needed to heat the gas by one
degree, y will approach 1 and the adiabatic gradient becomes very small.
In other words, if such a gas expands there is a large amount of energy
available for the expansion before the gas cools by one degree. Under those
circumstances, Vad becomes very small and convection sets in. We find
such a situation for temperatures and pressures for which an abundant
element like hydrogen or helium starts to ionize. In this case, a large
amount of energy is used to remove additional electrons from the hydrogen
or helium atom or ion as needed for a temperature increase according to
the Saha equation. This energy is not available to increase the temperature,
i.e., the kinetic energy of the average particle. Large amounts of energy
are necessary to heat the gas and at the same time supply the necessary
energy for the increased ionization.
Depending on the gas pressure, hydrogen starts to ionize for temperatures
between 6000 and 7000 K in the stellar atmospheres. For higher
temperatures Vad is therefore smaller than 0.4 and may be as small as 0.05.
182 14 Hydrogen convection zone
In Section 14.3.1 we found that for x ^ 1 roughly Vrad = Vr = (b + l)/4.
For constant K, i.e., b = 0, convection sets in for Vad ^ 0.25, which is easily
obtained in the hydrogen or helium ionization zones. This decrease in Vad
occurs in the same region where we find the increase in the absorption
coefficient which is coupled with the beginning ionization of hydrogen.
Large Vr and small Vad therefore occur in the same temperature range and
lead to one convection zone, the so-called hydrogen convection zone, because
it is due to the ionization of hydrogen. Vad also becomes small when helium
ionizes at somewhat higher temperatures than hydrogen. We call this
unstable region the helium convection zone. For most stars these zones
merge. On the main sequence we find only for early F stars separate
hydrogen and helium convection zones.
7
1.0
/
/
2.0
4000 4400 5000 6000 7000 8000
Fig. 14.2. The upper boundaries for the hydrogen convection zones are shown as
a function of the effective temperatures of the stars. The different curves refer to
stars with different gravitational accelerations g (and thus different luminosities).
14.5 Convective energy transport 183
convection sets in essentially at a given temperature, the convection sets
in at lower optical depths for stars with higher effective temperatures. When
the effective temperature is so high that hydrogen is already ionized at the
surface of the star, there is no longer a steep increase in k and hydrogen
convection stops. There are still the helium convection zones, but they are
of little importance for the stellar atmospheres.
Fig. 14.3. At a given horizontal layer at depth t0 bubbles coming from above fall
through this layer and bubbles coming from below rise through this layer. They
have started at different depths tu t2, *3, t4, t5 and therefore at t0 have different
temperature differences with respect to the average surroundings. For the
convective energy transport we consider the average temperature difference AT
corresponding to an average path length 1/2 for bubbles passing through f0, when
/ is the average total path length for the rising or falling gas.
Fig. 14.4. At a given horizontal layer gas is rising through half of the area and is
falling through the other half of the area. When calculating the convective energy
transport through 2 cm 2 of the horizontal layer we consider upward moving gas in
1 cm2 and downward moving gas in the other 1 cm 2 .
14.5 Convective energy transport 185
Here we have added the transport of kinetic energy. nFc is the convective
energy transport per cm 2 .
We now have to consider the mass transport, which has to be zero, as
we stated above. This means
which for pu~Pd~P yields vu = vd = v. 14.45
At this point we can assume that the densities are nearly the same, since
it is not the difference of the densities which is used here. Inserting this
into equation (14.44) and reordering, we obtain
pvcp(Tu - Td) + \pv{v2u -v2d) = 2nFc. 14.46
With vu vd the second term on the left-hand side becomes zero, which
means the kinetic energy transport does not contribute to the energy
transport because the contributions from the upwards and downwards
flowing material nearly cancel. This is due to the requirement of zero net
mass transport. We are then left with
_ nFe = pvcp(Tu-Td). 14.47
With AT being the average temperature difference between the upwards
flowing material and the average surrounding, we have
T M -T d = 2AT, 14.48
and
14.49
The fact that in equation (14.47) the difference TuTd enters and not only
AT tells us that the downward moving material carrying negative energy
downward contributes as much to the convective energy transport as the
upwards moving material carrying positive energy upward.
Including the convective energy transport in our thermal equilibrium
condition nF = const. = <xT*ff (see equation 6.9), we must now require that
nF = nFr + nFc = oTtff. 14.50
With nFr ^ 0 we find that
t. 14.51
This equation gives us an upper limit for the velocities v once we know
the relation between AT and v. That such a relation must exist becomes
obvious when we remember that the AT determine the buoyancy force
which gives the acceleration. If fb is the buoyancy force per cm 3 , we have
/b = Apfif, 14.52
where g is the gravitational acceleration. The acceleration a due to the
buoyancy force is then
Ap force
a= g= . 14.53
P P
186 14 Hydrogen convection zone
Assuming again pressure equilibrium between the bubble and the
surrounding gas, we find (assuming constant /LL)
An AT AP AT
fr+TF- l4M
and
~P9. 14-55
which now gives us an upper limit for y3. If at any point in the convection
zone essentially all the energy is transported by convection, then this upper
14.6 Importance of convective energy transport 187
3
limit for v does indeed give the true velocity at that layer in the atmosphere.
Equation (14.61) thus provides a simple way of calculating the maximum
convective velocity in a given convection zone.
In Figs. 14.5 and 14.6 we show temperature differences and velocities
calculated for stars with different effective temperatures and gravities.
0.6
Fig. 14.5. Maximum temperature differences AT between rising or falling gas and
the average temperature of the surrounding are shown as a function of the
effective temperatures of the stars for main sequence stars.
188 14 Hydrogen convection zone
1
1 1
7 \
-/Y
4 -
3 -
\ \ //
V
5800
2 -
i \
1
1
I
i
, 8000^
1.0 2.0 3.0 4.0
log g
Fig. 14.6. Average convective velocities of the rising or falling gas are shown as a
function of the effective temperatures of the stars for stars with different gravities.
log ^ = 4.5
logs'= 2.5-^?
i
8000 7000 6000 5000 4000
Fig. 14.7. The optical depths in the upper convection zones for which
nFJnFr % 0.01 are shown as a function of the effective temperatures for different
gravities log g.
14.6 Importance of convective energy transport 189
If convection transports a major fraction of energy, then there is less
energy to be transported by radiation, as is apparent from the energy
equation (14.50). Since the temperature gradient is proportional to the
radiative flux (see equation 14.10), the temperature gradient is then smaller
than in radiative equilibrium (see Fig. 14.8). The total energy transport
becomes easier; therefore, a smaller temperature gradient is sufficient in
order to transport the required flux.
We can easily convince ourselves that the actual temperature gradient
can never decrease below the adiabatic one, because then convection would
stop and all the energy would have to be transported again by radiation,
which would require a steep temperature gradient. The temperature
gradient adjusts in such a way that the stratification remains unstable to
convection and that convection transports enough energy to make up for
the reduced radiative flux. In Volume 3 we shall discuss how the temperature
stratification can be calculated.
So far we have also neglected the radiative energy exchange between the
rising or falling gas and the surroundings which reduces the temperature
differences and the velocities. Such effects will also be taken into account,
when numerical values are calculated in Volume 3.
Radiative equilibrium T(t)
T(t)
/ Adiabatic T(t)
Fig. 15.1. The solar corona during the sunspot minimum. (From Kuiper, 1953.)
190
15.1 Solar observations 191
only see the corona when the solar disc is occulted by the moon. On high
mountains, where the scattered light from the atmosphere is reduced (when
seen from an airplane the sky is much darker) we can also observe the
corona by means of the so-called coronagraph, which is a telescope with
a built-in disc to occult the light from the solar photosphere.
The solar corona actually looks quite different at different times (see
Figs. 15.1 and 15.2). During sunspot maximum the corona looks spherically
symmetric, while during sunspot minimum the corona is very extended
above the equatorial belt, but shrinks in the other direction. It then looks
quite oblate, with a lot of structure in it - bright rays or loops stick out
in the equatorial regions. These bright structures look especially impressive
on a solar image taken with an X-ray sensitive camera (see Fig. 15.3).
These X-ray bright structures are also very prominent on the disc because
the solar photosphere does not emit X-rays; only the corona does. Only
the corona is hot enough to emit such energetic rays as the X-rays, which
have wavelengths of a few angstroms to a few hundred.
Fig. 15.2. The solar corona during sunspot maximum. (From Kuiper, 1953.)
192 15 Stellar chromospheres, transition layers, and coronae
When the wavelengths for the coronal spectral lines were first measured,
there were fairly strong lines at 5303, 5694, and 6374 A. Their identification
was completely obscure. Only after B. Edlen in Lund had measured in the
laboratory wavelengths of ultraviolet lines of highly ionized lines like Fe IX,
which has lost eight electrons, and Fe X, which has lost nine electrons,
could he construct energy level diagrams for these ions. With these energy
level diagrams, W. Grotrian in Potsdam succeeded in 1939 in identifying
some of the coronal lines as being due to very highly ionized ions. Following
this discovery, B. Edlen in Lund was able to identify the remaining coronal
lines in the visible region as being due to other very highly ionized particles.
The line at 5303 A is due to an iron ion which has lost 13 electrons. The
6374 A line is due to an Fe ion which has lost 9 electrons, and the 5694 A
line is due to a Ca ion which has lost 14 electrons. For some of these ions,
energies of the order of 300 eV are needed to separate the last electron
from the next lower ionized particle. The radiation from the solar
photosphere certainly does not have such high energy photons. The only
way these particles can be ionized to such a high degree is by means of
Fig. 15.3. A solar image taken with an X-ray sensitive camera. (From NASA,
courtesy Golub and Harvard Smithsonian Observatory.)
15.1 Solar observations 193
collisions with very energetic particles. But to make particles with such
high kinetic energies the temperature in the corona has to be one or two
million degrees. The X-ray emission of the corona indicates similar
temperatures.
There had actually been earlier studies, apparently by several
astronomers, investigating the line widths of the coronal lines, which are
quite broad. When these widths were interpreted as being due to thermal
Doppler broadening, they also indicated temperatures around two million
degrees. Before Grotrian's discovery this seemed impossible. Therefore,
these studies were not accepted at the time and only became known after
Grotrian's and Edlen's coronal line identifications.
Even after those discoveries, it remained a puzzle how the corona could
be so hot. We have always emphasized that for the heat flow to go from
the inside out, the temperature has to decrease outwards. Now we find
this large temperature increase outwards. How can the radiation from the
relatively cool photosphere with 5800 K flow outwards to and through the
hot corona such that it reaches the Earth? Does it not contradict the second
law of thermodynamics when the cool photosphere heats the corona up
to a million degrees? Or what else is heating the corona? What keeps the
photosphere cool between the hot solar interior and the hot corona? All
these questions were hard to answer. We are now close to the answers to
many of the questions, but before we talk about the energy equilibrium
and the heating mechanism for the corona we need to discuss the
observations concerning the layers which are below the corona but above
the photosphere. Clearly, the temperature cannot jump abruptly from
5800 K in the photosphere to two million degrees in the corona. There
must be regions with intermediate temperatures. These are called the
chromosphere and the transition layer.
JVIoon
Sun
Sun
Fig. 15.4 explains the meaning of second and third contact during total solar
eclipses. The angular size of the moon is exaggerated relative to the sun.
1
-6371, F e X
. 0 ! &D2. N a l
5867. D3, He I
5303. FflXIV
-4SK, H e l l
J-45J?, T i l l
I 4554. Ba II
-4M0, HT
Hi
-40D7. SrII
W45. Fe I
Fig. 15.5. A chromospheric flash spectrum. Several line identifications are given.
The coronal lines are seen as whole circles, chromospheric lines as part of a circle.
(From Zirin, 1966.)
15.1 Solar observations 195
What about the hot corona you might ask, which is in front as well as
behind the chromospheric layer along the line of sight? The corona is so
optically thin, which means tA is so small in almost all of the wavelengths,
that the intensity emitted from the corona is much smaller than the one
emitted from the chromosphere. Therefore, we do not see it in the flash
spectrum, except in a few strong coronal lines seen in Fig. 15.5. The
chromospheric gas does indeed absorb some of the coronal light, but again,
the optical depth of the chromosphere is so small and the coronal intensity
is so small that we cannot see this absorption. Only in the chromospheric
lines is the optical depth large enough to cause any measurable absorption,
but since there is so little coronal light available to be absorbed, the emission
Fig. 15.6. For the solar chromosphere we see an emission line spectrum because
we see an optically thin (in the continuum) hot gas against the background of the
cool interstellar medium. (From Zirin, 1966.)
196 15 Stellar chromospheres, transition layers, and coronae
in the chromosphere is much stronger than the absorption of the coronal
light such that we do indeed see only emission lines.
In the far ultraviolet (X < 1700 A) the chromospheric emission lines can
be observed outside of solar eclipses. Of course, we cannot observe in these
wavelengths with ground-based observatories because UV light cannot
penetrate the Earth's atmosphere. The first UV observations were all done
by means of rockets. Now most of these observations are done by satellites.
In Fig. 4.12 we show an early ultraviolet spectrum of the sun. For
X > 1700 A, an absorption line spectrum from the photosphere is seen. At
these wavelengths we see fairly deep layers of the solar atmosphere. For
X < 1700 A, the continuous absorption coefficient increases steeply, mainly
due to the sharp increase of the Si I absorption coefficient, as we discussed
in Section 7.6. For shorter wavelengths therefore, we only see high layers
in the atmosphere. In fact, we see the outer layers where the temperature
increases; thus we see emission lines, even when we observe the solar disc.
Some people like to say that for these short wavelengths the photospheric
light has decreased so much because of the low photospheric temperatures;
that we only see the chromospheric emission lines. This, of course, also
means that we see the chromospheric emission lines because the
chromosphere is hotter than the photospheric layers which we see at these
wavelengths.
Fig. 15.7. The ultraviolet IUE spectrum of the bright giant /?Dra. The stronger
lines are identified with the elements and ionization states which give rise to them.
For example, O I is neutral oxygen and Si III is doubly ionized silicon. (From
Bohm-Vitense, 1979.)
198 15 Stellar chromospheres, transition layers, and coronae
(b)
a AOR G2 I b
i I I I 1 1
HR'4511 '
Ia-0 - + CEPHEIDS (not observed) O
o NO CLASSICAL
CHROMOSPHERIC
EMISSION SCaMa
O
o VERY WEAK EMISSION
o WEAK EMISSION + +
EMISSION a Per 45Dra yCyg ySAqr a AC
S^VARIABLE o o
8 +
uPer? S Vol /3Dro
++
-
^ p Pup?
n Peg?
m
7
69Tau 7Sc
EZ: - -
(
aCaMi
63 225
HR4I38 XMus aHyi aCae lOUMa
-O O O OO OO
| 1 1 1 I
O.I 0.2 0.3 0.4 05 06 0.7 08 0.9
- B-V
Fig. 15.9. Shown is the position of observed stars in the color magnitude diagram.
Stars with chromospheric emission are shown as filled symbols, stars without are
shown with open symbols. Half-filled symbols indicate uncertain cases. The
Cepheid instability strip is plotted. All filled symbols are on the cool side of the
instability strip. (From Bohm-Vitense and Dettmann, 1980.)
BOUNDARY LINE
FOR TRANSITION
LAYERS
102
10
10
x (Li-abundance) x 10 1
I A Rotation velocity (km s~
Ca* emission luminosity
0.1
0.01 0.1 1.0 10
T: age (gigayears)
r ac oc pviuYh/cs, 15.2
where cs is the speed of sound. These acoustic waves travel upwards from
the photosphere and enter layers of lower density, the energy flux being
given by \pv2cs (energy density = f/w2, where v is the particle velocity,
multiplied by propagation speed cs). If they do not lose energy their velocity
must increase because of the decreasing density. This means they steepen up
to shock waves. While the acoustic waves as such do not lose much energy
once they become shock waves they are damped very rapidly, which means
they transfer a lot of energy to their surroundings, as we know from
experience. This energy transfer to the surroundings, of course, means that
the surroundings experience an energy input, by which it gets heated.
In the presence of magnetic fields these sound waves and shock waves
are modified, becoming so-called magnetohydrodynamic waves (MHD
waves) of different kinds, but there is always one kind which behaves
similarly to sound waves and which can form shock waves. How quickly
are these acoustic waves damped? The answer depends on the period of
the waves. In the lower chromosphere we observe oscillations with periods
of about 300 seconds. If this number is used one finds that in the solar
chromosphere with a temperature of about 10 000 K the shock waves are
dissipating energy with a damping length X of about 1500 km. This means
that the change of the shock wave flux, F mech , with height is given by
dF u 1
_ j ^ = F w here X* 1500 km. 15.3
dh X
This damping length is comparable to the height of the solar chromosphere.
The energy amount Fmech/X is the amount dissipated into heat in each layer
and which heats that layer of the chromosphere. Integration of equations
(15.3) yields
17 T7 o~h/k 1^/1
10
1 1 1 1 1
\
1
(\Jy
1
10
II
- I1
\ N
\
\
\
10
I I
: i -
10
3
i . ..1 . . . . . ...i . ...1 . .
10 10 4 105 106 107 10
7-(K)
Fig. 15.12. The radiative loss function f(T) is shown as a function of temperature.
The different curves shown were calculated by different authors (from Rosner,
Tucker and Vaiana, 1978). Shown are the results of Pottasch (1965) [ ],
McWhirter et al. (1975) [ ], Raymond (1978) [ ], and an analytic fit to Raymond's
curve [].
15.3 Theory 207
This temperature dependence of the radiative energy losses is very
important for the understanding of the temperature stratification in the
outer layers of solar type stars.
~ , 15.7
dh
where rj is about 10~ 6 cgs units.
The minus sign says that the heat flux goes towards smaller temperatures.
Generally in stellar atmospheres conductive heat transport is unimportant,
except in white dwarfs and in stellar transition layers and coronae. Why?
The conductive heat flux carries energy from the transition layer and corona
back down to the chromosphere and photosphere. If through the transition
layer the conductive flux remains constant, then no heat energy is taken
out or put in, just the same as in radiative equilibrium, where no energy
is deposited when the radiative flux remains constant. If, however, the
conductive flux increases downwards, then this additional energy must be
taken out of the layer where the flux increases. An increasing conductive
flux downward then constitutes an energy loss to that layer. If the downward
flux becomes smaller, this requires an energy input into the layer where
208 15 Stellar chromospheres, transition layers, and coronae
the flux decreases. (We always mean here the absolute value of the flux.
We have to be careful, since in the mathematical sense the conductive flux
is negative because it goes to decreasing heights.)
Fig. 15.15. The gas pressure decreases with a scale height H, as we have
discussed in Chapter 9. The P2g therefore decreases as
which means it decreases with half the scale height as the gas pressure
-0.6 -
Uj
-0.7 -
o
-0.8 -
Fig. 15.13. The radiative energy losses in erg per cm 3 . The rad are plotted as a
function of temperature for seven different values of the gas pressure. For a given
energy input Ein in erg per cm 3 , higher equilibrium temperatures Te are needed for
lower gas pressures in order to balance the energy input Ein by the radiative energy losses.
-0.9
-1.0
4.4 4.8 4.9 5.0
Fi'g. 15.14. As Fig. 15.13, but the energy input Ein decreases with increasing
height. Even for an energy input which is decreasing with height, the temperature
still has to increase if the decrease of Ein is not too large.
210 15 Stellar chromospheres, transition layers, and coronae
itself. The energy input decreases with the damping length (or scale height)
L We therefore must conclude that the temperature does not increase
outward if k becomes less than half the pressure scale height, i.e., if
-0.5
log/ 3 ,
-0.6 - - 0 . 8 7 5 = log Pgy
^^^~^~^-0mQ = \ogPg2
-0.7 ^^^^~~ ^0.925 = logP,3
o
-0.8 -
*0~
-0.9 -
-1.0
T
4.4
) nth)
I i
4.5
i
4.6 4.7
i
4.8
i
4.9
i
5.0
i
Fig. 15.15. As Fig. 15.13, but with an energy input Ein which is strongly
decreasing with height. If Ein decreases faster than n2e, then the equilibrium
temperature does not increase with height. It may decrease.
15.3 Theory 211
part of the transition region, which means according to equation (15.13)
that
and
and if, for instance, X = X0Ta, i.e., if X increases with increasing height, then
we derive from equations (15.14) and (15.13)
f
dh dh dh
2 aqc
15.19
dh (/3 + a-2)RgTl 2X
As we saw when discussing the radiative energy losses, the emission line
intensities are also proportional to the electron densities n\. The values for
the electron densities can therefore be obtained from the measured emission
line intensities. Knowing n\ we can determine the ratio Fm/X from the
left-hand side of equation (15.13).
212 15 Stellar chromospheres, transition layers, and coronae
(ii) The upper part of the transition layer. Fig. 15.12 shows that
for temperatures larger than about 105 K the ratiative loss function f(T)
no longer increases with increasing temperature, which means /J ^ 0. With
decreasing density the radiative losses therefore decrease; an increase in
temperature cannot prevent this. If the energy input is larger than the
radiative losses for T > 105 K, then the transition layer cannot get rid of
it by means of radiation. There is then a surplus of energy input; the
atmospheric layer must heat up. It continues to heat up until some other
energy loss mechanism becomes effective. Due to the heating, a very steep
temperature gradient develops at this temperature which leads to a
relatively large conductive heat flux down into the lower layers. If the
temperature gradient is large enough this conductive heat flux is able to
take out the surplus energy. There can be an energy equilibrium between
the energy input dFm/dh and dFJdh, where Fc is the conductive heat flux.
There is conductive flux coming down from the corona. At each layer in
the transition zone there is some mechanical energy input and some
radiative energy losses. If the energy input is larger than the losses, then
the net energy input must also be transported down by the conductive
flux. This means at each layer some energy is added to the conductive flux,
which keeps increasing on its way down (see Fig. 15.16). The largest
conductive flux is then found at the bottom of this upper part of the
transition layer, that is, in the layer with T 105 K.
I'-
Fm
~A
Fig. 15.16. At each height h some energy Ein is put in. For an equilibrium this
energy has to be added to the conductive flux in order to get rid of this energy
and keep the temperature constant.
15.3 Theory 213
If there were no energy lost in these upper layers by radiation or by
some other means, then all the mechanical flux which entered at the bottom
of these high temperature regions must return in the form of conductive
flux. In reality, there is also some energy lost due to radiation and a
somewhat larger amount is lost due to the stellar wind, as will be discussed
in the next chapter. In fact, at each layer in the transition region the amount
of conductive flux coming down must equal the amount of mechanical flux
going up, minus the energy loss due to the wind and the radiation in the
layers above the one under consideration. As the stellar wind takes out
energy mainly at the top of the corona and the radiative losses are small,
we can say that at each layer
\Fc(h)\ = Fm(h) - AFwind - AFrad(fc), 15.20
2
where AFwind is the energy loss per cm s due to the stellar wind, and
AFrad(/i) the radiative losses per cm2 s above h.
Combining equations (15.7) and (15.20), we derive
" " It
an
15.21
I I I i
.
6.0
l d/7 d/7
5.5 -
-
5.0 -
4.5
I i i i
7.5 8.0 8.5 9.0 9.5 10.0
log/7
Fig. 15.17. The observed temperature stratifications in the two parts of the
transition region of Procyon are shown as a function of height (Jordan et al.,
1984). The difference in the temperature gradient between the lower and the upper
part of the transition layer is clearly seen, the transition occurring at T ~ 1 . 6 x 105 K.
214 15 Stellar chromospheres, transition layers, and coronae
Here, Fwind can be considered to be the same for all heights except in the
high corona.
The integration of the right-hand half of this equation between h = h0
and h yields
Emitted
photon
Photon
absorbed
c
Emitted
photon
Fig. 16.1. Each absorbed photon transmits a momentum hv/c to the absorbing
particle in the direction of propagation of the photon. The emitted photons
transmit a momentum in the direction opposite to the propagation direction. The
momenta transmitted by the isotropically re-emitted photons cancel on average.
The momenta of the absorbed photons add.
215
216 16 Stellar winds
first to suggest that there must be a particle stream with high velocities
transferring momentum radially away from the sun to the comet tail
particles when they hit them. These high velocity particles moving radially
outwards from the sun are called the solar wind.
In the meantime, the densities and velocities of these solar wind particles
have been measured by means of rockets and satellites. It has also been
detected that they carry magnetic fields with them. These solar wind
particles are also the ones which populate the van Allen belts in the Earth's
ionosphere. The particle density n in the wind is about five particles per
cm3 at the distance of one astronomical unit from the sun. The velocities
of the particles depend on the area on the sun from where they started.
Particles originating in regions of the so-called coronal holes have velocities
up to 700 km s" 1 , while particles originating from the average, so-called
quiet solar surface have velocities of the order of 200 to 300 km s" 1 . The
total mass flow from the sun can then be estimated to be
m = nmHv4nd2 where d = 1AU, 16.1
19 14
which turns out to be about 10 grams per year, or 10~ M 0 per year.
Even during its entire lifetime, which is about 1011 years, the total mass
loss of the sun, if it remains constant, is only 0.1% of its mass.
/?AQR
Jl1
GOIb
2 -
Mg11
1 -
0 -
61 km s 1 48 km s 1
i i i
-1
2795 2800 2805 2810
i i
6000
V
4000
" C11 \ \ \
(K) "I 57 km s' 1
/
2000
V
/
i i
Fig. 16.2. The profile of the Ca + 3933 A line in the supergiant p Aqr. An
absorption component is seen displaced to shorter wavelengths by 0.75 A,
corresponding to an outflow velocity of 57 km s " l . Similar displaced components
are seen in the Mg + lines at 2795.5 and 2802.7 A. The longer wavelength
component is due to an interstellar absorption line. (From Hartmann, Dupree and
Raymond, 1980.)
Fig. 16.3. The total number of absorbing atoms or ions in a column of 1 cm2
cross-section extending from the stellar surface to the observer is called the column density.
218 16 Stellar winds
comes out to be about 10~8 M o to 10" 5 M o per year. (The exact value
is still difficult to determine.) The lifetime for these stars as luminous red
giants is estimated to be about 105 to 106 years. They may therefore lose
a considerable fraction of their original mass during this stage of their
evolution.
1548.19 A
Fig. 16.4. The P Cygni profile of the C IV lines at 1550 A as seen in the spectrum
of the O4 V star 9 Sgr.
16.1 Observations 219
Such PCygni line profiles can be interpreted as being caused by a
spherically symmetric, expanding shell, as we explain in Figs. 16.5 and
16.6. When the observer is in the direction of the bottom of the page, the
material in the shaded area is in front of the star and contributes to an
9 9
w
The pressure decreases exponentially but never reaches the value zero. If
we take into account that the gravitational acceleration actually decreases
as l/r2 and therefore the scale height increases, the pressure decreases even
more slowly at large distances and we can imagine that at very large
distances from the star the gravitational force is no longer able to bind the
material to the star against the gradient of the gas pressure which has to
reach the very low interstellar value at large distances. At large distances
we also have to take into account the decrease in temperature because the
radiative flux Fr also decreases as l/r 2 . In radiative equilibrium the
temperature T is proportional to F r 1/4 (see equation 6.31), so the
temperature decreases at large distances roughly as Toe 1/^/V, which means
the decrease in gravity outweighs the decrease in temperature, and in a
rough approximation we can say that the scale height H increases as r 3/2 .
Chapman (1957) considered the case of a stellar corona in which the
temperature stratification is determined by heat conduction and found
16.5
222 16 Stellar winds
The temperature then decreases more slowly than in the case of radiative
equilibrium and the scale height increases somewhat faster. (In the presence
of winds the temperature has to decrease faster because of the expansion
of the wind.) If we insert Chapman's outer coronal temperature
stratification into the hydrostatic equilibrium equation and remember that
for a fully ionized hydrogen atmosphere with n = number of protons
p = nmH and Pg = 2nkT, 16.6
we find
dPg ~GM
r1 Ynm^ l6J
2
dr r
and
2k (nT) = 2kTor^ A (n/r2'7) = ^ - nmH, 16.8
dr dr r2
or
d
( n \\ =
=
GMnm
u _ lfiQ
dr\r2/1) llcTr2'1 r2'
ar \r / z/ci o r o r
A solution of this equation is
n (r) = n(r 0 )x 2/7 e C[x " 5/7 - 1] 16.10
with 6
x=^ and cJ- **"*. 16.11
r0 5 2kToro
For small distances one obtains with x = 1 + (r ro)/ro and (r ro)/ro 1
16.12
Toro r0
This is similar to equation (16.4), but the scale height H is now replaced by
GMmH GM
This new scale height is a constant, even though the dependence of T and
g on r have been taken into account. At the stellar surfaces, where r /?,
we find that
^ ^ =2^1. 16.14
H R2 M
At the stellar surface the isothermal scale height H describes the density
stratification rather well.
For the temperature of the solar corona To = 2 x 106 K we calculate
H' 60 000 km. Using the observed density of n(r 0 ) 106 cm 3 for the outer
corona, we find for the expected particle density at the distance of the
Earth, namely, d = 1.5 x 1013 cm = 1 AU or x = 2 x 102, n(d) 103.
16.2 Theory 223
This is a surprisingly large density - much larger than is actually observed
- telling us that the density in the far out layers of the solar corona decreases
much faster than in hydrostatic equilibrium. The actual pressure gradient
must therefore be much larger than in hydrostatic equilibrium and so must
be too large to be balanced by the gravitational forces. This means that
there is a net pressure force pushing the particles outward, away from the
sun. The question then is why does such a large pressure gradient develop?
Let us now look at the gas pressure as obtained from the hydrostatic
equilibrium density stratification equation (16.10), which comes out to be
^(x-5/7-l)). 16.15
Here we have also made use of equation (16.5).
For the pressure at very large distances we find
P,(ooHP, o e-^ / H ', 16.16
which means at infinity the pressure reaches a constant value. Because the
temperature decreases with increasing distance, this means also that the
density does not decrease to zero with increasing distance from the star in
hydrostatic equilibrium. This in turn means that the total mass contained
in this far out corona must be infinite.
All these derivations demonstrate that hydrostatic equilibrium is not
possible at very large distances from the stars because the gravitational
binding force is too small. For the solar coronal temperature of 2 x 106 K
density equation (16.16) gives as the pressure at very large distances
^ .... _- , 16.30
" M o R Re4Rg Mo R\
which tells us that for solar type coronae the temperature cannot be higher
1200 1 1 1 1 1 1 1 I
~ 4 x106K
1000 - 3 x106 K
^ -
800 2 x106K
1
" / / " "" 1.5 x 1 0 6 K
^
If
I 600 . -
1 x106 K
3 0.75 "x106J<
400 _ "
0.5 x 1 0 6 K
200
I earth
'3
1 1 I I I
0 20 40 60 80 100 120 140 160
/-(106km)
Fig. 16.7. The dependence of the wind velocity on the distance r from the solar
center is shown, where r is measured in units of 106 km. The velocities of these
isothermal coronae depend on the coronal temperature, given on each curve.
(From Hundhausen, 1972, p. 10.)
228 16 Stellar winds
than about 6 x 106 K. Otherwise, the sonic point would move into the star
and the mass loss would be tremendous. If the mass loss were that large
the energy loss would be so large that there would not be enough mechanical
energy available to make up for this loss and the corona would have to
cool down, thereby reducing the stellar wind.
Observations of so-called RS CVn stars show, however, much higher
coronal temperatures for these stars. We must conclude that in such stars
the stellar winds are suppressed, possibly due to magnetic fields with large
horizontal components.
^ = 5.7 x l O 6 ! ^ 16.31
R Tc R
where M and R are the mass and radius of the star in solar units.
For the solar values of Tc = 2 x 106 K and M = 1, R = 1, we calculate
(r c /*) 0 = 2.8. 16.32
For this coronal temperature in the sun the sonic point should be at
about three solar radii, but for a temperature of only 106 K it would be
at a distance of 5.5 R o . The distance increases rather rapidly with decreasing
temperature.
For very cool, luminous stars with Tc ~ 10 000 K and with M % 10 M o ,
R&l00RQ, we derive
^ = 57. 16.33
R
For these stars the sonic point is found to be very far away from the star.
With 0 = 30 the pressure scale height is roughly /J = 5.5xl0 1 0 cm, or
about 1% of the stellar radius. At a distance of 50 stellar radii the density
should have decreased to very low values indeed if this scale height were
correct. Observations show, however, that these stars have much larger
scale heights. The effective gravities must be much smaller than the purely
gravitational one. The reason for this is still debated. For lower effective
gravities the sonic point is moved closer to the stellar surface.
For hot, luminous stars with M/R = 2 and T(surface) 40 000 K we
would find
rJRx 2.8, 16.34
16.4 The Eddington limit 229
8
which is similar to the sun. However, with a scale height of 6.6 x 10 cm,
the distance corresponds to 2000 scale heights, which means the density
at this point should be extremely small. In fact, the interstellar pressure
could easily be matched for such a star. However, these stars are observed
to have a strong mass loss. The density at the sonic point must therefore
be much larger than calculated. This in turn means that the effective gravity
must be much smaller than the gravitational one. There must be additional
forces opposing the gravitational forces. For hot stars radiation pressure
is quite important and is probably the origin of the observed strong winds,
although it is still surprising that such large velocities can be achieved.
or
^ x ar J ^ v ar *-*
: 5500 K
T \ogPt log^ \ogn
logr (K) (dyn cm" ) (cm2 g 1 )
2
(cm 4 )
-3.0 4251 3.50 -1.95 11.56
-2.5 4380 3.86 -1.64 11.85
-2.0 4491 4.14 -1.40 12.15
-1.5 4630 4.41 -1.15 12.42
-1.0 4842 4.70 -0.86 12.75
-0.9 4894 4.73 -0.85 12.77
-0.8 4961 4.79 -0.80 12.82
-0.7 5040 4.84 -0.75 12.89
-0.6 5126 4.90 -0.71 12.96
-0.5 5224 4.95 -0.65 13.03
-0.4 5335 4.99 -0.60 13.10
-0.3 5465 5.04 -0.55 13.18
-0.2 5612 5.09 -0.47 13.28
-0.1 5770 5.14 -0.38 13.40
0 5975 5.18 -0.26 13.54
0.1 6218 5.22 -0.11 13.73
0.2 6465 5.25 + 0.04 13.93
0.3 6675 5.27 + 0.18 14.10
0.5 7098 5.31 + 0.44 14.41
0.7 7453 5.34 + 0.65 14.64
1.0 7913 5.39 + 0.92 14.95
241
242 Appendix
reff = 5500 K
= 1.5, log Z / Z 0 = <
rff = 5500
log 0 = 4.5, l o g Z / Z o = - 2
Tn = 10000 K
19 = 2.5, l o g Z / Z o = 0
T logP, \ogne
logr (K) (dyn cm 2 ) (cn/g- 1 ) (cm 4 )
Abell, G. O., 1982, Exploration of the Universe, Fourth Edition, CBS College Publishing.
Auer, L. and D. Mihalas, 1973, Astrophysical Journal Supplement, 24, 193.
Bhatnagar, P. L., M. Krook, D. H. Menzel, and K. N. Thomas, 1955, in Vistas of
Astronomy, ed. A. Beer, Vol. 1, 296.
Biermann, L., 1951, Zeitschrift fur Astrophysik, 29, 274.
Bohm-Vitense, E. and T. Dettmann, 1980, Astrophysical Journal, 236, 560.
Bohm-Vitense, E., 1979, Mercury, 8, 29.
Bohm-Vitense, E., 1986, Astrophysical Journal, 301, 297.
Chalonge, D. and V. Kourganoff, 1946, Annales d'Astrophysique, 9, 69.
Chapman, S., 1957, Smithsonian Contribution Astrophysics, 2, 1.
Crawford, D. L., 1958, Astrophysical Journal, 128, 185.
Davidson, C. R. and F. J. M. Stratton, 1927, Memoirs Royal Astronomical Society, 64, IV.
Deutsch, A., 1956, Astrophysical Journal, 123, 210.
Edlen, B., 1941, Arkiv for Matematik Astronomi och Fysik, 28, B, No. 1.
Edlen, B., 1942, Zeitschrift fur Astrophysik, 22, 30.
Grotrian, W., 1939, Naturwissenschaften, 27, 214.
Hartmann, L., A. K. Dupree, and J. C. Raymond, 1980, Astrophysical Journal
(Letters), 236, L143.
Hartmann, L. and K. B. McGregor, 1980, Astrophysical Journal, 242, 260.
Hundhausen, A. J., 1972, Coronal Expansion and Solar Wind, Springer Verlag, New
York, Heidelberg, Berlin.
Inglis, D. R. and E. Teller, 1939, Astrophysical Journal, 90, 439.
Johnson, H. L., 1965, Astrophysical Journal, 141, 940.
Jordan, C , A. Brown, F. M. Walter, and J. L. Linsky, 1985, Monthly Notices Royal
Astronomical Society London, 218, 465.
Keenan, P. C. and R. E. Pitts, 1980, Astrophysical Journal Supplement, 42, 541.
Kuiper, G. P., 1953, The Sun, University of Chicago Press, Chicago, Illinois.
Lighthill, M., 1952, Proceedings Royal Society London Ser. A, 211, 564.
Linsky, J. L. and B. M. Haisch, 1979, Astrophysical Journal (Letters), 229, L27.
Moore, C. E., 1959, A Multiplet Table of Astrophysical Interest, NBS Technical Note
No. 26, United States Department of Commerce.
Morgan, W. W., H. A. Abt, and J. W. Tapscott, 1978, Revised MK Spectral Atlas for
Stars Earlier than the Sun, Yerkes Observatory, University of Chicago and Kitt
Peak National Observatory.
Morgan, W. W., P. Keenan, and E. Kellman, 1943, An Atlas of Stellar Spectra,
University of Chicago Press, Chicago, Illinois.
245
246 References
Parker, E. N., 1958, Astrophysical Journal, 128, 664.
Proudman, I., 1952, Proceedings Royal Society London Ser. A, 214, 119.
Rosner, R., W. H. Tucker, and G. S. Vaiana, 1978, Astrophysical Journal, 220, 643.
Schmidt-Kaler, T., 1982, in Landolt-Bornstein: Numerical Data and Functional
Relationships in Science and Technology, New Series, Volume 2, Astronomy and
Astrophysics, ed. K. Schaifers and H. H. Voigt, Springer Verlag, Berlin,
Heidelberg, New York, p. 1.
Skumanich, A., 1972, Astrophysical Journal, 171, 565.
Stencel, R., D. J. Mullan, J. L. Linsky, G. S. Basri, and S. P. Worden, 1980,
Astrophysical Journal Supplement, 44, 383.
Stromgren, B., 1963, in Basic Astronomical Data, p. 123, ed. K. Strand, Vol. Ill of
Stars and Stellar Systems, gen. ed. G. P. Kuiper and B. M. Middlehurst, University
of Chicago Press.
Unsold, A., 1977, The New Cosmos, Springer Verlag, New York, Heidelberg, Berlin.
Vidal, C , J. Cooper, and E. Smith, 1973, Astrophysical Journal Supplement, 25, 37.
Wildt, R., 1939, Astrophysical Journal, 90, 611.
Zirin, H., 1966, The Solar Atmosphere, Blaisdell Publishing, Waltham, Massachusetts.
247
248 Index
equivalent height 102 magnetohydrodynamic waves (heating by) 203,
equivalent width 128 204
excitation energy, definition 68 magnitudes
excitation temperature, definition and absolute 11
determination 154 apparent Iff, 3
extinction, Earth's atmosphere 6ff bolometic 12
extinction correction 9 main sequence 13
maser 165
filter, sensitivity functions, UBV 6 mass loss
flux hot stars 218, 219
apparent 3 luminous, cool stars 217
surface 11, 43, 46 solar 216
Maxwell velocity distribution 119
Gaunt factor 71 metallic lines (dependence on abundances) 159
giants 13 mixing length (in convection theory) 186
gravitational acceleration 102 multiplets 148
effective 111
grey atmosphere, definition 46, 49 non-local thermodynamic equilibrium (NLTE)
163ff
height of stellar atmosphere, effective 116
helium convection zone 182 occupation numbers 73, 166ff
hydrogen convection zone 179, 182 optical depth, definition 7
hydrogen lines, dependence on abundances 159 optically thin, definition 31
hydrostatic equation 102 optically thick, definition 32
hydrostatic equilibrium lOlff oscillator strength 118
partition function 77
impact broadening 136
Paschen continuum 77
induced emission 165
path length (convection) 186
Inglis-Teller formula 139
Pauli principle 68
intensity, definition 20, 26
P Cygni, line profile 219
intensity (surface) 4Iff
photon energy 68
interstellar cloud 33
Planck formula 20
ionization energy, definition 68
planetary nebula, spectra 34
ionization temperature, definition and
pressure
determination 155
electron 74, 107
gas 102ff
Joule heating 203, 204 radiative 111, 112
stratification lOlff
kinetic temperature, determination 156 turbulence 109ff
Kirchhoff's law 29 principle quantum number 68
Chapter 1
With a photometer you measure the energy fluxes of a star whose unknown colors
you want to determine. You also measure the fluxes of the standard star Vega
(a Lyrae). For simplicity you measure both stars at the same zenith distance. For
each filter the photometer readings are proportional to the apparent fluxes. For
the different filters you have the following readings per unit of time (after
background subtraction):
U B V
2. You want to determine the unknown apparent blue magnitude of a star. You
have determined the photometer readings for Vega which are proportional to the
fluxes received when Vega was at a zenith distance of 20. You can measure the
apparent fluxes for your program star only at zenith distances of 3 = 30 and
6 = 45.
Photometer readings per unit of time (after background subtraction):
s Star Vega
20 2000
30 800 ?
45 710 ?
Determine the correction factors for the extinction in the Earth's atmosphere.
What is the apparent blue magnitude of the star?
231
232 Problems
3. The absolute visual magnitude of the sun if Mv = 4.82. The sun is at a distance
of 1 astronomical unit = 1.49 x 10 13 cm. What is the apparent visual magnitude
of the sun?
Chapter 3
1. The amount of energy we receive from the sun per cm 2 s
atmosphere is called the solar constant S. It is S nfQ = 1.38 x 106 erg c m ^ s " 1 .
Measure the angular radius of the sun and calculate nF, the surface flux on the
sun. Calculate the effective temperature of the sun.
2. (a) The apparent visual magnitude of the KO III star a Boo (Arcturus) is mv =
- 0 . 0 4 and B - V = 1.23. The bolometric correction is about BC = 0.30.
What is the apparent bolometric magnitude of the star?
(b) The apparent bolometric magnitude of the sun is m bolo = 26.82. The angular
diameter of a Boo was measured with the Michelson interferometer to be
a = 0.02 arcsec. Calculate the effective temperature of a Boo.
(c) The trigonometric parallax of a Boo was measured to be n = 0.09 arcsec. What
is the radius of a Boo?
Chapter 4
1. You have a well-insulated box filled with sodium vapor. The box has been left
for a long time. Suppose you punch a small hole into the box and observe the
energy distribution around the strong yellow sodium lines at X = 5896 A and
5890 A. What would the energy distribution look like? Would you see spectral
lines? If so, would you see absorption or emission lines?
2. Many early B stars have gaseous shells around them, which are optically thin.
Suppose this shell expands with a velocity of about 50 km s" 1 . How would the
line profile of a strong Si III line look in the stellar spectrum?
Problems 233
Chapter 5
1. Does the foreshortening factor cos S discussed in equation (4.2) for light leaving
1 cm 2 of the stellar surface under an angle S lead to a limb darkening which we
have omitted in our discussion?
2. We have been looking only at what happens in the light beam under consideration.
There are many photons scattered from all other directions into this beam. Did
we consider those too or were they omitted? If we did consider them, how did
we do this?
i= 0
j-,.
Jo
4. We approximate the source function by
Calculate Fx(0).
5. The center to limb variation for the solar radiation at 5010 A has been measured
to be
-^ = 0.2593 + 0.8724 cos S - 0.1336 cos2 3.
Chapter 6
1. The measured center to limit variation of the solar intensity is for the different
wavelengths
7A(0, i9)/7A(O, 0) = a0 + ax cos # + 2a2 cos2 <9.
2. (a) Verify the surface condition for a grey atmosphere, namely, J(0) = ^F(O),
assuming isotropic radiation for the outer half sphere and no incoming
radiation.
(b) Verify the Eddington approximation for this case of isotropy in a half sphere
and no radiation in the other half sphere.
Chapter 7
For a pure hydrogen gas with a gas pressure of Pg = 103 dyn cm" 2 and a
temperature T= 10 080 K, calculate the ratio H + / H and the electron pressure
Pe = nekT. Remember that ne = U+ and Pg = nkT, with n = e + H + + H .
2. For a pure helium gas with a gas pressure Pg = 103 dyn cm" 2 and T = 15 000 K,
calculate He 2 + /He + , He + /He, and Pe. Remember n = ne + H e 2 + + H e + + H e .
* ion (He) = 24.58 eV; xion (He + ) = 54.4 eV. Here *ion (He + ) is the energy needed in
order to remove the additional electron from the He + ion to make He 2 + .
3. The absorption coefficient per hydrogen atom in the level with main quantum
number n is given by { M ^ z W - >
v
where
v = c/X, c = velocity of light, c = 3 x 10 10 c m s " 1
m = electron mass = 9.105 x 10" 2 8 g
h = Planck's constant = 6.624 x 10" 2 7 erg s
e = charge of electron = 4.8024 x 10" 1 0 electrostatic units
Z = nuclear charge + 1 number of electrons in atom or ion
Z = 1 for H and He, Z = 2 for He + .
Calculate the hydrogen bound free absorption coefficient per atom (i.e.,
NH(n = 1) = 1) for the levels n ^ 4. Assume T = 5800 K, corresponding to the solar
photosphere.
4. For the sun compare the number of hydrogen atoms absorbing at k 3300 A
with the number of H " ions absorbing at that wavelength. Assume a temperature
T = 5800 K and an electron pressure log Pe = 1.3.
5. For the sun calculate the number of Si I atoms absorbing at 1600 A and compare
with the number of hydrogen atoms absorbing at this wavelength.
^.on (Si) = 8.15 V. For this estimate use u+ u and gx g0.
Chapter 8
1. Just for the fun of it assume that only the hydrogen absorption is important for
the sun, i.e., neglect H " , metals, electron scattering and Rayleigh scattering.
Calculate the emergent energy distribution in the continuum. For simplicity, use
Problems 235
a grey atmosphere temperature distribution and a depth independent absorption
coefficient, i.e., use /c(H) for Teff = 58OOK. Plot Fx(0) for the surface.
Chapter 9
1. Assume that the temperature and pressure dependence of the continuous
absorption coefficient /cgr can be approximately described by kgr = KOT. This is a
reasonably good approximation for temperatures between 6000 K and 10 000 K.
Calculate how the gas pressure increases with f in a given atmosphere.
Chapter 10
1. Explain why spectral lines in the red are generally weaker than the lines in the
blue and ultraviolet wavelengths.
2. For a grey atmosphere calculate the maximum line depth for a spectral line at
3000 A, at 6000 A, and at 10 000 A. Use Teff = 5800 K.
3. Derive how the line depth of an optically thin line of neutral iron, i.e., the Fe I
line depth, will change with decreasing metal abundances in solar type stars.
Derive how the line depth for an optically thin Fe + line will change with decreasing
abundances of the heavy elements. Assume that the abundances of all heavy
elements are reduced by the same factor.
Chapter 11
1. In Fig. 1 we show the spectra of the star a Lyr, which has a spectral type A0 V,
and of the star rj Leo, which has a spectral type A0 I. Determine the electron
densities for these two stars, making use of the Inglis-Teller formula.
236 Problems
2. How does the depth in the hydrogen line wings change with decreasing metal
abundances
(a) for solar type stars?
(b) for A stars?
Why are the hydrogen lines such a good means for the temperature
determination?
Fig. 1. Spectra of A stars with different luminosity classes are shown. For higher
luminosity stars the hydrogen lines are narrower. More Balmer lines can be seen
separately around 3800 A for the higher luminosity stars. (See Chapter 11,
problem 1. From Morgan, Abt and Tapscott, 1978.)
3. How does the depth in the hydrogen line wings change for increasing He
abundance. That is for decreasing hydrogen abundance,
(a) in solar type stars?
(b) in A stars?
4. How does the Balmer discontinuity change with decreasing abundances of heavy
elements
(a) for F stars?
(b) for B stars?
Table 1. Equivalent widths and oscillator strengths for some solar spectral lines
Chapter 13
The 21cm line of the hydrogen atom is a transition between two energy states
for which the magnetic moment orientation of the proton with respect to the
magnetic moment of the electron is different. The transition probability is extremely
small, namely, Aul = 2.84 x 10~ 15 s" 1 . The cross-section for collisional de-
excitation is Cul ~ 1 x 10" 1 5 cm2 .
Determine the excitation temperature of the upper level in the interstellar
medium with a kinetic temperature of 100 K and a density of about 1 atom per cm3 .
Chapter 14
1. Calculate the radiative gradient for the solar convection zone in a layer where
T = 10 000 K and log Pg = 5.33.
2. For any given layer in the hydrogen convection zone give the largest possible
ratio of F conv /F rad , or in other words, the smallest value for Frad/crTeff.
3. In the deep interior of stars the absorption coefficient kgr decreases with increasing
temperature approximately as /cgr oc T~ 3 - 5 (Kramers' opacity law). In layers where
there is no nuclear energy generation can there be convective instability?
Chapter 15
The height of the solar chromosphere is about 2000 km. After the second contact
during total solar eclipses this layer is slowly covered by the shadow of the moon.
Calculate how long some of the chromospheric spectrum can be seen. (We cannot
see the spectrum before second contact because the sunlight from the disc is too
bright.)
Problems 239
2. Calculate the temperature stratification in the lower part of a transition region
between chromosphere and corona, assuming a mechanical energy flux of
1 0 5 e r g c m " 2 s ' 1 and a damping length X = 1.5 x 1 0 6 T 0 5 cm. Use geff = g with
h
cm s~
4. In red giant chromospheres the electron densities are about ne = 108 cm" 3 . If the
temperature is 10 000 K and geff = g, how large would be the density at the sonic
point for such a star? How large would the mass loss be if the velocity were
100 km s "*? Assume that the red giant has a mass M = 1.5 M o and a radius of
200 R o .
6. (a) For the calculation of the Eddington luminosity limit for stars we have assumed
that GV + KV &GV = const. In reality there are a large number of absorption
lines. Assume that 20% of the spectrum is covered by strong absorption lines.
50
Fig. 2. (a) The frequency dependence of the absorption coefficient for our picket
fence model in Chapter 15, problem 6. (b) The assumed frequency dependence of
the flux in our picket fence model of problem 6.
240 Problems
For simplicity assume that these lines are evenly distributed over the spectrum
and that within these lines KL/GV = const. = 50. See Fig. 2. In such strong
absorption lines the flux is reduced, assume that the flux in the lines is 25%
of the continuum flux (see Fig. 2).
(i) By how much will the gravitational acceleration gr be increased due to
these lines?
(ii) What would be the luminosity limit for such stars for M = 1 M O and
M = 5OM0?
(b) In the Large Magellanic Cloud (LMC) stars the abundances of the heavy
elements are reduced by about a factor of 5 (the exact number is still uncertain)
as compared to galactic stars. Suppose the line absorption coefficients in the
above model (picket fence model) are reduced by this factor,
(i) By which factor is the radiative acceleration gr changed as compared to
galactic stars?
(ii) What would be the luminosity limit for a 50 M o star in the LMC?