Michelsen (UCSD) - Funky Electromagnetic Concepts
Michelsen (UCSD) - Funky Electromagnetic Concepts
Michelsen (UCSD) - Funky Electromagnetic Concepts
Concepts
The Anti-Textbook*
A Work in Progress. See physics.ucsd.edu/~emichels for the latest versions of the Funky Series.
Please send me comments.
Eric L. Michelsen
E
E
A
y
B
x
z
z
z
* Physical, conceptual, geometric, and pictorial physics that didnt fit in your textbook.
Instead of distributing this document, please link to physics.ucsd.edu/~emichels/FunkyElectromagneticConcepts.pdf.
Please cite as: Michelsen, Eric L., Funky Electromagnetic Concepts, physics.ucsd.edu/~emichels, 4/9/2013.
2006 values from NIST. For more physical constants, see http://physics.nist.gov/cuu/Constants/ .
Speed of light in vacuum
Boltzmann constant
Stefan-Boltzmann constant
Relative standard uncertainty
Avogadro constant
Relative standard uncertainty
calorie
4.184 J (exact)
Electron mass
Proton mass
Proton/electron mass ratio
Elementary charge
Electron g-factor
Proton g-factor
Neutron g-factor
Muon mass
Planck constant
Bohr radius
Bohr magneton
Other values:
Jansky (Jy), flux and spectral density
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Contents
1.
Introduction .............................................................................................................................. 5
Why Funky? ......................................................................................................................................... 5
How to Use This Document................................................................................................................... 5
Whats Wrong With Existing Electromagnetic Expositions?............................................................... 5
My Story ........................................................................................................................................... 5
Thank You ........................................................................................................................................ 5
Notation................................................................................................................................................ 6
2.
Circuits...................................................................................................................................... 7
Circuits Reference Desk ........................................................................................................................ 7
Brief Note on Phasor Analysis ........................................................................................................... 9
3.
Classical Electromagnetics...................................................................................................... 11
Just For Reference: Faradays Law ...................................................................................................... 11
Stunning Phasors and Fourier Space .................................................................................................... 11
Phasor Calculus ............................................................................................................................... 14
Time Averages ................................................................................................................................ 15
Polarization Vector ............................................................................................................................. 16
Sleepy Hollow: The Legend of the Headless Vector ......................................................................... 20
Poynting Vector For Linear Polarization.............................................................................................. 21
Beware of Solenoidal Poynting Vectors ........................................................................................... 21
Wave Packets...................................................................................................................................... 22
Phase Velocity and Group Velocity.................................................................................................. 23
Vector Potentials I Have Known.......................................................................................................... 24
Solving Laplaces Equation ................................................................................................................. 25
Two-D Laplace Solutions................................................................................................................. 26
Three-D Laplace Solutions............................................................................................................... 28
Boundary Conditions Determine Solutions....................................................................................... 33
Respecting Orthogonality .................................................................................................................... 35
Propagation In a Vacuum .................................................................................................................... 35
Waveguides ........................................................................................................................................ 37
Boundary Conditions and Propagation ............................................................................................. 38
Phase and Group Velocity In a Waveguide....................................................................................... 39
Cylindrical Hollow Waveguides....................................................................................................... 40
Multipoles: Dipoles and Quadrupoles .................................................................................................. 41
Quadrupoles .................................................................................................................................... 42
4.
Relativistic Electromagnetics.................................................................................................. 47
Construction of a Valid Frame of Reference ..................................................................................... 47
Time Dilation and Length Contraction ............................................................................................. 47
Transformation of E & B Fields ....................................................................................................... 48
Acceleration Without Force ............................................................................................................. 50
On-Axis Doppler ............................................................................................................................. 50
Transformation of Intensity (Power Density) .................................................................................... 51
How Big Is a Photon? ...................................................................................................................... 52
The Pressure of Light....................................................................................................................... 52
Example: Reflection Off a Moving Mirror........................................................................................ 53
Beaming.......................................................................................................................................... 54
Covariant Form of Maxwells Equations .......................................................................................... 54
5.
Shorts ...................................................................................................................................... 55
Coulomb Force Constant is Defined Exactly........................................................................................ 55
Charge Is the Fundamental Electrical Quantity .................................................................................... 55
Units of Measure: SI vs. Gaussian ....................................................................................................... 55
Bound and Gagged.............................................................................................................................. 56
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Appendices .............................................................................................................................. 64
References .......................................................................................................................................... 64
Glossary.............................................................................................................................................. 64
Formulas............................................................................................................................................. 64
Index .................................................................................................................................................. 72
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1. Introduction
Why Funky?
The purpose of the Funky series of documents is to help develop an accurate physical, conceptual,
geometric, and pictorial understanding of important physics topics. We focus on areas that dont seem to
be covered well in most texts. The Funky series attempts to clarify those neglected concepts, and others
that seem likely to be challenging and unexpected (funky?). The Funky documents are intended for serious
students of physics; they are not popularizations or oversimplifications.
Physics includes math, and were not shy about it, but we also dont hide behind it.
Without a conceptual understanding, math is gibberish.
This work is one of several aimed at graduate and advanced-undergraduate physics students. Go to
http://physics.ucsd.edu/~emichels for the latest versions of the Funky Series, and for contact information.
Were looking for feedback, so please let us know what you think.
My Story
The Funky series of notes is the result of my going to graduate school in physics after 20 years out of
school. Although I had been an engineer all that time, most of my work involved software and design
architectures that are far removed from fundamental science and mathematics. I expected to be a little
rusty, but I found that the rust ran deeper than I realized.
There are many things I wish I had understood better while taking my classes (first at San Diego State
University, then getting my PhD at University of California, San Diego). The Funky series is my attempt
to help other students acquire a deeper understanding of physics.
Thank You
I owe a big thank you to many professors at both SDSU and UCSD, for their generosity, even when I
wasnt a real student: Dr. Herbert Shore, Dr. Peter Salamon, Dr. Arlette Baljon, Dr. Andrew Cooksy, Dr.
George Fuller, Dr. Tom ONeil, Dr. Terry Hwa, and others.
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Notation
[Square brackets] in text indicates asides that can be skipped without loss of continuity. They are
included to help make connections with other areas of physics.
arg A
for a complex number A, arg A is the angle of A in the complex plane; i.e., A = |A|ei(arg A).
[Interesting points that you may skip are asides, shown in square brackets, or smaller font and narrowed
margins. Notes to myself may also be included as asides.]
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2. Circuits
Circuits Reference Desk
Which end of a resistor is positive? An inductor? A capacitor? A diode? A battery? It all comes
down to a few simple conventions. We start with the resistor, then proceed to the more complicated
devices.
These principles extend directly to AC analysis, using phasors and complex impedance.
dv
dt
cathode
i I 0 evq / kT 1
- v +
di
dt
vL
- v +
iC
v iR
- v +
- v +
- v +
anode
v constant
i arbitrary
For any circuit element, conventions define a reference polarity for the voltage, and the
reference direction for the current. We must write our I-V equations consistently with those
choices.
For a resistor: the current always flows from + to -, so we choose reference directions consistent with
that (diagram above, left). This allows us to write Ohms law without minus signs: v = iR.
For an arbitrary circuit, we may not know ahead of time which end of a resistor will end up being +.
For example:
For a capacitor: Things are a tad messier, because current does not always flow from + to . A
capacitor stores energy in its electric field. If we increase the voltage on the capacitor from zero, the
capacitor is drawing energy from the rest of the circuit (it is charging). In this case, it is qualitatively
similar to a resistor, and its current flows from + to . Therefore, to be consistent with resistors, we define
our reference directions for this case: reference current flows from reference + to reference (just like a
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resistor). But wait! The current through a capacitor is not related to the polarity of voltage across it; the
current is related to the rate of change of voltage:
iC
dv
dt
units: C / s C / V V / s
If the voltage is negative, but increasing (becoming less negative), the current through the capacitor is
still positive. Thus the I-V equation above is always valid: charging or discharging. When a capacitor is
discharging, either its voltage is + and its current is , or its voltage is and its current is +. Either way, the
capacitor is delivering energy to the circuit, and temporarily acts more like a battery than a resistor. But we
cant change our reference directions on the circuit based on the charging/discharging state of the capacitor
from moment to moment. Also, either way, the capacitors power consumed is
P VI
where
For an inductor: things are similar, but we exchange voltage and current, and replace electric
with magnetic. Again, the current does not always flow from + to . An inductor stores energy in its
magnetic field. If we increase the current from zero, the inductor is drawing energy from the rest of the
circuit (loosely, it is charging). In this case, it is qualitatively similar to a resistor, and its current flows
from + to . Again, to be consistent with resistors, we define our reference directions for this case:
reference current flows from reference + to reference (just like a resistor). But wait! The voltage across
an inductor is not related to direction of the current through it; the voltage is related to the rate of change of
the current. Therefore, if the current is negative, but increasing (becoming less negative), the voltage
across the inductor is still positive. This allows us to write a single I-V equation for both cases:
vL
di
dt
When an inductor is discharging, the current is decreasing, and the inductor is supplying energy to
the circuit. Now it is acting more like a battery than a resistor. But we made our reference choice for the
inductor, and we must stick with it. The above I-V equation is valid at all times. Again,
P VI
where
Note that when the inductor current is increasing (either becoming more positive or less negative), v is
positive. When the current is decreasing (either becoming less positive or more negative), v is negative,
which means the + reference terminal is really at negative voltage with respect to the terminal. In all
cases, the above equations are correct. We achieved that consistency by defining a single reference
polarity.
For a diode: Resistors, capacitors, and inductors are all symmetric or unpolarized devices: you
can reverse the two leads with no effect. Diodes, in contrast, are polarized: one lead is the anode; the
other is the cathode. You must connect them properly. The reference voltage is defined always with +
on the anode (and therefore on the cathode); reference current flows from + to . Diodes always
consume power (like resistors do). Consistently with that, these conventions require that
P VI
For a battery: Batteries usually supply energy to the circuit, so we define them as having positive
power when they do so (the opposite of all other devices here). This requires the opposite reference
directions:
Batteries use the opposite convention from other devices:
reference current flows through the battery from reference to reference +.
However, it is possible to force current backwards through a battery, and then it will consume energy
from the circuit (as demanded by the fundamental definitions of voltage and current). This is how we
recharge a rechargeable battery. Thus:
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large V,
forward
current,
positive
power
small V,
reverse
current,
negative
power
P = VI where
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i1
i2
v2 +
v1 +
For a transformer: Transformers are a new breed, because they have 4 terminals, rather than two:
v1 M
di2
dt
v2 Nv1
v2 M
i2
di1
dt
1
i1
N
d
,
dt
v2 N 2
d
dt
v2
N2
v1 Nv1
N1
where
N2
N1
Also, an ideal transformer has a highly magntizeable core such that the flux, , requires virtually no
current to create it. This means that the primary and secondary currents must cancel each other, leaving
nearly zero MMF (magneto-motive force). Therefore:
N1i1 N 2i2
or
i2
1
i1
N
where
N2
as before
N1
dv
dt
i i Cv
where
The phase of the derivative is positive, which means the derivative leads the original function. In this case,
the current leads the voltage, or equivalently, the voltage lags the current (below left).
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leading
current
lagging
voltage
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leading lagging
voltage current
90o
90o
(Left) In a capacitor, voltage lags current. (Right) In an inductor, current lags voltage.
For inductors, things are reversed:
v (t ) L
di
dt
v i Li
where
The voltage leads the current, or the current lags the voltage (above right).
You can remember the lead/lag relationships for circuit elements as follows:
Capacitors oppose a change in voltage, so the voltage lags the current.
Inductors oppose a change in current, so the current lags the voltage.
Similarly:
You can remember the lead/lag relationships for derivatives as follows:
Derivatives show where youre going before you get there, so they lead the function.
Integrals sum up where youve been, and so lag the function.
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3. Classical Electromagnetics
Just For Reference: Faradays Law
What is the meaning of the minus sign in Faradays law? Faradays law relates the induced voltage in
a circuit surrounding a magnetic field to the rate of change of that field:
V
d
dt
The minus sign can only be interpreted with respect to a standard set of reference directions and
polarities. As with reference directions in electric circuits (see Circuits Reference Desk elsewhere), a
reference direction (of a current, B-field, or E-field) is the direction which is called positive in the
equations. A reference polarity (of a voltage) is the polarity which is called positive in the equations. The
actual polarity may be the same as our reference choice, or it may be opposite. We may not know until we
solve some equations. But there is no problem either way, because if the actual polarity is opposite our
reference choice, it will simply have a negative value.
y
x
+
Figure 3.1 Reference directions for flux , current, and reference polarity for voltage.
Now, Faradays law: the reference direction for current is that which would produce the reference
direction for , and is in the same direction as B. In the diagram above, B and are positive in the zdirection (out of the page). The reference direction for the current is counter-clockwise, which produces
(by the right hand rule) a B-field out of the page. We can lump the resistance of the loop into a single
equivalent resistor. The reference polarity for the voltage must be consistent with the reference direction
for the current, and this forces the choice shown. We now have a consistent set of reference directions and
polarity for all four of the flux, B-field, current, and voltage. Positive voltage makes positive B-field.
This defines the meaning of the minus sign in Faradays law. Lenz law says that if the B-field
changes, the voltage induced will try to drive a current that produces a B-field which opposes the change.
This is the minus sign: decreasing B (negative d/dt) causes positive voltage, which boosts B. Increasing
B (positive d/dt) causes negative voltage, which reduces B.
The minus sign emphasizes that if the induced voltage reinforced the change in B-field, that would
induce more voltage, further changing the B-field, which induces more voltage, further changing the Bfield, in a never ending death spiral of infinite current and B-field.
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Working with phasors is also called working in Fourier space or in Fourier modes. A phasor can
also be called a complex amplitude [Gri E&M p??]. (Quantum mechanics uses the term complex
amplitude for a complex number whose relationship to a sinusoid is fairly abstract.)
This section requires that you understand complex numbers in both polar and rectangular form, and the
basic calculus of complex functions of real variables.
We start by noting that any real-valued sinusoid is fully characterized by 3 numbers: amplitude, phase,
and frequency:
c(t ) C cos( t ),
C real .
A phasor is a complex number that characterizes the amplitude and phase of a sinusoid.
A phasor says nothing about its frequency. You must know the frequency from some other condition.
Combining two phasors by simple addition only makes sense if they refer to sinusoids of the same
frequency. However, phasors of different frequencies are often combined by inserting the time dependence
explicitly before combining (quantum mechanics does this routinely).
A phasor A (a complex number) corresponds to the sinusoid (in the engineering time convention):
The sign of the times: For engineers (including electromagnetics) and classical physics [M&T, Tay],
and for AC circuit analysis, the time dependence is e+it. For most physicists in quantum mechanics and
electromagnetics, the time dependence is eit. You can remember this mnemonically by thinking that for
wave physicists, time goes backwards. The polarity of the exponent is purely conventional, and has no
physical significance. This work shows pictures for both time conventions.
The magnitude of the phasor is exactly the (real) amplitude of the sinusoid, and the complex angle of
the phasor is exactly the phase of the sinusoid, i.e. the angle of the cosine at t = 0. The geometric
interpretation of a phasor is that of a rotating stick, which casts a shadow on the horizontal axis. The
length of the shadow is a sinusoidal function of time, with amplitude |A|, and starting at an angle arg A:
t=0
|
|A
arg A
shadow
t1 > 0
t 2 > t1
shadow
shadow
Figure 3.2 NB: For engineers, the stick rotates counter-clockwise, as shown, per e+it..
For wave physicists, the stick usually rotates clockwise (opposite to that shown), per eit.
We can also view the rotation at frequency as complex multiplication by e+it (engineering), or eit
(physics). Recall that multiplication by a unit-magnitude complex number simply rotates another complex
number in the complex plane. Now imagine the unit-magnitude angle is not fixed, but changes linearly
with time, i.e. multiply not by a fixed complex angle , but by an increasing angle t. is the angular
frequency, in rad/s. When we multiply some complex number r ei by eit, we get a complex function of
time that rotates continuously around the origin in the complex plane. The magnitude of the result is fixed,
because |eit| = 1 at all times. But the angle of the result increases with time, at the rate .
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imaginary
imaginary
r
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(r ei)eit
real
(r ei)eit
real
Figure 3.3 Physics: (Left) Rotation in time at negative frequency < 0. (Right) Rotation in time
at frequency > 0. Engineering: (Left) Rotation in time at frequency > 0. (Right) Rotation in
time at negative frequency < 0.
The angular frequency is not constrained to be positive; it can just as well be negative. Rotation by a
negative frequency rotates in the clock-wise direction, rather than counter clockwise. Hence, both positive
and negative frequencies occur in complex rotations, and in signal analysis.
Recall that any linear combination of sinusoids, of arbitrary amplitude and phase (but identical
frequency, of course), is another sinusoid (at that frequency). The beauty of phasors is that the phasor
representing the sum of two sinusoids is simply the (complex) sum of the phasors representing the original
sinusoids (addends). The graphical demonstration of this is both simple, and illuminating:
A+B
t
t=0
A+B
t1 > 0
t2 > t1
t
t
A
shadow
shadow
shadow
Figure 3.4 Demonstration that the sum of any 2 sinusoids is a sinusoid, and its phasor is the sum
of the constituent phasors. Note that all 3 vectors rotate together. The sum of the shadows is the
shadow of the vector sum. NB: For physicists, the sticks usually rotate clockwise per eit.
Note that the phasor for cos t is 1; the phasor for sin t is i (physics, or -j for engineering). Thus, we
can represent the sine wave a(t) by its in-phase (cosine) and quadrature (sine) parts (in physics notation e
it
):
Let phasor
A Ar iAi .
Then
Thus we see that the real and imaginary parts of the phasor A are exactly the in-phase and quadrature
components of the sine wave.
We use complex numbers to represent sinusoids because the arithmetic (and some calculus) of
complex numbers (2D vectors) is the same as the arithmetic of adding sinusoids.
Direction of the wave-vector: For traveling waves, we can see that the wave-vector points in the
direction of propagation, by considering a point of constant phase on the wave. For constant phase, we
must have
k x t const
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because as t increases, so must kx, and therefore the point of constant phase (x x + dx) must move in
the same direction as k points.
Even in the engineering convention, with time evolution given by exp(+it), the wave-vector k still
points in the direction of propagation. The condition for propagation is t k x const , which can be
brought into the physics-convention form by absorbing a minus sign into the constant.
Phasor Calculus
We can easily see that phasors convert differential equations to algebraic equations. This is expected,
because phasors are a method of Fourier analysis, which is well known for converting differential to
algebraic equations. Lets take the first and second derivatives of a cosine, in both the old-fashioned realvalued way, and this new-fangled phasor way:
We can take time derivatives of phasors by noting that the time derivative of the real part of a complex
function equals the real part of the time derivative:
Let
z (t ) zr (t ) izi (t ).
Then
d
d
Re z (t ) Re z (t )
dt
dt
because
d
d
d
d
d
Re z (t ) Re zr (t ) i zi (t ) zr (t ) Re z (t ) .
dt
dt
dt
dt
dt
d
d
Then
Re Aeit Re Ae it Re i Ae it
dt
dt
d
In phasor notation:
A i A
dt
A 1
d
cos t sin t
dt
d
A i A
dt
d
dt 2
cos t 2 cos t
d2
dt
A i A 2 A
2
We can also have phasors defining both space and time sinusoidal variations.
These can be used for traveling waves.
Then the phasor carries the amplitude and phase of the traveling wave, but not its wave-vector k (spatial
frequency) or temporal frequency . We must be given k and separately. In 1-dimension, the spatial
derivative works like this:
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Let
z (t , x ) zr (t , x ) izi (t , x) .
Then
d
d
Re z (t , x) Re z (t , x )
dx
dx
d
d
d
In phasor notation:
A ikA .
dx
and
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d
,
dt
z (t , r ) zr (t , r ) izi (t , r ) .
Then
Re z (t , r ) Re z (t , r )
Then
d
.
dx
A ikA .
In phasor notation:
Time Averages
It is often useful to compute the time average of the product of 2 sinusoids. E.g., the time average of
the Poynting vector gives the effective radiation power density in W/m2 (or power per unit area). The time
average of two sinusoids does not depend on the absolute phase of either; it depends only on the relative
phase of the two. Specifically,
Let
iB b(t ) Re | B | e
A be a phasor
B be a phasor
B Br
a(t )b(t ) t
1
Re AB * .
2
it arg B
How so?
One way to see this is to decompose the phasors into the cosine (real) and sine (imaginary)
components.
a(t ) Ar cos t Ai sin t
a ( t )b ( t ) t
Ar Bi cos t sin t
Ai Br cos t sin t
Ai Bi sin 2 t
Only the cos-cos and sin-sin terms contribute to the time average, because the time average of cos-sin is
zero (they are orthogonal functions; sin is odd, cos is even). Therefore,
Use
cos 2
sin 2
1
2
Ai Bi sin 2 t
1
1
1
Ar Br Ai Bi Re AB *
2
2
2
[Notice that Re{AB*} is analogous to the dot-product of two spatial vectors: it is the component of B parallel
to A, times |A|, i.e. AB cos , where = arg B - arg A. Only the parallel components contribute to the time
average.]
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A+B
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t2 > t1
t1 > 0
t=0
A
B//
AB//
shadow
shadow
AB*
B*
a ( t ) b( t ) t
period
it arg B
AB
AB
The last equation is because the integral is over a full period, so shifting the starting and ending point
by a fixed time interval, or angle, doesnt change its value. This demonstrates that the time average
depends only on the phase difference between A and B, and not their absolute phases. Finally,
Use
a(t )b(t ) t AB
c t , d arg A arg B
period dt cos t cos arg A arg B sin t sin arg A arg B cos t
dt cos 2 t
AB cos arg A arg B
period
1
1
1
i arg A arg B
AB cos arg A arg B Re AB e
Re A ei arg A B ei arg B
2
2
2
1
Re AB *
2
QED
Polarization Vector
This section assumes you understand phasors, and the decomposition of a sinusoid into in-phase and
quadrature parts (see Phasors, above). We start with an overview, and then fill in the details.
Overview: A polarization vector in general applies to any vector (or other multi-component object)
which travels in space, and varies sinusoidally in time and space. It gives the relative amplitude and phase
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of each component of the vector. By convention for EM waves, the polarization vector, , gives the relative
amplitudes and phases of the E-field components (Ex, Ey, Ez) for the wave.
In applications where the A field is most relevant, we write a polarization vector for the A-field,
instead. In this case, is gaugedependent. In some gauges, it has unexpected components, including a
longitudinal component, which points along the propagation direction. This is a gauge artifact, and the
polarization vectors for E and B-fields (in vacuum) always have no component along the propagation
direction.
In principle, one could write a polarization vector for a B-field, but its not usually done since it is just
a right-hand rotation of the E-field polarization vector about the propagation direction by 90 o.
[Note that for gravity waves, where the propagating field is the 4 4 metric tensor perturbation field, h, the
polarization object is a 4 4 polarization tensor.]
2.
the fraction of the total EM wave intensity (power density) carried by each component of
polarization
The polarization vector is a set of 3 phasors, that describe the sinusoidal oscillations of each
component (Ex, Ey, Ez) of the E-field. So we can immediately write the form of the E-field at a point as a
function of time (in several different notations):
E Re e i t Re x x y y z z e it
Re ei t
x
x
it
i t
Re y e
Re y e
z
Re z e it
It is possible to represent any polarization (linear or elliptical) in any propagation direction with a 3D
complex vector. Since EM waves are transverse, the E-field, and thus the polarization vector, are
perpendicular to the wave-vector k, i.e., k = 0:
y
y
R
I
x
Figure 3.6 (Left) Wave vector k in an arbitrary direction, and some possible real polarization
vectors. (Right) Wave vector k, and the real (blue) and imaginary (red) parts of a complex
polarization vector.
To keep the x, y, and z intensities normalized with respect to total intensity, the polarization vector is
usually normalized to unit magnitude:
x , y , z x x y y z z ,
2
x y z
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Note, however, there is a subtle distinction between the amplitudes of spatially perpendicular components,
and the instantaneous maximum magnitude of the electric field, Emax. For a linearly polarized wave, say
halfway between the x and y-directions, Emax = (2)Ex, and the time averaged intensity <I> ~ Emax2/2.
However, for the same amplitude of perpendicular electric field components, but phase shifted to make
circular polarization, Emax = Ex = Ey, and <I> ~ Emax2. Note that the intensity is the same for both waves:
perpendicular component intensities always add; it is Emax which is different.
For linearly polarized waves, <I> ~ Emax2 / 2, whereas for circular waves, <I> ~ Emax2.
The real part of the polarization vector is the in-phase E-field parts. The imaginary part is the timequadrature E-field parts, i.e. is 90o out of phase with the real part. Any polarization vector for a
propagating wave in a given direction k can be written as a linear combination of two basis polarization
vectors for that k.
E-field
Circular polarization,
special case of elliptical
Linear polarization
Elliptical polarization
1 2
E
2
E 2
(gaussian)
c 2
E
4
In general:
E m P,
m 2
P
c
(SI)
[L&L p120]
or
(gaussian)
Consider an EM wave traveling in an arbitrary direction, and with arbitrary polarization. At any given
time, the wave has an E-field with components in the x, y, and z directions:
E(t ) E x (t )x E y (t )y E z (t )z
Each component is a sinusoid of frequency = c|k|, and so may be represented by a phasor (recall that
a phasor is a complex number which represents the amplitude and phase of a sinusoid; see Phasors above):
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E(t ) m Ptotal Re x e it x Re y e i t y Re z e it z
m Ptotal Re x x y y z z e it
where
For elliptical polarization, both the magnitude and direction of the E-field are functions of time.
The propagation vector is k (not to be confused with k z-hat), whose magnitude is the spatial
frequency in rad/m, and whose direction is the propagation direction. For a given E-field component j
{x, y, z}, the intensity (power/area) carried in the k direction is proportional to Ej2, where Ej is the (real)
amplitude:
2
Px m2 E x 2 x Ptotal ,
Py m2 E y 2 y Ptotal ,
Ptotal Px Py Pz W/m 2
Pz m2 Ez 2 z Ptotal
2
x y z
So |x|2 is the fraction of the total wave intensity carried by the Ex component, etc.
It is instructive to consider the real and imaginary parts of separately:
R i I
where
R is the direction of the E-field at the reference time, t = 0, and all integer oscillation periods after,
t = nT = 2n/ (Figure 3.6, right diagram, blue arrow). I is the direction of the E-field at quadrature times,
t = (n + )T (Figure 3.6, right diagram, red arrow). The E-field is always perpendicular to k, so both R
and I lie in the plane perpendicular to k.
R k I k 0
k 0 0i 0
Basis polarization vectors: Since EM waves are transverse, the E-field (and therefore polarization
vector) lies in a plane perpendicular to k. This plane is a 2-D space, and all vectors in that plane (for a
given direction of k) can be written as the linear combination of 2 basis vectors. For complex vectors like
polarization vectors, the basis vectors can be either real or complex. We always choose the basis vectors to
be orthonormal, i.e.
e1 e1 e 2 e2 1
e1 e2 0,
where
[The dot product of complex vectors is just like a quantum inner product.] The only two bases you
ever see are linear polarization bases, and circular polarization bases. To simplify the discussion, we now
focus on propagation in the z-direction (k = kez). There are 2 orthogonal directions of linear polarization, x
and y. Therefore, our linear polarization basis vectors are simply:
e x 1, 0, 0
e y 0,1, 0
For z propagation, the plane of the E-field is the x-y plane; thus every polarization vector for z
propagation can be written as a (possibly complex) combination of ex and ey. The polarization vector for
100% x polarization is just = e1 = (1, 0, 0). For 100% y polarization, = e2 = (0, 1, 0). For linear
polarization at 45o, = (1/2, 1/2, 0), which means the power is carried in Ex and half in Ey.
For right-hand-circular polarization (RHC), E(t) rotates counter clockwise (right hand rule applied to
k). This means E(t = 0) points in the x direction, so R ~ (1, 0, 0). E(t = T/4) points in the y direction, so I
~ (0, 1, 0). Then
~ R i I 1, 0, 0 i 0,1,0 1, i,0 .
Normalizing ,
1
2
1, i, 0
For LHC, E(t = T/4) points in the y direction, so = (1/2, i/2, 0). We have just derived the
circular polarization basis vectors:
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eR
1
1, i, 0
2
eL
emichels at physics.ucsd.edu
1
1, i, 0
2
[Beware that in optics, RHC is called left handed and LHC is called right handed. Beats me.]
Examples: Lets use our polarization basis vectors to find out how much power passes through some
polarization filters.
(1) Given an elliptical wave with = (3/5, i4/5, 0), what fraction f of the power will survive a ypolarizing filter? The filter passes only the part of the wave with = (0, 1, 0). Therefore, we project our
given polarization onto this, i.e., we see what fraction of our wave is Ey-polarized. This is just like finding
how much of a basis vector is in a quantum state. Then we square that, to get intensity:
2
f e y 0,1, 0 3/ 5, i 4 / 5,0 i4 / 5 16 / 25
2
A circular wave, = (1/2, i/2, 0), and a 45o linear wave, = (1/2, 1/2, 0), would both pass the
power.
(2) How much of an x polarized wave survives a chiral filter that passes only RHC?
f e R 1/ 2, i / 2,0 1,0, 0 1/ 2 1/ 2
2
The answer is the same for any direction of linear polarization (axial symmetry):
f e R 1/ 2, i / 2,0 a, b, 0
2
a ib
1/ 2,
since
1 a ib
Notice that in these cases, linear polarized light went in, but circular polarized light came out.
(3) How much of an LHC wave would pass the RHC filter?
f e R eL
1/ 2, i / 2, 0 1/ 2, i / 2, 0
1 1
0
2 2
Irrelevant point: Sometimes, you can find the direction of k from the polarization vector, sometimes
not. If R and I point in different directions, then because they are both perpendicular to k, R I points
along k. But if R and I are parallel, or if I = 0, then R I = 0, and you cant tell anything. So far as I
know, this fact is of no use at all.
TBS. Angular momentum of elliptically polarized waves.
Extension to other waves: For things like gravity waves, the field which varies sinusoidally is a rank2 tensor, which can be written as a 2D matrix (below, left):
tt
xt
(t , x) Re
yt
zt
tx
ty
xx
xy
yx yy
zx
zy
tz
xz it
e
yz
zz
1
0
Re
(t , x)
2
0
0 0 0
1 0 0 it
e
0 1 0
0 0 0
Above right is +, the polarization tensor for + polarization propagating in the z-direction.
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is reasonable, because the polarization is fully described by the direction of the E-field, at a given reference
time, and the E-field is a headful vector.
Recall every vector lies along an axis, and that the sense of a vector is simply which way the vector
points along the axis. In some cases, including most cases of polarization, we know the axis of the E-field,
but not its sense. Also, recall that the polarization vector specifies the complete (headful) direction and
phase of the E-field for a propagating wave.
Interference occurs between headful E-field vectors, including their sense. To illustrate, suppose we
superpose two equal-amplitude linearly polarized EM waves, with polarization vectors 1 and 2. Then at
the point of superposition, we get:
E1 (t ) Re E0 1e i t ,
E2 (t ) Re E0 2 e it ,
E1 E2 Re E0 1 2 eit .
Suppose that 1 and 2 are real and parallel: 1 = 2. Then we get constructive interference. Now
suppose we rotate in space the 2nd wave about its propagation direction by 180o. This is not a phase-shift;
there is no time delay involved. After rotation, the polarization vector gets negated: 2 1. The
interference is now destructive:
E1 E2 Re E0 1 2 e i t 0 .
Therefore, we see that 1 and 1 (or 2 and 2) are different polarizations. One is the negative of the other.
If 2 = (ex, 0, 0) then it is horizontal polarization. Then 2 is also horizontal polarization, but its a
different horizontal: it is the negative of 2s horizontal.
Thus the polarization vector describes not only the plane of polarization, but the sense of the E-field in
that plane. For linear polarization, you can write the polarization as a unit vector parallel to the E-field.
This is a simple case of a polarization vector.
Of course, the sense reverses every half-period, so the sense of the polarization vector is defined at
some reference time. In fact, the polarization vector describes the full 360o phase (in time) of the E-field,
and phase is always relative to some reference time (or phase), taken as zero. The phenomenon of
constructive or destructive interference is, as always, independent of our reference time (or phase).
Summary: Some experiments are not sensitive to the sense of the polarization, and therefore measure
only the plane of polarization. They often treat such a polarization measurement as a headless vector,
which is good enough for some applications. However, we have shown that interference reveals that
polarization is a headful vector, even if the sense is unknown.
S(t ) E(t ) H (t )
where
E E, H H .
Quite often, we are more interested in the time-averaged power density, rather than the instantaneous
power density. Then we use the fact that <sin2> = , over the long term (or an integral number of cycles):
S
1
,
EH E H
2
where
E, H
H are unit vectors in direction of E and H .
E
E
H
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This term seems to contribute to a Poynting vector, yet is said not to contribute to radiation. Some
references explanations are misleading: they say that the velocity term falls off as 1/r4, and so becomes
insignificant compared to the 1/r2 radiation term. While true, this is not the crux of the matter. The
Poynting vector describes the power density (= energy flux) of the propagating radiation: watts/m 2.
Regardless of how the magnitude of the Poynting vector falls off, a radial component to it represents real
energy flow. In the far field, we can imagine thick spherical shells around the radiation source. Since the
shells are in vacuum (or non-absorbing material), the power (energy/time) flowing into any shell (from the
source inside) must equal the power flowing out of that shell. Since the shell surface area increases as r2,
the power density must decrease as 1/r2, so that the product of the two, the total power, remains constant.
This is just conservation of energy. Therefore:
Conservation of energy requires that all power flows fall off exactly as 1/r2.
There cannot be any power flow that falls off as 1/r4, because it would violate conservation of energy.
So what of the velocity term? The real explanation is that the Poynting vector resulting from the velocity
term is solenoidal, i.e. the lines of power flow close on themselves. No power flows out; its almost as if
the power just flows around in circles. The radial component of the velocity term Poynting vector is 0:
S velocity r 0
So there is no outward power flow. The rate of fall off (1/r4 or anything else) is irrelevant.
TBS: Poynting Vector For Arbitrary Polarization
Wave Packets
Often, one sends finite-time messages or signals with electromagnetic waves (or some other linear
medium: wires, water, air, etc.). Such a signal can be decomposed into a sum of sinusoidal frequency
components (i.e., the Fourier Transform), each frequency component with its own amplitude and phase.
Since many propagation phenomena are frequency dependent, it is important to ask: What range of
frequencies are needed to add together to form a finite time signal? We will argue that a continuous band
of frequencies, of finite upper and lower bound, is roughly sufficient to construct a finite time signal. The
difference between the upper and lower frequency in a signal is called the bandwidth (BW). (This term is
much abused; in particular, it is frequently used (incorrectly) to mean data capacity.) We give a handwaving argument for a crude estimate of the bandwidth of a finite-time signal. When viewed as a sum of
sinusoidal waves, such a signal is called a wave-packet.
Fourier Transform:
Wave-packet in Frequency
amplitude
Wave-packet in Time
amplitude
time
t
frequency
BW
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sinusoids will have lots of positive and negative components, and will add up to approximately zero. To
have a significant contribution to our signal, most of the instantaneous phases need to be within a radian or
two of each other. And this has to be true throughout the width (in time or space) of the wave packet.
Since the signal comprises different frequencies, the instantaneous phase difference between frequency
components varies over time (or space). Components in-phase at one point will be out-of-phase at a
distant point. (Below) the low- and high- frequency components (red and blue) start out in-phase, but drift
to opposite phase over some time.
amplitude
upper
lower
time or
space
Wave components of different frequencies drift out of phase over time (or space).
The rate at which the phases drift apart is exactly the frequency difference between the two
components:
1 1t 1 ,
2 2 t 2
1 2 1t 1 2t 2 1 2 t const t const , or
k z const
The constant term is not important, and simply depends on the shape of the signal and the origin of our
t (or z) coordinate. What matters is that as t (or z) increases, so does the phase difference. Choose the
coordinate origin to be the start of the wave-packet; then the width of the packet is t (or z). Clearly, the
largest occurs when the frequency difference is maximum, i.e., at
or
Since must remain within 1 or 2 radians over the width of the signal, we have
~ 1 or 2 rad BW t
BW ~
1 or 2 rad
t
or
BW ~
1 or 2 rad
z
A more detailed analysis yields a theorem which states: a signal of finite time must (strictly speaking)
have infinite bandwidth, and conversely, a signal of finite bandwidth must extend over infinite time.
However, one can have an approximately finite-time and finite-bandwidth signal (for all practical
purposes), because the amplitudes outside a small interval are negligibly small.
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Phase Velocity
amplitude
emichels at physics.ucsd.edu
Constant Phase
Velocity
= vphasek
1 sec
time
1m
vphase = /k
We have 4 rad/s, and rad/m. The wave must then travel with
v phase
4 rad / s
4 m/ s
k rad / m
If all frequencies propagate at the same speed, then = vphasek (above right).
However, in most linear media, the propagation speed of a sinusoid depends on its frequency. Since a
wave-packet is made up of many frequencies, which frequency should we use to determine its speed? You
might guess the center frequency. Turns out, though, that what really matters is just how the speed varies
with frequency. The wave-packet, or group of waves travels with a speed (group-velocity)
vgroup
d
.
dk
Actually, since each component travels at a different speed, the wave packet does not truly travel as a
group. It gets smeared over time, because the relative phases of its components get misaligned. This
smearing is called dispersion, and in some cases is a serious limitation on a communication system. More
on phase and group velocity in Waveguides, later.
0
4
J (r ') 3
d r'
r r'
However, many problems start with a given magnetic field (rather than current distribution), but our
analysis tools use the vector potential. How can we find the vector potential for a given magnetic field?
There is no explicit inverse curl, but usually it is fairly easy to do by inspecting the formula for curl in the
coordinates of interest. For example, for a constant B-field, B A Bz z and rectangular coordinates
(chosen from other symmetries of the problem), we look at the formula for curl in those coordinates:
A Ay
Ax Az
A z
z
x
y
z
Ay Ax
y x y z
We are given only a Bz component, so we look at the coefficient of z above. We see that we can get it
from either a Ay/x, or a Ax/y term. Both terms are simple, so we choose one, and set
Ay
x
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Bz
Ay ( x )
dx Bz Bz x
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If other aspects of the problem favored an Ax instead, we could choose the other term in the curl. Note
that the curl operator is linear, so if there are more components of B, we can build them up from A terms,
and sum them for the final A.
Suppose we have a constant Bz in cylindrical coordinates,
1 Az A
1
A
Ar Az
A
rA r z
r
z
r
r r
z
r
1
1 Ar
rA and
. If the problem has axial
r r
r
symmetry, then the 2nd term must be zero. Also, the 2nd term cannot have a constant derivative, and be
single valued at = 0 = 2. So we choose the first term, and set
1
rA Bz
r r
rA rBz
r
rA
rBz dr
Bz 2
r
2
Bz
r
2
Spherical coordinates dont come up all that often in magnetics problems, and is a little more
complicated, but we can use the same method.
A
A
1
sin A
r sin
1 Ar 1
1
A
rA rA r
r
r r
r sin r r
For a constant Bz, we must first decompose z into its r and components:
1
r sin
1
r sin
A
sin A cos
Ar 1
rA sin
r r
z cos r sin
Choose : A 0
1
sin A cos
r sin
sin A r
d cos sin
r
sin 2
2
r
sin
2
Ar 1 r
r sin sin
r r 2
Ar
Ar
sin sin ,
0
Ar const , choose
Ar 0
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We consider both 2-D and 3-D BVPs, in all common coordinate systems: rectangular, polar,
cylindrical, and spherical. All problems start from Laplaces equation,
2 0 ,
plus some boundary conditions on . In all coordinate systems, the nontrivial solutions arise from
separations of variables. Separation of variables hopes that we can write the solution as a product of
functions, each of a single coordinate, such as
( r , , z ) R ( r )Q ( ) Z ( z ),
( x, y , z ) X ( x)Y ( y ) Z ( z ),
( r , , ) R( r)Q( ) P( ) .
or
We restrict ourselves to real-valued functions and parameters, since they are of most interest to E&M.
Finally, note that Laplaces equation is linear, so that the weighted sum (linear superposition) of any
number of solutions is also a solution. We use this property extensively in what follows.
We cover the various cases in order of difficulty. We do not consider orthogonal function expansions
for matching boundary conditions; see regular texts [refs??].
2
2
2,
2
x
y
or
1 1 2
r
.
r r r r 2 2
By inspection, this clearly satisfies Laplaces equation. Both terms of the Laplacian operator are zero.
It turns out, though, that this solution doesnt come up very often in E&M problems, because it doesnt
satisfy most physical boundary conditions. Nonetheless, it is a perfectly valid solution, and should be
considered in some cases. (It comes up in some 1D quantum mechanical scattering problems.)
More interestingly, we consider separations of variables, which requires
( x, y ) X ( x)Y ( y )
Consider the two terms of the Laplacian in rectangular coordinates, i.e., the partial derivatives w.r.t. x
and y:
2 X
2 x 2 x
Y ( y ),
2
2 y
X ( x)
2Y
2 y
2 X
2
x
2Y
Y ( y ) X ( x ) 2 0
y
If we could find X(x) and Y(y) with these properties, wed be set:
2 X
kX ( x ),
x 2
and
2Y
kY ( y ),
y 2
then
2 kX ( x )Y ( y ) kX ( x)Y ( y ) 0
This implies that X(x) and Y(y) are eigenfunctions of the 2nd derivative operator. In rectangular
coordinates, we can easily find such a solution. Recall that sinh(x) and cosh(x) (and equivalently exp(x)
and exp(x)) satisfy this, and their eigenvalues are always positive. Also, sin(x) and cos(x) satisfy this, and
their eigenvalues are always negative. The solutions are therefore
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x
x
X ( x ) Ae Be ,
Y ( y ) C sin y D cos y ,
OR
2 X
2 Ae x Be x 2 X ( x),
and
2
x
2Y
2
2
C sin y D cos y Y ( y )
y 2
X ( x ) C sin x D cos x,
Y ( y ) Ae y Be y ,
( x, y ) X ( x)Y ( y )
2 X
2 C sin x D cos x 2 X ( x),
x 2
2Y
2 Ae y Be y 2Y ( y )
2
y
and
Note that
sinh( x)
1 x 1 x
e e ,
2
2
cosh( x )
1 x 1 x
e e ,
2
2
where
and
tan 1 B / A
sinh(x) and cosh(x) and exp(x) and exp(x) are linear combinations of each other, and therefore equivalent.
However, sinh(x) and cosh(x) are often convenient in E&M problems because sinh has a zero, and cosh has
a zero derivative, which are useful for matching common boundary conditions. exp(x) have no zeros, and
no zero derivatives. Also, E cos(x + ) is equivalent to A sin(x) + B cos(x).
As always, the boundary conditions determine the coefficients A, B, C, and D (or A, B, E, and in
the alternate cos(x) form).
More on boundary conditions later.
Two-D Laplace Solutions in Polar Coordinates
In polar coordinates, (r, ), we might try the same trick for separation of variables: eigenfunctions of
the two Laplacian operator terms, with eigenvalues that sum to zero:
2
1 1 2
r
r r r r 2 2
1
r R (r ) kR( r ),
r r r
Let
1 2
r 2 2
(r , ) R ( r )Q( ).
Solve
Q( ) kQ( )
We see immediately that this wont work, because of the 1/r2 factor in front of the Q() term, and Q()
has no r in it, by definition. But we dont really need eigenfunctions of the Laplacian terms; we just need a
function f(r, ) such that:
1
r
R (r ) f (r , ) R (r ),
r r r
1 2
r 2 2
Q( ) f (r , )Q( ).
and
Then
2 f ( x, y ) R (r )Q( ) R (r ) f ( x, y )Q( ) 0
Given that the 2nd term of the Laplacian has 1/r2 before it, the simplest choice for f(r, ) that we could
hope for is:
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Let
f (r , )
Then
R (r ) Ar Br
r2
emichels at physics.ucsd.edu
1
1
r R (r )
r Ar 1 Br 1
r r r
r r
and
Q( ) C sin D cos
So
f (r , )
1
1
2
Ar Br 2 Ar 1 2 Br 1 2 R(r )
r r
r
r
1 2
r 2
r2
Q( )
2
C sin
2
D cos
2
Q( )
r2
2
,
r2
and we have found our solution to Laplaces equation in 2-D polar coordinates. [It happens now that Q()
is an eigenfunction of 2/2, but R(r) is not an eigenfunction of anything.] Note that, depending on the
boundary conditions, may or may not be an integer. [Jac sec 2.11 p75] covers this well. In particular, if
goes completely around the origin, then must be an integer, because it must be that Q[(2+)] = Q(),
so that Q is single-valued at every point. The solutions are thus
R (r ) Ar Br ,
Q( ) C sin D cos ,
(r , ) R( r )Q( ) .
Note that if the domain of includes the origin, then R(r) = Ar, because Br blows up there.
There is one more case: = 0. If this is allowed by boundary conditions and the domain of , then the
two Laplacian terms are both zero. There are separate solutions for this case:
1 2
Q( ) 0
r 2 2
1
r
R (r ) 0,
r r r
R (r ) B ln
r
A
(sometimes written R( r ) K B ln r )
1
1
r
R(r )
r r r
r r
and
Q( ) C D
B 1
r
B0
r r r
1 2
r 2 2
Q( ) 0
(r , ) R (r )Q( ) B ln C D
A
Note that this possibility exists only if the domain of excludes the origin, because (ln r) blows up there.
Also note that if the domain of allows to go all the way around the origin, then D = 0, so that Q() is
single valued, and hence Q() = 1 (constant).
In this case of the domain surrounding but excluding the origin, we get the most general solution by
combining the ( = 0, D = 0) case with the other integer [Jac 2.71 p77]:
(r , ) B0 ln
r
A r B r
A0 1
sin D cos .
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2
2
2
2 2,
2
x
y
z
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( x, y, z ) Ax By Cz D .
All 3 Laplacian terms are zero. Again, it doesnt come up much, but Jackson doesnt even mention it.
For less trivial solutions, we extend the separation of variables approach from the 2D rectangular
coordinate case:
( x, y , z ) X ( x )Y ( y )Z ( z )
2 X
aX ( x ),
x 2
2Y
bY ( y ),
y 2
2 Z
cZ ( z )
z 2
If any coordinate, say z, can be separated into the form Cz + D, its eigenvalue is zero, and we revert to the
2D solutions for the remaining two coordinates. In the 2D case, we had only two terms, so the eigenvalues
had to be negatives of each other, so that they would sum to zero, and satisfy Laplaces equation. Recall
that the eigenvalues of sinh(x) and cosh(x) are always positive, and the eigenvalues of sin(x) and cos(x) are
always negative. This meant that one coordinate had to be A sinh() + B cosh() (or equivalently exp( ) ),
and the other had to be C sin() + D cos().
In the 3-D case, there are more possibilities, but all we require is that the 3 eigenvalues sum to zero:
abc 0.
This means at least one function must be sinh() or cosh() (or equivalently exp( ) ), at least one
function must be sin or cos, and the last function can be either, subject to the constraint that the eigenvalues
of all 3 functions sum to zero. Some example solutions are:
X ( x ) sin 2 x,
Y ( y ) cos 3 y,
Z ( z ) sinh
13 z
Y ( y ) e 12 y ,
Z ( z ) 2 sin(13z ) 7 cos(13z )
2 3 13 0
3 4 5 0
5 12 13 0
2
1 1 2
2
r 2 2 2 ,
r r r r
z
(r , , z ) R (r )Q ( )Z ( z ) .
(3.1)
The sum of the 3 Laplacian terms must be zero. From our 2D work, we have a Q() = sin or cos()
solution that can offset an R(r)/r2 term, and a Z(z) solution that can offset either a + or constant factor
[Z(z) = sinh or cosh(kz) or Z(z) = sin or cos(kz)]. Wouldnt it be nice if we had an R(r) that would produce
the negative of the sum of those Q() and Z(z) terms? That is:
1 2
r 2
2 Z
z 2
Q( )
2
2
Q( ),
r2
Z ( z) k 2 Z ( z)
(3.2)
2 2
1
r R (r ) k 2 R ( r )
r r r
r
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For now, we call the eigenvalue of Z(z) k2, and it is positive. We dont have any pre-existing
functions that satisfy the R(r) equation, so we do the obvious: we make one! Using power series solutions
and some fancy math (that we dont need to know here), mathematicians define a family of functions which
satisfy the above R(r) equation for k2 > 0, and call them Bessel functions. Every value of gets its own
Bessel function, named J(kr). Recall that can be any real number, so there is an uncountably infinite set
of Bessel functions. Bessel functions have lots of magic properties (that we dont need to know here). We
dont need a separate Bessel function for each value of k, because a simple change of variable eliminates k
(w = kr) [I need to clarify this??]:
Let
w kr , r
w
.
k
Then
2 2
1
r R (r ) k 2 R (r )
r r r
r
Define
, and
w
w
w 2 k 2 2 w
k
k
k
R k 2 R
w w k w k
x k
w
J ( w) R . Then
k
2
1
w J ( w) 1 2 J ( w)
w w w
w
Then
w
R J ( w)
k
R (w) J (kw),
or
R (r ) J (kr )
Notice that since the eigenvalue of Z(z) > 0 (k2 > 0), Z(z) = sinh(kz) or cosh(kz). (Well get to the Z(z)
= sin or cos case in a minute.) But wait a minute, something is missing! We have 2 linearly independent
solutions for Q(), two linearly independent solutions for Z(z), but only one solution for R(r). Basic
differential equation theory tells us we need 2 solutions for each of the 3 variables. Well it turns out, for
non-integer , we have two Bessel functions: J and J. So the solution to Laplaces equation is really
non-integer
Finally, since Laplaces equation is linear, the linear combination of any two solutions is also a
solution:
(r , , z )
Note that each value of has its own set of A, B, C, D, E, and F. Also, the values of over which to
sum are determined by the problem from the allowed range of and the boundary conditions.
However, a very common case is = integer (solution valid for all around the axis). In that case, J
is a multiple of J, and the two solutions are not independent. So we do more math, and find a variant of
the Bessel function which satisfies the equation, and is linearly independent. Such a function is a Neumann
(pronounced noy-mon) function, also called a Bessel function of the second kind. Our solution is then
R (r ) AJ (kr ) BN (kr ),
(r , , z )
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Q( ),
r 2 2
r2
2Z
Z ( z ) k 2 Z ( z )
z 2
2 2
1
r R (r ) k 2 R (r )
r r r
r
This is a different differential equation for R(r), and the Bessel functions dont solve it. No problem; we
just make up a new set of functions for this case, do the fancy math/power series solutions, and call them
modified Bessel functions (not to be confused with Bessel functions of the 2nd or 3rd kinds). We use the
same change of variable (w = kr), and call the solutions I and K [Jac ??]:
2 2
1
r
R
(
r
)
k 2
r r r
r
R(r )
R(r ) AI ( kr ) BK ( kr ),
integer
As before, I is regular at the origin, and K blows up there. I assume for non-integer, we dont use K,
and instead use I and I, as in the Z(z) = sinh or cosh case.
Three-D Laplace Solutions in Spherical Coordinates with Axial Symmetry
The Laplacian in spherical coordinates is
2
1 2
1
r
sin
2
r r r r sin
Note:
1
2
2 2
2
r sin
1 2 1 2
2
2
r
2
r
2
2
r r r
r r r r r
We start with a common special case of spherical coordinates: axial symmetry. In this case, the whole
system is unchanged by any rotation about the z-axis. All of our results are thus independent of . We
now seek a 2D solution (r, ), embedded in a 3D physical space. Applying the simplified Laplacian, and
separation of variables, we seek something similar to the 2D polar case:
2
1 2
1
r
sin
r 2 r r r 2 sin
(r , , ) (r , ) R(r ) P ( )
As with the 2D polar case, the two operators must reduce to k/r2 for R(r) and P():
1
k
sin
P ( ) 2 P ( )
r sin
r
1 2
k
r
R ( r ) 2 R( r )
r 2 r r
r
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The difference from the 2D polar case is that the differential operators for the two Laplacian terms are
different. We dont have any standard functions that provide the desired results, so (as with cylindrical
coordinates), we invent them.
First, to find P(): Fancy math Legendre polynomials, Pl(cos ). These are a countably infinite set
of polynomials, labeled with their (integer) order l, which when applied to cos , satisfy the above equation
[Arf 11.37 p565]:
l l 1
1
sin
Pl (cos )
Pl (cos )
r sin
r2
2
The first few Legendre polynomials, and the corresponding P(), are:
P0 ( x) 1
P4 ( x)
P1 ( x ) x
P2 ( x)
1
35x 4 30 x 2 3 ,
8
P1 (cos ) cos
1
3x 2 1
2
P3 ( x )
1
5 x3 3x
2
or
P2 (cos )
1
3cos 2 1
2
P3 (cos )
1
5cos 3 3cos
2
1 2
1 2
l 2
l 1
r
R (r ) 2 r r Alr l 1 Br
r 2 r r
r
l l 1
1
1
l 1
Alr l 1 l 1 Br l 2 Al l 1 r l l l 1 Br
R (r )
2
r r
r
r2
Finally, since Laplaces equation is linear, the linear combination of any two solutions is also a
solution:
(r , , ) (r , ) R( r ) P( )
Al rl Bl r l 1 Pl cos
l 0
1 2
1
r
2
sin
2
r r r r sin
2
1
2
2
2
r sin
Note :
1 2 1 2
r
r
r 2 r r r r 2
(r , , ) R(r ) P ( )Q( )
Without axial symmetry, we need a real function of . Since the prefactor of 2/2 is 1/(r2 sin2 ),
which has no in it, we simply need the eigenfunction of 2/2, which is, as always, sin or cos(m), with
eigenvalue m2:
Q ( ) C sin m D cos m
2
Q( ) m2 Q( ) .
2
The Legendre polynomials no longer work for , because now we need to satisfy
l l 1
1
m 2 P( )
sin
P( )
2
r sin
r
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to cancel the new m2 term from the part above. For integer m, more fancy math creates a new set of
functions, the associated Legendre functions, Plm(cos ), which satisfy the above equation for given l and
m:
l l 1
1
m
m2 Pl m (cos )
sin
Pl (cos )
2
r sin
r
In other words, combining the and dependence, we see that (excluding r dependence):
l l 1
2 Pl m (cos )Q ( )
m 2 Plm (cos )Q( ) m2 Pl m (cos )Q( )
2
l l 1
r
Pl m (cos )Q ( )
This is exactly the same equation we had for the axial symmetry case, so R(r) is not affected. Note that
m is restricted to l m l. Also, when m = 0, the associated Legendre function is simply the original
(non-associated) Legendre polynomial of order l. [Associated Legendre functions are only polynomials for
even l; for odd l, they have a square root in them.]
The loss of axial symmetry introduced a dependence, which was cancelled by new functions of
. The radial function was not involved, and the radial function cannot see any dependence on
.
Now R(r): compare to the axially symmetric case (above): we added here a m2 eigenvalue for , but
cancelled all of it in the new Plm(cos ). Therefore, the radial function doesnt know anything about , and
our axially symmetric R(r) still works as before.
l 1
R (r ) Ar l Br
Finally, since Laplaces equation is linear, the linear combination of any set of solutions is also a
solution:
(r , , ) R( r ) P( )Q ( )
l 1
Al r l Bl r
l 0
P
l
m l
Note that each value of l has its own set of Al, and Bl, and each (l, m) pair has its own Clm, and Dlm.
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function
range
zeros at
derivative
derivative
range
sin
1 to 1
cos
1 to 1
cos
1 to 1
(n+1/2)
sin
1 to 1
to
cosh
1 to
none
sinh
to
> 0 to
none
exp(+x)
> 0 to
> 0 to
none
exp(x)
to < 0
sinh
+
cosh
exp(+x)
exp(x)
(green)
Recall that in rectangular coordinates, at least one coordinate must be sin/cos, and the other must be
sinh/cosh or e+x/ex. The following two cases often allow a simple choice for the coordinate which is
sin/cos:
(1)
(2)
In 3D, we might find two coordinates which are both sin/cos. Note, though, that other BCs can also be
solved with sin/cos, so the above choices are just two of many possible conditions.
Now having a good idea which coordinate(s) is/are sin/cos, we use the other BCs with the table above
to choose the simplest function for the other coordinate(s). These functions must be sinh/cosh or e+x/ex,
because they have the positive eigenvalues needed to complement the negative ones from the sin/cos
functions already identified in the solution. Some BCs allow us to write the solution as a single function,
instead of a linear combination. For example:
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(3)
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(4)
(5)
(6)
(7)
(8)
If the BCs do not allow any of the above simple solutions, then the solution must be a linear
combination of these functions, with coefficients that must be determined by solving simultaneous
equations, in the usual way.
Respecting Orthogonality
TBS: orthogonality wrt to linear operators. Orthogonality of interest of Bessel functions.
Propagation In a Vacuum
Maxwells equations imply that all vacuum EM fields, including inside waveguides, satisfy the wave
equations (in Gaussian units) [Jac sec. 8.2]:
2E
1 2
E,
c 2 t 2
2 Ex
Also for B:
2B
1 2
Ex ,
c 2 t 2
1 2
Ey ,
c 2 t 2
and
2 Ez
1 2
Ez .
c 2 t 2
2 By
1 2
By ,
c 2 t 2
and
2 Bz
1 2
Bz
c 2 t 2
1 2
B
c 2 t 2
2 Bx
1 2
Bx ,
c 2 t 2
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E
E
A
B
B
w
y
x
z
z
Linearly polarized EM wave in free space, propagating in the z direction. On the right, only the E
and B fields are shown, for simplicity. On the left, the A field is included. A is in the plane of E,
but leads E in time and space.
In the radiation gauge, the vector potential A for a linearly polarized EM wave is also a linearly
polarized transverse wave, in the plane of the E field. A leads E in time (A peaks 90o before E), and lags in
space (A peaks 90o after E). Consider the point w in space in the diagram above:
A Ay y,
Ay ( w)
A( w) 0,
Ay (w)
0,
0.
1 A
c t
At the same point, B is (negative, in this case) maximum at w, where A is 0 and increasing in space:
A
Ay
Ay
B A z
x
x
y
z
z
However, the A field decreases in magnitude with increasing frequency, for the same power density,
because E & B derive from A with derivative operators, which introduce factors of for E, or k for B:
E vs. A
Low Frequency
E vs. A
High Frequency
amplitude
amplitude
A
time
A larger
time
A smaller
Same magnitude E field and its associated A field at two different frequencies.
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Waveguides
We should first note that in the context of waveguides, the word mode is somewhat abused: in
classical mechanics, and in E&M cavities, a mode implies a specific frequency of oscillation. In
waveguides, the mode really refers to the eigenvalue of the waveguide transverse
eigenfunction/eigenvalue equation (described below), and covers a continuum of temporal frequencies, ,
and spatial frequencies (wave numbers), k.
This section relies heavily on the methods developed in Solving Laplaces Equation and
Propagation in a Vacuum, above. You should read those sections first. Those sections concern
electrostatics and magnetostatics. In the topic of waveguides, we extend those principles to electrodynamics, where the fields are time-dependent. We therefore use the full, time-dependent Maxwell
equations, including time derivatives.
Waveguides are concerned with the propagation of waves down a fixed (but arbitrary) cross sectioned
cylinder. The most common cross sections are rectangles, circles, and fractions of a circle. Since we are
concerned with wave propagation, prior experience strongly suggests decomposition of an arbitrary wave
into sinusoidal components. [More rigorously, sinusoids compose a basis set of functions for the
construction of any arbitrary function, and even better, are the eigenfunctions of linear integro-differential
equations (such as the wave equations resulting from Maxwells equations). This means we can study a
single sinusoid of arbitrary frequency, and know that an arbitrary function can be made from a linear
superposition of these sinusoids, and the sinusoids will not interact with each other. This is the standard
method of eigenfunction decomposition and analysis, familiar from quantum mechanics.]
Under what conditions can a sinusoidal wave propagate down the waveguide, but remain otherwise
unchanged? We assume an ideal conductor for the boundary of the waveguide, so there is no loss as the
wave propagates. (Loss is a small correction to the lossless case [Jac p353m]). The choice of propagating
sinusoids leads immediately to the use of phasors (complex numbers) for all components of E and B fields;
WOLOG, we assume propagation in the +z direction. Then all 3 components of both the E and B fields
vary sinusoidally in time and z-direction:
E( x, y, z, t ) Re E( x, y) eikz it ,
B( x, y, z, t ) Re B( x, y)eikz it
Note that is always real, and for propagation, k must be real. So far, these are still fully 3dimensional vector fields with components in all 3 directions. All the components propagate uniformly in
the z direction. Because we have now defined the variation of the fields in z and t, we can find the time and
z derivatives, and partially evaluate the Laplacian operator in the wave equations above:
2E
k 2 E,
2
z
2E
2 E,
2
t
2B
k 2 B,
2
z
2B
2 B
2
t
Since we know the variation in z, it becomes useful to separate the Laplacian operator into its z-part
and its transverse part (i.e., its x-y part), so we define
2 t 2
2
z 2
t 2 2
2
2
2 1 1 2
[Jac 8.20 p357]
r
z 2 x 2 y 2 r r r r 2 2
Note the subscript t means transverse here, not tangential. Also, we may use polar coordinates
(r, ) or (x, y) coordinates for the transverse part (or any other 2-D coordinates for that matter), whichever
is easier. With these definitions, keeping in mind that all vector components are complex-valued phasors,
the wave equations above become:
2 2
1 2
t 2 E 2 2 E
c t
z
i.e.
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2
t 2 Ex k 2 2 Ex
c
k2 E
2
t 2 E y k 2 2 E y ,
c
2
E,
c2
and
2
t 2 E k 2 2 E
c
2
t 2 E z k 2 2 E z
c
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Also for B:
2 2
1 2
t 2 B 2 2 B
z
c t
2
t 2 Bx k 2 2 Bx
c
i.e.
emichels at physics.ucsd.edu
2
t 2 B k 2 2 B
c
2
t 2 B y k 2 2 B y ,
c
and
2
t 2 Bz k 2 2 Bz
c
Thus we see that the conditions for propagation down a waveguide are that all 6 E & M components,
Ex, Ey, Ez, Bx, By, and Bz, must satisfy the same 2D eigenfunction/eigenvalue equation in the transverse
plane. Note, however, that some of the 6 components may be everywhere zero, in which case those
components identically satisfy any eigenfunction/eigenvalue equation. Each scalar field eigenfunction has
an eigenvalue, which determines the dispersion relation relating k and for that eigenfunction. Further, to
achieve propagation of such a mode, k and must be the same for all 6 E & M components. Therefore the
eigenvalues (and thus dispersion relation) must be the same for all non-zero E- and B-field components of a
single frequency. It is common to define the eigenvalue for a given waveguide frequency, and therefore the
dispersion relation, as
eigenvalue 2 k 2
2
,
c2
2
k2
c2
(k ) c k 2 2
Dispersion Relation
(k)
= ck
min= c
k
Dispersion relation for EM propagation
Note that for hollow cross-sectional waveguides, the eigenvalues are always negative (why??), so 2 >
0. For propagation, k must be real > 0. Thus, a given propagation mode has a minimum frequency for
propagation of
min c
In free space, = 0, the dispersion relation is linear: = ck, and there is no frequency dispersion. Note
that as , the dispersion relation for all waveguide modes is asymptotic to = ck, which is the free
space dispersion relation (of no actual dispersion).
Now that we have the basic equations of operation of a waveguide, we can consider the boundary
conditions that exist in a waveguide.
( NB : Bt Bx x B y y ) .
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TBS.
Phase Velocity in
Free Space
wave
crests
vp
ha
s
vgroup < c
vphase > c
vphase > c
In the direction of propagation (along the red arrow), the phase velocity is just c, the speed of light.
But suppose we set up electrometers to measure the electric field as it passes down the z-axis. The time
between wave crests is the same, but the distance between wave crests along the z-axis is greater. So the
wave crests are moving faster than light along the z-axis; this is required by the geometry. Hence, in the zdirection, vphase > c.
The propagation down a waveguide can be written as a superposition of two constituent waves
traveling at an angle to the waveguide axis (above right, blue and red). The constituent waves reflect off
the walls of the waveguide, and superpose to produce a net propagation in the z direction. The phase
velocity of both the blue and red constituents is the same as in the left diagram:
vphase = c/cos > c .
However, the net propagation speed in the z direction is just the z component of the red (or blue)
constituent:
vgroup = c(cos ) < c .
Therefore, in a waveguide, the phase velocity is always greater than c, and the group velocity is always < c.
Note that the constituents are not simply two uniform plane waves. The boundary conditions must be
met on each constituent wave: namely, that Ez = 0 at the walls, and that Bn = 0 (the normal component of
the B-field). So the constituent waves also taper to zero at the edges, just like the total wave does.
The angle can be determined from the z-component of the spatial frequency (the wave-vector), kz.
We find kz from solving Maxwell's equations, and from
kz
.
c cos
Multiple modes of one frequency: It is generally the case that a single frequency can propagate in
multiple modes. The modes typically have different group velocities, though, so they are dispersed in time
and space as they propagate. This is probably undesirable. To avoid this, the transmitter must excite only a
single desired mode. Methods of doing this are beyond our scope.
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2 Ez
2
c2
Ez .
1 1 2
2
r 2 2 2 .
r r r r
z
(3.3)
and
2
z 2
Z ( z) 2 Z ( z)
where
k z wave number .
Then substituting the separated Laplace equation (3.2) into the wave equation (3.3) above, we get:
2
2 2
2 E z k 2 2 2 2 E z 2 E z .
r
r
c
We can turn this into Laplaces equation by bringing the RHS over to the left:
2
2
2
2 E z k 2 2 2 2 2 E z 0 where
r
r
c
k2
k2
2
2 0 .
c2
Again we see that for propagation, must be large enough to make k2 > 0.
For a pipe of radius a, solutions of this equation must satisfy the boundary condition Ez(a) = 0. We can
choose arbitrarily, and then are allowed only discrete values of k which satisfy:
J v (ka) 0 .
(This is much like the rectangular waveguide which allows only discrete values of kx and ky.) Having an
acceptable (, k) pair, we find the z-direction wave-number:
2
k2 .
c2
You might wonder about modified Bessel function solutions that might work for < the cut-in
frequency. However, the required radial solutions dont work because one of them blows up at r = 0, so is
unphysical, and the other has no zeros for r > 0, and so cant satisfy Ez(a) = 0. Therefore, only the ordinary
Bessel equation and Bessel functions solve Maxwells equations for a propagating wave in the pipe.
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Q1
r'
Qn
Q2
r
Qi
Q3
reference point
for blob
r
Q4
observer
Q5
Q .
i
i 1
Qtotal
r
rr.
where
If all the point charges were exactly at r, this would be the exact potential. Now consider the effect of
moving one of the point charges, Qi, from the point r by some small amount, r'i. The potential at the origin
due to that charge changes by
Qi
1
Qi
Q
1
1
i Qi
Qi .
r r' r
r
r r' r
That is, the change in potential (seen by us at the origin) equals the charge times the change in 1/ r. The
function 1/r is a scalar field: it assigns a number to every point in space. Therefore, we can use its gradient
to approximate small changes in 1/r:
1
1
r'
r
r
1
Qi Qi r ' .
r
We can find the gradient direction with a simple geometric argument: the gradient points in the
direction of greatest rate of increase.
direction of fastest
increase in r
The greatest rate of increase of 1/r is opposite to r itself: r increases fastest along itself. Therefore, 1/r
decreases fastest along r, and 1/r increases fastest opposite to r: The magnitude of the gradient then
follows from a simple derivative:
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d 1
1
2
dr r
r
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1 1
2 r .
r r
We use this to approximate the change in potential at the origin due to moving Qi a small distance r' :
1
1
Qi Qi r ' Qi 2 r r ' .
r
r
Now lets move each of the charges by its own small displacement r'i.
superposition, we just add up the changes due to moving each charge:
total
Qr
i
i 1
r r 'i
n
1
1
r
r
'
Q
2 r p,
i
i
r
r 2 i 1
where
By the principle of
n
Qr' .
i
i 1
Note that p is independent of where the blob is located, i.e. independent of the relative position of observer
and blob. It is solely a function of the blob, i.e. the distribution of charges. p is called the dipole moment
of the charge distribution. We see that, to first order, the potential due to a net-neutral blob (I mean, charge
distribution) falls off as 1/r2, and is zero along a line through the blob perpendicular to p.
Note that often we consider the blob to be at the origin, and compute the field at some position r. This
simply reverses the sign of r above, to yield the more common formula:
total
1
r p
r2
Qtotal 1
2 r p ... .
r
r
A subtle point: the dipole moment of the blob may depend on the reference point within the blob from
which the r' are measured. However, if the total charge is zero, the dipole moment is then independent of
the reference point within the blob. Why??
Note that the dipole moment does not completely characterize the charge distribution, nor is it
necessarily the most significant component of such a characterization. It is simply the 1/r2 component of
the potential due to the charge distribution in the far field approximation.
Quadrupoles
The quadrupole moment extends the above approximation to 2 nd order. In short, we could say, Just
extend the dipole to the 3-variable 2nd order Taylor expansion of 1/r, and claim were done. But thats too
arcane, so lets see what it really means.
First, lets recall the meaning of the Taylor expansion of a simple function, f(x), about x0. To first
order, we approximate the first derivative as constant. Hence
f f '( x0 )x
f ( x x) f ( x0 ) f '( x0 )x .
To 2nd order, we approximate the 2nd derivative as constant, and therefore the first derivative changes
linearly with x:
f ' f ''( x0 )x
We can then find the average first derivative f ' over the interval x:
f 'avg f '( x0 )
1
1
f ' f '( x0 ) f ''( x0 )x .
2
2
This average f ' incorporates 2nd order effects. Now use this average f ' in the first order estimate for f(x +
x):
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1
1
2
f ( x x) f ( x0 ) f '( x0 )x
1
2
f ''( x0 ) x
2
Dipole moments are the 1st order Taylor coefficients in the expansion of potential due to a blob of
charges. Quadrupole moments are simply the 2nd order Taylor coefficients in the same expansion:
(r )
We follow a procedure similar to the 2nd order Taylor expansion of f(x) above, but with two additional
complications: (1) our function 1/r is a function of 3-D space, so its first derivative (gradient) is a vector,
not a number; and (2) the 2nd derivative of 1/r is the gradient of a gradient, which is the gradient of a vector,
which is a rank-2 tensor. We take these two issues in turn. As is traditional, well stick to Cartesian
coordinates.
[Aside: purists might object to the following explanation, but our goal here is to describe electromagnetics,
not differential geometry.]
First, the gradient of a scalar field at a point is a set of 3 numbers (a vector), which tell us how the
scalar field varies as we move in the x, y, and z directions:
f f f
f , , .
x y z
We usually write the gradient as a vector, but for now, you can think of it as just a set of 3 numbers. We
use the 3 numbers to approximate changes in f given small changes in position, dx, dy, dz:
f
f
dx,
x
f
dy ,
y
f
dz .
z
Further, for differentially small changes dx, dy, dz, the f are independent and can be summed, to get the
familiar total derivative rule:
f
f
f
f
dx
dy
dz .
x
y
z
This is what we used to compute the dipole moment. It is a first order approximation, because we take the
gradient to be constant over the intervals dx, dy, dz. To make the 2nd order approximation, we take the 2nd
derivative to be constant, thus making the gradient (first derivative) vary linearly over the intervals dx, dy,
dz.
So now we ask, what is the gradient of a vector field, g(r)? At a given point, it is simply a set of 3
vectors that tell us how the vector field varies as we move in the x, y, and z directions:
g g g
g , ,
x y z
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y g(x, y) + (g/y) dy
g (g/y) dy
(x, y)
g (g/x) dx + (g/y) dy
g(x, y) + (g/x) dx +
(g/y) dy
y+dy
dy
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(x+dx, y+dy)
g (g/x) dx
g(x, y)
dx
x+dx
g(x, y) + (g/x) dx
x
g
dx,
x
g
dy,
y
g
dz,
z
and
g
g
g
dx dy dz .
x
y
z
g yx
g yy
g yz
g zx
g zy ,
g zz
where
g xx
g
,
g xy
g x
xz
g yx
g
,
g yy
y
g yz
g zx
g
.
g zy
g z
zz
This is a rank-2 tensor. It can operate on a displacement vector to approximate the change in a vector
field over that displacement
g xx
g
g
g
g
dx
dy
dz g dr g xy
x
y
z
g xz
g yx
g yy
g yz
g zx dx
g zy dy
g zz dz
Note that a tensor is linear in its vector argument: g k dr1 dr2 k g dr1 g dr2 . [This implies
that if we transform coordinates (e.g., rotation), the columns and then rows of the tensor will transform like
vectors; but we dont really care about that right now.]
Now lets compute the gradient of the gradient of a scalar function f(x, y, z):
f
x
f f f f
f , ,
x y z y
f
z
2
f
f
f f
x 2
x
x
x 2
f
f
f , ,
x y y y z y
xy
f
f
f 2 f
z
z
z xz
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2 f
yx
2 f
y 2
2 f
yz
2 f
zx
2 f
zy
2 f
z 2
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Its just the matrix of all second derivatives. [It is called the Hessian matrix.]
Back to quadrupoles: We are the observer at the origin. We want the 2nd order change in potential
seen by us due to displacing a single charge Qi from the blob by a small amount r' = (x', y', z'). We follow
the same procedure as above for the 2nd order Taylor expansion of f(x). First, we compute the gradient of
the gradient of 1/r:
1
1
1
2 r 3 r x 2 y 2 z 2
r
r
r
x2 y 2 z 2
x
3/ 2
3/ 2
g g g
, ,
x y z
xx yy zz . Use g
xx yy zz
3 2
x y2 z2
5/ 2
2 x2 x2 y 2 z 2
3/ 2
3 2
2
2
x 2 x y z
5 / 2
2 x yy zz
3x 2 1
3xy
3 xz
5 3 x 5 y 5 z
r
r
r
r
3y2 1
3 yx
3 yz
[...] 5 x 5 3 y 5 z ,
y
r
r
r
r
Similarly:
x2
1 3
5 yx
r r
zx
xy
y2
zy
and
2 1 2
x 3 r
xz
1 0 0
1
3
yz 3 0 1 0 5
yx
r
r
2
z
0 0 1
zx
3z 2 1
3 zx
3 zy
[...] 5 x 5 y 5 3 z
z
r
r
r
r
xy
1
y2 r 2
3
zy
yz
1
z2 r 2
3
xz
1
r
avg
2 1 2
x 3 r
x'
1
3
1
yx
y ' 5
2
r 2r
'
z
zx
xy
1
y2 r2
3
zy
x'
yz
y'
z '
1
z2 r 2
3
xz
2 1 2
3
2 1 2
5 yxx ' y r y ' yzz '
3
2r
2 1 2
zxx ' zyy ' z r z '
3
1
3
1
2
Qi r ' Qi 5 x 2 r 2 x ' xyy ' x ' xzz ' x '
3
2r
r avg
1
1
2
2
yxx ' y ' y 2 r 2 y ' yzz ' y ' zxx ' z ' zyy ' z ' z 2 r 2 z '
3
3
Theres a trick here thats kind of complicated that I dont have time to write down yet, but it turns out
we can drop the (1/3)r2 terms in the first matrix by adding (1/3)r'2 terms to the quadrupole tensor. Briefly,
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its because grad-grad-1/r is traceless, so we can add any multiple of the identity matrix to the quadrupole
tensor D, and it wont affect . So we add (1/3)r'2[I] to make D traceless. This means we can now
add any multiple of the identity matrix to grad-grad-1/r, and it wont affect . So we add (r2/3)[I], to
eliminate those ugly terms.
Now notice that all the terms for separate cleanly into r factors and r' factors. We also use the
principle of superposition, to include the effect of all charges in the blob. This leaves us with
x
3
Qi 5 yx
2r
zx
2
Qi
xy
y2
zy
2 1 2
x' 3 r '
xz element by element
z 2 summed with
z'x'
x' y'
y'z'
1
z '2 r '2
3
x'z'
1
y ' 2 r '2
3
z' y'
So we define:
Da
i 1
2 1 2
xi ' 3 ri '
Qi yi ' xi '
zi ' xi '
xi ' yi '
1
yi '2 ri '2
3
zi ' yi '
yi ' zi ' ,
1
zi '2 ri '2
3
xi ' zi '
and then:
2r 5 , 1
D r r
2r 3 , 1
D r r ,
where
r th component of r .
D is the quadrupole tensor, which can simply be thought of as the matrix of numbers needed to
compute the 2nd order change in due to a blob of charges. Notice that when summed with components of
r-hat (the unit vector in the r direction), the change in varies as 1/r3.
[Notice that the quadrupole tensor is a rank-2 tensor, which acts directly on another rank-2 tensor, to
produce a scalar: . The quadrupole tensor acts on the tensor r r, which is the tensor product of the
displacement vector with itself. We have thus seen two ways rank-2 tensors can act: some can act on a
vector to produce a vector, and others can act on another rank-2 tensor to produce a scalar. It is not always
sensible for a single rank-2 tensor to act both ways. See Funky Mathematical Physics Concepts for details.]
[Aside: The above last few steps would be a lot easier if we recognize that
a b
x, y, z d e
g h
x2
yx
zx
element by element a b
c x
x
f y x, y, z y
multiplied and d e
i z
z summed with g h
xy
y
zy
xz element by element a b
yz multiplied and d e
z 2 summed with g h
c x
f yx
i zx
xy
y
zy
c
f
i
xz a b
yz d e
z 2 g h
c
f ,
i
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4. Relativistic Electromagnetics
Relativity is second only to statistics in conceptual difficulty, because it is so contrary to common
experience. However, when approached systematically, it can be understood, eventually. Here we not only
present valid reasoning, but also describe the failure of common invalid reasoning. Topics:
1.
2.
3.
4.
Doppler effect
5.
wall
Alice
light 1 lightbeam second
Alices view
y
x
Before
beam hits
wall
Bobs view
Bob
Alices view
Alice
As beam
hits wall
Bobs view
Alice shines a flashlight in the y direction. It travels 1 light-second (in, of course, 1 second), and hits a
wall. The two defining events are (1) Alice turns on the flashlight, and (2) the beam hits the far wall.
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Bob travels in the +x direction (parallel to the wall) at speed = v / c. What does Bob see? First, Bob
sees himself as stationary, and Alice as moving to the left. He sees that the light beam is not perpendicular
to the wall, so we conclude that the angle at which a light beam travels is relative; i.e., different observers
measure different angles. (This is true of baseballs or anything else that moves.) Also, Bob sees the light
travel a larger distance than Alice does; since the speed of light is the same for Alice and Bob, Bob clocks a
longer time between the 2 events. Call Bobs time (in seconds). During that time, Alice moves left a
distance (by his measure) of light seconds. Since distances perpendicular to the motion are the same for
Alice and Bob, Bob measures a y displacement of 1 (light-second). Therefore, Bobs time is
2 12 ,
2
2 2 2 1,
1
1 2
During this time, a clock stationary in Alices frame measures only 1 second. Therefore, Bob sees a
moving clock as ticking more slowly than his own (stationary) clocks. This is time dilation. By
symmetry, Alice (and everyone else in the universe) must see the same phenomenon.
Moving clocks tick more slowly than stationary ones.
This is not an illusion caused by the finite speed of light (see discussion of valid reference frames
above). This is real.
Do not be mislead by this erroneous argument: Alice sees one second on her clock, and sees Bob
time > 1 seconds on his clocks, therefore Alice thinks Bobs moving clocks run faster. The error
in this argument is that Bob used two clocks, at different x positions, to make his measurement. In
Bobs frame, those clocks are synchronized, but in Alices frame they are not.
If Alice looks at any one of Bobs clocks (not two clocks in two different x positions), then she will
indeed see Bobs clock run more slowly.
Length contraction: By symmetry, Alice and Bob agree that they are moving at a speed relative to
each other. When Alice turns on her flashlight, she simultaneously marks a green dot on Bobs moving
frame. When she detects the light on the wall, she marks a red dot on Bobs frame. During the interval,
Alice measures Bob move t = light-seconds across her frame; i.e. Alice measures the x distance
between two moving dots as light-seconds. The dots are stationary in Bobs frame, and as already noted,
Bob measures their separation as t = light-seconds. Therefore, Alice measures the length of moving
things as shorter than someone at rest with respect to those things, by the factor 1/.
Do not be mislead by this erroneous argument: Look at the diagram of Alices measurement of
the distance between the dots she painted on Bobs reference frame. As noted, Alice measures
that distance as . Suppose Alice has a ruler on the ground, with which she measures that
distance. One might (incorrectly) conclude that at that moment, Bob sees her moving ruler
aligned with his dots, and conclude that in his frame, the ruler measures , and thus moving rulers
grow instead of contract. This is wrong because in Bobs frame, the dots do not align with Alices
ruler at a single instant in time.
For the correct reasoning, see Bobs view of the ruler (above): In Bobs frame, Alice is moving to the
left, and her ruler is short. Bob measures the red dot aligns with the red end of Alices ruler before the
green dot aligns with the green end of Alices ruler.
Moving objects contract.
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y
E
z
x
sheet charge
density
v
Lorentz contraction:
= E= E
In the moving frame, the sheet charge is compressed to a smaller area, by the length contraction factor
. Therefore, E is increased by the same factor.
Now we add a B-field to the stationary frame:
y
E= (E + v B) (SI)
z
B
v B
Lorentz
contraction
E ' E
There is no contraction of the plane perpendicular to the direction of motion, and no Lorentz force
from B|| (parallel to the velocity). Hence, E|| = E||.
B is similar. TBS??
B ' B v E ,
B ' B
We can combine the E (and B) transformations above into a single vector equation for each. If we start
with the perpendicular formulas, and note that the cross-product term does not contribute anything to the
parallel component, then replacing B B would incorrectly inflate the parallel component by . We can
fix this by just subtracting off 1 E 1 E :
E ' E B 1 E
B ' B E 1 B
2
Jackson bizarrely complicates this formula with the identity 1 E
E
1
E ' E B
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2
2
E , B ' B E
B
1
1
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dp y
dt
Fx 0
dpx
dt
px mvx const.
increasing
vx decreases!
On-Axis Doppler
A
v dt
c dt
Enterprise v > 0
(observer)
stationary
light source
light
We first consider the simpler case of motion along the line of propagation. We are a stationary light
source, and there is a moving observer traveling directly toward us (above). We can deduce what the
observer sees from our own observations. Note that wave crests are countable things, and therefore the
same in all frames. In an interval of our time dt, the observer starts at A and ends at B, and simultaneously,
the wave train starts at C and also ends at B. During this interval, the observer sees all the crests between A
and C, i.e.
cv
v
n f dt
f dt 1
c
c
where
However, his clock advances only (dt / ) seconds, due to time dilation. Without loss of generality, let
dt = 1 second. Then in one of the observers seconds, he sees
v
n ' f ' f 1
c
This formula is valid for positive and negative v. And of course, only the relative motion of the source
and observer matter (after all, this is relativity, isnt it?)
Off-Axis Doppler
For variety, we now take the reverse (but equivalent) view that you are now the observer, you are
stationary, and the light source is moving. (I think this is a little harder to follow). You are in the
enterprise, and a star has an off-axis relative motion of v
v < 0 light emitting
(see diagram).
star
Let r = distance from enterprise to star
x = horizontal distance to star
v = speed of star in x direction (v < 0)
fs = frequency at which star emits light (in its
own frame)
fo = frequency at which enterprise sees light
+x
Enterprise
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r 2 x2 d 2
dr
dx
2x
2r cos v
dt
dt
dr v cos dt
2r
Because of time dilation, the emitted frequency in the Enterprise frame is multiplied by a factor of
(1/). In addition, the waves are compressed by the changing separation between the star and Enterprise.
In a time interval dt, the Enterprise sees the star emit (fs/) dt cycles of EM wave. But those cycles get
squeezed into a space of (c dt + dr) distance (dr < 0). Thus, the number of cycles the enterprise sees in a
full distance of (c dt) is (recalling again that v < 0):
f o dt
fo
fs
f s c dt f s
c dt
dt
dt
c dt dr
c dt v cos dt
c v cos
fs
1 cos
This reduces to the well known simpler formula for on-axis motion when = 0:
fo
fs
fs
1 cos 1
E' E B
c
E ' E B
c
v
E c E E 1 c
Doppler shift
Electric fields in propagating waves transform exactly like the Doppler frequency shift.
Then, for source and observer approaching:
IE
2
1 v / c
v
1 v / c
I ' 1 I
I
I
1
v/c
1 v / c 1 v / c
c
also square of
2
square of
Doppler shift
Doppler shit
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v dt
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c dt
1m
1m
Enterprise
observer
stationary
light source
v>0
Without loss of generality (wlog), the tube has cross-sectional area of 1m2. As seen by you
v
(stationary), the moving observer intercepts n1 f dt 1 photons in the time interval dt. But again,
c
his clock only advanced (dt / ) seconds. Therefore, the observers photons per unit time (the cross
v
sectional area for him is still 1m2) is n ' f ' f 1 . The flux transforms like Doppler shift. (Note
c
also that as the moving observer looks ahead at the photon density in space, his density is greater than
yours by a factor of , because the tube is shorter by a factor of due to length contraction.)
But each photons energy is proportional to its frequency, E , and each photons frequency is
Doppler shifted by (1 + v/c), as shown earlier. Ergo,
I ' n'E'
v v
v
I ' I 1 1 I 2 1
c c
c
v approach speed
as before.
Now force F = dp/dt, so for normal incidence light absorbed onto a surface:
F
dp 1 dE P
dt c dt c
where
P power of radiation
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Pressure
I
c
where
emichels at physics.ucsd.edu
transverse plane
reflected light, Pr
expanding
volume
mirror
E
B
v
v
vB
The intensity transformation way: In previous sections, we derived (in two different ways) that E
and B fields transform like the Doppler shift, and therefore intensity transforms like the square of Doppler
(the minus sign below is in the numerator because source and observer are receding):
1 v / c
I ' Ii
1 v / c
v recession speed
In this problem, we have two intensity transformations of the same factor: once into the mirror frame,
and once again from the mirror frame to the lab frame. The result is then simply
1 v / c
1 v / c
Ir I '
Ii
v
/
c
1 v / c
1 v / c
1 v / c
The energy way: It is perhaps instructive to consider conservation of energy in the lab frame: the
energy crossing a transverse plane (parallel to the mirror) goes into (1) filling up the increasing distance to
the mirror with electromagnetic energy, (2) doing work on the mirror from the pressure of light, and (3)
transmitting light across the boundary back into the laboratory. Without loss of generality, we can use an
incident power of 1 (in arbitrary units). In some time interval, the fraction of the incident energy that gets
stored in the expanding space up to the mirror is then v/c. The rest of the light hits the mirror, and reflects.
Similarly, the fraction of reflected energy that gets stored in the expanding space is v/c.
The work done on the mirror by the incident light is F x, where F = P/c, and x in a unit time is v.
Then W = Pv/c. However, P is not Pi = 1, but just the fraction of Pi that is not stored in the expanding
volume. We found above that the stored energy is v/c, so that which hits the mirror is (1 v/c), and the
work is W = (1 v/c)v/c.
Similarly for the work of the reflected light, but the total work is from the light stored in the expanding
volume plus the light sent back through the transverse plane into the lab. The stored energy is v/c, and the
reflected light power is . Hence the work is W = (1 + v/c)v/c.
Then by conservation of energy, the incident energy equals the sum of stored incident plus reflected
energy, plus the incident plus reflected work:
1
v
v
vv
vv
1 1
c
cc
c
c
reflected
stored
stored
in space
incident work
on mirror
1 v / c
1 v / c
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reflected work
on mirror
in space
2
2
v v
v v
1 2 1 2
c c
c c
into lab
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Beaming
Both Jackson and Griffiths have misleading diagrams for relativistic beaming at high . Both their
diagrams show no backward emission components at all. Inspection of the equation shows immediately
that this is not the case; the power is finite for all angles up to . Here are more suggestive diagrams:
dP/d
dP/d
~0
>0
dP/d
~1
x
Lorentz transformation matrix from s to s
x
x
Then
Note
In s :
F 0 J
In s :
F
F F 0 J 0 J
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x x x
0 J
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5. Shorts
Coulomb Force Constant is Defined Exactly
Recall that c and 0 are defined exactly:
c 299 792 458 m/s
k
0 4 10 7
1
is exact, and
c 2 0
c 2 0
1
SI
(r ) q / 4 0 r
(2)
c 0 0 1
(3)
F qv B
Gauss
(r ) q / r
0 0 1
v
F q B
c
0 1/ 4 .
0 4 / c 2 .
B B/c .
(1) This does not mean that 0 is 1/4 in Gaussian units; in fact, 0 = 1 in Gaussian units. It means that
when including the whole package of conversions from SI to Gaussian, including the unit of charge, the
final result is that 0 turns into 1/4.
(2) Note that we used #1 above (0 1/4) to eliminate 0.
(3) This means that the B-field in Gaussian units is inflated by the factor c, so when using it, you have
to deflate it to make it work in the equation. For example, thats why a propagating wave has:
(SI) E cB,
but
(gaussian) E B .
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0 Idl r
4 r 2
(gaussian)
dB
4 Idl r
c
4 c 2 r 2
dB
1 Idl r
.
c r2
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bound P n
[Gri ??]
The electric field from the dipoles is uniform throughout the slab.
However, this is clearly an unrealistic depiction. Real molecules arent that close together; their dipole
moments are not ideal point charges separated by vacuum; and adjacent charges dont exactly cancel.
More realistically (but still idealized), above right, the dipoles dont overlap (theyre shorter); each one
has a larger dipole moment, so the endpoint charges are larger; and there are more of them. However, the
dipole moment per unit volume is the same as the middle diagram. Therefore, these dipoles add up to the
same total dipole moment in the slab as the middle diagram, but they result in larger real surface charge
densities. In addition, there are sheet charges at other layers below the surface, which cause the E-field to
be large between the dipole charges, and zero in the gaps between layers of dipole. Thats because a real
dipole moment is spread throughout the volume of the dielectric. In fact, the standard definition of
polarization for a dielectric (in an E-field) is dipole moment density (dipole moment per unit volume). The
total dipole moment in the slab is the product of moment-density times volume:
p P (Volume)
where
Therefore:
The bound surface charge densities we compute for dielectrics are model densities which give
equivalent dipole moment to the real dielectric, but are not real.
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constant.) For example, in a pure magnetic dipole pointing up, all the magnetic field in the equatorial plane
points down:
dipole: B(r )
3 M r r M
r3
But in a finite loop, the field inside the loop points up. Hence, a finite loop cannot be a pure dipole.
M
B
B
equatorial
plane
Pure dipole
(Left)
Pure dipole B-field points down
everywhere
(Right) Finite-size current loop B-field points up inside loop.
I
Finite size
current loop
in
equatorial
plane.
dp
v
qE B
dt
c
This says that the E field acts on any charge, moving or not. The B field acts only on moving charges.
The E field can do work on a charge, but the B field cannot directly do work on a charge. The E and B
fields are directly measurable by their effects on a test charge. Physically, the sources of E fields are (1)
charges (moving or not), and (2) changing B fields. The source of B fields is charge currents (i.e., moving
charges).
Sometimes however, instead of working directly with E and B fields, it is more convenient to work
with potentials, which generate the E and B fields by a simple mathematical operation. Two such
potentials are needed: a scalar potential (t, x), and a vector potential A(t, x). These potential functions are
not unique: different choices can produce the exact same E and B fields, and thus the same physical
predictions. N
All known laws of physics, classical and quantum, are gauge invariant.
The statement of gauge invariance is nothing more than the mathematical fact that I can construct
multiple pairs of functions, (t, x) and A(t, x), that generate the same measurable E and B fields. The fact
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that many such pairs of functions exist is gauge invariance. There is no mathematical requirement that
these two functions be partitioned in such a way as to have any immediate physical meaning.
Static Fields
The defining property of the magnetic vector potential A(t, x) is that
B(t , x) A(t , x)
Static fields are not functions of time, so in the static case, the above definition can be reduced to:
B(x) A(x)
(static fields)
Then we can define an electric potential V(x) that directly relates to energy:
Define V (x) such that
E(x) V ( x)
V ( x)
x E(x) dx
0
This electric potential has no gauge freedom (only an arbitrary additive constant). The E-field is given
by the gradient of the electric potential, which in the static case also serves as a scalar potential:
E(x) V (x) (x)
where
(static fields)
V(x) has units of joules/coulomb, or volts. Both of these equations involve only derivatives of the
potentials, so adding a constant to or A doesnt change the fields. This is a trivial form of gauge
invariance, which we ignore.
More importantly, the A-field is not unique because there are many functions A(x) which have the
same curl. In particular, any irrotational (i.e., curl-free) vector field can be added to A, and it doesnt
change the curl.
Any irrotational field can be written as the gradient of a scalar field,
and the gradient of any scalar field is irrotational.
Our gauge freedom for static fields then means we can add the gradient of any scalar function to A,
and get another, physically equivalent, vector potential:
A '(x) A(x) (x)
B(t , x)
A (t , x )
t
t
A(t , x)
E(t , x)
0
t
Note that the curl and time derivatives commute, so we can bring the /t inside the curl operator.
Since the curl of E + A/t is zero, it can be written as the gradient of a scalar function which we call :
A(t , x)
(t , x)
t
A(t , x)
E(t , x) (t , x)
t
E (t , x)
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B(t , x) A(t , x)
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For dynamic fields, (t, x) is not the electric potential. It is the scalar potential, chosen for
mathematical convenience, which, along with a matching A(t, x), generates E(t, x).
In fact, the electric potential does not exist (in general) in the time varying case, since the electric field
has a non-zero curl. This means that the energy E required to move a charge from a point x1 to x2
depends on the path, and hence there cannot exist any function V(x) such that E = V(x2) V(x1).
Since both E and B are found from derivatives of A and , A and are not unique. Gauge freedom
(or gauge invariance) is the ability to impose additional mathematical constraints on A and , that dont
change the physical system. [It is sometimes said that E and B are the physical fields, and A is not; this
point is debatable, but the preceding definition of gauge invariance holds either way.]
Lorenz Family of Gauges
You may hear talk of the Lorenz gauge (often incorrectly referred to as Lorentz gauge [Jac p294]).
Lorenz gauges are actually a family of gauges, which satisfy
A (t , x )
or
1 (t , x)
0
t
c2
A (t , x)
Lorenz gauge
Relativistic form
[Aside: As a simple example of gauge invariance, look at the first equation above. We can generate another
Lorenz gauge vector potential A(t, x) by adding any static (i.e., time-independent) divergenceless vector field
G(x) to a given Lorenz gauge field A. Since the curl of G must be 0 (to preserve the B-field), G can be written as
the gradient of a scalar field:
G (t , x) (t , x)
Since we chose G = 0 (divergenceless), is a harmonic function (it satisfies Laplaces equation):
G (t , x ) 0
(t , x) 2 (t, x ) 0
For boundary conditions at infinity, this implies that (t, x) = constant. For restricted regions of space (and
therefore local boundary conditions), may be non-trivial.]
However, in general, a significant gauge transformation involves changing both A and . We can
generate another Lorenz-family gauge with any transformation of the following form:
Given that
2 1 2
2
2 2 (t , x) (t , x) 0,
c
A '(t , x) A(t , x) (t , x)
'(t , x) (t , x)
then
(t , x)
t
We can see that the new fields are also Lorenz gauge by simply plugging into the gauge definition
above:
A '
1 '
1
1 1 2
2
A 2
A 2
2 t
t
c
c t
c t c 2 t 2
1 2
1 2
A 2
2 2 0
c t
c t
Then
A ' A
and
4/9/2013 13:54
A ' A 0
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Lorenz gauges are Lorentz covariant: the 4-divergence A 0 . The Lorenz gauge makes the wave
equations nice and symmetric, and has a simple, covariant relativistic form [Jac p241].
Transverse, Radiation, or [gag] Coulomb Gauges
Another family of gauges is radiation gauge, aka transverse gauge, or (misleadingly) Coulomb
gauge, defined by
A(t , x) 0
(Gaussian units)
2 (t , x) 4 (t , x)
The term Coulomb gauge is a terrible misnomer. It suggests that (t, x) is the electric (so-called
Coulomb) potential, which it is not. Further, the relation of (t, x) to (t, x) looks the same as the
static field, but that is not physically significant.
[Jac p242b] says that Coulomb gauge indicates that Coulomb forces propagate instantly. This is
misleading, since is a mathematical field, not a physical one. However, [Jac] is quite clear that all EM
fields propagate at the finite speed of light, and all behavior is consistent with Special Relativity.
We can generate another radiation-family gauge by adding a static gradient of a harmonic function to
A:
A (t , x) A(t , x) (x)
where
Since is irrotational, it preserves B. Since ()/t = 0, it preserves E. [As before, for all of
space, harmonic implies that (x) = constant. For restricted regions of space, may be non-trivial.]
The radiation gauge is often used when no significant sources are nearby, such as the radiation fields
far from the source. Then 0 in the radiation gauge, and A satisfies the homogeneous (i.e. no-source)
wave-equation:
2 A (t , x )
1 2A
0
c 2 t 2
E(t , x)
A
,
t
B (t , x ) A
With this additional condition that = 0, the E-field is given only by E = (A/t). Note that:
Far from sources, when = 0, the radiation gauge is also a Lorenz gauge.
This follows immediately from the definitions:
A (t , x) 0,
radiation gauge
(t , x) 0
far field
1 (t , x)
A (t , x ) 2
0
t
c
Lorenz gauge
Note that the Lorenz gauge is Lorentz covariant, but neither of the radiation gauge conditions = 0,
nor A = 0 is covariant (i.e., they are frame dependent).
Even a static E-field can use a gauge in which = 0. This implies that A contains an irrotational part
which grows unbounded (linearly) with time. (This part has no curl, and so contributes nothing to the B
field.) Such an unbounded A-field, in this gauge, makes it hard to believe the A-field is a physical field
[but see the Bohm-Aharonov effect in quantum mechanics].
In general, for dynamic fields, the scalar potential (t, x) is not the electric potential. In fact, in
general, no time-varying electric potential is possible. The radiation gauge is no exception. (t, x) is a
mathematical convenience, whose relation to (t, x) looks the same as a static field, but whose meaning is
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very different. An electric potential cannot be constructed, and is certainly not simply the integral of the
static Coulomb potential over the charge distribution.
L T ( q1 ,... qn ) V ( q1,... qn )
or
e
c
q A (q ,... q )
j
j 1
q (q1 ,... qn ) are the set of generalized coordinates for the particle
e
L
or
mq A
c
q
where e is the charge of a partical.
p
pi mqi
e
Ai
c
Because of gauge freedom in choosing A(q), the canonical momentum is gauge dependent, and even
which components of momentum are conserved is gauge dependent. Consider a positive charge moving in
a uniform B-field directed out of the page. Here are 2 different A fields that describe this situation:
y
y
v
Bz
Ay
v
Bz
Ax
A(x, y) = xj
A(x, y) = yi
Two different gauges for A that describe the same physical situation.
left, py is conserved and px is not. On the right, px is conserved and py is not.
On the
dt
q j
q j
If
L
0, then
q j
L
d L
d
0
pj
dt q j dt
q j
In the situation above, the particle motion is a circle. In the left case,
A ( x, y ) xj,
L ( x, y )
L( x, y )
0
y
4/9/2013 13:54
1 2
1
mv v A m x 2 y 2 yx
2
2
L
e
e
p y const
my Ay my x
y
c
c
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and px is not conserved. You can see this on the left and right edges of the circle of motion. On the
left side, v and A point up, and add. On the right, v opposes A, and they subtract. The sum in both cases is
the same, i.e. py is conserved. But in the right case,
A ( x, y ) yi,
L
0
x
m x 2 y 2 xy
2
e
e
L
px const
mx Ax mx y
x
c
c
L ( x, y )
and py is not conserved. Here, the top and bottom edges of the circle are examples of v and A
combining to the same px value at both points.
Note that any rotation of the A field would produce exactly the same physics, and would be just
another gauge change. If we made the A field point at 45, then the linear combination px + py would be
conserved, but neither one separately would be conserved.
Reflection Symmetry
When considering magnetic systems, one might encounter reflection symmetry, which is essentially a
parity transformation. In such cases, we must distinguish between polar vectors and axial vectors. In
short, polar vectors flip direction under a parity transformation, and axial vectors do not. So position,
momentum, and E-fields, for example, are polar vectors. Angular momentum and B-fields are axial vectors
(some axial vectors point along an axis of rotation, e.g. angular momentum, hence the name axial). This
means that when you consider reflection symmetry (i.e., parity transformation) in E&M, you have to flip
the E-field, but not the B-field. This is a little weird. You have to be careful with reflection symmetry and
magnetic fields.
Bandwidth
[Taken from my original Simple English Wikipedia: Bandwidth page. As such, this section, and
only this section, is public domain.] Bandwidth is a measure of how much frequency space of a spectrum is
used. To clarify this, we must define some terms. Many systems work by means of vibrations, or
oscillations. Vibrations are something that goes back and forth at a regular rate. Each complete cycle of
back and forth is called, simply enough, a cycle. The number of cycles per second of a system is its
frequency.
Frequency is measured in cycles per second, usually called Hertz, and abbreviated Hz.
However, most systems dont operate at just a single frequency. They operate at many different
frequencies. For example, sound is vibrations. Therefore, it has at least one frequency, and usually many
different frequencies. People can hear sound frequencies as low as about 20 Hz, and as high as about
20,000 Hz. A band of frequencies is a continuous range of frequencies; in this example, the band of
frequencies people can hear is from 20 Hz to 20,000 Hz.
Finally, bandwidth is how wide the frequency band is, that is, the highest frequency minus the lowest
frequency. In the hearing example, the bandwidth of a persons ears is about 20,000 Hz - 20 Hz = 19,980
Hz.
Bandwidth is often applied to the electromagnetic spectrum: radio waves, light waves, X-rays, and so
on. Radio waves are oscillations of electric and magnetic fields. For example, the lowest United States
AM radio channel covers the band of frequencies from 535,000 Hz to 545,000 Hz. It therefore has a
bandwidth of 10,000 Hz (10 kHz). All United States AM radio stations have a bandwidth of 10,000 Hz.
The lowest United States FM radio channel covers the band from 88,000,000 Hz (88 MHz) to 88,200,000
Hz (88.2 MHz). It therefore has a bandwidth of 200,000 Hz (200 kHz).
The term bandwidth has been misappropriated into the field of digital data communication. It is
often incorrectly used to mean data carrying capacity. However, there is no such thing as digital
bandwidth. The proper term for the data carrying capacity of a communication channel is channel
capacity.
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It is true that, in general, the channel capacity of a system increases with the bandwidth used for
communication. However, many other factors come into play. As a result, in many (if not most) real
systems, the channel capacity is not easily related to the channel bandwidth.
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6. Appendices
References
Arfken, George B. and Hans J. Weber, Essential Mathematical Methods for Physicists, Elsevier Science,
2004. ISBN 0-12-059877-9.
Griffiths
Helliwell, Special Relativity, ??
[L&L]
Glossary
Definitions of common Electro-magnetic terms:
<x>
fact
A small piece of information backed by solid evidence (in hard science, usually
repeatable evidence). If someone disputes a fact, it is still a fact. If a thousand people
say a foolish thing, it is still a foolish thing. See also speculation, and theory.
linear
sense
Every vector lies along an axis. The sense of a vector is simply which way the vector
points along the axis. In some cases, we know the axis of a vector, but not its sense.
speculation
A guess, possibly hinted at by evidence, but not well supported. Every scientific fact and
theory started as a speculation. See also fact, and theory.
theory
WLOG or WOLOG
Formulas
Basics
SI
D
E
E DP
B
t
F qv B
H J
0
4
B H M H
B 0
A (r )
Gaussian
4/9/2013 13:54
D
t
D 4
E
1 B
c t
A(r )
B H 4 M H
B 0
H
J (r ') 3
d r ' [Jac 5.32 p181]
r r'
E D P
q
v B
c
4
1 D
J
c
c t
J (r ')
1
dr '
c
r r'
LL 43.5 p111
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physics.ucsd.edu/~emichels
(1)
SI
(r ) q / 4 0 r
(2)
c 0 0 1
(3)
F qv B
Gauss
(r ) q / r
0 0 1
v
F q B
c
emichels at physics.ucsd.edu
0 1/ 4 .
0 4 / c 2 .
B B/c .
Vector Stuff
er
A
1
1
e
e
r
r sin
r
1
1 A Az
A
z
1 2
1
1 A
sin A
r Ar
2
r sin
r sin
r r
A Ay
Ax Az
A ex z
ey
z
x
z
y
Ay Ax
e z x y
1 Az A
1
Ar
Ar Az
er
e
e z r r rA
r
z
z
r
er
2
1
r sin
A
sin A e
2
2
2
x 2 y 2 z 2
1 Ar 1
1
A
rA e rA r
sin
r
r
r
r
r
1 1 2
2
r 2
r r r r 2 z 2
1
2
1 2
1
,
sin
r 2 sin 2 2
r 2 r r r 2 sin
a ( a) 2a,
NB :
1 2 1 2
r
r
r 2 r r r r 2
a b c b a c c a b
or
( x, y ) Ax By D
( x, y ) X ( x)Y ( y)
2 0
1
1
X "
Y" 0
X ( x)
Y ( y)
X ( x) A sin x B cos x,
Y ( y ) C sinh y D cosh x
far-field lowest
X ( x) A sinh y B cosh x
Y ( y ) C sin x D cos x,
X ( x) A exp y B exp y
far-field lowest
4/9/2013 13:54
or
Y ( y ) C exp y D exp y
or
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(r , ) B0 ln
2D polar :
emichels at physics.ucsd.edu
C sin D cos
r
A r B r
A0 1
3D cylindrical : (r , , z ) R (r )Q( ) Z ( z )
R (r ) CJ m (kr ) DN m (kr )
f (r )
n1
A n J x n
a
3D rectangular :
or
( r a, ) 0
xmn
a
2
J21 ( x n ) 0
dr r f (r ) J x n
a
( x, y, z ) Ax By Cz D
2 0
( x, y, z ) X ( x )Y ( y )Z ( z )
Y ( y ) C sin x D cos x
4/9/2013 13:54
k mn
A n
X ( x ) A sin x B cos x
f ( x)
1
1
1
X "
Y "
Z"0
X ( x)
Y ( y)
Z ( z)
1
X " 2
X ( x)
Y " 2 2 2 2 ,
Y ( y)
1
Z " 2
Z ( z)
Am 2
Bm a
far-field lowest
cos 2 mx
dx f ( x)
a / 2
sin a
a/2
Page 66 of 72
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(r , , ) R(r )Q( ) P( )
3D spherical :
U (r )
R (r ) Ar l Br (l 1) ,
r
Q( ) eim
P1 ( x) x
P2 ( x )
Axial symmetry: ( r , )
A r
Y22
4
1 15
sin 2 e2i
4 2
g ,
P3 ( x)
Bl r (l 1) Pl (cos )
( r , , )
l 0
m l
1
35 x 4 30 x 2 3
8
V ( ) Pl (cos ) sin d
Y11
3
sin ei
8
Y10
3
cos
4
Y21
15
sin cos ei
8
Y20
5 3
1
2
cos
4 2
2
Alm
m l
P4 ( x )
lmYlm ( , )
1
5 x3 3x
2
2l 1
2
P0 ( x) 1
*
Yl , m , 1 Ylm
,
A
l 0
2l 1
Pl (r )
2
Al
2l 1 (l m)! m
Pl (cos )eim
4 (l m)!
Y00
U l (r )
1
3x 2 1
2
l 0
Ylm ( , )
emichels at physics.ucsd.edu
sphere
x x'
*
d Ylm
( , ) g ( , )
l 0
rl
rl 1
Pl (cos )
ds
c
1/ 1
2
t ' t vx / c2
x ' x vt
dxi
u
c, v
d
u ' v
1 2
dt
u ' v
c
1 2
2
2
dt '
c
1 v / c
( 1) 12
1
1
x ' Lx
( 1) 1 2
2 1
( 1) 1 3
3 1
2
vtotal
v1 v2
1 v1v2 / c 2
3/ 2
v2
1 2
c
a
a '
3
u ' v
1
c2
4/9/2013 13:54
2
1
( 1) 1 2
2
( 1) 22
2
( 1) 2 3
2
( 1) 1 3
1
( 1) 2 3 x
1
2
( 1) 3
1
vy v 'y
1 V 2 / c2
1 v 'x V / c2
v2
v a ' u '
c2
a
a '
3
c2
u ' v
2
c
[Jac p569]
Page 67 of 72
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E mc 2
p i E / c, p x , p y , p z E / c, p
E 2 mc 2 p 2 c 2
q
q
L
mv A p A
c
c
v
p mv Ev / c 2
' 1 cos
k i ck 0 , k
1 v
dv
2
/c
2 3/2
v
1 v / c
2
emichels at physics.ucsd.edu
1
1
2
sin 2
2
h ' h mc
2
e
e
1
L mc 2 1 v 2 / c 2 e A v [ Jac 12.12 p582] H
P A e [ LL 16.10 p 49]
2m
c
c
d
v
3 2
dv
c
E ' E
E ' E B
B ' B
Combined:
E ' E B 1 E
B ' B E 1 B
or :
E ' E B
vD c
EB
2
E
1
c a 2
2B 2
B B
vC c
2
B
1
v 2 R B0
m 2 R B0
v 2 2
c R RB0
e
R B0
Adiabatic invar: Ba 2 , v2 / B, p2 / B, , ec a 2 / 2c
d
dB
d m 2
mv B e
v
e
dt
ds
ds s 2
mv 2v2
s
s 2
B ' B E
B ' B E
1 2 1 2
mc 2 1 2
v v R B
v v R B
c R
2
2
eBR
mv
2B
NonRel :
vG
B2
vG vC
eB
ecB
E
mc
Ak Ai Ex
Fik i k
E y
x
x
Ez
Sx / c
w
S x / c xx
T ik
S y / c yx
S / c
zx
z
df dA T
f force
A area
4/9/2013 13:54
Ex
Ey
Bz
Bz
By
Bx
Sy /c
xy
yy
zy
Sz / c
xz
yz
zz
v0 B( z )
2B0 z
Ez
By
Bx
Rel :
mv v
mv
dt
s
F ik
Ex
Ey
Ez
Ex
E y
Bz
Bz
By
Bx
Ez
By
Bx
E 2 B2
w
[LL 31.5 p81] ?? Faraday?
1
1
2
2
4 E E B B 2 E B
2
2
2
2
2
2
xx 8 E y E z E x B y Bz Bx
1 E E B B
[LL 33.3 p87]
z y
xy 4 x y
Page 68 of 72
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j j c, J
j i
J 0
x i t
4 j
j i F ij
c
x
emichels at physics.ucsd.edu
p i E / c, p
A i , A
eF ik u
dpi
d
d
1
i
k
i
k
4-force: f F j
c mu mc, mv
c ik
d
ds
dt
c
F
ij
x k
jk
F ki F
0
j
xi
x
Scalars : F F
ij
P
2cos r sin
3
r
sn
dS 1
1
d 3r
8
2
2
1
1
1 Q2
Ciji j CV 2
2 i , j 1
2
2 C
1
0
r
(r )
K b cn M cM n
B H 4 M 1 4 m H H
M mH,
E 2 B2
d r
2W Q 2
V 2 2W
x x
q r d 1
Q 5 ...
r r 2 2 ,
r
2
1
x x
e ( r ) e (0) r e (0)
e (0) ...
x x
2 ,
Jac 4.22
d 3r (r )e ( r ) qe (0) p e (0)
4/9/2013 13:54
d r
B 2
8
1
1 F JB
J A (units ?? cf
)
2
c V
c
1
E
...
F qE (0) P E
Q
r 0 6
x
r 0
We
M Be
1
nQn
d 3r
2
iklm F F 4 E B
ik lm
I
M area
c
M r 3 M r r M M
2cos r sin
r3
r3
cr 3
B(r )
2 B2 E 2
1
1
ij
B2 E 2
F F
f 16 ij
8
2
Econductor 4
Lagrangian density:
ij
2
1
Q
(0) ...
x x e
6 ,
Page 69 of 72
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emichels at physics.ucsd.edu
1 B
c t
1
D da 4 Q E dl c t
E
2 A
1 A
c t
4
J
c
Ar
For B0 z :
B da 0
H dl
B A
LL 43.4 p110
A(r )
B0
or
2
4
1 D
J
c
c t
B 0
1 D
4
4
I
J
da
c
c t
c
q v r
E, 1
c r2
1
J (r ')
dr '
r r'
c
A xB0 y or
LL 43.5 p111
I
r
dB d 2
c
r
S
[LL 43.7]
c
E B
4
A yB0 x
4
1 E
4
B
J
0
J
t
c
c t
c
v
F J B
I
F e E B B
c
V
c
Radiation
SS
S
1 2
E
2
c
c 2
E n
EH
4
4
d P
d
E 2
d
d c 2
I
E
d
d 8
S
c
dielectric:
c 2 2
P sin
8 c
e 1 2 r
e r r
E (t )
1 r 3 R 2
c 1 r 3 R
t '
t t '
2 e2 a 2 2
p2
3 c3
3 c3
d P
Thompson scattering:
r r0 (t ')
d
1 2
k 4a6
Er
d
2
d
e4
2 4 E r
d m c
B (t ) n E t '
R (t ') r r0 (t ')
e n n dP (t )
c e2
e2
2
2
v
n
v sin 2 [Jac 14.21 p665]
Non-Rel : E
r 2 R 2 n S
4
3
c
R
d
c
4 c
t '
dP 2e 2 2
v
Q
Multipole: P (t )
3
3
3c
3c
180c5 ,
P (t )
4
2
2
P m
3c 3
4/9/2013 13:54
PL (t )
2e 2 6 2
3c
Relativistic: P(t )
PT
2e 2 6
3c
2e 2 4 2
,
3m 2 c
2 2
for fixed
dp
: PT 2 PL
dt
Page 70 of 72
physics.ucsd.edu/~emichels
dP(t ')
e 2 n n
d
4 c
1 n 5
t '
emichels at physics.ucsd.edu
sin v n
1 /
1
~
3
5
2
2
d
4 c 1 cos
1/ 2 / 2
2
v
dPT (t ')
e2
3
d
2 c 1 cos 3
sin 2 cos 2
1
, n to in 3
2
2 1 cos
Reflection/Refraction
2
EB
E
c
c
c
1
2
S
g
z
Re E B*
Re E E*
E
Re E B*
8
8
8
8 c
8 c
E'
2n1 cos
E n cos / n 2 n 2 sin 2
1
1
2
2
1
E :
2n1n2 cos
E'
E / n 2 cos n n 2 n 2 sin 2
1
2
2
1
2
1
S y ''
Sy
'
fne 2
c2
Power
area
2
Waveguides
min min c
v ph
4/9/2013 13:54
keff
n x
m y
sin
a
b
4 nfe 2
,
m 2
E( x ) z i 1
c
H0
4
c
H( x)
4
Cu 1017 s 1 ??
TE : Bz ( x, y) cos
n x
m y
cos
a
b
n m
2
a b
2
k2 2
c2
2 c2
c2 2
k
k
Et t eikz it
R 0
2
1 i
H ( x) H 0 exp
c
4 H 0
4 B
c 2 t
decay : 2 B
TM : Ez ( x, y ) sin
2
2
2
2
E '' 1 / 2 n2 cos n1 n2 n1 sin
E / n 2 cos n n 2 n 2 sin 2
1
2
2
1
2
1
0 ( 1/ ) 1 4 ne2 / m
n( ) ( ) ( )
2
2
2
E '' n1 cos 1 / 2 n2 n1 sin
E
n1 cos 1 / 2 n2 2 n12 sin 2
sin '
R
sin '
n n
E ''2
2 Rnormal 1 2 ,
E
n1 n2
tan '
,
R
tan '
Et , H t cont; Dn , Bn cont
vg
d
kc 2
vg v ph c 2
2
2
2
2
dk
c k c
t 2 0
Bt z Et
Page 71 of 72
physics.ucsd.edu/~emichels
emichels at physics.ucsd.edu
R
din
dout
R2
din
ln r 2 ??
Index
The index is not yet developed, so go to the web page on the front cover, and text-search in this document.
4/9/2013 13:54
Page 72 of 72