Phys411 - Electricity and Magnetism
of Maryland, College Park
Spring 2010, Professor:
Notes, Demos and Supplements
In these notes I'll try to mention things I talk about in class that
are not also in the textbook,
as well as supplementary material. Please do not assume that these notes
are even roughly complete. Tuesday May 11 review
And a student asked why it's called a "gauge transformation".
I explained its origin with Hermann Weyl's
failed unified field theory of electromagnetism
and gravity, in which the (4-dimensional) vector potential determined
the change of scale of an object as
it moves areound in spacetime. I said the theory was later
reinterpreted in terms of the phase of
the quantum wavefunction, rather than the scale of the object, by
Dirac. But I looked it up, and
it seems that it was Weyl himself, and also Fock, and London, who did
The lightlike line segments are limits of timelike ones
as well as spacelike ones.
On these grounds I argued that the proper time of the timelike lines
proper length of the spacelike lines must go to zero as the lines
common limit, the lightike line.
Proof that square of (c x proper time) is twice the
Euclidean area of the light rectangle.
Used two inertial observers, Alice and Bob, who exchange light signals
proper times for each of them. The light triangle (half the rectangle)
areas were compared
by writing each area as 1/2 base times height. The ratio of the bases
was seen to be
the reciprocal of the ratio of the heights, so the areas are the same.
This shows that the
square of the proper time is proportional to the area. The
proportionality constant can
be evaluated by a simple special case, the observer who bisects the 90
between the light rays. Note that these pictures break the symmetry:
there appears to
be a preferred "vertical" observer. But the spacetime structure is
independent of this
The time^2 = area easily demonstrates time dilation and the twin effect.
The twin effect is analogous to the fact, in Euclidean geometry, that
paths connecting two points have the same length.
The spacetime Pythagoras theorem expresses
time of a timelike displacement
in terms of the time and length measurements of a given observer:
(proper time)^2 = t^2 - (x/c)^2
I proved the theorem using the time^2 = area relation. Thursday May 6
- pertubative method to reduce the order of the equation: series
solution in powers of tau
- origin of self force: I didn't really go into this. But I mentioned
Griffiths' dumbbell model
of a charge q as a pair of charges q/2 separated by a distance d that
he takes to zero at the end.
The force of one end on the other gives the "interaction" part of the
self-force F_int(q), which
Griffiths points out is half of the total. To get the other half, in
problem 11.20 Griffiths changes
the model to a continuous charge distribution. But instead we can argue
Let F(q) be the TOTAL d-independent part of the self force on a charge
q in the limit as d goes to zero.
It breaks up into the force of the two q/2's on each other, plus the
force of each q/2 on itself:
F(q) = F_int(q) + 2 F(q/2).
The function F(q) must be proportional to q^2, since the field is
proportional to q and the force
to q times the field. Hence F(q/2) = F(q)/4, so the above equation
implies F(q) = 2 F_int(q). QED.
- Relativity: Maxwell's
equations imply electromagnetic waves which travel at the speed of
19th century physicists believed this speed was measured relative to
the rest frame of the aether.
This was strange however because Newtonian mechanics has no preferred
rest frame (it has
"Galilean symmetry"). Moreover, Maxwell's theory had an obvious element
of relativity of motion:
electromagnetic induction works the same whether it is the magnet that
moves or the wire that moves,
which strongly suggested (to Einstein, who began his first relativity
paper commenting on
exactly this) that in fact Maxwell's theory has no preferred rest
frame, and in particular that the
aether frame is undetectable. Theoretical work by Lorentz and Poincare
showed this to be the case,
and the Michelson-Morely experiment demonstrated that light propagates
at the same speed in
all directions relative to the (moving) earth. So the conclusion was
that there was in fact no preferred
frame...However, this was incompatible with the usual Galilean relation
between relatively moving
observers, who should see waves as traveling at different speeds.
Einstein's resolution of this conundrum
was to physicalize the process by which coordinate systems are defined,
using light rays to synchronize
clocks. This produced relations between coordinates that were identical
to those which Lorentz had
previously shown to be symmetries of Maxwell's equations. But what for
Lorentz was just a mathematical
symmetry, was for Einstein a relation between different physical
To start things off, I took the Lorentz point of view and just asked
for which changes of coordinates
(t,x,y,z) -> (t',x',y',z') is the wave operator unchanged, box =
box'? Because I want to show you something else,
we don't have time to show that the entire set of Maxwell equations has
this symmetry, when in addition the
fields are transformed appropriately. However it should perhaps at
least be plausible, since the retarded
potentials are exactly solutions to the wave equation in Lorenz gauge.
The primed observer moves at speed v relative to the unprimed one in
the x direction, hence x' = 0 = y = z
corresponds to the line x = vt, y = z = 0. That is x' = 0 is x - vt =
0. Thus the most general linear transformation
(I won't try to motivate the linearity here) that could relate the
coordinates is (with a convenient parametrization)
t' = g(at + b/c2 x)
x' = g(x - vt)
where g, a, and b are some constants. We expressed the partial
derivatives wrt t and x in terms of the partial
derivates wrt t' and x', and required dx2 - c-2
dt2 = dx'2
- c-2 dt'2.
= 1, b = -v, and g = (1- v2/c2)-1/2.
This is the Lorentz transformations.
Rather than going further down this algebraic path, I want to
show you a coordinate independent, geometrical
way of understanding special relativity...
Simultaneity: The Einstein
radar convention for establishing simultaneity yields different
simultaneity for inertial obesrvers in relative motion. I drew a
spacetime diagram illustrating this.
Proper time: I claimed that in
the spacetime diagram, the area of the light rectangle is proportional
square of the proper time interval. I will demonstrate this by a
geometrical argument on Monday. But given this
fact, it is easy to see that there is time dilation and that the twin
effect occurs... Tuesday May 4
- Radiation from point charges
- Lienard's generalization of Larmor's formula
- special case: linear acceleration; radiation pattern, "beaming"
- special case: circular motion: synchrotron radiation; example of
charged particles in the Crab nebula
which emit synchrotron radition from electrons and
positrons cycling in a magnetic field, with a spectrum
ranging all the way from radio frequencies up to 100 GeV, the latter
being generated by particles with
kinetic energies of order 1500 TeV!
- radiation relaction: motivation/"derivation" for the Abraham-Lorentz
force Monday May 3
- Something I neglected to point out last Thursday: when we
compute the retarded vector potential, the
leading order contribution to the radiation comes just from
(µ0/4π)(1/r)∫ J(r', t-r/c) dtau'.
Note that this integral is taken over the value of the current density
at one time, t-r/c.
For a closed current loop this vanishes, but if div J is nonzero it does not. Griffiths
invokes problem 5.7
to show that the integral is the time derivative of the dipole moment.
A simpler way to see this is to write
the integral as a sum over charges times velocities:
∫ J = ∑ qivi = (d/dt)∑ qiri = dp/dt
Dimensional analysis/physical picture
of radiated power P = (µ0/6πc)pddot2.
The dipole radiation comes from the dipole field "detaching", which
happens at a distance of order the light
travel time in one cycle, l = c/f, which is also the wavelength, where
f is the frequency. The power should scale as
(energy in the dipole at the distance l)/(period of a cycle). The
energy density in the dipole goes as epsilon_0 times the
E^2, where E is the dipole field strength at the distance l, which goes
as (1/epsilon_0)p/l^3. The total energy goes like
this density times l^3. Thus we have
Magnetic dipole and electric
This comes from the next order term in the expansion from the distance
dependence of the retarded time
in rho(r', t-|r-r'|/c)
t-|r-r'|/c). This arises from the first
derivative term in the Taylor expansion of
the time dependence, and goes in the case of the current as (d/dt)J(r', t-r/c) rhat.r'/c. Compared to the
electric dipole contribution to the vector potential this is suppressed
by the time derivative times r'/c, i.e. by
fd/c ~ d/l, where, as below, d is the size of the source. When
computing the energy, the fields are squared,
so that the power in this radiation is suppressed relative to the
electric dipole by (d/l)^2. The contribution
from rho at this order just goes into the electric quadrupole, but that
from J sources both the
quadrupole and the magnetic dipole radiation.
- comment on fields of uniformly moving point charge: I pointed
out what is happening in (10.69), where
the unit vector from the retarded position winds up relaced by the
velocity vector divided by c.
- by the way, I remarked on what Griffiths calls the "extraordinary
coincidence" that the field of a
uniformly moving point charge is directed from the present position of the charge
rather than the retarded
position. This can be understand in a different way using
the fact that in the rest frame
of the charge the field is just the Coulomb field, which is directed
from the charge, then making
a Lorentz transformation on the fields. See Example 12.13 for a
discussion of this. I wonder if there
is an even simpler argument along these lines...
- continuing with
radiation: using these approximations (see Tuesday) in the
retarded potential leads
to an expression containing the monopole potential, the retarded dipole
potential, and the dipole radiation field,
etc, the last being from the time derivative of the dipole moment. The
radiation field drops with one
inverse power of r rather than two, and dominates as long as lambda
The electric and magnetic radiation fields must also fall of as 1/r,
When taking the derivatives of the potentials to obtain the fields, the
arise from the terms in which the derivatives act on the t or r in the
argument of rho.
- No monopole radiation because of charge conservation. I made the
comment that there are no dipole
gravitational waves because the time derivative of the mass dipole
moment is the momentum, which is
also conserved. So the leading order gravitational waves are
- Electric and magnetic dipole radiation fields: The electric field is
proportional to minus (1/r) times
the projection of the second time derivative of the electric dipole
moment orthogonal to rhat. The
field is perpendicular to both E and rhat, with magnitude E/c and
direction such that E x B (and hence the Poynting
vector) is in the direction of rhat.
dipole radiation: zero along the
direction fo the dipole derivative, maximal perpendicular to that.
- The total power radiated is given by the simple formula P = (µ0/6πc)pddot2,
derivative of the dipole moment, evaluated at the retarded time.
- For a point charge, pddot is the charge q times the acceleration, so
we obtain the famous
Larmor formula P = (µ0/6πc)(qa)2.
- Decay of classical atomic orbits: it takes roughly a
nanosecond. This is on the homework. Tuesday April 27
- Point charges, Lienard-Wiechert potentials, electric and
- field of a uniformly moving point charge
- relativistic "pancake" field of rapidly moving charge
- radiation: the multipole
expansion exploits the hierarchy of length scales often present in
systems: d << lambda << r, where d is the source size,
lambda is the wavelength, and r is the distance
from the source to the field point. Lambda is c/frequency. The first
inequality thus says that the system
is must smaller than the distance light travels in one cycle. The key
approximation following from
d << r is
|r - r'| = r - rhat.r'
and the key approximation from omega << c/d is
rho(r', t - |r - r'|/c)
rhat.r'/c + O(rho (omega d/c)2)
Monday April 27
Lorenz gauge and retarded
potentials: in Problem 10.8 Griffiths asks to confirm that the
retarded potentials (10.19)
satisfy the Lorenz gauge condition which was assumed in deriving those
potentials. I proposed a much simpler
way of doing this, but it's not quite conclusive. Namely, apply the
d'Alembertian oeprator (box) to the Lorenz
gauge condition itself. To simplify notation I'll work with units in
which mu_0 and epsilon_0 are unity. Then we have
box(div A + ∂tV) = div
A + ∂t boxV = -
div J - ∂t
rho = 0,
where the last equality follows from the cotinuity equation (local
charge conservation). So the Lorenz gauge condition
expression div A
+ ∂tV satisfies the wave equation. This equation has the
property that solutions are determined by initial
values and time derivatives of the function at a given time. So if the
Lorenz gauge condition and its time derivative are
satisfied at one time, it will be satisfied for all time.
- Wire example
- Jefimenko's equations Thursday April 22
Exam 2 Tuesday April 20
review for exam 2 Monday April 19
potentials in electrodynamics:
B = curl A and E = - grad V - ∂tA is possible because
div B = 0 and curl E = - ∂tB. Conversely, the latter equations
are implied by writing E and B
in terms of the potentials. The potentials are not uniquely determined:
For any function
f of space and time, A + grad f and V - ∂tf
give the same E and B as do A and V. This change
of the potentials is called a "gauge transformation".
The other Maxwell eqns involve the sources: charge and current
They can be simplified by choosing the Lorentz (Lorenz) gauge
condition, div A + (1/c2)∂tV
after which they take the simple form
box V = -rho/epsilon0 , box
A = - µ0J,
where box = d'Alembertian = Laplacian - (1/c2)∂t2
is the wave operator.
and V' don't satisfy the Lorentz gauge condition, then A' + grad f and V' - ∂tf
gauge transformation function f is chosen to satisfy the equation box f
= s, where the source s
is s = - div
A' + (1/c2)∂tV'.
all satisfy a similar equation.
The eqn box f = s has no unique solution,
because we can add to f any solution of box f = 0.
This means that there is residual gauge freedom. That is, even after we
impose the Lorentz
gauge conditions, there remains some freedom to change the potentials
and still satisfy this
condition. As for the potentials, we need to supplement the equation by
to select a unique solution. The "retarded" boundary condition imposes
that the potentials are
determined only by the values of the sources to the past. (The "advanced"
is the opposite, with dependence only on the future.) We can write the
general retarded solution:
see (10.19). I drew a spacetime
diagram showing how this formula is interpreted: to find the
value of the field at a point in space and time, one integrates over
the past light cone of that
I showed that the integral (10.19) indeed satisfies the equation, by
evaluating box [r-1s(t-r/c)].
Using the Laplacian in spherical coordinates, and ignoring the trouble
at the point r = 0,
this is zero (it is because s(t - r/c) satisfies the one dimensional
wave equation (∂r2
If we do worry about r = 0
then we pick up a delta function from Laplacian r-1
= - 4π delta3(r).
So we have box [r-1s(t-r/c)] = -
4π delta3(r) s(t).
We can now shift this so the origin is at any other
position r', and using that we
can apply the box operator to the integral in (10.19). It comes inside
the integral, acting on the r
and t dependence, and produces a delta function. The integral can then
be carried out, verifying that the equations (10.16) are
Thursday April 15
- A bit more on RealD glasses: 1) If you place two pairs face to
face, so the left eye of one pair
faces the right eye of the other, then the two quarter wave plates have
their fast axes aligned,
so it becomes a half-wave plate, which has the effect of rotating the
polarization to a vertical one. This is then blocked by the horizontal
polarizer on the back side,
so the result is dark...but not completely, since there is chromatic
dispersion. What you actually
see is a dark violet, and if you look at an angle you can see different
color mixtures depending
on the angle. 2) If you place the two pairs face to face, and then
rotate one by 90 degrees, it then
becomes the opposite quarter wave plate, and the two quarter wave
plates cancel. They form
a sandwich between the now crossed linear polarizers on the backs of
the glasses, so now the light
is blocked for all colors.
- Waveguides. I gave the
highlights of Griffith's treatment, skipping some of the algebra.
A student asked about the limit of a large waveguide being equivalent
to empty space, admitting
ordinary transverse plane waves. We can address that as follows.
Consider the limit where the lengths
of the waveguide a and b are "very large". More specifically, consider
a mode for which m/a and n/b are
very small compared to k, the wavenumber for the z-direction. Then the
dispersion relation becomes
approximately omega^2 = c^2 k^2. But what about the z-components of the
Refer to eqn (i) of (9.179), which comes from Faraday's law. This tells
us that the magnitude of the
z-component of B scales as (m/a)/omega = (1/c)k_x/k_z times the
transverse components of E.
It is thus suppressed by the ratio k_x/k_z compared to the transverse
components of B in a plane wave.
A similar analysis works for the z-component of E, using (iv), which
comes from the Maxwell-Ampere
One student pointed out something neat after class: The zig-zag
bouncing waves model discussed in and
around Fig. 9.25 can also be exploited to "explain" why there is a z
component of the magnetic or electric field,
and it even explains how large that component is: For example, if a
transverse wave is travelling
"diagonally", with transverse B field (B_x,B_z), then similar triangles
show that B_z/B_x = k_x/k_z.
This is just the relation inferred in the previous paragraph from
- The potential formulation of electrodynamics: I covered parts of
Tuesday April 13
- explained the notion of group velocity, and why it is equal to
- frequency dependence of the refractive index -
covered the model in Griffiths
- how RealD
3D projection and glasses work. A digital signal with 24 frames per
for each eye is tripled in frequency and alternated RLRLRL ...
by a light projector, so each eye
sees the same image three times, each for 1/144 th of a second. The
light from the projector goes
though a "Z-screen", which is an electro-optical liquid crystal
modulator, that puts out the light
as right circular or left circular polarized, let's say for the right
or left eye. This Z-screen must be
switchable at this high frequency, and synchronized with the incoming
light sequence. Half the
light goes to each eye, and half of that is lost when it is circularly
polarized, and the Z-screen
has some finite switching time, so less than 25% of the light makes it
out of the projection
camera for each eye. The screen in the theatre is a "silver screen"
that preserves the polarization
when it reflects the light, and it has a "gain" of brightness in the
direction of the audience of
a factor of about 2.2-2.4 relative to a normal matte screen.
The glasses worn by the viewer pass one circular polarization through
the right eyepiece, blocking the
opposite circular polarization. (Only 80% of the light of the correct
polarization is transmitted, further
decreasing the intensity.) The left eyepiece passes the opposite
How do the eyepieces accomplish this? Each one consists of two layers:
1) a quarter wave plate
in front and 2) a horizontal polarizer in back. The quarter wave plate
converts the incoming
circular polarizations into vertical and horizontal polarizations. The
fast and slow axes for linear
polarizations of the quarter waveplates lie at 45 degrees, and are
opposite for the two eyes,
so that opposite circular polarizations emerge with horizontal linear
I'm guessing the quarter
plate is about 0.5 mm thick (but this could be totally wrong),
and the wavelength of the visible light is around 0.5 microns, so 1000
In class I remarked that in order to produce the quarter cycle phase
the fast and slow indices of refraction must differ by one part in 4000.
A student asked about chromatic dispersion, i.e. the fact that
different colors would have
different indices of refraction, so the quarter wave plate wouldn't
work. I answered that the
difference of indices would be of order 1%, which shouldn't present a
problem. Well I think my
answer was confused! The wavelengths of different colors will differ by
something of order unity,
so a quite different number of cycles will be completed for the
different colors. Moreover,
If n_fast and n_slow differ from 1 by something of order unity, which
probably they do, then
if their difference has to be tuned to one part in 4000 that tuning
cannot possibly work for all
At the above link a design is mentioned that may be what is used, that
seems to address part
of the tuning problem but not all: take two wave plates of the same
birefringent material but slightly different thicknesses L and L + ∆L,
with their fast and slow axes
rotated 90 degrees. Then the second plate almost undoes what the first
plate did, and the net phase shift
just comes from the difference in thickness. One polarization has a net
phase change accumulation
k_slow L + k_fast (L + ∆L), and the other has k_fast
∆L). These differ by π/2 if
(k_slow - k_fast)∆L = π/2. Now let's look at the frequency
dependence. We have k = n(w)w/c,
where I use w in place of the angular frequency omega. So the condition
[n_slow(w) - n_fast(w)]w ∆L/c = π/2. Problem:
∆L that works for all frequencies w in the optical range, so we need [n_slow(w)
constant in this range! That's a significant demand since w changes by
almost a factor of 2. So are there
materials whose indices of refraction behave this way? Maybe not. But
the Wikipedia article also
mentions combining plates of different materials in order to cancel out
the frequnecy dependence...
I don't know how it's actually done for these RealD glasses. If you
find out please let me know! Monday April 12
more on em waves in conductors
skin depth in silver at 6 E14 Hz (blue/green, wavelength 5000 A) is
about 26 Angstroms,
whereas at 60 Hz it is about 8.5 mm.
skin depth in a wire: The
electric field in an AC current carrying wire, and the current, are
on the surface of the wire at high frequencies. This is analogous to
what we found for the plane wave
in a conductor. Another way to see it is to note that the oscillating
B-field in the wire induces an E field
whose emf adds to the current on the outside of the wire and subtracts
on the inside.
relation between skin depth and
absorption? I did this in class but messed up with numerical
but also with the energy dependence. The simplest way to compute the
fraction absorbed is to compute the
fraction reflected, since there the ratio R is just the ratio of the
squares of the electric field amplitudes.
Here is the correct argument:
According to (9.147) R = |(1-b)/(1+b)|^2, where (for typographical
reasons) I write b for betatilde.
We are talking about the limit |b| >> 1, so we get approximately
R = 1 - 4/|b| (get this by factoring
out a b from the numerator and denominator, then using the expansion (1
+ 1/b)^n = 1 + n/b +... a
couple of times), hence T = 4/|b|. This transmitted part is absorbed in
Now for vacuum/conductor interface, with µ approximately
µ_0, we have from (9.146)
|b| = (c/omega)|ktilde| = |ktilde|/k_vac = (√2/2π)(lambda_vac/d_skin),
T = (8π/√2) d_skin/lambda_vac. We learn that the ratio of skin depth to
controls the amount absorbed, but the numerical factor is pretty large,
With d_skin = 25A and lambda_vac = 5000 A we find T = 4π/√2 % = 9%.
This is pretty close
to the 7% figure mentioned in the book.
An interesting question came up in class: how does the fraction
depend on the
index of refraction of the transparent medium? For example, with a
silver mirror covered with
glass, how does the index of refraction of the glass change the
The answer is that lambda_vac is replaced by lambda_glass, which is
around 1.5 times smaller,
so there is 1.5 times as much absorbed, it seems... I think this can be
understood as due to a better
impedance match. The wave speed in the conductor is very low. The speed
in glass is lower than
in vacuum, so there is a closer match, hence more energy
transferred. Thursday April 8
oblique reflection and transmission:
Note that Fig. 9.14 and 9.15 are inconsistent with the fact (9.96,
9.97) that the components
of all three wavevectors in the plane of the interface are equal.
One point not made by Griffiths is that when µ1 = µ2
, the condition for Brewster's angle
becomes equivalent to the statement
that the reflected and refracted wavevectors are perpendicular.
configuration, the polarization vectors of these waves are also
perpendicular, and are parallel
other wavevector. I have heard this given as an explanation of why
there can be no
at this angle of incidence, i.e. because there is no motion of the
the surface that can
produce both of these waves. This explanation sounds like it has some
truth in it, but it can't
be really correct, because the true condition for no reflected wave is
different if the magnetic
permeabilities of the two media are different. Also, one can ask why it
it the reflected wave
rather than the transmitted one that is extinguished. Also, the
transmitted wave is really a combination
of the incoming wave plus the wave generated by the charges in the
material. Since the
incoming and transmitted waves are not parallel, the generated wave
must not be parallel to
the transmitted wave. This appears to weaken the argument based on the
Perhaps there is a way to clarify and support
this simple argument however.
em waves in conductors: I
followed Griffiths, but here are a couple of notes:
The demonstration that the free charge density vanishes contains a
subtlety. The argument goes like this:
the free charge density rhof is separately conserved, so
∂t rhof = -div Jf
= - sigma div E = - (sigma/epsilon) rhof
so rhof decays exponentially in
time. But if rhof refers to the
conduction electrons, then surely their
density does not decay exponentially! So what gives? The first two
equalities are unassailable.
The problem lies in the last one: epsilon takes into account the
"bound" charge, but what this
really means is the charge density arising from polarization. In a
conductor, the ion charge is not
included in this polarization charge, so must be included in the "free"
charge density on the
right hand side of the last term. Since it is constant in time, it can
also be added to rhof
time derivative term. So, what decays away exponentially is the sum of
the ion and conduction
Another point not made explicitly in the book is that, for a good
conductor, the imaginary
part of the square of the complex wavenumber (9.124) is much larger
than the real part, so
on the complex plane ktilde^2 points almost along the imaginary axis.
Therefore, its square
root, ktilde, lies at nearly a 45 deree angle, and has nearly equal
real and imaginary parts, both
nearly sqrt(µ sigma omega/2) Tuesday April 6
Perfectly conducting plasma and "Frozen-in
theorem" a.k.a. "frozen field theorem" a.k.a. "Alfven's theorem"
(see problem 7.59)
Implications: stellar collapse produces strong magnetic fields,
folding/mixing fluid produces strong fields with rapid spatial variation
in direction. Magnetic reconnection.
theorem, electic and magnetic field energy density, Poynting
magnetic flux tubes: Consider a tube of flux lines with cross
sectional area A and length L. If the plasma has approximately
constant density, then AL=const. as the tube is stretched. The
frozen-in theorem says BA =constant, so B ~ L. The energy density
the field is ~ B^2 _ L^2, and the volume is constant, so the field
energy scales as L^2. This means that the tube acts like a spring
with a constant spring constant (I said this wrong in class). It takes
energy to stretch it, and that energy is stored in the magnetic field.
It can be released when magnetic reconnection occurs.
Flow of energy into a resistor.
Momentum. rather than derive this as in the textbook, I appealed
to the fact the momentum = energy/c for a massless particle.
(E^2 = p^2 c^2 + m^2 c^4, so if m=0, E = pc). This relation between
momentum, energy, and speed, can be generalized:
Non-relativistically, momentum is mass times velocity, mv.
Relativistically, the mass is replaced by (E/c^2), and the momentum
is p = (E/c^2)v. When v=c this becomes p = E/c. Thus
the momentum density is the energy flux density divided by c^2,
Examples of the charged hoop around a solenoid: the em field has
circulating momentum, i.e. angular momentum. When the
electric or magnetic field is turned off, a torque is applied
internally to the system and something starts rotating, so that
the angular momentum is conserved.
Discharge of coil: field strength needed to tear charges from surface
is very high, probably
too high for the discharge to be initiated by that alone. Instead, I
think that ions in the air
play a critical role...but I'm not positive (or negative).
Perfect conductors: in order
not to have an infinite current the force must vanish, i.e. E + v
x B = 0.
This means that the induced emf in a perfectly conducting circuit
vanishes, so that the flux in the
circuit must be constant in time.
Example: square loop partially in a magnetic field (Problem 7.51).
Example: neutron star, treated as a conducting sphere spinning in a
constant magnetic field along
the axis (Problem 7.45).
Maxwell's equations in matter. The current has a free part, a
magnetization part that goes into H, and a polarization part that
goes into D.
Maxwell's conception of the displacement current: even the "vacuum" was
a medium, the aether,
and the presence of an electric field entailed a displacement of some
kind, such that when the
electric field is changing there is a changing displacement, hence a
(In quantum field theory, the vacuum, while Lorentz invariant, is a
sort of medium, and can be
polarized, etc.) Monday March 29
Maxwell's displacement current term added to Ampere's law. First, in
vacuum, then, in matter. Thursday March 25
Example: H for a permanent magnet. Recall H = B/µ0 - M. While curl H is zero, div H is not.
I discussed the example of a cylindrical bar magnet with uniform
magnetization M. Then div H = div M,
which is a delta function at the "caps" of the cylinder. So the H field of the magnet is identical
to the E field of a capacitor with two
parallel circular plates. For a magnet much longer than it is
wide, the H inside is then much smaller*
than M, so to a good
approximation we have B = µ0M inside. Outside,
B = µ0H,
H is the field generated
by the div M of the caps. A
different way to think about the
problem is to identify the bound current, and see what B that generates. The bound current
is Jb= curl M,
which is zero except for a surface contribution Kb=
M x n, which corresponds to current
the cylinder like a solenoid, and the magnetic field is thus that of
such a solenoid.
* Proof : The H from each cap
is (1/4π) ∫ [(div M)/z2] dV =~ MA/4πz2
where z is the distance from the cap
to the field point. At the midpoint of the cylinder, a distance z = h/2
from the cap, the contributions for the two
caps add to M(2 r2/h2),
Magnetization and saturation:
The permanent magnet on the lecture table is marked as having a field of
3.5 kilogauss, which is about 1/3 Tesla. A typical ferromagnetic
material has a magnetic field that saturates
at a field of the order of a Tesla with 100% domain alignment. So it
seems that magnet would be a few tens
of percent aligned. A small neodymium magnet by contrast has a field at
the cap of order one Tesla.
on Faraday's law curl E =
- The dimensions match. (The combination E + v
x B in the Lorentz force law
shows that the ratio
of the dimensions of E and B is velocity.)
- The parities match: both sides depend on the arbitrary right hand
- The divergence of curl E is
identically zero, which matches with the fact that div B = 0.
- In terms of the vector potential A,
differently, the curl part
of E is -∂tA. A gauge change of A changes only the grad part
- The sign is essential for stability, otherwise there are runaway
effects, whereby a changing flux induces more
of the same type of change, rather than opposing the change.
Lenz's law: the effects of
electromagnetic induction always oppose the change that produced them.
The induced electric field generates a current in the can that is
opposite to the current in the coil. The opposite
currents repel, and the can is pinched and repelled outwards. To
understand the outwards force you have to
consider the flaring out of the magentic field of the coil: a field
component perpendicular to the symmetry axis
is needed to produce the outward force along the axial direction.
Here a motional emf (in the rest frame of the magnet) generates
currents in the pendulum. Once the current
begins to flow it feels a repulsive force opposing the change. When the
penduum begins to exit the magnet
then it feels an attractive force, again opposing the change. In the
(instantaneous) rest frame of the moving
pendulum, there is no velocity, but the magnetic field is changing, so
there is an induced electric field, which
causes the current to flow.
The primary coil has an AC current, and surrounds an iron core. The
induced current in the secondary ring
opposes the changing flux. This thing can produce a steady nonzero
average force, and therefore "floats" the ring.
I used to think this was simply explained just by analogy with the can
smasher: when the flux increases, the induced
current opposes the change, hence is opposite to the primary current,
hence repelled, and when the primary current reverses,
increasing the flux in the opposite direction again the induced current
is opposite, hence still repels. This explanation
is bogus however, since it only covers two of the four quarters of a
complete cycle. For instance in the second
quarter cycle, the flux is decreasing because the current in the
primary is decreasing, so the induced current in the secondary
is in the same direction,
opposing the decrease, and hence is attracted.
So something is amiss in this explanation. The missing thing is the
contribution to the flux coming from the current in
the ring itself, i.e. the self-inductance!
current in the secondary lags that in the primary by 90 degrees in
phase; with the self-inductance it lags by more, which
results in avoiding total zeroing of the average force. If the
self-inductance dominates over the resistance, then the
secondary lags by 180 degrees, hence is always repelled. See Problem S7 of
HW7 for details.
Tuesday March 23
Ferromagnetism - Started by
"melting" the ferromagnetic domains of Nickel with this demo:
J7-13 CURIE POINT OF NICKEL
Some Curie point temperatures:
Nickel: 358 C
Cobalt: 1400 K
Iron: 1043 K (770 C)
Gadolinium: 16 C
Dysprosium: 85K (liquid nitrogen boils at 77 K)
Explained the nature of ferromagnetic domains and domain walls. The
walls are several hundred
atoms thick. I can't find consistent statements about domain sizes. I
read microns, but also millimeters
or fractions thereof. In any case, a domain is a part of the material
where neighboring electron
spins are aligned, resulting in a mesoscopic magnetic moment. Only some
materials have the
needed crystal structure. In those, the spin alignment occurs because
that way the spin states are symmetric,
which implies that the spatial wave function is antisymmetric under
electron intechange, which helps minimize
the electron-electron Coulomb electrostatic repulsion energy. If the
energy is sufficiently lowered by this
arrangement, then the spins will beheld in alignment. This depends on
temperature: if the termal energy
is too large, then the spin order will be melted. The transition where
this happens is the Curie point.
Different domains form, oriented randomly with respect to each other,
rather than one giant domain, because the
randomized domains have lower total energy. The cost is the domain
walls where the spins don't get to
align perfectly. The balance of this effect with the overall magnetic
energy determines the minimum total
energy domain size. In an external magnetic field, the domains aligned
with the field grow at the expense
of other domains, and some domains flip, so the material gets a
macroscopic magnetization. Some materials
can hold that magnetization for a long time, after then external field
is removed. Those are called
"permanent" magnets. Lodestone,
magnetized by lightning strikes.
Ferromagnetic material when magnetized can provide a huge amplification
of the original magnetic
field that triggered the magnetization. We demonstrated this with a
fromed from a coil with a 1.5 V battery, and an iron core inside: J6-04 LOW
POWER - HIGH FORCE ELECTROMAGNET.
Einstein de Haas effect:
Quite remarkably, Einstein, together with de Haas (who was Lorentz' son
in law whom Einstein had
helped get a temporary position), demonstrated experimentally in 1915
is due to circulating currents that carry angular momentum. The method
was to suspend an iron cylinder from a
fiber. Around the cylinder was a solenoid, in which an alternating
current was driven, alternately magnetizing the
cylinder in opposite directions. The magnetization arises from
circulating currents of charge...but also of mass.
So the magnetization also implies an angular momentum. Since there is
no external torque on the cylinder, the
bulk material must aquire an equal and opposite angular momentum. By
tuning the alternating current
frequency to the natural vibrational frequency of the torsional
oscillations of the suspended mass, the resonance
can be used to enhance the effect. Einstein assumed the microscopic
currents were due to cirulating electrons,
hence he used the gyromagnetic ratio obtained from the mass and charge
of the electrons. They thought they
confirmed the predictions to within 10%, but that resulted from
selective choice of data and massaging the
theoretical calculations. In fact, the prediction was off by a factor
of 2, since it is electron spin, not orbital
motion that accounts for ferromagnetism, and the gyromagnetic ratio of
electron spin is twice that for
orbital motion. Later experients saw this factor of two, even before
electron spin and its associated
gyromagnetic ratio was discovered, but these experiments played no role
in that discovery, which was
instead entirely based on atomic spectra.
Regarding this experiement, Einstein told a friend: "In my old age I am
developing a passion for experimentation."
(He was 36 years old!) In fact, his physics, though fundamental, was
always grounded in physical reasoning.
Perhaps his work at the patent office played a role in this. A case has
been made for an important role for that
work in his discovery of relativity theory: Einstein's
Galison. But an expert
historian of Einstein's work tells me that the case doesn't stand up to
scrutiny. I think it's an interesting idea,
but the book is very repetitive.
At first magnetism and electricity were disjoint phenomena. Then
Oersted and Ampere showed that an electric
current (produced by a voltaic pile) produces magnetic effects. A
natural question then was whether, conversely,
magnetic effects can produce an electric current. This was tried
without success using static configurations,
but eventually Faraday
found that a changing
magnetic field, or moving a wire in a static magnetic field,
produces a current. I demonstrated this with K2-02
INDUCTION IN A SINGLE WIRE.
A useful quantity in characterizing the effects that produce a current
in a circuit is the electromotive
or emf. I covered the definition and properties of emf as described by
A "motional emf" is produced by moving a wire in a time independent
magnetic field. Quite generally
the motional emf is minus the rate of change of magnetic flux through
the circiut loop. I proved this in the
way shown by Griffiths. If the loop is stationary but the magnet moves,
the same effects result, since all
that matters is the relative motion. But if the wire does not move, it
cannot be a magnetic force that makes
the current flow...so what can it be? It is an electric force...but
unlike electrostatic forces, it can produce
an emf around a closed loop. So, it is a new kind of electric field,
not derivable from the gradient of a potential,
whose integral ∫ E.dl around a loop bounding a surface
S is equal to -(d/dt)∫ B.da, minus the rate of change of
flux through S. The loop integral of E
can be written as a flux integral of curl E, and since this holds for
any surface S, we infer the differential form of Faraday's law,
curl E = - ∂tB. Monday March
(link to a video demonstration). The difference comes down
to electronic structure:
N has 7 electrons in the orbitals 1s (two), 2s (three), and 2p (three).
The spins on the 1s
and 2s electrons are paired, but those on the 2p electrons are not.
However, when two
nitrogen atoms bond to make N2, the six 2p electrons are shared, and
the spins pair,
yeilding no net spin or orbital magnetic moment. In contrast, O has 8
one more 2p electron. It turns out that the energy is minimized if the
spins are NOT
all paired, so O2 has a net spin magnetic moment, hence is
paramagnetic. The susceptibility
is mentioned in the table in Griffiths: 3.9 E-3 at -200C (= 73K,
boiling point is 90K).
Diamagnetism of water is easily demonstrated using a dish of water,
or with a stream of water.
rod is probably due to induced negative charge in the water stream,
pulled from ground.
In any case, what I'm linking it for here is the last part, showing the
repulsion from a magnet.)
A spectacular demonstration is levitating a water droplet, a
and even a frog,
in a 10 Tesla field. Pyrolytic carbon
is similar to graphite but has a parallel planar structure
with macroscopically large planes. It has the largest diamagnetic
susceptibility by weight
of any room temperature material, and can be levitated above
small neodymium magnets.
Discussed the classical model of diamagnetism presented by Griffiths.
Actually, it can
only be understood using quantum mechanics.
Ferromagnetism introduced by demo in which dysprosium (element number
66) is cooled to
liquid nitrogen temperature and undergoes a ferromagnetic phase
transition, becoming strongly
attracted to a large permanent magnet:
J7-14 CURIE POINT OF DYSPROSIUM
11 Exam 1 Tuesday March
9 Review for exam. Monday March 8
The auxilliary field quantity H
= (1/mu_0)B - M is introduced in analogy with the
Since curl M is the bound
volume current density we have curl H
= Jf. Note
the div H = div M ≠ 0 in general.
Jump conditions on H.
Demonstration of paramagnetism of copper sulfate (see above).
Linear, isotropic magnetic media: M
= chi_m H. (N.B. we use
the auxialliary quantity H
here, unlike in
defining the dielectric susceptibility where we use P = chi_e epsilon_0 E.) The magnetic susceptibility
chi_m is a dimensionless number, positive for paramagnetic materials
and negative for diamagnetic ones.
Thursday March 4
Ampere, after Oersted's and his own work showing that electric
currents deflect a compass needle,
proposed that the magnetic field of the earth must arise from electric
"...we could suppose that before we knew about the South-North
orientation of a magnetic needle,
we already knew the needle's property of taking a perpendicular
position to an electric current [...].
Then for one who tries to explain the South-North orientation, would it
not be the simplest idea to
assume in the earth an electric current?"
Magnetic monopole moment
vanishes (proved this), and dipole
moment is 1/2 ∫ r
x J dtau
(talked about what goes into proving this). For a planar loop showed
the dipole moment is the
current times the area times a unit vector normal to the loop. For an
arbitrary loop it is I ∫ da.
Proved that the integral of the area element is independent of the
surface spanning the loop.
Jump condition for vector potential in Coulomb gauge.
Force grad(m.B) and torque m x B
on a magnetic dipole. Illustrated with
and permanent magnets. A dipole will be flipped by the torque until it
aligns with the magnetic
field, then the force will pull it into the strong field region, where m.B
Looked at example of a pair of parallel current loops. The are physical
(not pure) magnetic dipoles.
They attract. You can see this thinking about the parallel currents
attracting. Or you can see it in
terms of the magnetic poles, or the magnetic dipole. As explained also
on p. 257, I explained that
in a uniform magnetic field there would be no attraction, and it is
only the flaring out of the field
sourced by one loop that can produce the net attractive force acting on
the other current loop.
Magnetization: magnetic dipole
moment per unit volume. Magnetized material has a curculating
bound surface current Kb= M
x n, where n is the unit normal, and a bound
volume current Jb= curl M.
Tuesday March 2
Jump conditions for the
magnetic field at a planar interface: div B = 0 implies the perpendicular
of the magnetic field is continuous, and curl B = mu_0 J implies any jump parallel to the
surface must be
perpendicular to the surface current K.
B must be
perpendicular to both the surface
normal n and the surface
current, i.e. it must be parallel to the cross product K x n.
shows that the coefficient is just mu_0: B^above - B^below = mu_0 K x n.
Vector potential, gauge freedom,
In Coulomb gauge the vector potential generated by an element of
current is parallel to that current.
Meaning of Coulomb gauge in the Fourier transform: div A = 0 implies k.a(k) = 0,
where lower case a is the
Fourier transform of A, which
is thus transverse.
Example of a spinning charged shell: constant magnetic field
inside, pure dipole field outside.
The radius if the sphere is R and the angular velocity omega and
surface charge density sigma.
Then the surface current density is K
= sigma v phihat = sigma omega
R sin(theta) phihat.
(Example 5.11) gets the vector potential by integrating over the
surface current. The magnetic field
is then obtained by taking the curl of the vector potential.
An alternate way to solve for magnetic field of the spinning charged
shell: introduce a magnetic scalar
U, so that B = - grad U. This
is possible as long as curl B is
one function U^out, and inside with another U^in inside. Then div B = 0 implies that these each
satisfy Laplace's equation,
and they are axisymmetric, so can be expanded in Legendre polynomials.
Only positive powers of r occur in U^in,
only negative powers in U^out. The matching conditions come from the
jump conditions for B, which
imply (i) continuity
of the radial derivative, and a jump in the theta derivative
proportional to the surface current. We can now virtually solve
this by inspection: the theta dependence of the surface current is
sin(theta) = -(d/dtheta)P_1(theta). So only the P_1
term has a jump. This means that there can only be a P_1 term, since
the absence of a jump in the r derivative and the
absence in the jump of the theta derivative are incompatible (check
this). So we have
for some A, where the coefficient -1/2 was chosen so the r derivatives
would match. U^in is proportional to z, hence
the interior magnetic field is constant in the z-direction. U^out has
the form of a pure dipole potential.
To determine A we need to impose the condition that the negative of the
gradient of U in the theta direction jumps
by mu_0 K = mu_0 sigma omega
R sin(theta). This yields -1/2 A/R sin(theta)
- A/R sin(theta) = mu_0 sigma
so A = -2/3 mu_0 sigma
expansion of the vector potential Monday March 1
Force on a wire.
To what extent do the div and curl of
a vector field determine the vector field?
One answer to this is the Helmholtz
theorem, as described in Appendix B of Griffiths.
If div F = D and curl F = C,
C fall to zero
faster than 1/r2 as r goes to infinity,
then there is a unique F that
satisfies these equations and vanishes at infinity, which is given
F = -grad U + curl W,
= (1/4π)∫ d3r' D(r')/|r-r'|
(1/4π)∫ d3r' C(r')/|r-r'|. As a special case, if
= 0 and curl F = 0, and F vanishes at infinity, then F vanishes everywhere.
A different approach is to assume the functions can all be Fourier transformed. Use the
lower case letters f, d, c,
F = C, it must be that div C = 0, so k.c = 0.)
As I explained in class, div F = D and curl F = C
are equivalent to ik.f = d and ik x
f = c.
The first equation determines the "longitudinal" component of f and the second determines the
We can write the solution for f:
f = (-i/k2) (d k + c x k)
This determines all but the k
= 0 component, i.e. the constant part of F.
Solenoid with arbitary
cross-section: Take the solenoid to run along the
z-direction. If the transverse
components B_x and B_y are zero, then the loop integral form of
Ampere's law immediately yields
B_z = mu_0 n I, where I is the current in the wire and n is the number
of turns per unit length.
For the analogous toroidal problem, Griffiths uses the Biot-Savart law
to show that the transverse
field is zero. But here is a nicer way to show it: split B into the components parallel and
to the z-direction, B = B_z + B_perp, and note that by the
symmetry neither of these depend on z.
Thus by itself div B_z = 0, hence since div
B = 0,
div B_perp = 0. Also curl B_z has only
x & y components, and curl B_perp
curl B = mu_0 J implies
curl B_perp = 0. We have shown
that both the div and the curl of B_perp vanish
everywhere. As explained above, this
implies that B_perp itself
must vanish everywhere (except for a
possible constant part that is unrelated to the solenoid.)
A similar construction using cylindrical coordinates shows that the
magnetic field of a toroidal solenoid
purely in the angular direction, and is hence easily determined using
the integral form of Ampere's law.
I suggest you verify this yourself. Thursday Feb. 25
More on cycloid motion in crossed electric & magnetic
fields. To understand the motion,
imagine the charge is moving perpendicular to the crossed fields at
speed v_0 = E/B. Then the
electric and magnetic forces balance, so the particle is unaccelerated.
In a proper relativistic
treatment, the speed must be less than c, the speed of light, so this
undeflected motion is
possible only if E/B < c.
Next consider a more general motion, r(t)
v_0 t + s(t), which has acceleration equal
where the ' means time derivative. This should be equal to the force
divided by the mass,
(q/m)(E + (v_0 + s') x B). We choose v_0 so that E + v_0
B = 0, hence s'' = (q/m) s'
So the motion described by s(t) is a cyclotron motion, which is
added to the uniform velocity v_0.
Now suppose the particle is initially at rest. Then evidently s' starts out equal and oposite to v_0.
This motion is a cycloid.
Comment on magnetic forces do no work Example 5.3.
Oersted first showed that a magnetic compass needle is deflected by an
This showed that magnetism may be due to electric currents. A couple of
Ampere established the law of force between currents.
Magnetostatics: constant charge density and current density. Continuity
equation then implies also div J
Griffiths derives div B = 0,
and curl B = µ0J from the Biot-Savart law. But
these are more basic
equations, and I will skip that "derivation" (you are welcome to read
it). Notice the second equation
makes sense only in magnetostatics, since div curl B is automatically zero, so we'd
better have div J = 0.
Given enough symmetry, these equations can be used to find the magnetic
field given the current.
Example 5.7: an infinite straight line current. Rather than appealing
to the Bio-Savart law to get the
direction, cylindrical symmetry tells us all three cylindrical
components of the magnetic field depend
only on the distance s from the axis. The curl equation then gives the
phi component of B.
The curl eqn tells us there is no circulation of B around a rectangular loop in the
since no current flows through that loop. By symmetry the top and
bottom edges cancel, so the
inner and our edges must cancel as well, so the z-component must be
constant. A constant field
in the z direction can always be added but is not sourced by the line
div equation tells us there is no magnetic flux through a cylinder
centered on the axis. The top and
bottom caps cancel since the z-component of the field is constant (or
vanishes), so the
radial component must vanish as well. Tuesday Feb. 23
Started out with this demo: J4-13
MATCHSTICK ON NICKLE UNDER GLASS
which illustrates how an electric field can polarize a neutral material
(in this case the wooden matchstick), and thus exert a force on it.
Then I charged a balloon by rubbing it with fur, after which it nicely
to the door of the classroom, illustrating the same principle as above.
Examples of finding electric field given a polarization, or given a
linear dielectric material
and an external electric field.
Jump conditions for displacement, constrasted with those for electric
Energy stored in system of linear dielectrics.
Magnetic (Lorentz) force on a moving charge, F = q v x B. This is orthogonal to the
hence does no work. It only redirects velocity, but does not add
Example: cyclotron motion. Mentioned that these circular orbits are
quantized in quantum physics,
and the corresponding energies are called Landau levels.
Example: Cycloid motion in crossed electric and magnetic fields. Monday Feb. 22
A permanent dipole moment p
in an electric field E has
energy W = - p.E. But what if the dipole
moment is induced, say p =
alpha E. Then we must
integrate up from zero field to find the energy,
W = ∫ dW = ∫ -alpha E.dE = -1/2 alpha E^2. I used this in
the discussion about computing alpha
from quantum perturbation theory in the 2/18 notes.
J4-22 PARALLEL PLATE CAPACITOR WITH DIELECTRIC
Discussed this demo. How to compute the field? I introdced
here the electric displacement
D = epsilon_0 E + P. Approximate the dielectric using
the linear, isotropic
relation between P = chi_e
epsilon_0 E, in which case D = epsilon E, where
epsilon = epsilon_0(1+chi_e).
Terminology: chi_e is the electric
susceptibility, while epsilon is the permittivity. epsilon_r = epsilon/epsilon_0 = 1 + chi_e is the relative permittivity or dielectric constant.
Discussed uniformly polarized sphere.
Solved parallel plate capacitor with dielectric. Noted how capacitance
is increased by a
factor of the dielectric constant. Thursday Feb. 18
Examples of multipole moments, physical and "pure" dipoles,
conducting sphere in an
external field produces a pure dipole field, electric field of a dipole.
separation) of matter, polarizability, polarization density P
An "electret" is a
material with a permanent electric polarization, analogous to a
magnet. These can be fabricated by heating a material containing polar
an external electric field, then cooling the material until the
"freezes in". I read in Wikipedia that quartz can be in a naturally
state, but I didn't find anywhere the magnitude of polarizations that
can occur that way.
The textbook mentions in Problem 4.11 that barium titanate is the "most
electret that holds its own polarization (as a reset of its intrinsic
structure). If quartz does
as well, this is puzzling, since quartz seems more familiar! Note that
an electret would
tend to neutralize its field, by accumulating opposite charges from the
environment on its ends.
The dipole moment of
from Example 3.8. Dividing by the field strength yields the
polarizability, which is
R^3(4π epsilon_0). It's interesting to compare this to
the table 4.1 of atomic
Here I'm getting a bit carried away...
through by 4π epsilon_0, they are, in units of cubic
Angstroms (10-30 m3),
0.667 for H, 0.205 for He, and 24.3 for Li.
I was wondering how to account for the differences.
One can calculate the polarizability using quantum mechanics. The
energy of a permanent
dipole moment p aligned in an electric field of strength E is -pE. If
the dipole moment is
induced by the electric field, p = alpha E, then the energy is the
integral of -alpha E dE,
i.e. it is -1/2 alpha E^2. So the polarizability is minus twice the
energy shift divided by the squared
field strength. For the ground states of the atoms in the table, I
think the first order shift
vanishes by symmetry, so the leading order comes from second order
The result for H is 9/2 (a_B)^2, where a_B = 0.53 A(ngstroms) is the
Bohr radius, which
yields 0.670 A^3. This differs from 0.667 by only 3/667 0r about half a
percent. So for this
simple atom, lowest (second) order perturbation theory gives an
For He, despite having two electrons, the result is less than a
third as large.
I mumbled in class something about the Pauli exclusion principle, but
this is nonsense,
since in any case the electron spins are different so there is no
exclusion of their spatial states.
I think the real issue is that the jump to the lowest excited state in
Helium is roughly twice
as large as in H, because of the greater nuclear charge (although once
the electron jumps
outside the ground state orbital the other electron screens the nuclear
charge). In second order
perturbation theory this energy jump is squared, so this presumably
accounts for the different
For Li, on the other hand, as pointed out by a student in class, there
are 2p spatial states available
that lie rather nearby in energy. These lie only about 1.5 eV above the
ground state, compared
with the jump of about 10 eV for H. (They would be degenerate were it
not for the screening effect
of the two 1s electrons.) The contribution from these is thus
= 44 times larger than
for H. Applying this factor to the polarizability of H (0.667) we get
about 30. The actual
polarizability of Li is 24.3. The fact that it is not as high as 30
makes sense, because the energy
shift in second order perturbation theory comes from summing over all the states, and as we
go to the higher levels (3s,p,d, etc) the energy jumps approach those
of H, since the two
1s electrons screen the nuclear charge.
Tuesday Feb. 16
Legendre polynomials, multipole expansion
Note that a single charge at the origin has only a monopole moment, but
the same single charge
displaced to another point has nonzero multipole moments. The
to 1/|r - r'|, whichis a solution to Laplace's
equation, hence can surely be expanded in the general form of (3.65),
∑(Alrl + Blr-l-1) Pl(cos
To find the Bl, consider the case where r' = r' zhat. Then on the positive z axis we
1/|r - r'| = 1/(r-r') =
so we read off that in that case, Bl = (r')l.
When r' is not on the
z axis, we can infer the result by rotating the z axis,
i.e. we replace theta by theta', where theta' is the angle between r and r'. Thus
1/|r - r'| = (1/r) ∑(r'/r)l Pl(cos
In deriving the general form of the multipole expansion, (3.95),
Griffiths expands 1/|r - r'|,
remarkable result" that it is equal to the expression above. But since
he already derived the general form (3.56), it shouldn't seem so
Monday Feb. 15
Separation of variables, cartesian and spherical Monday Feb. 8, Tuesday Feb.
9, Thursday Feb. 11: SNOWED OUT Thursday Feb. 4
Charge distribution on a conducting
disk or needle: I found an extremely nice, two page paper on
this: R. H. Good, American Journal of Physics 65, 155 (1997). He shows
that the charge distribution which
produces a zero field on the disk is obtained by projecting a uniform
charge distribution on a spherical shell
onto the equatorial plane. This yields a non-uniform charge density on
the disk, divergent at the edge.
Similarly the distribution on a needle is obtained by projecting the
spherical shell onto an axis, and this
yields a uniform charge
density.... which seems paradoxical, since it seems as if the field at
except the midpoint could not vanish. See the paper for the explanation!
Method of images (cf. section
3.2 Griffiths) - Covered as in Griffiths. Regarding the plane, he says
the induced charge
is -q, as "you can perhaps convince yourself it had to be." The best argument for
this that I could find so far is to consider
a Gaussian surface that is a large disk under the conducting plane,
then follows the field lines around behind the charge
q, and then jumps across. In the limit that this surface goes to
infinity in all directions, the flux will vanish, so the total induced
charge must cancel the charge q. For a picture of the field lines for
the plane and sphere problems see http://www.scielo.br/img/revistas/rbef/v31n3/916fig01.gif Tuesday Feb. 2
- Computational note:
to find the leading order behavior of a function of z, when z >>
express the function in terms of the small dimensionless ratio R/z, and
make a Taylor expansion about R/z = 0.
- Conductors (cf. section 2.5
Griffiths). Note that charges are bound to a conductor by some
so charges are confined on the conductor unless the electric field is
really large. Surface charge density
is determined by the jump condition (since the field inside vanishes)
to be epsilon_0 times the outward
normal component of the electric field at the surface (there is no
[Here is the talk I mentioned in class by David Griffiths on the charge
distribution on a conductor in
- Uniqueness theorem for solutions to
Suppose div grad V1 = 4π rho and
div grad V2 = 4π rho in a region R.
Then the difference W =
V1 - V2 satisfies
Laplace's equation, div grad W = 0.
Then div(W grad W) = (grad W).(grad W). Now integrate both sides of
this equality over R.
Using the divergence theorem, the left hand side becomes the flux of W
grad W through the
boundary of R. If the boundary conditions imply that this vanishes,
then it follows that grad W =0,
i.e. W is constant. We consider three versions of this theorem:
1) Suppose V1
and V2 are equal to the same fixed
function on the boundary (Dirichlet
Then W = 0 on the boundary. So W = constant in the region, and since W
= 0 on the boundary, it must
actually be zero everywhere in R. Thus
V1 = V2,
2) Suppose the boundary condition is that grad V1
and grad V2 are perpendicular to the boundary
(Neumann boundary condition). Then
W = constant, i.e. any two solutions (if they exist) differ at most by
3) Suppose the boundary is the surface of a conductor or multiple
conductors, so that V1
are each constant on all components of the boundary,
and by Gauss' law the flux of grad V1
and grad V2
over any component of the boundary are equal to each other and equal to
-Q/epsilon_0, where Q is the total
charge on that conductor. Then the flux
- V2) grad (V1
- V2) through the boundary vanishes,
so again we conclude that W = constant, so the potentials differ at
most by a constant.
- Solutions to
Laplace's eqn have no local minima or maxima: If div grad V = 0
then the flux of grad V through
any closed surface is zero. Suppose V has a local minimum at a point p.
Then grad V points away from
p in all directions, so the flux of V through a small sphere
surrounding p is positive. This contradicts the
fact that it must be zero. Hence there cannot be a local minimum.
Similarly there cannot be a local maximum.
This gives an alternate proof of the fact that if a solution V to
Laplace's equation is constant on the boundary
of a compact region R, then it is constant everywhere throughout R,
otherwise it would have a local maximum
or minimum inside.
- Examples: Applied the
uniqueness theorem(s) to the examples discussed in section 2.5.2. As
pointed out to me after
class, I overcomplicated the argument for example 2.9. The potential is
constant on the conductor, hence
constant on the spherical outer surface. There is a spherical solution
to Laplace's equation outside the conductor
satisfying this boundary condition, and by the uniqueness theorem this
must be the solution. Thus
the charge density
at the surface is spherically symmetric as well, i.e. the charge q is
spread uniformly over the sphere, and contributes
nothing to canceling off the field in the interior of the conductor.
- Capacitors: given an
arrangement of two conductors, carrying charge Q and - Q, the potential
between the conductors must be proportional to Q. Proof using the
uniqueness theorem: given a charge distribution
and potential satisfying Laplace's equation and conductor boundary
condition for one value of Q, a solution is
obtained for 2Q simply by scaling the charge density and the potential
everywhere by a factor of 2. By the uniqueness
theorem, this must be the actual solution. Thus Q = C ∆V for some
constant C, the capacitance,
determined by the
conductor geometry. The SI unit of capacitance is the farad (F), 1 F =
1 C/V. A single conductor is assigned a capacitance
using for ∆V the potential difference between the conductor and
Monday Feb. 1
- Finding V from E: the
potential change from a to b is minus the line integral of E along any curve from a to b.
For example, taking V = 0 at infinity, V(r) is minus the line integral of the
electric field from infinity to r.
- Work and energy in electrostatics - cf section 2.4 of Griffiths.
Thursday Jan. 28
Applications of Gauss' law:
charge distribution and showed that the field outside is the same as if
all the charge were
concentrated at the center. We also showed that the field inside a
spherical shell is zero.
The field of an infinite plane is constant, independent of the distance
from the plane.
It's magnitude is sigma/2epsilon_0, and direction is away from the
plane if sigma > 0.
Jump Conditions: The field
flips direction from one side of an infinite plane to the other,
so the field itself jumps by sigma/epsilon_0 in the direction
perpendicular to the plane.
This jump condition holds across any
surface charge. For example, across a spherical
shell, the field jumps from 0 inside to sigma/epsilon_0 in the radial
Spherical shell versus infinite plane:
sparked by my erroneous statement that the field very close to any
surface charge is
the same as the field very close to an infinite plane, i.e.
sigma/2epsilon_0. The example
of a spherical shell, which has a field twice as large as this just
outside, shows this to be wrong.
One student suggested that this discrepancy could be understood by
thinking of a contribution
sigma/2epsilon_0 coming from the nearby cap of the sphere, and an
coming from the rest of the sphere. This is right in the following
sense. Consider a point
at a distance d
from the north pole of a sphere of radius R, and consider a cap of
If you take
the limit in which both d and s tend to zero, and d goes to zero faster
also tends to zero, then the contribution from the cap indeed limits to
Electrostatic potential: A
Coulomb field r/r2
can be written as (minus) the gradient of
a function, -grad(1/r). A superposition of Coulomb fields
is thus minus the gradient of
a sum of 1/r functions centered at the different charges. That is, E = -grad V for some
function V, called the electrostatic
potential. The SI unit of potential
is the volt (V):
1 V = 1 Nm/C = 1 J/C.
I was asked in class about the expression for the electric field as an
integral over Coulomb
fields for a continuous charge distribution. This has a script r
squared in the denominator,
which blows up when r = r'. I said this does not lead to an
infinite result, because it is multiplied
by the script r unit vector,
which points in all directions near r = r'
While this is true, there is another reason why no infinity occurs: the
volume element in the
neighborhood of that point also goes as script r squared, which cancels
the former one.
One could ask a similar question about the scalar potential. Suppose we
the function 1/r over a volume including the origin r = 0. Do we get
infinity? No: the volume
element in spherical coordinates is r2 sin(theta) dr dtheta
dphi, so the integrand including the
volume element goes as r, not 1/r. Tuesday Jan. 26
the electric fields from multiple charges add vectorially.
This simplifies things a lot! It is not exactly true: due to the
nature of the vacuum in quantum field theory there are corrections,
normal circumstances. But in strong enough fields these
corrections can become
important, even dominant. An example is the scattering of light by
two photons each split into electron-positron pairs, which then
each other pairwise to produce two new photons. This becomes important
when the energy of the field in a volume of the electron Compton
cubed, times the square of the fine structure constant, is comparable
rest mass of an electron times the squared speed of light. This turns
out to be
an electric field of about E18 V/m, or a magnetic field of about E9 T.
Magnetic fields of order E10 T are believed to exist around certain
stars, called magnetars.
centered on a point charge q, and found it is equal to q/epsilon_0,
of the radius of the sphere. Then we argued it is also independent of
the shape of
the surface. Therefore the flux through any surface is the total charge
divided by epsilon_0. Using the divergence theorem, and the fact that
this is true
for any surface, we
infer that div E =
rho/epsilon_0, where rho is the charge density.
We should be able to derive Gauss' law instead by just computing the
of the electric field of a charge distribution, given the integral
expression for that.
This involved computing div(r/r3)
- 3r-4(grad r).r = 3r-3 -
But this can't be right at r = 0, since integrating this divergence
over an arbitrarily
small sphere around r = 0 yields, with the help of the divergence
The resolution: div(r/r3) = 4π
0.Monday Jan. 25
Supplementary books on vector calculus: Div, Grad, Curl and all That: An
Informal Text on Vector Calculus, H.M. Schey A Student's Guide to Maxwell's
Equations, Daniel Fleisch
I will essentially skip Chapter 1. Please review the math and make sure
you know it,
and learn it if you don't know it. I am happy to answer any questions
you have about it.
Electromagnetism is now understood as part of a (partially) unified
based on a "gauge theory" of the group SU(2)xU(1). This symmetry is
broken by the Higgs
field, and except for the electromagnetic force all the others become
short ranged. We have
a subject of classical electromagnetism because the force is long
ranged, and macroscopic fields
due to multiple sources add together coherently.