Phys411 - Electricity and Magnetism
of Maryland, College Park
Spring 2011, Professor:
Notes, Demos and Supplements
In these notes I'll try to just indicate the topics covered in class.
I'll also mention things I talk about in class that
are not also in the textbook,
as well as supplementary material, if they are not in last years notes.
Please do not assume that these notes
are even roughly complete.
Material strongly deformed
by a magnetic field, http://focus.aps.org/story/v27/st9
Physics Today article from Sept.1998, about diamagnetic levitation of
frogs and other things:
Everyone's Magnetism, Andrey Geim
(the guy who won the Nobel Prize last year for fabricating graphene
with adhesive tape)
European Journal of Physics article from 1997 with soime math details
about magnetic levitation:
Of flying frogs and levitrons, MV Berry
and AK Geim
On the history of the Einstein-de Haas
effect, V. Ya. Frenkel
Monday May 9
- 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
part of the
self-force F_int(q), which Griffiths points out is half of the total.
To get the other half,
problem 11.20 Griffiths changes the model to a continuous charge
distribution. But instead we
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.
- Regarding the runaway solutions, I mentioned that you can get rid of
them by imposing a final boundary
condition, but then there is "pre-acceleration", i.e. acceleration
before the external force acts. If you're
curious, here are some notes I wrote last year that explain this:
solutions of Abraham-Lorentz equation
- pertubative method to reduce the order of the equation:
make a series
solution in powers of tau.
The Abraham-Lorentz equation: a = (F_ext)/m + tau adot. Treat the
adot term as a small
correction, and therefore replace it by tau (F_ext)dot/m. Can
systematically expand in a power series in tau...
- Transformation of the fields when the reference frame changes. I was
running out of time, so didn't
write the transformation, but mentioned the invariants: E.B
and E^2 - c^2 B^2.
- I made an attempt to say something about special relativity in a few
minutes, but there was really no time.
See last year's notes for some flavor. In particular, you might be
interested in this article about
relativity and spacetime diagrams:
Spacetime and Euclidean
Geometry, by Dieter Brill and Ted Jacobson
But there's some more in last year's notes.
Thursday May 5
+ Absence of monopole radiation is linked to charge
conservation, and to the fact that the field outside a spherical charge
distribution depends only on the total charge. Gravitational radiation
also has no monopole contribution, because the source
of gravity is energy, and that's conserved, and the field outside a
spherical mass/energy depends only on the total energy.
Also it has no dipole contribution, because the time derivative of the
mass/energy dipole moment is the momentum, and
momentum of an isolated source is conserved. So gravitational radiation
is sourced at leading order by the quadrupole moment.
Specifically, it's the third time derivative of the quandrupole moment.
That's the pattern: one more derivative for each higher
multipole. And each derivative brings another factor of wa/c =
+ We briefly discussed whether charge could appear out of a wormhole,
in effect violating charge conservation and thus
somehow producing monopole radiation. However this won't work: a sphere
surrounding the wormhole will have an electric
flux detemined by the total charge enclosed, whether the charge is
outside the wormhole or inside or beyond.
+ What it means to say electromagnetic radiation has "spin-1", and
gravitational radiation "spin-2", etc: if the minimum rotation
under which the field returns to it's original configuration is 2π/s,
then the field has "spin-s". An
electromagnetic wave is characterized
by the electric field vector, which roates fully under 2π, hence has
spin-1. A gravitatonal wave stretches in one direction and compresses
in the perpendicular direction, so returns to it's original pattern
under a π rotation, hence s=2.
+ Radiation beaming: examples
of velocity instantaneously parallel to acceleration, and perpendicular
to the acceleration.
I showed a mathematica notebook with a graph
having a know to crank up the v/c factor. The perpendicular case is
for circular motion, which produces what's called synchrotron radiation. A common
source is fast moving charged particles
in a magnetic field. I talked about the example of the Crab nebula, which
emits synchrotron radiation from radio frequencies up
to 100 MeV (I mistakenly said GeV in calss). At the upper end, the
corresponding electrons have energies around
1500 TeV = 1.5 10^15 eV. These have a relativistic gamma factor of
around 3 x 10^9, hence a velocity within 1 part in 10^19 of the
speed of light. (To understand what sets the scale of the cutoff
frequency for a a given charge velocity,
one can esimate the angular width of the beam, hence the time it takes
for the beam to sweep past the viewer. The reciprocal of this
time sets the scale of the maximum Fourier frequency component in the
+ Radiation reaction force: I
discussed this more or less along the lines in the textbook. One
additional comment was how to
think about the "derivation", given all it's shortcomings. Why should
it work? Well, presumably there is some simple formula for
the radiation reaction force, so once you've found something that
"works" to preserve energy conservation in certain limited
you can guess that this works more generally. There are independent
arguments that it does. In fact, a student suggested that one
just use the Poynting flux through a small sphere surrounding the
particle, and this is one way I've seen the radiation reaction force
derived (but in the fully relativistic setting).
+ runaway solutions: the Abraham-Lorentz radiation reaction force is
third order in the time derivatives of the position. There are
exponentially growing solutions. I'll explain next time how to get rid
Tuesday May 3
+ At what distance do the radiation fields become much larger
than the near-zone
fields? The electric dipole radiation 1/r
divided by the (time-dependent) electric
dipole 1/r^2 potential is of order rw/c = 2π r/wavelength, so when r
>> wavelength the radiation
field is dominant.
+ Poynting flux of radiation, angular dependence, and total power
+ Note that the radiation field at a given large radius "looks like" a
plane wave propagating in the radial direction, in the sense that
E and B are perpendicular to each other
and to the radial direction.
+ Next order corrections to electric dipole: electric quadrupole and magnetic dipole. Examples of
magnetic dipole radiation are:
oscillating current in a fixed loop, fixed current in a rotating loop,
rotating magnetized neutron star, and the hyperfine spin-flip
transition (which produces the famous 21 cm line) in the ground state
+ The origin of the electric quadrupole radiation is found in the next
order terms in the Taylor expansion of the charge and current density:
rho(r', t0 + r'.rhat/c + ...) = rho(r', t0) +
rhodot(r', t0) r'.rhat/c +
(1/2) rhoddot(r', t0) (r'.rhat/c)^2 + ...
the integral of which over r'
Q + pdot.rhat/c +
(1/(2c^2)) Qddot_ij rhat^i rhat^j,
where everything is evaluated at the time t_0, and Q_ij is the integral
of rho x^i x^j over the source.
This is called the second moment of the charge distribution, and its
tracefree part is the quadrupole moment tensor.
The size of the quadrupole term compared to the dipole term is
determined by (i) the extra derivative, (ii) the extra r',
and the extra 1/c, i.e. by wa/c = a/wavelength, where a is the typical
source dimension. The power radiated by the
quadrupole will be suppressed by the square
of this ratio.
To get the full electric quadrupole field, as well as to find the
magnetic dipole field, we need to also expand the
current density in the vector potential to the next order in the time
dependence. This involves
J(r', t0 + r'.rhat/c
+ ...) = J(r',
t0) + Jdot(r',
r'.rhat/c + ...
The integral of this over all space is
pdot + mdot x rhat/c + 1/2 Qdot.rhat/c
is the tensor defined above [in class I omitted this term]. The
from the integral of J,
and yields the
vector potential contribution to the electric dipole radiation. The
second and third terms come from the integral of Jdot, and
to derive. The details are explained in the index notation notes. The
first, m term, produces the
magnetic dipole radiation.
The Q term
contributes to the electric quadripole. As you can see, these come in
at the same order. The electric field of the magnetic
dipole radiation is minus the time derivative of the m term,
E_(mag dip rad) =
-(µ0/4π) (1/rc) mddot x rhat
Note that this has exactly the same form as the electric dipole
-(µ0/4π) (1/r) pddot_perp
with the replacement of p by m/c, and rotated by π/2 about rhat. Note, again, that the ratio
between the magnetic and electric
dipole amplitudes is (m/c)/p ~ qwa^2/qac = a/wavelength (assuming the
current is I ~ qw and the area is ~ a^2).
+ Point charges: we can apply
the retarded potentials to point
particles, using delta function sources. The result is the
Lienard-Wiechert potentials. From those we can find the exact electric
and magnetic fields. These and the potentials are
written out in the book. Then from the fields we can identify the
radiation part, the power radiated into a given solid angle,
and the total radiated power. I refer you to the book for all the
formulae. The calculations are complicated, and we won't
go through them. An important final result is that, at relativistic
velocities, the radiation is beamed into the forward direction
and amplified in intensity by a huge amount.
Monday May 2
+ derived electric dipole
radiation fields from the retarded potentials
Thursday April 27
+ example of retarded potential from a wire carrying a time
dependent current (see example 10.2)
+ Jefimenko's equations (section 10.2.2)
RADIOWAVES - ENERGY AND DIPOLE PATTERN
+ Most of the class I devoted to trying to develop intuition about
radiation: the 1/r dependence of the fields, the
kinks of field lines determining the radiation pattern and the
polarization, and the dipole approximation.
+ electric dipole raidiation:
the length scales involved in the dipole approximation: a <<
lambda << r.
I wrote the potentials (11.51, 11.54) and fields (11.56,
11.57) for electric dipole radiation , and discussed some examples
like a charge oscillating in one direction, and a rotating
charge (or pair of charges). In the latter case, we saw how that
could produce circular polarization viewed perpendicular to the
rotation, and linear when viewed in the plane of the rotation.
Next time I'll explain the derivation of these formulae.
Tuesday April 25
More on potentials:
+ Can we access Lorentz gauge? This is equivalent to the question: can
we solve the inhomogeneous wave equation box f = s?
+ Note the time-independent version of this: Laplacian f = s. We know
well how to solve this using the Coulomb potential.
A guess for the time-dependent case is that in the
integral over s, the time should be replaced by the retarded time, t_r
= t - |r-r'|/c.
I showed that this indeed gives a solution, so Lorentz
gauge can be accessed. This is called the "retarded solution".
+ I drew a spacetime diagram illustrating the concept of the retarded
solution: f(r,t) is determined
by the values of s on the past
light cone of the spacetime point (r,t).
+ Similarly we can write the retarded solution for the potentials.
+ We must check that the retarded solution for the potentials satisfies
the Lorentz gauge condition, since we assumed that condition
in deriving the equation for the potentials. I showed that
this is true at all times if it is true at one time. I didn't
explicitly show it is
true at one time. I suggest you do problem 10.8 and show
it explicitly for yourself.
+ One could also write the advanced solutions, reversing the roles of
future and past, since the equations themselves have time reversal
symmetry. One could even take a linear combination of the
retarded and advanced solutions. It is the initial conditions that tell
which solution to use. It seems that we inherit the vacuum
state from the past (in a cosmological sense), so that evidently it is
retarded solutions we should use.
Here are some detailed notes copied from what I wrote last year:
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
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 continuity 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.
Monday April 25
waveguide modes: A
cylindrical cavity of arbitrary cross section admits modes that have no
E_z (TE modes)
or no B_z (TM modes), but no TEM modes. I showed a great falstad applet of waveguide
modes, and we
looked at a few quantitites and modes. I suggest you play with it to
see if it all makes sense with respect
to the boundary conditions, charge, and current.
I highlighted the steps used in finding these modes (you can read all
about it in the textbook):
+ Maxwell's eqns determin E_x,y and B_x,y from E_z and B_z.
+ Assume B_z = B_z0(x,y) exp(ikz - iwt). For a TE mode this will
+ The wave equation implies that B_z0 satisfies Poisson's equation. Use
separation of variables to
solve this for B_z0 of the form X(x)Y(y). The boundary
conditions for TE modes imply X and Y
are cosine functions with wavevectors mπ/a and nπ/b. The wave
equation implies then that
the dispersion relation
takes the form w2 = c2k2
wmn = c
[(mπ/a)2 + (nπ/b)2]1/2
+ I explained, with reference to a graph, how the phase
velocity of these modes is > c for all k,
but the group velocity is < c for all k.
potentials for electrodynamics:
Previously the scalar and vector potential were introduced for
fields. Now we extend this to electrodynamic fields. This is worth
doing because it is often more convenient
to use the potentials to find the fields. Also, in quantum mechanics,
in the Schrodinger picture, and in quantum field
theory, charges couple directly to the potentials, not to the fields.
(In quantum mechanics it is possible to use the
fields, in the Heisenberg picture.) I covered this so far pretty
much as in the book.
Thursday April 21 Exam
Tuesday April 19
Monday April 18
waveguide modes - A
cylindrical cavity of arbitrary cross section doesn't admit purely
transverse waves, whereas
a coaxial cable does. The dispersion relation for the lowest mode coax
waves (the one with no angular dependence)
is w2 = c2 k2, i.e. the same as in
vacuum. See textbook for details. I think modes with angular dependence
and have different dispersion relations.
Thursday April 14
- explained the evanescent wave that runs along the boundary
when there is total internal reflection
- finished waves in conductors
- discussed the role of skin depth in current carrying wires: the
induced back emf pushes the current to the surface
of the wire. Clint's remark prompted me to admit that the argument
isn't convincing unless, as in the Thomson
coil, there is a an extra phase lag. I asked Clint to work this out for
us and explain it!
- dispersion: origin of the frequency dependence of the index of
refraction. I covered the essence of what Griffiths has in 9.4.
Tuesday April 12
- reviewed 9.2.2 (reflection and transmission at normal
- covered 9.3.3: oblique incidence (didn't do the jump condns in
detail, just quoted the result
- discussed the point that when µ1=µ2, Brewster's angle
occurs when reflected and refracted wavevectors are perpendicular,
and mentioned that this means those waves have orthogonal
- a student asked why isn't the amplitude of
the transmitted wave maximum when there is no reflected wave, i.e. at
I mumbled something about the direction of the energy flux. Here's the
clear answer: when there is no reflected
wave, the the energy flux
perpendicular to the interface, in the z-direction, should be equal for
the incident and transmitted waves. As the angle of incidence grows,
at fixed incoming amplitude, the incident energy flux in the
z-direction drops, so the transmitted flux in the z-direction need not
maximum. In fact, we can check explicitly that the z-flux of
transmitted wave is equal to that of the incident wave at Bresters
incoming z-flux: eps_1 v_1 cos(theta_I) E_I^2
transmitted z-flux: eps_2 v_2 cos(theta_T)
At Brewster's angle we have cos(theta_T)/cos(theta_I) =
µ1v1/µ2v2, and E_T/E_I = (µ2 v2)/(µ1v1), which
together with eps µ v^2 = 1
shows the the incoming and transmitted z-fluxes are equal.
- discussed how RealD 3d glasses work
(quarter wave plate on top of a horizontal polarizer)
- section 9.4.1: em waves in conductors, up through eqn (9.124).
- a student asked how can it be that the charge density in a conductor
always decays exponentially at each point? If there
is an initial charge concentration somewhere, and zero charge
elsewhere, won't the charge spread out to the surface and temproarily
become nonzero at other points inside the conductor, as the charge
flows past? I think the answer is that charge can flow past without
the charge density being nonzero. For example, in a current carrying
wire, of course, the wire can be neutral.
Monday April 11
OPTICAL BOARD - BREWSTER’S ANGLE
- Maxwell's equations in linear, isotropic medium
- jump conditions at a plane interface between two media
- reflection and transmission at normal incidence
Thursday April 7
Ruhmkorff coil works.
- how Hertz generated and detected electromagnetic waves: he useda n
coil and a resonating circuit to generate the waves, and a loop, tuned
to be resonant,
with a tiny gap across which a spark would jump when an emf if
generated. He dummed the
lights and looked carefully for the spark...
- linear & circular polarization - a
applet demonstrates this. Horizontal plus vertical
combines to 45 degrees polarization when added in phase, and combines
to circular polarization when added
one quarter of a cycle out of phase. In between yields elliptical
polarization. To see this in the applet, set the
phase difference to 0 (45 degrees), 0.5 pi (clockwise circular), -0.5
pi (counter clockwise circular).
- Electromagnetic monochromatic plane waves derived from Maxwell's
- time averaged energy density, energy flux, intensity, momentum
density, and radiation pressure
Tuesday April 5
- Maxwell's equations in matter; the bound part of the displacement
- local energy conservation: energy density and Poynting vector
- field momentum and angular momentum
- decoupled, 2nd order 3d wave equations for the components of electric
and magnetic fields in vacuum
- nature of the wave equation in 1d, right moving and left moving
Monday April 4
- estimate of self-inductance of a ring & exact formula for
a ring of circular cross section
- skin effect: at very high freqency current is on the surface of a
conductor, which changes the self inductance. At
60 Hz, for our Thompson coil, this is not an issue. (More on this later
in the semester.)
- perfect conductor: force per uit charge must vanish, so the total emf
vanishes. How this happens in a perfect
conductor is that E + v x B
= 0, where the E field is due, I suppose, to both Coulomb
repulsion of charges and induced
electric field from changing current. Anyway, you don't have to think
about that, as they key thing for the problem
in the homework is that to total emf vanishes, which means that the
back emf cancels the external contribution to the emf.
This links the time derivative of the current to the velocity of the
- Maxwell's displacement current
term: First consider the consistency of Faraday's law for the induced E
curl E = - ∂tB
div curl E =
div B = 0 since there is
no magnetic charge, indeed div ∂tB = 0, so Faraday's law is
Now consider the analogous question for Ampere's law:
= µ0 J
It's not consistent, since div J is
div J + ∂trho
So we must fix Ampere's law by adding to J something whose divergence is ∂trho.
we see that div(ϵ0∂tE) = ∂trho.
ϵ0∂tE to J the sum will be divergence-free.
This is Maxwell's modification of
Ampere's law, and the extra term in the current is called the displacement current:
curl B = µ0 J + µ0ϵ0
If you recall that µ0ϵ0
= 1/c2, where c is the speed of light, you can see why this
term would not be noticed unless you
observe very rapidly changing electric fields. Note that now, if you
set the current to zero, rather than having zero magnetic
field, the changing electric field can support a changing magnetic
field, which by Faraday's law supports a changing
electric field, so one can have propagating solutions to Maxwell's
equations in vacuum, i.e. electromagnetic waves.
Thursday Mar 31
Lenz' law: Effects of induced
emf always oppose the change that produced them.
Lenz' law follows from the other laws, but it is a convenint way to
infer the direction of induced emf effects.
It must hold otherwise physics would be unstable: a small perturbation
would amplify itself. This would mean
that there is negative energy around; I think the magnetic field would
carry negtaive energy.
Applied to the K2-44 EDDY
CURRENT PENDULUM. As the pendulum is entering the poles of the
the induced currents make a magnetic dipole that is anti parallel to
the dipole it's entering, and side by side
opposite dipoles repel.
Applied to K2-62
ELECTROMAGNETIC: the rapid rise of current in the
induces an opposite current around the aluminum soda can. The opposite
currents repel, pinching and
tearing apart the can.
Applied to K2-61
COIL: The induced current in the ring lags the primary
coil by 90 degrees, so the
currents are parallel every other quarter cycle and anti-parallel every
other quarter cycle, so the time averaged
force would vanish, if were not for the self-inductance of the ring.
Because of the self inductance, the phase lag is
more that 90 degrees. See the homework.
Magnetic energy and inductance:
Showed that the work to put a current I in an inductor is W = 1/2 L
I^2. In fact, this
is really the more general definition of self-inductance, since the
loop associated with the current is not always well-defined.
This energy is really the energy stored in the magnetic field,
(1/(2µ_0))∫ B^2 dtau. The self inductance is analogous to
mass, with I analogous to velocity, and W kinetic energy. Put
differently, L is a kind of inertia that resists establishing
Tuesday Mar 29
motional emf: example of the pulled rectangular loop
motional emf: general loop in general field
emf and voltage
differential form of Faraday's law
& properties thereof (see last year's notes for some details)
work and emf: I said that any
power deliverd in a circuit as a result of motional emf must come from
done by the agent moving the loop. I talked about showing that in a
particular case of the rectangular pulled
loop in a constant magnetic field in the half-space. But Clint pointed
out that we can show it quite generally,
once and for all: the magnetic force on an element of current-carrying
loop is dF = I dl x B.
then something must supply an equal and opposite force, and this force
delivers power at the rate
dP = - dF.v = - I (dl x B).v = I (v x B).dl = I f.dl.
For example, if the power is dissipated in a resistor, ℇ = IR, then the Ohmic power loss
is indeed ℇI = I2
induced electric field: Example 7.9, using differential form of
Monday Mar 28
INDUCTION IN A SINGLE WIRE
Henry and Faraday in 1831 found that a changing magnetic field induces
Faraday found that moving a conductor in a constant magnetic field also
a current. To characterize this current, we use the concept of
"electromotive force", emf.
conductivity & resistivity
"intensive" version of Ohm's law
relation between resistance and resistivity
Thursday Mar 17
Tuesday Mar 15
Monday Mar 14
I think I showed how Laplace's equation is equivalent to the
statement that the
average over any sphere is equal to the value at the center. This means
function can't have a local minimum or maximum. You can also see this
just by integrating
the Lapacian of the function over a small sphere surrounding a point,
and using the
divergence theorem. I also showed that the square of the electric or
in vacuum can have a local minimum, but not a local maximum.
I'm not sure, but I may have also shown this day
how to solve for the Fourier transform
of a vector field from the Fourier transform of its divergence and curl.
Thursday Mar 10
Tuesday Mar 8
for links to the video of pouring liquid nitrogen and liquid oxygen
between the poles of a magnet,
and the video of levitating a frog and a strawberry, etc, see last
also those notes have good discussion of much of what we covered.
Monday Mar 7
J7-11 PARAMAGNETISM AND DIAMAGNETISM
Thursday Mar 3
- Math identities: curl grad f = 0, div curl W = 0.
- Math theorems:
if curl W = 0, then W = grad f for some f.
- Physics uses these theorems:
if div F = 0, then F = curl G for some G.
= 0, so E = -grad V for some
scalar potential V, defined up to addition of a constant (the minus
sign is conventional)
- Coulomb gauge: div A = 0. Even with this freedom fixed,
A remains ambiguous up
to the addition of grad f such that
div B = 0, so B = curl A for some vector potential A, defined up to addition of grad f,
div grad f = 0, i.e. Laplacian f = 0. There are many such f, but they
all blow up at spatial infinity.
- Why vector potential is useful: 1) Exploit gauge freedom to simplify
equations, e.g. with div A =
0, Ampere's law becomes
Lapacian A = -µ0J, which is just three Poisson
equations, with the components of A
decoupled. 2) In describing a
particle by a wave, the effect of a magnetic field is directly
manifested through A. 3) When
applying quantum mechanics to the
electromagnetic field itself, it is most straigthforward to use A.
- mulipole exapansion of
the vector potential;
Magnetic monopole moment vanishes (proved this), and dipole moment is
1/2 ∫ r x J dtau
(didn't prove this). For a planar current loop the dipole moment is the
current times the area times a
unit vector normal to the loop. For an arbitrary current loop, J dtau = I dl, so the magnetic moment
is 1/2 I ∫ r x dl
= I ∫ da
(see Problem 1.61d of hw1).
- example of the magnetic field of the earth, 90% dipole at the surface
(by some measure)
- finite sized current loop, infinitesimal current loop and pure
dipole, magnetic pure dipole field same form as electric dipole
field (showed that curl (m x rhat)/r2 = grad (m.rhat)/r2).
Tuesday Mar 1
- jump conditions on magnetic field
- vector potential
Monday Feb 28
- Ampere's law in differential and integral form
- consistency: curl B = µ0J : divergence of the left hand side
vanishes identically, while according to the continuity equation for
charge, divergence of the right hand side is equal to -∂t
rho. Hence the equation is consistent only with time independent charge
- Using Ampere's law to find the magnetic field in
situations with symmetry.
Thursday Feb 24
- Showed and explained J4-22 PARALLEL
PLATE CAPACITOR WITH DIELECTRIC
- Showed Oersted experiment J5-20 OERSTED
EXPERIMENT - LARGE COIL AND COMPASS
and Ampere experiment K1-01 FORCE
BETWEEN CURRENT-CARRYING WIRES
- Lorentz force: F = q v x B,
- motion in a magnetic field: straight lines, cyclotron, and spirals.
- motion in crossed electric and magnetic fields: cycloid orbit. I
explained the cycloid orbit
as the sum of a cyclotron with the uniform linear motion with velocity v = E
x B (see 2010 notes).
CYCLOID - LIGHT BULB ON WHEEL to illustrate what a cycloid is.
- current; current density; relation between current, charge density,
and charge velocity, for
line, surface, and volume currents.
- force on a current element: dF
= I dl x B
- Charge continuity equation: div J
= -∂t rho.
Tuesday Feb 22
- jump conditions for D
- linear, isotropic dielectrics
- example of dielectric sphere in an
external field: The polarization is determined by the total
field, which includes the contribution from the polarization. How to
escape this loop?
The way I suggested to do this in class was not so useful. I said that
we can find the surface charge density
by using the fact that it both determines the jump of the perpendicular
component of the electric field,
and is equal to the perpendicular component of the polarization, which
is determined by the field inside.
sigma_b/eps_0 = Eperp^out -
sigma_b = Pperp = chi_e eps_0 Eperp^in (2)
and we can eliminate Eperp^in, to express sigma_b in terms of
Eperp_out. But then it's still not a straight
shot to solve the problem. A more direct way would be to eliminate
sigma_b, which yields
Eperp^out = (1+chi_e)Eperp^in (3)
which can be used as a boundary condition for matching solutions to
Laplace's equation for the potential
inside and outside the sphere, as Griffiths shows in example (4.7).
Note that in that example, Griffiths obtains
this jump condition in a simpler way: there is no jump of Dperp, since
no free charge. But D = eps E, so there is no
jump of the product (eps Eperp), so eps_0 Eperp^out = eps
Eperp^in. This is the same as Eq. (3) above.
This is a cleaner way to think of it.
I also explained that another way to solve the problem is to try the
guess that the polarization inside is uniform
and see if it's consistent. If so, we can solve the problem using what
we already found for the field of a uniformly
polarized sphere: - P/3eps_0 inside, dipole
outside. In this case, the net field inside is E
- P/3eps_0, where
the first term is the applied field. But the polarization is determined
by the net field, P = chi_e
eps_0 E. Putting
these together we find that, inside, E
= E_0/(1+chi_e/3). (Check that
this is the same as (4.49).
- example of the point charge next
to a dielectric slab: This is example 4.8 in Griffiths. The
volume bound charge
density is always zero in a linear isotropic dielectric (see (4.39)).
So we just need the surface charge density. This case
is different from the sphere case above, since here it seems easier to
find an expression for the surface charge density in terms of
the field of the point charge. I think the reason this is relatively
easy is that the field of a planar surface charge is known
to be - sigma_b/2eps_0 zhat,
even if sigma_b is not constant (the Gaussian pillbox argument plus
plane reflection symmetry
this). The total z-component of the field inside is that from the point
charge q plus
that from the surface charge:
Eperp^in = Eperpq^in - sigma_b/2eps_0 (4)
Together with (2) this yields
sigma_b = eps_0 Eperpq^in (2chi_e)/(2+chi_e)
which is the same as (4.50) in Griffiths.
- Example of capacitor with
dielectric: I explained that since the dielectric screens the
electric field, you can have
the same charge on a capacitor with less potential difference. Since Q
= CV, this means that the capacitance is increased
by filling the space between the plates with a dielectric material. I
started to compute the effect of the dielectric
on the field and was incoherent and ran out of time. Here is a very
simple way to do it: Use the relation for linear dielectrics,
D = eps E. The only free charge is on the
capacitor places, and since we have planar symmetry and D is perpendicular
to the plane, we have curl D =
0. So the usual Gaussian
pillbox argument tells us that D is
uniform between the plates
and equal in magnitude to sigma_f. Hence E is equal in magnitude to
sigma_f/eps. Without the dielectric it would have
been sigma_f/eps_0, so E is decreased by a factor of the dielectric
constant eps_r, and the potential difference is decreased
by the same factor. Thus the capacitance is increased by the same
factor, C = eps_r C_0.
- Regarding capacitors, I did get a chance to properly explain this
previously, so I'll place here my notes on that from
last year (it's also explained by Griffiths):
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 21
- polarization density, bound charge, and free charge
- why rho_b = -div P implies
sigma_b = P_perp
- total bound charge = 0
- field of a uniformly polarized cylinder
- field of a uniformly polarized sphere (Example 4.2)
- the electric displacement vector D
= eps_0 E + P
- div D = rho_f, curl D = curl P
- thick spherical shell with polarization P = k
rhat/r (problem 4.15). Method (a): find the bound surface
charge on the inner and outer surfaces, and the bound volume
charge, and find the electric field they generate
inside the inner surface, inside the shell, and outside the out
surface. Method (b): note that by spherical symmetry,
D = D(r) rhat, and since there is no free
charge, div D = 0 everywhere.
Together these imply that D =
= eps_0 E + P = 0, so
E = -P/eps_0. In
particular, the electric field is zero except in the interior of the
(you should also be able to infer this directly from Gauss' law
and spherical symmetry), and inside it is opposite to P.
Thursday Feb. 17
- Rubbed a small balloon with fur and it stuck to the wood door
of the classroom.
Why? Although the door remains neutral, the negatively charged balloon
polarization, i.e. separation
of positive and negative charges, in the door. It pushes the
electrons back a bit, leaving unbalanced positive charge in front.
Since this positive charge is
closer to the balloon than the negative
charge that is pushed away there is a net attractive
force. Note that the wood is not a conductor, so the polarization
is not due to charges flowing
anywhere, rather only a small rearrangement of the charges on
- Torque and force on a dipole. I used the index
notation method to relate two ways
of writing the force on the dipole.
- Polarization P: dipole
moment density. Potential generated by a polarization P is equivalent
to that generated by a "bound" volume charge density rho_b = -div P and a "bound" surface charge
density sigma_b = P_perp, the outward normal component of P at the
Tuesday Feb. 15
- More on charge at distance a from origin: for r > a can
expand in a/r, i.e. in negative
powers of r. For r < a can expand in r/a, i.e. in positive powers of
r. These meet the boundary
conditions at infinity and at the origin, respectively.
- Generalized this to the potential of a charge at any location a. See (3.94). I got this by a
route than Griffiths: for the case of a charge on the z axis, we just
matched coefficients in the expansion
of 1/(r-a) to the coefficients in the general solution ∑ B_l r^-l-1
P_l(cos theta), putting theta=0. This
yields B_l = a^l. Then for a charge at location a, just replace theta by theta', the
angle between a and r.
- Used the previous result to write a formula for the potential of any localized charge distribtion, as
a sum over inverse powers of r. This is the multipole expansion. The
1/r term is the monopole term,
determiend by the total charge. The 1/r^2 term is the dipole term,
determined by the dipole moment.
I showed that the dipole moment is independent of the origin only if
the total charge vanishes.
- Looked at a few examples of dipole moments.
- Discussed the pure dipole as a limit of physical dipoles, pd, with d going to zero,
p going to infinity, with pd held fixed.
- Computed the electric field of a pure dipole, in two ways. (i) Using
as did Griffiths, on p. 153, eqn (3.103). (ii) Using index notation for
I've posted a separate pdf document about the index
method, as it's hard
to read with ascii
characters, and it's worth learning because the techniques generalize...
- A student asked if we could use three point charges to make a
zero total charge and zero dipole moment, but non-zero quadrupole
moment. The answer is yes...
I may assign a homework problem on this question!
Monday Feb. 14
- more on conducting sphere in external electric field
- potential of point charge away from the origin, expanded in powers of
r and Legendre polynomials.
How to choose the separation
I've had several questions about this, so let me spell this out.
When you apply the boundary conditions, you find conditions on the
You can solve these for the constants. But in the examples we've seen,
it's possible to anticipate
some or all of the solution, rather than writing out all the conditions
First, don't forget that Poisson's equation implies that the constants
must add to zero,
C_x + C_y + C_z = 0. So, for example, if you know that C_x and C_y are
then you know C_z must be positive.
Now consider an example where the boundary condition says that V is
zero at both x=0 and x=a.
Could C_x be positive, say C_x = k^2? Well, can A sinh(kx) + B cosh(kx)
= 0 at two points?
The answer is no: if you think about the shape of the functions sinh
and cosh you can see that this
sum can cross zero at most at one value of x. for instance, suppose A
and B are positive. Then
this function is positive for x > 0, and if A is greater than B it
will pass through zero somewhere
on the negative x axis, but only once. So C_x must be negative, say
-k^2, and V(x) = A sin(kx) + B cos(kx).
Now since V(0)=0, it follows that B=0. And since V(a)=0, it follows
that k = n pi/a, for some integer n.
Alternatively, you can just turn the algebra crank: If V(x) = A
sinh(kx) + B cosh(kx), then
0 = A sinh(0) + B cosh(0) = B and 0 = A sinh(ka) + B sinh(ka),
which since B=0 implies also A=0.
Could this function be zero at some other two places neither of them
being 0, say at x=c and x=d?
It takes a bit more algebra, but you can show this. A slick, indirect
way to show it is to notice that
the second derivative is k^2 times the function itself, so has the same
sign as the function. But
if the function is zero at two places with a maximum in between, it
must be positive at the maximum
with a negative second derivative. And if it has a minimum in between,
then it must be negative there
with a positive second derivative. In neither case do the
function and its second derivative have the same
sign. So the function can't be zero at two places.
Thursday Feb. 10
FARADAY CAGE - ELECTROSCOPE
- finished example 3.4
- separation of variables using spherical coordinates, Legendre
- example of conducting sphere in external electric field
Tuesday Feb. 8
- Force on a charge next to a
conducting plane: the force is the one produced by the
part of the field taht comes from the image charge, so goes like
z is the distance from the charge to the plane. This blows up as z goes
and fast enough so that the work done by this force is infinite in
charge into the plane. In effect, there is a potential -kq/(4z), that's
at z = 0. The infinity is unphysical, it's cut off at some scale,
determined by a combination
of the atomic structure of the conductor and the quantum spread of the
wavefunction of the
charge. Together these determine the work function, i.e. the energy it
takes to remove an
electron from the conductor.
[For those who know quantum mechanics, the wave function for an
electron in this potential
satisfies a Schrodinger equation that is identical to the equation for
u(r), where Psi(r) = u(r)/r
is a spherically symmetric (zero angular momentum) state in a Coulomb
momentum transverse to the conducting plane is conserved, and just
produces a shift in the energy eigenvalue.]
- Visualizing the field of charges
and conductors: here's the link for the applet I showed today:
If you look under the "Full Directions" link, he specifies that the
default setting for the
conducting sphere is that it is grounded, i.e. at zero potential. Then
a net charge is induced
on it, and indeed the electric field is pointing outward everywhere.
As pointed out in class by a student, if the net charge is zero on the
sphere then the field
must be pointing outward at some places and inward at others.
It would be nice if the potential slider on the applet showed the total
charge on the sphere,
so we could see when the sphere is neutral (zero total charge).
I was curious to know at what angle, measured relative to the direction
from the center of the
sphere to the actual charge at a, the field changes sign at the surface
of the sphere, in the
neutral case. The answer I found is that this happens at the angle
cos(theta) = [1 + a2(1 - (1 - a-2)2/3]/(2a)
and where I've used length units with R=1. That is, a is actually a/R.
When a=1 (i.e. a=R) this
yields cos(theta)=1, so theta=0. That is, we get an image charge -q at
theta=0, and everywhere
else on the sphere we have a negative charge density. When a goes to
infinity, we get cos(theta)=0,
so theta = pi/2, which makes sense: the charge q induces a symmetric
dipole on a distant sphere.
- Charge distribution on a conducting
disk or needle: This came up so I mentioned a nice two page
this: R. H. Good, American Journal of
Physics 65, 155 (1997),
that the charge
distribution that produces a zero field on the disk is obtained by
projecting a uniform
on a spherical shell onto the equatorial plane. This yields a
non-uniform charge density on
divergent at the edge. Similarly the distribution on a needle is
obtained by projecting the
onto an axis, and this yields a uniform
density.... which seems paradoxical, since it seems as if the
any point except the midpoint could not vanish. The explanation, given
in the paper, is that
the field due to the nearby charge is infinite, so the imbalance makes
a negligible contribution.
Also, one can carve up the needle into contributions that do balance!
See the paper for details.
- Separation of variables: The
idea is that Laplace's equation is linear, so the solutions form a
space. Various special bases for this space can be found, each suitable
for a different symmetry
situation. The basis vectors are solutions formed from a product of
functions of a single variabe.
I started with the Cartesian case, and explained it more or less as in
the text, and applied it to the
case in Example 3.4 Be sure to read all the text and examples.
Monday Feb. 7
- finished discussing the uniqueness theorem for solutions to Poisson's
- the method of images
Thursday Feb. 3
- electrostatic energy of a charge distribution
- uniqueness theorem for solutions to Poisson's equation. I covered
this a little bit differently than does
Griffths. See last year's Feb. 2 notes for a concise summary of the
reasoning I gave. Today I didn't get to
the third reason the boundary term might vanish (conductors with fixed
total charges). I'll start with that
Tuesday Feb. 1
- electrostatic potential
- jump conditions at a surface charge
- electrostatic potential and potential energy per unit charge
Monday Jan. 31
- integral and differential forms of Gauss' law
- 3d Dirac delta function and Coulomb field of point charge
- applications of Gauss' law
Thusday Jan. 27 snow
Tuesday Jan. 25
- Coulomb's law
- electric field
- Gauss' law
Monday Jan. 24
- review of vector claculus.
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