Phys411 Notes, Demos and Supplements

Phys411 - Electricity and Magnetism
University of Maryland, College Park
Spring 2011, Professor: Ted Jacobson

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.

Index notation

Article about stessing materials by magnetic domain alignment:
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 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 like this:

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:
runaway and pre-acceleration 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 = a/wavelength.

+ 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 relevant
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 signal.)

+ 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 circumstances,
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 of these...

Tuesday May 3

+ At what distance do the radiation fields become much larger than the near-zone fields? The electric dipole radiation 1/r potential
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 radiated.

+ 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 of hydrogen.

+ 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' yields

   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', t0) r'.rhat/c + ...

The integral of this over all space is

   pdot + mdot x rhat/c + 1/2 Qdot.rhat/c + ...

where Q is the tensor defined above [in class I omitted this term]. The first term comes 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
are tricky 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 radiation,

E_(elec dip rad) = -(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)

+ applet showing radiation from accelerating charge

+ K8-42 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 us
   which solution to use. It seems that we inherit the vacuum state from the past (in a cosmological sense), so that evidently it is the
   retarded solutions we should use.

Here are some detailed notes copied from what I wrote last year:

potentials in electrodynamics: writing 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 densities.
They can be simplified by choosing the Lorentz (Lorenz) gauge condition, div A + (1/c²)∂_tV = 0,
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.

If A' and V' don't satisfy the Lorentz gauge condition, then A' + grad f and V' - ∂_tf do if the
gauge transformation function f is chosen to satisfy the equation box f = s, where the source s
is s = - div A' + (1/c²)∂_tV'. So both the needed gauge transformation function, and the potentials
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 boundary conditions
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" boundary condition
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
point.

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- (1/c²)∂_t²)s(t - r/c) = 0).
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 satisfied.

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 box 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 determine everything.

+ 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+ w_mn², with w_mn = 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 electrostatic
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 2

Tuesday April 19 review

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 exist
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 incidence)

- 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 polarizations...

- a student asked why isn't the amplitude of the transmitted wave maximum when there is no reflected wave, i.e. at Brewster's angle.
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 have a
maximum. In fact, we can check explicitly that the z-flux of transmitted wave is equal to that of the incident wave at Bresters angle:

incoming z-flux: eps_1 v_1 cos(theta_I) E_I^2

transmitted z-flux: eps_2 v_2 cos(theta_T) E_T^2

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
M7-11 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

M9-03 CIRCULAR POLARIZATION - STICK MODEL

K8-01 ELECTROMAGNETIC WAVE - MODEL

K2-21 RUHMKORFF INDUCTION COIL

How the Ruhmkorff coil works.

- how Hertz generated and detected electromagnetic waves: he useda n induction
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 very nice applet demonstrates this. Horizontal plus vertical polarization
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 equations

- time averaged energy density, energy flux, intensity, momentum density, and radiation pressure

Tuesday April 5

K2-61 THOMSON’S COIL

- Maxwell's equations in matter; the bound part of the displacement current: ∂_tP

- 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 solutions

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 conducting rectangle.

- Maxwell's displacement current term: First consider the consistency of Faraday's law for the induced E field:

curl E = - ∂_tB

div curl E = 0 for any vector field, so div ∂_tB had better be zero. Since div commutes with ∂_t, and div B = 0 since there is
no magnetic charge, indeed div ∂_tB = 0, so Faraday's law is consistent.

Now consider the analogous question for Ampere's law:

curl B = µ₀J

It's not consistent, since div J is not zero when the charge density is changing: the charge continuity equation gives

div J + ∂_trho = 0

So we must fix Ampere's law by adding to J something whose divergence is ∂_trho. Using Coulomb's law, div E = rho/ϵ₀,
we see that div(ϵ₀∂_tE) = ∂_trho. So if we add 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 = µ₀J + µ₀ϵ₀∂_tE

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

K2-44 EDDY CURRENT PENDULUM

K2-61 THOMSON’S COIL

K2-62 CAN SMASHER - ELECTROMAGNETIC

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 magnet,
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 CAN SMASHER - ELECTROMAGNETIC: the rapid rise of current in the primary coil
induces an opposite current around the aluminum soda can. The opposite currents repel, pinching and
tearing apart the can.

Applied to K2-61 THOMSON’S 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
a current.

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 work
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. To pull the wire
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. Integrating this around the whole loop yields Iℇ.
For example, if the power is dissipated in a resistor, ℇ = IR, then the Ohmic power loss is indeed ℇI = I² R.

induced electric field: Example 7.9, using differential form of Faraday's law

Monday Mar 28

Electrodynamics

K2-02 INDUCTION IN A SINGLE WIRE

Henry and Faraday in 1831 found that a changing magnetic field induces a current.
Faraday found that moving a conductor in a constant magnetic field also induces
a current. To characterize this current, we use the concept of "electromotive force", emf.

emf

conductivity & resistivity

"intensive" version of Ohm's law

relation between resistance and resistivity

Faraday's law

SPRING BREAK

Thursday Mar 17

Exam 1

Tuesday Mar 15

review

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 a harmonic
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 magnetic field
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

J6-04 LOW POWER - HIGH FORCE ELECTROMAGNET

J7-23 MAGNETIC DOMAINS - MODEL

J7-01 LODESTONE

J7-13 CURIE POINT OF NICKEL

J7-14 CURIE POINT OF DYSPROSIUM

Ch. 6

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 year's notes.

also those notes have good discussion of much of what we covered.

Ch. 6

Monday Mar 7

J7-11 PARAMAGNETISM AND DIAMAGNETISM

Ch. 6

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.
if div F = 0, then F = curl G for some G.

- Physics uses these theorems:

curl E = 0, so E = -grad V for some scalar potential V, defined up to addition of a constant (the minus sign is conventional)
div B = 0, so B = curl A for some vector potential A, defined up to addition of grad f, any f.

- Coulomb gauge: div A = 0. Even with this freedom fixed, A remains ambiguous up to the addition of grad f such that
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 quantum mechanical
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 density.

- 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, redirects velocity, does no work.

- 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).
Used D1-11 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 electric
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.
That is,

sigma_b/eps_0 = Eperp^out - Eperp^in (1)

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 = E_0 - 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
tells us 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):

- Capacitors: given an arrangement of two conductors, carrying charge Q and - Q, the potential difference ∆V
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 infinity.

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 = 0 everywhere.
So D = eps_0 E + P = 0, so E = -P/eps_0. In particular, the electric field is zero except in the interior of the shell
(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 induces a
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 indicidual atoms.

- 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 surface.

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 different
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 spherical coordinates,
as did Griffiths, on p. 153, eqn (3.103). (ii) Using index notation for Cartesian coordinates.
I've posted a separate pdf document about the index notation 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 configuration with
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.

Supplement:
How to choose the separation constants:

I've had several questions about this, so let me spell this out.

When you apply the boundary conditions, you find conditions on the separation constants.
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 explicitly.

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 negative,
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

DEMOS:
J1-13 ELECTROSTATIC INDUCTION
J3-22 FARADAY CAGE - ELECTROSCOPE

- finished example 3.4

- separation of variables using spherical coordinates, Legendre polynomials

- 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 -kq/(2z)^2, where
z is the distance from the charge to the plane. This blows up as z goes to zero,
and fast enough so that the work done by this force is infinite in accelerating a
charge into the plane. In effect, there is a potential -kq/(4z), that's negatively infinite
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 potential. Any
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:
http://www.falstad.com/vector3de/index.html

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 theta, where

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
paper on this: R. H. Good, American Journal of Physics 65, 155 (1997), which shows that the charge
distribution that 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 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 vector
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 eqns.

- the method of images

Thursday Feb. 3

- electrostatic energy of a charge distribution

- conductors

- 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
in Monday.

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 day

Tuesday Jan. 25

- Coulomb's law

- superposition

- 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