Meaning of the Maxwell Tensor

We define the Maxwell Stress Tensor as

 1  1  1  T = ε E E − δ E E + B B − δ B B ij 0  i j ij k k  µ  i j ij k k  2 0 2

While it looks scary, the Maxwell tensor is just a different way of writing the fact that electric fields exert forces on charges, and that magnetic fields exert forces on cur- rents. Dimensionally,

Momentum T = (Area)(Time)

This is the dimensionality of a pressure:

Force = dP dt = Momentum Area Area (Area)(Time) Meaning of the Maxwell Tensor (2)

If we have a uniform electric field in the x direction, what are the components of the tensor?

E = (E,0,0) ; B = (0,0,0)

    2   1 0 0 1 0 0 ε 1 0 0 2    1    1 0 E   = ε  0 0 0  −  0 1 0  + ( ) =  0 −1 0  Tij 0 E 0       µ  −  0 0 0  2  0 0 1 0 2  0 0 1

This says that there is negative x momentum flowing in the x direction, and positive y momentum flowing in the y direction, and positive z momentum flowing in the z direction.

But it’s a constant, with no divergence, so if we integrate the flux of any component out of (not through!) a closed volume, it’s zero Meaning of the Maxwell Tensor (3)

But what if the Maxwell tensor is not uniform? Consider a parallel-plate capacitor, and do the surface integral with part of the surface between the plates, and part outside of the plates

Now the x-flux integral will be finite on one end face, and zero on the other. So x-momentum is flowing though the ε 2 closed surface, and the pressure is 0 Ex 2.

The force per unit area is the charge per unit area times the electric field at the charges. With this field, the = ε charge per unit area on the plates is Q A 0 Ex .

ε 2 If we used Ex for that field, we would get 0 Ex , which is twice what we calculated from the Maxwell stress tensor, which is wrong.

I wasted an hour last night worrying about this, until I re- membered the solution (Griffiths 3d section 2.5.3). Meaning of the Maxwell Tensor (4)

Charges reside at the surface of a conductor. So the electric field is finite on the exterior side, and zero on the interior side. The proper electric field to use turns out to be the average of the full field and the zero field.

One way to see this is to make the charge layer a finite thickness, so the electric field ramps up linearly from zero to the full value, and compute the integrated force on the distributed charge.

Another way to see it is to consider a uniform charge sheet not in a conductor, which makes equal and op- ε ( ) posite electric fields of 0Q 2A on opposite sides, but with a superposed uniform electric field of the same strength, which doubles the field on one side and cancels it on the other. If we isolate a patch of the charge, the charge sheet exerts no net force on the patch by symme- try, and all the force is from the uniform external field.

With either explanation, the actual net force on the ca- ε 2 pacitor plate is 0 Ex 2, in agreement with the Maxwell stress tensor calculation.

The moral is, sometimes the equations are smarter than me! Meaning of Field Momentum Density

Consider two positive point charges moving at equal speeds along lines that cross at right angles, but timed so they don’t actually collide:

V V B F

At the instant show, the horizontally moving particle makes a magnetic field that is into the screen, and the V × B force on the vertically moving particle is then hori- zontal and backwards.

But the vertically moving particle cannot possibly make an equal and opposite backwards force on the other particle, either electrically or magnetically! So we have a violation of Newton’s Third Law, i.e., momentum is not conserved!

It will get even worse later, when we will find that the electric forces between particles in motion aren’t instan- taneously balanced!

The solution to these paradoxes is that the fields between the particles must also carry momentum, and only the sum of the particle momentum and field momentum (and net Maxwell stress flux) is conserved. Angular Momentum in Statics

If a wire with charge Q per length is parallel to the axis of a solenoid with magnetic field B, and we move it radi- ally from the edge to the center with velocity v, there is a Lorentz force per unit length given by

Q Q QB F = v × B = (−rˆ v) × (zˆ B) = vφˆ

To keep the path radial, we have to apply a torque. Rela- tive to the center of the solenoid, the torque is

QBr QBr τ = r × F = v(rˆ × φˆ ) = vzˆ

A torque causes a change in angular momentum

dL dL dr dL τ = = = v dt dr dt dr

Equating these we have

QBr dL τ = vzˆ = v dr which we can solve for

QBr dL = zˆ dr Angular Momentum in Statics (2)

The integral of the angular momentum change from radius r = R to r = 0 is

0 0  2  0 = QBr = QB = QB r L zˆ ∫ rdr zˆ ∫ rdr zˆ   R R 2 R QB  R2  QBR2 = zˆ  0 −  = − zˆ  2  2

When we are done moving the charge to the center of the solenoid, we have applied a torque sufficient to trans- mit this much angular momentum to something.

Where does the angular momentum go? Angular Momentum in Statics (3)

The charged wire has a radial electric field, so E × B is non-zero. The linear momentum density stored in the fields is p = ε E × B volume 0

The angular momentum density then should be L r × S = = ε r × (E × B ) volume c2 0

The electric field around a line charge by Gauss’ Law is Q 1 E = rˆ πε 2 0r

When the charge is at the center of the solenoid, the angular momentum density is L  Q 1  QB = ε r × rˆ × zˆ B = −zˆ 0  πε  π volume 2 0 r 2

The total field angular momentum per length is thus

L QB QBR2 = −zˆ πR2 = −zˆ 2π 2

So the the integrated mechanical torque from the Lor- entz force on the charged wire agrees exactly with the angular momentum density stored in the E × B fields. Angular Momentum in Dynamics

Consider two cylinders of radius a and b with charges +Q and −Q, concentric with each other and with a solenoid of with magnetic field B and radius R, a < R < b.

If we turn off the solenoid in time T , we generate electric fields, inside and outside, in the cirumferential direction. By the integral form of Faraday’s Law

−B Br 2πrE = −πr 2 ⇒ E = φˆ inside T inside 2T −B BR2 2πrE = −πR2 ⇒ E = φˆ outside T outside 2rT

These fields generate torques on the charged cylinders for time T and impart angular momenta

L = (r × QE )T Ba QBa2 L = aQ Tzˆ = zˆ inside 2T 2 BR2 QBR2 L = b(−Q) Tzˆ = − zˆ outside 2bT 2 QB L = (a 2 − R2 )zˆ total 2

Where did the angular momentum come from? Angular Momentum in Dynamics (2)

If the two cylinders have length , the electric field between them is

Q rˆ Q 2πr E = ⇒ E = ε πε 0 2 0 r

The angular momentum density is

L  rˆ Q  −1 QB = r × (ε E × B ) = r ×  ε × B = zˆ 0  0 πε  π volume 2 0 r 2

If we multiply by volume we have

 −1 QB  QB L =  zˆ  [π (R2 − a2 )] = (a2 − R2 )zˆ  2π  2

So again, the angular momentum in the fields equals the mechanical angular momentum.

There’s something fishy, though.

The electric field doesn’t actually depend on the outer cylinder radius. If we have no outer cylinder, we still get the same electric field, and the same angular momentum stored in the field.

But angular momentum doesn’t balance if we only have the inner cylinder. Angular Momentum in Dynamics (3)

When we turn off the solenoid we generate a circumferential electric field outside, which is what exerts the torque on the outer cylinder.

That electric field is itself time-varying, so it will generate its own magnetic fields, and propagate as a wave. The wave carries angular momentum to the outer cylinder.

If the outer cylinder is 10 light years away, for 10 years the angular momentum will be in the wave. It will eventually be absorbed by the outer cylinder. But it could just go on forever...

But what about the case of no inner cylinder, just the outer one. Now there is zero electric field inside the solenoid, so there is no E × B momentum stored, so no angular momentum stored. But when we turn the solenoid off, there is still an external electric field that makes the outer cylinder start to rotate.

Where did the angular momentum come from this time? Angular Momentum in Dynamics (4)

Well, instead of starting with the solenoid on and turning it off, start with it off and turn it on. And instead of mak- ing it a coil of wire, make it another charged cylinder. Rotating the cylinder generates a magnetic field just like a current does.

When we start rotating the cylinder, the dB dt generates an electric field that opposes the motion. We have to do work against this force, and this is responsible for the energy stored in the magnetic field.

But is also corresponds to a torque on the inner cylinder, so we must be putting angular momentum someplace. But where is it?

There is E × B momentum inside the cylinder, but only while we are changing the rotation speed. If we stop changing the speed, the electric field goes away, so it can’t be stored there.

I think the angular momentum must be radiated away in the external circumferential electric field wave. When we stop the cylinder, we radiate away angular momentum with the opposite sign.

These two waves also should carry some energy, so there must be some additional “radiation inductance” asso- ciated with turning on and off the solenoid, not account- ed for in the B2 magnetostatic energy. Where Are We?

We reviewed electrostatics (chapters 1-4)

We reviewed magnetostatics (chapters 5-6)

We reviewed Faraday’s Law of induction (chapter 7) consistent with Lorentz force law + simple relativity also works for time-dependent B with no motion

We reviewed Maxwell’s displacement term (ch. 7) required for consistency with charge conservation

Energy and momentum of fields (chapter 8) = ε 2 + 2 µ Energy density is U 0E B 0 = × µ Energy flux is S E B 0 = ε × = ε µ = 2 Momentum density is P 0 E B 0 0S S c Momentum flux is   1  1  1   −T = − ε  E E − δ E2  +  B B − δ B2   0  i j ij  µ  i j ij   2 0 2 Momentum flux is closely related to pressure

Angular momentum (chapter 8) Also stored in static and dynamic fields Lots of neat paradoxes if you’re not careful.

Second Edition chapter 7.5 covers 3d edition chapter 8 Second Edition chapter 7.4 starts 3d edition chapter 10, which we’ll do later. Fields and Waves

A field is defined over a continuous region of space •it could be a single real function (or even complex) •it could be a vector (N numbers, rotatable) •it could be a function of 1, 2, 3, N coordinates •it could be a function of time as well

Many fields obey equations that have travelling-wave solutions. Examples:

Transverse waves on a string •Two transverse functions of 1 coordinate + time

Waves on a water surface •1 height function of 2 coordinates + time

Sound waves in air •1 pressure function of 3 coordinates + time

Electrostatics •3 functions, each of 3 coordinates, no time dep.

Electromagnetic waves •6 functions (E,B), each of 3 coordinates + time

Schroedinger equation •1 complex function of 3 coordinates + time Field Equations vs Oscillator Equations vs Wave Equations

A field equation is describes some constraint on on the space or time dependence of the field, usually in the form of a partial differential equation.

The differential equation for a mass on a spring isn’t a field equation (time is the only independent variable). If we had many masses on springs, with coupling between them, we could describe the system by a field equation.

Wave equation is a rather loose term that implies that there are solutions that “travel,” particularly if they can get far away from their source.

A time-dependent field equation does not necessarily have wavelike solutions. The diffusion equation

∂f (x,t) ∂2 f (x,t) = −D ∂t ∂x 2 has solutions that spread out with time, but don’t travel far from the source. It would not be considered a wave equation. Field Equations, etc. (2)

Some wave equations allow nearly arbitrary functions of the coordinates to be solutions, which travel with no dis- tortion. There is nothing intrinsically sinusoidal or har- monic about the solutions to these equations. Maxwell’s equations in vacuum are in this class.

Many equations in physics have travelling wave solutions, but the wave gets more or less distorted as it travels •Attenuation (amplitude change) •Dispersion (shape change)

Maxwell’s equations outside of vacuum can have dispersion and attenuation, and we’ll do some cases.

Schrodinger’s equation for a free particle

∂Ψ   2 ∂2 Ψ i =   ∂t  i  ∂x 2 has travelling wave solutions with wave velocities that depend on frequency or wavelength (dispersion), but no attenuation.

While it is often useful to consider sinusoidal time dependence when working with field equations, it isn’t manditory in the same way that it is with oscillator equations. Inventing a Wave Equation

We can view a wave equation as a prescription for the change with time of the value of a function at a point x, to be calculated from the values of the function near x.

If an arbitrary function is moving past with velocity v, how is the time-derivative related to the function?

v dt

f(x,t) f(x,t+dt) ∂f vdt ∂x

motion

∂f f (x,t + dt) − f (x,t) = −vdt ∂x ∂f ∂f = −v ∂t ∂x

This has solution f (x,t) = F(x − vt) for arbitrary F(x).

Note that the wave can only go to the right for this equation, not the left. Inventing a Wave Equation (2) ∂ ∂ If we view this as an operator equation = −v ∂t ∂x and square both sides before applying it to f , we get

∂ ∂ ∂  ∂  ∂2 f ∂2 f f = −v  −v  f ⇒ = v2 ∂t ∂t ∂x  ∂x ∂t 2 ∂x 2

This has solutions f (x,t) = F(x − vt)+ G(x + vt).

In the first-order case, if someone tells us the function at all points (so we can calculate its derivative), we can pre- dict it’s value at any future time.

In the second order case, the f value at all x points only tells us the second time derivative. We also need to be told df dt at all points. In other words, we need not just one arbitrary function of position as the initial condition, but a second arbitrary function of position df dt t =0 .

Not surprising, since we need to specify both F and G... Inventing a 3-D Wave Equation

What function describes a travelling wave in 3 dimen- sions?

( ) = − − − = ( − ) f x, y,z,t f (x vx t , y vy t , z vzt) f x v t

What is it’s time derivative?

∂f ∂f ∂f ∂f = −v − v − v = −(v ⋅ ∇ ) f ∂t x ∂x y ∂y z ∂z

So we could write down a wave equation

∂f = −(v ⋅∇ ) f ∂t but it has some problems. We can have an arbitrary function f , but the velocity v in a fixed direction is built into the equation.

Can we make an equation that only has the velocity mag- nitude built in, the same in any direction? Inventing a 3-D Wave Equation (2)

The trick of squaring the operators that we used in 1-D ∂2 creates terms like v v , not just rotationally invari- x y ∂x∂y ant terms like v ⋅v .

The time derivative of the above function depends on the components of the velocity, and we need a time derivative that is does not depend on the velocity direction. But the function still needs to depend on x − v t to travel properly.

Let’s try f (x,y,z,t) = f (vˆ ⋅(x − v t)) = f (vˆ ⋅x − v t).

The time derivatives are

∂f ∂f = −(vˆ ⋅v ) ∂t ∂(arg) ∂2 f ∂2 f = v2 ∂t 2 ∂(arg)2

The space derivatives are ∂ ∂ ∂ ∂ ∇ = + f + f + f = f f vˆ x vˆ y vˆ z vˆ ∂(arg)2 ∂(arg)2 ∂(arg) ∂(arg) ∂2 f ∂2 f ∇2 f = vˆ 2 = ∂(arg)2 ∂(arg)2 Inventing a 3-D Wave Equation (3)

So the function

f (x,y,z,t) = f (vˆ ⋅(x − v t)) = f (vˆ ⋅x − v t) satisfies an isotropic wave equation

∂2 f = v2 ∇2 f ∂t 2 or equivalently

1 ∂ 2 f ∇2 f = v 2 ∂t2

Note that f can be an arbitrary function of its argument, and travel in an arbitrary direction, but it is constant in planes normal to the velocity vector.

That’s why we call these solutions to this equation plane waves...

The equation is linear, so the sum of two solutions is also a solution. 3-D Spherical Waves

∂2 f The wave equation = v2 ∇2 f also has solutions like ∂t 2 g(r − vt) f (r,t) = r

We can write the Laplacian in spherical coordinates

1 ∂  ∂  1 ∂  ∂  1 ∂2 ∇2 =  r2  +  sinθ  + r2 ∂r  ∂r r 2 sinθ ∂θ  ∂θ r2 sin 2 θ ∂φ2

Let’s work out the Laplacian of the claimed solution. There is no φ or θ dependence, only r dependence. The term in parenthesis is

 ∂f   r ∂g ∂arg − g  ∂g  r 2 = r 2 =  r − g  ∂r   r 2   ∂arg 

Now we take the r-derivative of this

1 ∂  ∂g  1  ∂g ∂2 g ∂g   r − g =  + r −  r 2 ∂r  ∂arg  r 2  ∂arg ∂arg2 ∂arg 1 ∂2 g 1 ∂2 g = r = r 2 ∂arg 2 r ∂arg2 3-D Spherical Waves (2)

g(r − vt) The time derivatives of f (r,t) = are r

∂f ∂g = −v ∂t ∂arg ∂2 f ∂2 g = v2 ∂t 2 ∂arg 2

Thus we have

1 ∂2 g 1 ∂2 g 1 ∂2 f ∇2 f = = v 2 = r ∂arg2 v2 ∂arg2 v 2 ∂t 2

This is a spherical wave, but in a single scalar field, not a vector field like Maxwell’s Equations. There are spherical solutions to Maxwell too, but they involve multiple components of E and B and are somewhat complicated.

Note that the solution falls off as 1 r , not 1 r 2 . The integral of such a field over the surface of a sphere will grow with distance, not stay constant or fall off. But things like energy go like the square of the field, so the surface integral will be constant independent of distance.