## PHY450H1S. Relativistic Electrodynamics Lecture 14 (Taught by Simon Freedman). Wave equation in Coulomb and Lorentz gauges.

Posted by peeterjoot on February 17, 2011

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# Reading.

Covering chapter 4 material from the text [1].

Covering lecture notes pp.103-114: the wave equation in the relativistic Lorentz gauge (114-114) [Tuesday, Feb. 15; Wednesday, Feb.16]…

Covering lecture notes pp. 114-127: reminder on wave equations (114); reminder on Fourier series and integral (115-117); Fourier expansion of the EM potential in Coulomb gauge and equation of motion for the spatial Fourier components (118-119); the general solution of Maxwell’s equations in vacuum (120-121) [Tuesday, Mar. 1]; properties of monochromatic plane EM waves (122-124); energy and energy flux of the EM field and energy conservation from the equations of motion (125-127) [Wednesday, Mar. 2]

# Trying to understand “c”

Maxwell’s equations in a vacuum were

There’s a redundancy here since we can change and without changing the EOM

with

which gives

Maxwell’s equations are now

Can we make , while .

We need

How about

We want the divergence of to be zero, which means

So we want

Can we solve this?

Recall that in electrostatics we have

and

which meant that we had

This has the identical form (with , and ).

While we aren’t trying to actually solve this (just show that it can be solved). One way to look at this problem is that it is just a Laplace equation, and we could utilize a Green’s function solution if desired.

## On the Green’s function.

Recall that the Green’s function for the Laplacian was

with the property

Our LDE to solve by Green’s method is

We let this equation (after switching to primed coordinates) operate on the Green’s function

Assuming that the left action of the Green’s function on the test function is the same as the right action (i.e. and commute), we have for the LHS

Substitution of on the RHS then gives us the general solution

## Back to Maxwell’s equations in vacuum.

What are the Maxwell’s vacuum equations now?

With the second gauge substitution we have

but we can utilize

to reduce Maxwell’s equations (after dropping primes) to just

where

Note that for this to be correct we have to also explicitly include the gauge condition used. This particular gauge is called the \underline{Coulomb gauge}.

# Claim: EM waves propagate with speed and are transverse.

\paragraph{Note:} Is the Coulomb gauge Lorentz invariant?

\paragraph{No.} We can boost which will introduce a non-zero .

The gauge that is Lorentz Invariant is the “Lorentz gauge”. This one uses

Recall that Maxwell’s equations are

where

Writing out the equations in terms of potentials we have

So, if we pick the gauge condition , we are left with just

Can we choose such that ?

Our gauge condition is

Hit it with a derivative for

If we want , then we have

This is the physicist proof. Yes, it can be solved. To really solve this, we’d want to use Green’s functions. I seem to recall the Green’s function is a retarded time version of the Laplacian Green’s function, and we can figure that exact form out by switching to a Fourier frequency domain representation.

Anyways. Returning to Maxwell’s equations we have

where the first is Maxwell’s equation, and the second is our gauge condition.

Observe that the gauge condition is now a Lorentz scalar.

But the Lorentz transform matrices multiply out to identity, in the same way that they do for the transformation of a plain old four vector dot product .

# What happens with a Massive vector field?

## An aside on units

“Note that this action is expressed in dimensions where , making the action is unit-less (energy and time are inverse units of each other). The has units of (since ), so has units of , and then has units of mass. Therefore has units of and therefore you need something that has units of to make the action unit-less. When you don’t take , then you’ve got to worry about those factors, but I think you’ll see it works out fine.”

For what it’s worth, I can adjust the units of this action to those that we’ve used in class with,

## Back to the problem.

The variation of the field invariant is

Variation of the term gives us

so we have

The last integral vanishes on the boundary with the assumption that on that boundary.

Since this must be true for all variations, this leaves us with

The RHS can be expanded into wave equation and divergence parts

With for the wave equation operator

we can manipulate the EOM to pull out an factor

If we hit this with a derivative we get

Since is presumed to be non-zero here, this means that the Lorentz gauge is already chosen for us by the equations of motion.

# References

[1] L.D. Landau and E.M. Lifshitz. *The classical theory of fields*. Butterworth-Heinemann, 1980.

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