## PHY450H1S. Relativistic Electrodynamics Lecture 15 (Taught by Prof. Erich Poppitz). Fourier solution of Maxwell’s vacuum wave equation in the Coulomb gauge.

Posted by peeterjoot on March 2, 2011

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# Reading.

Covering chapter 6 material from the text [1].

Covering lecture notes pp. 115-127: reminder on wave equations (115); 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]

# Review of wave equation results obtained.

Maxwell’s equations in vacuum lead to Coulomb gauge and the Lorentz gauge.

\paragraph{Coulomb gauge}

\paragraph{Lorentz gauge}

Note that is invariant under gauge transformations

where

So if one uses the Lorentz gauge, this has to be fixed.

However, in both cases we have

where

is the wave operator.

Consider

where we are looking for a solution that is independent of . Recall that the general solution for this equation has the form

PICTURE: superposition of two waves with moving along the x-axis in the positive direction, and in the negative x direction.

It is notable that the text derives 2.11 in a particularly slick way. It’s still black magic, since one has to know the solution to find it, but very very cool.

# Review of Fourier methods.

It is often convienent to impose periodic boundary conditions

## In one dimension

When is real we also have

which implies

We introduce a wave number

allowing a slightly simpler expression of the Fourier decomposition

The inverse transform is obtained by integration over some length interval

\paragraph{Verify:}

We should be able to recover the Fourier coefficient by utilizing the above

where we use the easily verifyable fact that

latex m \ne n$} \\ 1 & \quad \mbox{if } \\ \end{array}.\end{aligned} \hspace{\stretch{1}}(3.20)$

It is conventional to absorb for

To take notice

when changes by , changes by

Using this

With , and

\paragraph{Verify:}

A loose verification of the inversion relationship (the most important bit) is possible by substitution

Now we employ the old physics ploy where we identify

With that we see that we recover the function above as desired.

## In three dimensions

## Application to the wave equation

Now operate with

Since this is true for all we have

For every value of momentum we have a harmonic osciallator!

Fourier modes of EM potential in vacuum obey

Because we are operating in the Coulomb gauge we must also have zero divergence. Let’s see how that translates to our Fourier representation

implies

The chain rule for the divergence in this case takes the form

But since our vector function is not a function of spatial coordinates we have

This has two immediate consequences. The first is that our momentum space potential is perpendicular to the wave number vector at all points in momentum space, and the second gives us a conjugate relation (substitute after taking conjugates for that one)

Since out system is essentially a harmonic oscillator at each point in momentum space

our general solution is of the form

Define

so that

Our solution now takes the form

\paragraph{Claim:}

This is now manifestly real. To see this, consider the first term with , noting that with

Dropping primes this is the conjugate of the second term.

\paragraph{Claim:}

We have .

Since we have , 3.40 implies that we have . With each of these vector integration constants being perpendicular to at that point in momentum space, so must be the linear combination of these constants .

# References

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

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