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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}

\begin{aligned}A^0 &= 0 \\ \boldsymbol{\nabla} \cdot \mathbf{A} &= 0 \\ \left( \frac{1}{{c^2}} \frac{\partial^2 {{}}}{\partial {{t}}^2} - \Delta \right) \mathbf{A} &= 0\end{aligned} \hspace{\stretch{1}}(2.1)

\paragraph{Lorentz gauge}

\begin{aligned}\partial_i A^i &= 0 \\ \left( \frac{1}{{c^2}} \frac{\partial^2 {{}}}{\partial {{t}}^2} - \Delta \right) A^i &= 0\end{aligned} \hspace{\stretch{1}}(2.4)

Note that \partial_i A^i = 0 is invariant under gauge transformations

\begin{aligned}A^i \rightarrow A^i + \partial^i \chi\end{aligned} \hspace{\stretch{1}}(2.6)

where

\begin{aligned}\partial_i \partial^i \chi = 0,\end{aligned} \hspace{\stretch{1}}(2.7)

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

However, in both cases we have

\begin{aligned}\left( \frac{1}{{c^2}} \frac{\partial^2 {{}}}{\partial {{t}}^2} - \Delta \right) f = 0\end{aligned} \hspace{\stretch{1}}(2.8)

where

\begin{aligned}\frac{1}{{c^2}} \frac{\partial^2 {{}}}{\partial {{t}}^2} - \Delta \end{aligned} \hspace{\stretch{1}}(2.9)

is the wave operator.

Consider

\begin{aligned}\Delta = \frac{\partial^2 {{}}}{\partial {{x}}^2}\end{aligned} \hspace{\stretch{1}}(2.10)

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

\begin{aligned}f(t, x) = F_1 \left(t - \frac{x}{c}\right)+F_2 \left(t + \frac{x}{c}\right)\end{aligned} \hspace{\stretch{1}}(2.11)

PICTURE: superposition of two waves with F_1 moving along the x-axis in the positive direction, and F_2 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

\begin{aligned}\mathbf{A}(\mathbf{x} + \mathbf{e}_i L) = \mathbf{A}(\mathbf{x}), i = 1,2,3\end{aligned} \hspace{\stretch{1}}(3.12)

In one dimension

\begin{aligned}f(x + L) = f(x)\end{aligned} \hspace{\stretch{1}}(3.13)

\begin{aligned}f(x) = \sum_{n=-\infty}^\infty e^{i \frac{2 \pi n}{L} x} \tilde{f}_n\end{aligned} \hspace{\stretch{1}}(3.14)

When f(x) is real we also have

\begin{aligned}f^{*}(x) = \sum_{n = -\infty}^\infty e^{-i \frac{2 \pi n}{L} x} (\tilde{f}_n)^{*}\end{aligned} \hspace{\stretch{1}}(3.15)

which implies

\begin{aligned}{\tilde{f}^{*}}_{n} = \tilde{f}_{-n}.\end{aligned} \hspace{\stretch{1}}(3.16)

We introduce a wave number

\begin{aligned}k_n = \frac{2 \pi n}{L},\end{aligned} \hspace{\stretch{1}}(3.17)

allowing a slightly simpler expression of the Fourier decomposition

\begin{aligned}f(x) = \sum_{n=-\infty}^\infty e^{i k_n x} \tilde{f}_{k_n}.\end{aligned} \hspace{\stretch{1}}(3.18)

The inverse transform is obtained by integration over some length L interval

\begin{aligned}\tilde{f}_{k_n} = \frac{1}{{L}} \int_{-L/2}^{L/2} dx e^{-i k_n x} f(x)\end{aligned} \hspace{\stretch{1}}(3.19)

\paragraph{Verify:}

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

\begin{aligned}\frac{1}{{L}} \int_{-L/2}^{L/2} dx e^{-i k_n x} \sum_{m=-\infty}^\infty e^{i k_m x} \tilde{f}_{k_m} \\ &= \sum_{m = -\infty}^\infty \tilde{f}_{k_m} \delta_{mn} = \tilde{f}_{k_n},\end{aligned}

where we use the easily verifyable fact that

\begin{aligned}\frac{1}{{L}} \int_{-L/2}^{L/2} dx e^{i (k_m - k_n) x} = \begin{array}{l l}0 & \quad \mbox{if latex m \ne n$} \\ 1 & \quad \mbox{if m = n} \\ \end{array}.\end{aligned} \hspace{\stretch{1}}(3.20)$

It is conventional to absorb \tilde{f}_{k_n} = \tilde{f}(k_n) for

\begin{aligned}f(x) &= \frac{1}{{L}} \sum_n \tilde{f}(k_n) e^{i k_n x} \\ \tilde{f}(k_n) &= \int_{-L/2}^{L/2} dx f(x) e^{-i k_n x}\end{aligned} \hspace{\stretch{1}}(3.21)

To take L \rightarrow \infty notice

\begin{aligned}k_n = \frac{2 \pi}{L} n\end{aligned} \hspace{\stretch{1}}(3.23)

when n changes by \Delta n = 1, k_n changes by \Delta k_n = \frac{2 \pi}{L} \Delta n

Using this

\begin{aligned}f(x) = \frac{1}{{2\pi}} \sum_n \left( \frac{2\pi}{L} \Delta n \right) \tilde{f}(k_n) e^{i k_n x}\end{aligned} \hspace{\stretch{1}}(3.24)

With L \rightarrow \infty, and \Delta k_n \rightarrow 0

\begin{aligned}f(x) &= \int_{-\infty}^\infty \frac{dk}{2\pi} \tilde{f}(k) e^{i k x} \\ \tilde{f}(k) &= \int_{-\infty}^\infty dx f(x) e^{-i k x}\end{aligned} \hspace{\stretch{1}}(3.25)

\paragraph{Verify:}

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

\begin{aligned}\int \frac{dk}{2\pi} e^{i k x} \tilde{f}(k) &= \iint \frac{dk}{2\pi} e^{i k x} dx' f(x') e^{-i k x'} \\ &= \int dx' f(x') \frac{1}{{2\pi}} \int dk e^{i k (x - x')}\end{aligned}

Now we employ the old physics ploy where we identify

\begin{aligned}\frac{1}{{2\pi}} \int dk e^{i k (x - x')} = \delta(x - x').\end{aligned} \hspace{\stretch{1}}(3.27)

With that we see that we recover the function f(x) above as desired.

In three dimensions

\begin{aligned}\mathbf{A}(\mathbf{x}, t) &= \int \frac{d^3 \mathbf{k}}{(2\pi)^3} \tilde{\mathbf{A}}(\mathbf{k}, t) e^{i \mathbf{k} \cdot \mathbf{x}} \\ \tilde{\mathbf{A}}(\mathbf{x}, t) &= \int d^3 \mathbf{x} \mathbf{A}(\mathbf{x}, t) e^{-i \mathbf{k} \cdot \mathbf{x}}\end{aligned} \hspace{\stretch{1}}(3.28)

Application to the wave equation

\begin{aligned}0 &= \left( \frac{1}{{c^2}} \frac{\partial^2 {{}}}{\partial {{t}}^2} - \Delta \right) \mathbf{A}(\mathbf{x}, t) \\ &=\left( \frac{1}{{c^2}} \frac{\partial^2 {{}}}{\partial {{t}}^2} - \Delta \right) \int \frac{d^3 \mathbf{k}}{(2\pi)^3} \tilde{\mathbf{A}}(\mathbf{k}, t) e^{i \mathbf{k} \cdot \mathbf{x}} \\ &=\int \frac{d^3 \mathbf{k}}{(2\pi)^3} \left( \frac{1}{{c^2}} \partial_{tt} \tilde{\mathbf{A}}(\mathbf{k}, t) + \mathbf{k}^2 \mathbf{A}(\mathbf{k}, t)\right)e^{i \mathbf{k} \cdot \mathbf{x}} \end{aligned}

Now operate with \int d^3 \mathbf{x} e^{-i \mathbf{p} \cdot \mathbf{x} }

\begin{aligned}0 &=\int d^3 \mathbf{x} e^{-i \mathbf{p} \cdot \mathbf{x} }\int \frac{d^3 \mathbf{k}}{(2\pi)^3} \left( \frac{1}{{c^2}} \partial_{tt} \tilde{\mathbf{A}}(\mathbf{k}, t) + \mathbf{k}^2 \mathbf{A}(\mathbf{k}, t)\right)e^{i \mathbf{k} \cdot \mathbf{x}}  \\ &=\int d^3 \mathbf{k}\delta^3(\mathbf{p} -\mathbf{k}) \left( \frac{1}{{c^2}} \partial_{tt} \tilde{\mathbf{A}}(\mathbf{k}, t) + \mathbf{k}^2 \mathbf{A}(\mathbf{k}, t)\right)\end{aligned}

Since this is true for all \mathbf{p} we have

\begin{aligned}\partial_{tt} \tilde{\mathbf{A}}(\mathbf{p}, t) = -c^2 \mathbf{p}^2 \tilde{\mathbf{A}}(\mathbf{p}, t) \end{aligned} \hspace{\stretch{1}}(3.30)

For every value of momentum we have a harmonic osciallator!

\begin{aligned}\dot{d}{x} = -\omega^2 x\end{aligned} \hspace{\stretch{1}}(3.31)

Fourier modes of EM potential in vacuum obey

\begin{aligned}\partial_{tt} \tilde{\mathbf{A}}(\mathbf{k}, t) = -c^2 \mathbf{k}^2 \tilde{\mathbf{A}}(\mathbf{k}, t)\end{aligned} \hspace{\stretch{1}}(3.32)

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

\begin{aligned}0 &= \boldsymbol{\nabla} \cdot \mathbf{A}(\mathbf{x}, t) \\ &= \int \frac{d^3 \mathbf{k} }{(2 \pi)^3} \boldsymbol{\nabla} \cdot \left( e^{i \mathbf{k} \cdot \mathbf{x}} \cdot \tilde{\mathbf{A}}(\mathbf{k}, t) \right)\end{aligned}

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

\begin{aligned}\boldsymbol{\nabla} \cdot (\phi \mathbf{B}) = (\boldsymbol{\nabla} \phi) \cdot \mathbf{B} + \phi \boldsymbol{\nabla} \cdot \mathbf{B}.\end{aligned} \hspace{\stretch{1}}(3.33)

But since our vector function \tilde{\mathbf{A}} is not a function of spatial coordinates we have

\begin{aligned}0 = \int \frac{d^3 \mathbf{k} }{(2 \pi)^3} e^{i \mathbf{k} \cdot \mathbf{x}} (i \mathbf{k} \cdot \tilde{\mathbf{A}}(\mathbf{k}, t)).\end{aligned} \hspace{\stretch{1}}(3.34)

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 \mathbf{k} \rightarrow -\mathbf{k}' after taking conjugates for that one)

\begin{aligned}\mathbf{k} \cdot \tilde{\mathbf{A}}(\mathbf{k}, t) &= 0 \\ \tilde{\mathbf{A}}(-\mathbf{k}, t) &= \tilde{\mathbf{A}}^{*}(\mathbf{k}, t).\end{aligned} \hspace{\stretch{1}}(3.35)

\begin{aligned}\mathbf{A}(\mathbf{x}, t) = \int \frac{d^3 \mathbf{k}}{(2\pi)^3} e^{i \mathbf{k} \cdot \mathbf{x}} \left( \frac{1}{{2}} \tilde{\mathbf{A}}(\mathbf{k}, t) + \frac{1}{{2}} \tilde{\mathbf{A}}^{*}(- \mathbf{k}, t) \right)\end{aligned} \hspace{\stretch{1}}(3.37)

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

\begin{aligned}\partial_{tt} \tilde{\mathbf{A}}(\mathbf{k}, t) &= - \omega_k^2 \tilde{\mathbf{A}}(\mathbf{k}, t) \\ \omega_k^2 &= c^2 \mathbf{k}^2\end{aligned} \hspace{\stretch{1}}(3.38)

our general solution is of the form

\begin{aligned}\tilde{\mathbf{A}}(\mathbf{k}, t) &= e^{i \omega_k t} \mathbf{a}_{+}(\mathbf{k}) +e^{-i \omega_k t} \mathbf{a}_{-}(\mathbf{k}) \\ \tilde{\mathbf{A}}^{*}(\mathbf{k}, t) &= e^{-i \omega_k t} \mathbf{a}_{+}^{*}(\mathbf{k}) +e^{i \omega_k t} \mathbf{a}_{-}^{*}(\mathbf{k})\end{aligned} \hspace{\stretch{1}}(3.40)

\begin{aligned}\mathbf{A}(\mathbf{x}, t) = \int \frac{d^3 \mathbf{k}}{(2 pi)^3} e^{i \mathbf{k} \cdot \mathbf{x}} \frac{1}{{2}} \left( e^{i \omega_k t} (\mathbf{a}_{+}(\mathbf{k}) + \mathbf{a}_{-}^{*}(-\mathbf{k})) +e^{-i \omega_k t} (\mathbf{a}_{-}(\mathbf{k}) + \mathbf{a}_{+}^{*}(-\mathbf{k})) \right)\end{aligned} \hspace{\stretch{1}}(3.42)

Define

\begin{aligned}\boldsymbol{\beta}(\mathbf{k}) \equiv \frac{1}{{2}} (\mathbf{a}_{-}(\mathbf{k}) + \mathbf{a}_{+}^{*}(-\mathbf{k}) )\end{aligned} \hspace{\stretch{1}}(3.43)

so that

\begin{aligned}\boldsymbol{\beta}(-\mathbf{k}) = \frac{1}{{2}} (\mathbf{a}_{+}^{*}(\mathbf{k}) + \mathbf{a}_{-}(-\mathbf{k}))\end{aligned} \hspace{\stretch{1}}(3.44)

Our solution now takes the form

\begin{aligned}\mathbf{A}(\mathbf{x}, t) = \int \frac{d^3\mathbf{k}}{(2 \pi)^3} \left( e^{i (\mathbf{k} \cdot \mathbf{x} + \omega_k t)} \boldsymbol{\beta}^{*}(-\mathbf{k})+e^{i (\mathbf{k} \cdot \mathbf{x} - \omega_k t)} \boldsymbol{\beta}(\mathbf{k})\right)\end{aligned} \hspace{\stretch{1}}(3.45)

\paragraph{Claim:}

This is now manifestly real. To see this, consider the first term with \mathbf{k} = -\mathbf{k}', noting that \int_{-\infty}^\infty dk = \int_{\infty}^\infty -dk' = \int_{-\infty}^\infty dk' with dk = -dk'

\begin{aligned}\int \frac{d^3\mathbf{k}'}{(2 \pi)^3} e^{i (-\mathbf{k}' \cdot \mathbf{x} + \omega_k t)} \boldsymbol{\beta}^{*}(\mathbf{k}')\end{aligned} \hspace{\stretch{1}}(3.46)

Dropping primes this is the conjugate of the second term.

\paragraph{Claim:}

We have \mathbf{k} \cdot \boldsymbol{\beta}(\mathbf{k})  = 0.

Since we have \mathbf{k} \cdot \tilde{\mathbf{A}}(\mathbf{k}, t) = 0, 3.40 implies that we have \mathbf{k} \cdot \mathbf{a}_{\pm}(\mathbf{k}) = 0. With each of these vector integration constants being perpendicular to \mathbf{k} at that point in momentum space, so must be the linear combination of these constants \boldsymbol{\beta}(\mathbf{k}).

References

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

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