# I Deriving the E and B fields in a plane wave from the four-potential

#### SiennaTheGr8

Summary
Having trouble deriving the electric field of a plane wave from the four-potential.
I'm trying to derive the electric and magnetic fields of a plane wave from the four-potential $\mathbf{A} = (A^t , \mathbf{a})$ in the Lorenz gauge. Given:

$\mathbf{A}(\mathbf{R}) = \Re \left( \mathbf{C} e^{i \mathbf{K} \cdot \mathbf{R}} \right)$

for constant future-pointing lightlike $\mathbf{K} = (K^t, \mathbf{k})$ and constant $\mathbf{C} = (C^t , \mathbf{c})$ orthogonal to $\mathbf{K}$, I think I correctly get the magnetic field by taking the real part of this:

$\nabla \times \mathbf{a} = \nabla \times \left( \mathbf{c} e^{i \mathbf{K} \cdot \mathbf{R}} \right) = \nabla e^{i \mathbf{K} \cdot \mathbf{R}} \times \mathbf{c}$,

which gives me $\mathbf{B} = \sin (\mathbf{K} \cdot \mathbf{R}) (\mathbf{k} \times \mathbf{c})$. That seems reasonable, and if it's right then I expect $\mathbf{E}$ to have magnitude $\sin (\mathbf{K} \cdot \mathbf{R}) \Vert \mathbf{k} \times \mathbf{c} \Vert$ and be perpendicular to both $\mathbf{B}$ and $\mathbf{k}$ (and perhaps parallel to $\mathbf{c}$?). But I'm having trouble:

$- \nabla A^t - \partial^t \mathbf{a} = - \nabla \left( C^t e^{i \mathbf{K} \cdot \mathbf{R}} \right) - \partial^t \left( \mathbf{c} e^{i \mathbf{K} \cdot \mathbf{R}} \right) = -C^t \nabla e^{i \mathbf{K} \cdot \mathbf{R}} - \mathbf{c} \partial^t e^{i \mathbf{K} \cdot \mathbf{R}}$.

If the first term vanishes, taking the real part of the second term I get $\mathbf{E} = \sin (\mathbf{K} \cdot \mathbf{R}) k \mathbf{c}$, which works if it's true that $\mathbf{c} \parallel \mathbf{E}$ (correct magnitude, and correct right-handed set for $\mathbf{E}$, $\mathbf{B}$, and $\mathbf{k}$). Otherwise, for the first term I get:

$-C^t \nabla e^{i \mathbf{K} \cdot \mathbf{R}} = -i e^{i \mathbf{K} \cdot \mathbf{R}} C^t \mathbf{k}$,

whose real part I think is $\sin (\mathbf{K} \cdot \mathbf{R}) C^t \mathbf{k}$, giving:

$\mathbf{E} = \sin (\mathbf{K} \cdot \mathbf{R}) \left( C^t \mathbf{k} + k \mathbf{c} \right) = \sin (\mathbf{K} \cdot \mathbf{R}) \left( (\mathbf{c} \cdot {\hat{\mathbf{k}}}) \mathbf{k} + k \mathbf{c} \right)$,

which doesn't seem right at all, unless it's true that $\mathbf{c} \perp \mathbf{k}$ in all frames (the condition under which the first term drops out). But I haven't been able to prove to myself that $\mathbf{c} \perp \mathbf{k}$ is a Lorentz-invariant statement. In fact, the notion seems silly, because unless I'm missing something it's equivalent to saying that I can't boost to a frame in which $C^t \neq 0$.

Given that $\mathbf{K}$ is lightlike and $\mathbf{C}$ is orthogonal to it (and obviously spacelike), does it follow that their spatial three-vector components $\mathbf{k}$ and $\mathbf{c}$ are perpendicular in all Lorentz frames? If so, is it easily demonstrated? And if not, can you spot where I've gone off the rails?

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#### SiennaTheGr8

It occurs to me that I probably made a sign error (or errors), but that otherwise I was on the right track. If instead of:

$\mathbf{E} = \sin (\mathbf{K} \cdot \mathbf{R}) \left( C^t \mathbf{k} + K^t \mathbf{c} \right)$

it were:

$\mathbf{E} = \sin (\mathbf{K} \cdot \mathbf{R}) \left( C^t \mathbf{k} - K^t \mathbf{c} \right)$,

then $\mathbf{E} \perp \mathbf{k}$ as desired, since $C^t \mathbf{k} \cdot \mathbf{k} - K^t \mathbf{c} \cdot \mathbf{k} = C^t K^{t \, 2} - C^t K^{t \, 2}$ (and of course $\mathbf{E} \perp \mathbf{B}$, since the former is coplanar with $\mathbf{c}$ and $\mathbf{k}$ while the latter is perpendicular to them).

Additionally, I think this sign correction gives $E = B \propto ck \sin \theta_{[ \mathbf{c}, \mathbf{k} ]}$:

$\left( C^t \mathbf{k} - K^t \mathbf{c} \right) = ck \left( (\mathbf{\hat c} \cdot \mathbf{\hat k}) \mathbf{\hat k} - \mathbf{\hat c} \right) = ck \left( \mathbf{\hat k} \cos \theta - \mathbf{\hat c} \right)$,

where $\Vert \mathbf{\hat k} \cos \theta - \mathbf{\hat c} \Vert ^2 = \left( \mathbf{\hat k} \cos \theta - \mathbf{\hat c} \right) \cdot \left( \mathbf{\hat k} \cos \theta - \mathbf{\hat c} \right) = 1 - \cos^2 \theta = \sin^2 \theta$.

I'll still have to find my sign error(s), but this looks promising (in case anyone was wondering...).

#### Orodruin

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You can always choose your polarisation vector to have time-component of zero by subtracting from it a vector proportional to the wave vector. This will not change the EM fields.

#### SiennaTheGr8

I have a related question—

In some sources I see a negative sign on the exponent like $e^{-i \mathbf{K} \cdot \mathbf{R}}$, and in others I don't. Is this an arbitrary choice, or does it correspond to the chosen sign convention for the Minkowski dot product? That is, must one make the exponent negative if the Minkowski product gives $K^t ct - \mathbf{k} \cdot \mathbf{r}$, so that the sinusoidal argument becomes $\mathbf{k} \cdot \mathbf{r} - K^t ct$?

#### Orodruin

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The real part of $e^{-ik\cdot x}$ and $e^{ik\cdot x}$ are the same. You will also get a redefinition of the polarisation vector. Of course, you will also get a difference in terms of the metric convention. Always check the metric convention.

#### SiennaTheGr8

Ah, I think I get it: cosine is obviously an even function, and although after differentiation (of the exponential) the real part will be an odd function (sine), you also inherit the correct sign from the exponent.

So it really is just a matter of preference whether to slap a negative sign up there (to get a classic "$x - t$" argument instead of "$t - x$"). Either way, you end up with a right-handed set of $\mathbf{E}, \mathbf{B}, \mathbf{k}$, yeah?

"Deriving the E and B fields in a plane wave from the four-potential"

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