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Consider a simple 3D wave solution of Maxwell's equations: $$\vec B=\vec e_z \cos\left(\frac{2\pi}{\lambda}(ct-x)\right),$$ $$\vec E=\vec e_y\cos\left(\frac{2\pi}{\lambda}(ct-x)\right),$$ and propagation in the direction of $\vec e_x$.

I'd like to find some vector potential $\vec A$ and scalar potential $\phi$ for such a wave. I've tried using the known expression for static uniform magnetic field: $\vec A=\vec e_y B x$, which satisfies $\vec B=\nabla\times \vec A$ and multiplying it by the cosine factor: $$\vec A=\vec e_y B x\cos\left(\frac{2\pi}{\lambda}(ct-x)\right),$$ but although it does satisfy $\vec B=\nabla\times \vec A$, it appears that I can't have the correct result for $\vec E=-\nabla\phi-\frac{\partial\vec A}{\partial t}$ (at least if I use $\phi=\mathrm{const}$). Using another expression for the static part of $\vec A$, $\vec A=\frac{B}2\left(\vec e_y x-\vec e_x y\right)$, gives an even worse result. It either I am using the wrong expression for $\vec A$, or I have to add a non-constant $\phi$.

What would be the correct way of determining the potentials?

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  • $\begingroup$ Trial and error, plus remembering calculus (specifically the Leibniz rule that $\frac{d}{dx}\left[f(x)\cdot g(x)\right]=\frac{df}{dx}g+f\frac{dg}{dx}$ $\endgroup$
    – Kyle Kanos
    Commented Dec 28, 2013 at 15:34
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    $\begingroup$ Set $\phi = 0$ and $\mathbf A = -\mathbf e_y E_0/\Omega \sin (\Omega t - kx)$. This gives electric and magnetic field of plane wave. $\endgroup$ Commented Dec 28, 2013 at 18:36

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There are actually an infinite number of possible answers. The E- and B-field do not uniquely specify the potentials - you have gauge freedom. That is, you can specify some $\vec{A}$, $\phi$, which will give you $\vec{E}$ and $\vec{B}$, but you could equally add the gradient of any scalar function to $\vec{A}$ and subtract the time derivative of the same scalar function from $\phi$ and you would get the same result.

So you need to specify what gauge you are working in. Typically for a plane electromagnetic wave you would choose $\phi=0$ and then all you need to do is $$ \vec{A} = -\int \vec{E}\ dt = -\vec{e}_y\frac{\lambda}{2\pi c}\sin\left(\frac{2\pi}{\lambda}(ct-x)\right) + \vec{A}_0(\vec{r}), $$ where $\vec{A}_0$ is some time-independent vector field with a zero curl (see below).

If you take the curl of this A-field you get $$\nabla \times \vec{A} = \vec{e}_k \frac{1}{c} \cos\left( \frac{2\pi}{\lambda}(ct-x)\right) + \nabla \times \vec{A}_0$$

This is (or should be) your magnetic field, providing that $\vec{A}_0$ is curl-free (or zero for convenience). I say should be, because judging from your expression for the E-field in terms of the potentials, you are using SI units. In which case the amplitude of the B-field should be $c$ times less than the E-field amplitude.

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  • $\begingroup$ What do you mean by "typically for a plane electromagnetic wave"? A plane wave seems to be too specific to have any "typical" approach. $\endgroup$
    – Ruslan
    Commented Mar 6, 2016 at 14:24
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    $\begingroup$ @Ruslan I mean that you could choose $\phi$ to be anything you like. Typically, we choose zero. In other situations we might choose something else. $\endgroup$
    – ProfRob
    Commented Mar 6, 2016 at 14:25

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