# Vector and scalar potentials of plane wave

Consider a simplest 3D 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 is in direction of $$\vec e_x$$.

I'd like to find some vector potential $$\vec A$$ and scalar potential $$\phi$$ for such wave. I've tried using 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 despite it does satisfy $$\vec B=\nabla\times \vec A$$, it appears that I can't have 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 static part of $$\vec A$$, $$\vec A=\frac{B}2\left(\vec e_y x-\vec e_x y\right)$$, gave even worse result. Seems either I use wrong expression for $$\vec A$$, or I have to add non-const $$\phi$$.

What would be the correct way of determining the potentials?

• Trial and error, plus remembering calculus (specifically the Leibniz rule that $\frac{d}{dx}\left[f(x)\cdot g(x)\right]=\frac{df}{dx}g+f\frac{dg}{dx}$ – Kyle Kanos Dec 28 '13 at 15:34
• 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. – Ján Lalinský Dec 28 '13 at 18:36

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.
• @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. – Rob Jeffries Mar 6 '16 at 14:25