Energy in Field for a Given Vector Potential

We have the vector potential

\begin{displaymath}\bgroup\color{black} \vec{A}(\vec{r},t)\equiv 2\vec{A}_0\cos(\vec{k}\cdot\vec{r}-\omega t). \egroup\end{displaymath}

First find the fields.

\vec{E}&=&-{1\over c}{\partial \vec{A}\over \partial t}=2{\ome...
...A}=2\vec{k}\times\vec{A}_0\sin(\vec{k}\cdot\vec{r}-\omega t) \\

Note that, for an EM wave, the vector potential is transverse to the wave vector. The energy density in the field is

\begin{displaymath}\bgroup\color{black} U={1\over 8\pi}\left(E^2+B^2\right)={1\o...
...over 2\pi c^2}A_0^2\sin^2(\vec{k}\cdot\vec{r}-\omega t) \egroup\end{displaymath}

Averaging the sine square gives one half, so, the energy in a volume \bgroup\color{black}$V$\egroup is

\begin{displaymath}\bgroup\color{black} Energy={\omega^2A_0^2V\over 2\pi c^2} \egroup\end{displaymath}

Jim Branson 2013-04-22