Birkhoff's theorem (electromagnetism)

Derivation using Lie derivatives

Defining the 1-form ${\displaystyle E}$ and 2-form ${\displaystyle B}$ in ${\displaystyle {\mathbb{R}}^3}$ as:

${\displaystyle \begin{array}{cc}E& =E_{i}d{r}_i\\ B& ={ϵ}_{ijk}{B}_{i}d{r}_j\wedge d{r}_k\end{array}}$

Using the Hodge star operator, we can rewrite Maxwell's Equations with these forms as

${\displaystyle \begin{array}{cc}dB& =0\\ d\star E& =0\\ \star d\star E& =\dot{E}\\ dE& =-\dot{B}\end{array}}$.

The spherical symmetry condition requires that the Lie derivatives of ${\displaystyle E}$ and ${\displaystyle B}$ with respect to the vector field ${\displaystyle V}$ that represents their rotations are zero

${\displaystyle \begin{array}{cc}{V}_i& ={ϵ}_{ijk}{r}_j\frac{\mathrm{\partial }}{\mathrm{\partial }{r}_k}\\ {\mathcal{L}}_V(E)& =0\\ {\mathcal{L}}_V(B)& =0.\end{array}}$

By the definition of the Lie derivative as the directional derivative along ${\displaystyle V}$

${\displaystyle \begin{array}{cc}{V}_j(r_i)& ={ϵ}_{jki}{r}_k\\ 0& ={\mathcal{L}}_V{E}_{i}d{r}_i+E_i{\mathcal{L}}_{V}d{r}_i\\ & =V_j(E_i)d{r}_i+E_i{V}_j(d{r}_i)\\ & =V_j(E_i)d{r}_i+E_{i}d(V_j(r_i))\\ & =V_j(E_i)d{r}_i+E_i{ϵ}_{jki}d{r}_k\\ & =V_j(E_i)d{r}_i+{ϵ}_{jki}{E}_{k}d{r}_i\\ V_j(E_i)& =-{ϵ}_{jki}{E}_k\end{array}}$.

Therefore, ${\displaystyle E}$ is equivalent to ${\displaystyle r}$ under rotation and we can write for some function ${\displaystyle g}$

${\displaystyle E=g(r^2,t)r_{i}d{r}_i}$.

Because the product of the components of the vector are just its length

${\displaystyle r_i{r}_i=r^2}$.

And substituting back into our equation and rewriting for a function ${\displaystyle f}$

${\displaystyle E=df(r^2,t)}$.

Taking the exterior derivative of ${\displaystyle E}$, we find by definition that,

${\displaystyle dE=0}$.

And using our Maxwell equation that ${\displaystyle dE=-\dot{B}}$,

${\displaystyle \dot{B}=0}$.

Thus, we find that the magnetic field is static. Similarly, using the second rotational invariance equation, we can find that the electric field is static. Therefore, the solution must be static.