I've recently played around with 1+1D-Electrodynamics a little. In this theory, the potential $\mathbf A(t,x)$ looks like $$ \mathbf A = \phi(t,x)dt + A(t,x)dx $$ The field strength tensor in this theory only has one independent component: $$ \mathbf{F} = d\mathbf A = \frac{\partial\phi}{\partial x}dt\wedge dx + \frac{\partial A}{\partial t}dx\wedge dt \equiv E(t,x)dx\wedge dt $$ Which, in vacuum, satisfies the sole "Maxwell equation" $$ \frac{\partial E}{\partial x} = 0 $$ But this means that the only electric field which can appear in 1+1D is constant, and we get no radiation. Another argument goes along the lines that since there are no directions orthogonal to the $E$ and $B$ (which is entirely absent in 1+1D) fields, electromagnetic radiation has no direction available to propagate. See also the answers to this question.
Birkhoff's theorem now states that the vacuum solution of a classical field theory is unique. In the case of electromagnetism, this solution is given by the Coulomb potential. Since the Coulomb potential does not radiate, and there are no other spherically symmetric solutions to Maxwell's equations, there can be no spherically symmetric electromagnetic radiation. Now consider a spherically symmetric charge distribution: $$ \rho = \rho(t,r) $$ All the relevant physics is now happening on the $t,r$-plane. Is it therefore reasonable to assume that we can use the above formalism to accurately describe this charge distribution, and is Birkhoff's theorem simply a consequence of the absence of radiation in 1+1 dimensions?
