# Accelerating charges giving rise to electromagnetic waves from Maxwell's laws?

For a long time I have been wondering how accelerating charges give rise to electromagnetic radiation. I have now seen 'graphical' reasoning in terms of requiring continuity of electric field lines, such as in this post.

However I like to see if things can be derived from the mathematics, and I have been trying to see how this comes out of Maxwell's equations. Each time I have just gotten very stuck. For example, considering it case by case and initially when in a frame in which the accelerating particle is at rest (and assuming there is initially no changing electric field, I get that the curl of a magnetic field from this motion is zero, so the magnetic field, if there is one, is conservative. So we can write $\textbf{B}=-\nabla \phi$ and $\frac{\partial\textbf{B}}{\partial t}=-\nabla \frac{\partial\phi}{\partial t}$, so from one of the equations $\nabla \times \textbf{E} =\nabla \frac{\partial\phi}{\partial t}$ and this doesn't seem to get me anywhere further.

One of the stack exchange posts I looked at led me to the Larmor formula, but looking at the derivation I didn't see Maxwell's equations creep in? (This may just be me being dumb!)

Anyhow, having spent such a long time trying to figure this out, I remembered that when we studied Maxwell's equations we plugged in a sine/cosine solution and found that they work. The equations didn't just 'spit out' the solution (perhaps there are more?). I just wondered whether a similar thing might be happening here, and I could be playing with the equations for a long time but would never see that the magnetic and electric fields are zero, or unchanging etc. I would have to plug these into the equations along with the conditions of non-accelerating charge etc, and see that they work. But the equations themselves would not 'show' me the result?

If you want a rigorous derivation, the best way is probably via the Liénard-Wiechert potentials, which give the solutions of the Maxwell equations in response to a moving point charge, i.e. with sources given by \begin{align} \rho(\mathbf{r}', t') & = q \delta^3(\mathbf{r'} - \mathbf{r}_s(t'))\\ \mathbf {J} (\mathbf {r} ',t') & =q\mathbf {v} _{s}(t')\delta ^{3}(\mathbf {r'} -\mathbf {r} _{s}(t')). \end{align} Once you do that development, then the electric field produced by the particle is given by $${\mathbf {E}}({\mathbf {r}},t)={\frac {1}{4\pi \epsilon _{0}}}\left({\frac {q({\mathbf {n}}-{\boldsymbol {\beta }})}{\gamma ^{2}(1-{\mathbf {n}}\cdot {\boldsymbol {\beta }})^{3}|{\mathbf {r}}-{\mathbf {r}}_{s}|^{2}}}+{\frac {q{\mathbf {n}}\times {\big (}({\mathbf {n}}-{\boldsymbol {\beta }})\times {\dot {{\boldsymbol {\beta }}}}{\big )}}{c(1-{\mathbf {n}}\cdot {\boldsymbol {\beta }})^{3}|{\mathbf {r}}-{\mathbf {r}}_{s}|}}\right)_{{t_{r}}},$$ where the second term is the radiative term, as it is proportional to the inverse of the distance to the charge. That radiative term is directly proportional to the time derivative of $\boldsymbol\beta=\mathbf v/c$, i.e. to the acceleration of the particle.