# Vanishing dipole radiation of equidistant point charges on a ring

I learned that if I distribute point charges uniformly on a circle and let them run along the circle with the same constant speed, the dipole radiation of that configuration vanishes.

However, moving along the circle, the charges still experience an accelerated motion and should thus emit radiation. Why is this not the case?

Surely, when you consider remote point charge $e$ circling on the ring it emits radiation. That means it creates fields $\mathbf{E}$ and $\mathbf{H}$ such that they decade as $1/R$. That field is discribed by Lienard--Wiechert potentials $$\mathbf{E} = \frac{e}{c^2R} \frac{[\mathbf{n}\times[(\mathbf{n}-\mathbf{v}/c) \times \mathbf{w}]]}{(1-\left(\mathbf{n}\cdot\mathbf{v})/c\right)^3}, \quad \mathbf{H}=[\mathbf{n}\times\mathbf{E}].$$ where all values are taken at retarded time moment $t'$. For simplicity we can consider non--relativistic particles $v\ll c$. Then $t'\approx t - R/c$ and $$\mathbf{E} \approx e\frac{\mathbf{n}(\mathbf{n}\cdot\mathbf{w})-\mathbf{w}}{c^2R}, \quad \mathbf{H}\approx[\mathbf{n}\times\mathbf{E}].$$ From that formula it is easy to see that for distances $R\gg r$, where $r$ is the radius of the circle fields from diametrically opposite charges eliminate each other. It also take place in relativistic case. Therefore integral radiation equals to zero.
Your question was about dipole radiation. So then you should know that intensity of radiation can be written in form $$I = \frac{2\ddot{\mathbf{d}}^2}{3c^3} + \frac{2\ddot{\boldsymbol{\mu}}^2}{3c^3} + \frac{\dddot{D}_{\mu\nu}\dddot{D}^{\mu\nu}}{180c^5} + \dots$$ where $\mathbf{d}$ is dipole moment, $\boldsymbol{\mu}$ is magnet dipole moment and $D^{\mu\nu}$ is quadrupole moment tensor. But your system is so symmetric so all moments are constants and hereby all summands vanish. In particular, dipole moment $$\mathbf{d} = \sum_i e_i \mathbf{r}_i = \mathrm{const}$$ and equals to zero if you put the origin of coordinate system in the center of the ring.