How to prove that if a particle performs cyclical motion then its energy loss rate averaged over the period equals averaged radiation intensity?
The energy loss rate is the quantity of energy that the particle loses in unit time. And radiation intensity is an energy which observer detects some time later than it was emitted (due to finitude of light speed).
$$\frac{d E}{d t} = - \int d\Omega \left( 1 - \frac{(\vec n, \vec V(t))}{c}\right)\frac{dI(t)}{d\Omega}$$
where $\vec V(t)$ is the particle's speed, and $\vec n$ is unit vector on sphere $d \Omega$.
One can integrate it over time period $\int_{0}^{T}$ and get averaged energy loss rate $\langle\frac{d E}{d t}\rangle$, averaged intensity $\langle I\rangle$ and $$\int dt \int d\Omega \frac{(\vec n, \vec V(t))}{c}\frac{dI(t)}{d\Omega} $$ which should equals zero in order the initial sentence to be true.
