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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.

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It seems that when you traverse the cyclical motion, the velocity vector cancels in pairs over the period (point in opposite directions along antipodal points on the sphere), so that your last term disappears. –  Chris Gerig Dec 27 '11 at 20:09
    
@ChrisGerig , I still don't understand how to easily prove it. $\frac{dI(t)}{d\Omega}$ depends on $\vec v$ and $\dot{\vec{v}}$ so one can't just integrate over time and substitute $\vec v(t_1) = \vec v(t_2)$ –  0x2207 Dec 30 '11 at 17:06

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