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I am very curious if an easy calculable formula for the bremstrahlung radiation of deeply relativistic, charged particles exists, if they are moving on circular orbit:

$P(E,m_0,Q,r)=?$

...where

  • $P$ is the power of the Bremstrahlung radiation;
  • $E$ is the total kinetic energy of the particles (we are in deeply relativistic case, thus $E\gg{m_0}c^2$);
  • $m_0$ is the total rest mass of the particles;
  • $Q$ is the total charge of the particles;
  • and $r$ is the radius of the orbit.

If a such clean, trivial formula doesn't exist, a link were also okay, where it can be found.

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1 Answer 1

up vote 2 down vote accepted

The radiation from an ultrarelativistic ($v \approx c$) particle on a circular path is called synchotron radiation. The total power radiated from such a particle is $$P = \frac{e^2 a^2}{6\pi \epsilon_0 c}\gamma^4$$ where $a$ is the acceleration and $\gamma$ is the Lorentz factor, $\gamma^2 = 1/(1-v^2/c^2)$.

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1  
And what is the ultrarelativistic acceleration? $\omega^2r$? –  Peter Horvath May 26 at 19:18
    
good question: physics.stackexchange.com/q/66839/10531 –  diffeomorphism May 26 at 19:42
    
For circular motion $a = v^2/r = \omega^2 r$. –  Robin Ekman May 26 at 20:08

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