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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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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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    $\begingroup$ And what is the ultrarelativistic acceleration? $\omega^2r$? $\endgroup$
    – peterh
    Commented May 26, 2014 at 19:18
  • $\begingroup$ good question: physics.stackexchange.com/q/66839/10531 $\endgroup$ Commented May 26, 2014 at 19:42
  • $\begingroup$ For circular motion $a = v^2/r = \omega^2 r$. $\endgroup$ Commented May 26, 2014 at 20:08
  • $\begingroup$ This is not taking into account magnetic field or the particles charge (both important factors), this formula does not look correct to me... $\endgroup$
    – MoonKnight
    Commented Jan 8, 2015 at 12:42

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