# Total Power Radiated by Ultra-relativistic Particle

by jameson2
Tags: radiated power, relativistic power, ultrarelativistic
 P: 53 1. The problem statement, all variables and given/known data Given the formula for power radiated into a solid angle, evaluate the total power radiated to all angles by an ultra relativistic particle, keeping the leading power of $$\gamma$$ only. 2. Relevant equations The power formula: $$\frac{dP'}{d\Omega}=\frac{q^2 \alpha^2}{\pi^2 c}\frac{2\gamma^{10}\theta^2}{(\gamma^2 \theta^2 +1)^5}$$ 3. The attempt at a solution Basically, I can't integrate this: $$P'= \int^{2\pi}_0 \int^\pi_0 \frac{q^2 \alpha^2}{\pi^2 c}\frac{2\gamma^{10}\theta^2}{(\gamma^2 \theta^2 +1)^5} sin(\theta)d\theta d\phi$$ I was thinking that since gamma will be large, you can ignore the 1 on the bottom line, but that doesn't get me anywhere. Possibly I'm just missing how I can use the fact that the question says "keeping only the leading power of gamma", but it's not clear to me at all. Thanks for any hints.
 HW Helper P: 3,309 how about re-arrange as follows $$Psingle-quote = \int^{2\pi}_0 \int^\pi_0 \frac{q^2 \alpha^2}{\pi^2 c}\frac{2\gamma^{10}\theta^2}{(\gamma^2 \theta^2 +1)^5} sin(\theta)d\theta d\phi = \int^{2\pi}_0 \int^\pi_0 \frac{q^2 \alpha^2}{\pi^2 c}\frac{2\theta^2}{( \theta^2 +\frac{1}{\gamma^2})^5} sin(\theta)d\theta d\phi$$ then consider a power expansion
 P: 53 So just get $$\theta^{10} + 5\theta^8\frac{1}{\gamma^2}$$ on the bottom line? Is this what it means by only keep the leading power of gamma? I'm still not sure how to go about integrating this though.
HW Helper
P: 3,309

## Total Power Radiated by Ultra-relativistic Particle

as gamma, gets large, 1/gamma gets small, so I would try to expand in power series in terms of 1/gamma and only keep the lowest order terms. geometric series may be handy here, though you will need to be careful on how you manipulate theta...

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