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Total Power Radiated by Ultrarelativistic Particle 
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#1
Mar611, 07:38 AM

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 [tex] \gamma [/tex] only. 2. Relevant equations The power formula: [tex] \frac{dP'}{d\Omega}=\frac{q^2 \alpha^2}{\pi^2 c}\frac{2\gamma^{10}\theta^2}{(\gamma^2 \theta^2 +1)^5} [/tex] 3. The attempt at a solution Basically, I can't integrate this: [tex] 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 [/tex] 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. 


#2
Mar611, 12:00 PM

HW Helper
P: 3,307

how about rearrange as follows
[tex] Psinglequote = \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 [/tex] then consider a power expansion 


#3
Mar611, 01:47 PM

P: 53

So just get
[tex] \theta^{10} + 5\theta^8\frac{1}{\gamma^2} [/tex] 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. 


#4
Mar711, 09:29 AM

HW Helper
P: 3,307

Total Power Radiated by Ultrarelativistic 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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