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Total Power Radiated by Ultra-relativistic Particle

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jameson2
#1
Mar6-11, 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.
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lanedance
#2
Mar6-11, 12:00 PM
HW Helper
P: 3,307
how about re-arrange as follows
[tex]
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
[/tex]

then consider a power expansion
jameson2
#3
Mar6-11, 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.

lanedance
#4
Mar7-11, 09:29 AM
HW Helper
P: 3,307
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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