Total Power Radiated by Ultra-relativistic Particle

[jameson2]

Homework Statement

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.

Homework Equations

The power formula:
$$\frac{dP'}{d\Omega}=\frac{q^2 \alpha^2}{\pi^2 c}\frac{2\gamma^{10}\theta^2}{(\gamma^2 \theta^2 +1)^5}$$

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.

 

[lanedance]

how about re-arrange as follows
$$Psingle-quote<br /> = \int^{2\pi}_0 \int^\pi_0 \frac{q^2 \alpha^2}{\pi^2 c}\frac{2\gamma^{10}\theta^2}{(\gamma^2 \theta^2 +1)^5} sin(\theta)d\theta d\phi <br /> <br /> = \int^{2\pi}_0 \int^\pi_0 \frac{q^2 \alpha^2}{\pi^2 c}\frac{2\theta^2}{( \theta^2 +\frac{1}{\gamma^2})^5} sin(\theta)d\theta d\phi$$

then consider a power expansion

 

[jameson2]

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.

 

[lanedance]

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...