# Total Power Radiated by Ultra-relativistic Particle

1. Mar 6, 2011

### jameson2

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.

Last edited: Mar 6, 2011
2. Mar 6, 2011

### lanedance

$$Psingle-quote = \int^{2\pi}_0 \int^\pi_0 \frac{q^2 \alpha^2}{\pi^2 c}\frac{2\gamma^{10}\theta^2}{(\gamma^2 \theta^2 +1)^5} sin(\theta)d\theta d\phi = \int^{2\pi}_0 \int^\pi_0 \frac{q^2 \alpha^2}{\pi^2 c}\frac{2\theta^2}{( \theta^2 +\frac{1}{\gamma^2})^5} sin(\theta)d\theta d\phi$$

then consider a power expansion

3. Mar 6, 2011

### 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.

4. Mar 7, 2011

### 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...