How to Perform a Time Average of Radiation from a Moving Charge?

[venatorr]

1. The problem statement
The problem is from the textbook Mathematics for Physicist by S.M. Lea. it's problem 2.35

The power radiated per unit solid angle by a charge undergoing simple harmonic motion is

$\frac{dP}{dΩ} = K \sin^{2}θ \frac{cos^{2}(ωt)}{(1+β \cosθ \sin(ωt))^{5}}$​

where the constant K is

$K=\frac{e^{2}~c~β^{4}}{4~\pi~a^{2}}$​

and

$β=\frac{a\omega}{c}$​

is the amplitude/c. Using the Residue Theorem, perform the time average over one period to show that

$\left\langle \frac{dP}{dΩ} \right\rangle = \frac{K}{8}\sin^{2}\theta \frac{4+\beta ^{2}\cos^{2}\theta}{(1-\beta^{2}\cos^{2}\theta)^{10}}$
​

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Homework Equations

$\cos\theta=\frac{1}{2}\left( z +\frac{1}{z}\right)$​

$\sin\theta=\frac{1}{2i}\left( z -\frac{1}{z}\right)$​

where $z=e^{i\theta}$

Residue theorem:

$\oint_{C}f~dz = 2\pi i \sum Resf(z_{n})$​

The Attempt at a Solution

the time average of a function f(t) is

$\frac{1}{T}\int_{0}^{T}f(t)$​

My guess is that I have to convert the integral over time to an integral over θ and than convert it to an integral on the complex unit circle using the formula converting cosθ and sinθ to z. I can then do the integral using the Residue theorem

So i first converted it into an integral over theta

since $\omega = \stackrel{~.}{\theta}$

$K \sin^{2}θ \frac{cos^{2}(ωt)}{(1+β \cosθ \sin(ωt))^{5}}dt$

$= K \sin^{2}θ \frac{cos^{2}(ωt)}{(1+β \cosθ \sin(ωt))^{5}} \frac{d\theta}{\omega}$​

However I'm stuck with the term ωt in the sin and cos and I don't know how to convert them into theta. Since the speed is not constant in simple harmonic motion, I can't simply say ωt = θ. I also don't think I can say that ωt is constant with respect to theta.

 

[gabbagabbahey]

My guess is that I have to convert the integral over time to an integral over θ and than convert it to an integral on the complex unit circle using the formula converting cosθ and sinθ to z. I can then do the integral using the Residue theorem

No, in the given radiation law, $\theta$ is the angle between the axis of the charged particles motion and the point at which the radiation is measured (the field point), while $\omega$ is the angular frequency of its motion. The two quantities are completely independent.

Instead, just use the substitution $z\equiv e^{i\omega t}$.

 

[venatorr]

Thanks a lot i got it !

 

[TSny]

Using the Residue Theorem, perform the time average over one period to show that

$\left\langle \frac{dP}{dΩ} \right\rangle = \frac{K}{8}\sin^{2}\theta \frac{4+\beta ^{2}\cos^{2}\theta}{(1-\beta^{2}\cos^{2}\theta)^{10}}$
​

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In case this thread is referred to in the future, I believe the power of 10 written above should be corrected to 7/2 (see for example Jackson's text, 2nd edition, problem 14.5). So

$\left\langle \frac{dP}{dΩ} \right\rangle = \frac{K}{8}\sin^{2}\theta \frac{4+\beta ^{2}\cos^{2}\theta}{(1-\beta^{2}\cos^{2}\theta)^{7/2}}$
​

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