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}}
​

\textit{}

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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