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

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

where the constant K is

[itex]K=\frac{e^{2}~c~β^{4}}{4~\pi~a^{2}}[/itex]​

and

[itex]β=\frac{a\omega}{c}[/itex]​

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

[itex]\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}}[/itex]
​

[itex] \textit{}[/itex]

Homework Equations

[itex]\cos\theta=\frac{1}{2}\left( z +\frac{1}{z}\right)[/itex]​

[itex]\sin\theta=\frac{1}{2i}\left( z -\frac{1}{z}\right)[/itex]​

where [itex]z=e^{i\theta}[/itex]

Residue theorem:

[itex]\oint_{C}f~dz = 2\pi i \sum Resf(z_{n})[/itex]​

The Attempt at a Solution

the time average of a function f(t) is

[itex]\frac{1}{T}\int_{0}^{T}f(t)[/itex]​

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 [itex]\omega = \stackrel{~.}{\theta}[/itex]

[itex] K \sin^{2}θ \frac{cos^{2}(ωt)}{(1+β \cosθ \sin(ωt))^{5}}dt [/itex]

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

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, [itex]\theta[/itex] is the angle between the axis of the charged particles motion and the point at which the radiation is measured (the field point), while [itex]\omega[/itex] is the angular frequency of its motion. The two quantities are completely independent.

Instead, just use the substitution [itex]z\equiv e^{i\omega t}[/itex].

 

[venatorr]

Thanks a lot i got it !

 

[TSny]

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

[itex]\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}}[/itex]
​

[itex] \textit{}[/itex]

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

[itex]\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}}[/itex]
​

[itex] \textit{}[/itex]

 

[paranormal]

Therefore, I am not sure how to proceed with the time average over one period. I would appreciate any guidance or corrections on my approach.