1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Radiation from a moving charge

  1. Oct 25, 2012 #1
    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]

    2. Relevant 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]​


    3. 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.
     
  2. jcsd
  3. Oct 25, 2012 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    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].
     
  4. Oct 26, 2012 #3
    Thanks a lot i got it !
     
  5. Oct 26, 2012 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook