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Homework Help: 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

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    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

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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

    [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]
     
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