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Homework Help: Electrodynamics of a moving particle

  1. Sep 10, 2014 #1
    1. The problem statement, all variables and given/known data
    There is a charged particle (charge=q) moving on the x axis such that [itex]x(t)=A\,sin(\omega t)[/itex]. Prove that:
    [tex]\int <\rho>\,dV=q [/tex]

    2. Relevant equations
    We have the following equations:
    [tex]<\rho>=\frac{1}{T} \int_0^T \rho\, dt [/tex]

    Where [itex]T=\frac{2 \pi}{\omega}[/itex]

    3. The attempt at a solution

    So for this particle we have the following charge density [itex]\rho=q\,\delta(z)\delta(y)\delta(x-A\,sin(\omega t))[/itex] therefore:

    [tex]<\rho>=\frac{\omega}{2\pi}q\,\delta(z)\delta(y) \int_0^T \delta(x-A\,sin(\omega t))\,dt[/tex]

    However, we know that [itex]\delta(f(t))=\sum \frac{\delta(t-t_i)}{|f^\prime (t_i)|}[/itex] for every root of [itex]f(t)[/itex]. So we have [itex]\delta(x-A\,sin(\omega t))=\delta(x/A-sin(\omega t))/|A|[/itex] where [itex]f(t)=x/A-sin(\omega t)[/itex].

    Considering the integral is over one period I'll only take one root into account:

    [tex]\delta(x-A\,sin(\omega t))=\frac{1}{|A|}\frac{\delta(t-sin^{-1}(x/A)/\omega)}{\omega cos(sin^{-1}(x/A))}=\frac{\sqrt{A^2-x^2}}{\omega}\delta(t-sin^{-1}(x/A)/\omega)[/tex]

    Then, the average value is:

    [tex]<\rho>=\frac{\sqrt{A^2-x^2}}{\omega}\frac{\omega}{2\pi}q\,\delta(z)\delta(y) \int_0^T \delta(t-sin^{-1}(x/A)/\omega)\,dt=\frac{\sqrt{A^2-x^2}}{2\pi}q\,\delta(z)\delta(y) [/tex]

    Finally we have:

    [tex]<\rho>=\int \frac{\sqrt{A^2-x^2}}{2\pi}q\,\delta(z)\delta(y)\, dV =\frac{q}{2\pi}\int \sqrt{A^2-x^2}\,dx[/tex]

    This is as far as I go, please help me if there is any mistake and how can I get the final result. Thanks a lot.
  2. jcsd
  3. Sep 10, 2014 #2


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    But [itex] x[/itex] is itself a function of time. In the following steps you proceed as if x was a time independent parameter.

    The easy (and only way, I think) to do it is to do the volume integral before doing the time integral. Then the calculation is trivial.
  4. Sep 10, 2014 #3
    Thanks a lot! It's far easier that way. I have one final question, does this hold:

    [tex]\int f(x) \delta(x^\prime-x(t)) dx^\prime=f(x(t))[/tex]

    Thanks again.
  5. Sep 10, 2014 #4


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    Yes, that's correct. To see this, just imagine picking a certain value of t. Then it is obviously true at that time. But it must be true for any value of t so it is a valid relation even given that x(t) is a function of time.

    You are welcome:cool:
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