1. Limited time only! Sign up for a free 30min personal 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!

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

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

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

Have something to add?
Draft saved Draft deleted



Similar Discussions: Electrodynamics of a moving particle
  1. Moving particle decay (Replies: 1)

Loading...