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Verify about the solution of wave equation of potential.

  1. Dec 12, 2012 #1
    I read in the book regarding a point charge at the origin where [itex]Q(t)= \rho_{(t)}Δv'\;[/itex]. The wave eq is.

    [tex]\nabla^2V-\mu\epsilon\frac{\partial^2 V}{\partial t^2}= -\frac {\rho_v}{\epsilon}[/tex]

    For point charge at origin, spherical coordinates are used where:

    [tex] \nabla^2V=\frac 1 {R^2}\frac {\partial}{\partial R}\left( R^2 \frac {\partial V}{\partial R}\right)[/tex]

    This is because point charge at origin, [itex]\frac {\partial}{\partial \theta} \hbox{ and }\; \frac {\partial}{\partial \phi}[/itex] are all zero.

    My question is this:

    The book then said EXCEPT AT THE ORIGIN, V satisfies the following homogeneous equation:

    [tex]\frac 1 {R^2}\frac {\partial}{\partial R}\left( R^2 \frac {\partial V}{\partial R}\right)-\mu\epsilon \frac {\partial^2 V}{\partial t^2}=0[/tex]

    The only reason I can think of why this equation has to exclude origin is because R=0 and origin and this won't work. Am I correct or there's another reason?

    Thanks

    Alan
     
  2. jcsd
  3. Dec 12, 2012 #2
    Yep, you're right. Nothing complicated here.
     
  4. Dec 12, 2012 #3
    Thanks

    Alan
     
  5. Dec 13, 2012 #4
    I have another question follow up with the original wave equation:

    [tex]\frac 1 {R^2}\frac {\partial}{\partial R}\left( R^2 \frac {\partial V}{\partial R}\right)-\mu\epsilon \frac {\partial^2 V}{\partial t^2}=0[/tex]

    According to the book, to simplify this equation, Let:

    [tex]V_{(R,t)}=\frac 1 R U_{(R,t)}[/tex]

    This will reduce the wave equation to:

    [tex]\frac {\partial^2U_{(R,t)}}{\partial R^2}-\mu\epsilon\frac {\partial^2U_{(R,t)}}{\partial t^2}=0[/tex]

    One of the solution is [itex]U_{(R,t)}=f\left(t-\frac R {\mu\epsilon}\right)[/itex]

    My question is I want to find the solution if U(R,t) is time harmonic which is a continuous sine wave. The book does not show the anything. So I am using the regular solution of time harmonic wave equation and using direction to be [itex]\hat R[/itex]. So the formula become:

    [tex]\frac {\partial^2U_{(R,t)}}{\partial R^2}-\mu\epsilon\frac {\partial^2U_{(R,t)}}{\partial t^2}= \frac {\partial^2U_{(R,t)}}{\partial R^2}+\omega^2\mu\epsilon U= 0.[/tex]
    [tex] δ^2=-\omega^2\mu\epsilon\;\Rightarrow\; \frac {\partial^2U_{(R,t)}}{\partial R^2}-δ^2 U= 0.[/tex]

    This is a 2nd order ODE with constant coef and the solution is:

    [tex]U_{(R,t)}= V_0^+ e^{-δ R}+V_0^- e^{δ R}[/tex]

    For potential that reach to infinity space, there will be no reflection so the second term disappeared, whereby:

    [tex]U_{(R,t)}= V_0^+ e^{-δ R}[/tex]

    Do I get this right? This might look very obvious, but I am working on retarded potential and I am interpreting what the book don't say, I have to be careful to make sure I get everything right.

    Thanks

    Alan
     
    Last edited: Dec 13, 2012
  6. Dec 13, 2012 #5
    [itex]\frac {\partial^2U_{(R,t)}}{\partial R^2}-\mu\epsilon\frac {\partial^2U_{(R,t)}}{\partial t^2}= \frac {\partial^2U_{(R,t)}}{\partial R^2}+\omega^2\mu\epsilon U= 0.[/itex]
    just from the last in it you can write the periodic solution.Don't substitute any δ after.
     
  7. Dec 13, 2012 #6
    Thanks for the reply, I just finished updating the last post, please take a look again. The reason I use δ is because this will be used in phasor form which is going to be in form of:

    [tex]U_{(R,t)}= V_0^+ e^{\alpha R} e^{j\beta R}\; \hbox { where } δ=\alpha + j\beta[/tex]
     
  8. Dec 13, 2012 #7
    a plus sign before ω2 in eqn is necessary for periodic solution.First write solution in terms of ω and then introduce other definitions.
     
  9. Dec 13, 2012 #8
    This is because for lossy dielectric medium ε=ε'+jε'' therefore [itex] δ=j\omega \sqrt{\mu(ε'+jε'')}[/itex] is a complex number where [itex]δ=\alpha + j \beta[/itex]. Solution of:

    [tex]\frac {\partial ^2 U}{\partial R^2} -δ^2U=0 \hbox { is }\; U_{(R,t)}= U_0^+ e^{-δR}\;=\; U_0^+ e^{-\alpha R} e^{-j\beta R}[/tex]

    This is a decay periodic function.

    For lossless, we use [itex] k=\omega\sqrt{\mu\epsilon}\hbox { where ε is real.}[/itex]

    Then

    [tex]\frac {\partial ^2 U}{\partial R^2} +k^2U=0 \hbox { is }\; U_{(R,t)}= U_0^+ e^{-jkR}[/tex]

    Actually because you brought this up, I dig deep into this and answer the question why the books use δ for lossy media and k for lossless media. I never quite get this all these years until now.
     
    Last edited: Dec 13, 2012
  10. Dec 13, 2012 #9
    that is why it is better to go with + sign.but it is ok,now.In case,ε is complex it is still possible to define things after.But you already got it.
     
  11. Dec 13, 2012 #10
    Thanks.
     
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