Verify about the solution of wave equation of potential.

  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?


  2. jcsd
  3. Yep, you're right. Nothing complicated here.
  4. Thanks

  5. 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.


    Last edited: Dec 13, 2012
  6. [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. 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. 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. 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]


    [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. 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. Thanks.
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?