# Verify about the solution of wave equation of potential.

1. Dec 12, 2012

### yungman

I read in the book regarding a point charge at the origin where $Q(t)= \rho_{(t)}Δv'\;$. The wave eq is.

$$\nabla^2V-\mu\epsilon\frac{\partial^2 V}{\partial t^2}= -\frac {\rho_v}{\epsilon}$$

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

$$\nabla^2V=\frac 1 {R^2}\frac {\partial}{\partial R}\left( R^2 \frac {\partial V}{\partial R}\right)$$

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

My question is this:

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

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

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. Dec 12, 2012

### elfmotat

Yep, you're right. Nothing complicated here.

3. Dec 12, 2012

### yungman

Thanks

Alan

4. Dec 13, 2012

### yungman

I have another question follow up with the original wave equation:

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

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

$$V_{(R,t)}=\frac 1 R U_{(R,t)}$$

This will reduce the wave equation to:

$$\frac {\partial^2U_{(R,t)}}{\partial R^2}-\mu\epsilon\frac {\partial^2U_{(R,t)}}{\partial t^2}=0$$

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

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 $\hat R$. So the formula become:

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

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

$$U_{(R,t)}= V_0^+ e^{-δ R}+V_0^- e^{δ R}$$

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

$$U_{(R,t)}= V_0^+ e^{-δ R}$$

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
5. Dec 13, 2012

### andrien

$\frac {\partial^2U_{(R,t)}}{\partial R^2}-\mu\epsilon\frac {\partial^2U_{(R,t)}}{\partial t^2}= \frac {\partial^2U_{(R,t)}}{\partial R^2}+\omega^2\mu\epsilon U= 0.$
just from the last in it you can write the periodic solution.Don't substitute any δ after.

6. Dec 13, 2012

### yungman

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:

$$U_{(R,t)}= V_0^+ e^{\alpha R} e^{j\beta R}\; \hbox { where } δ=\alpha + j\beta$$

7. Dec 13, 2012

### andrien

a plus sign before ω2 in eqn is necessary for periodic solution.First write solution in terms of ω and then introduce other definitions.

8. Dec 13, 2012

### yungman

This is because for lossy dielectric medium ε=ε'+jε'' therefore $δ=j\omega \sqrt{\mu(ε'+jε'')}$ is a complex number where $δ=\alpha + j \beta$. Solution of:

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

This is a decay periodic function.

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

Then

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

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
9. Dec 13, 2012

### andrien

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.

10. Dec 13, 2012

Thanks.