# Complex vector potential of solid sphere and Heaviside function

1. Feb 19, 2013

### Shinobii

1. The problem statement, all variables and given/known data

I have done this problem for the case of a spherical shell, however, I am not understanding how to go about this for a solid sphere.

2. Relevant equations

$$\vec{A} = \frac{1}{4 \pi} \int_{\phi' = 0}^{2 \pi} \int_{-1}^1 \int_0^R \rho_o \Theta(R-r) \sum_{l=0}^\infty \sum_{m=-l}^l \frac{4 \pi}{2l +1} \frac{r_<^l}{r_>^{l+1}}Y_{l,m}^*(\theta',\phi')Y_{l,m}(\theta,\phi) r'^2 d(\cos(\theta'))d\phi'$$

3. The attempt at a solution

For the case of a shell, there is a delta function which makes life easy.

My question is, what do I do with this Heaviside function? Do I treat it differently for r < R and r > R? This is my first time encountering this function.

Any hints would be greatly appreciated.

2. Feb 20, 2013

### vela

Staff Emeritus
You're missing a dr in your integral. Given the limits on your integral, you always have $r\le R$, so the step function is equal to 1.

3. Feb 20, 2013

### Shinobii

Hi, sorry I meant to put in the dr'.

I guess when I am integrating from 0 to r > R, I would split the integral up and only the integral from 0 to R would contribute while the other integral (R to r) would be zero.

4. Feb 20, 2013

### vela

Staff Emeritus
What are the limits on your r' integral supposed to be? Are they 0 and R as you've written? Should the r inside the step function be r'?

5. Feb 20, 2013

### Shinobii

Yes sorry, the r is r' inside the Heaviside function.

Do we split up the integral like the case of the electric potential? If so,

For r<R:

The integral, like solving for the electric potential, would have to broken up into two components, that is $\int_{\infty}^r \rightarrow \int_{\infty}^R + \int_R^r.$ Would the first integral = 0 by definition of the Heaviside function?

For r>R:

The integral is from $\int_{\infty}^r$ and therefore 0?

Again solving it for the shell case none of this was an issue...

6. Feb 20, 2013

### vela

Staff Emeritus
I'm still unclear on what you're calculating here. The righthand side doesn't appear to be a vector. Where are you getting $r$ and $\infty$ for your limits?

7. Feb 20, 2013

### Shinobii

Yeah, I messed up real good. I forgot the cross product $(\omega \times r')$ from the velocity.

Using the complex current density,

$$J_x + iJ_y = -i \sqrt{\frac{3}{2\pi}}\frac{q \omega r'}{ R^3} \Theta(R-r')Y_{1,1}.$$

And using this to find the complex vector potential I get,

$$A_x + iA_y = -i\sqrt{\frac{3}{2\pi}}\frac{q \omega}{ R^3} \int_{\phi' = 0}^{2 \pi} \int_{-1}^1 \int_0^r \Theta(R-r')Y_{1,1}(\theta',\phi') \sum_{l=0}^\infty \sum_{m=-l}^l \frac{4 \pi}{2l +1} \frac{r_<^l}{r_>^{l+1}}Y_{l,m}^*(\theta',\phi')Y_{l,m}(\theta,\phi) r'^3 dr'd(\cos(\theta'))d\phi'$$

From here I believe I can use orthogonality to kill off some terms setting l=m=1, and I get

$$A_x + iA_y = \sqrt{\frac{3}{2\pi}}\frac{q \omega}{ R^3} \int_0^r \Theta(R-r')Y_{1,1}(\theta',\phi') \frac{4 \pi}{2(1) +1} \frac{r_<^{(1)}}{r_>^{(1)+1}} r'^3 dr'$$

I hope this is right thus far. . . And yes I am clearly not understanding what I am integrating over for the r' integration.

8. Feb 22, 2013

### vela

Staff Emeritus
You want to integrate over all space, so the limits should be 0 to infinity. Since the step function vanishes for r'>R, you can simply integrate from 0 to R.

9. Feb 23, 2013

### Shinobii

You are correct, thanks! Sorry again for my sloppy mathematics, it's my pet peeve yet I am sloppy myself...

10. Feb 23, 2013

### Shinobii

One more quick question on notation, would this be correct?

$$\int_0^{\infty} r' \, \Theta(R-r') dr' \rightarrow \int_0^{R} r' \, dr'.$$

Would I simply change the limits of integration and remove the Heaviside function from the equation?

11. Mar 4, 2013

### vela

Staff Emeritus
Sorry for the late response. Yes, that's correct.