Delta function potential

1. Oct 24, 2009

realcomfy

I am trying to integrate a charge density over a volume in order to obtain a total charge, but there is a delta function involved and I am not entirely sure how to treat it.

$$\rho = q* \delta (\textbf{r})- \frac {q\mu^{2} Exp(- \mu r)} {4 \pi r}$$

$$Q = \int \rho (\textbf{r})d^{3}r$$

$$d^{3}r = r^{2} dr d(cos \theta )d \phi$$

Plugging the density above in the the Q equation should give me the integral over a delta function times the differential volume element minus the integral of the uglier function times the same volume element.

All this is good and fine except that I get that $$\int q \delta (\textbf{r}) r^{2} drd(cos (\theta)d \phi$$ and I'm not totally sure how to evaluate it. I am pretty sure it should be something like 4 $$\pi$$ times the function r evaluated at zero, but I can't find the right answer with this approach. Any help would be appreciated.

2. Oct 24, 2009

Bill Foster

Are you familiar with the properties of the delta function? In particular, this one?

$$\int_{-\infty}^{+\infty}f\left(x\right)\delta\left(x-x'\right)dx = f\left(x'\right)$$

3. Oct 24, 2009

realcomfy

Ya, I got that down. So I was thinking that this integral should be equal to zero because we are evaluating r at 0. But I wasn't sure if that was right or if since we are integrating over a delta function, the whole thing should be equal to q. or $$4 \pi q$$. Ya, I'm a little confused

4. Oct 24, 2009

Bill Foster

It looks like you'll get two integrals. One that has the delta function in it. That one, when you integrate it, will be zero. But the other one should give you some value.

5. Oct 24, 2009

Avodyne

The delta function is a three dimensional delta function

$$\delta(\textbf{r})=\delta(x)\delta(y)\delta(z)$$

which obeys

$$\int d^3r\,\delta(\textbf{r}) = 1.$$

6. Oct 24, 2009

Winzer

What volume are you exactly integrating over? Why $$d^{3}r = r^{2} dr d(cos \theta )d \phi$$.
Were you thinking spherical coordinates?
$$d^{3}r = r^{2} sin \phi dr d\theta d\phi$$
That would make things simpler.

7. Oct 24, 2009

jdwood983

Actually, using $$d(\cos\theta)$$ is more simple than $$\sin\theta d\theta$$ when there are cosine terms in the integrand. They do actually work out to the same thing, but in electric potential theory, due to the prevalence of cosine terms in the integrands, using $$d(\cos\theta)$$ is the better choice.

8. Nov 16, 2009

realcomfy

Thanks, I think I get it now