3-D Dirac Delta Fnc (DDF) Evaluated

[Living_Dog]

Hi,

After reading DJGriffiths sections on the DDF I have a question about evaluating it in regards Prob. 1.46 (a), to wit:

"Write an expression for the electric charge density $$\rho(\vec{r})$$ of a point charge q at $$\vec{r}$$'. Make sure that the volume integral of $$\rho$$ equals q."

This is easily done in rectangular coordinates, namely, let $$\rho(\vec{r}) = q\delta(\vec{r}-\vec{r}')$$.

Checking that this resultsin a volume integral of q I did:

$$\int_V \rho(\vec{r}) d\tau = \int_V q\delta^3(\vec{r}-\vec{r}') d\tau = q \int_V \delta^3(\vec{r}-\vec{r}') d\tau = q \cdot 1 = q.$$

Here is my question: doing the integral as is I get the correct answer. However, when I wrote it out in spherical coord's. It seems that I should rewrite $$\rho(\vec{r})$$ as $$\frac{q}{4\pi}$$?

Why? Because:

$$\int_V \rho(\vec{r}) d\tau = \frac{q}{4\pi} \int_V \delta^3(\vec{r}-\vec{r}') dr d\Omega = \frac{q}{4\pi} \cdot \int_0^R \delta^3(\vec{r}-\vec{r}') dr \cdot \int_0^{4\pi} d\Omega = \frac{q}{4\pi} \cdot 4\pi \cdot \int_0^R \delta^3(\vec{r}-\vec{r}') dr = q \int_0^R \delta^3(\vec{r}-\vec{r}') dr = q$$.

So my question is: is this second form of evaluation correct? And if it is, then why does $$\rho$$ have two different expressions? Also, what about other curvilinear coordinates? Shouldn't his expression be independent of the geometry being used? So I suspect there is something wrong with my second evaluation. Can you not integrate over $$\delta^3$$ over one dimension?

tia!
-LD

EDIT: I see it now. I was missing the fact that $$\delta^3(\vec{r}-\vec{r}') = \delta(r-r')\delta(\theta-\theta')\delta(\phi-\phi')$$ and so the integral over $$\Omega$$ is still 1 and not $$4\pi$$.

PS: I wasn't sure if I should have deleted this post having figured out the correct answer, or leave it and post what I found out. Let me know and I will do whatever is the correct procedure here.

thx!
-LD

 

[HallsofIvy]

You are integrating the delta function incorrectly in your last formula. Since it is a 3 dimensional delta, its integral over all 3 dimensions, not just r, is equal to 1.

 

[Living_Dog]

You are integrating the delta function incorrectly in your last formula. Since it is a 3 dimensional delta, its integral over all 3 dimensions, not just r, is equal to 1.

:) You caught me mid-edit! :)

but thanks for the confirmation! :!)

-LD