- #1
Living_Dog
- 98
- 0
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
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
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