Proving Divergence Theorem using Gauss' Theorem

[latentcorpse]

I need to show that, using Gauss' Theorem (Divergence Theorem), i.e. integration by parts, that:

$\int_V dV e^{-r} \nabla \cdot (\frac{\vec{\hat{r}}}{r^2}) = \int_V dV \frac{e^{-r}}{r^2}$

any ideas on where to start?

 

[gabbagabbahey]

Hint:The following vector identity should be useful:

$$\vec{\nabla}\cdot{f\vec{A}}=f(\vec{\nabla}\cdot\vec{A})+\vec{A}\cdot(\vec{\nabla}f)$$

 

[latentcorpse]

thanks.
1, do i set $f=\frac{1}{r^2}$?
2, how did you know to do that so quickly?

 

[gabbagabbahey]

1. No, set $$f=e^{-r}$$

2. It's a common theme in many EM derivations.

 

[latentcorpse]

ok so it then equals
$\int_V dV \nabla \cdot (\frac{e^{-r} \vec{\hat{r}}}{r^2}) - \int_V dV (\frac{\vec{\hat{r}}}{r^2}) \cdot (\nabla e^{-r}) = \int_S (\frac{e^{-r} \vec{\hat{r}}}{r^2}) \cdot \vec{n} dS - \int_V dV (\frac{\vec{\hat{r}}}{r^2}) \cdot (\nabla e^{-r})$
My instinct is to do the integral over the surface S using S = sphere of radius a as then $\vec{n}=\vec{\hat{r}}$ and i can get rid of that dot product. but then $dS=4 \pi a^2$ which messes stuff up and also what do I do on the right?

 

[gabbagabbahey]

I'm assuming that the original integral was over all of space, correct?

If so, then your surface integral is over the boundary of all space which is at $$r\to\infty$$...what is the value of the integrand there?

For the second term; Take the gradient of $$e^{-r}$$ using spherical coords...

 

[latentcorpse]

ok. might be getting somewhere now. on the right i took the grad of the exponential term and got $\nabla e^{-r} = -e^{-r} \vec{\hat{r}}$ is that right?
seems to be so because that causes the integral on the right to simplify substantially to $-\int_V dV (\frac{\vec{\hat{r}}}{r^2}) \cdot (\nabla e^{-r}) = + \int_V dV \frac{e^{-r}}{r^2}$ which is what we're looking for - therefore it appears to just be a matter of showing the integral on the left vanishes?

 

[latentcorpse]

indeed it is over all space. i would then imagine it would zero because $e^{-r} \rightarrow 0$ as $r \rightarrow \infty$. If this is the case, how owuld you recommend I write that in order to be an acceptable answer? Also, I took the grad of the exponential using Cartesians - does that matter?

 

[gabbagabbahey]

indeed it is over all space. i would then imagine it would zero because $e^{-r} \rightarrow 0$ as $r \rightarrow \infty$. If this is the case, how owuld you recommend I write that in order to be an acceptable answer? Also, I took the grad of the exponential using Cartesians - does that matter?

Formally, you could start by assuming that your volume is a sphere of radius $a$ and then take the limit as $a$ approaches infinity.

And it doesn't matter that you used Cartesian coords to take the gradient; although it would probably have been quicker to use spherical coords.

 

[latentcorpse]

ok
but then
$\int_S (\frac{e^{-r} \vec{\hat{r}}}{r^2}) \cdot \vec{n} dS = \int_S (\frac{e^{-r} \vec{\hat{r}} \cdot \vec{\hat{r}}}{r^2}) dS= \frac{e^{-r}}{r^2} 4 \pi a^2$
if i take the limit as $a \rightarrow \infty$ i get $\infty$

 

[latentcorpse]

unless i take $dS=4 \pi r^2$ and then the intergal becomes $4 \pi e^{-r}$ and then the limit as $r \rightarrow \infty$ would be 0. Any advice? Cheers.

 

[gabbagabbahey]

ok
but then
$\int_S (\frac{e^{-r} \vec{\hat{r}}}{r^2}) \cdot \vec{n} dS = \int_S (\frac{e^{-r} \vec{\hat{r}} \cdot \vec{\hat{r}}}{r^2}) dS= \frac{e^{-r}}{r^2} 4 \pi a^2$
if i take the limit as $a \rightarrow \infty$ i get $\infty$

Along the surface, $r=a$.

 

[latentcorpse]

i don't understand what you mean there sorry

 

[latentcorpse]

actually surely if i integrate over a sphere of radius a, $dS=a^2 \sin{\theta} d\theta d\phi$?

in which case this integral becomes $\int_S \frac{e^{-r} a^2}{r^2} \sin{\theta} d\theta d\phi = 4 \pi \frac{a^2}{r^2} e^{-r}$

would i then let a or r go to infinity, or is this all complete bollocks? HELP I am really confused now! arghhhh!

 

[gabbagabbahey]

$r$ is the distance from the origin to the infinitesimal area element $dS$. Since the entire surface is a distance $a$ from the origin, that means that $r=a$ everywhere on the surface.

$$\implies \int_S \frac{e^{-r}}{r^2}dS=\int_S \frac{e^{-a}}{a^2} dS =\int_S \frac{e^{-a} a^2}{a^2} \sin{\theta} d\theta d\phi = 4 \pi e^{-a}$$

 

[latentcorpse]

cheers buddy!