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!