Proving Divergence Theorem using Gauss' Theorem

[latentcorpse]

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

[itex]\int_V dV e^{-r} \nabla \cdot (\frac{\vec{\hat{r}}}{r^2}) = \int_V dV \frac{e^{-r}}{r^2}[/itex]

any ideas on where to start?

 

[gabbagabbahey]

Hint:The following vector identity should be useful:

[tex]\vec{\nabla}\cdot{f\vec{A}}=f(\vec{\nabla}\cdot\vec{A})+\vec{A}\cdot(\vec{\nabla}f)[/tex]

 

[latentcorpse]

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

 

[gabbagabbahey]

1. No, set [tex]f=e^{-r}[/tex]

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

 

[latentcorpse]

ok so it then equals
[itex] \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}) [/itex]
My instinct is to do the integral over the surface S using S = sphere of radius a as then [itex]\vec{n}=\vec{\hat{r}} [/itex] and i can get rid of that dot product. but then [itex]dS=4 \pi a^2[/itex] 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 [tex]r\to\infty[/tex]...what is the value of the integrand there?

For the second term; Take the gradient of [tex]e^{-r}[/tex] using spherical coords...

 

[latentcorpse]

ok. might be getting somewhere now. on the right i took the grad of the exponential term and got [itex] \nabla e^{-r} = -e^{-r} \vec{\hat{r}}[/itex] is that right?
seems to be so because that causes the integral on the right to simplify substantially to [itex]-\int_V dV (\frac{\vec{\hat{r}}}{r^2}) \cdot (\nabla e^{-r}) = + \int_V dV \frac{e^{-r}}{r^2}[/itex] 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 [itex] e^{-r} \rightarrow 0[/itex] as [itex] r \rightarrow \infty[/itex]. 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 [itex] e^{-r} \rightarrow 0[/itex] as [itex] r \rightarrow \infty[/itex]. 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 [itex]a[/itex] and then take the limit as [itex]a[/itex] 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
[itex]\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 [/itex]
if i take the limit as [itex]a \rightarrow \infty [/itex] i get [itex]\infty[/itex]

 

[latentcorpse]

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

 

[gabbagabbahey]

ok
but then
[itex]\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 [/itex]
if i take the limit as [itex]a \rightarrow \infty [/itex] i get [itex]\infty[/itex]

Along the surface, [itex]r=a[/itex].

 

[latentcorpse]

i don't understand what you mean there sorry

 

[latentcorpse]

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

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

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

 

[gabbagabbahey]

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

[tex]\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} [/tex]

 

[latentcorpse]

cheers buddy!