# Divergence Theorem Question

1. Jan 13, 2009

### latentcorpse

This may well be the wrong place to post this so apologies for that if it's the case.

Anyway, I'm stuck on this question, any help appreciated

Use Gauss' Theorem to show that:
(i) If $\psi(\mathbf{r}) ~ \frac{1}{r}$ as $r \rightarrow \infty$,

then,
$\int_V {\psi \nabla^{2} \psi} dV \less 0$ where the integral is over all space.

Hint: take $\mathbf{V}=\psi \nabla \psi$

My initial plan of attack was to write the integral as

$\int_V {\nabla \cdot \mathbf{V}} dV$

which then becomes using the divergence theorem

$\oint_S {\mathbf{V} \cdot \mathbf{n}} dS$

This probably isn't right because I then have no clue how to show this is less than 0

2. Jan 13, 2009

### gabbagabbahey

Careful;

$$\nabla \cdot \mathbf{V}=\nabla \cdot (\psi \nabla \psi)=\psi (\nabla \cdot \nabla \psi)+\nabla\psi \cdot \nabla \psi \neq \psi\nabla^2\psi$$

3. Jan 13, 2009

### latentcorpse

Thanks,
ok so could i make it
$\int_V {\psi \nabla^{2} \psi} dV = \int_V {\nabla \cdot \mathbf{V} - \nabla \psi \cdot \nabla \psi} dV = \oint_S {V \cdot \mathbf{n}} dS - \int_V {\nabla \psi \cdot \nabla \psi} dV$

Now I guess I need to show $\oint_S {V \cdot \mathbf{n}} dS < \int_V {\nabla \psi \cdot \nabla \psi} dV$ in order to prove that $\int_V {\psi \nabla^{2} \psi} dV <0$

Any advice on how to do that ....

4. Jan 13, 2009

### gabbagabbahey

First note that $\nabla \psi \cdot \nabla \psi=||\nabla \psi||^2\geq 0$

Then use the fact that your surface integral is at the boundary of all space (i.e. at $r\to\infty$), where $$\psi\sim\frac{1}{r}$$ and hence $$\nabla\psi\sim\frac{1}{r^2}$$

What does that make your surface integral for $r\to\infty$?

5. Jan 13, 2009

### latentcorpse

ahh ok. how's this then
$\oint_S \mathbf{V} \cdot \mathbf{n} dS = \oint_S (\frac{1}{r} \frac{1}{r^2}) \cdot \mathbf{n} dS =\oint_S \frac{1}{r^3} \cdot \mathbf{n} dS$

and
$\int_V {|| \nabla \psi ||}^{2} dV=\int_V {\frac{1}{r^4} dV$

sorry but i still don't see how this is going to show it's less than 0.

Also if $\psi = \frac{1}{r} = (r_j r_j)^{-\frac{1}{2}}$ i get that $\nabla \psi =-\frac{1}{2} (r_j r_j)^{-\frac{3}{2}} 2r_i = -\frac{r_i}{r^3}$ as opposed to $\nabla \psi = \frac{1}{r^2}$. I assume this has something to do with it being at the boundary of all space. Can you explain this?

6. Jan 13, 2009

### gabbagabbahey

Sure, and since $dS\sim r^2$, the integrand $\sim \frac{1}{r}$ And since$r\to\infty$ for the entire surface, that means the integrand (and hence the integral) is zero doesn't it?

All you need to know is that $$\int_V {|| \nabla \psi ||}^{2} dV>0$$ since the integrand is positive definite.

That gives you zero minus some positive number, which is of course less than zero.

7. Jan 13, 2009

### latentcorpse

cheers. could you explain that bit about how you got $\nabla \phi$ to be $\frac{1}{r^2}$ though. still a bit confused there.

8. Jan 13, 2009

### gabbagabbahey

Another way to write $\psi \sim \frac{1}{r}$ is $\psi=\frac{k}{r}$ for some unknown constant 'k'...then just take the gradient in spherical coordinates: you should end up with $$\nabla\psi=\frac{-k}{r^2}\sim\frac{1}{r^2}$$.

9. Jan 13, 2009

### latentcorpse

in spehrical polars the gradient is $\nabla \chi = (\frac{\partial \chi}{\partial r},\frac{1}{r} \frac{\partial \chi}{\partial \theta}, \frac{1}{\sin{\theta}} \frac{\partial \chi}{\partial \phi})$

This doesn't give me what I want when I let $\chi = \frac{1}{r}$

10. Jan 13, 2009

### gabbagabbahey

Why not?

$$\frac{\partial}{\partial\theta} \left(\frac{1}{r}\right)=0$$

$$\frac{\partial}{\partial\phi} \left(\frac{1}{r}\right)=0$$

and

$$\frac{\partial}{\partial r} \left(\frac{1}{r}\right)=\frac{-1}{r^2}$$

11. Jan 13, 2009

### latentcorpse

ok so then $\nabla \psi = (-\frac{1}{r^2},0,0) = -\frac{1}{r^2} \vec{e_r}$
Problems I have with this are threefold:
(i)first of all this is negative and you said it was $\frac{1}{r^2}$ which is positive
(ii)also $\frac{1}{r^2}$ doesn't specify a direction whereas it should be in the $\vec{e_r}$ direction, no?

(iii)furthermore this answer disagrees with the one i got in post 5 in this thread when I took the grad in cartesians - this is the most worrying part!!!

12. Jan 13, 2009

### gabbagabbahey

It doesn't matter whether its positive or negative: $$\lim_{r\to \infty} \frac{k}{r}=0$$ whether k is positive or negative and that's all the final conclusion rests on.

Okay, so to be more accurate you could say that $$\nabla \psi \sim \frac{-1}{r^2} \vec{e_r}$$ But the surface normal is also radial so you still end up with the integrand being proportional to 1/r, and hence being zero on the boundary of all space.

No it doesn't. Remember,

$$\vec{e_r}=\frac{x\vec{e_x}+y\vec{e_y}+z\vec{e_z}}{\sqrt{x^2+y^2+z^2}}=\sum_i \frac{\vec{r_i}}{r}$$

And so,

$$\nabla\psi= \frac{-1}{r^2} \vec{e_r}=\sum_i \frac{-\vec{r_i}}{r^3}$$

13. Jan 13, 2009

### latentcorpse

thanks a lot mate

14. Jan 13, 2009

Welcome