Using Gauss' Theorem to Show Integral Convergence in Divergence Theorem

[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

 

[gabbagabbahey]

My initial plan of attack was to write the integral as

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

Careful;

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

 

[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 ...

 

[gabbagabbahey]

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 ...

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 $r\to\infty$?

 

[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?

 

[gabbagabbahey]

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

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?

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

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.

 

[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.

 

[gabbagabbahey]

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

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 $$\nabla\psi=\frac{-k}{r^2}\sim\frac{1}{r^2}$$.

 

[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 $\chi = \frac{1}{r}$

 

[gabbagabbahey]

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 $\chi = \frac{1}{r}$

Why not?

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

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

and

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

 

[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!

 

[gabbagabbahey]

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

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.

(ii)also $\frac{1}{r^2}$ doesn't specify a direction whereas it should be in the {latex] \vec{e_r} [/itex] direction, no?

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.

(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!

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,

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

 

[latentcorpse]

thanks a lot mate

 

[gabbagabbahey]

Welcome