Hi,

I have two energy potentials:

$\displaystyle \phi = \int_{V}\frac{1}{|\mathbf{r}-\mathbf{r}'|}dv$

$\displaystyle \psi = \int_{V}|\mathbf{r}-\mathbf{r}'|dv$

where $\displaystyle \mathbf{r}'$ is a constant vector.

Apparently the following is true:

$\displaystyle \nabla^2 \psi = 2\phi$

$\displaystyle \nabla^4 \psi = 2\nabla^2\phi$

I did this a few years back and since then my math has become rather rusty but if anyone could explain how I got the first equality I could work on the next one.

This is my working without including the constant vector:

Attempt 1:

$\displaystyle \frac{d|\mathbf{r}|}{dx} = \frac{d(\mathbf{r} \cdot \mathbf{r})^{1/2}}{dx} = \frac{\mathbf{r}_{,x}\mathbf{r}}{|\mathbf{r}|}$

where (,x) refers to the partial derivative with respect to x.

But then this doesn't give the desired result when applying it for all three directions.

Attempt 2 (Just blindly differentiate with respect to the magnitude r):

$\displaystyle \frac{d|\mathbf{r}|}{dr} = \frac{d(rr)^{1/2}}{dr} = \frac{r}{|r|}$

differentiating again with respect to r:

$\displaystyle \frac{|r|-\frac{r}{|r|}r}{|r|^2}=0$

I've also tried the vector laplacian but it still didn't work.

Thanks for any help.

K

EDIT: I don't know why I did attempt 2 that is just 1 right off the bat.