How can we get this formula for Gravitational potential?

Question

I just read the wikipedia page

http://en.wikipedia.org/wiki/Gravitational_potential

But I don't understand how to get this formula:

$$\rho(\mathbf{x}) = \frac{1}{4\pi G}\Delta V(\mathbf{x})$$

Can anyone tell me why?

Answer 1

This equation comes most simply from the following two: $$ {\bf g}=-\nabla V, $$ $$ \nabla\cdot{\bf g}=-4\pi G\rho. $$ Plut the first one into the second, and you're done.

Of course, the natural next question is where those two equations come from. The first one is the relationship between the gravitational field ${\bf g}$ and the gravitational potential $V$. It's pretty much the definition of gravitational potential.

The second one is Gauss's Law for gravitational fields. It follows from the inverse-square law. Wikipedia gives a derivation. (That derivation is for electric fields and potentials, rather than gravitational ones, but if you just replace ${\bf E}$ with $-{\bf g}$ and "charge" with "mass," it's exactly the same.)

Answer 2

$\Delta=\nabla^2$ is the Laplace operator. So from $V(\textbf{x})=-G\int_{\mathbb{R}^3}\frac{1}{|\textbf{x}-\textbf{r}|}\rho(\textbf{r})dv(\textbf{r})$ we have:

$\Delta V(\textbf{x})=-G\int_{\mathbb{R}^3}\left(\nabla\cdot\nabla\frac{1}{|\textbf{x}-\textbf{r}|}\right)\rho(\textbf{r})dv(\textbf{r})$

But it's easy to show that $\nabla\frac{1}{|\textbf{x}-\textbf{r}|}=-\frac{\textbf{x}-\textbf{r}}{|\textbf{x}-\textbf{r}|^3}$ and $\nabla\cdot\left(\frac{\textbf{x}-\textbf{r}}{|\textbf{x}-\textbf{r}|^3}\right)=4\pi\delta^3(\textbf{x}-\textbf{r})$

(tell me if you need help with this)

Therefore

$\Delta V(\textbf{x})=G\int_{\mathbb{R}^3}4\pi\delta^3(\textbf{x}-\textbf{r})\rho(\textbf{r})dv(\textbf{r})=4\pi G \rho(\textbf{x})$