I just read the wikipedia page


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?


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.)


$\Delta=\nabla^2$ 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)


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

  • $\begingroup$ How to get $\nabla\cdot\left(\frac{\textbf{x}-\textbf{r}}{|\textbf{x}-\textbf{r}|^3}\right)=4\pi\delta^3(\textbf{x}-\textbf{r})$ ? $\endgroup$ – NGY Jul 8 '11 at 0:58
  • $\begingroup$ @NGY This document gives a semi-rigorous proof. $\endgroup$ – mmc Jul 8 '11 at 1:41
  • $\begingroup$ OK, thanks. And I have another question: mathematically, how to ensure the integral $V(\textbf{x})=-G\int_{\mathbb{R}^3}\frac{1}{|\textbf{x}-\textbf{r}|}\rho(\textbf{r})dv(\textbf{r})$ converges? For example, if $\rho(\textbf{r})\equiv\text{Constant}$(when $\textbf{r}$ near $\textbf{x}$), it diverges. $\endgroup$ – NGY Jul 8 '11 at 2:44
  • $\begingroup$ @NGY I don't understand your example. Are you speaking about a density $\rho$ that is constant over all space? $\endgroup$ – mmc Jul 8 '11 at 3:37
  • $\begingroup$ @mmc Sorry, I misunderstood the meaning of $\rho$(Ignore the physical background, in mathematics, if $\rho(\mathbf{r})$ is constant when $\mathbf{r}$ is near $\mathbf{x}$, and vanishes when $\mathbf{r}$ is far from $\mathbf{x}$, then the integral diverge.). Now I understand it. $\endgroup$ – NGY Jul 8 '11 at 3:47

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