3
$\begingroup$

How can it be shown that $$\nabla^2\phi = 4\pi G \rho(\vec{r})$$ ?

$\endgroup$
2

2 Answers 2

4
$\begingroup$

This is Poisson's equation and it derives from Gauss's law:

Since the gravitational field has zero curl (equivalently, gravity is a conservative force) as mentioned above, it can be written as the gradient of a scalar potential, called the gravitational potential:

$$g =-\nabla \phi$$

Then the differential form of Gauss's law for gravity becomes Poisson's equation:

$$\nabla ^{2}\phi =4\pi G\rho$$

This provides an alternate means of calculating the gravitational potential and gravitational field. Although computing g via Poisson's equation is mathematically equivalent to computing g directly from Gauss's law, one or the other approach may be an easier computation in a given situation.

$\endgroup$
0
2
$\begingroup$

Poisson's law is ultimately based on Newton's hypothesis for a law of universal gravitation. Mathematically, the acceleration $\vec{g}(\vec{r})$ caused by some distribution of mass, characterised by a density $\rho(\vec{r}')$ is written $$ \vec{g}(\vec{r}) = -G\int \rho \frac{\left(\vec{r}'-\vec{r}\right)}{|\vec{r}'-\vec{r}|^3}\ dr'^{3}, $$ where $\vec{r}'$ is a position in the source density distribution and the integral is over the entire volume of the source.

The above equation is a hypothesis - it cannot be proved (and turns out to be wrong in detail, which is why is must be replaced with Einstein's field equations).

Taking the divergence of both sides of this equation, and using the vector calculus identity that $$\nabla \cdot \frac{\vec{r}'-\vec{r}}{|\vec{r}'-\vec{r}|^3} = 4\pi \delta(\vec{r}'-\vec{r}),$$ where $\delta$ is the 3D Dirac delta function, we have $$\nabla \cdot \vec{g}(\vec{r}) = -\int 4\pi G \rho \delta(\vec{r}'-\vec{r})\ dr'^3 = -4\pi G \rho$$

If we then define a scalar potential such that $\vec{g}=-\nabla \Phi$ and insert this into the equation above, we get $$\nabla \cdot \nabla \Phi = \nabla^2\Phi = 4\pi G \rho\ .$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .