Gauss’s Flux Theorem for Gravity and Newton’s Law

Gauss’s flux theorem for gravity (also known as Gauss’s law for gravity) in differential form says that

\nabla \cdot \mathbf{g} = -4 \pi G \rho

In this note I want to show that one can get quite far towards deriving Gauss’s law for gravity without knowing Newton’s law of universal gravitation, but not all the way. To explore this, suppose that all we know is that the gravitational force depends on mass and radial distance:

\mathbf{g} = k(M, r) \cdot \mathbf{e_r}

Here, k is an unspecified function, M is a mass which can be taken as being located at the origin, r is the radial distance from the origin, and \mathbf{e_r} is a radial unit vector.

Now we imagine a closed spherical surface \delta V of radius r centered at the origin. The total flux of the gravitational field \mathbf{g} over the closed surface \delta V is

\oint_{\delta V} \mathbf{g} \cdot d\mathbf{A} = \oint_{\delta V} k(M, r) \cdot \mathbf{e_r} \cdot d\mathbf{A}

= k(M, r) \oint_{\delta V} \mathbf{e_r} \cdot \mathbf{e_r} \cdot dA

= k(M, r) \oint_{\delta V} dA

= k(M, r) \cdot 4 \pi r^2 (this explains where the 4 \pi comes from).

The total flux is independent of r so to eliminate r^2 we must have k(M, r) = \frac{k^*(M)}{r^2} where k^*(M) is some unspecified function of M, and therefore

\oint_{\delta V} \mathbf{g} \cdot d\mathbf{A} = k^*(M) \cdot 4 \pi

By the divergence theorem we can write this as

\oint_{V} \nabla \cdot \mathbf{g} dV = k^*(M) \cdot 4 \pi

and therefore differentiating both sides with respect to V we get

\nabla \cdot \mathbf{g} = k^{* \prime}(M)\frac{dM}{dV} 4 \pi

If we set \frac{dM}{dV} = \rho we see that this is *nearly* Gauss’s law:

\nabla \cdot \mathbf{g} = k^{* \prime}(M) 4 \pi \rho

We only need Newton’s law to tell us that k^{* \prime}(M) = -G at this final stage.

The link with the scalar potential comes through \mathbf{g} = -\nabla \phi which gives us

\nabla^2 \phi = 4 \pi G \rho

(a well known type of partial differential equation known as Poisson’s equation).