# Proof of Gauss' Law of gravitation without reference to Newton?

Gauss' Law of gravity is:

$$\bigtriangledown \cdot \mathbf{g}= 4\pi G\rho$$

This can be shown to be equivalent to Newton's Law of gravity via the divergence theorem. However, this does not really constitute a proof. Where does the $4\pi$ come from? I would like to derive Gauss' Law from the notion of solid angle and/or the definition of the scalar potential.

Specifically, just using the following facts:

$$\mathbf{g}=\bigtriangledown\cdot\phi$$

Where $\phi$ is the scalar potential, and the definition of the scalar potential (from the wikipedia article):

$$4\pi\phi=\int_V \frac{\bigtriangledown\cdot\mathbf{g}}{r}\:\:dV$$

(Where the RHS is a volume integral, and the necessary assumptions about asymptotic vanishing towards infinity are made).

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You're trying to take the divergence of a scalar field; In general, for any conservative force $\vec{F}$ there is a scalar potential $V$ such that $\vec{F}=-\nabla V$. I would imagine the $4\pi$ comes from some evaluation of a surface integral over a sphere at some point. –  FireGarden Jan 30 at 9:11
Why "this does not really constitute a proof"? –  Martín-Blas Pérez Pinilla Jan 30 at 9:13
I think you mean $\textbf{g} = \nabla \phi$, not $\nabla \cdot \phi$. –  Paul Siegel Jan 30 at 9:13
Aside from that, I too am curious why you don't like the divergence theorem argument? –  Paul Siegel Jan 30 at 9:14
@PaulSiegel I guess, he even means $\textbf{g} = - \nabla \phi$, which is the usual convention in physics. –  jibe Jan 30 at 10:12