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$$\Phi=\iint_{\partial V}\mathbf{g} \cdot d \mathbf{A}=-4 \pi G M$$

Essentially, why is $\Phi$ independent of the distribution of mass inside the surface $\partial V$, and the shape of surface $\partial V$? That is, I'm looking for a mathematical justification for the characterization of flux as a kind of "flow" of gravitational field through a surface.

The reason I worry about that is this. I have encountered the following proof of Gauss' Law for gravity: $$\iint_{\partial V}\frac{-MG}{r^2}\mathbf{e_r} \cdot \mathbf{e_r}d A=\frac{-MG}{r^2}\iint_{\partial V}d A$$ $$=-MG4 \pi$$ However, two assumptions are made in the above proof that I've not convinced myself hold true for the general case:

  • The Gaussian surface is a sphere.

  • It is assumed that all mass is concentrated at the centre of the sphere.

Returning to my original question:

  • Why is it true that when the Gaussian surface is deformed (not adding or removing any mass to inside the surface, of course), the amount of flux, $\Phi$, through the surface is equal to the flux in the spherical case?
  • Why is it true that the positions of the masses inside the surface doesn't affect $\Phi$?

Edit: removed reference to Poisson's equation, changed much of question to make it clearer.

Further edit: I am happy that $$\iint_{\partial V}\mathbf{F}\cdot d \mathbf{A}=\iiint_{ V}( \nabla \cdot \mathbf{F})dV$$

However, my problem is with the following equation: $$\nabla \cdot \mathbf{g}=-\rho G 4 \pi $$

Which can be used to get $$\iint_{\partial V}\mathbf{g}\cdot d \mathbf{A}=\iiint_{ V}(-\rho G 4 \pi)dV=-M G 4 \pi$$

Meow
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2 Answers2

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Maxwell's equations say that the divergence of E is the charge density. By Stokes' theorem, the flux through a surface is equal to the volume integral of the divergence over the volume enclosed. This completes the proof of Gauss' Law.

To show that Coulomb's Law follows, take a point charge and calculate the flux through a spherical shell surrounding it. By symmetry, the E-field must be the same everywhere on the shell and so the flux is proportional to $r^2$ times the E-field strength. By Gauss' Law the surface integral of the E-field must equal the charge enclosed, so $E r^2 \propto q$.

Mark Eichenlaub
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Its because the divergence is undefined at the origin and 0 everywhere else. Thus as long as the volume you integrate over contains the origin, it doesn't matter what region you integrate over.

Ben
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