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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 $$\iint_{\partial V}\mathbf{g}\cdot d \mathbf{A}=\iiint_{ V}(-\rho G 4 \pi)dV=-M G 4 \pi$$

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I'm a bit confused about what your question is. Poisson's equation essentially directly follows from empirical facts about electric fields (such as the expression for the electric field of a point charge, and the principle of superposition). Given Poisson's equation, one can mathematically prove Gauss's law, and Gauss's law immediately implies electric flux through a closed surface depends only on the total enclosed charge. If you knew all of this, then are looking for a mathematical justification for the characterization of flux as a kind of "flow" of electric field through a surface? – joshphysics Mar 10 '13 at 22:54
Related electrostatic analog: How to prove Gauss' law from Coulomb's law: – Qmechanic Mar 10 '13 at 22:55
What do you mean "proof of Poisson's equation"? Poisson's equation is... a differential equation. Like the diffusion equation. It's just there in some Platonic sense without any need for proof. What you could mean conceivably is "proof of the uniqueness of solutions to Poisson's equation given boundary conditions," or "derivation of Poisson's equation for the electric potential given something more basic (which you might take as Coulomb's law)," or "experimental evidence that Poisson's equation accurately describes electrostatics," or.... – Michael Brown Mar 11 '13 at 0:09
@Alyosha Do you understand the proof of the divergence theorem? (They build it up in steps - starting from rectangular shapes and pasting them together to make arbitary volumes.) – Michael Brown Mar 11 '13 at 10:18
Are you looking for a justification of $\nabla\cdot\mathbf g = -4\pi G\rho$ from some other, more basic, physical facts? At some point, you have to accept something physical here that cannot be proven but which leads to this (Poisson's) equation. The related post linked by @Qmechanic for example shows how to obtain Poisson's equation from Coulomb's Law. Is this somehow not satisfying to you? – joshphysics Mar 12 '13 at 5:03
up vote 2 down vote accepted

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 $E r^2 \propto q$.

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You mean Divergence Theorem, Not Stokes' Theorem. – cspirou Mar 11 '13 at 3:54
@cspirou Actually, the name Stokes' Theorem is more widely used in mathematics to refer to a theorem on the integration of differential forms over manifolds and has, as a subcase, both the classical Stokes' Theorem from vector calculus and the Divergence Theorem. – joshphysics Mar 11 '13 at 5:27

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.

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This is only true for point charges, not the general case. Also, the part of your answer after the comma essentially restates the question. – Meow May 18 '14 at 19:30

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