# The shell theorem and divergence in Susskind's GR lecture

In the lecture available here, at around 1:36:50, Susskind seems to argue that you can derive the fact that the gravitational field inside a spherical shell (of constant density) is 0 from the divergence theorem. I don't see how this argument works at all. All you get is that the divergence of the gravitational field inside the shell is 0, but that doesn't imply that the field itself is 0 of course.

Am I missing something here?

He doesn't derive that the divergence is 0. He uses the fact that the divergence is 0 to show the gravitational field is also vanishing. The argument goes as follows. The divergence theorem states that

$$\iiint_{\text{ball of radius R}} \left( \vec{\nabla} \cdot \vec{F}_{\text{grav}} \right)d^3x= \iint_{\text{sphere of radius R}}\vec{F}_{\text{grav}}\cdot d\vec{S}.$$

Now if the divergence is 0, then the left hand side is 0. Then we evaluate the right hand side on the spherical shell. This yields

$$4\pi R F_r=0,$$

where $$F_r$$ is the radial component of the gravitational force. Since $$R\neq 0$$, the only way this equation could be satisfied is if $$F_r$$=0.

To show that the divergence is $$0$$, one uses Gauss' law for the gravitational field and recognizes that there is no mass within the shell

$$\vec{\nabla} \cdot \vec{F}_{\text{grav}} = -4 \pi Gm =0.$$

I hope this helps.

• I see, but why does $F_r = 0$ imply $F = 0$? I have some thoughts about this but maybe there's an easy way to do it Jul 9 '20 at 9:21
• Well, in principle, it doesn't. But you can take the integral to be over any shape and size as long as it is inside the region of zero mass density and you can show that the components of the gravitational field vanish this way. Or you can recognize that the problem has a spherical symmetry and say that the only possible non-zero component of the force can be the radial and then you apply the argument from above. Either way works, but the latter is quicker and more elegant, in my opinion. Jul 9 '20 at 9:24
• I think you need to use the spherical symmetry somehow. Otherwise, imagine just a point particle somewhere. Fixing also some other point, for any neighborhood of this point you choose there will be no mass, and so zero divergence. But the gravitational field is not $0$ of course. The spherical symmetry is crucial here. Jul 9 '20 at 9:28
• But I don't think it's so immediate to actually show that it implies the gravitational field points radially. In principle all it implies is that the field is preserved under rotation. I believe you can then argue this means it has no tangential component essentially due to the Hairy ball theorem in topology. But I'm curious if that's really necessary. Jul 9 '20 at 9:30
• I made a different post about that here physics.stackexchange.com/questions/564698/… Jul 9 '20 at 17:04