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I was doing a little thought experiment and ran into a paradox. How to compute the gravitational field inside a spherical hole in a universe that is otherwise filled with uniformly distributed mass?

If we start with just a spherical shell of mass in an otherwise empty universe, then it is well-known that inside the shell, there is zero gravitation, independent of the thickness of the shell. If we then subsequently keep the inside of the shell the same size, but make the shell thicker and thicker, we should be able to make it fill the whole universe (assuming it is a spherically curved universe). By that reasoning, there should not be any gravitation inside a spherical hole inside a universe of equally distributed mass. So a small test mass near the border of the hole should not feel any gravitational force.

Now we subsequently fill the hole with a sphere of mass. The small test mass near the border of the hole should feel gravitation from this sphere of mass, that is just normal gravitation laws for spherical masses. If the test mass feels gravitation from the sphere, but not from the rest of the universe, then there should be a net gravitational force on it.

But if the hole is filled with a spherical mass of the same mass density as the rest of the universe, then effectively, there is no hole, and we just have a universe filled with uniformly distributed mass. In such a universe, the test mass should not feel any gravitation where ever it is placed.

Where does my reasoning go wrong, and how can the gravitation inside a spherical hole be calculated?

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    $\begingroup$ I'm a little confused by your description. If you fill the hole, there's no room left for the test particle. As for Newtonian gravity in an infinite homogeneous universe, see physics.stackexchange.com/a/430442/123208 $\endgroup$ – PM 2Ring Feb 27 at 4:14
  • $\begingroup$ @PM2Ring It does not need to be a dense distribution of mass in the universe. Think of a mass density comparable to our universe, but more equally distributed. You can surely find a spot for an infinitesimal small test mass in that. Your link is interesting, but mainly just says that such a universe is unstable and will start to collapse. That is fine by me, I am just interested in the initial case before the collapse starts. $\endgroup$ – fishinear Feb 27 at 12:06
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You don't even need a shell, I think this is a simpler description of your problem:

Consider a sphere with radius R, constant mass density $\rho$ inside and zero mass density outside. The gravitational force inside the sphere at a distance r from the center is $$F=r\frac{GmM}{R^3}=rGm\frac{4}{3}\pi\rho \quad\text{for}\quad r<R.$$

If we now let R go to infinity, then we have a universe filled with uniform mass density, yet the force is not zero everywhere.

In such a universe, the test mass should not feel any gravitation where ever it is placed.

I think this is where your reasoning goes wrong. If we start with a cube instead of a sphere and let the borders go to infinity, then we will also get a universe with infinite uniform mass density, but $F$ will look completely different. This is because by choosing such a shape, we define different boundary conditions.

There is a good analogy given by knzhou in this thread.

Here's an analogous question: suppose a function $f$ obeys $f''(x)=1$ and we want to solve for $f(x)$. Since every point on the real line is the same as every other point, we might think that by symmetry, $f(x)=\text{constant}$. But this is completely wrong, because the second derivative of a constant is zero. And stepping back, the whole question doesn't make any sense, because there isn't enough information. To solve a general differential equation, you need boundary conditions.

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  • $\begingroup$ I see what you mean, you cannot extrapolate to infinity in that way. And I assume that applies to my case as well with a universe that is not infinite, but spherically curved. The curvature will start to play a role, and the mass that is on "the other side" of the universe will start to counter-act the gravity at "this side" of the universe. $\endgroup$ – fishinear Feb 27 at 20:12
  • $\begingroup$ The question still remains: how do I calculate the gravitational field inside a spherical empty space in the universe? $\endgroup$ – fishinear Feb 27 at 20:13
  • $\begingroup$ @fishinear I'm not sure, but you are right that in a closed finite universe, the "mass that is on "the other side" of the universe will start to counter-act the gravity at "this side" of the universe." Maybe this means that gravity inside a shell is not zero everywhere in such a universe, if the shell gets big enough? The result that it is zero inside a shell assumed a flat universe I think. So in a closed universe, that result might only be approximately true for small enough shells. $\endgroup$ – Azzinoth Feb 27 at 20:52

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