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?

  • 1
    $\begingroup$ 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, 2019 at 4:14
  • 1
    $\begingroup$ @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, 2019 at 12:06

2 Answers 2


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.

  • $\begingroup$ 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, 2019 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, 2019 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, 2019 at 20:52

You should sum up contribution of shells centered at the test particle, not at the center of the cavity. The shell theorem is not applicable in this case.

Let's consider a simpler 1D case.

enter image description here

Suppose a rope pulling (tug of war) game with infinite number of players in the both teams with a test particle at zero.

All players produce the same pull and are located at all integer distances from zero, but at zero itself and at $1$ the players are missing.

Which team will win? Here we step into the area of divergent series and integrals. The cardinalities of the sets {-1,-2,-3,...} and {2,3,4,...} are the same, but they have different numerocities. By definition, numerocity is the sum of the indicator function over the set: $\sum_{k=-\infty}^\infty p(k)$, where $p(k)$ is $1$ if $k$ is in the set and $0$ otherwise.

So, the numerocity of the left team is $\sum_{k=-\infty}^{-1} 1$ while the numerocity of the right team is $\sum_{k=2}^\infty 1$. The both sums are divergent, but their regularized values are different.

The regularized value of the left-hand sum is $-1/2$ while the regularized value of the right-hand sum is $-3/2$.

They also can be expressed with integrals as $\int_{-\infty}^{-1/2}dx$ and $\int_{3/2}^\infty dx$.

If the infinite parts cancel each other, the finite parts give victory to the left-hand-side team.

In a more complicated 3D case one should account for situations, where the infinite parts may not cancel each other (this is not the case of your original question, but may arise if at the sides of a test particle are infinite cones, cylinders, paraboloids, sponges or other infinite shapes, which need to be compared).

Due to Fourier or Laplace transform we can formally denote the divergent integral $\int_0^\infty dx$ as $\pi\delta(0)$ or, for simplicity, $\tau$.

This way, the strength of the left-hand team in our example is $\int_{-\infty}^{-1/2}=\tau-1/2$ and the strength of the right-hand team is $\int_{2/2}^\infty=\tau-3/2$

The integrals of monomials, needed in multi-dimensional complicated cases will follow the following law:

$\int_0^\infty x^n dx=\frac{\left(\tau +\frac{1}{2}\right)^{n+2}-\left(\tau -\frac{1}{2}\right)^{n+2}}{(n+1)(n+2)}$

For more information on divergent integrals, see here (need MathML-capable browser (Firefox, PaleMoon).

  • $\begingroup$ As far as I know, and as Azzinoth notes in his answer, you cannot compare infinities that way. You state that the cardinalities of the left and right side are the same, but with just as much justification, you can instead state that the cardinality of the left side is twice that of the right side. Just like the cardinality of all even integers is the same as the cardinality of all integers. $\endgroup$
    – fishinear
    Aug 11, 2021 at 8:51
  • $\begingroup$ @fishinear the thing is, cardinality is absolutely irrelevant here, because cardinality is all about possibility of bijection. What is relevant is numerocity, density, integrals. Cardinality of even numbers and all of integers is the same, numerocity is twice less. $\endgroup$
    – Anixx
    Aug 11, 2021 at 9:11
  • $\begingroup$ The point is still that you cannot compare an infinity on one side with an infinity on the other side that way. I see nothing in your answer that discredits my original reasoning of the hollow sphere that grows to infinity. Apart from that, I assumed a finite Universe, so I see no reason to talk about infinities at all. $\endgroup$
    – fishinear
    Aug 11, 2021 at 12:47
  • $\begingroup$ @fishinear "The point is still that you cannot compare an infinity on one side with an infinity on the other side that way." - who said so? This answer shows how you can do it. "Apart from that, I assumed a finite Universe" - if the universe is finite, you should define its shape. If it is a sphere, where is its center? This will affect the answer. $\endgroup$
    – Anixx
    Aug 11, 2021 at 12:53
  • $\begingroup$ I still don’t see how your answer differs from my empty shell argument in the question. And I assumed a spherically curved Universe, as stated in the question, and is common in physics. So it is a uniform finite universe that has no boundary and no center. $\endgroup$
    – fishinear
    Aug 11, 2021 at 13:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.