Suppose infinite 3D Euclidean space is uniformly filled with dense matter except a spherical cavity. Will a particle in this cavity experience gravitational acceleration towards the nearest wall?

  • $\begingroup$ @CountTo10 Because if the space is uniformly filled with matter, there is no force. Now if we remove matter non-symmetrically around the test particle, the balance is broken. I think the cavity will pull as the matter of the same form in empty space but into opposite direction. $\endgroup$ – Anixx Nov 6 '16 at 13:40
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    $\begingroup$ Hmmmm. ...I guess I will have to think about that aspect, and I apologise for writing before thinking :) $\endgroup$ – user108787 Nov 6 '16 at 13:48

Either there is no force, or the situation is ill-defined.

That "no force" is the only consistent answer is evident because we may divide the infinite matter into infinitely many spherical shells of finite thickness, and for each of those shells we know that there is no force exerted on something within.

However, we may also imagine dividing the matter up in some other asymmetrical way, and it is not evident that the limit of all such divisions should agree with each other. For any finite distribution of mass, this would be the case because $\int_{\mathbb{R}^3} \rho(x) \mathrm{d}V = \sum_i \int_{V_i}\rho(x)\mathrm{d}x$ for any collection of disjoint sets $V_i$ with $\bigcup_i V_i = \mathbb{R}^3$, but in this case the integral $\int_{\mathbb{R}^3}\rho(x)\mathrm{d}V$ doesn't exist. That is, we might produce different answers depending on the limiting procedure used to model "infinite space filled with matter".

  • $\begingroup$ If you divide it into shells centered at the cavity's center. What about shells centered at the test particle? $\endgroup$ – Anixx Nov 6 '16 at 13:42

Well, after reading other answers, I say different. Point me out where I am wrong and I would be glad to understand my mistake.

Suppose the entire region was filled with matter. So, using any kind of symmetry, there would be no force any particle. Now we remove a sphere, which contains that particle that we're observing. Isolating the particle and the sphere, were know that the force due to the sphere is same as in the case of force on particle inside a sphere with uniform mass distributed.

Thus, by removing the sphere, we tend to create some sort of 'negative' mass, with causes a force on the particle on the opposite direction.

Also note that there is symmetry only when the particle we consider is at the center of the sphere, wherein my answer also stands in agreement with the no force result.

Hope this helped!


I think that the result is undeterminate. Let me explain why.

At first, one could think that the answer is that there is no force inside the cavity because of Newton's shell theorem, which is just an application of Gauss's law plus the divergence theorem. The usual proof is this:

We start from the differential form of Gauss's law (proof):

$$\tag{1} \nabla \cdot \vec g (\vec r) = - 4 \pi G \rho (\vec r)$$

Then we use the divergence theorem:

$$\tag{2} \int_V \nabla \cdot \vec g \ d \vec r = \int_S \vec g \cdot d\vec S$$

where $S$ is an arbitrary closed surface and $V$ the volume enclosed by it. For simplicity, we will take a spherical surface centered at the center of our cavity.

The left side is easy to evaluate using $(1)$:

$$\int_V \nabla \cdot \vec g \ d \vec r =- 4 \pi G \int _V \rho(\vec r) \ d \vec r = -4 \pi G M$$

where $M$ is the mass contained in $V$. We therefore obtain from $(2)$:

$$\int_S \vec g \cdot d\vec S = - 4 \pi G M$$

If the chosen surface contains no mass, then

$$\tag{3} \int_S \vec g \cdot d\vec S = 0$$

Now the crucial passage: usually, we say that since we are considering a uniform spherical shell, then by symmetry the force must be directed radially outwards and must be constant over $S$, so that we have

$$\int_S \vec g \cdot d\vec S = \int_S | \vec g | \ \hat r \cdot d \vec S = | \vec g | \int_S dS = |\vec g | 4 \pi R^2 = 0$$

where $R$ is the radius of the spherical surface and $|\vec g |$ is the magnitude of $\vec g$ evaluated over the surface. We therefore conclude that

$$|\vec g|=0$$

for every spherical surface inside the shell, and therefore

$$\vec g = \vec 0$$

inside the shell.

So what is the problem with an "infinitely thick" shell, like in the problem you propose? The problem is in the passage where we say that by symmetry the force must be radial and constant over the surface. This is certainly true for every shell of uniform density and finite thickness, but if the thickness is infinite the whole concept of spherical symmetry is in my opinion undefined and we cannot proceed from $(3)$. Therefore, I think that the problem is undeterminate.

To put it simply, think at the problem this way: imagine, instead of a uniform shell, a uniform spherical distribution of separate point masses (picture below). Now, place your test mass closer to the "left side" of this (picture below...). For such a finite set of point masses, you can see intuitively why the shell theorem works: on the "left side" the test mass interacts strongly with a small number of masses, while on the "right side" it interacts with many more masses, but more weakly (of course, the definition of "left side" and "right side" is completely arbitrary). But if you have an infinitely tick set of points both on the "right side" and on the "left side", then what is going to happen...? enter image description here

  • $\begingroup$ Is it possible to show the problem is indeterminate by coming up with a diverging improper integral or something like this? $\endgroup$ – Anixx Nov 6 '16 at 15:05
  • $\begingroup$ Not really: the problem is with the symmetry argument. What is a spherically symmetric infinite mass distribution? I will update the answer to try to put it in a simpler way. $\endgroup$ – valerio Nov 6 '16 at 15:12
  • $\begingroup$ I think we can assume infinite uniformly filled space is symmetric, and starting from this postulate evaluate only the contribution of non-uniform areas. $\endgroup$ – Anixx Nov 6 '16 at 15:14
  • $\begingroup$ Right now I am busy (sorry), but as soon as possible I will try to update my answer and try to explain in a more intuitive way what the problem of an infinite mass distribution is. $\endgroup$ – valerio Nov 6 '16 at 15:17

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