I was just reading a question about the gravity inside a hollow neutron star. It was a trivial question, obviously there is no force felt. But then it got me thinking.

Suppose you had a hollow sphere that was massive enough that outside of it, it became a black hole. It's not hard to imagine a situation where the magnitude of the force of gravity at the surface (while strong) is equal to the magnitude of the force on the particles of the sphere due to some phenomenon involving one of the other 3 forces (just so I can allow this hollow sphere to continue being hollow).

Obviously, I don't have to do the integral to know that any object inside the hollow would experience no net gravitational force due to the shell. However, being a few months short of General Relativity, I am curious to know whether there would be any effects from Gravity observed.

Would an observer inside experience any time dilation? There may not be any net force due to gravity, but one cannot deny that there are still some fairly powerful gravitational fields present that happen to superimpose to zero. Furthermore, (assuming a relatively massless observer) would they have access to any of the information that was "lost" in the black hole? If so, then if they were long-lived enough for the black hole to be able to evaporate around them, does this mean that not only would the black hole radiate that information, but it would also store a copy of it in the hollow?

I want to apologize in advance. I know this question seems a bit more like a discussion than a Q&A thing. To be clear, I am asking: In my scenario, would there be any effects of gravity on things in the hollow? And would information lost in the black hole be accessible to the hollow (which is essentially its own isolated bubble of the universe)? A good answer can address these two questions. If you want to add in your own opinions/contribute to a discussion about it, that'd be a bonus.

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    $\begingroup$ Possible duplicate: physics.stackexchange.com/q/52021/2451 $\endgroup$
    – Qmechanic
    Jul 2, 2013 at 16:19
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    $\begingroup$ You can't have a spherical shell inside an event horizon - see physics.stackexchange.com/q/47828 $\endgroup$ Jul 2, 2013 at 17:10
  • $\begingroup$ @JohnRennie Good question, not the same concept though. That one was about a shell around the gravity-generating singularity. In my scenario, it is the shell itself that generates the gravity. This distinction changes the gravitational force acting on the particles of the shell $\endgroup$
    – Jim
    Jul 2, 2013 at 17:15
  • $\begingroup$ As for the duplicate comment, I hadn't even seen that question. It's definitely related, but it isn't a duplicate. That question asks about the measurable gravity. While both ask about the possible effects of GR. Mine is more specific and also asks about the accessibility of "lost" information to the inside observer $\endgroup$
    – Jim
    Jul 2, 2013 at 17:18
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    $\begingroup$ @Jim: you can't have any static structure inside an event horizon, and that applies to all event horizons regardless of how the matter is (temporarily) organised inside the horizon. For example if you took a spherical shell of mass $M$ and compressed it to a radius less than $2GM/c^2$ then a horizon would form around it and that shell would inevitably and rapidly collapse into a singularity. What an observer inside the shell would (briefly!) experience is an interesting question but one that I suspect would be hard to answer. $\endgroup$ Jul 2, 2013 at 17:31

2 Answers 2


After I commented on the question I started wondering what an observer inside a collapsing shell would experience.

If you construct a spherical shell then an observer inside it feels no gravity. This is true in Newtonian gravity, and is also true in General Relativity as a consequence of Birkhoff's theorem i.e. the metric inside the shell is the Minkowski metric.

In principle we can take the shell and compress it until it's external radius falls below the Schwarzschild radius $r = 2GM/c^2$, at which point the shell will start collapsing inwards and form a singularity in a finite time. In fact it's a very short time indeed. Calculating the lapsed time to fall from the horizon to the singularity of an existing black hole is a standard exercise in GR, and the result is:

$$ \tau \approx 6.57 \frac{M}{M_{Sun}} \mu s $$

That is, for a black hole of 10 solar masses the fall takes 65.7 microseconds! I would have to indulge in some head scratching to work out if the same time would be measured by an observer riding on the collapsing shell, but if the time isn't the same it will be of a similar order of magnitude. This means much of the question doesn't apply, since the shell cannot be stable long enough for the black hole to evaporate. However it leaves open the interesting question of what the observer inside the shell experiences.

Curious as it seems, Birkhoff's theorem implies the observer experiences absolutely nothing until the collapsing shell hits them and sweeps them, along with the shell, to an untimely end (a few microseconds later!).

Response to comment: time dilation

The infall time I calculated above is the proper time, that is the time measured by the freely falling observer on their wristwatch. You need to tread carefully when talking about time in relativity, but the proper time is usually easy to understand.

Re time dilation: again we need to be careful to define exactly what we mean. In the context of black holes we usually take an observer far from the black hole (strictly speaking at an infinite distance) as a reference and compare their clock to a clock near the black hole. By time dilation we mean that the observer at infinity sees the clock near the black hole running slowly.

A clock in a gravitational potential well runs slowly compared to the clock at infinity. This was discussed in the higher you go the slower is ageing (and also in Gravitational time dilation at the earth's center). It's important to understand that it's the potential that matters, not the gravitational acceleration, so even though the observer inside the shell feels no gravitational acceleration they are still time dilated compared to the observer at infinity.

Note that the time dilation relative to the observer at infinity goes to infinity at the event horizon, so it makes no sense to compare times inside the event horizon to anything outside.

  • $\begingroup$ Very interesting, I like this answer. However, it still leaves me wondering whether or not information about what has fallen into the black hole is available within the shell for its brief existence. I mean, would an active scan of the shell from inside result in any information about its composition? Would there be radiation emissions from the shell's particles to the interior? Also, I assume your 65.7 microseconds is time dilated for an observer on the surface. Does this mean an observer inside also experiences time dilation even without gravity? If not, how long is it in that frame? $\endgroup$
    – Jim
    Jul 2, 2013 at 18:46
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    $\begingroup$ A collapsing shell introduces time dependence, so how can one continue to apply Birkhoff's theorem? $\endgroup$
    – Will
    Jul 2, 2013 at 23:48
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    $\begingroup$ @will: because the system is spherically symmetric $\endgroup$ Jul 3, 2013 at 5:35
  • $\begingroup$ @Jim : Assuming a flat space-time in the hollow, all physical properties in the hollow are these of a standard Minkowski space-time, so I don't see any reason why an inside observer cannot acess inside information. $\endgroup$
    – Trimok
    Jul 3, 2013 at 7:49
  • $\begingroup$ Re the time dilation update: This is perfect! absolutely marvelous. I see now that it is the potential and not the net gravitational force that dilates time. As for the access to information; the best I can expect anyone to do is speculate. But anyone willing to contribute to a discussion on the hypotheticals of it is welcome. I hope this problem was as fun for everyone else to think about as it was for me :-) $\endgroup$
    – Jim
    Jul 3, 2013 at 14:50

Take a shell with outer radius $R$ and inner radius $r$. If the density of the ring material is $d$, The formula to calculate the mass of the shell is $$\frac{4}{3}d\pi(R^3-r^3)$$ The Schwarzschild radius of this shell is $$\frac{\frac{8}{3}dG\pi(R^3-r^3)}{c^2}$$ It follows that if this equals $R$, the shell should collapse into a black hole. Solving for $d$: $$d=\frac{3c^2R}{8G\pi (R^3-r^3)}$$ A sphere of density $d$ and radius $R-r$ has a Schwazschild radius of $$\frac{8\pi dG(R-r)^3}{3c^2}$$ Substituting $d$ gives us: $$\frac{R(R-r)^3}{R^3-r^3}$$ If the Schwarzschild radius of a sphere inside the ring of the shell is less than the radius of the shell at collapse, the shell breaks apart into infinite black holes: $$\frac{R(R-r)^3}{R^3-r^3}<\frac{\frac{8}{3}\frac{3c^2R}{8G\pi (R^3-r^3)}G\pi(R^3-r^3)}{c^2}$$ The radius of the shell at collapse is R anyway, so $$\frac{R(R-r)^3}{R^3-r^3}<R$$ This inequality holds true only if both $R$ and $r$ are positive.

This solution seems to suggest, that any shell with a thin radius will collapse into local black holes at $r + \frac{(R-r)}{2}$ before the shell reaches a density high enough to collapse into the black hole at a smaller radius.

I am unsure whether any of this is true. It may only work for very small values of $R-r$. Even so, I don't know how black holes would form in that region, or if they dissapear due to hawking radiation. If the latter is the case, this would be a way of converting mass into energy in the form of gravitational waves 100% efficiently.

  • $\begingroup$ I realise this is a non-sensical answer, and I wrote it at 7am, but I'll keep it up $\endgroup$
    – Max
    Feb 1, 2021 at 15:10

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