As a thought experiment imagine an incredibly strong spherical shell with a diameter a bit smaller than the event horizon of a particular large black hole. The shell is split into two hemispheres, and the two hemispheres are released from either side of the black hole such that they will meet and join inside the event horizon, forming a stationary uniform shell around the singularity. Of course, we consider that the shell is flawless, and exactly centred around a stationary singularity - clearly this is an unstable equilibrium but we imagine it is possible.

Is is possible that the structural forces provided by such a shell could be strong enough to resist the extreme gravity conditions inside the event horizon, and hence avoid being sucked into the singularity?

(For the purposes of this question we can consider taking the shell to extreme levels of strength, e.g. consider that the shell is made from a theoretical maximally densely packed neutronium-like material that is strongly resistant to any further compression due to the Pauli exclusion principle).

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    $\begingroup$ No more than a strong spherical shell elsewhere could resist time passing. What happens at the event horizon of a black hole is much more profound than just strong gravity. $\endgroup$ Commented Dec 29, 2012 at 5:41
  • $\begingroup$ in addition to @dmckee 's comment this neutronium shell will be sitting on a very metastable point. One neutron decay at the surface might destroy the symmetry and start kinematic effects that will destroy it. $\endgroup$
    – anna v
    Commented Dec 29, 2012 at 5:56
  • $\begingroup$ anna v: yes I agree it would be incredibly unstable, clearly not practical but as a thought experiment it is interesting... (well to me anyway) $\endgroup$
    – roblev
    Commented Dec 29, 2012 at 18:28
  • $\begingroup$ A bit of history: the idea of black hole formation was discovered when considering the breaking strength of such a ball of packed neutrons. By definition, the shell can't be any stronger. $\endgroup$
    – Joshua
    Commented Jan 7, 2015 at 20:27
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    $\begingroup$ Good luck. I learned this one pre-interent by reading physics books in the library. What happens is the breaking strength of a solid material is proportional to the speed of sound in the material. As the speed of sound can't be greater than c, the breaking strength must have a maximum. $\endgroup$
    – Joshua
    Commented Feb 24, 2015 at 17:27

3 Answers 3


As dmckee says in his comment, the answer is no, a stationary spherical shell isn't possible. This is because not even the interparticle forces in neutronium are strong enough to support it.

The problem is that once inside the event horizon there is no way to travel away from the singularity, or even maintain your distance from it, without travelling faster than light. However the fundamental forces like electromagnetism and the weak and strong forces have a maximum propagation speed of $c$, and that means the three forces cannot propagate outwards. A neutron at the outer edge of the spherical shell can't be held there because none of the three forces can move outwards to hold it there, so it must fall inwards. This applies to all the neutrons in the shell, so they and the shell must inevitably fall inwards and hit the singularity.

Proving this is a bit involved. see my answer to Why is a black hole black?, but even though I've tried to simplify the maths it's still a bit intimidating.


I've just spotted Can we have a black hole without a singularity?, which is not exactly a duplicate but is related.

Later still:

You might have noticed that dmckee said in his comment:

No more than a strong spherical shell elsewhere could resist time passing

and you might wonder what this means. Well when we talk about objects falling into black holes we generally (often without realising it) choose a set of co-ordinates called Schwarzschild co-ordinates. These are time, $t$, radial distance from the centre of the black hole, $r$, and two angular coordinates that I'll ignore for now.

As long as you're outside the event horizon $t$ and $r$ make intuitive sense. It's true that time slows down for an object near the event horizon, but we're already used to time changing for fast moving objects. Time still behaves like time and distance still behaves like distance. Specifically, we can move in either $r$ direction but we can only move forward in the $t$ direction i.e. we can't move back in time, only forward.

However, if we use our Schwarzschild coordinate system to describe events inside the event horizon we find something very odd. The $r$ coordinate that describes distance from the centre of the black hole now behaves like time, and just like time it's now only possible to move forward (towards the singularity) in $r$ and it's impossible to move backwards. This is why dmckee made the reference to the shell being unable to resist time passing. The spherical shell inside the event horizon cannot resist moving forward in $r$ any more than a spherical shell outside the event horizon can resist moving forward in $t$.

This sounds weird (it's used over and over in science fiction), but it's perfectly good physics. However I'm not fond of this viewpoint because the apparent weirdness is the result of a poor choice of coordinates. Since in the Schwarzschild coordinates an object takes infinite time to reach the event horizon you shouldn't be surprised to find the Schwarzschild coordinates do a poor job of describing what happens inside the event horizon. It's important to emphasise that if you were falling into a black hole you wouldn't experience distance behaving like time or vice versa. In your (brief!) journey to destruction at the singularity you'd find time and distance behaving just as you'd expect them to.

  • $\begingroup$ Thanks, it does make sense to me that the fundamental forces cannot act on the neutrons to stop them falling further, so the shell would fall. But I wonder if there is any mileage in considering the Pauli exclusion principle further, i.e. if gravity is trying to force neutrons to occupy the same space then there could be a resistance to that movement that is different in nature to electomagnetism, strong and weak forces? $\endgroup$
    – roblev
    Commented Dec 29, 2012 at 18:24
  • $\begingroup$ One other thought... re: "the three forces cannot propagate outwards" if the shell is a thin shell then the forces do not need to propagate outwards, they need to propagate sideways. A sideways force on a collapsing atom in a shell could provide a component of resistance to the downward motion. $\endgroup$
    – roblev
    Commented Jan 1, 2013 at 3:54
  • $\begingroup$ Sadly not. A sideways force obviously has no component towards the centre. The reason arches stand up is because the forces on the blocks in them are not sideways. If you look closely at the blocks you'll find they are wedge shaped i.e. their sides are not parallel, so there is an outwards component of the apparently sideways forces on them. $\endgroup$ Commented Jan 1, 2013 at 7:24
  • $\begingroup$ The Schwarzschild solution only applies outside a sphere where there is no energy-matter, so how can you argue that the particles will have to fall inwards? I was thinking that the natural way would be solving Einstein equations where the energy-momentum is non-zero and see that all geodesics reach the singularity, but this probably isn't possible. $\endgroup$
    – jinawee
    Commented Dec 24, 2015 at 15:03
  • $\begingroup$ For realistic rotating black holes in Kerr and Kerr-Newman spacetimes, it is not at all necessary for the object within the event horizon to fall into the singularity. There are many different orbits and ranges of motion to avoid a ring singularity. If two such black holes merge, this will still be the case - indeed, the larger the black hole, the easier it is not to fall into the singularity. $\endgroup$ Commented Dec 11, 2023 at 1:26

A shell made from "stiff" matter has minimum $r$-coordinate $\approx 5M/2$, where $M$ is the mass of the black hole. This is between the event horizon at $r=2M$ and the photon sphere at $r=3M$. The approximation is only valid when the shell is much lighter than the black hole.

This is from a concise paper by Frauendiener et al, "A shell around a black hole" (1990), who give more results and details. Unfortunately there is no free arXiv version.


No, but that does not mean that a spherical shell could not be used as a means of keeping a black hole from expanding. Obviously, the shell would need to be quite large for most stellar black holes. The durability of material required would be inversely proportional to the durability of material involved. Meaning that, for example, a sphere of steel around a solar-mass object would need to be AT LEAST 750000000000 KM across, which is not remotely practical, even if it itself.

Even a nano-diamond sphere would need to be 750000000, 5 earth radii, across. If it were a millimeter thick, it would weigh multiple Jupiter-masses.

The only material that would have a better strength-to-weight ratio would probably be a whatever neutron stars have to support mini-mountains on their crust.

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    $\begingroup$ Could expand a bit on where your numbers come from to give your answer more general value?! $\endgroup$
    – Michiel
    Commented Apr 15, 2013 at 6:24

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