I have read this question (unfortunately this mentions supernova and not black hole):

If it it the latter, then the instabilities that lead to the collapse of a neutron star would begin near the centre of the star at the highest densities. Collapse timescales go as the free-fall timescale, which is ∝(Gρ)−1/2 where ρ is the density. Thus dense regions collapse quicker and the collapse would proceed on an inside-out basis.

Does the neutron star collapse from the surface to the center or from the center to the surface?

And this one (in the comments you can see it says the event horizon forms at the center and expands outwards):

What happens when a neutron star becomes a black hole?

Would a black hole instantly form when a neutron star slips below the phantom event horizon?

Now let's say for the sake of argument that you could survive at the center of the neutron star (where you are weightless), and the neutron star collapses into a black hole. This collapse would (based on the above) start at the center, and the event horizon might do the same (extend from the center outwards), but there are contradictory statements about this on this site.

Thus, if I am at the center, would I be at the singularity right away, or would I still experience a journey to the singularity?


  1. What would the collapse of a neutron star into a black hole look like from the center?
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    $\begingroup$ As observed from outside, the internal observer will be pushed to just outside the horizon and remain there forever. He himself would not necessarily be aware of this. Any answer or comment below referring to the density gradient is misleading, because a black hole does not have a center (either Schwarzschild or OS) - all directions inside equally lead to the singularity just like any roads in your city equally lead to tomorrow. Also the TOV equation is not a black hole solution, but just a condition for the collapse to start. Stan’s answer is correct for OS (and other solutions are similar). $\endgroup$
    – safesphere
    Commented Feb 25 at 5:32

3 Answers 3


From the interior, collapse into a black hole is equivalent to the Universe collapsing in a Big Crunch, at least in the simplest approximation. Specifically, the Oppenheimer-Snyder model of the collapse of a uniform sphere of pressureless matter is equivalent to a collapsing matter-dominated FLRW cosmological model. As the star collapses, you would experience density increasing without bound as you approach the singularity (which is in the future, not at a point in space). In this approximation, your experience is the same no matter where you are within the star -- you do not need to be at the center.

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    $\begingroup$ I did not expect this question to get a good answer within 30 mins of being asked. Nice. $\endgroup$
    – AXensen
    Commented Feb 23 at 0:14
  • $\begingroup$ Thank you so much! $\endgroup$ Commented Feb 23 at 7:09
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    $\begingroup$ This is a good (partial) answer, it would be even better if it would comment on how the simplifications in the Oppenheimer-Snyder model affect the final statement. $\endgroup$
    – TimRias
    Commented Feb 23 at 8:54
  • $\begingroup$ Is there any kind of theoretical higher order approximation which would affirm one of the OP's questions? That is, if you were not at the exact center, then the black hole would form first without you and then (a picopicopico...second) later your world line would come to an end at the black hole? $\endgroup$
    – Lee Mosher
    Commented Feb 23 at 14:34
  • $\begingroup$ Does not it at least matter whether I am placed within the Schwarzschild radius or outside it? $\endgroup$ Commented Feb 23 at 15:30

The answer given by Sten is good, but does depend on particular approximation of the matter in the star. Specifically, it assumes there is no pressure, which allows for the homogeneity of the interior. Of course, this is not realistic for a neutron star just prior to collapse, since the neutron star is supposed to be kept stable by the internal pressure. The solution for a relativistic spherically symmetric star is given by the Tolman-Oppenheimer-Volkoff solution.

In the slightly less unrealistic idealized case of a star consisting of an incompressible fluid (i.e. constant density), the TOV equation gives the following density profile:

$$p(r) = \rho_* c^2 \frac{R \sqrt{R-2GM/c^2}-\sqrt{R^3-2GMr^2/c^2}}{\sqrt{R^3-2GMr^2/c^2}-3R\sqrt{R-2GM/c^2}}, $$

where $p$ is the pressure as a function of the radial position $r$, $\rho_*$ is the density of the fluid, $c$ is the speed of light, $R$ is the radius of the star, $G$ is gravitational constant, and $M$ is the mass of the star. Note that an observer inside the star can always tell whether they are at the center or not by looking at the gradient of the pressure. The density at the center $p(0)$ approaches infinity as the total mass $M$ approaches the Buchdahl limit $4GM/(9c^2)$, at which point the metric becomes singular at that point and the star must collapse.

Of course, a realistic collapse lies somewhere in between these extremes.

  • $\begingroup$ Thank you so much! $\endgroup$ Commented Feb 23 at 18:58
  • $\begingroup$ "the singularity spreads through the star" sounds kind of weird, as if the singularity would get a nonzero radius and volume, but multiply infinite density with any volume and you get infinite mass. $\endgroup$
    – Yukterez
    Commented Feb 23 at 19:08
  • $\begingroup$ @Yukterez I think it meant "the event horizon spreads through the star"? $\endgroup$ Commented Feb 23 at 21:14
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    $\begingroup$ @Yukterez I actually thought I’d removed that part of the sentence. Statements about how the collapse proceeds simply do not make sense in the of a non compressible fluid, which has an infinite speed of sound and breaks relativity. $\endgroup$
    – TimRias
    Commented Feb 23 at 21:26
  • $\begingroup$ I’ve removed that part now. $\endgroup$
    – TimRias
    Commented Feb 23 at 21:26

A calculation based on the FLWR metric assumes a universe that is basically a uniform distribution of matter and energy throughout the universe and so a uniform average density. The question assumes starting with a neutron star which in simplified form is a solid sphere surrounded by vacuum. Even if we disregard the vacuum outside the spherical solid body (as per the valid comment by Sten) the density inside the body will not be uniform, but greater nearer the centre. I think a more appropriate solution would be based on matching the Schwarzschild interior metric with the Schwarzschild exterior metric. The interior solution also assumes the density is constant throughout the body and is a static solution, so it would have to be modified to be dynamic perhaps by incorporating an FLWR type metric. Either way a valid solution would have to incorporate an allowance for density varying in radial distribution and time. I do not think there is currently a single analytical solution to this question at the moment and the solution would perhaps require some very sophisticated finite element simulation program. The answer would be very enlightening.

The 'phantom' Schwarzschild radius grows from the centre, but is not an event horizon until the phantom horizon reaches the surface of the body. For example, the Earth has a theoretical Schwarzschild radius that is a tiny fraction of the radius of the Earth. If we kept racing down matter on the surface of the Earth, the increase in mass will cause the theoretical Schwarzschild radius to grow and the same time the internal pressure of the Earth would grow as it compresses under its own weight. Once the theoretical Schwarzschild radius is equal to the radius of heavy Earth the Earth is officially a black hole. The collapse was probably underway before this occurs and it likely that due to the density and pressure being greatest near the centre the collapse starts nearest the centre and works it way outwards heuristically.

  • $\begingroup$ Oppenheimer-Snyder collapse is the analytic solution for a sphere of dust surrounded by vacuum (assuming only uniformity within the sphere). Within the sphere, it is locally equivalent to FLRW, even though the global conditions are not the same. $\endgroup$
    – Sten
    Commented Feb 23 at 19:49
  • $\begingroup$ My argument is that there is not uniformity within the sphere and more sophistication is required. There is also the simple fact that the infinite density of the singularity at the centre is unphysical, implies we do not fully understand black holes. $\endgroup$
    – KDP
    Commented Feb 23 at 19:54
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    $\begingroup$ That is fair. I read "solid sphere surrounded by vacuum, so not a uniform distribution of matter" as emphasizing the vacuum surroundings and wanted to point out that those do not present a difficulty. $\endgroup$
    – Sten
    Commented Feb 23 at 19:57
  • $\begingroup$ and your comment is also fair ;-) $\endgroup$
    – KDP
    Commented Feb 23 at 19:58
  • $\begingroup$ @Sten Modified my answer to try and incorporate the correction you suggested. $\endgroup$
    – KDP
    Commented Feb 23 at 20:08

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