Something that's always confused me. How large is a black hole's physical size - not mass?

From descriptions, it would seem that the 'singularity' is a single point, but is it really?

Say for arguments sake, a 110 solar mass black hole. Obviously it's not going to be the same massive size as a 110 solar mass star (such as Cygnus OB2-12 at a physical size of 246 R☉), but would it really be a tiny point smaller than say a pin head?


2 Answers 2


A few comments:

1) We don't know what happens with objects of volume smaller than $\ell_{p}^{3}$, where $\ell_{p}$ is the Planck length, which is approximately $10^{-35}$ m. The known laws of physics do not hold for anything that small.

2) General relativity, in its classical glory, predicts that objects will collapse down to a shape with zero volume. A spinning black hole will take the shape of a ring, with a radius determined by the angular momentum and mass of the black hole. Since it is virtually impossible to tune down the angular momentum of physical objects to exactly zero, we will expect most singularities to be ring-shaped, perhaps having a $\ell_{p}$ thickness.

3) the singularity is not expected to be observable to outside observers, so most astrophysicists will instead talk about the size of the black hole horizon, which is a macroscopic size.

  • $\begingroup$ General relativity, in its classical glory, predicts that objects will collapse down to a shape with zero volume. Not true: physics.stackexchange.com/questions/144447/… $\endgroup$
    – user4552
    Apr 11, 2019 at 20:59
  • $\begingroup$ @BenCrowell: it's kind of weird that that answer doesn't discuss the well-known Oppenheimer-snyder collapse, in which a matter distribution does produce a schwarzschild black hole at late times. Also, a Kerr singularity is known to be a zero-volume ring in the spacetime, and since both the kerr and schwarzschild spacetimes are known to be locally stable against small perturbations, one would expect 'near hole" matter collapse distributions to produce them after ringdown. $\endgroup$ Apr 11, 2019 at 22:59

Usually the size of the black hole is considered to be the point known as the Schwarzschild radius. This is the point at which time slows down to the extent that matter is never seen crossing it - one becomes frozen in time approaching this point. It is given by $2 G M/c^2$. As you can see the black hole can in principle be as small or as large as one desires, although interactions other than gravitational might prevent very tiny black holes from forming (it is an interesting question what is the smallest possible size of a black hole, to think of it). I suggest you read more about it here: http://en.wikipedia.org/wiki/Schwarzschild_radius

  • $\begingroup$ Both of these are great, thanks! I'll have to give the answer to Jerry though as that goes into a little more of the specifics of the object itself $\endgroup$ Oct 5, 2012 at 0:04
  • $\begingroup$ this answer might be a bit misleading. Incoming matter may appear to be frozen at the Schwarzschild radius, never seen crossing it, but in reality incoming matter does cross the Schwarzschild radius and keeps on going into the singularity at r=0. Only the outgoing light rays are frozen. So the real size of any black hole is 0, as stated in the first answer. Of course this does not take into account any quantum effect at Planck length. $\endgroup$
    – magma
    Oct 6, 2012 at 15:26
  • $\begingroup$ Is this really true though? I was wondering about this myself. If you look at the coordinates of the objects as the asymptotic observer sees it, the object approaching the event horizon actually slows down and does not cross the horizon. I am happy to rewrite my answer if you can convince me that it is only the light rays that make the object appear as if it had not crossed the horizon but that it actually did. $\endgroup$
    – SMeznaric
    Oct 7, 2012 at 19:28
  • $\begingroup$ BH is usually defined to be a region of spacetime where nothing can come out of. In that sense, an event horizon (for stationary BH) is the boundary of a BH. $\endgroup$
    – Demian Cho
    Oct 11, 2012 at 21:52
  • $\begingroup$ In-fall object crosses horizon in in its own frame and hit the singularity in finite proper time in BH. $\endgroup$
    – Demian Cho
    Oct 11, 2012 at 21:54

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