# Can a massive object have an escape velocity close to $c$ and not become a black hole?

It is well-documented that a given body has a well-defined Schwarzschild radius, defining a minimum radius for its volume, given its mass, before it becomes a black hole. It is clearly true, also, that the escape velocity for black holes is greater than $c$ - however, would a mass with an escape velocity 0.999$c$ also be a black hole?

My belief is that such an object would be a black hole (or on the way to being one) since such escape velocities might only be achievable in a situation where a mass is going to or is becoming one (for example, it might be the escape velocity of a collapsing star). My question, then, is: what is the lower-bound for the escape velocity of an object before we can know that it is or will be a black hole?

• You should probably be more precise with your question in the title of your question. At the moment it just says "Escape Velocity for Black Holes" which is not a valid question. – Alf Mar 14 '18 at 20:00
• “Escape velocity” is not sufficient to explain an event horizon at all. Being slower than the escape velocity means that the object will eventually fall back instead of escaping, but it doesn’t define an impassable border. With Newtonian mechanics, you could still travel through the entire universe while being slightly slower than the escape velocity. Or, as others already put it, Newtonian mechanics can’t explain a black hole… – Holger Mar 15 '18 at 9:17
• There is also the matter of Hawking radiation. Your question is too simple. – Klaws Mar 15 '18 at 13:09
• @Klaws: hawking radiation is not relevant to this question. – Jerry Schirmer Mar 15 '18 at 18:33
• You are speaking of speed as though it were absolute or relative to light-You are already going 99.996ish% the speed of light relative to some particles entering the atmosphere! You, me, the sun and every other large chunk of matter is going exactly 0.000% the speed of light relative to LIGHT. We don't move or accelerate relative to light or in any absolute way. If you accelerate to 99.9%C relative to earth, you can still call that a "Standstill" and accelerate by another 99.9% of C in the same direction and do it again and never move relative to "light", only to the objects left behind. – Bill K Mar 15 '18 at 20:28

Whether the black hole could form at all among other thing is determined by the distribution of angular momentum in the system. So if we allow rotating bodies then the answer is: escape velocity of a stationary body could be arbitrarily close to $c$ without forming a black hole.

The only other answer that mentions rotations is Rob Jeffries' who seems to believe that rotation could only change situation only slightly. This may be true for astrophysical bodies such as neutron stars which shed excessive angular momentum during collapse. However, why limit ourselves with only natural bodies? If we can construct stationary rotating structures, then we in principle could produce extended bodies with rather extreme metrics, for example metrics arbitrarily close to extreme Kerr metric, yet not being a black hole. And so an escape velocity for such an object would be arbitrarily close to $c$ (at least in some parts of this object). The matter for the structure could be quite ordinary (no need for degenerate neutron matter) or dust-like (which simply means that internal tensions or pressures everywhere would be arbitrarily small compared to relativistic energy density $\rho c^2$).

An example of such construction could be seen here:

Neugebauer, G., & Meinel, R. (1993). The Einsteinian gravitational field of the rigidly rotating disk of dust. The Astrophysical Journal, 414, L97-L99, full text at adsabs.

There the authors construct a family of solutions for thin rigidly rotating disk (with varying density) that has an extreme Kerr metric as a limiting case. And of course, such configuration is not the only one possible. Quick googling produced another one:

Ansorg, M., Kleinwächter, A., & Meinel, R. (2002). Relativistic Dyson rings and their black hole limit. The Astrophysical Journal Letters, 582(2), L87, arXiv.

• Maximally rotating neutron stars also have escape velocities that must be less than $c$. – Rob Jeffries Mar 15 '18 at 23:08
• I never suggested otherwise. From your updated answer (and Cipolletta et al) I see that this maximum rotation still has a finite 'gap' from a black hole. The constructions that I dug up do not require stiff equations of state and would approach extremal Kerr (from the 'other side' to black hole's) arbitrarily close. – A.V.S. Mar 16 '18 at 4:13

Unlike with the other answer, my presumption is that you're asking about the largest possible escape velocity from a stable object that's not a black hole. I.e., the circumstances you're considering don't include a situation in which the "object" involved includes some matter that's in the process of being rapidly expelled. The best candidate for such an object that's moderately well understood is likely a neutron star.

The relationship between density and mass of a neutron star isn't very precisely understood; there are a number of somewhat differing models for a neutron star's equations of state. There may well even exist quark matter that's even denser than the neutron degenerate matter that's thought to form the bulk of a neutron star, but that's venturing off even further into the limits of what's currently known.

However, according to some equations of state, there can exist "ultracompact" neutron stars which are so dense that they have a photon sphere. A photon sphere occurs at the radius at which a photon can orbit the object in question in a circle. In Schwarzschild coordinates, the photon sphere occurs at 1.5 times the Schwarzschild radius. I.e., the photon sphere occurs at a radius given by

$$r_{ps}=\frac{3GM}{c^2}\ \ .$$

Using $r_{ps}$ as the radius in the equation for escape velocity,

$$v_e=\sqrt{\frac{2GM}{r}}\ \ ,$$

which is valid even relativistically, gives that the escape velocity from a photon sphere is

$$v_e=\sqrt{\frac{2}{3}}c \approx 0.816c\ \ .$$

So it appears that there can exist stable objects (ultracompact neutron stars) which are not black holes, but which require an escape velocity of more than 0.8c to escape from the surface.

As a preliminary, you're using a Newtonian explanation of what is a black hole, which doesn't really work. A black hole is not defined as an object from which the escape velocity is $c$. If the Newtonian idea of escape velocity were the only thing at work, then we could hoist matter out of a black hole using a bucket on a rope.

Neither Newtonian mechanics nor general relativity prevents us from having materials with exotic properties that would allow such a collapse to halt at any stage. However, in GR we have things called energy conditions, which state some bounds on our typical expectations for the behavior of most ordinary forms of matter. Given an appropriate energy condition, there is a theorem called the Penrose singularity theorem that states that under a certain condition (the formation of a trapped surface), a singularity is guaranteed to exist somewhere in the spacetime. This singularity does not necessarily have to be a black hole singularity, and it does not necessarily have to be surrounded by an event horizon.

So your Newtonian intuition was not completely off. There is something like the thing you imagined, but the details play out much differently in GR than in Newtonian gravity.

• Ben I find material in Shapiro & Teukolsky that seems to contradict this answer. Even if infinite pressure were allowed, collapse will occur at radii greater than $r_s$. For crazy equations of state like $P = \rho c^2$, the limit is even larger. – Rob Jeffries Mar 15 '18 at 18:33
• @RobJeffries: I can't really tell what you're describing. Would it be possible to post this as a separate question, quoting the relevant material from Shapiro? – Ben Crowell Mar 15 '18 at 21:17
• I don't have the book on me at the moment. The relevant pages are referred to in my answer. 260-261. What i am describing is that if a mass is compressed to less than some multiple of its Schwarzschild radius then it inevitably collapses, whatever the EOS. That multiple is like 1.2-1.3. – Rob Jeffries Mar 15 '18 at 22:31
• @RobJeffries: I located a copy of the book (1st ed., which seems to be the only edition?). I don't see anything like what you're describing on pp. 260-261, which contain exercises 9.6-9.10 and figure 9.7. There is a note in the figure about the idea that the speed of sound should be less than or equal to the speed of light. Note that that is equivalent to an energy condition. Without an energy condition, I don't see how one could ever make any statement like the one you're describing. – Ben Crowell Mar 17 '18 at 16:10
• I still have no access to my copy. The argument is in the section on neutron star stability. This specifically rules out any kind of stability for configurations with radii less than about 1.3 $r_s$, even if $P= \rho c^2$. The limit is larger for more realistic EOS. What exotic properties would allow you to have a smaller neutron star? – Rob Jeffries Mar 17 '18 at 16:24

You are correct, that there is a mass/radius ratio that makes it inevitable that an object will collapse to form a black hole and that this mass/radius ratio has a corresponding "escape velocity" (NB. it is a speed in Newtonian physics, but in GR it is a velocity because I think the direction matters) that is less than $c$. If an object of a given mass shrinks below this critical radius, which is larger than the Schwarzschild radius, then it will collapse to form a black hole.

The structure of a General Relativistic object is controlled by the Tolman-Oppenheimer-Volkhoff equation of hydrostatic equilibrium. This has the pressure gradient on the LHS but also features the pressure on the right hand side, because pressure is a source of space-time curvature in GR. As the object gets smaller and approaches the Schwarzschild radius, the central pressure must increase to provide the necessary pressure gradient to support the increasing weight. However, this pressure also contributes to the requirement for an increased pressure gradient and the whole thing becomes self-defeating and the object will collapse.

The details depend on the specifics of the equation of state for material at ultra-high densities, thought to exist in neutron stars, which is highly uncertain. However, there is a limit. In "Black Holes, White Dwarfs and Neutron Stars" by Shapiro & Teukolsky, (pp.260-263), it is shown, approximately, that even if the equation of state hardens to the point where the speed of sound equals the speed of light, that instability sets in if $(GM/Rc^2)<0.405$. [NB. This is for non-rotating objects, which might change things slightly, but even if causality were abandoned and you allow $P \rightarrow \infty$ then $(GM/Rc^2)<0.444$ (this the so-called Buchdahl limit).]

The Schwarzschild radius is $R_s=2GM/c^2$ and therefore $R > 1.23 R_s$ for stability. This limit is reached for a neutron star with $M \simeq 3.5 M_{\odot}$ using this equation of state. The "radial escape velocty" (according to a "shell observer" stationary at that radius) for such an object is $$v = \left( \frac{2GM}{1.23R_s} \right)^{1/2} = \frac{c}{\sqrt{1.23}}$$

A more accurate treatment in Lattimer (2013) suggests that a maximally compact neutron star has $R\geq 1.41R_s$, which leads to an escape speed of $c/\sqrt{1.41}$.

In practice the maximum escape speed will be smaller than this because the real equation of state is unlikely to be as extreme as supposed above.

The picture below (from Demorest et al. 2010) shows the mass-radius relations for a wide variety of equations of state. The limits in the top-left of the diagram indicate the limits imposed by (most stringently) the speed of sound being the speed of light (labelled "causality" and which gives radii slightly larger than Shapiro & Teukolsky's approximate result) and then in the very top left, the border marked by "GR" coincides with the Schwarzschild radius. Real neutron stars become unstable where their mass-radius curves peak, so their radii are always significantly greater than $R_s$ at all masses and the escape speed will be given by $c$ divided by the square root of their smallest possible radius as a multiple of the Schwarzschild radius. EDIT: Just to address the point on rotation. I found a paper that adopt the "causal" equation of state and allows neutron stars to rotate as fast as they possibly can (Friedman & Ipser 1987; see also more modern work by Cipolleta et al. 2015). These configurations do allow more massive neutron stars to exist (by 30% or more), but they also have larger radii. The net outcome is almost identical - the minimum stable radius is about $1.3R_s$. What I am unsure about is what the relationship between escape velocity and radius is in the Kerr metric. (Or even how that would be defined).

As long as the escape velocity is less than $c$, then the matter composing the body in question could be moving radially outward at a larger velocity which would prevent it from collapsing into a black hole. Therefore there is no threshold above the Shwarzschild radius at which something must become a black hole in the future.

Note this answer is theoretical and does not deal with what may or may not be considered "realistic" velocities or situations.

• An electron moving at a speed of .999999999999c will have an energy of merely 361 GeV. The LHC could realistically create plenty of such particles. – probably_someone Mar 14 '18 at 19:56
• I was answering the question as though it were purely theoretical. Of course you can always add enough nines and say nothing has ever or will ever move this fast and then the statement "If something has this escape velocity, then in all cases it will become a black hole." – dylnan Mar 14 '18 at 19:57
• @probably_someone I love the image of throwing the LHC into an almost-black hole, just to demonstrate that we can throw electrons out of it. – Cort Ammon - Reinstate Monica Mar 15 '18 at 0:31
• @CortAmmon I mean, what else are you going to do with it when the Future Circular Collider gets built? – probably_someone Mar 15 '18 at 0:32
• I think this answer obfuscates the true intent of the question. The question talks about an object, not an explosive event. For an "object" there is a radius (greater than the Schwarzschild radius) for which collapse to a black hole is inevitable. Even then, the problem you have not considered is that by having matter with this sort of energy you are vastly increasing the gravitational mass of what is there and thus vastly increasing (in energetic terms) the escape speed. The process becomes self-defeating. – Rob Jeffries Mar 15 '18 at 18:29

Since time stops inside a black hole, an observer far away from a collapsing star would observe that it takes an infinitely long time for the star to completely collapse. It would only be 99.9999% collapsed. This is similar to walking in a race and the closer you get to the finish line, the slower you go to where you never reach the finish line. This has the effect of allowing a fast space ship to go into the black hole and then leave as if it were still a neutron star (from the perspective of the ship). Although from our perspective, the ship went into the black hole and never came out because it took them billions of our years to leave. If what's preventing the star from collapsing is time itself being very slow due to gravitational time dilation, it would be very close to being a black hole and light would take a very long time to leave the star. This would make the black hole seem black and at the same time the laws of physics don't break down. This would explain why when black holes collide, they form a bigger black hole. It could be seen as two neutron stars colliding with the collision taking an almost infinitely long time to happen from our perspective. So to answer your question: A star with an escape velocity of .9999 c would seem like a black hole because light would take a very long time to leave and the number of photons released per second would decrease to very close to zero.