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My understanding is that light can not escape from within a black hole (within the event horizon). I've also heard that information cannot propagate faster than the speed of light. It would seem to me that the gravitational attraction caused by a black hole carries information about the amount of mass within the black hole. So, how does this information escape? Looking at it from a particle point of view: do the gravitons (should they exist) travel faster than the photons?

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The no-hair theorem says black holes are not completely hairless. They have five hairs: mass-energy $M$, linear momentum $P$ (three components), angular momentum $J$ (three components), position $X$ (three components), electric charge $Q$. –  Ali Jul 3 '13 at 16:50
If you think of gravity as a repelling forces then it makes sense that gravity doesn't need to "escape" from a black hole. What fails to escape is the force that repels matter which makes the gravity seem stronger. –  Derek Tomes Aug 14 at 3:08

13 Answers 13

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Well, the information doesn't have to escape from inside the horizon, because it is not inside. The information is on the horizon.

One way to see that is from the fact that from the perspective of an observer outside the horizon of a black hole, nothing ever crosses the horizon. It asymptotically gets to the horizon in infinite time (as it is measured from the perspective of an observer at infinity).

An other way to see that is the fact that from the boundary conditions on the horizon you can get all the information you need to describe the space-time outside but that is something more technical.

Finally, since classical GR is a geometrical theory and not a quantum field theory*, gravitons is not the appropriate way to describe it.

*To clarify this point, GR can admit a description in the framework of gauge theories like the theory of electromagnetism. But even though electromagnetism can admit a second quantization (and be described as a QFT), GR can't.

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Good answer, this addresses many of the issues. –  Noldorin Nov 16 '10 at 18:05
Last paragraph is just wrong, please remove it: GR is very similar to other (classical) gauge theories and there are reformulations where it is completely equivalent. Also gravitons are precisely what describes gravity in any reasonable quantum theory of gravity. And there is also no problem in quantizing e.g. gravitational waves (the quanta of which are gravitons) on curved background in the very same manner we quantize EM waves to get photons. –  Marek Nov 16 '10 at 18:59
I didn't say it is not a gauge theory. It is not a quantum field theory. Electromagnetism is also a gauge theory, one that admits a second quantization. I am not aware of a similar process for GR. –  Vagelford Nov 17 '10 at 10:50
Even with the clarification the last statement is not correct. Gravity is a field theory and you can think of the gravitational force in terms of graviton exchange. You can also quantize gravity perturbatively, see for instance arxiv.org/abs/gr-qc/9405057. Sure, it is a non-renormalizable QFT, but so are many other QFTs that are widely used in physics. –  Daniel Grumiller Feb 10 '11 at 1:35
Yes. GR is an effective quantum field theory. It is non-renormalizable, like many other effective quantum field theories. However, I think for the current context quantum considerations are actually irrelevant. The static 1/r potential comes from classical considerations. –  Daniel Grumiller Feb 10 '11 at 13:00

There are some good answers here already but I hope this is a nice short summary:

Electromagnetic radiation cannot escape a black hole, because it travels at the speed of light. Similarly, gravitational radiation cannot escape a black hole either, because it too travels at the speed of light. If gravitational radiation could escape, you could theoretically use it to send a signal from the inside of the black hole to the outside, which is forbidden.

A black hole, however, can have an electric charge, which means there is an electric field around it. This is not a paradox because a static electric field is different from electromagnetic radiation. Similarly, a black hole has a mass, so it has a gravitational field around it. This is not a paradox either because a gravitational field is different from gravitational radiation.

You say the gravitational field carries information about the amount of mass (actually energy) inside, but that does not give a way for someone inside to send a signal to the outside, because to do so they would have to create or destroy energy, which is impossible. Thus there is no paradox.

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Note that there is NO NEED to indroduce any quantum mechanics AT ALL into this discussion. That's why I specifically said "electromagnetic radiation" and "gravitational radiation", not "photons" or "gravitons". –  Keenan Pepper Nov 17 '10 at 2:47
Hello Keenan, what about static magnetic field? –  Georg Jan 18 '11 at 11:29
+1 for being the only answer a non-physics major can understand. –  Anonymous Type Mar 10 '11 at 0:02

Let's get something out of the way: let's agree not to bring gravitons into this answer. The rationale is simple: when you talk about gravitons you imply a whole lot of things about quantum phenomena, none of which is really necessary to answer your main question. In any case, gravitons propagate with the very same speed as photons: the speed of light, $c$. This way we can focus simply in Classical GR, ie, the Differential Geometry of Spacetime: this is more than enough to address your question.

In this setting, GR is a theory that says how much curvature a space "suffers" given a certain amount of mass (or energy, cf Stress-Energy Tensor).

A Black Hole is a region of spacetime that has such an intense curvature that it "pinches out" a certain region of spacetime.

In this sense, it's not too bad to understand what's going on: if you can measure the curvature of spacetime, you can definitely tell whether or not you're moving towards a region of increasing curvature (ie, towards a block hole).

This is exactly what's done: one measures the curvature of spacetime and that's enough: at some point, the curvature is so intense that the light-cones are "flipped". At that exact point, you define the Event Horizon, ie, that region of spacetime where causality is affected by the curvature of spacetime.

This is how you make a map of spacetime and can chart black holes. Given that curvature is proportional to gravitational attraction, this sequence of ideas completely addresses your doubt: you don't have anything coming out of the black hole, nor anything like that. All you need is to chart the curvature of spacetime, measuring what happens to your light-cone structure. Then, you find your Event Horizon and, thus, your black hole. This way you got all the information you need, without having anything coming out of the black hole.

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Suppose, very hypothetically of course, that some extra mass were suddenly created inside the black hole. Would the spacetime curvature outside the black hole change? I realize this is an unphysical process, but if the hand of God reached down and created a large lump of stuff just inside the event horizon, what do the equations of GR tell us about whether we would we be able to tell about the event from outside the event horizon? –  Mark Eichenlaub Nov 16 '10 at 19:40
The thing to note is that curvature is not something that lives only inside the Black Hole: this is a property of spacetime as a whole, and that's what counts. Global, topological, properties are very non-intuitive things. ;-) –  Daniel Nov 16 '10 at 19:50
@MarkE: only if you were God. Look, the bottom-line is that we're dealing with classical GR, and not Quantum Gravity nor its effects. And, within the framework of classical GR, it's simply not possible for you to change any of the properties (charge, mass, angular momentum) of a black hole from the inside of it. A black hole is simply a "sink" of gravitational fields. –  Daniel Nov 16 '10 at 20:25
@MarkE: I could open my magic toolbox and talk about holonomies and their relation with orbits in GR (ie, with closed geodesics). And, by mapping the holonomies of a space you can get information about its curvature. So, if you're orbiting a black hole, you can gather all the information about its curvature. (That's why i made that comment about global properties of spaces: they are very non-intuitive.) –  Daniel Nov 16 '10 at 20:27
@Nogwater: As far as we know, the mass itself is at the singularity, for some definition of "is" (namely that the proper time between crossing the event horizon and reaching the singularity is finite). But, speaking in vague terms, the information of how much mass there is gets "imprinted" on the horizon, it does not "fall" down to the singularity along with the mass. In more precise form, this is called the holographic principle. –  David Z Nov 17 '10 at 18:30

The problem here is a misunderstanding of what a particle is in QFT.

A particle is an excitation of a field, not the field itself. In QED, if you set up a static central charge, and leave it there a very long time, it sets up a field $E=k{q \over r^2}$. No photons. When another charge enters that region, it feels that force. Now, that second charge will scatter and accelerate, and there, you will have a $e^{-}->e^{-}+\gamma$ reaction due to that acceleration, (classically, the waves created by having a disturbance in the EM field) but you will not have a photon exchange with the central charge, at least not until it feels the field set up by our first charge, which will happen at some later time.

Now, consider the black hole. It is a static solution of Einstein's equations, sitting there happily. When it is intruded upon by a test mass, it already has set up its field. So, when something scatters off of it, it moves along the field set up by the black hole. Now, it will accelerate, and perhaps, "radiate a graviton", but the black hole will only feel that after the test particle's radiation field enters the black hole horizon, which it may do freely. But nowhere in this process, does a particle leave the black hole horizon.

Another example of why the naïve notion of all forces coming from a Feynman diagram with two pairs of legs is the Higgs boson—the entire universe is immersed in a nonzero Higgs field. But we only talk about the 'creation' of Higgs 'particles' when we disturb the Higgs field enough to create ripples in the Higgs field—Higgs waves. Those are the Higgs particles we're looking for in the LHC. You don't need ripples in the gravitational field to explain why a planet orbits a black hole. You just need the field to have a certain distribution.

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I think it's helpful to think about the related question of how the electric field gets out of a charged black hole. That question came up in the (now-defunct) Q&A section of the American Journal of Physics back in the 1990s. Matt McIrvin and I wrote up an answer that was published in the journal. You can see it at https://facultystaff.richmond.edu/~ebunn/ajpans/ajpans.html .

As others have pointed out, it's easier to think about the question in purely classical terms (avoiding any mention of photons or gravitons), although in the case of the electric field of a charged black hole the question is perfectly well-posed even in quantum terms: we don't have a theory of quantum gravity at the moment, but we do think we understand quantum electrodynamics in curved spacetime.

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While in many ways the question was already answered, I think it should be empasized that on the classical level, the question is in some sense backwards. The prior discussion of static and dynamic properties especially comes very close.

Let's first examine a toy model of a spherically-symmetric thin shell of dust particles collapsing into a Schwarzschild black hole. The spacetime outside of the the shell will then also be Schwarzschild, but with a larger mass parameter than the original black hole (if the shell starts at rest at infinity, then just the sum of the two). Intuitively, the situation is analogous to Newton's shell theorem, which a more limited analogue in GTR. At some point, it crosses the horizon and eventually gets crushed out of existence at the singularity, the black hole now gaining mass.

So we have the following picture: as the shell collapses, the external graviational field takes on some value, and as it crossed the horizon, the information about what it's doing can't get out the horizon. Therefore, he gravitational field can't change in response to shell's further behavior, for this would send a signal across the horizon, e.g., a person riding along with the shell would be able to communicate across it by manipulating the shell.

Therefore, rather than gravity having a special property that enables it to cross the horizon, in a certain sense gravity can't cross the horizon, and it is that very property that forces gravity outside of it to remain the same.

Although the above answer assumed a black hole already, that doesn't matter at all, as for a spherically collapsing star the event horizon begins at the center and stretches out during the collapse (for the prior situation, it also expands to meet the shell). It also assumes that the situation has spherical symmetry, but this also turns out to not be conceptually important, although for far more complicated and unobvious reasons. Most notably, the theorems of Penrose and Hawking, as it was initially thought by some (or perhaps I should say hoped) that any perturbation from spherical symmetry would prevent black hole formation.

You may also be wondering about a related question: if the Schwarzschild solution of GTR is a vacuum, does it make sense for a vacuum to bend spacetime? The situation is somewhat analogous to a simpler one from classical electromagnetism. Maxwell's equations dictate how the electric and magnetic fields change in response to the presence and motion of electric charges, but the charges alone do not determine the field, as you can always have a wave come in from infinity without any contradictions (or something more exotic, like an everywhere-constant magnetic field), and in practice these things are dictated by boundary conditions. The situation is similar in GTR, where the Einstein field equation that dictates how geometry are connected only fixes half of the twenty degrees of freedom of spacetime curvature.

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Gravitation doesn't work the way light does (which is why quantum gravity is hard).

A massive body "dents" space and time, so that, figuratively speaking, light has a hard time running uphill. But the hill itself (i.e. the curved spacetime) has to be there in the first place.

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A bit vague, but on its way to being a good answer. –  Noldorin Nov 16 '10 at 18:06
But, you could also ask "How does electric force escape a charged black hole", which would be an equally valid question. –  Jerry Schirmer Feb 10 '11 at 15:10

I think the best explanation that can be given is this: you have to discern between statical and dynamical properties of the space-time. What do I mean by that?

Well, there are certain space-times that are static. This is for example the case of the prototypical black-hole solution of GTR. Now, this space-time exists a priori (by definition of static: it always was there and always will be), so the gravity doesn't really need to propagate. As GTR tells us gravity is only an illusion left on us by the curved space-time. So there is no paradox here: black holes appear to be gravitating (as in producing some force and being dynamical) but in fact they are completely static and no propagation of information is needed. In reality we know that black-holes are not completely static but this is a correct first approximation to that picture.

Now, to address the dynamical part, two different things can be meant by this:

  • Actual global change of space-time as can be seen e.g. in the expansion of the universe. This expansion need not obey the speed of light but this is in no contradiction with any known law. In particular you cannot send any superluminal signals. In fact, opposite is true: by too quick an expansion parts of universe might go too far away for even their light to ever reach us. They will get causally disconnected from our sector of space-time and to us it will appear as if it never existed. So it shouldn't be surprising that no information can be communicated.
  • Gravitational waves, which is a just fancy name for the disturbances in the underlying space-time. They obey the speed of light and the corresponding quantum particles are called gravitons. Now these waves/particles indeed wouldn't be able escape from underneath the horizon (in the precisely the same way as any other particle, except for Hawking radiation, but this is a special quantum effect).
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In my opinion this is an excellent question, which manages to puzzle also some accomplished physicists. So I do not hesitate to provide another, a bit more detailed, answer, even though several good answers exist already.

I think that at least part of this question is based upon an incomplete understanding on what it means to mediate a static force from a particle physics point of view. As others have mentioned in their answers already, you encounter a similar issue in the Coulomb problem in electrodynamics.

Let me answer your question from a field theory point of view, since I believe this concurs best with your intuition about particles being exchanged (as apparent from the way you phrased the question).

First, no gravitational waves can escape from inside the black hole, as you hinted already in your question.

Second, no gravitational waves have to escape from inside the black hole (or from the horizon) in order to mediate a static gravitational force.

Gravity waves do not mediate the static gravitational force, but only quadrupole or higher moments.

If you want to think about forces in terms of particles being exchanged you can view the static gravitational force (the monopole moment, if you wish) as being mediated by "Coulomb-gravitons" (see below for the analogy with electrodynamics). Coulomb-gravitons are gauge degrees of freedom (so one may hesitate to call them "particles"), and thus no information is mediated by their "escape" from the black hole.

This is quite analog to what happens in electrodynamics: photon exchange is responsible for the electromagnetic force, but photon waves are not responsible for the Coulomb force.

Photon waves do not mediate the static electromagnetic force, but only dipole or higher moments.

You can view the static electromagnetic force (the monopole moment, if you wish) as being mediated by Coulomb-photons. Coulomb-photons are gauge degrees of freedom (so one may hesitate to call them "particles"), and thus no information is mediated by their "instantaneous" transmission.

Actually, this is precisely how you deal with the Coulomb force in the QFT context. In so-called Bethe-Salpeter perturbation theory you sum all ladder graphs with Coulomb-photon exchanges and obtain in this way the 1/r potential to leading order and various quantum corrections (Lamb shift etc.) to sub-leading order in the electromagnetic fine structure constant.

In summary, it is possible to think about the Schwarzschild and Coulomb force in terms of some (virtual) particles (Coulomb-gravitons or -photons) being exchanged, but as these "particles" are actually gauge degrees of freedom no conflict arises with their "escape" from the black hole or their instantaneous transmission in electrodynamics.

An elegant (but perhaps less intuitive) way to arrive at the same answer is to observe that (given some conditions) the ADM mass - for stationary black hole space-times this is what you would call the "black hole mass" - is conserved. Thus, this information is provided by boundary conditions "from the very beginning", i.e., even before a black hole is formed. Therefore, this information never has to "escape" from the black hole.

On a side-note, in one of his lectures Roberto Emparan posed your question (phrased a bit differently) as an exercise for his students, and we discussed it for at least an hour before everyone was satisfied with the answer - or gave up ;-)

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Could you provide a link to the lecture? –  Ehryk Aug 19 '12 at 2:22
@Ehryk: no, sorry - there were no lecture notes, just a series of blackborad lectures –  Daniel Grumiller Aug 22 '12 at 17:24

The black hole does "leak" information, but it is not due to "gravitions, but in the form of the Hawking radiation. It has its basis in quantum mechanics, and is a thermal sort of radiation with extremely low rate. This also means that the black hole is slowly evaporates, but on a time scale that is comparable to the age of the universe.

The origin of this radiation can be described in a little bit hand-waving way as such: due to quantum fluctuations, there's particle-antiparticle pair creation going on in the vacuum. If such a pair-creating happens on the horizon, one of the pair can fall into the black hole while the other can escape. To preserve the total energy (since the vacuum fluctuations are around 0) with a particle now flying away, its fallen pair has to have a negative energy from the black hole's point of view, thus it is effectively losing mass. The outside observer perceives this whole process as "evaporation".

This radiation has a distribution as described by a "temperature", which is inversely proportional to the black hole's mass.

Might want to check out http://en.wikipedia.org/wiki/Hawking_radiation and other sources for more details...

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The various theories - QED, GTR, classical electromagnetism, quantum loop gravity, etc - are all different ways to describe nature. Nature is what it is; theories all have defects. So far as saying whether gravity resembles electromagnetism in some way or not, is just blowing warm air about how humans think and not saying anything substantial about physical reality.

So what if we don't have a full grasp of quantum gravity? Gravitons are a sensible concept, and a key part in some unified (or semi-unified) field theories. It might get tricky because unlike other quantum particles, gravitons are a part of the curvature of spacetime and the relations of nearby lightcones, as they fly through said spacetime. We can sort of ignorer that for now. The question is good, and can be answered in terms of quantum theory and gravitons. We just don't know, given the existing state of physics knowledge, how far we can push the idea.

When charged particle attract or repel, the force is due to virtual photons. Photons like to travel at the universal speed c, but they don't have to. Heisenberg says so! You can break the laws of conservation of energy and momentum as much as you like, but the more you deviate, the shorter the time span and smaller the bit of space in which you violate these laws. For the virtual photons connecting two charged particles, they've got the room between the two particles, and a time span matching that at lightspeed. These not running waves with a well-defined wavelength, period or phase velocity. This ill-defined velocity can be faster than c or less equally well. In QED, the photon propagator - the wavefunction giving the probability amplitude of a virtual photon connecting (x1, t1) to (x2, t2) is nonzero everywhere - inside and outside the past and future light cones, though becoming unlimited in magnitude on the light cones.

So gravitons, if they are that much like photons, can exist just fine outside the horizon and inside. They are, in a rough sense, as big as the space between the black hole and whatever is orbiting or falling into it. Don't picture them as little energy pellets flying from the black hole center (singularity or whatever) - even with Heisenberg's indulgence, it's just not a matter of small particles trying to get through the horizon the wrong way. A graviton is probably already on both sides!

For a more satisfying answer, I suspect it takes knowing the math,Fourier transforms, Riemann tensors and all that.

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The holographic principle gives a clue, as pointed out by David Zaslavsky. The Schwarzschild metric element $g_{tt}~=~1 – r_0/r$, for $r_0~=~2GM/c^2$ gives a proper distance called the delay coordinate $$ r^*~=~r~+~r_0 ln[(r-r_00)/r_0] $$ which diverges $r^*~\rightarrow~-\infty$ as you approach the horizon. What this means is that all the stuff which makes up the black hole is never seen to cross the horizon from the perspective of a distant outside observer. The clock on anything falling into a black hole is observed to slow to a near stop and never cross the horizon. This means nothing goes in or out of the black hole, at least classically. So there really is not problem of gravity escaping from a black hole, for as observed from the exterior nothing actually ever went in.

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"Proper distance" means taking the length of a curve in a spatial hyperslice, but $r^*$ is not produced in the surface of constant Schwarzschild time, so it's unclear what you're referring to. For radial light rays, $\Delta t = \pm\Delta r^*$, which is relevant to "not seeing", but this comes from $g_{rr}$, not $g_{tt}$. "Nothing goes in or out of the black hole" is just wrong, although it was the view before the mid-1960s, with falling objects slowing and stopping at the infinite redshift surface. (EG: acceleration in Minkowski; things obviously cross the horizon without been seen to do so.) –  Stan Liou Jan 18 '11 at 5:00
I should have said proper interval. The toroise coordinate indicates that to see something from the horizon it is seen from the "infinite past." Nothing can be directly observed to actually reach the event horizon. –  Lawrence B. Crowell Jan 18 '11 at 19:19
There are observational evidences of matter crossing the horizon of black holes and simply increasing the BH mass, in binary systems. Because the typical spectral fingerprint of the shock wave heating in similar systems but with a white dwarf as accreting object instead of a BH, is absent. Whatever happens to proper time of the accreting matter, it crosses the horizon. –  Eduardo Guerras Valera Dec 5 '12 at 21:40

No escape is necessary (a slightly diifferent perspective).

A lot of nice answers so far but a couple of things need mentioning. It's not clear where, exactly, the mass of the black hole is supposed to be. Where does the mass reside? That's one thing. The other thing is, how does the mass/energy in the gravitational field, itself, fit into this picture?

I think (and I'll no doubt get hammered mercilessly for this) that the mass of a black hole resides spread out through its external gravitational field and nowhere else. The mass of a black hole resides, wholly and solely, in the gravitational field outside the hole. Fortunately for me, I'm not completely alone here.

The calculation of the total gravitational field energy of a black hole (or any spherical object) was made in 1985 by the Cambridge astrophysicist Donald Lynden-Bell and Professor Emeritus J. Katz of the Racah Institute Of Physics. http://adsabs.harvard.edu/full/1985MNRAS.213P..21L, Their conclusion was that the total energy in the field is ... (drum-roll here) ... mc^2 !!!

The total mass of the BH must reside, completely, and only, in the self-energy of the curvature of spacetime around the hole!

Here are a couple of quotes from the paper: "... the field energy outside a Schwarzschild black hole totals Mc^2." and, " ... all these formulae lead to all the black hole's mass being accounted for by field energy outside the hole."

The answer to your question, then, is this: information about the mass of a black hole doesn't have to escape from within the black hole because there is no mass inside the black hole. All the mass is distributed in the field outside the hole. Therefore, no information needs to escape from inside.

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protected by David Z Jan 13 '11 at 22:20

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