Reconciling graviton exchange with a black hole

In summary, the conversation discusses the concept of gravity in both classical general relativity and quantum gravity. In classical general relativity, gravity is seen as a geometric manifestation and the orbits of test particles are described as geodesics. However, in quantum gravity, gravitational interactions are mediated by graviton exchange between two masses. The conversation also delves into the issue of graviton sources and whether they are emitted from the horizon of a black hole. In addition, the concept of virtual particles and their role in explaining non-locality is also brought up. The conversation ends with a discussion on how this non-locality may have wider implications for black hole entropy and event horizons.
  • #1
A/4
56
3
Perhaps this is a bit of a naive question, but I'm having trouble with the following issue. It is easy to describe the motion of a test particle near a black hole using classical general relativity. The underlying assumption is that gravity is a geometric manifestation, and the orbits are geodesics. The test particle does not directly interact with the black hole, only the spacetime curvature.

From the quantum gravity perspective, however, gravitational interactions are mediated by graviton exchange between two masses. In the case of black holes, what is the graviton source? Are they emitted from the horizon? (since clearly they cannot come from the interior).

Any insight or pointers to appropriate literature would be appreciated.
 
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  • #2
Ok, I think the if you think of the horizon as a horizon in space-time and not just space it becomes more obvious what is going on in terms of gravitons. Consider a mass M collapsing to form a black hole; as the radius containing the mass approaches the Schwarzschild radius the time as observed from a distant observer tends to infinity. This means that according to the distant observer the black hole really takes an infinitude time to form an event horizon. In other words the event horizon has a time coordinate at t=infinity. This means that all the mass is still outside the horizon and the gravitons escape from all the matter even that within the Schwarzschild radius because it is not until all the mass is within the radius that the event horizon is formed.

Of coarse when an observer falls into the black hole they will cross this horizon in a finite amount of time as measured by their own clock.

These arguments are really just classical ones as the matter is producing the gravitational field. The gravitational simple being the metric.

In QG a graviton would be quanta of the metric.
 
  • #3
Finbar said:
as the radius containing the mass approaches the Schwarzschild radius the time as observed from a distant observer tends to infinity.
This is valid only at the classical level though, so how can you explain a quantum effect ?

Take it that way : forget about gravitons since we do not have a consensus on quantum gravity. The black hole can still carry an electric charge. How can you account for the photon exchange ?

This is exactly what Baez uses to approach this FAQ.
 
  • #4
humanino said:
This is valid only at the classical level though, so how can you explain a quantum effect ?

Take it that way : forget about gravitons since we do not have a consensus on quantum gravity. The black hole can still carry an electric charge. How can you account for the photon exchange ?

This is exactly what Baez uses to approach this FAQ.

Does Baez have a specific FAQ / "This Week's Find" on the subject? If so, can you please direct me to it.

Edit: Never mind. Google is your friend:

http://math.ucr.edu/home/baez/physics/

That being said, does anyone else have input on the original question?
 
  • #5
humanino said:
This is valid only at the classical level though, so how can you explain a quantum effect ?

Take it that way : forget about gravitons since we do not have a consensus on quantum gravity. The black hole can still carry an electric charge. How can you account for the photon exchange ?

This is exactly what Baez uses to approach this FAQ.

Well as u say there is no consensus on QG. But in the classical limit what i said applies. Gravitons are still being exchanged in the classical limit so the conceptual problem expressed in the OP still applies there.

I would account for photon exchange the same way. All the mass and charge remains outside the event horizon, as observed by a distant observer, so the photons never have to cross the horizon.
 
  • #6
Finbar said:
...
I agree that one can calculate classical results using quantum field theory methods. But I have never actually done or seen such a calculation for charged BH so I can not claim certainty. It seems to me however that if one wanted to attack the problem, most likely one would take a semi-classical approach where the gravitational field is classical, the BH is stationary, the charge has definitely fallen into the BH, far after the remaining amount of energy trapped on the surface is negligible enough so that no single real photon can be detected from there. My point was that this situation is not so unrealistic at all, and virtual photons are not confined to be inside a light cone, so they can escape the BH. This is what Baez claims. I would appreciate if someone has reference showing calculation of what part comes from the past infalling charge and what part actually comes from inside. I believe if one waits long enough, nothing comes from the past infalling matter at all.
 
  • #7
If you think gravitons are "emitted", you don't understand field theory. That should be your first step, before trying to do any reconciliation. I would start with QED, as it's a relatively simple theory.
 
  • #8
Baez explanation of it as being due to a non-locality mechanism via virtual particles is neat. It ties in with Feynman sum-over-histories models where even the faster than light trajectories are explored before a collapse back to a least action mean path.

But does this nonlocal story have any wider implications for black hole entropy and event horizons?
 
  • #9
Vanadium 50 said:
If you think gravitons are "emitted", you don't understand field theory. That should be your first step, before trying to do any reconciliation. I would start with QED, as it's a relatively simple theory.
I think I understand your point, but just thought (jokingly) that real gravitons are emitted in the Hawking radiation :biggrin:
 
  • #10
Vanadium 50 said:
If you think gravitons are "emitted", you don't understand field theory. That should be your first step, before trying to do any reconciliation. I would start with QED, as it's a relatively simple theory.

The semantics aren't the issue, though your point is taken. If you draw a diagram for the exchange, the graviton will propagate between two particles will mass (say fermion lines). One of them will be the test particle. The other will be the "black hole" -- but what is that in this case?
 
  • #13
A/4 said:
The semantics aren't the issue, though your point is taken. If you draw a diagram for the exchange, the graviton will propagate between two particles will mass (say fermion lines). One of them will be the test particle. The other will be the "black hole" -- but what is that in this case?

The gravitons are coming from the matter fields (so yes the fermions lines, the photon lines all...particles in fact). And these matter fields remain outside the horizon as observed by a distant stationary observer. The gravitons from the matter that has fallen into the black hole are never felt by the observer since the occur after t= infinity as recorded by his clock.

Of course QG will spoil this picture but these effects will only become important when the mass of the black hole is very small and we don't know what will happen then. In QG one would expect all possible curvatures of space time to be included in the path integral. In this sense any virtual gravitons must be exchanged in a background independent way so one cannot assume the existence of a horizon during the graviton exchange it is only when the path integral is performed over all possible graviton exchanges that an expectation value of the space time geometry can be given. Remember the gravitons are the gravitational field and the gravitational field is the space time geometry.

I think the key to understanding the original post is to understand that the all the matter that constitutes the black hole must first of been just normal matter which then started to collapse. If one thinks about the matter already being behind the horizon then one is considering events that cannot effect observers outside the black hole.
 
  • #14
So I'll repeat again a little more explicitly... If at lowest order in a non-relativistic single graviton exchange, your test particle scatters "from infinity" at initial given momentum to a final momentum "at infinity", the scattering taking place on a static, infinitely massive BH, I claim that it is easy to show that you get a space-like graviton exchanged.

It is actually the same story if you consider a single photon exchanged in the scattering from a charged BH.

So my basic point is : showing this does not even involve calculating a matrix element. Can you do this ? If you can not, do you realize that you are questioning Feynman diagram calculations without being able to perform the most simple special relativity elementary calculation ? If you can, and you get a space-like exchange, why do you worry that it gets out of a light-cone !? It was never inside a light-cone to begin with !
 
  • #15
humanino said:
So I'll repeat again a little more explicitly... If at lowest order in a non-relativistic single graviton exchange, your test particle scatters "from infinity" at initial given momentum to a final momentum "at infinity", the scattering taking place on a static, infinitely massive BH, I claim that it is easy to show that you get a space-like graviton exchanged.

It is actually the same story if you consider a single photon exchanged in the scattering from a charged BH.

So my basic point is : showing this does not even involve calculating a matrix element. Can you do this ? If you can not, do you realize that you are questioning Feynman diagram calculations without being able to perform the most simple special relativity elementary calculation ? If you can, and you get a space-like exchange, why do you worry that it gets out of a light-cone !? It was never inside a light-cone to begin with !


eh? at lowest order there would be no loop corrections. Hence it must be on-shell so its following a null geodestic.

Are you claiming that if you turned on an electromagnetic field from behind the event horizon this could be observed by an observer outside the event horizon? Because that's what u seem to be claiming by saying a photon can travel out of the black hole?
 
  • #16
Finbar said:
eh? at lowest order there would be no loop corrections. Hence it must be on-shell so its following a null geodestic.
Certainly not. I think I gave you enough details for you to do the calculation. External legs are on-shell obviously, and this is exactly what forces a single graviton (or photon) exchanged (obviously in the t-channel) to be space-like.
Finbar said:
Are you claiming that if you turned on an electromagnetic field from behind the event horizon this could be observed by an observer outside the event horizon? Because that's what u seem to be claiming by saying a photon can travel out of the black hole?
No that's not what I am claiming.

A BH is characterized by just 3 numbers and one direction : its mass, electric charge and angular momentum. The charge is well inside the BH behind the horizon if you wait long enough for it to be stationary, or static, or whatever you want to call it. You are welcome to provide a reference or display the calculation of a dynamical situation as I mentioned earlier.
 
  • #17
The problem lies in trying to make the black hole act like a 'point' mass or charge and then use a test particle to probe the location of the horizon.

Clearly something goes wrong, since the horizon is a global property of spacetime and no local measurement can determine where it is (otherwise you violate the equivalence principle)

Instead you have to deal with the full nonlinear gravitational field directly and its influence on a test particle, only then do you recover the desired physics.

Even if you linearize everything the geometry will show up in the physics at infinity if you do the calculation properly.
 
  • #18
humanino said:
Certainly not. I think I gave you enough details for you to do the calculation. External legs are on-shell obviously, and this is exactly what forces a single graviton (or photon) exchanged (obviously in the t-channel) to be space-like.No that's not what I am claiming.

A BH is characterized by just 3 numbers and one direction : its mass, electric charge and angular momentum. The charge is well inside the BH behind the horizon if you wait long enough for it to be stationary, or static, or whatever you want to call it. You are welcome to provide a reference or display the calculation of a dynamical situation as I mentioned earlier.

Ok so if i do the calculation of a lowest order scattering amplitude in QED in flat space are you saying the exchanged photon would be space-like? say like page 153 in peskin and schroeder? here the q^2 = -2k^2 (1-cos theta). so q^2 is negative hence its space-like? I then this carrys over to the black hole in some limit? Is this your point?

If so ok I'll have a think.

Does this imply that if i have two electrons separated by a large distance that they can exchange photons quicker than the speed of light? This would be a virtual photon moving out side its light cone aswell. Something seems very wrong here.
 
  • #19
The truth is that I have very limited competence in those questions, and Haelfix could elaborate much more. My only very elementary point is the exchanged (boson) in the t-channel at lowest order is certainly space-like. I think Peskin and Schroeder explain very well how this does not violate causality or any other pathology.
Finbar said:
Does this imply that if i have two electrons separated by a large distance that they can exchange photons quicker than the speed of light? This would be a virtual photon moving out side its light cone aswell. Something seems very wrong here.
Note that you are confusing here the (momentum) virtuality of the exchanged (boson) with the (spatial) interval between the two scattering particles.
 
  • #20
Forget about doing a qft calculation, it just confuses the issue.

The problem as I understand what's being asked in this thread, remains the same at a purely classical (tree) level and a textbook on GR or black hole physics will resolve the confusion, preferably one that uses the same 'exchange carrier' lingo that particle physicists use. So maybe something like Weinberg.
 
  • #21
Finbar said:
This means that according to the distant observer the black hole really takes an infinitude time to form an event horizon.

So essentially, there are no black holes except on blackboards, right?
 
  • #22
Finbar said:
Ok, I think the if you think of the horizon as a horizon in space-time and not just space it becomes more obvious what is going on in terms of gravitons. Consider a mass M collapsing to form a black hole; as the radius containing the mass approaches the Schwarzschild radius the time as observed from a distant observer tends to infinity. This means that according to the distant observer the black hole really takes an infinitude time to form an event horizon.

In which case, an infalling observer will place the event horizon at some smaller radius than one at assymtotic infinity, so there no black holes except on blackboards, right?
 
  • #23
Haelfix said:
Forget about doing a qft calculation, it just confuses the issue.

The problem as I understand what's being asked in this thread, remains the same at a purely classical (tree) level and a textbook on GR or black hole physics will resolve the confusion, preferably one that uses the same 'exchange carrier' lingo that particle physicists use. So maybe something like Weinberg.

Agreed. The problem in the original post is the same for classical GR. That's why i think my original resolution of the problem may be along the right lines.
 
  • #24
Phrak said:
In which case, an infalling observer will place the event horizon at some smaller radius than one at assymtotic infinity, so there no black holes except on blackboards, right?

No the event horizon is at the same radius. The in falling observer will cross the horizon in a finite amount of time.
 
  • #25
Finbar said:
No the event horizon is at the same radius. The in falling observer will cross the horizon in a finite amount of time.

Ok.. you are assuming a coordinate singularity independent of observer, right? But there will not be a coordinate singularity for the observer at asymptotic infinity without assuming infinite time will be physically meaningful.
 
  • #26
A/4 posts:
The test particle does not directly interact with the black hole, only the spacetime curvature...From the quantum gravity perspective, however, gravitational interactions are mediated by graviton exchange between two masses.

I wonder why you accept spacetime curvature from a black hole but not gravitons...how does either "emerge"?? And how do we observe/measure mass, angular momentum and charge "from behind the horizon"...we don't...
 
  • #27
Naty1 said:
A/4 posts:

I wonder why you accept spacetime curvature from a black hole but not gravitons...how does either "emerge"?? And how do we observe/measure mass, angular momentum and charge "from behind the horizon"...we don't...

I don't think I said any of those things. I accept both curvature and gravitons.

I'm seeing two schools of thought emerging here. One is the "no hair" school -- all measurable quantities that would influence a test particle are encoded on the horizon (e.g. via entropy, etc...). The second is that this information is behind the event horizon, but can "communicate" with a test particle via spacelike gauge bosons.

Obviously this isn't as simple an issue as some make it out to be.
 
  • #28
Phrak said:
Ok.. you are assuming a coordinate singularity independent of observer, right? But there will not be a coordinate singularity for the observer at asymptotic infinity without assuming infinite time will be physically meaningful.

I assume the einstein equations. When i say 'event horizon' i mean the surface in space time that once information crosses it, it cannot travel back to the region of space it was in before. This is a global property independant of coordinates. A soloution to the eintein equations with such a property i call a 'black hole soloution'. If the einstein equations are correct such a soloution will be present in the universe; not just on blackboards.
 
  • #29
Naty1 said:
A/4 posts:

I wonder why you accept spacetime curvature from a black hole but not gravitons...how does either "emerge"?? And how do we observe/measure mass, angular momentum and charge "from behind the horizon"...we don't...


Yeah i agree. The problem exsists classically just as it does with gravitons. The conceptual error as i see it is to view soloutions such as the Schwarzschild soloution as physical. They are not because they assume that the matter has intially present behind the horizon. But one cannot assume this; the black hole must of collapsed to form a black hole. I won't repeat myself again so you can follow the logic.
 
  • #30
I posted:
I wonder why you accept spacetime curvature from a black hole but not gravitons...how does either "emerge"?? And how do we observe/measure mass, angular momentum and charge "from behind the horizon"...we don't...

and A/4 replied:
I don't think I said any of those things. I accept both curvature and gravitons


Yet your original post which started this thread seems to question the source for gravitons not curvature:

The test particle does not directly interact with the black hole, only the spacetime curvature. From the quantum gravity perspective, however, gravitational interactions are mediated by graviton exchange between two masses. In the case of black holes, what is the graviton source?

My only point is that it appears you have accepted the curvature of spacetime without any source...but question the source of gravitons...Remember gravitions are the quantum (discrete) version of contininuous classical waves...so I am simply pointing out that both the curvature, and separately gravitons, do have a cause, a source...and it's not from within the spacetime horizon...

In other words, why single out gravitons from all the other effects I list (angular momentum, curvature,etc) ?? All are rather mysterious since the event horizon censures (isolates) all within.
 
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