Tags:
1. Jan 25, 2017

### stevendaryl

Staff Emeritus
I'm sure that there are limits to the analogy between the event horizon of black holes and the "Rindler horizon" for an accelerated observer, but there are a number of similarities:

For Schwarzschild spacetime as described in Schwarzschild coordinates:
1. Spacetime is static, and a rocket must exert a constant thrust to remain stationary.
2. There is a horizon such that no information can be sent from below that horizon to a stationary receiver above the horizon.
3. If you drop an object from a stationary rocket above the horizon it will fall away toward the horizon, and approaches it asymptotically as coordinate time goes to infinity.
4. On the other hand, the object will cross the horizon in a finite amount of the object's proper time.
5. Taking into account quantum mechanics, the stationary rocket receives radiation (Hawking radiation) from the general direction of the horizon.
For Minkowsky spacetime as described in Rindler coordinates:
1. Spacetime is static, and a rocket must exert a constant thrust to remain stationary.
2. There is a horizon such that no information can be sent from below that horizon to a stationary receiver above the horizon.
3. If you drop an object from a stationary rocket above the horizon it will fall away toward the horizon, and approaches it asymptotically as coordinate time goes to infinity.
4. On the other hand, the object will cross the horizon in a finite amount of the object's proper time.
5. Taking into account quantum mechanics, the stationary rocket receives radiation (Unruh radiation) from the general direction of the horizon.
So there are lots of similarities. On the other hand, there is one stark difference: a black hole is a real thing, while the Rindler horizon is just an effect of choosing noninertial coordinates. My question, though, is: To what extent is the "black hole information paradox" analogous to something in Rindler spacetime (or rather, Minkowsky spacetime described in Rindler coordinates)? Or is that pushing the analogy past the breaking point?

In spite of the similarities, there seems to be a big difference between Hawking radiation and Rindler radiation: In the case of Hawking radiation, you can interpret the radiation as coming from the black hole--if not in the sense of literally traveling from the black hole, then at least in the sense of energy balance: the black hole mass shrinks to balance out the energy lost due to radiation. Objects dropping into a black hole increase its mass, and radiation shrinks its mass. The black hole information paradox is that if you consider a black hole as a box that you put energy into and get energy back out of, information is lost in that transaction--the energy that comes out contains no clue as to the information that went in. So the entire history of the black hole, from its initial creation to its demise due to Hawking radiation represents a loss of information, in apparent contradiction to the microscopic reversibility of all the fundamental laws of physics.

In contrast, there is no mass associated with the Rindler horizon. The horizon is not changed by dropping objects into it, nor by radiation coming "out" of it. So maybe there is no paradox in that case: The "stationary" observer can just assume that the information dropped into the horizon just remains forever below the horizon. So maybe the analogy breaks down. However, it seems to me likely that any proposed mechanism for information loss in the case of black holes would have an analogy with Rindler horizons. If not, then it would be insightful to know how the two cases differ when it comes to information loss.

Last edited: Jan 25, 2017
2. Jan 25, 2017

### Demystifier

I think the main difference is the following. In the Hawking case, inertial observers far from the black-hole horizon see radiation. In the Rindler case, inertial observers far from the Rindler horizon don't see the radiation. Nevertheless, it is a very interesting idea that correct resolution of the information paradox should exploit the analogy with the Rindler case. The current approaches do not use this analogy very seriously.

3. Jan 25, 2017

### martinbn

Just a side comment about the terminology. I don't think that the way you phrase points 1 is the standard way. Schwarzschild is static only outside the horizon, inside it isn't even stationary. In Rindler you say that to remain stationary the observer needs to exert a constant thrust, but here by stationary you mean non-changing Rindler coordinates, not that that the world line of the observer is an integral line of the timelike Killing field.

4. Jan 25, 2017

### stevendaryl

Staff Emeritus
I'm specifically talking about the exterior region, though.

Yes, I'm using a coordinate-dependent notion of "stationary": You're stationary if your spatial coordinates are not changing with respect to your time coordinate. Of course, that terminology depends on having a coordinate system with 3 spacelike coordinates and one timelike coordinate.

5. Jan 25, 2017

### Staff: Mentor

The worldlines of Rindler observers are integral curves of a timelike KVF. It is just not the one that is obvious in Minkowski coordinates.

But it can easily be converted to an invariant notion, the one martinbn referred to (though he might not have realized it): "stationary" means "following an integral curve of a timelike Killing vector field". That applies to both of the observers you describe, and it does so regardless of the coordinates you choose. (For example, you can show that the hyperbolas followed by Rindler observers are integral curves of a timelike KVF using Minkowski coordinates.)

6. Jan 27, 2017

### stevendaryl

Staff Emeritus
You could certainly define things that way, but it's pretty far removed from the informal notion of "stationary". The biggest thing lacking is the criterion that if two objects are both stationary, then the distance between them is constant. Of course, you could drop that criterion, but then the notion of "stationary" would have so little connection with the informal notion that it's not appropriate to use the same word. But if you keep the criterion, then that makes it a coordinate-dependent concept.

The business about timelike Killing vector fields is relevant, though, because only when there is such a field is it possible to come up with a coordinate system such that there is a sensible notion of "stationary" for that coordinate system.

Here's a technical GR question: Is "There exists a timelike Killing vector field" true if and only if "There exists a coordinate system such that the metric components are time-independent"?

7. Jan 27, 2017

### Staff: Mentor

In the sense that the "informal notion" is coordinate-dependent, yes. But to me that's a reason for discarding the informal notion and teaching people a notion of "stationary" that works better.

If both objects are following integral curves of the same timelike Killing vector field, this will be true if we define "constant distance" to mean "constant round-trip light travel time", i.e., "radar distance". (If the KVF is also hypersurface orthogonal, we can also define a "ruler distance" in the orthogonal hypersurfaces which will be constant.)

No, it doesn't. See above.

That depends on what you mean by "sensible". Comoving observers in cosmology are "stationary" in the coordinate-dependent sense in FRW coordinates. I think many cosmologists would say that is a "sensible" notion of stationary for that coordinate system, but it certainly doesn't meet the timelike KVF definition of "stationary".

If you modify the claim a bit, yes. Here is a modified version: There is a timelike KVF iff there exists a chart which has exactly one timelike coordinate, and none of the metric coefficients depend on it.

8. Jan 28, 2017

### stevendaryl

Staff Emeritus
I don't agree. It's possible that the word "stationary" is just not useful, and it should be avoided. But giving it a meaning that has no connection to its informal meaning doesn't seem very useful, either.

"...of the same timelike Killing vector field"

That's the point. If you say that an object is stationary with respect to a timelike Killing vector field, then that's a sensible notion of "stationary" (to me). But if you say that an object is stationary if there EXISTS a timelike Killing vector field, then two stationary objects will not be at a constant distance relative to each other.

"Stationary" is a relative concept. You can make it relative to a choice of a timelike Killing vector field, rather than relative to a coordinate system. That might be an improvement.

Thanks.

9. Jan 28, 2017

### Haelfix

There are a lot of similarities between the Rindler case and the black hole case. The formal similarities are among the key points in the modern treatment of the information paradox, and the role entanglement plays.

For a good discussion of the semiclassical treatment start on p17 of the following and it goes through the two cases.
https://arxiv.org/abs/1409.1231

10. Jan 28, 2017

### Staff: Mentor

That would come as a great surprise to the authors of all the GR textbooks, such as Hawking & Ellis, that explicitly define "stationary" in terms of a timelike Killing vector field. They certainly appear to find it useful.

That's not what I said. Here are the technical definitions:

(1) A spacetime is stationary if a timelike Killing vector field exists in the spacetime. (If the KVF only exists or is only timelike in a portion of the spacetime, then only that portion is stationary. For example, Schwarzschild spacetime is only stationary outside the event horizon.)

(Also, Schwarzschild spacetime obeys an additional constraint, that the timelike KVF is hypersurface orthogonal. The term for that is "static". There are spacetimes which are stationary but not static, e.g., Kerr spacetime.)

(2) An object is stationary if its worldline is an integral curve of a timelike Killing vector field. (If more than one timelike KVF exists in a spacetime, as in Minkowski spacetime, then you have to specify which KVF, so "stationary" is relative in the sense that an object can only be stationary with respect to one timelike KVF.)

11. Jan 28, 2017

### stevendaryl

Staff Emeritus
That's what I suspected. It is not useful to say that an object is stationary without saying relative to what.

12. Jan 31, 2017

### Ilja

The analogy between Rindler and Hawking radiation fails already much earlier. The easiest way to see that it is misleading is to look, instead of a BH, at a stable star. For a stable star, as described outside as a Schwarzschild spacetime in Schwarzschild coordinates:
1. Spacetime is static, and a rocket must exert a constant thrust to remain stationary.
2. Taking into account quantum mechanics, the stationary rocket receives no radiation (Hawking radiation) from the general direction of the horizon.
So the analogy fails. And it fails consistently even if the difference between the BH and the stable star, described by the difference between horizon radius and the radius of the star, or, if one likes, the surface time dilation factor, is trans-Planckian.

This is not a merely technical problem, but unavoidable. Semiclassical gravity does not have the power to derive consistently Hawking radiation. What prevents this is a conceptual problem: The failure of a well-defined energy-momentum density of the gravitational field, caused by diffeomorphism invariance. Energy conservation needs a preferred time to apply the Noether theorem, and without energy conservation you have nothing not even a good justification for writing down a Schroedinger equation.

Once this is recognized, one can see that it depends on how you define the preferred time, within a preferred frame, if you obtain Hawking radiation or not. To identify that preferred frame correctly, you need some physics beyond GR.

13. Jan 31, 2017

### stevendaryl

Staff Emeritus
That definitely puts an obstacle into any attempt to come up with an intuitive understanding of Unruh and Hawking radiation:
1. In empty space, constant proper acceleration implies Unruh radiation.
2. Outside a black hole, constant proper acceleration implies Hawking radiation.
3. Outside a star, constant proper acceleration does not imply Hawking radiation.
On the one hand, you can understand them all from the point of view that 1& 2 have event horizons: there are regions of spacetime that cannot send messages to the accelerated observer.

On the other hand, the discrepancy between 2&3 seems very strange, because the spacetime metric outside a spherically symmetric star is indistinguishable from that outside a black hole. So what mechanism can make the difference in the presence/absence of Hawking radiation?

14. Jan 31, 2017

### Ilja

Hawking radiation comes from that arbitrary small, trans-Planckian environment of the BH. This is the trans-Planckian problem of Hawking radiation.

It is a quite serious problem, given that it is an exponential effect. Unruh describes it in the following words: "Thus one second after a solar mass black hole forms,the radiation, produces by whatever the process is that produces Hawking radiation, originated from frequencies in the initial state of the uncollapsed system of order $e^{10^5}$, a number so absurdly large that any imaginable units would simply produce an insignificant change in that exponent. And the later the radiation one is considering is emitted, the larger and more absurd this factor becomes".

Beyond this, in principle, if you ignore this exponential redshift, what you see if you look at a black hole is the surface of the collapsing star, in the process of the collapse. But a collapsing star radiates.

15. Jan 31, 2017

### Jilang

Ilja, are you saying that if such an effect existed it would be so huge that the star/black hole would fizzle away quickly?

16. Jan 31, 2017

### Ilja

No. The effect would require that the semiclassical approximation remains valid for time dilation factors so extremal as $e^{10^5}$ to allow to predict Hawking radiation lasting even one second after the collapse, and would require to multiply it again with the same factor $e^{10^5}$ for lasting it one second more. But nobody believes that the theory is applicable below Planck time, so that the very idea to apply this approximation to predict something is absurd.

The size of the effect is extremely small, and is what it is. But to believe that existing theory allows to predict Hawking radiation except for the time of the collapse itself is nonsense.

17. Jan 31, 2017

### Haelfix

The way I understand all the different effects heuristically is to also include a 4th case. So include the case of a star that supernovaes or something which leads to mass loss. In that case there is time variation within the metric as the star shrinks from a mass M to a mass M'. We know from QFT in curved spacetime that the gravitational analogue of the Schwinger effect for gravity must involve inhomogenous fields and only under such conditions will you have particle creation that take places, thus one expects some particle creation to take place during this physical process. The second the star settles down, you are back to something that looks like the stationary Schwarschild exterior solution and the effect turns off. This particle creation is a tiny exponentially small effect, and completely swamped by the massive stress energy flow from the usual radiation of the supernovae, nevertheless it is predicted to be there, and a detector at infinity should be able to measure it.

Ok, so it seems like it's time dependence of the gravitational field that leads to whether there is really radiation or not. So whats going on in the case of Rindler space and the Blackhole case (with a physical interior region). The answer is that in those two cases there is a presence of a Killing horizon. So consider a family of spacelike surfaces that straddle the horizon of a blackhole, modes that are on the left hand side under the flow of a Killing vector are pushed towards the singularity, whereas modes that are on the right hand side are pushed towards Scri+. It is this warping and stretching that leads to an 'effective' time dependence and produces Hawking radiation near the horizon.

Why does a blackhole register the presence of a particle at infinity and the Rindler case does not? This requires a subtle analysis, and is conceptually and qualitatively very similar to a perhaps more familiar story: Namely the question of whether an electron sitting on a table radiates or not. Heuristically this can be seen by noting that the Hawking temperature explicitly depends upon the presence of the surface gravity in the numerator whereas the Unruh temperature depends on the proper acceleration, so the analogy must be similar. Anyway, when proper normalization is taken into account, the upshot is the Hawking temperature asymptotes to a finite value and the Unruh effect goes to zero at infinity.

18. Feb 3, 2017

### martinbn

That's a very good point, I definitely didn't think about that. On the other hand I still don't see it. It seems to me that they are not the integral lines of a KVF. But it is very likely that I am confusing something about Rindler observers or worse I am confused about basic geometry. Anyway this is probably of topic.

19. Feb 3, 2017

### stevendaryl

Staff Emeritus
As @PeterDonis clarified: "There is a timelike KVF iff there exists a chart which has exactly one timelike coordinate, and none of the metric coefficients depend on it." So the constantly-accelerated observer can pick a coordinate system in which he is at rest and the metric components are unchanging. So his time coordinate is a scalar field whose gradient is a KVF.

This thread should actually be in the relativity section, since it has very little QM in it. Maybe I'll move it, using my superpowers.

20. Feb 3, 2017

### Staff: Mentor

It can get confusing because the KVF is not actually the same vector field as the vector field of 4-velocities of Rindler observers. The two just happen to have the same integral curves. (The same thing happens for static observers in Schwarzschild spacetime.) In Rindler coordinates, the KVF is $\partial / \partial t$; but the 4-velocities of Rindler observers are $(1 / x) \partial / \partial t$ (assuming we've chosen units appropriately), because 4-velocities have to be unit vectors. Try plugging $\partial / \partial t$ in Rindler coordinates into Killing's equation; you will see that it satisfies it (but $(1 / x) \partial / \partial t$ does not).