Does entropy change on horizons

[friend]

As I understand it, horizons, such as black hole or cosmological event horizons is a place where time seems to stop. So if nothing changes at the horizon, then can there be entropy there? Does entropy require things to change with time? If not energy dissipates because nothing moves, then what happens to entropy?

 

[PeterDonis]

As I understand it, horizons, such as black hole or cosmological event horizons is a place where time seems to stop.

That's not a correct definition. So your question is not well posed, because it's based on an incorrect premise.

 

[mathman]

To a distant observer, time seems to stop. For an object in the vicinity of a black hole time proceeds.

 

[kimbyd]

To a distant observer, time seems to stop. For an object in the vicinity of a black hole time proceeds.

To try to explain this a little bit, once the object has passed the horizon, no new photons can ever reach an external observer. The external observer only ever sees the photons emitted while the object was still outside the horizon. But these are stretched out so that over time the outside observer sees the last few moments before the object crossed the horizon happen more and more slowly, with the light shifting to longer and longer wavelengths. The crossing itself is never witnessed.

Imagine, for example, that according to the object's clock, it took 60 seconds to reach the horizon of the black hole. All of the photons the external observer will ever see were emitted in those 60 seconds, and the time at which those photons manage to leave the black hole will be stretched out over the rest of time.

 

[friend]

outside observer sees the last few moments before the object crossed the horizon happen more and more slowly, with the light shifting to longer and longer wavelengths.

So not only does time seem to stop at the horizon (as viewed from outside), but those objects get red-shifted to the point of disappearing. If nothing can be observed at the horizon, then it would seem entropy would have to be zero there, right?

I also wonder, if two particles are entangled with one of them far outside the horizon and the other very near the horizon, what happens to the entanglement? Is the outside particle entangled with a particle that can never be measured? What would that mean?

 

[PeterDonis]

not only does time seem to stop at the horizon (as viewed from outside), but those objects get red-shifted to the point of disappearing

No. Neither of those things are true in the sense in which you are interpreting them.

If nothing can be observed at the horizon

Wrong.

it would seem entropy would have to be zero there, right?

Wrong.

The fundamental mistake you are making is to confuse what light signals reach the distant observer, with what is actually happening at the horizon. To an observer free-falling through the horizon, the horizon is a perfectly normal place: time does not "slow down" there, nothing is redshifted, the second law works just fine. The horizon itself appears as a surface of outgoing light rays to this observer. The fact that light emitted outward by this observer at or below the horizon can never reach an observer very far away from the black hole is irrelevant to understanding what is happening at and below the horizon and how entropy works there.

if two particles are entangled with one of them far outside the horizon and the other very near the horizon, what happens to the entanglement?

Nothing. The two particles are spacelike separated, but there is no issue with spacelike separated particles being entangled. It happens all the time.

Is the outside particle entangled with a particle that can never be measured?

No. Anyone who free-falls into the black hole can measure the other particle.

 

[kimbyd]

So not only does time seem to stop at the horizon (as viewed from outside), but those objects get red-shifted to the point of disappearing. If nothing can be observed at the horizon, then it would seem entropy would have to be zero there, right?

Nope.

Entropy is a tricky concept. It's not directly-observable (though its effects are). Entropy is a measure of the number of ways you can rearrange the microscopic components of the system and leave the large-scale behavior of the system unchanged. Whether or not something is observable simply plays no role in this.

There's also the problem that we don't have a firm grasp on what the entropy of a self-gravitating system is. At least, not in general. We do know the entropy of a black hole, and the entropy of an empty universe with a positive cosmological constant. But those are very particular, contrived cases. We don't know the entropy of a star, or of a planetary system, or of an object falling into a black hole.

The entropy of a black hole is directly proportional to the area of its event horizon (de Sitter space, or empty space with a positive cosmological constant, also has a horizon and entropy proportional to said horizon). When an object enters a black hole, its horizon increases in area, which causes the entropy of the black hole to increase. We can describe the increase in entropy of the black hole very precisely, but it's not possible (yet) to describe the entropy of the infalling object.

I also wonder, if two particles are entangled with one of them far outside the horizon and the other very near the horizon, what happens to the entanglement? Is the outside particle entangled with a particle that can never be measured? What would that mean?

Entanglement doesn't transmit information, so even if the particles do remain entangled, you can't use the entanglement to measure anything about the black hole.

 

[kimbyd]

No. Anyone who free-falls into the black hole can measure the other particle.

Is that true? I had thought that the worldlines of objects entering the black hole at different times never intersect.

 

[George Jones]

Is that true? I had thought that the worldlines of objects entering the black hole at different times never intersect.

They do not have to coincide physically in order for a measurement to be made.

Suppose observers A and B are inside the event horizon. If B is inside A's future lightcone, then A can send a signal to B's worldline to start an experiment on B. The apparatus then sends a signal (the result of the measurement on B) that A can receive when A is inside B's future lightcone.

I have not thought much about for how much of the interior this is possible.

 

[friend]

Entropy is a tricky concept. It's not directly-observable (though its effects are). Entropy is a measure of the number of ways you can rearrange the microscopic components of the system and leave the large-scale behavior of the system unchanged. Whether or not something is observable simply plays no role in this.

Just a moment. I thought an horizon (of any kind) was by definition a boundary beyond which no information can be known at least to a distant observer. And information as I recall has to do with the way things are distributed if it were possible to observe them. And information is closely related to entropy, right? So from the point of view of a distant observer, if an horizon is a boundary of no information, then the entropy is also zero there, right?

 

[PeterDonis]

I had thought that the worldlines of objects entering the black hole at different times never intersect.

For some worldlines (see below), this is true. But there are plenty of pairs of worldlines that fall in at different times but do intersect.

What you might be thinking of is a more restricted claim: any two radially ingoing freely falling objects that pass a given radial coordinate at different times will never intersect. But that is only a small subset of all the possible worldlines that enter the black hole at different times.

 

[PeterDonis]

I thought an horizon (of any kind) was by definition a boundary beyond which no information can be known at least to a distant observer. And information as I recall has to do with the way things are distributed if it were possible to observe them. And information is closely related to entropy, right? So from the point of view of a distant observer, if an horizon is a boundary of no information, then the entropy is also zero there, right?

No. You can't reason about this subject by using vague ordinary language terms and trying to string them together. You have to actually dig down into the details--the mathematical models. And, as @kimbyd has pointed out, we don't know exactly what the right mathematical model is that describes entropy in general in the presence of gravity. But we do know that, whatever it is, it won't say that a black hole's entropy is zero, either at the horizon or in general.

Also, even as vague ordinary language, some of your statements are incorrect. A horizon is not a "boundary of no information" period. It only restricts one very particular kind of information flow: classical information cannot go from inside the horizon to outside the horizon. But that does not prevent, as already noted, quantum entanglement between particles inside and outside the horizon. Nor does it prevent other quantum processes, such as Hawking radiation, from potentially revealing information that fell into the hole. (Figuring that out is one of the primary open questions being researched in this area.)

 

[kimbyd]

Just a moment. I thought an horizon (of any kind) was by definition a boundary beyond which no information can be known at least to a distant observer.

Not quite. The statement about information with regards to black hole horizons comes from what is known as the "no hair theorem". It states that the only parameters that are needed to fully describe a black hole in General Relativity are:
1. Mass
2. Rotation
3. Electric charge

That's it. Anything else we might describe about a black hole must be a function of those properties, and those properties alone.

It turns out that the entropy of a black hole is proportional to the area of its event horizon, which in tern depends upon the above parameters. Wikipedia actually has a pretty good article on black-hole thermodynamics.