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How reasonable it it to conclude that, from a remote observer’s frame, matter falling towards a black hole never crosses the event horizon, because ∆ t → 0 as v → c (according to the Lorentz transform)?

If that is a tenable view, then it seems to solve the information paradox because infalling matter is never actually lost from the observable universe. Also, because the infalling matter, from the observer’s frame, remains at the event horizon, it represents negative (gravitational) energy, which balances the energy in the underlying Hawking radiation, thus preserving conservation of matter/energy.

Existing threads on StackExchange point to widely divergent answers to this (e.g. black hole event horizon). These divergences seem largely determined by the incompatibility of quantum mechanics and general relativity.

So, I guess I’m asking whether this classical view of black holes can be considered correct or incorrect, as opposed to “simply” incompatible with quantum mechanics.

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  • $\begingroup$ The information paradox is a lot more subtle than often presented. In particular, even if you can account for all the information at the end and the beginning of the process, I think there is an issue in the middle. Most of the Hawking radiation comes out at the end of the black hole's life, when the hole itself is already too small to contain much information. Perhaps someone can fill in the details in this line of thinking. $\endgroup$ – user10851 Apr 15 '15 at 22:50
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/21319/2451 , physics.stackexchange.com/q/5031/2451 , physics.stackexchange.com/q/160060/2451 and links therein. $\endgroup$ – Qmechanic Aug 6 '15 at 13:41
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My interpretation is that you are raising the following objection to the black hole information paradox:

According to observers distant from the hole, causal lines take infinite coordinate time to cross the event horizon. To these observers, infalling information is thus never lost, but only very strongly redshifted; in essence it remains "painted" on the horizon, very dimly, but forever. Therefore, there is no paradox, just a misunderstanding of how coordinates work.

There are several responses to this standard but often-undiscussed objection.

  1. Real matter is quantized. The exponential redshift thus eventually leads to a sitatuation where there is a "last quantum" to fall into the hole. Eventually, it does fall in, and the matter is truly gone.

  2. Fate of the hole after decay. Black holes emit Hawking radiation, which reduces their mass. Eventually it reduces their mass to zero and the holes vanish. After this point only the Hawking radiation remains. But the radiation is exactly thermal according to theory, so information is eventually lost, even according to distant observers: we started off with a bunch of matter, and ended up with a radiation field whose temperature depends only on the matter's total mass.

One might then offer the following also-standard response to objection 2:

This objection shows only that something strange must be happening during the actual destruction of the hole. But this is obviously a quantum gravity effect. Thus there is no need to modify our understanding of what happens to the hole before the decay: it just stays painted on the horizon until the hole is destroyed.

Some canonical responses are:

  1. Remants seem absurd. If this response were taken seriously, it would essentially imply that all the information about the black hole - an object of potentially arbitrary mass! - can somehow be contained within a Planck-scale volume. This would be very odd.

  2. Page timescale. It can be shown that about the first half of the black-hole information must be emitted over the same "Page" timescale as it takes to emit about half of the mass. This seems to imply that something poorly-understood is going on even while the hole is large.

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A first problem is that there is in GR no such thing as "an observers frame", except in sloppy speech. There are various systems of coordinates. Two systems of coordinates may agree for an observer as much as one likes but differ elsewhere. And all the systems of coordinates are on equal foot, none is preferred.

What could replace the "observer's frame"? The most plausible candidate seems to be the use of harmonic coordinates - they really essentially simplify the Einstein equations, are used in the Newtonian limit and in the PPN formalism (see here) or in Choquet-Bruhat's local existence and uniqueness proof, roughly the only reasonable candidate for preferred coordinates. To define harmonic coordinates, one need some initial values, because the harmonic condition is only an evolution equation $\square X^\mu = 0$ for them, but this is given by the Minkowski coordinates before the collapse. This gives, roughly, the Schwarzschild time coordinate, thus, it would be such a system of coordinates where the material of the collapsing star never reaches the horizon.

From point of view of classical GR, the only objection is that these coordinates would not cover the complete solution.

If one includes Hawking radiation into the consideration, one should care about the trans-Planckian problem (to derive that Hawking radiation lasts more than a second after the collapse, one has to presuppose that semiclassical theory remains valid for distances of $10^{-1000}$ of Planck length or so). The mainstream way to solve this problem is to rely on some results of Boulware and others that some Hawking radiation remains if one considers various regularizations. The problem is that these regularizations break covariance, thus, require preferred coordinates, and they obtain Hawking radiation only if the preferred coordinates are those related with the infalling observers. If they are stationary, there will be no Hawking radiation. And if one wants to rely on modifications of GR with preferred coordinates, see above for candidates.

If one ignores this and assumes Hawking radiation, then one can publish the idea that the BH evaporates before being formed even in Phys.Rev., as done by Gerlach, PRD 14, 1479 (1976), and get citations by standard textbooks like Birrell, Davies, quantum fields in curved spacetime, so it is close enough to being tenable.

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    $\begingroup$ Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files, e.g., arxiv.org/abs/1102.5192 $\endgroup$ – Qmechanic May 19 '16 at 13:46
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Note that the property "black hole" is relative to an observer. A freely falling observer will not notice anything strange like an event horizon etc. Those observers will not have an information paradox. But similar, an observer at rest relative to the black hole will have no information paradox too, since nothing ever falls into it.

For a calculation approach see this: https://physics.stackexchange.com/a/171596/75518

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  • $\begingroup$ Why won't a freely falling observer notice the event horizon? Won't they see a black sphere blocking the star light behind it and slowly filling their view as they fall towards it? $\endgroup$ – Quantumplate Apr 20 '15 at 1:28
  • $\begingroup$ In what sense is "black hole" relative to an observer? An observer may not have an easy way of knowing they have passed the horizon, but the definition of an event being in a black hole is observer-independent, namely that there are no more causal paths connecting the event to future null infinity. $\endgroup$ – JohnnyMo1 Jun 28 '15 at 20:51
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    $\begingroup$ It is an observer-independent property of the manifold whether a black hole - a region of spacetime which is causally disconnected from the rest - exists. $\endgroup$ – AGML Nov 16 '15 at 22:31
  • $\begingroup$ @Quantumplate That surface is not the event horizon, and, if the hole's mass changes over time, need not have anything to do with it. For example, if the hole expands, light initially outside the shadow could later be enveloped by it. Despite being outside the shadow at first, that light was always inside the event horizon (which is the surface bounding light which never escapes, not light that initially moves inward). $\endgroup$ – AGML Nov 16 '15 at 23:01
  • $\begingroup$ @AGML: can you point me to a proof of that statement? $\endgroup$ – image Nov 17 '15 at 23:09
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The thing to consider is not $t$ the 'coordinate time', but $\tau$ the 'proper time' measured by the clock of the in-falling particle.

$$ \Delta\tau = \int \sqrt{-ds^2}, $$

where $ds^2$ is the spacetime interval along a time-like path. All in-falling particles take finite proper time to cross the horizon. Further, all in-falling, massive particles will have $v/c<1$ at the horizon.

The problem is the coordinate singularity in the Schwarzschild metric when it is expressed in 'Schwarzschild coordinates'. This is nice a coordinate system for trajectories outside a black hole, because they remind us of spherical coordinates. Unfortunately, they break at $r=2M$, the event horizon.

Changing coordinates to Eddington-Finkelstein or Kruskal–Szekeres coordinates (or others) will eliminate the coordinate singularity at the horizon. Particles fall in just fine.

Edit: my $\Delta\tau$ is defined for a metric with signature $(-,+,+,+)$, so $ds^2<0$ corresponds to time-like separated events.

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  • $\begingroup$ Let's stick with the information paradox. If the issue here is the loss of information to the observable universe, then the proper frame to consider is that of a remote observer (in the observable universe), and not the proper frame of the infalling particle. $\endgroup$ – odyssoma Apr 15 '15 at 22:33
  • $\begingroup$ It still takes finite time for a particle to fall into a black hole. What we call time $t$ for a distant observer doesn't make any sense at the horizon. Inside the horizon $t$ points in a space-like direction and $r$ is the time-like coordinate. The problem is that $t$ is not 'time' at the horizon. $\endgroup$ – Paul T. Apr 15 '15 at 22:51
  • $\begingroup$ While the particle certainly crosses the event horizon in finite proper time, it is nevertheless true that according to the distant observer it never falls into the hole. Those events are not in the distant observer's past. $\endgroup$ – AGML Nov 16 '15 at 22:22
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If you consider a spaceship as v → c then someone in a stationary frame observing the spaceship will see ∆ t → 0 as well (on the spaceship). The thing to note is that the spaceship is still moving forwards at close to c. Time is only frozen for things moving on the spaceship (as observed by someone in a stationary frame) but the spaceship itself is still moving forwards at a very fast rate.

The same thing happens with matter falling into a black hole. Time slows down for that matter (as observed by a remote observer) but the matter as a whole (let's say a spaceship) is still travelling towards the black hole at speeds approaching c.

You are correct that there are widely divergent views on this (and a lot of confusion). This arises (in my opinion) from the fact that light escaping a black hole takes longer to escape than it does to fall in. e.g. a photon exactly on the event horizon travelling directly away from the black hole cannot escape and can theoretically be stationary. A photon just outside the black hole will escape very slowly and so on.

This has the effect that a remote observer (who is say observing the light reflect off the matter falling in) will never see the object fall in. This is because the light reflecting off the matter takes longer and longer to get back to the observer the closer it gets to the event horizon and in essence the image they see slowly fades out. The actual matter has long since crossed the event horizon but the outside observer can never see this occur.

Now you can take the point of view that things actually happen, even if no one observes them, or you can take the point of view that with Relativity everything is relative to a frame of reference so that you can only describe something from a frame of reference (e.g. the remote observer never see's the matter cross the event horizon (and no other frames of reference do either) so it doesn't).

So let's agree that matter falls into a black hole. This doesn't solve the Information Paradox and as far as I know there is no satisfactory explanation. We could say that either we are missing something in the solution OR one of the premises it rests on is wrong (and so there is no paradox).

One of the premises is that a black hole evaporates (due to Hawking Radiation) and information about particles falling in is therefore lost. Hawking Radiation is caused by vacuum fluctuations around the event horizon which give rise to virtual particles, one of which escapes (with positive energy (not to be confused with charge)) and one of which falls into the black hole (which by the conservation of energy has negative energy and therefore causes the black hole to evaporate).

My question would be, if the energy for these virtual particles came from outside the black hole (e.g. entered the small space as unseen particles, say collisions of dark matter or something else) then would there be no need for negative energy particles, black holes wouldn't evaporate and there would be no information paradox?

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    $\begingroup$ The events inside the hole are not in the past of the distant observers. Such observers are therefore perfectly correct in saying that the matter never falls in, unless they attempt to chase after it through the horizon. This is not an optical illusion and can be seen very clearly from the black hole's Penrose diagram. $\endgroup$ – AGML Nov 16 '15 at 22:57

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