# Hawking radiation prevents event horizon crossing?

1. Mar 24, 2015

### Asher Weinerman

No one seems to be bothered by this except me:

Black holes have a finite lifetime measured in Schwartzchild time due to Hawking radiation. Similarly, the universe probably has a finite lifetime measured in Schwartzchild time. In that case, nothing ever falls through the event horizon of a black hole because it takes infinite Schwartzchild time so the black hole would disappear and the universe would come to an end before you crossed the event horizon. You could even calculate dr/dt of the event horizon of a black hole shrinking due to Hawking radiation which would presumably be non-zero and presumably negative. Since an infalling observer's dr/dt approaches zero at the event horizon, as you fall towards the event horizon it will shrink away from you as you approach and you will never cross it. So why does everyone spend time analyzing what happens after you cross the event horizon, lured by the fact that proper time is finite? I really just don't understand - I'm not trying to be clever - I just want to know what I am missing here.

2. Mar 24, 2015

### phinds

This is all wrong. You should read some more (and not pop-sci stuff) about black holes.

Things always fall through the EH of a BH, it just doesn't look that way to a remote observer. The in-falling object never notices that there even IS such a thing as an EH.

3. Mar 24, 2015

### Staff: Mentor

No, this is not correct. First of all, Schwarzschild coordinates are not good ones to use to describe an object falling through the event horizon even for a purely classical black hole, which does not emit Hawking radiation, because those coordinates are singular at the horizon. Painleve or Eddington-Finkelstein coordinates, which are nonsingular at the horizon, are much better. In these coordinates, it only takes a finite time for an object to fall through the horizon.

If we add Hawking radiation to the mix, and use coordinates that are nonsingular at the horizon, like Painleve or Eddington-Finkelstein (more precisely, if you use the analogues of these coordinates for a spacetime with an evaporating black hole, which is different from a spacetime with a purely classical black hole), then we find that, as before, it only takes a finite time for an object to fall through the horizon, and this time will be shorter than the time (in the same coordinates) that it takes the hole to evaporate via Hawking radiation. If, however, you insist on using Schwarzschild coordinates (more precisely, coordinates that are analogous to them), then you find that, while the coordinate time it takes for an object to reach the horizon is infinite, the time it takes for the hole to evaporate is also infinite. So these coordinates can't even represent the evaporation process properly, and you can't use them to answer the question of whether objects can reach the horizon before the hole evaporates.

4. Mar 24, 2015

### Asher Weinerman

But it takes finite time in Schwartzchild coordinates for a black hole to evaporate if the universe is open. If the universe is closed I guess we all fall into the same black hole. I say finite time if the universe is open because the Hawking radiation will dissolve the black hole until it is in thermal equilibrium with the cosmic background radiation. This cosmic background radiation will become less and less powerful until it disappears entirely if the universe is open, which means all black holes will eventually evaporate. Apparently, according to wiki, at the moment only a black hole the size of the moon or less will evaporate before it is in thermal equilibrium with the current state of the cosmic background radiation.

So again, if the universe is open, all black holes will evaporate before anything falls into them. Right?

5. Mar 24, 2015

### Staff: Mentor

No, it doesn't. See my previous post. Please become familiar with the actual math and theory before making claims about it.

If the universe is closed, then strictly speaking there is no such thing as a black hole, because there can't be any event horizons. Event horizons can only exist in a spacetime that has a future null infinity, and the spacetime of a closed universe does not have one.

Yes, this is true, but the "time" being used here cannot be Schwarzschild coordinate time (see my previous post), even though it will be very close to Schwarzschild coordinate time far away from the black hole (this is true of both of the alternate coordinate charts I mentioned in my previous post). Close to the horizon, the "time" implicitly being used here will differ greatly from Schwarzschild coordinate time; at the horizon, it will differ infinitely greatly, since Schwarzschild coordinates assign an infinite time coordinate to the horizon.

Wrong. See above.

6. Mar 24, 2015

### PAllen

Wrong. Let's describe the physics observed by two different observers, and forget coordinates which are just labels attached to events.

Bob freely falls into BH, reaching singularity in quite finite time on their watch. Later, the BH completes evaporating, including the energy contributed by Bob.

Distant observer watches Bob disappear, progressively, on approach to horizon. 'Disappear' means that any light from Bob is zillions of orders of magnitude less than either CMB or the prior Hawking radiation from the hole (which, for a stellar mass BH order of magnitude less than CMB). The BH continues to radiate and shrink until a final burst. The total emitted includes radiation equivalent to all matter that has fallen into the BH - including that of Bob and all else that has reached the singularity. There is complete consistency between this and Bob's observation. It is true that distant observer cannot see Bob crossing horizon but since when has what some observers sees been taken as the definition of the universe?

As for coordinates, SC coordinates simply don't cover events at or inside the horizon, any more than Rindler coordinates cover the boundary or beyond the Rindler wedge. Thus, the complete picture of matter forming the BH, more matter falling in, and all evaporating is simply not covered by SC coordinates. Instead, all events at or inside the (evolving) horizon are simply not labeled by these coordinates. However, the subset of events covered by SC coordinates are completely consistent with the more complete description provided by other coordinates - it is just that some part of spacetime is not labeled.

7. Mar 24, 2015

### PAllen

It is worth noting that your overall argument is common one that has been refuted repeatedly.

8. Mar 24, 2015

### Asher Weinerman

I appreciate you helping me to understand things, since I'm not an expert in this subject like you are.
Every reference I've checked says that black holes will evaporate via Hawking radiation in finite far-away time on the order of 10^60 times the age of the universe (wiki for example). What reference are you using - can you give me a link to it so I can become familiar with the actual math and theory?

9. Mar 24, 2015

### Asher Weinerman

Hi PAllen and PeterDonis - thank you for helping me understand.
PAllen - my question is that doesn't BH evaporate before Bob falls through the horizon in quite finite proper time, since I'm saying black holes evaporate in finite far-away time, which occurs before Bob enters BH.
Maybe you could give me a link that states black holes take infinite far-away time to evaporate by Hawking radiation if the universe is open?

10. Mar 24, 2015

### Asher Weinerman

Oops - I meant to say above "before Bob enters EH, not BH"

11. Mar 25, 2015

### pervect

Staff Emeritus
Ted Bunn's cosmology FAQ is one resource on this topic, see for instance http://cosmology.berkeley.edu/Education/BHfaq.html#q9
The sci.physics.faq at http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/fall_in.html also covers the issue. Both FAQ's come to the same conclusion, the black hole will not evaporate before you fall in.

On the more techincal side, T Vachaspati, D Stojkovic and L M Krauss, Phys. Rev. D76, 024005 (2007) [arXiv:gr-qc/0609024] do make claims which could be regarded as saying that black holes evaporate before you fall in. Rebutting this paper we have:

"Radiation from collapsing shells, semiclassical backreaction and black hole formation" which is a response to the Krauss paper that can be found on arxiv at http://arxiv.org/abs/0906.1768

1) Paranjape and Padmanabahn are aware of Krauss' et al work - they even cite it

2) P&P cite over a dozen papers that they claim support the view (that they describe as a consensus view) that black holes don't evaporate before you fall in.

3) I haven't seen a published response by Krauss, et al, to P&P's comments. (It's possible I could have missed it).

I recall some other comments on the topic that were much simpler than Paranjape and Padmanababhn's, and came to the same conclusion, but not exactly where they were or who made them. It might have been Unruh. It was a rather simple argument about the implausibility of any terms in the stress-energy tensor due to evaporation that could halt collapse, as I recall.

12. Mar 25, 2015

### Staff: Mentor

"Far-away time" is not the same as "Schwarzschild time". Do any of those references actually say that the hole will evaporate before anyone falls in? If so, please give an explicit quote and link. It looks to me like you are misinterpreting what the references are telling you.

We can't because that statement is false. The hole does evaporate in finite "far-away time". But people can fall into the hole in a finite "far-away time" that is much shorter than the far-away time it takes for the hole to evaporate. Again, this has already been stated.

The question you should be asking is, how can someone fall into the hole in finite "far-away time"? The answer is that you need to use appropriate coordinates, such as the ones I mentioned previously (Painleve or Eddington-Finkelstein), which are not singular at the horizon, so they can be used to assign a well-defined "far-away time" to events near or at or inside the horizon. Schwarzschild coordinates cannot do this because they are singular at the horizon, so you can't use Schwarzschild coordinates as your "far-away time" if you are trying to analyze events near or at or inside the horizon.

13. Mar 25, 2015

### stevendaryl

Staff Emeritus
You have two events: (1) The infalling observer crosses the event horizon. (2) The black hole evaporates. Obviously, event (1) must happen before event (2), if it happens at all.

If there were no Hawking radiation, then for events taking place outside the event horizon, you can use Schwarzschild coordinates, and event (1) never happens. But "never happens" just means that there is no $t$ coordinate for that event. The Schwarzschild coordinates just don't cover that event. It happens, but just not at a finite $t$ coordinate. To see that this doesn't really mean that it doesn't happen, you can compare the Schwarzshild coordinates with the Rindler coordinates:

$X = \sqrt{x^2 - c^2 t^2}$
$T = tanh^{-1}(\frac{gt}{cx})$

where $x$ and $t$ are the usual inertial coordinates.

An object dropped from the point $X_0, T_0$ will approach the point $X=0$ as $T \rightarrow \infty$, but it will never reach it in Rindler coordinates. But it will reach it at time $t=X_0/c$ in inertial coordinates.

So getting back to black holes, event (1) does happen, it's just that it isn't covered by Schwarzschild coordinates. What about event (2)? Well, Schwarzschild coordinates are appropriate for eternal black holes. For evaporating black holes, you have to use some other coordinate system. If you pick a coordinate system such that the evaporation takes place at a finite time, then event (1) will also take place at a finite time. What'll happen is that the event horizon will shrink with time, and the infalling observer will hover just outside. The black hole will disappear when the event horizon shrinks to nothing, and at that moment, the infalling observer will be squashed into zero volume, also.

14. Mar 25, 2015

### Staff: Mentor

If you mean this is what the infalling observer's worldline will actually be, this is not correct. The infalling observer will fall through the horizon long before the hole evaporates, and will be destroyed in the singularity. (At least, this is what will happen in the original model of a hole evaporating by Hawking radiation, the one Hawking originally worked with. There are lots of other proposed models where different things happen, but I don't think we need to open that can of worms in this thread. ) If we use appropriate coordinates, such as ingoing Eddington-Finkelstein coordinates, the infalling observer will fall through the horizon at an Eddington-Finkelstein time that is long before the Eddington-Finkelstein time at which the hole evaporates, and will hit the singularity at an E-F time that is somewhat later, but still long before the E-F time at which the hole evaporates.

One could, however, set up a sort of "Schwarzschild-like" coordinate chart in which the "time" coordinate was assigned based on when light signals are seen by a far-away observer. (This would actually be more like outgoing Eddington-Finkelstein coordinates.) In such a chart, the infalling observer would appear to "hover" outside the horizon until the hole evaporated, because the light signals he emitted just before reaching the horizon would be extremely delayed getting out to the far-away observer; they would reach that observer just before the light from the hole's final evaporation does. Light emitted by the infalling observer at the instant he crossed the horizon would reach the far-away observer at the same instant as the light from the hole's final evaporation, and so would be assigned the same "time" coordinate in this chart.

However, in this latter chart, no time could be assigned to any event inside the horizon, including the event at which the infalling observer hits the singularity. The far-away observer will never see any of those events, because the light from them will have hit the singularity. The event horizon, while it exists, is still an event horizon; light from events inside it can never get out. When the far-away observer sees the light emitted by the infalling observer as he crosses the horizon, those signals will show that observer in the (presumably) normal condition he was in when he crossed the horizon; the far-away observer will never see him being crushed in the singularity. It will just look like the infalling observer vanished at the instant he crossed the horizon.

15. Mar 25, 2015

### stevendaryl

Staff Emeritus
I am talking about from the point of view of someone observing the saga from a safe distance. I realize that that doesn't uniquely determine a coordinate system. However, during the time when the black hole is shrinking slowly, I would think that a distant observer could use, throughout most of the lifetime of the black hole, use a coordinate system that is Schwarzschild coordinates with a slow time-dependence in the $M$ parameter. In such a coordinate system, the infalling observer would be hovering outside the black hole.

That's what I mean.

Yes, the importance of the "Schwarzschild-like" coordinates is just to understand that there is no contradiction between the black hole evaporating and the infalling observer never crossing the event horizon. From the point of view of this coordinate system, the evaporation and the crushing of the infalling observer are simultaneous.

16. Mar 25, 2015

### PAllen

It is not so clear to me what 'SC equivalent coordinates' would mean for an evaporating BH. Classical models are usually based on the Vaidya metric, and the P&P paper Pervect mentions states that despite quantum issues near the horizon, the distant region must be described by a Vaidya metric.

SC coordinates for the SC spacetime are similar to radar coordinates, and cover the same spacetime region. In radar coordinates defined by a distant observer, it seems to me that something interesting happens to the labeling of an infaller world line in the case of an evaporating BH. Light trapped on the horizon would be released simultaneous to the final BH evaporation. This means (following the way radar assign coordinate time), the actual event of infaller reaching the horizon would be given a finite coordinate time before the event final evaporation (because the inward light reaching infaller at horizon would start much earlier than an inward light path reaching the horizon at moment of evaporation). There would still be no labeling events inside the horizon. The world line of the infaller would just cease at some finite coordinate time corresponding to horizon crossing, at the horizon radius at the time of crossing. The energy associated with the infaller would then come out later in the Hawking radiation.

Last edited: Mar 25, 2015
17. Mar 25, 2015

### Staff: Mentor

Far away from the hole, you can do this; but if you try to extend such a chart close to the horizon, you will find that it doesn't work; you can't have $M$ depend on just the $t$ coordinate. It has to depend on both $t$ and $r$. If you want $M$ to just be a function of the "time" coordinate, you have to use something like Eddington-Finkelstein coordinates, as I said before.

If you're ok with $M$ depending on $r$ as well as $t$, then you can use these "Schwarzschild-type" coordinates, but in these coordinates, the $r$ coordinate of the horizon changes with time, which makes a difference. See below.

Also, remember that this spacetime is not a vacuum solution; the Hawking radiation has a nonzero stress-energy tensor. This sort of spacetime is actually more like the outgoing Vaidya solution than the Schwarzschild black hole solution. (PAllen has already mentioned this.)

Not quite. In these coordinates, as I said above, the $r$ coordinate of the horizon decreases with time; so while the infalling observer might appear to "slow down" as he falls, he won't "hover", because, roughly speaking, the horizon is "falling" too.

No, they aren't. From the point of view of this coordinate system, the evaporation and the falling through the horizon of the infalling observer are simultaneous. This coordinate system cannot assign a time to the event where the infalling observer hits the singularity and is crushed (or to any event inside the horizon).

18. Mar 25, 2015

### PAllen

Following this up, the key point I intended is that contrary to the SC spacetime, radar coordinates for a distant observer of an evaporating BH do not give infinite coordinate time to horizon crossing or horizon formation. Both coordinate times are finite [due to final evaporation 'releasing' frozen outward null geodesics], and causal ordering is as expected: horizon formation precedes later infaller crossing horizon, which precedes final evaporation.

19. Mar 25, 2015

### stevendaryl

Staff Emeritus
Right, but after evaporation, the infalling observer is effectively gone from the universe of the external observer. Whether or not you assign a time to his demise, there is a time after which he's beyond interacting with.

20. Mar 25, 2015

### stevendaryl

Staff Emeritus
Okay, here's a question. If we temporarily forget about Hawking radiation, we can use the Schwarzschild metric to compute two numbers of an infalling observer:
1. The time $\tau_1$ on his watch when he crosses the event horizon.
2. The time $\tau_2$ on his watch when he smashes into the singularity. (Ignoring the fact that his watch will be smashed before he reaches the singularity)
So if we assume that Hawking radiation is a tiny perturbation, then those two numbers are not changed much by evaporation.

The question for a distant observer is this: Does evaporation allow the light from these two events reach a distant observer?