Hawking radiation prevents event horizon crossing?

[m4r35n357]

OK, thanks for clearing up a couple of points. Where my mind is leading me is along these lines. I was watching some numerical simulations of merging masses (black holes) and it occurred that this model of merging "horizons" is much nicer to visualize than infinite red-shifting/smearing at a static "horizon". I'm quoting the word horizon because I've always associated it with the infinite future, and I'm not sure how or even whether the concept of future infinity arises in a simulation. I suppose it must because they plotted it. So I've replaced one uncertainty with another!
I suppose I'll have to look into collapsing dust equations . . . feel free to close this if I've said nothing of interest!

 

[Asher Weinerman]

The Schwarzschild solution (SC) is static, and is also a single body solution. Both these aspects prevent proper consideration of "horizon crossing", and we should not be surprised to see "paradoxes" like this appear. In numerical simulations the infalling particle eventually "merges" with the black hole and the event horizon expands to engulf both.

Maybe the above quote contains the answer I'm looking for. In numerical simulations, will the infalling particle merge with the black hole and the event horizon expand to engulf both IN A FINITE AMOUNT OF FAR_AWAY TIME?

 

[m4r35n357]

Maybe the above quote contains the answer I'm looking for. In numerical simulations, will the infalling particle merge with the black hole and the event horizon expand to engulf both IN A FINITE AMOUNT OF FAR_AWAY TIME?

According to these simulations, yes. The relevance/accuracy of the computer models and initial/boundary conditions is another matter though.

 

[Asher Weinerman]

According to these simulations, yes.

Thank-you. I now understand that the infinite (far-away) time calculation is for a single body, and produces a paradoxical result. A correct two body treatment requires a computer simulation and results in a FINITE (far-away) time for the horizon crossing because the horizon itself expands to engulf the infalling particle.

I knew the answer should be simple, but I couldn't find it anywhere!

Thanks again,
Asher

 

[PeterDonis]

In numerical simulations, will the infalling particle merge with the black hole and the event horizon expand to engulf both IN A FINITE AMOUNT OF FAR_AWAY TIME?

The ingoing Vaidya metric is a closed form solution that shows the same effect; it models spherically symmetric radiation falling into a black hole, and the hole's horizon radius increases as the radiation falls in.

(The outgoing Vaidya metric, which has been mentioned in this thread as being a suitable idealized model for a black hole emitting Hawking radiation, is just the time reverse of the above; as the radiation goes out, the horizon radius of the hole decreases.)

 

[sshai45]

I was reading these posts and am wondering about this. It seems that it is being said here that it does not, in fact, take "infinite" time to fall into the black hole. So then, how much time does it take? If I'm watching something falling into the hole from a spaceship orbiting some good distance from that hole, at what time on my watch will I register the object has crossed the horizon?

 

[Nugatory]

If I'm watching something falling into the hole from a spaceship orbiting some good distance from that hole, at what time on my watch will I register the object has crossed the horizon?

You will never register it because light from the object crossing the horizon will never reach your eyes. But that doesn't mean it didn't happen, just that you didn't see it.

 

[Asher Weinerman]

You can put stationary clocks synchronized to some far-away time at all radii up to but not including r = 2M, and if the infalling observer stops each clock as he passes it, all clocks of course will read a finite time regardless of whether or not you actually "see" the observer at these locations. However it is not possible to put a stationary clock at r = 2M (or somewhat more because the horizon expands to engulf the observer), so I'm not sure if it is possible to attach a meaning to "when" the observer falls through the horizon in far-away time. If we can't measure it, the question has no meaning. But if the black hole eventually meets some fate such as through Hawking Radiation or otherwise, does it become important in some way or measurable if our observer has actually fallen through r = 2M before the fate of the black hole is realized. I know I'm not answering your question sshai45, I'm just adding to it. I'm not sure if anyone knows the answer to this.

 

[PeterDonis]

it is not possible to put a stationary clock at r = 2M

Correct. But the reason for this is that the horizon is a null surface, and no object moving on a null worldline can ever be "stationary". That means you have to be careful drawing further conclusions from this fact; see below.

(or somewhat more because the horizon expands to engulf the observer)

That doesn't make the horizon non-null or change the relation r = 2M for the horizon; it just means that M increases when an object falls in.

I'm not sure if it is possible to attach a meaning to "when" the observer falls through the horizon in far-away time.

Sure it is. You just have to use coordinates that aren't singular on the horizon. Surfaces of constant time in these coordinates will be different from the surfaces of constant time in standard Schwarzschild coordinates (the ones in which all the stationary clocks for r > 2M are synchronized--though it's worth noting that they run at different rates depending on r); but coordinates are just conventions, so this is fine. There is no "right" answer to the question "when did the object cross the horizon in far-away time"; you can adopt coordinates that make the answer "infinity" or you can adopt coordinates that make the answer some finite number. If your intuitions are telling you that there ought to be some unique "right" answer to the question, that means your intuitions are wrong and need to be re-trained.

if the black hole eventually meets some fate such as through Hawking Radiation or otherwise, does it become important in some way or measurable if our observer has actually fallen through r = 2M before the fate of the black hole is realized

Yes. That question does have a unique "right" answer. See below.

I'm not sure if anyone knows the answer to this

Yes, we do. The answer is that the observer does fall through the horizon before the fate of the black hole is realized. This will be true regardless of what coordinates you adopt--except that if you use coordinates that are singular on the horizon, you won't be able to compute the answer at all. But "unable to compute the answer in those coordinates" is not the same as "we don't know the answer"; we do know the answer, because we can adopt coordinates in which it can be computed.

 

[Asher Weinerman]

I was reading these posts and am wondering about this. It seems that it is being said here that it does not, in fact, take "infinite" time to fall into the black hole. So then, how much time does it take? If I'm watching something falling into the hole from a spaceship orbiting some good distance from that hole, at what time on my watch will I register the object has crossed the horizon?

The above quote is a good question. Although an outside observer cannot measure what happens at the event horizon they can position clocks synchronized to some far-away clock as close to the event horizon as they want in principle, and take a limit as r approaches the event horizon. So what time does a far-away clock register for an object of mass m to reach the event horizon of a black hole of mass M, starting out at position 3M, for example. Is there a computer simulation that can do this two-body problem as mentioned before? What is the answer? Does anyone know? When does the event horizon reach out and grab the infalling object before Hawking Radiation or something else destroys the black hole as time marches toward infinity?

 

[PeterDonis]

how much time does it take?

There is no invariant answer to this question, at least not from the viewpoint of the faraway observer. It depends on how that observer chooses coordinates. This is just a special case of the general rule for curved spacetimes that there is no invariant way for an observer to assign a "time" to events that are not on his worldline. (Strictly speaking, this is true even in flat spacetime, but in flat spacetime there are coordinates--global inertial frames--that are physically "picked out" as having special properties. In curved spacetimes, there are no such global coordinates at all.)

From the viewpoint of the observer falling in, there is an invariant answer to the question: the amount of proper time it takes to free-fall to the horizon is determined by the altitude at which the observer starts his fall and the initial velocity of fall at that altitude (usually taken to be zero, i.e., the observer falls from rest at some altitude). The exact formula is somewhat complicated, but a reasonable rule of thumb is that the proper time it takes to fall goes like the 3/2 power of the altitude (in units where $c = 1$).

 

[PeterDonis]

When does the event horizon reach out and grab the infalling object before Hawking Radiation or something else destroys the black hole as time marches toward infinity?

There is no invariant answer to the "when" question; it depends on the choice of coordinates. There is, however, an invariant answer to the question of ordering, i.e., does the infalling object reach the horizon before the hole evaporates away. The answer to that question is "yes". The fact that there is no invariant way to assign an exact "time" to the two events does not change the fact that their ordering is invariant.

 

[PAllen]

The above quote is a good question. Although an outside observer cannot measure what happens at the event horizon they can position clocks synchronized to some far-away clock as close to the event horizon as they want in principle, and take a limit as r approaches the event horizon. So what time does a far-away clock register for an object of mass m to reach the event horizon of a black hole of mass M, starting out at position 3M, for example. Is there a computer simulation that can do this two-body problem as mentioned before? What is the answer? Does anyone know? When does the event horizon reach out and grab the infalling object before Hawking Radiation or something else destroys the black hole as time marches toward infinity?

The answer is: whatever time you want, within limits. The problem is your statement "clocks synchronized to some far-away clock". No objective meaning can be attached to this, only convention. The convention chosen determines the answer.

Other parts of your question have objective answers for the classical case without Hawking radiation However, with Hawking radiation, I am not aware of an explicit simulation. Pervect, in post #11, provided links (some collected by me), to the general question, but they do not include detailed simulations of BH + infalling body in the presence of Hawking radiation.

 

[PeterDonis]

with Hawking radiation, I am not aware of an explicit simulation

I should clarify that my statements about the invariant ordering of events (the infalling object crosses horizon before the hole evaporates) are based on an idealized model, not on simulations. A Penrose diagram of this idealized model is given here:

http://en.wikipedia.org/wiki/Black_hole_information_paradox#Hawking_radiation

Of course this idealized model may not actually be the right one; that depends on how the black hole information paradox is resolved. But for this discussion, that idealized model appears to be the one being assumed.