Remote observer sees black hole evaporate

[phinds]

In a couple of recent threads, I have attempted, unsuccessfully, to elicit a response to the following, which is my belief about why an observer standing well off from a black hole will in fact see the black hole evaporate (over a LONG period of time, of course, so the observer has to have an unrealistic lifespan, but that's not the point).

My logic is this: assuming Hawking Radiation is real then in the far far future the black hole will evaporate down to nothing. As it finishes evaporating, the view seen by the incredibly long-lived observer will change from not seeing anything falling into the event horizon, to seeing it because the event horizon will shrink to nothing as the BH evaporates, and the photons showing the in-falling object actually falling in will be released.

SO ... I contend that an observer WILL see a black hole evaporate, he just has to wait a while. QUITE a while. And so the oft-heard statement that an observer will never see anything fall into the black hole seems wrong to me.

Comments?

P.S. I have no idea whether this thread should be "I" or "A" but the forum requires a choice so I flipped a coin.

 

[PeterDonis]

I contend that an observer WILL see a black hole evaporate, he just has to wait a while.

Yes, this is true. But "a while" can be a pretty long time. For a black hole of one solar mass, for example, "a while" is about $10^{67}$ years. The time goes up as the cube of the mass, so for a 10 solar mass hole it would be $10^{70}$ years, and so on.

the oft-heard statement that an observer will never see anything fall into the black hole seems wrong to me.

It depends on the context. In the context of classical GR, there is no black hole evaporation, and the statement is true.

In the context of quantum gravity, our best understanding is that black holes can evaporate, but it can take a very, very long time, as above. But nobody has actually confirmed this by observation, nor is such confirmation expected any time in the near future, because the Hawking radiation predicted to be emitted by any black hole we are likely to observe is so faint. So while the statement is false on our best understanding, that understanding doesn't have the kind of experimental support that classical GR does.

 

[PeterDonis]

I have no idea whether this thread should be "I" or "A"

"I" seems appropriate to me.

 

[martinbn]

I have a question about this. I suppose that the answer depends a lot on the exact nature of the evaporation and whether there is a remnant or not, but assuming the ussual explanation of the evaporation, as in the Penrose digaram bellow, my question is how much will the observer see? Is he going to see all things that fell into the hole and the event ot the hole disappearing? From the diagram it seems that his world line will cross the null surface, that contains the event horizon, at one point, so he will be a wittness of only one event. I'd say he will see the event horizon at only one stage, not the entire history of the horizon.

Edit: I removed the diagram. It is apparently too big. Here is link to it.
http://i.stack.imgur.com/Qtjrx.png

 

[phinds]

Yes, this is true. But "a while" can be a pretty long time. For a black hole of one solar mass, for example, "a while" is about $10^{67}$ years. The time goes up as the cube of the mass, so for a 10 solar mass hole it would be $10^{70}$ years, and so on.
It depends on the context. In the context of classical GR, there is no black hole evaporation, and the statement is true.

In the context of quantum gravity, our best understanding is that black holes can evaporate, but it can take a very, very long time, as above. But nobody has actually confirmed this by observation, nor is such confirmation expected any time in the near future, because the Hawking radiation predicted to be emitted by any black hole we are likely to observe is so faint. So while the statement is false on our best understanding, that understanding doesn't have the kind of experimental support that classical GR does.

Thank you Peter. Everything thing you say is exactly what I thought was the case and I appreciate the confirmation

 

[phinds]

I have a question about this. .

Could you PLEASE not use such a GIGANTIC image? When clicked on, it covers the reply box

 

[anorlunda]

If we interpret the OP as a thought experiment rather than a question about real life BHs, then one could imagine a micro BH with a mass of only 1 kg instead of N solar masses. Its lifetime before complete evaporation would be very short; micro? nano? pico? seconds.

One could also think of the OP question in the context of speculation about the ultimate heat death of the universe N years in the future (is a googleplex of years enough?). In that case, I would expect that all BHs had evaporated but the existence of an observer would be the fantasy hypothesis.

 

[phinds]

If we interpret the OP as a thought experiment rather than a question about real life BHs, then one could imagine a micro BH with a mass of only 1 kg instead of N solar masses. Its lifetime before complete evaporation would be very short; micro? nano? pico? seconds.

One could also think of the OP question in the context of speculation about the ultimate heat death of the universe N years in the future (is a googleplex of years enough?). In that case, I would expect that all BHs had evaporated but the existence of an observer would be the fantasy hypothesis.

I had considered a micro BH and agree w/ you. Whether or not a distant-time observer is a fantasy or might exist is pure speculation for either of us but I tend to agree w/ you. I cannot image how it could be possible but I've learned not to trust my "common sense" in cosmology.

 

[PeterDonis]

how much will the observer see?

In the model shown in the diagram, an external observer can, in principle, eventually see every event that happened on the event horizon. But he can't see anything that happened inside the horizon (i.e,. in the upper-left triangle bounded by the horizon and the singularity).

From the diagram it seems that his world line will cross the null surface, that contains the event horizon, at one point, so he will be a wittness of only one event.

No--he will witness (as in receive light signals from), in principle, an infinite number of events--all the ones that took place on the horizon line--because the light signals from all those events overlap, so to speak--they all travel along the same outgoing null worldline. But he will receive all those light signals at one event on his own worldline.

Of course this is a classical argument, so it can't quite be correct since the whole scenario requires quantum physics. Quantum mechanically, I would expect there to be some kind of "cutoff" such that the observer can only receive a finite amount of total energy from all the events that happened on the horizon. But that's just a heuristic guess. I don't know that anyone has tried to compute this in detail.

 

[martinbn]

@PeterDonis : Ah, I see, you're right.

 

[Chronos]

The latest ideas along this line suggest Hawking radiation does not orginate from the event horizon, but, at some distance away from the EH. It is true there remains the suspicion some kind of cuttoff limits emissions that can reach a remote observer. For discussion, see http://arxiv.org/abs/1511.08221, Hawking radiation, the Stefan-Boltzmann law, and unitarization. The calculations are not trivial and it is easy to misinterpret coordinate transformation consequences for remote obsevers.

 

[phinds]

@PeterDonis : Ah, I see, you're right.

Hey, that's a given. He's ALWAYS right

 

[martinbn]

Hey, that's a given. He's ALWAYS right

Well, yes, but I meant that I understood how I got myself into a pickle.

 

[phinds]

Well, yes, but I meant that I understood how I got myself into a pickle.

Unpickling people is Peter's specialty, he does it for me all the time.

 

[PeterDonis]

Unpickling people

As a Python programmer and a food enthusiast, there are just so many ways I could go with this...

 

[phinds]

As a Python programmer and a food enthusiast, there are just so many ways I could go with this...

If it's a pun you have in mind, please don't. The only good puns are the ones *I* make.

 

[CalcNerd]

an observer standing well off from a black hole will in fact see the black hole evaporate (over a LONG period of time, of course, so the observer has to have an unrealistic lifespan, but that's not the point).

.
Well, if you had more courage and got a bit closer, the BH would evaporate MUCH faster! Just get into an orbit just outside the horizon (Hey, you found the BH, so obviously you have an advanced spaceship). Let Relativity do the rest! Relativistic speeds and gravity fields should shorten those time periods for you down to whatever time you want. I would suggest exiting a bit before the hole evaporated completely. And sadly, you might be in a lonely universe when you do (or become the local caveman, if you're lucky).

 

[Darryl]

Why would the event horizon shrink? Is not the event horizon due to the gravity 'curving' space back onto itself, so if the BH lost matter that 'curve' would be less and less, making the EH grow, until the curve was such that there was no even horizon at all, and the light/energy would just radiate like anybody such as the Sun..

 

[phinds]

Why would the event horizon shrink? Is not the event horizon due to the gravity 'curving' space back onto itself, so if the BH lost matter that 'curve' would be less and less, making the EH grow, until the curve was such that there was no even horizon at all, and the light/energy would just radiate like anybody such as the Sun..

The radius of the EH is a function of the mass. As the BH loses mass, the EH gets smaller. Do you seriously think that a 5 solar mass BH has a larger EH than a supermassive BH?

 

[Darryl]

The radius of the EH is a function of the mass. As the BH loses mass, the EH gets smaller. Do you seriously think that a 5 solar mass BH has a larger EH than a supermassive BH?

I was referring to the amount of curvature, as the mass gets lower the curvature is less, it would then take more distance for light (that now has mass!) to curve back onto itself at an angle of 180 degrees or more, will lower mass that curve will be bigger it will take more space for light to be curved.

So as you BH loses mass that point of curvature will become larger, and would continue to become larger until the angle was less than 180 degrees at which point the event horizon would cease to exist at all. Unless you seriously think that a BH that is losing mass would get more dense.

[Mentor's note: Some text that violated the Physics Forums rule against posting personal theories and alternative explanations not supported by publication in appropriate peer-reviewed journals or standard texts has been removed]

 

[Bandersnatch]

I was referring to the amount of curvature, as the mass gets lower the curvature is less, it would then take more distance for light (that now has mass!) to curve back onto itself at an angle of 180 degrees or more, will lower mass that curve will be bigger it will take more space for light to be curved.

That's not how this works. Remember that EH is the equivalent of a surface from which escape velocity is c. That is, an object traveling at c needs an infinite 'distance' of space to escape the gravitational field. You can't have more than infinite distance for the light to travel. In terms of curvature, what you get when summing from EH to infinity is the minimum curvature needed to keep light from escaping the black hole.
As you reduce mass, the escape velocity at a given distance is reduced, i.e., there's not enough curvature to keep sufficiently fast objects from escaping, i.e., the event horizon shrinks.