# Falling into a black hole, time dilation and Hawking radiation

• I
I know that has been discussed elsewhere but never could find for a satisfying answer, so I try this here again.

Let us not take into account that an observer (an astronaut or a clock or just let us take both: an astronaut with a clock) falling into a black hole (BH) will be killed and torn apart by tidal forces. Just imagine what an observer falling towards the BH and an external observer will report.

If someone falls into a black hole (BH), then, as I understand it, due to gravitational time dilation the external observer should see the infalling clock ticking slower and slower until it almost stops ticking when nearing the event horizon (EH). But, as I understand it, this implies that also the speed with which I see the astronaut/clock nearing the EH should slow down after a certain point and then remain 'frozen' in time forever almost at the Schwarzschild radius. Some object that due to gravitational redshift the external observer would see nothing because light signals are becoming weaker and redshifted and nothing can be said. I don't find that a good objection that means much. Because in principle on can follow at it in the infrared and radio wave until to a certain point. One can follow the descend into a BH until a certain point 'asymptotically' so to speak, in principle until the EH.

On the other side, so goes the theory, the infalling observer with the clock (proper time) would notice nothing by crossing the EH and could not observe anything of the external universe and would see how he/she would fall towards the singularity very quickly. My understanding however is that the infalling observer, as long he/she will be able to still receive light form the external universe, will see it evolve faster and faster until it nears to the EH where (if equipped to observe the extreme blueshifted light with the same reasoning above there is in principle no limitation until the EH) and would see the entire history of the universe to play out in almost an instant.

I saw several explaining away the discrepancy by saying that nothing can be said because, for the infalling observer the external universe is outside the light cone and, since there is no meaning of a present moment in this context, these speculations are therefore meaningless. The two points of view are said to be compatible despite its intrinsic logical contradiction. This however, sounds incorrect to me as well. We can in principle avoid any consideration about proper or coordinate time and consider Hawking radiation. After an extreme long but still not infinite time, the BH will evaporate completely and the infalling observer who still did not cross the EH will indeed be able to observe again the universe trillions and trillions of years later (depending from the mass of the BH).

The bottom line is that nothing can really cross an EH. Not only, an EH can not form in the first place because the formation of a BH is an asymptotic process in time which however never will be completed.

Please tell me why this is wrong?

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phinds
Gold Member
2019 Award
The bottom line is that nothing can really cross an EH. Not only, an EH can not form in the first place because the formation of a BH is an asymptotic process in time which however never will be completed.
That looks at things from the point of view of a remote observer. The infalling object and the EH do not care what they look like to a remote observer, they just do their thing.

Dale
Demystifier
Gold Member
an EH can not form in the first place because the formation of a BH is an asymptotic process in time which however never will be completed.
The event horizon cannot be completed in a finite time, but the apparent horizon can. Most of the peculiar properties of the "horizon", including Hawking radiation, should really be interpreted as properties of the apparent horizon.

PeterDonis
Mentor
2019 Award
The event horizon cannot be completed in a finite time
What is your basis for this claim? There are known solutions of the Einstein Field Equation in which an event horizon does form in a finite time.

PeterDonis
Mentor
2019 Award
My understanding however is that the infalling observer, as long he/she will be able to still receive light form the external universe, will see it evolve faster and faster until it nears to the EH where (if equipped to observe the extreme blueshifted light with the same reasoning above there is in principle no limitation until the EH) and would see the entire history of the universe to play out in almost an instant.
This is not correct. The infalling observer will see the light from the external universe as more and more redshifted, and will not see the entire history of the universe play out. You are confusing the infalling observer with an observer who is hovering at a very low altitude just above the horizon; this observer will see light from the external universe as highly blueshifted, and will see the external universe's history evolving very fast by his own clock.

Pencilvester
Nugatory
Mentor
Let us not take into account that an observer (an astronaut or a clock or just let us take both: an astronaut with a clock) falling into a black hole (BH) will be killed and torn apart by tidal forces.
Not necessarily.... The more massive the black hole, the less the gravitational gradient and hence the weaker the tidal force on the observer. If the black hole is large enough, you could fall through the event horizon without feeling anything except weightless. (Yes, this is a both a quibble and a digression from your main question - I'm only mentioning it because you might want to consider the possibility that whatever you've learned about black holes from has left some important stuff out).

My understanding however is that the infalling observer, as long he/she will be able to still receive light form the external universe, will see it evolve faster and faster until it nears to the EH where (if equipped to observe the extreme blueshifted light with the same reasoning above there is in principle no limitation until the EH) and would see the entire history of the universe to play out in almost an instant.
That is not correct. There's a good explanation of what happens (along with discussion of some other common misconceptions) here: https://arxiv.org/abs/0804.3619

For a more intuitive explanation of what's going on: The infaller reaches the horizon, passes through it, and dies at the "central" singularity in a finite amount of time according to his wristwatch. The last thing he sees before his death is light that has passed through the horizon just after him and has been able to catch up with him before he reaches the singularity. Because it passes through the horizon just after he did, it has to have been emitted very shortly after he started his fall.

This will make more sense if you start with a Kruskal diagram. Plot on it the worldine of someone falling into the black hole from outside (the details of the trajectory don't matter as long is it is timelike - moving upwards at angle of more than 45 degrees on the diagram), the worldline of a Schwarzschild stationary outside observer (this will be a hyperbola of constant ##r##) and the paths of light signals sent from the outside observer to the infaller (these will be straight lines going up and left at an angle of 45 degrees). You will pretty quickly see how it is that the infaller does not see the entire future of the universe, just the little bit that he can reasonably say "happens before" he dies at the singularity.
Next, draw the path of light signals sent from the infaller to the outside observer (these will be straight lines going up and right at an angle of 45 degrees) to see what it really means to say that it takes infinite Schwarzschild coordinate time to fall through the horizon

I really strongly recommend that you try this exercise next - it is one of the most clarifyng things you can do.

I saw several explaining away the discrepancy by saying that nothing can be said because, for the infalling observer the external universe is outside the light cone and, since there is no meaning of a present moment in this context, these speculations are therefore meaningless.
Either you've found some bad explanations or you've misunderstood something in them, because that's simply not correct. The external observer is outside the light cone, but that doesn't mean what you're saying it does. try the exercise with the Kruskal diagram I suggest above and you'll see what it does mean.
After an extreme long but still not infinite time, the BH will evaporate completely and the infalling observer who still did not cross the EH will indeed be able to observe again the universe trillions and trillions of years later (depending from the mass of the BH).
That is not correct. The most that we can say is that if and when the black hole evaporates, the light that left the infaller as he passes the horizon will finally escape so that in principle an outside observer will be able to see it. But by then the infaller is long gone; you'll see this in the Kruskal diagram.
The bottom line is that nothing can really cross an EH. Not only, an EH can not form in the first place because the formation of a BH is an asymptotic process in time which however never will be completed.
There are several things going on here. First, the formation of a black hole is not an asymptotic process, it's described by the Oppenheimer-Snyder solution of the Einstein Field equations.

Second, the apparently asymptotic process by which an outside observer sees the infaller getting closer and closer to the horizon without reaching it comes from a calculation in which the mass of the infaller is assumed to be zero. For astronauts being dropped into a stellar-mass black hole this is a really excellent approximation, but it is an approximation and it breaks down when very near to the horizon. Even Schwarzschild coordinates predict that an infaller with non-zero mass will cross the horizon in a finite time according to the outside observer.....I wrote a post about this a while back, I'll see if I can dig it up.

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Dale
Demystifier
Gold Member
What is your basis for this claim? There are known solutions of the Einstein Field Equation in which an event horizon does form in a finite time.
Yes, but these solutions assume that all matter arrives during a finite time. In other words, it assumes that no matter will arrive later, after the horizon has been formed. Only if one knows that no matter will arrive later, one can call this horizon the event horizon. Otherwise, this is only an apparent horizon. That's why it if often (correctly) said that event horizon is a teleological concept. See e.g. https://physics.stackexchange.com/questions/277720/teleological-property-of-event-horizon

PeterDonis
Mentor
2019 Award
these solutions assume that all matter arrives during a finite time. In other words, it assumes that no matter will arrive later, after the horizon has been formed. Only if one knows that no matter will arrive later, one can call this horizon the event horizon.
Yes, agreed. But this doesn't mean there isn't an actual event horizon. If there is one at all, it's there as soon as the first blob of matter collapses to be inside a 2-sphere with a small enough area (actually, it's there even earlier than that, since an event horizon will form at the center of the blob and gradually expand outward to meet the blob's surface). So it's not correct to say the event horizon doesn't form in a finite time; it does. It's just that we don't know where the true event horizon actually is, since it can't be locally detected, whereas an apparent horizon can be.

phinds and Demystifier