# Infalling observer could never cross Black Hole event horizon? [duplicate]

I have an interest in cosmology, but I would no way consider myself an expert. It would be great if the more knowledgeable contributors would be able to answer the question I have posed.

The intense gravitational field near a black hole causes time dilation. This means, to an observer close to its event horizon, time runs more slowly. It can mathematical be expressed as:

Time dilation = $$\frac{\delta t_f}{\delta t_0} = \frac{1}{\sqrt{1-\frac{R_s}{R}}}$$.

Where $$\delta t_f$$ is an interval of time a long distance from the black hole, $$\delta t_0$$ is the time interval a distance $$R$$ from the black hole’s centre and $$R_s$$ is the radius of the event horizon.

This formula means that time dilation becomes infinite at the event horizon so to a distant observer a falling object would take an infinite time to reach the event horizon and would never cross it. However, in the frame of an observer falling into the black hole it should them take a finite time to reach the event horizon. To me the complexity is that all black holes are believed to ‘evapourate’ and eventually disappear in a finite time, by emitting Hawking radiation. Although this an incredibly slow process. For example, a solar mass hole would have a lifetime of $$10^{66}$$ years (in a time-frame measured a distance from the black hole).

My Question Assuming Hawking radiation exists and black hole evaporation occurs, does this mean that an infalling observer would never actually reach the event horizon, because the black hole has finite lifetime and would evaporate before an infalling observer would reach it?

Further Note Since posing this question I have had one further thought, which may or may be relevant, since my knowledge of GR and quantum theory is pretty sketchy.

The way that the time dilation formula works is that when the observer is a distance of one Planck length ( ~ $$10^{-35}$$ m ) away from the event horizon of a solar mass black hole (which has radius ~3km), the time dilation would be still only be ~ $$10^{19}$$. However, the lifetime of a solar mass black hole is ~$$10^{66}$$ years. Therefore, even when they are within 1 Planck length of the event horizon, the time dilation near the black hole is not sufficient for the observer to see the black hole evaporate.

It is often pointed out that the infinite time dilation at the EH is what is known as a "coordinate singularity", rather than an essential singularity that says something is physically singular there (like never reaching the EH for the infaller). A coordinate singularity is essentially a hiccup in the language we are using, like the way a north-south-east-west coordinate system is of no use at the North pole. The problem is that if you want to talk about "whether something has crossed the EH yet", you must match up your own concept of "now" with some time at the EH, and that matching up is the job of a coordinate system. So saying "it hasn't reached the EH yet," and repeating that every day for infinity of time here at Earth, is not the same thing as a physical statement about what is happening at the EH, it's just a reflection of the way we are matching up our "nows" with theirs-- it is purely a matter of the chosen coordinates.

To know what is physically happening to the observer at the EH, the ideal solution would be to simply ask them, and they would say "yes I am crossing the EH now". Here it gets tricky, because the normal coordinates we use to talk about the EH (the one you were using also), says that we could never receive that communication. But those coordinates make physical assumptions that are violated by having something falling into the black hole. At this point it would be best to have a GR expert, but I believe what happens is that the very act of having mass fall into the black hole in effect moves the EH out to swallow that mass, and we should therefore be able to receive the communication that the observer has crossed the EH in a finite time. This must be true because, as pointed out, the very existence of the black hole is like information coming to us that stuff has crossed its EH, an EH that formed and moved out as the black hole was created by mass infall.

• "I believe what happens is that the very act of having mass fall into the black hole in effect moves the EH out to swallow that mass". I think it's something like that, Ken. The black hole was originally known as a frozen star, and I think it grows something like a hailstone. You're a water molecule, you alight upon its surface. You can't pass through this surface, but you're surrounded and buried by other water molecules. So the surface passes through you. I think it's similar for the event horizon. See Friedwardt Winterberg's 2001 paper gamma ray bursters and Lorentzian relativity. – John Duffield Jan 29 '19 at 22:05
• In relation to John Duffield comment on Ken's reply What would happen if there the black Hole were not surrounded by much matter? This is certainly possible. in this case, I assume the event horizon wouldn't gradually grow . – John Davies Jan 30 '19 at 17:52
• @John Davies : why don't you ask a question on that? – John Duffield Jan 31 '19 at 8:24
• Thanks for the interesting reply. To me there is still something of a paradox. For any observer outside the event horizon nothing appears to enter the black hole. This is because of the way the time dilation formula works, Infalling objects “freeze” outside the event horizon. BUT, assuming Hawking radiation exists, to the observer a black holes would only last for a finite time (albeit very long time) – John Davies Jan 31 '19 at 12:01
• I was giving my reasoning that your statement, that nothing appears to enter the black hole, is incorrect. The time dilation formula you used assumes nothing falls into the black hole, it's a static solution. So if you assume that, you get what you assume! – Ken G Jan 31 '19 at 13:43

Assuming Hawking radiation exists and black hole evaporation occurs, does this mean that an infalling observer would never actually reach the event horizon, because the black hole has finite lifetime and would evaporate before an infalling observer would reach it?

The formulas I know which say that a black hole has a finite lifetime because of evaporation were done in a free-falling inertial frame. As an example see formula (11.108) from the book A First Course in General Relativity by Bernard Schutz; it says,

$$\tau \sim M^3$$where $$\tau$$ is the time in a freely falling system.

So, in order to answer your question. If you want to compare two times you must be sure that both times were calculated in the same coordinate system. Then to know what happens first (the whole evaporation or the body crossing the Schwarzschild limit) you can take both times in a freely falling system, and compare if the proper time for the particle is more than $$M^3$$ or not.

It makes no sense to compare times measured in different systems.

The intense gravitational field near a black hole causes time dilation. This means, to an observer close to its event horizon, time runs more slowly. It can mathematical be expressed as:
Time dilation $$= \frac{\Delta t_f}{\Delta t_0} = \frac{1}{\sqrt{1-\frac{R_s}{R}}}$$...

Actually, time dilation is associated with gravitational potential rather than field. The strength of the field at some elevation is proportional to the gradient in the potential at that location. This is important. If there's a bigger difference in the gravitational time dilation between two elevations A and B, then when I drop you at A, you will be falling faster when you fall past B.

This formula means that time dilation becomes infinite at the event horizon

Yes, the gravitational time dilation becomes infinite at the event horizon.

so to a distant observer a falling object would take an infinite time to reach the event horizon and would never cross it.

That's what people say. That's what Einstein said too. He wrote a paper in 1939. It was on a stationary system with spherical symmetry consisting of many gravitating masses. He said this

$$g_{44}=\left(\frac{1-\frac{\mu}{2r}}{1+\frac{\mu}{2r}}\right)^2$$
vanishes for $$r = \mu/2$$. This means that a clock kept at this place would go at the rate zero. Further it is easy to show that both light rays and material particles take an infinitely long time (measured in “coordinate time”) in order to reach the point $$r = \mu/2$$ when originating from a point $$r > \mu/2$$
.

Now, I'm a big fan of Einstein. But there's two problems with this. One is that falling bodies don't slow down. The bigger the difference in gravitational time dilation between elevations A and B, the faster they fall past B. In fact, if there was no difference in gravitational time dilation, a body wouldn't fall down at all. The other problem is that Einstein concluded that black holes cannot form, and we have great evidence that there are black holes out there. So I have to sigh and say my hero Einstein got this one wrong, more's the pity.

However, in the frame of an observer falling into the black hole it should them take a finite time to reach the event horizon.

That's right. There is said to be some "paradox" here. You can read about it in the New Scientist article the elephant and the event horizon. The article talks about Leonard Susskind, and says this: "If his calculations are correct, the elephant must be in more than one place at the same time".

To me the complexity is that all black holes are believed to ‘evaporate’ and eventually disappear in a finite time, by emitting Hawking radiation. Although this an incredibly slow process. For example, a solar mass hole would have a lifetime of $$10^{66}$$ years ( in a time-frame measured a distance from the black hole).

I think you should sideline the Hawking radiation actually. We have a major problem here even without it. How can the infalling body fall faster and faster, and yet take forever to reach the event horizon? And how can it be in two places at once? It can't. Something's wrong, so how do we resolve it? That's perhaps another question you could ask.

My Question Assuming Hawking radiation exists and black hole evaporation occurs, does this mean that an infalling observer would never actually reach the event horizon, because the black hole has finite lifetime and would evaporate before an infalling observer would reach it?

No. Black holes exist. If the infalling observer never reached the event horizon, black holes wouldn't exist. And they do.

• If you have to consider relativity, there is no "at the same time". Given two events, where one can't cause each other because a light ray from one can't hit the other, different observers will see them as in either order or at the same time. There is no privileged observer. – David Thornley Jan 29 '19 at 18:19
• I was unaware we were still in a debate about what's at the center of our galaxy. Frame of reference is everything (and I'm not a mathematician but if you're talking about formulas and infinity, all you can say is that it approaches infinity). Stand on the surface of the black hole and be an observer. Oh wait, you can't. But if you could you wouldn't "receive that communication" back from them anyway, until it turned back into a pulsar, or if you can decipher signals out of Hawking radiation. There's nothing special about an EH, other than not being able to garner information from beyond it. – Mazura Jan 29 '19 at 22:50
• "Falling bodies don't slow down". According to who? Until you sort out your problems with coordinate systems and frames of reference you are never going to "get" GR. Ditto with your comment about the coordinate speed of light being zero. True for Schwarzschild coordinates, not true for other coordinate systems and certainly not true for an observer moving through the event horizon. Note that an observer cannot be stationary at the event horizon in order to observe anything that lasts of any duration. – ProfRob Jan 30 '19 at 15:30
• @JohnDuffield No, the poor sap falling into the black hole doesn't notice time dilation. (It is possible to avoid spaghettification at the event horizon given a big enough black hole, say a million solar masses.) – David Thornley Jan 31 '19 at 4:04
• @JohnDuffield "I'm confident of my understanding of GR" - it's not confidence you're lacking. Please stop posting confusing and erroneus "answers" here. You're not in a place where you can explain GR to others. – Florin Andrei Jan 31 '19 at 6:19