<p>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.</p>

<p>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.</p>

CC BY-SA 4.0
2019-01-29T22:05:33.887Z
_"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
CC BY-SA 4.0
2019-01-30T17:52:35.360Z
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
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2019-01-31T08:24:32.797Z
@John Davies : why don't you ask a question on that?
John Duffield
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2019-01-31T12:01:17.013Z
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
CC BY-SA 4.0
2019-01-31T13:43:57.640Z
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
CC BY-SA 4.0
2019-01-29T16:39:26.427Z
2019-01-29T16:39:26.427Z
Ken G
<blockquote>
  <p>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?</p>
</blockquote>

<p>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 <em>A First Course in General Relativity</em> by Bernard Schutz; it says,</p>

<p><span class="math-container">$$\tau \sim M^3$$</span>where <span class="math-container">$\tau $</span> is the time in a freely falling system. </p>

<p>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 <span class="math-container">$M^3$</span> or not.</p>

<p>It makes no sense to compare times measured in different systems.</p>

CC BY-SA 4.0
2019-01-29T14:17:36.787Z
2019-01-31T07:01:56.653Z
2019-01-31T07:01:56.653Z
Gabriel Palau
<blockquote>
  <p>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:<br>
  Time dilation <span class="math-container">$= \frac{\Delta t_f}{\Delta t_0} = \frac{1}{\sqrt{1-\frac{R_s}{R}}}$</span>... </p>
</blockquote>

<p>Actually, time dilation is associated with gravitational <em>potential</em> 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.   </p>

<blockquote>
  <p>This formula means that time dilation becomes <strong>infinite</strong> at the event horizon </p>
</blockquote>

<p>Yes, the gravitational time dilation becomes infinite at the event horizon. </p>

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

<p>That's what people say. That's what Einstein said too. He wrote a paper in 1939. It was <a href="http://www.cscamm.umd.edu/tiglio/GR2012/Syllabus_files/EinsteinSchwarzschild.pdf" rel="nofollow noreferrer">on a stationary system with spherical symmetry consisting of many gravitating masses</a>. He said this</p>

<p><em>“<span class="math-container">$g_{44}=\left(\frac{1-\frac{\mu}{2r}}{1+\frac{\mu}{2r}}\right)^2$</span><br>
vanishes for <span class="math-container">$r = \mu/2$</span>. 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 <span class="math-container">$r = \mu/2$</span> when originating from a point <span class="math-container">$r &gt; \mu/2$</span>”</em>.</p>

<p>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.         </p>

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

<p>That's right. There is said to be some "paradox" here. You can read about it in the New Scientist article <a href="https://www.newscientist.com/article/mg19225751-200-the-elephant-and-the-event-horizon/" rel="nofollow noreferrer">the elephant and the event horizon</a>. The article talks about Leonard Susskind, and says this: <em>"If his calculations are correct, the elephant must be in more than one place at the same time"</em>.  </p>

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

<p>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.</p>

<blockquote>
  <p><strong>My Question</strong>
  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?</p>
</blockquote>

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

CC BY-SA 4.0
2019-01-29T18:19:02.597Z
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
CC BY-SA 4.0
2019-01-29T21:11:56.797Z
" If the infalling observer never reached the event horizon, black holes wouldn't exist." Unless by "black hole" we mean "the phenomenon of mass asymptotically approaching a location", which you have not established is not what physicists mean by "black hole". The rest of your answer has further problematic argument-by-assertion.
Acccumulation
CC BY-SA 4.0
2019-01-29T21:16:05.490Z
"And how can it be in two places at once? It can't." According to standard interpretations of quantum mechanics, nothing is at a particular place. Everything has a probability distribution over space. Everything is in some sense in an infinite number of places at once. Normally, the probability distribution for macroscopic objects is highly localized, but the effects of a black hole spread the probability distribution over the whole event horizon.
Acccumulation
CC BY-SA 4.0
2019-01-29T21:48:12.820Z
@Acccumulation : Check out Sagittarius A*. There’s something there with a mass that’s circa 4.28 million times the mass of the Sun. But it’s at most thirty times bigger than the Sun in terms of spatial extent. There’s only one thing it can be, and that’s a black hole. The problematic stuff doesn't come from me. I'm making the simple point that falling bodies don't slow down.
John Duffield
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2019-01-29T21:49:21.013Z
@David Thornley : all observers agree that the falling body is indeed falling down, and that gravitational time dilation is greater at the lower elevation. .
John Duffield
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2019-01-29T22:50:32.583Z
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
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2019-01-30T12:26:32.427Z
@Mazura : if you were at the event horizon, you'd be at a place where the "coordinate" speed of light is zero. See [Wikipedia](https://en.wikipedia.org/wiki/Propagation_of_light_in_non-inertial_reference_frames). So it would take you forever to observe anything. People tend to say the local speed of light is still c, but I think that's a mistake myself. How can anybody say that if it takes forever to measure the local speed of light?
John Duffield
CC BY-SA 4.0
2019-01-30T15:30:21.900Z
"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
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2019-01-30T18:35:47.373Z
@Rob Jeffries : according to me. See my answer. The bigger the difference in gravitational time dilation between elevations A and B, the faster the body dropped from A falls past B. Are you disputing that? If so please say so, and I'll ask a question on the subject so you can give details. Also see my link to Wikipedia above. It says this: _"For example, at the event horizon of a black hole the coordinate speed of light is zero"_. It's assuming the coordinate system of the distant observer, eg you or me, which I think is perfectly reasonable. I'm confident of my understanding of GR, thanks.
John Duffield
CC BY-SA 4.0
2019-01-31T04:04:46.747Z
@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
CC BY-SA 4.0
2019-01-31T06:19:17.837Z
@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
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2019-01-31T08:19:33.197Z
@Florin Andrei : please explain what's confusing or erroneous about this  answer, or provide an answer of your own. Alternatively, I suggested to Rob that I could pose a new question on the infalling issue. Perhaps you could answer that? As for explaining GR, that's something I've been trying to do elsewhere. You might like to read [this](http://physicsdetective.com/black-holes/) and [this](http://physicsdetective.com/firewall/). I welcome any comments or correction - I'd hate it if I'd said anything that was misleading or wrong.
John Duffield
CC BY-SA 4.0
2019-01-31T08:52:15.617Z
@Mazura There is still some room for doubt regarding the existence of actual black holes. Sure, we've found lots of BH *candidates*, and for most of them, if they aren't BHs they must be something even more exotic. It's not easy to get irrefutable observational evidence of BHs, especially from a distance, and we really need a quantum gravity theory to describe them fully.
PM 2Ring
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2019-01-29T13:52:59.747Z
2019-01-31T07:02:09.470Z
2019-01-31T07:02:09.470Z
John Duffield
<p>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.</p>

<p>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:</p>

<p>Time dilation = <span class="math-container">$\frac{\delta t_f}{\delta t_0} = \frac{1}{\sqrt{1-\frac{R_s}{R}}}$</span>. </p>

<p>Where <span class="math-container">$\delta t_f$</span> is an interval of time a long distance from the black hole, <span class="math-container">$\delta t_0$</span> is the time interval a distance <span class="math-container">$R$</span> from the black hole’s centre and <span class="math-container">$R_s$</span> is the radius of the event horizon. </p>

<p>This formula means that time dilation becomes <strong>infinite</strong> 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 <strong>‘evapourate’ and eventually disappear in a finite time</strong>, by emitting Hawking radiation. Although this an incredibly slow process. For example, a solar mass hole would have a lifetime of <span class="math-container">$10^{66}$</span> years (in a time-frame measured a distance from the black hole).  </p>

<p><strong>My Question</strong>
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?</p>

<p><strong>Further Note</strong>
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. </p>

<p>The way that the time dilation formula works is that when the observer is a distance of one Planck length ( ~ <span class="math-container">$10^{-35}$</span> m ) away from the event horizon of a solar mass black hole (which has radius ~3km), the time dilation would be still only be  ~ <span class="math-container">$10^{19}$</span>. However, the lifetime of a solar mass black hole is ~<span class="math-container">$10^{66}$</span> 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. </p>

CC BY-SA 4.0
2019-01-29T21:52:18.070Z
Good question John.
John Duffield
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2019-01-29T22:24:40.740Z
[Zeno's parachute](https://i.imgur.com/nh6o29v.gif)
Dennis Williamson
CC BY-SA 4.0
2019-01-29T23:45:29.993Z
You might be interested in this mostly logically equivalent [question about whether or not matter accumulates just outside of the event horizon](https://astronomy.stackexchange.com/questions/2441/does-matter-accumulate-just-outside-the-event-horizon-of-a-black-hole).
mtraceur
<black-hole>
Infalling observer could never cross Black Hole event horizon?
John Davies
# Infalling observer could never cross Black Hole event horizon?

(Asked by: username_0 on 2019-01-29)

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.

**Tags:** black-hole
> **Comments:**
> **username_1**: Good question John.  
> **username_2**: [Zeno's parachute](https://i.imgur.com/nh6o29v.gif)  
> **username_3**: You might be interested in this mostly logically equivalent [question about whether or not matter accumulates just outside of the event horizon](https://astronomy.stackexchange.com/questions/2441/does-matter-accumulate-just-outside-the-event-horizon-of-a-black-hole).  

---

## 3 answers

### Answer by username_4 on 2019-01-29. Score: 10

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.

> **Comments:**
> **username_1**: _"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_.  
> **username_0**: 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 .  
> **username_1**: @John Davies : why don't you ask a question on that?  
> **username_0**: 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)  
> **username_4**: 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!  

---

### Answer by username_5 on 2019-01-29. Score: 2

> 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.


---

### Answer by username_1 on 2019-01-29. Score: 0

> 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](http://www.cscamm.umd.edu/tiglio/GR2012/Syllabus_files/EinsteinSchwarzschild.pdf). 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](https://www.newscientist.com/article/mg19225751-200-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.

> **Comments:**
> **username_6**: 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.  
> **username_7**: " If the infalling observer never reached the event horizon, black holes wouldn't exist." Unless by "black hole" we mean "the phenomenon of mass asymptotically approaching a location", which you have not established is not what physicists mean by "black hole". The rest of your answer has further problematic argument-by-assertion.  
> **username_7**: "And how can it be in two places at once? It can't." According to standard interpretations of quantum mechanics, nothing is at a particular place. Everything has a probability distribution over space. Everything is in some sense in an infinite number of places at once. Normally, the probability distribution for macroscopic objects is highly localized, but the effects of a black hole spread the probability distribution over the whole event horizon.  
> **username_1**: @Acccumulation : Check out Sagittarius A*. There’s something there with a mass that’s circa 4.28 million times the mass of the Sun. But it’s at most thirty times bigger than the Sun in terms of spatial extent. There’s only one thing it can be, and that’s a black hole. The problematic stuff doesn't come from me. I'm making the simple point that falling bodies don't slow down.  
> **username_1**: @David Thornley : all observers agree that the falling body is indeed falling down, and that gravitational time dilation is greater at the lower elevation. .  
> **username_8**: 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.  
> **username_1**: @Mazura : if you were at the event horizon, you'd be at a place where the "coordinate" speed of light is zero. See [Wikipedia](https://en.wikipedia.org/wiki/Propagation_of_light_in_non-inertial_reference_frames). So it would take you forever to observe anything. People tend to say the local speed of light is still c, but I think that's a mistake myself. How can anybody say that if it takes forever to measure the local speed of light?  
> **username_9**: "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.  
> **username_1**: @Rob Jeffries : according to me. See my answer. The bigger the difference in gravitational time dilation between elevations A and B, the faster the body dropped from A falls past B. Are you disputing that? If so please say so, and I'll ask a question on the subject so you can give details. Also see my link to Wikipedia above. It says this: _"For example, at the event horizon of a black hole the coordinate speed of light is zero"_. It's assuming the coordinate system of the distant observer, eg you or me, which I think is perfectly reasonable. I'm confident of my understanding of GR, thanks.  
> **username_6**: @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.)  
> **username_10**: @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.  
> **username_1**: @Florin Andrei : please explain what's confusing or erroneous about this  answer, or provide an answer of your own. Alternatively, I suggested to Rob that I could pose a new question on the infalling issue. Perhaps you could answer that? As for explaining GR, that's something I've been trying to do elsewhere. You might like to read [this](http://physicsdetective.com/black-holes/) and [this](http://physicsdetective.com/firewall/). I welcome any comments or correction - I'd hate it if I'd said anything that was misleading or wrong.  
> **username_11**: @Mazura There is still some room for doubt regarding the existence of actual black holes. Sure, we've found lots of BH *candidates*, and for most of them, if they aren't BHs they must be something even more exotic. It's not easy to get irrefutable observational evidence of BHs, especially from a distance, and we really need a quantum gravity theory to describe them fully.  

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