<p>You are essentially asking the following: if someone falls from the Earth from some way beyond the event horizon of a black hole, how long after they have left can an observer on Earth still signal to them with a light beam?</p>

<p>The answer of course depends on exactly how far the Earth is from the black hole. It is also often forgotten that it is not just light reaching them before the falling observer reaches the event horizon, but there is also a little bit more to add in the (proper) time it takes the falling observer to travel from the event horizon to the singularity.</p>

<p>The answer in all cases is that there is a finite time available, so no, the infalling person does not look back and see the cosmic aeons flying by.</p>

<p>The handwaving reason is that from the perspective of Schwarzschild coordinate time both the falling person and light beams approach the event horizon along an exponential asymptote. The light beams are always travelling faster, so their <span class="math-container">$dt/dr$</span> is always shallower and hence it is possible for them to catch the falling observer as long as the delay in sending the light beam does not exceed a finite critical value. Or to put it another way; you can contruct a past light cone from any point on the event horizon (or the space-like singularity) and it encompasses only a finite time anywhere outside the black hole.</p>

<p>The full, very gritty, details were worked out (for a Schwarzschild black hole) at <a href="https://physics.stackexchange.com/a/396157/43351">https://physics.stackexchange.com/a/396157/43351</a></p>

<p>For the case of the person free-falling from a position <span class="math-container">$r_0$</span>, then the span of time seen by the falling person when looking back towards <span class="math-container">$r_0$</span>, before they hit the singularity, is given by
<span class="math-container">$$ c(\Delta t)_{\rm singularity} = r_s \ln \left(\frac{r_s}{r_0-r_s}\right) + \pi r_s\left(\frac{r_0}{r_s} -1\right)^{1/2}\left(1 + \frac{r_0}{2r_s}\right) -r_0,$$</span>
where <span class="math-container">$r_s$</span> is the Schwarzschild radius.</p>

<p>Note that as <span class="math-container">$r_0$</span> gets bigger then of course <span class="math-container">$\Delta t$</span> get bigger and approaches <span class="math-container">$\Delta t \simeq r_s \pi(r_0/r_s)^{3/2}/2c$</span>, which is just the freefall <em>proper</em> time (i.e. measured on a clock carried by the falling object) for an object to fall into a black hole. i.e. You could see that length of time into Earth's future, but only because you had spent that long falling into the black hole !</p>

<p>Edit: To address Jonathan's question in comments. Take a 10 solar mass BH with <span class="math-container">$r_s=30$</span> km. If it takes 1 hour (proper time) to "free fall" radially (formulae above are for radial infall), then <span class="math-container">$r_0 = 2.42\times 10^{9}$</span> m. The formula above then shows that <span class="math-container">$\Delta t$</span> is 1 hour to three sig figs. The reason for this is that on such a trajectory, the falling object spends a negligible fraction of its infall time anywhere near the black hole, so the GR effects are negligible.</p>

CC BY-SA 4.0
2020-06-06T13:03:18.913Z
I say this with respect for you, Rob. And I predict my comment will be deleted. But I am regularly impressed with the eloquence you display in your consistent, determined attempts to validate most-likely incorrect mainstream theories.
White Prime
CC BY-SA 4.0
2020-06-06T13:21:33.740Z
And it is an honour coming from someone who's top network post is "Is the Hulk a coward?" @WhitePrime
ProfRob
CC BY-SA 4.0
2020-06-06T16:29:43.393Z
So are you saying that if it took you an hour to fall into the black hole, you would only see approximately an hour of Earth's future?  I thought time dilation would kick in and change things significantly, allowing you to see a significant amount of time into the future (e.g. perhaps a number of years).
Jonathan
CC BY-SA 4.0
2020-06-06T18:28:51.903Z
@Jonathan for a plausible set of parameters, yes it is almost exactly 1 hour. The falling object spends a negligible part of that hour anywhere near enough the black hole for GR effects to be significant. Now if it was *in orbit* that is different.
ProfRob
CC BY-SA 4.0
2020-06-06T11:58:52.593Z
2020-06-06T18:26:27.257Z
2020-06-06T18:26:27.257Z
ProfRob
<p>If someone were to jump into a black hole, looking back at Earth, how far into the future would they see of Earth?  As I understand it, time would dilate approaching infinity as you approach the event horizon.  As I understand, time dilation would reach infinity for an infinitely small amount of time, but Planck time would probably prevent that infinity from arising.  How much time would pass in the outside universe as we fall in?  How far would we see into Earth's future, if we were looking back at it as we fell in?  This is further pondering my prior question and the answers received for:</p>

<p><a href="https://astronomy.stackexchange.com/questions/2524/would-time-go-by-infinitely-fast-when-crossing-the-event-horizon-of-a-black-hole">Would time go by infinitely fast when crossing the event horizon of a black hole?</a></p>

CC BY-SA 4.0
2020-06-06T18:48:46.233Z
Don't try this.
Strawberry
<black-hole><general-relativity><time><special-relativity><time-dilation>
How much time passes in the outside universe when falling into a black hole?
Jonathan
# How much time passes in the outside universe when falling into a black hole?

(Asked by: username_0 on 2020-06-06)

If someone were to jump into a black hole, looking back at Earth, how far into the future would they see of Earth? As I understand it, time would dilate approaching infinity as you approach the event horizon. As I understand, time dilation would reach infinity for an infinitely small amount of time, but Planck time would probably prevent that infinity from arising. How much time would pass in the outside universe as we fall in? How far would we see into Earth's future, if we were looking back at it as we fell in? This is further pondering my prior question and the answers received for:

[Would time go by infinitely fast when crossing the event horizon of a black hole?](https://astronomy.stackexchange.com/questions/2524/would-time-go-by-infinitely-fast-when-crossing-the-event-horizon-of-a-black-hole)

**Tags:** black-hole, general-relativity, time, special-relativity, time-dilation
> **Comments:**
> **username_1**: Don't try this.  

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## 1 answers

### Answer by username_2 on 2020-06-06. Score: 8 | Accepted answer

You are essentially asking the following: if someone falls from the Earth from some way beyond the event horizon of a black hole, how long after they have left can an observer on Earth still signal to them with a light beam?

The answer of course depends on exactly how far the Earth is from the black hole. It is also often forgotten that it is not just light reaching them before the falling observer reaches the event horizon, but there is also a little bit more to add in the (proper) time it takes the falling observer to travel from the event horizon to the singularity.

The answer in all cases is that there is a finite time available, so no, the infalling person does not look back and see the cosmic aeons flying by.

The handwaving reason is that from the perspective of Schwarzschild coordinate time both the falling person and light beams approach the event horizon along an exponential asymptote. The light beams are always travelling faster, so their $dt/dr$ is always shallower and hence it is possible for them to catch the falling observer as long as the delay in sending the light beam does not exceed a finite critical value. Or to put it another way; you can contruct a past light cone from any point on the event horizon (or the space-like singularity) and it encompasses only a finite time anywhere outside the black hole.

The full, very gritty, details were worked out (for a Schwarzschild black hole) at <https://physics.stackexchange.com/a/396157/43351>

For the case of the person free-falling from a position $r\_0$, then the span of time seen by the falling person when looking back towards $r\_0$, before they hit the singularity, is given by
$$ c(\Delta t)\_{\rm singularity} = r\_s \ln \left(\frac{r\_s}{r\_0-r\_s}\right) + \pi r\_s\left(\frac{r\_0}{r\_s} -1\right)^{1/2}\left(1 + \frac{r\_0}{2r\_s}\right) -r\_0,$$
where $r\_s$ is the Schwarzschild radius.

Note that as $r\_0$ gets bigger then of course $\Delta t$ get bigger and approaches $\Delta t \simeq r\_s \pi(r\_0/r\_s)^{3/2}/2c$, which is just the freefall *proper* time (i.e. measured on a clock carried by the falling object) for an object to fall into a black hole. i.e. You could see that length of time into Earth's future, but only because you had spent that long falling into the black hole !

Edit: To address Jonathan's question in comments. Take a 10 solar mass BH with $r\_s=30$ km. If it takes 1 hour (proper time) to "free fall" radially (formulae above are for radial infall), then $r\_0 = 2.42\times 10^{9}$ m. The formula above then shows that $\Delta t$ is 1 hour to three sig figs. The reason for this is that on such a trajectory, the falling object spends a negligible fraction of its infall time anywhere near the black hole, so the GR effects are negligible.

> **Comments:**
> **username_3**: I say this with respect for you, Rob. And I predict my comment will be deleted. But I am regularly impressed with the eloquence you display in your consistent, determined attempts to validate most-likely incorrect mainstream theories.  
> **username_2**: And it is an honour coming from someone who's top network post is "Is the Hulk a coward?" @WhitePrime  
> **username_0**: So are you saying that if it took you an hour to fall into the black hole, you would only see approximately an hour of Earth's future?  I thought time dilation would kick in and change things significantly, allowing you to see a significant amount of time into the future (e.g. perhaps a number of years).  
> **username_2**: @Jonathan for a plausible set of parameters, yes it is almost exactly 1 hour. The falling object spends a negligible part of that hour anywhere near enough the black hole for GR effects to be significant. Now if it was *in orbit* that is different.  

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