<p>As far as we know, once beyond the event horizon, someone might experience all of time in the blink of an eye until the hawking radiation makes the black hole disappear and the person comes out again. More likely though you'd just get spaghettified..</p>

<p>In all seriousness though, there is no real limitation, although they aren't really <em>transported</em> to the future, the outside world just moves faster.</p>

CC BY-SA 3.0
2016-04-21T23:36:37.400Z
Does this mean time is essentially slowed down dramatically giving the illusion that time has moved faster in the outside world?
NuWin
CC BY-SA 3.0
2016-04-22T01:32:41.643Z
Basically, yes, time does slow down
Tanenthor
CC BY-SA 3.0
2016-04-22T01:35:23.100Z
And to whomever down voted my answer, could you please explain *why*? So I can correct the issue
Tanenthor
CC BY-SA 3.0
2016-04-22T06:32:16.093Z
I downvoted it. The first sentence is not a true reflection of black holes in GR. A person falling into a black hole does not experience all of time in the blink of an eye unless they can hover just above the event horizon. Beyond the event horizon, nothing can remain stationary and falls to the singularity on a timescale of $\pi GM/c^3$ and is intercepted by light from only a little way into the future.
ProfRob
CC BY-SA 3.0
2016-04-22T12:31:06.823Z
This isn't an answer to the question; the question specifically mentions going close to the event horizon without going past it, while the answer is about going beyond the event horizon.
DylanSp
CC BY-SA 3.0
2016-04-23T00:45:12.747Z
@Rob Jeffries - In an eternal rotating black hole, you would see the entire history of the universe compressed to a finite time as you crossed the *inner* horizon, though as you say this doesn't happen at the outer event horizon (see the last paragraph on p. 153 of [this book](http://books.google.com/books?id=GJT4g1hNUGIC&lpg=PP1&pg=PA153)). In a more realistic rotating black hole that evaporates, presumably you only see the entire finite future history of the black hole, though I believe there is still an infinite blueshift of the signal so the speedup factor should still go to infinity.
Hypnosifl
CC BY-SA 3.0
2016-04-23T08:34:49.970Z
@Hypnosifl It is true that my comment holds only for Schwarzschild BHs. A brief glance at some other resources suggest that the idea outlined in your comment is probably unrealisable in practice because light arriving from well into the future changes the nature of the black hole and its singularity. I.e. I think what the book you reference says is true, so long as nothing is falling into the black hole. It appears this is still a topic of contemporary debate and research.
ProfRob
CC BY-SA 3.0
2016-04-23T17:56:10.980Z
@Rob Jeffries - You're right that the Kerr black hole solution is altered if you take into account the infalling electromagnetic and gravitational waves, especially near the neighborhood of the inner horizon where they should be infinitely blueshifted, along with taking into account the idea that the black hole formed from a collapsing star rather than existing eternally. But although the spacelike singularity at the center is probably eliminated, calculations still end up concluding there's an internal horizon where blueshift approaches infinity--in fact, there turn out to be two of them.
Hypnosifl
CC BY-SA 3.0
2016-04-23T17:58:28.100Z
(continued) Physicist Kip Thorne discusses this a bit in his book *The Science of Interstellar*--I summed up his discussion of the two inner horizons of a realistic rotating black hole in [this answer to an *Interstellar* question on the sci fi stack exchange](http://scifi.stackexchange.com/a/86187/22250), and the paper which Thorne references for this info can be found [here](http://arxiv.org/abs/1109.5139).
Hypnosifl
CC BY-SA 3.0
2016-04-21T16:53:47.370Z
2016-04-21T16:53:47.370Z
Tanenthor
<p>The way that you have specified the question, the answer is as far as you like. You simply put your spaceship into any orbit around the black hole and wait.</p>
<p>A more sensible question is what is the largest time dilation factor that can be accomplished - i.e. that maximises your travel time into the future for a given amount of proper time experienced on the spaceship.</p>
<p>This in turn is governed by how close to the black hole you can come and still tolerate the tidal forces. If you don't put a limit on this (your first case), then the answer is again infinite; you can hover as close to the event horizon as you like, using an enormous amount of rocket fuel, and the time dilation (see below) can be arbitrarily large.</p>
<p>Your second case is more realistic. Roughly we can say that the tidal acceleration across a body of length <span class="math-container">$l$</span> is given by <span class="math-container">$2GMl/r^3$</span>, where <span class="math-container">$M$</span> is the black hole mass and <span class="math-container">$r$</span> is the distance from the black hole. If we make this acceleration equal to say <span class="math-container">$1 g$</span>, and your body length <span class="math-container">$l \sim 1$</span>m, then for a <span class="math-container">$5M_{\odot}$</span> black hole <span class="math-container">$r \simeq 5000$</span> km (well outside the Schwarzschild radius of 15 km).</p>
<p>If you could &quot;hover&quot; at this radius, then the time dilation factor would be
<span class="math-container">$$\frac{\tau}{\tau_0} = \left( 1 - \frac{2GM}{rc^2}\right)^{1/2},$$</span>
where <span class="math-container">$\tau$</span> is the time interval on a clock on the spaceship and <span class="math-container">$\tau_0$</span> is the time interval well away from the black hole.</p>
<p>For <span class="math-container">$M=5M_{\odot}$</span> and <span class="math-container">$r = 5000$</span> km, this factor is 0.9985.</p>
<p>If the spaceship is in a <a href="https://en.wikipedia.org/wiki/Gravitational_time_dilation#Circular_orbits" rel="nofollow noreferrer">circular orbit</a> at this radius, the factor is <span class="math-container">$(1 - 3GM/rc^2)^{1/2} = 0.9978$</span>.</p>
<p>If you ignore tidal forces ripping you and your ship apart then the smallest stable orbit you can accomplish is at <span class="math-container">$r=6GM/c^2$</span> - the so-called innermost stable circular orbit. Using the formula for the circular orbit above, then the time dilation factor becomes 0.816.</p>
<p>These factors are perhaps not as big as you might have imagined! If you want to improve on that then you must consider rapidly rotating <a href="https://en.wikipedia.org/wiki/Rotating_black_hole" rel="nofollow noreferrer">Kerr black holes</a>. The innermost prograde stable circular orbit (i.e. in the same direction as the black hole spin) can be much closer - approaching <span class="math-container">$r= GM/c^2$</span> and the time dilation factor calculated above can become arbitrarily small.</p>
<p>Of course the tidal forces are still there, so the way to get around this is to be in orbit around a much more massive black hole <span class="math-container">$&gt;10^6$</span> solar masses, where it turns out the tidal forces at these radii might be tolerable to a human.</p>

CC BY-SA 3.0
2016-04-24T04:32:53.820Z
Even for the "realistic" case, I think it makes sense to consider a much larger black hole--after all, there seem to be plenty of [supermassive black holes](https://en.wikipedia.org/wiki/Supermassive_black_hole) at the centers of many galaxies. Sagittarius A at the center of our own galaxy is about 4.3 million times the mass of the Sun, while the [largest ones known](http://en.wikipedia.org/wiki/List_of_most_massive_black_holes) are thought to have masses in excess of 10 *billion* times the mass of the Sun. And the tidal forces outside the horizon are negligible for very large black holes.
Hypnosifl
CC BY-SA 3.0
2016-04-24T04:38:43.647Z
Also, [this answer](http://physics.stackexchange.com/a/146683/59406) calculates an approximate answer for the time dilation at the innermost stable circular orbit (ISCO) for a rotating black hole...I noted in a comment that there are theoretical reasons to think the maximum value for the rotation rate $a = 1 - \epsilon$ (expressed as a fraction of the 'extremal' rotation rate where the black hole loses its event horizon and becomes a 'naked singularity') is about $a = 0.998$, or $\epsilon = 0.002$, giving a time dilation factor of 1 second experienced for every 10 seconds for distant clocks.
Hypnosifl
CC BY-SA 4.0
2022-05-29T10:56:25.057Z
@Hypnosifl a 5 solar mass black hole is specified in the question. The tidal forces are essentially the same for even a maximally rotating black hole.
ProfRob
CC BY-SA 4.0
2016-04-23T20:14:20.870Z
2022-05-29T15:48:25.397Z
2022-05-29T15:48:25.397Z
ProfRob
<p>If we sent someone on a path that passed as near as possible to a black hole without getting pulled in, how far into the future would they go due to time dilation?  Let's assume the black hole is 5 solar masses (I am assuming the mass will affect the calculation).  I would like to know from two considerations:</p>

<ul>
<li><p>An observer that can survive anything (e.g. ideal case)</p></li>
<li><p>The forces that a human can tolerate (e.g. a human would not be able to get as close to a black hole as would be ideal) </p></li>
</ul>

CC BY-SA 3.0
2016-03-12T19:18:05.193Z
You can come arbitrarily far into the the future. How far depends on the mass $M$ of the BH, the distance $r$ to the BH, and how long time $t$ you spend there. If you only give us $M$, you don't even need to go near a black hole; you can come $x$ years into the future by waiting $x$ years.
pela
CC BY-SA 3.0
2016-03-12T20:50:37.700Z
Good point, although I am talking about the amount of time advanced due to time dilation :-)
Jonathan
CC BY-SA 3.0
2016-03-12T21:39:46.180Z
For a 5 solar mass black hole there would be enormous tidal issues that would tear a space ship and/or person in the ship apart long before the time dilation got anywhere close to interesting.      For a larger black hole, the question is asked and answered here:   http://physics.stackexchange.com/questions/146105/what-is-the-maximum-time-dilation-factor-when-orbiting-a-rotating-black-hole
userLTK
CC BY-SA 3.0
2016-03-14T08:09:41.623Z
Jonathan, you can find the appropriate equation [here](http://astronomy.stackexchange.com/questions/10467/time-according-to-the-gravity-of-sagittarius-a/10470#10470). If you plug in your favorite numbers and find a time dilation factor of, say, $t=0.5$, this means that for every hour you spend at the BH, two hours will pass for an external observer, i.e. when you get back to civilization, you will experience it as having traveled one hour into the future.
pela
CC BY-SA 3.0
2016-03-17T22:15:30.357Z
They would go as far into the future as they lived. They would not experience time dilation - time would run at a constant rate in their frame of reference.
adrianmcmenamin
CC BY-SA 3.0
2016-04-23T00:50:53.390Z
@Jonathan - Do you want a semi-practical answer that takes into account that your whole body can't get arbitrarily close to the event horizon since it has a finite thickness, along with the fact that the G-force you'd feel hovering above the horizon (your [proper acceleration](http://en.wikipedia.org/wiki/Proper_acceleration)) would get too large to survive if the distance got small enough? Or just an ideal answer for a pointlike observer who can withstand any G-force?
Hypnosifl
CC BY-SA 3.0
2016-04-23T14:43:33.900Z
@Hypnosifl - Actually an answer that considers both aspects would be good.
Jonathan
<black-hole><space-time><general-relativity><special-relativity><time-dilation>
How far into the future can we go by traveling close to a black hole?
Jonathan
# How far into the future can we go by traveling close to a black hole?

(Asked by: username_0 on 2016-03-12)

If we sent someone on a path that passed as near as possible to a black hole without getting pulled in, how far into the future would they go due to time dilation? Let's assume the black hole is 5 solar masses (I am assuming the mass will affect the calculation). I would like to know from two considerations:

* An observer that can survive anything (e.g. ideal case)
* The forces that a human can tolerate (e.g. a human would not be able to get as close to a black hole as would be ideal)

**Tags:** black-hole, space-time, general-relativity, special-relativity, time-dilation
> **Comments:**
> **username_1**: You can come arbitrarily far into the the future. How far depends on the mass $M$ of the BH, the distance $r$ to the BH, and how long time $t$ you spend there. If you only give us $M$, you don't even need to go near a black hole; you can come $x$ years into the future by waiting $x$ years.  
> **username_0**: Good point, although I am talking about the amount of time advanced due to time dilation :-)  
> **username_2**: For a 5 solar mass black hole there would be enormous tidal issues that would tear a space ship and/or person in the ship apart long before the time dilation got anywhere close to interesting.      For a larger black hole, the question is asked and answered here:   http://physics.stackexchange.com/questions/146105/what-is-the-maximum-time-dilation-factor-when-orbiting-a-rotating-black-hole  
> **username_1**: Jonathan, you can find the appropriate equation [here](http://astronomy.stackexchange.com/questions/10467/time-according-to-the-gravity-of-sagittarius-a/10470#10470). If you plug in your favorite numbers and find a time dilation factor of, say, $t=0.5$, this means that for every hour you spend at the BH, two hours will pass for an external observer, i.e. when you get back to civilization, you will experience it as having traveled one hour into the future.  
> **username_3**: They would go as far into the future as they lived. They would not experience time dilation - time would run at a constant rate in their frame of reference.  
> **username_4**: @Jonathan - Do you want a semi-practical answer that takes into account that your whole body can't get arbitrarily close to the event horizon since it has a finite thickness, along with the fact that the G-force you'd feel hovering above the horizon (your [proper acceleration](http://en.wikipedia.org/wiki/Proper_acceleration)) would get too large to survive if the distance got small enough? Or just an ideal answer for a pointlike observer who can withstand any G-force?  
> **username_0**: @Hypnosifl - Actually an answer that considers both aspects would be good.  

---

## 2 answers

### Answer by username_7 on 2016-04-23. Score: 5 | Accepted answer

The way that you have specified the question, the answer is as far as you like. You simply put your spaceship into any orbit around the black hole and wait.

A more sensible question is what is the largest time dilation factor that can be accomplished - i.e. that maximises your travel time into the future for a given amount of proper time experienced on the spaceship.

This in turn is governed by how close to the black hole you can come and still tolerate the tidal forces. If you don't put a limit on this (your first case), then the answer is again infinite; you can hover as close to the event horizon as you like, using an enormous amount of rocket fuel, and the time dilation (see below) can be arbitrarily large.

Your second case is more realistic. Roughly we can say that the tidal acceleration across a body of length $l$ is given by $2GMl/r^3$, where $M$ is the black hole mass and $r$ is the distance from the black hole. If we make this acceleration equal to say $1 g$, and your body length $l \sim 1$m, then for a $5M\_{\odot}$ black hole $r \simeq 5000$ km (well outside the Schwarzschild radius of 15 km).

If you could "hover" at this radius, then the time dilation factor would be
$$\frac{\tau}{\tau\_0} = \left( 1 - \frac{2GM}{rc^2}\right)^{1/2},$$
where $\tau$ is the time interval on a clock on the spaceship and $\tau\_0$ is the time interval well away from the black hole.

For $M=5M\_{\odot}$ and $r = 5000$ km, this factor is 0.9985.

If the spaceship is in a [circular orbit](https://en.wikipedia.org/wiki/Gravitational_time_dilation#Circular_orbits) at this radius, the factor is $(1 - 3GM/rc^2)^{1/2} = 0.9978$.

If you ignore tidal forces ripping you and your ship apart then the smallest stable orbit you can accomplish is at $r=6GM/c^2$ - the so-called innermost stable circular orbit. Using the formula for the circular orbit above, then the time dilation factor becomes 0.816.

These factors are perhaps not as big as you might have imagined! If you want to improve on that then you must consider rapidly rotating [Kerr black holes](https://en.wikipedia.org/wiki/Rotating_black_hole). The innermost prograde stable circular orbit (i.e. in the same direction as the black hole spin) can be much closer - approaching $r= GM/c^2$ and the time dilation factor calculated above can become arbitrarily small.

Of course the tidal forces are still there, so the way to get around this is to be in orbit around a much more massive black hole $>10^6$ solar masses, where it turns out the tidal forces at these radii might be tolerable to a human.

> **Comments:**
> **username_4**: Even for the "realistic" case, I think it makes sense to consider a much larger black hole--after all, there seem to be plenty of [supermassive black holes](https://en.wikipedia.org/wiki/Supermassive_black_hole) at the centers of many galaxies. Sagittarius A at the center of our own galaxy is about 4.3 million times the mass of the Sun, while the [largest ones known](http://en.wikipedia.org/wiki/List_of_most_massive_black_holes) are thought to have masses in excess of 10 *billion* times the mass of the Sun. And the tidal forces outside the horizon are negligible for very large black holes.  
> **username_4**: Also, [this answer](http://physics.stackexchange.com/a/146683/59406) calculates an approximate answer for the time dilation at the innermost stable circular orbit (ISCO) for a rotating black hole...I noted in a comment that there are theoretical reasons to think the maximum value for the rotation rate $a = 1 - \epsilon$ (expressed as a fraction of the 'extremal' rotation rate where the black hole loses its event horizon and becomes a 'naked singularity') is about $a = 0.998$, or $\epsilon = 0.002$, giving a time dilation factor of 1 second experienced for every 10 seconds for distant clocks.  
> **username_7**: @Hypnosifl a 5 solar mass black hole is specified in the question. The tidal forces are essentially the same for even a maximally rotating black hole.  

---

### Answer by username_5 on 2016-04-21. Score: 0

As far as we know, once beyond the event horizon, someone might experience all of time in the blink of an eye until the hawking radiation makes the black hole disappear and the person comes out again. More likely though you'd just get spaghettified..

In all seriousness though, there is no real limitation, although they aren't really *transported* to the future, the outside world just moves faster.

> **Comments:**
> **username_6**: Does this mean time is essentially slowed down dramatically giving the illusion that time has moved faster in the outside world?  
> **username_5**: Basically, yes, time does slow down  
> **username_5**: And to whomever down voted my answer, could you please explain *why*? So I can correct the issue  
> **username_7**: I downvoted it. The first sentence is not a true reflection of black holes in GR. A person falling into a black hole does not experience all of time in the blink of an eye unless they can hover just above the event horizon. Beyond the event horizon, nothing can remain stationary and falls to the singularity on a timescale of $\pi GM/c^3$ and is intercepted by light from only a little way into the future.  
> **username_8**: This isn't an answer to the question; the question specifically mentions going close to the event horizon without going past it, while the answer is about going beyond the event horizon.  
> **username_4**: @Rob Jeffries - In an eternal rotating black hole, you would see the entire history of the universe compressed to a finite time as you crossed the *inner* horizon, though as you say this doesn't happen at the outer event horizon (see the last paragraph on p. 153 of [this book](http://books.google.com/books?id=GJT4g1hNUGIC&lpg=PP1&pg=PA153)). In a more realistic rotating black hole that evaporates, presumably you only see the entire finite future history of the black hole, though I believe there is still an infinite blueshift of the signal so the speedup factor should still go to infinity.  
> **username_7**: @Hypnosifl It is true that my comment holds only for Schwarzschild BHs. A brief glance at some other resources suggest that the idea outlined in your comment is probably unrealisable in practice because light arriving from well into the future changes the nature of the black hole and its singularity. I.e. I think what the book you reference says is true, so long as nothing is falling into the black hole. It appears this is still a topic of contemporary debate and research.  
> **username_4**: @Rob Jeffries - You're right that the Kerr black hole solution is altered if you take into account the infalling electromagnetic and gravitational waves, especially near the neighborhood of the inner horizon where they should be infinitely blueshifted, along with taking into account the idea that the black hole formed from a collapsing star rather than existing eternally. But although the spacelike singularity at the center is probably eliminated, calculations still end up concluding there's an internal horizon where blueshift approaches infinity--in fact, there turn out to be two of them.  
> **username_4**: (continued) Physicist Kip Thorne discusses this a bit in his book *The Science of Interstellar*--I summed up his discussion of the two inner horizons of a realistic rotating black hole in [this answer to an *Interstellar* question on the sci fi stack exchange](http://scifi.stackexchange.com/a/86187/22250), and the paper which Thorne references for this info can be found [here](http://arxiv.org/abs/1109.5139).  

---

