Everyone says that time stops in the black hole. It's a "fact". However, I have never heard everyone explaining that.

Of course, I know that observer in weaker gravitational field sees that something in stronger gravitational field is experiencing slower time. However, slower and no at all is quite different.

I have no idea what equation is used to calculate dime dilatation, but it will use gamma and therefore division. And the only time division of non-zero constant yields zero is when you divide by infinity.

And although black holes are super heavy, super badass and super black, they posses finite energy and therefore finite gravitational acceleration (even at event horizon).

So shouldn't just the time be very slow, rather than just stop from our point of view?


3 Answers 3


Why does time stop in black holes?

Time according to whom?

The fact is that, in special and general relativity, there is no universal time. Indeed, time is a coordinate in relativity so one must be careful to specify the coordinate system when asking questions like this.

Now, every entity also has an associated proper time which is not a coordinate which means that it is coordinate independent (invariant). Think of your proper time as the time according to your 'wrist watch'.

In the context of the static black hole (Schwarzschild black hole) solution, there is a coordinate system (Schwarzschild coordinates) that we can associate with the observer at infinity. That is to say, the coordinate time corresponds to the proper time of a hypothetical entity arbitrarily far from the black hole.

In this coordinate system, we can roughly say that the coordinate time 'stops' at the event horizon (in fact, there is no finite value of this coordinate time to assign to events on the horizon).

However, there are coordinate systems with finite coordinate time at the horizon, e.g., Kruskal-Szekeres coordinates.

Moreover, for any entity falling freely towards the horizon, the proper time does not 'stop'. Indeed, the entity simply continues through the horizon towards the 'center' of the black hole and then ceases to exist at the singularity.

We interpret the fact that the Schwarzschild coordinate time does not extend to the horizon as follows: no observer outside the horizon can see an entity reach (or fall through) the horizon in finite time. This is simply understood as the fact that light emitted from (or inside) the horizon cannot propagate to any event outside the horizon.

Why? Because the spacetime curvature at the horizon is so great that there is no light-like world line the extends beyond the horizon. Indeed, the horizon is light-like. A photon emitted 'outward' at the horizon simply remains on the horizon.

Within the horizon, the spacetime curvature is such that there are no world lines that do not terminate on the singularity - the curvature is so great within the horizon that the future is in the direction of the singularity.

  • $\begingroup$ I like to imagine effects as being somewhat similar to a Doppler shift. If a radio transmitter is transmitting at 1.00MHz as it approaches the event horizon, outside observers will see the wavelength of the signal get shorter and shorter, implying that every second it will see more and more cycles. From the transmitter's perspective, every 1,000,000 cycles will still be one second. Is that a fair view? $\endgroup$
    – supercat
    Jul 10, 2014 at 15:53
  • $\begingroup$ I have pointed out that I understand relativity. $\endgroup$ Jul 10, 2014 at 18:49

Short answer: It doesn't stop.

Slightly longer answer: The case of a non-rotating, non-charged black hole is described by the Schwarzschild solution. It is now the case that, if you draw the worldline of a particle falling into a black hole, you will find that the coordinate time in the Schwarzschild metric grows infinite as the particle approaches the event horizon. Naively, this would seem to imply that a particle takes forever to fall into a black hole, which would mean that it becomes slower and slower as it approaches the event horizon. And as it would seem to imply that the particle comes to stop, some people say that "time stops at the event horizon". But this is just an artifact of the coordinates. The Schwarzschild coordinates are simply chosen badly. The proper time, i.e. the time the falling particle/observer would perceive, is finite, and there are other coordinates in which there is also no singularity at the event horizon, so that all coordinates stay finite. Nothing particulary terrible happens at the event horizon from the view of the falling particle, it is just that no light-like paths connect the interior of the horizon the the exterior, so that nothing can cross the horizon from the inside.

Inside the horizon, some weird stuff happens when looked at from the Schwarzschild coordinates, like the former time-coordinate becoming space-like, but this is again rather an artifact of the coordinate system than a property of the true black hole. The are coordinates which cover the whole of the spacetime except for the center of the hole, where the is a true singularity. All bets are off as to what happens there.

  • $\begingroup$ I was trying to make sure answerers know that I'm aware or the observer relative stuff. I was wondering why from my point of view the stuff is stuck at the event horizon (which must be really crowded from our point of view...) $\endgroup$ Jul 9, 2014 at 22:40
  • $\begingroup$ @TomasZato: Well...actually, from your point of view, all the stuff stuck at the horizon is also black due to gravitational redshift, so you don't see it. As for why the coordinate time in the Schwarzschild metric (which is the time of an observer at rest at infinity) grows infinite, just observe that the factor between $t$ and $\tau$ in the metric goes to zero as the radius approaches the horizon. $\endgroup$
    – ACuriousMind
    Jul 9, 2014 at 22:43
  • $\begingroup$ So it's currently just a mathematical thing which can eventually have some constant added to it as soon as we find out that things are a little bit different? $\endgroup$ Jul 9, 2014 at 22:46
  • $\begingroup$ @TomasZato: Not really. If it is wrong that we cannot see things falling into a black hole in a finite amount of our time, then GR is wrong. [But note that black holes are unfortunately in short supply for experiments.] $\endgroup$
    – ACuriousMind
    Jul 9, 2014 at 22:48
  • $\begingroup$ I thought you didn't know any GR physics. $\endgroup$
    – Ryan Unger
    Feb 19, 2015 at 0:49

If someone is falling in a black hole, the nearer he/she gets to black hole the slower time will pass and when he/she reaches the edge of event horison, time it would take for an observer to see him/her to cross event horison will be infinite (in other words if their friend was watching him/her he would never see him/her crossing the event horison). Gravitational time dilation can be derived from Einstein field equations: $$ G_{\mu\nu}+\Lambda g_{\mu\nu}=\frac{8\pi G}{c^4}T_{\mu\nu} $$ Where $G_{\mu\nu}$ is Einstein Tensor, $\Lambda$ is cosmological constant, $g_{\mu\nu}$ is metric tensor and $T_{\mu\nu}$ is Stress energy momentum tensor, (for more info visit: Einstein Field Equations).

or you can use most famous solution of Einstein field equations called Schwarzschild solution: $$ c^2d\tau^2=\left(1-\frac{r_s}{r}\right)c^2dt^2-\left(1-\frac{r_s}{r}\right)^{-1}dr^2-r^2(d\theta^2+\sin^2\theta\text{ }d\phi^2) $$ and difference in proper time and time will be: $$ \frac{\tau}{t} = \sqrt{\left(1-\frac{r_s}{r}\right)} \Rightarrow \tau = t\sqrt{1-\frac{2GM}{rc^2}} $$ where $r_s$ is Schwarzschild radius and $r$ is distance of observer from a (in this case) black hole: $$ r_s = \frac{2GM}{c^2} $$

For more info see: This page.


So as observer gets closer and closer to a black hole time passes more slowly relative to other observer and that other observer will never see him to cross event horison because it would take infinite amount of time and gravitational time dilation is calculated using this equation: $\tau = t\sqrt{1-\frac{2GM}{rc^2}}$ where $\tau$ is proper time

So shouldn't just the time be very slow, rather than just stop from our point of view?

If you will use time dilation equation for a black hole you will see that as you go to black hole ($r\to r_s$) time for an observer that is watching someone falling in black hole would be infinite (in other words observer will never see that someone crossing event horison) $t = \frac{\tau}{\sqrt{1-1}}=\frac{\tau}{0} = \infty$

  • $\begingroup$ Could you read the answer in the conclusion and just write up the question it answers? $\endgroup$ Jul 9, 2014 at 22:37
  • $\begingroup$ Also I've actually looked up the equations however I'm not as good to know what to put into them to get 0. $\endgroup$ Jul 9, 2014 at 22:41
  • $\begingroup$ "So as observer gets closer and closer to a black hole time passes more slowly"; in which sense? It's true that external observers will see someone freefalling into a Schwarzschild black hole asymptotically approaching the event horizon, but for the observer in freefall, the event horizon is just as any other point in spacetime and is reached in a finite amount of time. $\endgroup$
    – Alex A
    Jul 9, 2014 at 22:42
  • $\begingroup$ @TomášZato when someone gets to the edge of event horison or in other words distance between someone and black hole $r$ has same value as $r_s$ (Schwarzschild radius) ($r\to r_s$) when you will use this values in gravitational time dilation $t= \frac{\tau}{\sqrt{1-\frac{r_s}{r}}}$ you will get $t = \frac{\tau}{0} = \infty$ which means that observer that is watching someone falling into black hole will have to wait $\infty$ seconds (never) to see him/her crossing event horison $\endgroup$
    – user21420
    Jul 9, 2014 at 22:53

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