At the event horizon, time essentially stops. An outside observer would never see you cross the event horizon, as my understanding goes.

But in that case, how would this work out if I imagine the object in terms of each atom individually? Would the constituent atoms appear to be 'frozen' once they reach the event horizon, but the others that have not yet reached the event horizon and would still appear to be traveling towards the event horizon? As such, would any object falling into a black hole appear to become 'flattened' into a 1 atom thick silhouette on the event horizon to an outside observer?

If not, why not?

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    $\begingroup$ See Does someone falling into a black hole see the end of the universe? Spoiler: no they don't :-) $\endgroup$ Commented Jun 23, 2023 at 4:19
  • $\begingroup$ Also, are you asking if the object is flattened i.e. if you fell across an event horizon would you be flattened? Or are you asking is someone watching from afar would see the falling object appear to flatten out? The latter involves the travel time of the light rays from the object to your eye, so it's a complicated question. Light takes longer to reach your eye from the parts of the object nearer the horizon. $\endgroup$ Commented Jun 23, 2023 at 4:22
  • $\begingroup$ From famous Mathpages by Kevin Brown: “One common question is whether a man falling (feet first) through an even horizon of a black hole would see his feet pass through the event horizon below him. As should be apparent from the schematics above, this kind of question is based on a misunderstanding. Everything that falls into a black hole falls in at the same local time, although spatially separated, just as everything in our city is going to enter tomorrow at the same time.” - mathpages.com/rr/s7-03/7-03.htm $\endgroup$
    – safesphere
    Commented Jun 24, 2023 at 4:10
  • $\begingroup$ @JohnRennie While a falling observer cannot see it, the event of him crossing the horizon is not in the past light cone of any external event in the universe ever. And since the Schwarzschild spacetime is globally hyperbolic thus preserving the causal order of events, it is true that the universe dies in the infinite future before he crosses the horizon. $\endgroup$
    – safesphere
    Commented Jun 24, 2023 at 4:31

1 Answer 1


It is true that if you fall into a black hole an outside observer would never see you cross the horizon. It is also true that, according to a stationary observer just outside the event horizon, you would appear to be flattened. This is essentially Lorentz contraction which also happens to fast-moving (relative to some observer) objects in flat spacetime. In order to remain stationary just outside the horizon of a black hole, an observer would need to continually accelerate (the same way we continually accelerate to stay on the surface of the earth). This results in a relative velocity between the stationary observer and the in-falling object that changes as the object approaches the horizon. It is this relative velocity that makes the free-falling object appear to be contracted as it approaches the horizon.

It is not correct that an in-falling observer witnesses the entire history of the universe as they cross the event horizon. This misconception stems from treating the proper time of an observer far from the black hole as physically meaningful for an observer passing the horizon. Light can still freely pass from outside the event horizon to the inside (just not the other way around). So, when the observer crosses the horizon, they don't see anything crazy. As far as they are concerned the rest of spacetime continues on normally. It's just that after they cross the horizon, there is some finite time in their future where they are terminated at (or close to) the singularity.

Here is a picture that shows what I mean more precisely, Blue: Stationary observer outside the horizon. Red: observer that falls through the horizon.

If you are familiar with spacetime diagrams this is roughly a spacetime diagram for a black hole (called a Kruskal diagram). Time increases upwards and the $V$ axis is the event horizon. The red line is the worldline of someone that falls through the event horizon and the blue line is the worldline of someone staying stationary, outside the black hole. The last ray of light the red line can receive is the curvy line. You see that this is by no means the "last light every emitted in the history of the universe". There are plenty of events along the blue worldline that the red observer can never see.

  • $\begingroup$ This answer is incorrect. It describes the velocity length contraction and neglects the gravitational length contraction -1 $\endgroup$
    – safesphere
    Commented Jun 23, 2023 at 14:45
  • $\begingroup$ @safesphere The answer looks fine to me. What you called gravitational length contraction in another recent comment is not really a thing in GR. $\endgroup$
    – benrg
    Commented Jun 23, 2023 at 19:42
  • $\begingroup$ @benrg It is indeed, see my response there. If you don’t like the term, it is your opinion, but relativistic length contraction and time dilation exist in both SR (due to velocity) and GR (due to gravity). This is elementary and the fact that you are objecting the obvious is surprising. $\endgroup$
    – safesphere
    Commented Jun 23, 2023 at 21:00
  • $\begingroup$ @safesphere Sorry if this is a repeat! I added a comment here yesterday but it seems to have vanished. In any case, there is nothing inherently gravitational about this "squashing" effect of event horizons. This effect also occurs in flat spacetime in the frame of a uniformly accelerated observer (see Rindler metric and Rindler horizon). The effect of gravity in the Schwarzschild metric is that it changes what me mean by a stationary observer. An observer stationary with respect to Schwarzschild coordinates is actually uniformly accelerated (i.e. they have a positive proper acceleration). $\endgroup$ Commented Jun 24, 2023 at 0:10
  • $\begingroup$ I am glad you are admitting the existence of the “squashing” effect of the horizon better known as the gravitational length contraction. The Rindler metric is gravitational per the equivalence principle of gravity and acceleration being equivalent. The fact that the Rindler spacetime is flat is irrelevant, because locally the Schwarzschild spacetime is also flat. In both cases the length contraction (or time dilation) depends on the observer, because these are the relative effects of perspective. I hope you understand that the Rindler case you mentioned proves that your answer is incorrect. $\endgroup$
    – safesphere
    Commented Jun 24, 2023 at 3:47

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