![]() All bets are off as to what happens there. The are coordinates which cover the whole of the spacetime except for the center of the hole, where the is a true singularity. 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. 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. 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. The Schwarzschild coordinates are simply chosen badly. But this is just an artifact of the coordinates. And as it would seem to imply that the particle comes to stop, some people say that "time stops at 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. 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. Slightly longer answer: The case of a non-rotating, non-charged black hole is described by the Schwarzschild solution. 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. A photon emitted 'outward' at the horizon simply remains on 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. This is simply understood as the fact that light emitted from (or inside) the horizon cannot propagate to any event outside the horizon. 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. Indeed, the entity simply continues through the horizon towards the 'center' of the black hole and then ceases to exist at the singularity. Moreover, for any entity falling freely towards the horizon, the proper time does not 'stop'. However, there are coordinate systems with finite coordinate time at the horizon, e.g., Kruskal-Szekeres coordinates. 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). ![]() ![]() That is to say, the coordinate time corresponds to the proper time of a hypothetical entity arbitrarily far from the black hole. 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. Think of your proper time as the time according to your 'wrist watch'. Now, every entity also has an associated proper time which is not a coordinate which means that it is coordinate independent (invariant). Indeed, time is a coordinate in relativity so one must be careful to specify the coordinate system when asking questions like this. ![]() The fact is that, in special and general relativity, there is no universal time. ![]()
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