Thanks for the link LeeE. I notice that the Schwarzchild radius is part of the equation. It is not immediately obvious to me that t = 0 at that radius. I don't doubt that it might, I just notice that many folks think that t = 0 closer in toward the singularity. I'll have to do some arithmetic. []

This is the equation:

cutting straight to the shortened form, to save time, when the distance *r* from the center of the object is equal to the Schwarzchild radius *r*_{0}, we get 1 - 1 = 0.

LeeE, You are right in that Schwarzchild metric defines time as being zero as seen from a 'stationary observer' being at the event horizon. But to me that seems more of a theoretical limit? As I understand it, this is a situation that won't exist as there is no way to place an observer stationary at the EV, the definition of a EV is a 'point of no return'. When you're there you won't be able to 'hover' without expending a lot of energy in the opposite direction, and from the point of the in-falling observer there is nothing stopping him from falling further in towards that black hole. If I was observing him I expect that there would be a 'last moment' where light was reflected from that person, falling in.

That light I would expect to be both distorted and redshifted, but, it would still hit my retina at 'c' so that image wouldn't, as some imply, 'freeze' at that EV (event horizon). There is no 'hovering' allowed there, neither for the person falling in, nor for any light obeying spacetimes geodesics. So that last 'reflection' will arrive to the 'stationary observer' outside and then there would be no more reflected light emitted from that in-falling person.

From the situation of the person falling in it seems trickier though, one might argue that, as seen from the point of the in-falling observer, time would would 'slow down' allowing the universe to 'die' before he ever reached that EV. But that won't hold from our stationary observers point of view I think. He will observe this person disappearing from our view and so draw the conclusion that he have passed what he saw as the EV.

Why I think so goes back to my thought experiment with a 'stationary' relative Earth super-telescope watching a our spacecraft leave earth toward a star and then come back. That spacecraft, even if being extremly close to light speed in space, will have the same 'time dilation' as we observe at a black hole, you could say that the ship represents the in-falling observer and our telescope represents the observer outside the gravity well (black hole). To say that the observer of the in-falling person would observe, from his frame of reference, that this person would freeze in space due to the time dilation caused by the black hole seems to me to be equivalent to expecting that our spacecraft, as observed from our super telescope, would 'freeze' in space.

If that was a fact, I believe that this ship never would be able to make that journey, as seen from the telescopes (stationary observers) point of view. And all discussions about length contraction seems then to be a exercise in futility as any 'stationary observer' only would experience that spacecraft passing it at near 'c' as being frozen in time. So I expect the in-falling 'observer' to pass, what the 'staionary observer' outside the gravity well will observe to be, the Event Horizon.

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When it comes to the Kerr metric for a spinning black hole there will be framedragging added to the situation though. But we're not talking about that as I understands it? Anyway

I might be wrong in my 'comparison' between those two situations, but even without that 'comparison' I see no real problem for the guy at the EV to keep falling as that is the only way he can take, as seen from his frame of reference.