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clipka wrote:
> Thus, in the observer's reference frame (I'm deliberately avoiding the phrase
> "as seen from the observer", because the formula doesn't say anything about
> perception, but plain hard facts), as soon as the "victim" would hit the EH,
> its time would come to a standstill, so it couldn't move *any* distance in
> finite time.
I'm not sure that follows. Photons move as fast as possible, and in their
own reference frame, time has come to a standstill. Why does the victim's
clock influence the velocity of the victim in the observer's timeframe?
This is like saying I stick a giant engine on a spaceship, turn it on, and
just go, and after a while you see me slowing down, because my clock is
getting asymptotically close to stopped from your reference frame.
> (Note however that time slows down so dramatically already very
> close to the EH that the "victim" will not even *reach* EH in finite time.)
*That* is why I think the victim will not find himself inside the EH - the
universe (including himself) will have decayed to nothingness or some such
before he gets there.
> So, the equation does *not* state that the EH has a radius of r[s] = 2GM/c^2
> after all. Instead, it states that it has a *surface* of 4 pi (2GM/c^2)^2.
That difference is exactly the warped space. It happens with the sun and the
earth, too, because of their gravity.
IIRC, I read where the circumference of the (ideal) earth is something like
a cm shorter than what it should be given the radius, and the sun is a
kilometer or ten kilometers or some such different.
> Given that spacetime is notoriously distorted at the EH, this makes *no*
> statement whatsoever about its radial distance from the singularity. It could
> be - ta-ding! - zero after all...
More like infinity, methinks.
> (Good enough to quality for a world-class crackpot this time, huh? =B))
I don't think real crackpots go "Mwa ha ha ha!"
--
Darren New, San Diego CA, USA (PST)
"Ouch ouch ouch!"
"What's wrong? Noodles too hot?"
"No, I have Chopstick Tunnel Syndrome."
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