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I Don't Know About The Reduced Circumference..

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Posted by Bruce on February 22, 2001 01:41:01 UTC

The far away observer measures the distance to the event horizon to be infinite
drshell=dr/(1-2M/r)1/2
so when r=2M the expression results in an infinity.
At the same time the far away observer can take the integral of that expression and show that the distance is actually finite
=[r1/2(r-2M)1/2+2M ln{r1/2+(r-2M)1/2}]r2,r1
The far away observer also measures the velocity of the freely falling object to -> 0 as the object approaches r=2M (event horizon) and this is even true for light
dr/dt=-((1-2M/r)(2M/r)1/2 [from rest at infinity]
dr/dt=-(1-2M/r) [light]
The local shell observer very close to r=2M measures the freely falling object to -> c as it crosses the event horizon
drshell/dtshell=-(2M/r)1/2
also for light
drshell/dtshell=-1
Now to get to what you are talking about
the far away observer measures the total energy in Schwarzchild geometry to decrease with decreasing r-coordinate for the freely falling object.
E/m=(1-2M/r)dt/dt and as r->2M the total energy -> 0. This is an amazing result. So where is this energy conserved as it grows smaller with decreasing r-coordinate? It is conserved in the gravitational field. So as you said all the gravitational energy for the black hole is in the field outside the event horizon with respect to the far away observer who is the GR bookkeeper. This is only one perspective. I think the local explanation requires a quantum solution. Especially if we are talking about gravitons. There is nothing in physics which precludes virtual gravitons from tunneling the horizon with respect to limits set by the uncertainty relationship for energy and time. This means that within this relationship they can exceed c during the tunnel. What do you think about this?

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