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 Be the first pioneers to continue the Astronomy Discussions at our new Astronomy meeting place...The Space and Astronomy Agora Re: Relativity Of Event Horizon Forum List | Follow Ups | Post Message | Back to Thread Topics | In Response ToPosted by Bruce on February 25, 2001 01:40:23 UTC

In GR the ruler is ridgid. There is another paradigm that gets the same results where clocks and rulers are 'rubbery'. 'rubbery' clocks and rulers means no curvature. Same results to measurements.
This is wrong
"Now, those photons lose their energy as they ascend from EHFO and lose ALL their energy at infinity."
This is a thought experiment which shows how energy is conserved in Schwarzchild geometry.
A object starting from rest at infinity with rest energy (m=1) begins freefalling towards the black hole.
At a given radius r=2M+dr outside the event horizon (any radius you choose as long as r>2M) there is a machine which will catch the object and convert all its kinetic energy into a pulse of light which is sent radially back to a receiving station where the object started freefall with m=1.
Using SR the shell observer at the machine measures the objects total energy
vshell=(2M/r)
Eshell=m/(1-2M/r)1/2
And the kinetic energy converted to a pulse of light
Eflash=[m/(1-2M/r)1/2]-m
The far away observer at the receiving station measures the energy of the pulse of light
Efaraway=Eflash(1-2M/r)1/2
So
Efaraway=m-m(1-2M/r)1/2
There are no paradox in relativity. You can't have a black hole and have it disappear just by continuing to approach the event horizon. The distance to the event horizon is not infinite. The reason we need the integral for
drshell=dr(1-2M/r)-1/2 is because the Schwarzchild result for r=2M doesn't work. The other metric I posted doesn't have an infinity at r=2M but is derived from SR and Schwarzchild metric. Local observers use SR to do the physics.