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Jan 30

## What is the temperature of black hole and cosmic horizons really?

Posted by: JackSarfatti
Tagged in: Untagged

However, it looks as though we need H ~ Gm/c^2 ~ 10^-56 cm to get T > 2mc^2 and this is problematical since it's much smaller than the Planck length Lp ~ 10^-33 cm.

It depends how close you are to it as a static LNIF.

Here is one important idea that Nick got hung up on.
The Newtonian surface gravity is
gNewton = c^2rs/r^2 ---> c^2/rs
rs = 2GM/c^2
However, that is not correct for GR we need the time-dilation factor g00^-1/2
g(r) = (c^2/rs)(1 - rs/r)^-1/2   ----> infinity at the black hole horizon for static LNIFs hovering outside it.
we see c^2/rs as rs/r ---> 0
The temperature of the Hawking thermal radiation that the static LNIF detects T = hc/rskB = hc^3/2GMkB
that Unruh cites is what we see far from the black hole where the static LNIF merge to LIFs as rs/r ---> 0
Similarly for the observer-dependent cosmic horizon
g(r) = c^2/^1/2(1 - / ^2)^-1/2
we are at r = 0, so we see T = hc/^1/2/kB
but a static LNIF distance H << /^-1/2 from the horizon sees
T ~ hc/^1/4H^-1/2kB^-1
Of course a geodesic LIF falling through the horizon sees T = 0 because its covariant acceleration is zero.
The photon is a null geodesic LIF, but the virtual electron-photon pairs clamped to the relative horizon of that retarded photon emitted from r = 0 are static LNIF with covariant acceleration c^2/^1/4H^-1/2.
On Jan 29, 2011, at 11:09 PM, JACK SARFATTI wrote:
On Jan 29, 2011, at 10:49 PM, nick herbert wrote:
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Dear Abbie--
To whom should I turn for reliable information