On 5/6/2018 6:56 PM, Bruce Kellett wrote:
From: *spinozalens via Free Thinkers Physics Discussion Group*
<atvoi...@googlegroups.com>
Δ
No, Susskind makes clear that the Hawking radiation is blue shifted
near the Horizon, and every other source I have on this agrees. You
can't get a detectable photon for the outside observer if the photon
wasn't at a very much higher energy near the horizon.
We have argued this back and forth many times. But the answer is very
clear. Hawking radiation is produced just above the BH horizon with
exactly the energy with which it is observed by the stationary
observer at infinity. The apparent divergence in energy near the
horizon occurs *only* for a fiducial observer, held at rest there. The
photon does not lose energy climbing through the gravitational field.
There is no gravitational potential energy. All that changes with
distance from the horizon is the clock rate.
This is explained very clearly in MTW. Einstein used energy
conservation to deduce the red shift, but Schild improved this
argument to show that the red shift is in fact caused by spacetime
curvature (MTW, pp187-189). In their discussion of the
Pound-Rebka-Snider red shift experiment, MTW make an even clearer
explanation. On page 1058 they explain in detail that if one views a
photon as a sequence of wave crests, then each successive wave crest
sees exactly the same gravitational field, "...therefore the crest of
each electromagnetic wave that climbs upward must follow a world line
t(z) identical in form to the world lines of the crests before and
after it.... Hence, expressed in /coordinate/ time, the interval
between reception of successive wave crests is the same as the
interval between emission. Both are ΔΔ/t/." (MTW p1058). Atomic clocks
(and stationary observers) measure proper time, not coordinate time.
Hence the difference as given by the Killing factor.
Which means that an atomic clock lowered to near the event horizon will
measure the frequency of a photon that is a few ev far from the black to
have very high energy. So what looks like low temperature Hawking
radiation at infinity will look like high temperature for the object
suspended near the horizon, because that objects internal "clocks" run
slower than when it was at infinity.
So for Susskind's argument does it matter whether the photons are hotter
close to the event horizon or the thermometers run slower?
I think he is wrong to assume he can reason about the Hawking the
radiation by taking the limit of going to the event horizon.
Brent
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