From: *Brent Meeker* <[email protected] <mailto:[email protected]>>
On 5/6/2018 6:56 PM, Bruce Kellett wrote:
From: *spinozalens via Free Thinkers Physics Discussion Group*
<[email protected] <mailto:[email protected]>>
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 Δ/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.
Right
So for Susskind's argument does it matter whether the photons are
hotter close to the event horizon or the thermometers run slower?
What was Susskind's argument, exactly? But in general, I think it is
better to be physically accurate rather than to rely on dubious heuristics.
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.
You are probably right. He is wrong about Hawking radiation in most
respects: particularly because he think it comes from vacuum loops near
the horizon -- there are no such loops containing only photons. Hawking
radiation comes about because the horizon mixes the positive and energy
frequency parts of the field, so the definition of the vacuum (zero
particle state) is different in the presence of the horizon. Nothing to
do with virtual loops.
Bruce
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