On Monday, May 7, 2018 at 12:01:14 AM UTC-5, Brent wrote: > > > > On 5/6/2018 6:56 PM, Bruce Kellett wrote: > > From: spinozalens via Free Thinkers Physics Discussion Group < > [email protected] <javascript:>> > > 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 >
Hawking radiation does not necessarily appear right above the horizon and then climb out of the gravitational potential well. The black hole does emit a Planck unit of mass-energy and its appearance as Hawking radiation away from the BH takes into account this energy balance. The Hawking radiation appears largely around 4M and is a spontaneous tunneling or nonlocal event. For an observer right above the horizon there is a different perspective. From that observer's perspective the radiation does occur close to the event horizon. We do not have a locality of quantum fields at locations in spacetime. The accelerated observer Bob witnesses radiation emerging from the horizon very rapidly and near the Planck scale in energy. It is a horrific situation, and as I have illustrated the near horizon condition of a black hole is locally equivalent to a frame in an anti-de Sitter spacetime. This spacetime has negative vacuum energy and so quantum particles will spontaneously appear since the quantum field is not bounded below. The inertial and accelerated observers, Alice and Bob, witness this process from complementary frames. Nature is not observable from both of these perspectives simultaneously, but Alice and Bob witness events in a manner that reflects a general nonlocality of quantum field in curved spacetime. We are familiar with nonlocality in quantum mechanics, but with quantum field theory much of this is swept under the rug. This is permitted because nonlocal physics is only observable on a tiny scale before entanglement phase is randomly lost in a shower of daughter particles. So to avoid problems with understanding causality quantum fields are defined to have zero commutators on spatial surfaces. This works well enough, but for black holes the lensing of observed events and the contraction of systems along the radial direction recovers Schrodinger type of nonrelativistic physics. Nonlocality is in effect amplified by gravitation. LC -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To post to this group, send email to [email protected]. Visit this group at https://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
