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| Tags: bekensteinhawking, emit, horizons, radiation, rindler |
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This question came up in discussion with some colleaugues. There are several papers that claim that Rindler horizons have entropy (For example gr-qc/0302099). Thus Rindler horizons must radiate. But if they do, their infinite size would mean that they radiate an infinite amount of energy. Is that possible? Frank |
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Ben Rudiak-Gould wrote:
wrote: There are several papers that claim that Rindler horizons have entropy (For example gr-qc/0302099). Thus Rindler horizons must radiate. They do, and it's called Unruh radiation. I think you can derive the Unruh formula from the Hawking formula if you take the limit correctly (which is a bit tricky). Yes. But if they do, their infinite size would mean that they radiate an infinite amount of energy. Even a finite positive amount seems like too much to me. I'm afraid I don't understand how the energy accounting works in this case (and I wish I did). Remember that energy is the Noether current corresponding to translation in time. For Unruh radiation and a Rindler horizon to occur, one must use accelerating coordinates. In such coordinates there is no time translation symmetry (i.e. the time coordinate is NOT a Killing vector) and energy is not conserved. Without conservation, energy becomes pretty useless.... Use inertial coordinates on the same manifold: there is no Unruh radiation, no Rindler horizon, and energy is conserved. Basically this "infinite energy" is merely a mathematical artifact that appears in this unphysical situation (flat manifold, acceleration for infinitely long time over infinite volume). I'm not sure anyone understands it completely. I think the above gives the basics. Tom Roberts |
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Tom Roberts wrote:
For Unruh radiation and a Rindler horizon to occur, one must use accelerating coordinates. In such coordinates there is no time translation symmetry (i.e. the time coordinate is NOT a Killing vector) and energy is not conserved. Actually Rindler time translation is a Killing vector field; that's what makes this problem interesting. The analogy to the Schwarzschild geometry (and coordinates) is very close. Hawking radiation can't be just an artifact of the coordinates, if black holes really do evaporate (after emitting exactly as much energy as they consumed). But in the Unruh case there's nothing to evaporate or even shrink, is there? I seem to recall hearing that emission of Unruh radiation actually is associated with a recession of the Rindler horizon, but I don't understand how that works in either Rindler or Minkowski coordinates. It would make sense, in that the recession would correspond to an infinite loss of mass which might be expected to balance the infinite amount of radiation emitted. -- Ben |
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#6
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"Tom Roberts" wrote in message .com... Ben Rudiak-Gould wrote: wrote: There are several papers that claim that Rindler horizons have entropy (For example gr-qc/0302099). Thus Rindler horizons must radiate. They do, and it's called Unruh radiation. I think you can derive the Unruh formula from the Hawking formula if you take the limit correctly (which is a bit tricky). Yes. ... Here's the problem I have with a Rindler *horizon* radiating (not to be confused in my view with the accelerating *object* experiencing an effective radiation bath where it sits based on interaction with *local* virtual particles): The RH is the plane in space corresponding to the convergence of lines of simultaneity for hyperbolic motion (each part of an accelerating system lorentz-contracts according to its momentary velocity, leading to different velocities for each part of such a "Born-rigid" system. Their hyperplanes of simultaneity intersect at the RH.) The motion is given as a = c^2/X, where X is the Rindler coordinate zeroed on the RH at the moment everything is at "rest" in a given frame. So, the RH is a plane in space perhaps far away from a given accelerating body. The body's co-moving observer O considers the RH as a sort of black hole horizon, since light coming from the RH would be infinitely red-shifted (the shift follows the rule nu'/nu = X_emit /X_recv.) To us, this is due to the velocity differences between emitter and receiver, and to O, this is due to the light climbing the gravity potential. Well, the trouble is, the RH is just "space" identical to the surrounding vacuum and doesn't "know" that there is an object accelerating somewhere to which it can be "referenced" as a horizon (unlike - ? - a real black hole.) It is no more physically real as an interface or structure or operational entity apart from ordinary space than the earth's equator, or better yet, the locus sphere one light-year from the sun etc. It can no more properly emit radiation than any other conceptually defined region. The rest of us sure as hell won't find any "radiation" coming from there, any more at least than from any other plane in space. Well, maybe in some snarky sophistic way, the background flux can be "considered" by O to contain photons specifically emitted there yada yada, but that sounds like sophistry to me (but it might not be.) We have to get this issue hashed out before we can think about infinite energy "radiated" by such a weird notional somesuch. I raised a similar point on sci.physics.research and don't remember just what they said - some kind of mushy compromise perhaps. N. Bates http://tyrannogenius.blogspot.com |
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"Ben Rudiak-Gould" schrieb I'm afraid I don't understand how the energy accounting works in this case (and I wish I did). I'm not sure anyone understands it completely. There are analogous classical problems (with the radiation backreaction) that remain unresolved. I think this is one of the many things which cannot be defined without a preferred frame. It is the same thing which happens with the energy-momentum tensor of the gravitational field itself: All we have here is the Landau pseudotensor, which gives an integrable conservation law of type partial_m (T^m_n + t^m_n) = 0 in classical GR in some preferred set of coordinates. A similar freedom is that of the choice of vacuum. Indeed, in Minkowski space we have, together with natural Minkowski coordinates, also a natural choice of the vacuum. And a similar choice is also that of regularization of the energy-momentum tensor. Again, in the Minkowski space we have a natural regularization: Let it be zero for the vacuum. In a similar way we need preferred coordinates for the definition of a Hamilton formalism (ADM decomposition). All this, taken together, gives a quite natural concept: For QG, we need an additional structure, which defines all these things - energies, back-reactions and so on. And, the simplest choice is a classical preferred frame. I have worked out this choice in gr-qc/0205035. Ilja |
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"Neil" wrote in message ... "Tom Roberts" wrote in message .com... Ben Rudiak-Gould wrote: wrote: There are several papers that claim that Rindler horizons have entropy (For example gr-qc/0302099). Thus Rindler horizons must radiate. They do, and it's called Unruh radiation. I think you can derive the Unruh formula from the Hawking formula if you take the limit correctly (which is a bit tricky). Yes. ... Just a slight adjustment for clarity (and per the confusion of using "a" as a variable with word wrap and without italic font): Here's the problem I have with a Rindler *horizon* radiating (not to be confused in my view with the accelerating *object* experiencing an effective radiation bath where it sits based on interaction with *local* virtual particles): The RH is the plane in space corresponding to the convergence of lines of simultaneity for hyperbolic motion (each part of an accelerating system lorentz-contracts according to its momentary velocity, leading to different velocities for each part of such a "Born-rigid" system. Their hyperplanes of simultaneity intersect at the RH.) The motion is given as Corrected lines: a_0 = c^2/X, where X is the Rindler coordinate zeroed on the RH at the moment everything is at "rest" in a given frame, and a_0 is the "proper acceleration" (co-moving value.) .... http://tyrannogenius.blogspot.com |
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