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Do Rindler horizons emit Bekenstein-Hawking radiation?
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Neil1
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Joined: 24 May 2005
Posts: 16

PostPosted: Thu Jul 13, 2006 12:50 am    Post subject: Re: Do Rindler horizons emit Bekenstein-Hawking radiation? Reply with quote

"Neil" <neil_delver@caloricmail.com> wrote in message
news:12b0nqc3pbv9e93@corp.supernews.com...
Quote:

"Tom Roberts" <tjroberts137@sbcglobal.net> wrote in message
news:kmPrg.128556$dW3.51142@newssvr21.news.prodigy.com...
Ben Rudiak-Gould wrote:
frank_k_sheldon@yahoo.co.uk 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):
Quote:
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
Quote:
everything is at "rest" in a given frame, and a_0 is the "proper
acceleration" (co-moving value.)

....
Quote:
http://tyrannogenius.blogspot.com
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Ilja Schmelzer
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Joined: 11 May 2005
Posts: 377

PostPosted: Tue Jul 11, 2006 11:34 am    Post subject: Re: Do Rindler horizons emit Bekenstein-Hawking radiation? Reply with quote

"Ben Rudiak-Gould" <br276deleteme@cam.ac.uk> schrieb
Quote:
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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Neil1
science forum beginner


Joined: 24 May 2005
Posts: 16

PostPosted: Sun Jul 09, 2006 1:43 am    Post subject: Re: Do Rindler horizons emit Bekenstein-Hawking radiation? Reply with quote

"Tom Roberts" <tjroberts137@sbcglobal.net> wrote in message
news:kmPrg.128556$dW3.51142@newssvr21.news.prodigy.com...
Quote:
Ben Rudiak-Gould wrote:
frank_k_sheldon@yahoo.co.uk 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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Edward Green
science forum addict


Joined: 21 May 2005
Posts: 95

PostPosted: Sat Jul 08, 2006 10:56 pm    Post subject: Re: Do Rindler horizons emit Bekenstein-Hawking radiation? Reply with quote

Ben Rudiak-Gould wrote:
Quote:
frank_k_sheldon@yahoo.co.uk 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).

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).
I'm not sure anyone understands it completely. There are analogous classical
problems (with the radiation backreaction) that remain unresolved.

Short sketch, please?
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Ben Rudiak-Gould
science forum Guru


Joined: 04 May 2005
Posts: 382

PostPosted: Sat Jul 08, 2006 3:38 pm    Post subject: Re: Do Rindler horizons emit Bekenstein-Hawking radiation? Reply with quote

Tom Roberts wrote:
Quote:
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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Tom Roberts
science forum Guru


Joined: 24 Mar 2005
Posts: 1399

PostPosted: Sat Jul 08, 2006 2:32 pm    Post subject: Re: Do Rindler horizons emit Bekenstein-Hawking radiation? Reply with quote

Ben Rudiak-Gould wrote:
Quote:
frank_k_sheldon@yahoo.co.uk 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.


Quote:
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).


Quote:
I'm not sure anyone understands it completely.

I think the above gives the basics.


Tom Roberts
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Ben Rudiak-Gould
science forum Guru


Joined: 04 May 2005
Posts: 382

PostPosted: Sat Jul 08, 2006 12:44 pm    Post subject: Re: Do Rindler horizons emit Bekenstein-Hawking radiation? Reply with quote

frank_k_sheldon@yahoo.co.uk wrote:
Quote:
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).

Quote:
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).
I'm not sure anyone understands it completely. There are analogous classical
problems (with the radiation backreaction) that remain unresolved.

-- Ben
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frank_k_sheldon@yahoo.co.
science forum beginner


Joined: 28 Jan 2006
Posts: 8

PostPosted: Sat Jul 08, 2006 9:57 am    Post subject: Do Rindler horizons emit Bekenstein-Hawking radiation? Reply with quote

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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