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Unruh effect and Entropy

  1. Aug 22, 2014 #1
    Do different observers, say, inertial and accelerated, moving thru the same point in space, agree on the entropy of the same isolated system they observe?

    I am interested in it in the context of Unruh effect. If we switch between inertial and accelerated frames, we switch Unruh particles ON/OFF (or switch them between being "real" and "virtual"), and there are states associated with these particles

    Thank you
     
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  3. Aug 22, 2014 #2

    marcus

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    Personally I think this raises some non-trivial issues and I'm hoping some others will respond. My view (which others may wish to challenge) is that entropy is observer-dependent. IOW not absolute and intrinsic to the system. There has to be an entity interacting with the system (via some coarse-grained variables that affect that entity) in order for macrostates to be defined.

    So I would respond kind of reflexively and say "no, the two observers, inertial and accelerated, do not necessarily agree." I think your example of Unruh temperature is pertinent. They can't even agree about the temperature! :biggrin:

    This does raise some really interesting issues. There's an amazing paper from 1995 by Ted Jacobson, which you may know of. He derives the Einstein GR equation by probing using accelerated observers in every possible direction from a given point. I have to run an errand so I will give the link and check what I'm saying later.

    It would be on arxiv. with a title with words like "Einstein equation of state". Back later.
     
  4. Aug 22, 2014 #3

    marcus

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    Back again, I googled "Jacobson equation of state" and got

    http://arxiv.org/abs/gr-qc/9504004
    Thermodynamics of Spacetime: The Einstein Equation of State
    Ted Jacobson
    (Submitted on 4 Apr 1995)
    The Einstein equation is derived from the proportionality of entropy and horizon area together with the fundamental relation δQ=TdS connecting heat, entropy, and temperature. The key idea is to demand that this relation hold for all the local Rindler causal horizons through each spacetime point, with δQ and T interpreted as the energy flux and Unruh temperature seen by an accelerated observer just inside the horizon. This requires that gravitational lensing by matter energy distorts the causal structure of spacetime in just such a way that the Einstein equation holds. Viewed in this way, the Einstein equation is an equation of state. This perspective suggests that it may be no more appropriate to canonically quantize the Einstein equation than it would be to quantize the wave equation for sound in air.
    8 pages, 1 figure.

    You may know of it. Really interesting. Over the almost 20 years various other researchers have responded to the idea. I think the most interesting recent response was by Chirco et al, this year.
    They showed how you could carry thru Jacobsons derivation of the GR equation in the QG context without postulating some additional underlying microscopic degrees of freedom.

    http://arxiv.org/abs/1401.5262
    Spacetime thermodynamics without hidden degrees of freedom
    Goffredo Chirco, Hal M. Haggard, Aldo Riello, Carlo Rovelli
    (Submitted on 21 Jan 2014)
    A celebrated result by Jacobson is the derivation of Einstein's equations from Unruh's temperature, the Bekenstein-Hawking entropy and the Clausius relation. This has been repeatedly taken as evidence for an interpretation of Einstein's equations as equations of state for unknown degrees of freedom underlying the metric. We show that a different interpretation of Jacobson result is possible, which does not imply the existence of additional degrees of freedom, and follows only from the quantum properties of gravity. We introduce the notion of quantum gravitational Hadamard states, which give rise to the full local thermodynamics of gravity.
    12 pages, 1 figure

    One useful thing about the Chirco paper is that it analyzes how Jacobsons thermodynamic derivation of GR works and what it actually depends on. One gets a better understanding of the 1995 paper by reading the 2014 one.
     
  5. Aug 22, 2014 #4

    atyy

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    From Chirco's discussion around Eq 18-21, I don't know whether the different observers observe "the same" system. The accelerated observer "sees" the Rindler wedge, which is why its density matrix (which gives the entropy) is different.
     
    Last edited: Aug 22, 2014
  6. May 13, 2015 #5

    naima

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    This raises a curious thing. The non geodesic observer meet particles which heats its thermometer. For the geodesic observer these particle are not thermic but virtual. One often read that virtual particles are just a mathematical trick. It seems that one can manage continuously their reality fron 0 to 1 by tuning the geodicity.
    It is the same near a black hole , the free falling observer sees no radiation while another can cook his eggs near the boundary. This changes our notion of event. We cannot describe anymore an event in the space time as observer independent. A crucial issue for science Does the falling observer see raw eggs?
     
  7. May 13, 2015 #6

    wabbit

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    The fact that two observers see different things seem quite natural to me, but in the Unruh effect, I am not sure how to interpret the readings of the inertial observer : when he sees a thermometer accelerating away in the vacuum, he can read the Unruh temperature. What does he attribute it to ?
    There is no Unruh radiation from his viewpoint. Perhaps he sees the accelerated observer splitting pairs of virtual particles ? If so, there must be radiation emitted by the accelerated observer ?
    It looks like a kind of friction the accelerated observer encounters, a "resistance of the vacuum" reminiscent of air friction heating up a moving observer... (No, I am not saying there is a friction from moving in the aether :wink: )

    Edit : actually this is the same question naima asked - only I am convinced the infaller sees cooked eggs (the hoverer might even throw him an omelette as he passes by, and I don't think eggs get uncooked when they stop accelerating), but the question as I see it, is how does he explain the cooking ?
     
    Last edited: May 13, 2015
  8. May 13, 2015 #7
    If there was a process by which space-time had to "emerge" for any causal event and that process had some (as yet undiscovered) characteristics that naturally projected into the rates and/or shape of that emergence, then I can imagine emergence "shaped" by acceleration (converting some length to - t) being different than emergence "at rest". That's the cartoon I've been toting for Unruh temperature.

    What is a "Hadamard" State? I would love to know, if there was an easy version...

    And I'm very confused when I think about how that notion of Unruh Temperature is tangled up with SR and GR, though this was part of what appealed from Verlinde's Entropy Holographic Gravity. These are all characteristics of the space-time emergence process, whatever it is.

    [Edit] Gonna try to read as far as I can into these. Discipline!!
    But to @wabbit's comment analogy w/friction, a weak selection process would have a natural "governor" I believe. Someone mentioned in a different thread, what if there was a complex plane representation of the state of that emergence process, it's rate limiting. I was picturing the Ashtekar thing, detangling n planck-holes, dump, onto the next n planck-holes. The clock on the complex plane timing the dump, counting the n, corkscrewing from one "event" to the next.
     
    Last edited: May 13, 2015
  9. May 14, 2015 #8

    Demystifier

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    The acceleration itself does not produce Unruh effect. What produces Unruh effect is the interaction of the accelerating detector with the environment, even when the environment is initially in the ground "vacuum" state. Of course, the Unruh temperature does not depend on the coupling constant of interaction, but the measured effect does. In particular, for a zero coupling constant there is no entanglement between the detector and the environment.

    What is misleading here is a classical way of thinking that the thermal environment should be there even if nobody observes it, i.e. even when the coupling constant is zero. But that's wrong in quantum mechanics. Quantum mechanics is contextual. The interaction with the measuring apparatus, or more precisely the entanglement created by this interaction, creates properties which did not exist before.
     
  10. May 14, 2015 #9

    haushofer

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    I understood the unruheffect as due to the fact that the numberoperator in qft is only poincare-covariant. If entropy is not a general-covariant quantity, like the connectionn GR, then the idea of entropic gravity also seems plausible :P
     
  11. May 14, 2015 #10
    I had a sentence in there about how you need some kind of event, that "you can't have an event without an object". The idea of accelerating a vacuum doesn't seem meaningful, but then I got confused. So, your statement feels relevant to my confusion...

    I am still just hardly anywhere on QM or QFT at all. I've been told a number of things it means, and that those are not what it means. I have the books inc Susskind, but they just stay at the same place in the stack. So pardon my ignorant question. So for there to be any non zero coupling constant there must be a particle? What is it the field made of? Is the coupling constant "carried" by the particle or the field? I'm sure I'm being too literal minded about it.
     
  12. May 14, 2015 #11

    Demystifier

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    No.

    Nobody knows.

    No, neither.

    More detailed answers would be highly off-topic.
     
  13. May 14, 2015 #12

    naima

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    When an observer stays near a BH he can measure a temperature. This temperature is more linked to the BH surface than to his trajectory.
    Rovelli defines a "diamond" temperature for bounded trajectories
    If the temperature is observer dependent can we say that it is boudary dependent?
     
  14. May 27, 2015 #13

    naima

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    I am trying now to undestand this paper

    There is a formula (A8) that I do not understand. Please skip to page 10.
    Just before A8 we have an integral on time then in A8 we still have an integral on time but a ## e^{i \epsilon t}## term appears. I think that it comes from the time evolution of the operator A but how?

    Edit: I can see now that it comes from the product of ## e^{-iHt} ## and ## e^{iHt} ## on |0> and |1>
    But what is A in the interaction hamiltonian?
    Thank you.
     
    Last edited: May 27, 2015
  15. May 28, 2015 #14
    Since none of the guys who know anything are jumping to your aid...

    Isn't A the QM "observable".

    "Consider a quantum system described by an observable algebra [itex]A[/itex] with observables A,B,..., which is in a state [itex]\sigma :A\rightarrow C[/itex]..."

    with some assumptions then place on the Algebra [itex]A[/itex], like it has flow and is "realized by operators on a Hilbert space H where a Hamiltonian... with eigenstates.. and Eigenvalues... is defined, the flow and the state are defined in terms of the evolution operator..."

    You question was probably over my head but just in case...
     
  16. May 28, 2015 #15

    naima

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    Thank you.

    We have not so many possible observables for the system : position, (angular) momentum energy spin in current matter. So in the interaction hamiltonian
    [g(|0 X 1| + |1 X 0|) A] Can A be sometimes a momentum sometimes a spin etc?
    Let us take a black hole and a no falling thermometer near the Schwarzchild area. It can feel the heat of the BH
    What can be A? Is it a the distance between the BH or and the thermometer?
    the "motionless" thermometer shows the temperature T: is the flow null?

    Many questions...
     
  17. May 28, 2015 #16
    Yeah, that appendix - I would like to understand it better (meaning a little). Clearly it's critical to the whole paper.

    Seems to be saying that you can define a "temperature" for a QM system that meets some criteria, that the transition probability per unit time, or the time rate of state change is in a sense the "temperature" - and the referent of "transition" or "state change" A, could be "any observable". In fact my guess is that in this sense of QM "clock" == QM thermometer, you could collect any and/or all observables...

    IOW it seems to be highlighting the idea of "temperature" is only post calibrated against some specific observable,... at this fundamental level of definition, it is just in units of pure change, however you want to count them and however you want to aggregate or bucket them up), pure information as a property of time - is sufficient for the definition.

    In fact it seems to be suggesting that time is temperature? Which I kindof buy... I think. Isn't that consistent with a(t) in FLRW.

    Yeah, many questions...

    My guess in the BH case is.... uh....

    "... something something entanglement entropy?"

    Thanks for taking me down closer to that... very interesting.
     
  18. May 28, 2015 #17

    wabbit

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    As I read it there are not so many possibilities, it's a two-state thermometer. Now the actual physical process involved in reading the temperature can be anything of course, but those details don't matter.

    Then again I can't follow the details so I probably don't even know what you are talking about...
     
  19. May 28, 2015 #18

    naima

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    I found the same formulas here but with the Unruh effect as an example.
    Here A is the field operator!
    This is what i was looking for.
     
  20. May 28, 2015 #19

    wabbit

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    Oh OK, but that is mentionned already in the beginning of the appendix, they define A as an observable and quote the example of a field operator.
     
  21. May 29, 2015 #20
    I kindof read it as the QM thermometer was two state but the QM system it is coupled to had states i-f. I may be reading that wrong.

    So I took the thermometer detecting "Two states" to be in the sense of "I saw more than one state"? So QM system S with states I-f could be any multi-state observable.

    If you stick Crook's fluctuation theorem, Jarzynski -> England I think the notion of "temperature" being the probability of state change (a violation of equilibrium) in QM observables is nicely continuous from the thought experiment here up to something like a regular thermometer. Still doesn't "explain" why the QM system S has a time probability to change state... But then that's just the initial conditions of the universe right.

    And... one sentence that I completely don't get...
    "for instance these equations permit the treatment of thermal quantum field theory, where the Hamiltonian is ill-defined, because of the infinite energy of a thermal state in infinite space"

    This makes it sound like the simple two state thermometer coupled cleanly to a QM system S via some simple observable is questionable...
     
    Last edited: May 29, 2015
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