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I Hawking radiation from de Sitter horizon

  1. Aug 18, 2017 #1
    I have some questions regarding the temperature of empty space in a de Sitter universe or to say it better - the Hawking radiation emmited from the cosmological horizon:

    1) Do particles that make up the radiation get produced by the empty space inside the patch (the Bunch Davies vacuum) or by the horizon which then 'sends them' into the interior?

    When it is said that the observer would detect a temperature, would he detect the temperature of the empty space or the temperature of the radiation emmited by the horizon which takes time to travel to the observer?

    2) I watched a video from prof Lenny Susskind in which he talks about this situation with Hawking radiation and he says that most of the particles would be concentrated near the horizon. How would the temperature of empty space then be constant in any region since some would contain more particles than others?

    Furthermore, if there is another observer located differently, he would have a different horizon and he would describe the location of the 'many particles' of the first observer as a region with less particles, since his density would be greater near his own horizon. What is wrong with this argument?

    3) Sean Carroll has made an argument that the temperature of de Sitter isn't real if nobody is measuring it, and that fluctuations can only arise when matter is present? Did he mean that fluctuations arise in normal matter in radiation before it redshifts away or that the fluctuations don't arise because the matter that redshifts away acts like a measuring device on the vacuum which is the thing that really produces particles?

    4) If the horizon produces particles that become real like in the black hole case, what happens to the second pair which is left behind the horizon like falling into a black hole? And how can an observer in the center of the patch measure the creation of the particles, since by this logic they were created way before they came to the observer?

    As you can see I am very confused by this topic and I hope that it is not too nonsensical for some good answers.

    Thanks in advance.
     
  2. jcsd
  3. Aug 18, 2017 #2

    jambaugh

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    They are the same but better expressed the second way you said it. Temperature is defined by the distribution of energy across the system modes. In this case that will be (mostly) photon frequencies. So not "empty space".

    But be aware that event horizons are all through space-time. Snap your fingers and the future lightcone of that space-time event is an event horizon. No action inside it can communicate outside. What is special about BH horizons and deSitter horizons is that space-time curvature allows them to be stationary w.r.t. some observer frame.

    As to your other questions, they relate to the tricky task of understanding the relativistic vacuum. I won't try to cover the whole topic (not least because it is beyond my abilities and education) but here are the key points for this query.

    • The gauge field vacuum is not lorentz invariant. If I prepair a chamber with 0 photons in it, you, moving at non-zero velocity to me will observe a non-vacuum mixed state.
    • If you are accelerating you are thus consantly redefining your vacuum and thereby are able to absorb a flux of particles out of what I might view as empty T=0 space.
    MORE LATER gotta go meet someone.
     
  4. Aug 18, 2017 #3
    Regarding my question about present day Hawking radiation, one user posted this:

    !Theoretically, yes, there is a temperature associated with the event horizon of an accelerating spacetime. You can call it a "Hawking" (although see caveat below) or "de Sitter" temperature. Since one observer's position -- say yours -- is at the event horizon of some other observer, you will measure this temperature at your location; and likewise he'll measure radiation at your event horizon. Extending this construction across the isotropic universe results in the conclusion that the spacetime is uniformly bathed in thermal background of radiation. The temperature associated with this radiation is proportional to the Hubble scale (i.e. the horizon size) and so is very much lower than the CMB temperature. Hence, it is currently undetectable."

    Is this true - that Hawking radiation already exists everywhere?
     
  5. Aug 18, 2017 #4

    bapowell

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    Yes, but it's due to the existence of a cosmological rather than black hole event horizon.
     
  6. Aug 18, 2017 #5

    bapowell

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    The popular account of Hawking radiation involves a particle/antiparticle pair that pops out of the vacuum, with one particle getting captured by the black hole while the other flies off. A similar cartoon can be imagined in de Sitter space. Here you have a particle/antiparticle pair that get separated by the rapidly inflating spacetime, eventually getting pulled outside each other's event horizons.

    Having said this, it's generally not advisable to think about Hawking/de Sitter radiation in terms of particles. In fact, at no point in Hawking's derivation of his effect are there ever particles. He instead works at the level of the quantum field and studies how it evolves within the vicinity of a black hole horizon. His treatment makes an interpretation in terms of physical particle states murky at best, and I prefer simply not to strive for such a picture. The same goes for the derivation of the de Sitter temperature: it's an analysis of quantum fields evolving in an inflating spacetime. One finds in both cases that the quantum vacuum changes---from zero particles to nonzero particle count---as a result of the gravitational physics, but exactly where individual particle pairs are or what they're doing is not elucidated in the mathematics.
     
  7. Aug 19, 2017 #6
    But is this only a far future effect when we have de Sitter space or does there exist Hawking radiation right now and during the era with matter?

    I've mostly heard that it is a far future effect, since the radiation cannot reach the oberver until the end of his wordline.
     
  8. Aug 20, 2017 #7

    bapowell

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    No, it isn't some sort of asymptotic effect afaik. The de Sitter radiation is generated uniformly throughout space.
     
  9. Aug 21, 2017 #8
    But then, why does Sean Carroll in this paper: https://arxiv.org/abs/1405.0298 talk about the radiation as if it were only an effect of empty de Sitter phase, when all matter is diluted by expansion - when the universe is empty with just a cosmological constant?

    Does the Hawking radiation (the particles that make it up) still get diluted by expansion so that at the end we have only empty space with a cosmological constant?
     
  10. Aug 21, 2017 #9

    bapowell

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    I haven't read the paper, but your statement cannot be correct: during inflation, quantum fluctuations about the vacuum get amplified by the expansion and become classical curvature perturbations in the post-inflationary universe. The generation of these quantum fluctuations is a manifestation of the de Sitter "effect", and it occurs in space that is not pure vacuum (the inflationary period was "vacuum dominated", but surely not purely vacuum).

    All radiation gets redshifted by the expansion. Eventually, all blackholes will evaporate via the Hawking effect and cease to exist.
     
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