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Local Versus Remote Views of Temperature

  1. Aug 14, 2013 #1

    anorlunda

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    I struggled long and hard to resist the idea that Alice observes nothing special as she falls into the Schwarzschild horizon of a BH. Now, I accept it. The thing that did it for me was professor Susskind's description of deSitter space, and how the math of the metric for the cosmological horizon in de Sitter space was the same as the math of the Schwarzschild horizon of a BH. The same effects of the remote observer Bob seeing Alice dim and red shift apply to both kinds of horizons.

    Since every point in spacetime lies on the horizon of some hypothetical observer 46 billion light years away. If Alice is at Bob's cosmological horizon, then Bob is at Alice's horizon. That means that Alice, Bob, and you, and I all sit on someone's horizon. Yet I observe nothing special.

    But then Susskind continued discussing the temperature just above the BH horizon. He specified temperature in the sense of average kinetic energy. Whereas the Hawking temperature of a large BH is low, the temperature just above the horizon is very hot. Close to the horizon, it is described by 1/2∏ρ (where ρ is the distance from the horizon, say from one plank length up to 1 cm above the horizon). Then, Susskind said that the same applies to the temperature just short of the cosmological horizon in de Sitter space.

    Susskind illustrated with the imaginary experiment in which the remote observer lowers a thermometer on a string to just above the horizon, then reels it back up to read the recorded temperature. Note that the thermometer does not transmit its reading; it records a reading and is then reeled in. Schwarzschild horizon, cosmological horizon; same thing.

    Now, I'm struggling again to understand. How can my local sense of temperature differ from my temperature as seen by a remote observer? I imagine the remote observer's thermometer hovering in front of my eyes and I wonder how it can record a temperature different than that I perceive.
     
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  3. Aug 14, 2013 #2

    PeterDonis

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    I don't think that's quite what Susskind meant to convey. I think he meant to convey that there are two ways a remote observer can record the BH's temperature:

    (1) He can measure the temperature of the Hawking radiation as it passes him, far away from the BH.

    (2) He can lower a thermometer to just above the horizon, have it measure the temperature of the Hawking radiation there and record what it measures, then pull it up again to take the reading.

    What Susskind is saying (I think--you haven't linked to a specific source) is that the reading from #1 will give a much lower temperature than the reading from #2. This is because of gravitational redshift: the Hawking radiation is greatly redshifted as it rises up from just outside the horizon to where the remote observer is. #1 measures the radiation after it has redshifted; #2 measures it before it has redshifted. So of course the two measurements will give different answers.
     
  4. Aug 15, 2013 #3

    anorlunda

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    Citation and a clue

    Whoops, sorry to have left out the citation. It was the video course Topics in String Theory, lecture 9, viewable on Youtube here.

    I think Susskind deliberately chose his words to exclude redshift as the explanation. He said the thermometer records (I think of it as a digital recording) and is then reeled in. The thermometer does not lose energy to the gravity well; the string reeling it in supplies the energy.

    But in researching this question, I think I found the clue. The Unruth Effect. I hadn't realized that was what Susskind was talking about.

    The Wikipedia article says, "The hypothetical Unruh effect (or sometimes Fulling–Davies–Unruh effect) is the prediction that an accelerating observer will observe black-body radiation where an inertial observer would observe none."

    Applying that to the cosmological horizon case, I guess I am the inertial observer and the thermometer is the accelerated observer. While the thermometer hovers before my eyes, it is not accelerating and it sees the same temperature I do; but the instant it is reeled in it accelerates away from me it will observe and record a higher temperature. When it gets back to its origin, it can report the highest temperature recorded.

    So, I guess the answer to my quest to understand is to go away and learn the logic behind The Unruth Effect. Sorry to have bothered everyone, I should have dug deeper before posting my question.
     
  5. Aug 15, 2013 #4

    PeterDonis

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    Yes, but the measurement of Hawking radiation by the remote observer at his own location (i.e., *not* by lowering the thermometer) *is* affected by the redshift. Basically Susskind is saying that the usual calculation of a black hole's Hawking temperature is a calculation of its "redshifted" temperature, the temperature that would be measured far away from the hole; it is *not* the temperature that would be measured by a static observer just above the hole's horizon.

    I don' think it is. The Hawking effect is related to the Unruh effect, but that relationship is not what Susskind is talking about. See below.

    Yes. However, I think you've misidentified the observers in the horizon scenario.

    No. If you and the thermometer are both "hovering" close to the horizon, you are both accelerated, and you will both see a high temperature. That's the temperature the thermometer records. Yanking the thermometer away changes its acceleration, but only after it has already recorded its reading of the high temperature while hovering close to the horizon.

    An observer who is freely falling towards the horizon will *not* observe a high temperature; but if you and the thermometer are hovering close to the horizon, you will see such an observer free-falling past you at close to the speed of light. To that observer, it seems like you are accelerating outward very, very hard (and thus moving outward at close to the speed of light).

    That can help to understand how the prediction of Hawking radiation is derived, but by itself I don't think it will clear up the question you posed. See above.
     
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