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Is temperature relative?

  1. Oct 17, 2012 #1
    If Einstein's relativity tells us that movement is relative to the observer, and if temperature is determined in the following way, movement creates friction and friction creates heat, then is temperature not relative to the observer also?
     
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  3. Oct 17, 2012 #2

    bcrowell

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    Temperature isn't defined that way. Temperature is fundamentally defined by [itex]1/T=\partial S/\partial E[/itex], where S is the entropy and E is the internal energy (not counting energy that's locked up in collective motion).

    Some papers that may be relevant:

    http://arxiv.org/abs/0804.3827
    http://arxiv.org/abs/0904.4628
     
  4. Oct 17, 2012 #3

    tom.stoer

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    The transformation law I know is

    [tex]T_\beta = \frac{T_0}{\sqrt{1-\beta^2}}[/tex]

    where the T0 is the temperature of a system as observed in the rest frame, whereas Tβ is the temperature of the same system measured by an observer moving with β = v/c.

    http://www.sdu.dk/media/bibpdf/Bind%2030-39%5CBind%5Cmfm-36-1.pdf [Broken]

    But there was (is?) a debate how to define temperature for relativistiv motion, therefore I am not sure whether the above mentioned formula is widely accepted and correct; have a look this discussion https://www.physicsforums.com/showthread.php?t=228758
     
    Last edited by a moderator: May 6, 2017
  5. Oct 17, 2012 #4

    pervect

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    There is apparently some debate on the best way to describe relativistic thermodynamics. Most of my texts handle the situation by defining a fluid and only working in its rest frame, which sidesteps the question.

    My personal opinion agrees with that of Nakamura, http://arxiv.org/abs/physics/0505004, which seems to be in agreement with Ben's (bcrowell's) references as well. The approach, originating with Van Kampen and Israel, treats inverse temperature as one component of an 4-vector.

    Nakumura has this to say about Moeller's treatment:

     
    Last edited: Oct 17, 2012
  6. Oct 17, 2012 #5

    bcrowell

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    My naive question is -- what the heck is an inverse four-vector?
     
  7. Oct 17, 2012 #6

    pervect

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    A typo for "inverse temperature is one component of a 4-vector" (fixed now)
     
  8. Oct 18, 2012 #7
    Just a super-naive question: if that would be true, didn't it imply that the system would transfer heat passing near a stationary object with temperature T0, so the first system's mass would decrease while the object's mass would increase, after the passage, and the other way round, from the system's frame of reference, generating a paradox?
     
  9. Oct 18, 2012 #8

    tom.stoer

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    You are talking about two systems SA and SB, each system in thermodynamic equilibrium, each system with temperature TA = TB = T0, measured in their rest frames fA and fB

    No two things happen: you allow for a contact (allowing heat transfer between the systems), and you introduce a relative speed vB = v = βc w.r.t to an observer having the same rest frame as SA.

    This questions goes beyond the scope of the original question b/c there we had only one system SA but two observers with rest frames fA and fB; in addition you allow heat transfer between the two systems. If you do that you have to use something like an adiabatic approximation (as usual when you are no longer in thermodynamic equilibrium but want to use thermodynamics).

    I do not say that this is not a valid questions - it's a very interesting question - but first we have to discuss the far more simple but still not completely understood question from post #1
     
  10. Oct 18, 2012 #9

    Vanadium 50

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    Non-relativistically, there are several equivalent ways to define temperature. Relativistically, these are no longer equivalent.

    In most physical systems, this can be "sidestepped" (thanks, pervect) by working in one frame of reference. The reason is that it is difficult to simultaneously have two systems with a relative and relativistic velocity that also are in contact long enough to thermalize.
     
  11. Oct 18, 2012 #10
    Suppose an environment, containing some mass A, is in thermal equilibrium at temp T0 in lab frame S. Rapidly spin-up A to some constant circular orbital speed u about a fixed axis in S - radius can be arbitrarily large so as to avoid any issues related to centripetal acceleration/stresses. This avoids Twin Paradox style ambiguities. Unambiguously, averaged over a per-revolution basis a clock co-moving with A's instantaneous proper frame S' runs slow as seen in S by factor 1/γ = √(1-u2/c2). It also answers the problem of obtaining sufficient time to reestablish thermal equilibrium. In S' A has initially the same proper temp T' = T0 as when formerly at rest in S. In S net thermal energy W of A has been boosted by the factor γ. But A's clock is running in S at a rate the inverse of that same factor. Consequently A is radiating into S an invariant average quantity of thermal power dW'/dt' = γdW/dγt' = dW/dt. Surely an unavoidable conclusion. If the net radiated power is invariant, so surely is A's 'average' temperature T = T' = T0 seen in S[2]. Which conclusion disagrees with both Moller and Planck; sitting between as the geometric mean of those two's contradictory relations.

    As has already been pointed out, there is an angular redshift/blueshift spectral spread of environmental radiation seen by A in S'. Clearly if A has a finite aspect ratio (i.e. does not form a circular loop spinning about it's major axis) then the leading edge will heat, trailing edge will cool, and an axial heat flow ensues until a proper equilibrium temp distribution is reached. At least to first order, seems reasonable this new distribution will not alter that T = T' = T0[1]. Prior to equilibrium, A was experiencing a retardation force opposing circular speed u owing to the bias of blueshifted (increased) radiation pressure on it's leading face vs redshifted pressure on the trailing face. After equilibrium, that net force should increase not reduce. Which should also be reflected in the blueshift/redshift angular bias in A's radiant output seen in S.

    [1]: But some more thought and this conclusion fails to properly account for the inherent asymmetry owing to circular motion. In S', A must determine that the per-revolution mean clock-rate of thermal oscillators in S has increased, not decreased by factor γ. Implying that as determined in S', radiant input into A from environment has increased by factor γ2. One γ for clock-rate, another for boosted number density of oscillators seen in S'. Hence an initial net inflow over outflow into A by a factor γ2-1. Transformed back into S, this net inflow is reduced by clock-rate factor 1/γ, leaving a net initial inflow factor (γ2-1)/γ. Final result then is a boosted temperature of A seen in S. Think that's now about right. Which sort of complicates things and makes circular motion somewhat special or at least distinct from uniform rectilinear motion case where clock-rate asymmetries do not enter. Then again it can be well argued uniform rectilinear motion is an unreal scenario.]

    [2]: Forgot something: Net power flow (initially - neglecting subsequent cumulative effect of radiant imbalance as per [1] above) is invariant, but there is length contraction of A in S by factor γ, so radiating surface area is reduced. Hence power density is boosted, and that directly implies a boosted temperature of A seen in S. However this cannot be a simple γ-factor affair as geometry of A strongly enters - for a given γ factor, change in net surface area of an already squat-shaped A (say a thin disc) is much less affected than say for a slender rod shape.
     
    Last edited: Oct 18, 2012
  12. Oct 18, 2012 #11
    Wouldn't the color temperature of the radiation of the hot body be blue shifted if it is approaching you and red shifted if receding from you? So an approaching object would appear hotter and a receding object appear colder? Isn't it that simple?
     
  13. Oct 18, 2012 #12
    Simple relative to what question? Sure there is the Doppler type effect you mention, but is that what OP is meaning to ask? Honestly I cannot quite make heads or tails of #1 - friction and relativity are mixed up in some kind of 'relative motion' soup. Maybe Stu21 should clarify quick-smart. Anyway, the subsequent focus seems to be on how to relate/define temperature between frames, and that entails some concept of averaging over an angular distribution. I have made an additional edit in #10 btw.
     
  14. Oct 18, 2012 #13

    pervect

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    The formulations I've seen associate entropy with the matter that the light is emitted from, and the temperature as well, rather than the light.

    There might be a way of regarding the light as having it's own entropy of it's own, but I"m not aware of the details of such any such formulation.
     
  15. Oct 19, 2012 #14

    tom.stoer

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    Yes.

    But we are not talking about the doppler effect for the frequencies but about the temperature. There are two transformations, namely

    [tex]T_0\;\stackrel{\beta}{\longrightarrow}\;T_\beta[/tex]

    and

    [tex]\nu_0 \;\stackrel{\beta}{\longrightarrow}\;\nu_\beta[/tex]

    The first one is the on which bothers us. Once we have solved this problem the next step would be to derive a new radiation formula (Planck's formula transformed, but in order to do that we already need the transformed temperatrure) and apply the second transformation.
     
  16. Oct 19, 2012 #15

    DrDu

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    From a phenomenolocigal point of view, temperature (not necessarily absolute temperature) is defined for systems being in thermal equilibrium. As two objects of different speed cannot be in thermal equilibrium relativistically, it is not so clear how to extend temperature to relativistic situations. Specifically, consider an object flying in a black body cavity. The thermal radiation will not appear isotropic to an observer on the bullet and will transfer momentum so as to slow down the bullet.
     
  17. Oct 19, 2012 #16
    Something already pointed out in #10. I should add a correction here to the indirect inference in #10 that a circular hoop of matter spinning about it's major axis would not spin down owing to spectral angular bias. In fact there is no need for leading and trailing surfaces in order for a net momentum/angular momentum transfer. Annular surface of a spinning hoop is perfectly capable of emitting and receiving a relativistically biased radiant flux having net component along direction of local relative motion. Resulting retarding force will simply be reduced considerably over that for an otherwise equivalent body of compact shape, because SR bias of the locally Lambertian radiant distribution will be greater when leading and trailing surfaces exist.
     
  18. Oct 20, 2012 #17
    As I recall, temperature can be expressed as a function of the average 3D velocities. Then we can use the fact that only one of the three velocity components is changed by relative motion. If we make this change in the T(Vavg), we should get a relative temperature formula, right?
     
  19. Oct 20, 2012 #18

    tom.stoer

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    This should be correct for a rather specific system, namely when we look at an (ideal) gas. For other systems this will not work. And then there's another problem, namely the fact that the Maxwell-Boltzmann distribution of velocities uses Newtonian mechanics. So this will not work in general

    Have a look at http://en.wikipedia.org/wiki/Maxwell–Boltzmann_distribution#Distribution_for_relativistic_speeds
     
  20. Oct 20, 2012 #19

    pervect

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    If you use the statistical mechanics definition of entropy, the entropy of light would be related to the density of states by

    [tex]
    S = -k \sum_{i} P_i ln P_i
    [/tex]

    See for instance http://en.wikipedia.org/w/index.php?title=Entropy&oldid=518903651

    Is energy per unit photon a "good enough" defintion of state of a photon for the counting we need to define entropy? I don't really know. It should be close-ish.

    Temperature is generally related to entropy by T = dS/dQ, where ds is the change in entropy and dQ is the change in energy.

    So I would predict, semi-confidently (thermo isn't really my thing) that the entropy of a monocrhomatic laser beam would be zero or low, and since Q is fairly high, the temperature of a laser beam would be low.

    The fact that lasers are routinely used to cool things down to ultra-low temperatures makes me think that this idea is probably true.

    http://en.wikipedia.org/w/index.php?title=Laser_cooling&oldid=510230895

    IF laser beams were "hot", interacting with them wouldn't be able to cool down atoms and molecules.

    So you can see the "red-shift, blue-shift" argument doesn't necessarily have much to do with the temperature of a particular distribution of frequencies.

    You should probably try the thermo forum first to see if the concept of the "temperature" of a distribution of light makes sense, and some idea of how it's calculated non-relativistically first, before moving on to relativity.
     
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