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Relativity and temperature question

  1. Feb 17, 2005 #1
    Since the other thread got moved for a reason that I do not understand, let me try to ask a question which belongs in this category.

    Suppose that the temperature of some object is T(t), where t is the time coordinate in the objects rest frame. Now, in reality nothing has a perfectly constant temperature, rather there are tiny fluctuations about some average temperature Te.

    Suppose the temperature of some object in its rest frame is given by

    [tex] T(t) = T_e + T_0 sin(\omega t) [/tex]

    Where Te is the average temperature of the object, and T0 is the maximum amplitude of the temperature flux, and [tex] \frac{dT_0}{dt} = 0 [/tex], and [tex] \frac{d\omega}{dt} = 0 [/tex]

    What does relativity say that the temperature is in a reference frame moving at a relative speed of v?

    Regards,

    Guru
     
    Last edited: Feb 17, 2005
  2. jcsd
  3. Feb 17, 2005 #2

    dextercioby

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    Ignore relativity:what does that [itex] T=T(t) [/itex] mean,PHYSICALLY SPEAKING...?

    Daniel.
     
  4. Feb 17, 2005 #3
    That the temperature is a function of time.

    Regards,

    Guru
     
  5. Feb 17, 2005 #4

    ZapperZ

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    Considering that a bunch of your threads got moved to the TD section, you appear to be a very slow learner, or you simply refuse to learn.

    Can you point out one, JUST ONE, example of such time variation of temperature that is in a thermal equilibrium? And if you point out that this occurs in a superconductor once again via a supposition, I will repeat my recommendation.

    Zz.
     
  6. Feb 17, 2005 #5

    dextercioby

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    Sorry,this the mathematical interpretation... :wink: PHYSICS,please...

    Daniel.
     
  7. Feb 17, 2005 #6
    I want to watch someone do a Lorentz transformation on a very simple formula.
     
  8. Feb 17, 2005 #7

    russ_watters

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    Is this another one of those 'if time dilation slows down motion, doesn't that decrease the temperature of an object?' misunderstandings? Hint: Temperature is measured in your frame. Time dilation is something that happens in the other guy's frame. Time dilation does not affect temperature. (measurements of radiated energy of moving objects, however...)
     
  9. Feb 17, 2005 #8
    Mathematically to say that something is a function of time t, means many things, the main one being that if we differentiate that function with respect to t, that we will get a formula, rather than zero.

    Physically, to say that temperature is a function of time, would be to take out statistical analysis, and say that there is some precise reason why the temperature must fluctuate. Is that the kind of answer you were looking for?

    Regards,

    Guru
     
  10. Feb 17, 2005 #9

    dextercioby

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    And to what kind of "fluctuations" would you refer to...?

    Daniel.
     
  11. Feb 17, 2005 #10

    ZapperZ

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    Fine. However, if you start invoking any "magical" superconductivity stuff without bothering to actually learn what it is, you can bet that I'll be all over it like a cheap suit.

    Zz.
     
  12. Feb 17, 2005 #11
    Well, a body can emit photons, and absorb photons. Those photons either carry away energy (in the form of heat) or supply energy (in the form of increased speed of electrons). So to say that there is a "temperature" fluctuation, it suffices to say that there is a "thermal energy" fluctuation.

    Regards,

    Guru
     
  13. Feb 17, 2005 #12

    dextercioby

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    So what chapter of physics could account for your formula...?

    Assume i know nothing and i ask you what discipline of (more or less) modern physics explains [itex] T=T_{equil}+Amplit\cdot\sin\omega t [/itex]

    Daniel.

    P.S.And promiss Zapper you won't bring superconductivity into discussion...
     
  14. Feb 17, 2005 #13

    russ_watters

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    No, see, since electrons are moving "fast," they get cold due to time dilation, thus: superconductivity!
     
  15. Feb 17, 2005 #14
    Quantum mechanics covers the emission and absorption of photons from a hydrogen atom. For atoms of higher atomic number, the mathematics is too complex to find a precise formula for the wavefunction. This would be the 'modern physics' answer.

    According to quantum physics, a photon has an energy given by:

    [tex] E = hf [/tex]

    Where f is the frequency of a photon, and h is planck's constant, which is presumed to be a fundamental constant of nature. In other words, h is NOT a function of time, so that dh/dt=0.

    In quantum mechanics, the frequency f of a photon is proportional to omega we have:

    [tex] \omega = 2\pi f [/tex]

    Which appears in the quantum mechanical wavefunction:

    [tex] \psi (x,y,z,t) = e^{i(k\vec r - \omega t)} [/tex]


    Less modern, would be to try and use thermodynamics/statistical mechanics, founded by Boltzmann in the late 1800's. One could also try to derive h from classical electromagnetism, thermodynamics, and statistical mechanics, by trying to connect stephan's constant [tex] \sigma [/tex] to quantum mechanics, and electrodynamics, but that would be difficult, if not impossible.

    Regards,

    Guru
     
    Last edited: Feb 17, 2005
  16. Feb 17, 2005 #15

    pervect

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    I've always seen temperature defined in the rest frame of the fluid when speaking relativistically.

    For an ideal gas/ideal fluid, temperature is (physically), proportional to the kinetic energy of the particles in the fluid in the fluid's rest frame, as dexter was hinting.

    For more complex fluids such as actual gasses, one has to consider factors such as rotational and vibrational degrees of freedom -- every degree of freedom has an energy of kT/2 because of the equipartition theorem.
     
  17. Feb 17, 2005 #16

    dextercioby

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    Incorrect.

    An incorrect answer... :rolleyes:

    Yes.

    Incorrect.

    [tex] \omega = 2\pi f [/tex]

    Incorrect.


    Why would you say that...?

    "h" cannot be derived."h" is postulated.Period.

    Daniel.
     
  18. Feb 17, 2005 #17
    Then by all means, suggest to me your corrections, and lets see if i correct them or not.

    Kind regards,

    Guru
     
  19. Feb 17, 2005 #18
    Hi all, im slightly interested in this question so I thought I'd jump aboard.
    Questions I have before further thought on this:
    By temperature do you mean the moving objects rate at which it emits radiation?
    Or do you mean it has a temperature that, if it weren't moving, would remain constant. It emits nothing?

    Are you asking what the emmitted radiation would do under Lorentzian transformation?

    Why does temperature have to be a function of time? Can it not be constant?

    Are you trying to imply that the rate of change of temperature(emmitted radiation) is a function of velocity?

    Thanks.
     
  20. Feb 17, 2005 #19
    Mmm hmm, U = 3/2 kT, in the rest frame of the fluid, using the Mawellian speed distribution formula. The derivation of the formula quite long, but rather interesting.

    Regards,

    Guru
     
  21. Feb 17, 2005 #20

    dextercioby

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    The reason why i put that "incorrect" word after almost every phrase of yours is because you don't seem to grasp the object of quantum mechanics...Specifically,you do not know that in the field of quantum mechanics,the word photon doesn't have any meaning...And Planck's constant is postulated...

    Do you insinuate that,by performing a LT over the formula giving internal energy vs.equilibrium temperature,the time dependence of temperature would become explicit...?

    Daniel.
     
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