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Zeroth law of thermodynamics

  1. Nov 29, 2015 #1
    if two bodies are in thermal equilibrium in one frame, will they be in thermal equilibrium in all frames? also, does the temperature of a body depend on the frame from which it is observed?
     
  2. jcsd
  3. Nov 29, 2015 #2
    It's exactly the transitive property of mathematics. If a=b and b=c, then a=c.

    The temperature is defined as the average kinetic energy of the particles. The frame is taken to be one instance of measurement. The relative temperatures depend on the equilibrium state of the system. Is it closed or open? Is the ambient temperature ~close to the temperature of the system of particles?
     
  4. Nov 29, 2015 #3
    Temperature is the macroscopic performance of the motion of molecules, so I believe it may rely on the frame the observers stay, but I'm not sure if the zeroth law of thermodynamics still works anytime...
     
  5. Nov 29, 2015 #4
    Yes temperature is calculated as a macroscopic quantity but there are microscopic corollaries hence statistical mechanics. The zeroth law should hold as often as the 1st,2nd,3rd laws do. The rotation and velocity of particles in their micro and macro configurations are determined in probability by the partition function.
     
  6. Nov 29, 2015 #5

    mfb

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    It is not.

    Temperature in special relativity is actually a tricky concept, and there is more than one approach to define it. Thermal equilibrium of co-moving objects (zero relative velocity) is the same in all reference frames, however.
     
  7. Nov 29, 2015 #6

    Vanadium 50

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    Geez, there's a lot of misinformation in this thread, and it's only a couple hours old.

    Mfb is right. Relativistic thermodynamics is tricky. Classically, there are multiple ways to define thermodynamic quantities, and they are equivalent. Relativistically, they are no longer equivalent.

    That said, this question has an answer. If I have an extended object at uniform but not constant temperature T(t), we would say it's in thermal equilibrium with itself (but not the outside world). If it is moving, Relativity of Simultaneity implies that it is no longer at a uniform temperature, and we would not say it's in thermal equilibrium with itself.
     
  8. Nov 29, 2015 #7

    vanhees71

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    It is very important to be precise here, particularly if it comes to thermodynamics in the relativistic realm. Temperature is by definition a scalar (or more precisely a scalar field if you generalize it to local thermal equilibrium). Historically, this was not always the case. That's why you find in the older literature other ideas about the thermodynamic quantities. Here, I refer to the modern definition. I'm not sure, when it was precisely established. A lot has been done by van Kampen. I can only recommend to learn the modern definition, because it is pretty confusing in the old way (the same holds true for various old-fashioned concepts about relativistic mass or even transverse and longitudinal mass).

    Temperature is defined in the local rest frame of the fluid, i.e., it is a measure for thermal energy in a local inertial frame of reference, where the fluid element where the temperature is measured is at rest (or it's total three-momentum vanishes).

    This also implies that the phase-space distribution functions in statistical mechanics are scalars (which is not so trivial, because one has to define the single-particle phase-space volume element ##\mathrm{d}^3 \vec{x} \mathrm{d}^3 \vec{p}## in a covariant way).

    In local thermal equilibrium the phase-space distribution function takes the manifestly covariant form
    $$f(t,\vec{x},\vec{p})=\frac{g}{\exp[\beta(x) [p \cdot u(x)-\mu(x)]] \pm 1}.$$
    Here ##\beta(x)=1/k_{\text{B}} T(x)## is the inverse temperature field, ##u(x)## the flow-velocity-four-vector field, and ##\mu(x)## the chemical potential of some conserved charge, if applicable, and ##g## a degeneracy factor for intrinsic quantum numbers of the constituents like spin, flavor, color,... The energy is determined by the on-shell condition, i.e., ##p^0=E(\vec{p})=\sqrt{m^2 c^4+\vec{p}^2 c^2}##. The upper (lower) sign is for fermions (bosons).

    The particle-number four-density is given by
    $$n^{\mu}(x)=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi \hbar)^3} \frac{1}{E} f(x,\vec{p}).$$
    It is immideately clear that this is a four-vector, because ##f## is a scalar field and ##\mathrm{d}^3 \vec{p}/E## is a scalar too.

    For details, see my lecture notes on relativistic kinetic theory:

    http://fias.uni-frankfurt.de/~hees/publ/kolkata.pdf
     
  9. Dec 1, 2015 #8
    Thank you everyone for clearing this doubt, thanks for the generous help. :)
     
  10. Dec 1, 2015 #9
    True. I agree to this.
     
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