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Thermodynamics and intertial frame

  1. Sep 15, 2014 #1
    Is thermodynamics states like temperature, pressure and material properties like specific heat, thermal conductivity etc. invariant with inertial reference frame?.
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  3. Sep 15, 2014 #2

    Vanadium 50

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    There is no generally accepted version of relativistic thermodynamics. Classically, there are several equivalent ways tp define temperature. Relativistically, they are no longer the same.
  4. Sep 15, 2014 #3
    Yes, classily they are invariant with respect to the motion of the observer.

  5. Sep 17, 2014 #4
    Then sqrt(gamma*R*T) should be invariant with respect to inertial frame. (At least for speed in order of 100). Then can I claim speed of sound is invariant in all inertial frame of reference. I expect reply to be NO, but why it is no?
  6. Sep 17, 2014 #5


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    I think temperature is Lorentz invariant. Invariance under spatial rotation is trivial.About the boosts, when you change coordinates from one inertial one to another, you're actually changing the energy of each particle by a constant amount which doesn't contribute to the temperature.
    But pressure doesn't seem to be Lorentz invariant to me. Because it depends on the average number of particles hitting the container's wall in a given amount of time. That definition has a concept of simultaneity hidden in it! and in addition it depends on time and space intervals which are not Lorentz invariant.
    About material properties, I'm not sure but I think they're Lorentz invariant.
    I should also say that its not trivial to say [itex] \sqrt{\gamma R T} [/itex] is Lorentz invariant.
    As far as I remember, people replace things with other things in that formula to get speed of sound in terms of different things. So you should go to the actual derivation of this formula and see whether there is an assumption of taking a particular reference frame or not in between of the derivation. Because people aren't careful about SR when they talk about things as classical as speed of sound!!!
    Last edited: Sep 17, 2014
  7. Sep 17, 2014 #6
    The speed of sound is a quantity that is inherently derived for a specific dynamic problem that assumes that the gas is stationary relative to the observer, and the wave is traveling through the gas. If the observer were moving at the speed of sound relative to the gas when he passed the point at which a pulse was applied, he would observe that the wave would be stationary, but the gas would be moving backwards. This would not be consistent with the premise that the observer is stationary relative to the gas. So the speed of sound is the solution to a dynamic problem with a specific set of boundary conditions. On the other hand, transport and intensive thermodynamic properties such as temperature, pressure, heat capacity, thermal conductivity, and viscosity are more fundamental, and are not tied to the solution of any specific dynalmic problem.

  8. Sep 17, 2014 #7


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    In my opinion, the main problem is that two thermodynamic systems which move at different speed cannot be in thermal equilibrium as e.g. their blackbody radiation spectrum will be blue or redshifted which will tend to equalize their velocity. So it is not possible to define relative temperature via the zeroth law for systems of different speed.
  9. Sep 17, 2014 #8
    In my opinion entropy is scalar under Lorentz transformation.
    Energy has coefficient [tex]\gamma[/tex] in Lorentz transformation.
    So temperature dE/dS has coefficient [tex]\gamma[/tex] in Lorentz transformation as well.
  10. Sep 18, 2014 #9
    I may be misunderstanding the question. When the wind blows, pressure becomes a function of the wind direction and velocity. Equilibrium gaseous equations of state are invalid under non-equilibrium conditions (wind, evaporation, condensation, etc.).

    This is a common concern in the atmospheric sciences, since the global atmosphere is never in a state of equilibrium. Even small volumes of the free atmosphere are rarely (if ever) in a state of equilibrium.
  11. Sep 26, 2014 #10


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    Ok, but dE/dS isn't a scalar quantity as T is.
  12. Sep 27, 2014 #11
    Thanks. I am almost sure S is scalar but not sure T is. Could you tell me proof or reason ?
  13. Sep 27, 2014 #12


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    temperature is defined by the zeroth law of thermodynamics, i.e. the transitivity of thermal equilibrium. Two systems are in thermal equilibrium if they have the same temperature. So temperature is a scalar. But E is part of the energy momentum four vector, so if S is scalar, dE/dS can't be a scalar.
  14. Sep 27, 2014 #13
    Thanks. Scalar is number that does not change under Lorentz transformation. Two system A and B are in thermal equilibrium. In IFR #0
    In IFR #1
    How can we conclude that
    in order T to be a scalar?
  15. Sep 27, 2014 #14
    PS I consider
    [tex]T_{1A}=T_{0A},\\T_{1A}=\gamma T_{0A},\\T_{1A}=\frac{T_{0A}}{\gamma}[/tex]
    are the three major controverse ideas.
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