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Zeroth law of thermodynamics and empirical temperature

  1. Oct 4, 2014 #1
    In these lecture notes (http://ocw.mit.edu/courses/physics/8-333-statistical-mechanics-i-statistical-mechanics-of-particles-fall-2007/lecture-notes/lec1.pdf [Broken]), they get the equation:

    [tex] F_{AC} (A_1, A_2, \ldots; C_2, C_3, \ldots) = F_{BC} (B_1, B_2, \ldots; C_2, C_3, \ldots) [/tex]

    Then they claim that the additional contraint

    [tex] f_{AB}(A_1, A_2, \ldots; B_1, B_2, \ldots) = 0 [/tex]

    means that the first equation is independent of [itex] C_i [/itex], and so there are functions [itex] \Theta_A [/itex] and [itex] \Theta_B [/itex] such that

    [tex] \Theta_A(A_1, A_2, \ldots) = \Theta_B (B_1, B_2, \ldots ) [/tex]

    Maybe I'm missing something, but the whole thing feels a little hand-wavy to me, and I'm having trouble seeing a more mathematically rigorous justification for this step. Can anyone help me fill in the gaps and justify this step in a little more detail?
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Oct 6, 2014 #2
    It's rigorous because it's impossible to contradict! Forget the Zeroth Law for a minute and consider just arbitrary functions.

    Let f1 = f1(x1, x2, x3, x4) = 0 and f2 = f2(x1, x2, x3, x4) = 0.

    Furthermore, suppose that we know whenever f1 and f2 are 0, so is some third function f3 = f3(x1, x2, x3) = 0. Then, we must be able to derive the fact that f3 = 0 from the equality f1 = f2 = 0. For this to be true, x4 must have been extraneous as far as f3 is concerned. In other words, although f1 and f2 individually depended on x4, in reducing f1 = f2 = 0 to f3 = 0, it is immaterial what the value of x4 is. It must have cancelled out in some intermediate step of the reduction.


    For a more concrete example, consider systems a, b, and c consisting of Boyle gases separated by movable walls (in order of |--c--|--a--|--b--|).

    From experience, we know when system a is in mechanical equilibrium with b, and a is also in mechanical equilibrium with c, b must be in mechanical equilibrium with c. In other words, 'being in mechanical equilibrium' is reflexive, symmetric, as well as transitive. (Search 'equivalence relation' for more on this.)

    Because of the interaction between a and b, their pressures and volumes are related: PaVa - PbVb = 0. Or more generally, f_1(Pa, Va, Pb, Vb) = 0.
    Because of the interaction between a and c, their pressures and volumes are related: PaVa - PcVc = 0. Or more generally, f_2(Pa, Va, Pc, Vc) = 0.

    Because of the transitivity of mechanical equilibrium we previously established, b and c must also be in equilibrium. Thus, we claim the two equations above must directly imply a function of the form: f_3(Pb, Vb, Pc, Vc) = 0. Furthermore, we are claiming that, because Pa and Va appear in both f_1 and f_2 yet not in f_3, this means we could have also found two simpler functions independent of Pa and Va (call them g_1(Pb, Vb) and g_2(Pc, Vc) ) whose values also must have been equal under those conditions.

    Now verify; is this true? In this case, it's trivial! PaVa - PbVb = 0 combined with PaVa - PcVc = 0 clearly implies that PbVb - PcVc = 0. As we claimed, the extraneous variables Pa and Va cancelled out. Thus, our g_1 and g_2 in this case would have been g_1 = PbVb and g_2 = PcVc and these two functions, as we claimed, were also equal at equilibrium.

    I study more thermodynamics than mathematics, but I can see why this sounds like a hand-waving argument. This is the only way I've seen this topic introduced in all the thermodynamics textbooks I have read. Perhaps a mathematician could chime in with more thoughts on the rigor of this explanation.
    Last edited: Oct 6, 2014
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