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This question refers to the following set of lecture notes: http://ocw.mit.edu/courses/physics/8-333-statistical-mechanics-i-statistical-mechanics-of-particles-fall-2013/lecture-notes/MIT8_333F13_Lec1.pdf

The notes state the zeroth law of thermodynamics as follows: "If two systems, A and B, are separately in equilibrium with a third system C, then they are also in equilibrium with one another." The notes then assert that despite the simplicity of the zeroth law it implies the existence of an "empirical temperature, such that systems in equilibrium are at the same temperature." To prove this assertion, the following argument is made.

Equilibrium between systems A and C implies a constraint on the thermodynamic coordinates describing A and C. Mathematically, the constraint can be given as follows:

$$f_{AC}(A_1,A_2,...;C_1,C_2,...)=0.$$

Similarly, equilibrium between B and C can be mathematically stated as follows:

$$f_{BC}(B_1,B_2,...;C_1,C_2,...)=0.$$

These equations can be solved for $C_1$ and then recast to

$$C_1=F_{AC}(A_1,A_2,...;C_2,...),$$ $$C_1=F_{BC}(B_1,B_2,...;C_2,...),$$

where the $F$ equations have a different functional form than the $f$ equations. Since C is simultaneously in equilibrium with A and B we have

$$F_{AC}(A_1,A_2,...;C_2,...)=F_{BC}(B_1,B_2,...;C_2,...).$$

The zeroth law implies that A and B are also in equilibrium with each other, therefore we must have

$$f_{AB}(A_1,A_2,...;B_1,B_2,...)=0.$$

Now we get to the part I didn't really understand. From here, the notes say "Therefore it must be possible to simplify [the second-to-last equation] by cancelling the coordinates of C. Hence, the condition [given in the previous equation] for equilibrium of A and B must be expressible as

$$\Phi_A(A_1, A_2,...)=\Phi_B(B_1, B_2,...).$$

I needed a little help with the last step. I attempted to answer my own question, which is given below, but I'm not sure the answer is correct. Namely, I'm not convinced that the decomposition I've given for my function is the most general decomposition. In fact, I'm fairly convinced that it's not the most general decomposition. However, I don't see what alternative argument can be made, so even if I'm wrong in the particulars, I do believe the general form of my argument is correct. Comments would be appreciated, though.

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  • $\begingroup$ There is not just one system C - but as many as you like. If we could not cancel out the variables of C, it would be the case that the equilibrium between A and B is dependent on the details of C - however since we claim it is true for any system C, this can not be. Hence, the equilibrium condition of A and B can be reduced to a function of system A variables being equal to a function of system B variables. $\endgroup$
    – safkan
    Jan 15, 2018 at 7:56

2 Answers 2

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The following argument is a variation of one expounded in C. J. Adkins "Equilibrium Thermodynamics".

Begin with the equation with excluded $C_1$: $$ F_{AC}(A_1,A_2,\ldots;C_2,\ldots)=F_{BC}(B_1,B_2,\ldots;C_2,\ldots). \tag{1} $$ Solve for $A_1$: $$ A_1=G_{ABC}(A_2,\ldots;B_1,B_2,\ldots;C_2,\ldots). \tag{2} $$ By the zeroth law A and B are in equilibrium: $$ f_{AB}(A_1,A_2,\ldots;B_1,B_2,\ldots)=0. \tag{3} $$ Now again, solve for $A_1$: $$ A_1=G_{AB}(A_2,\ldots;B_1,B_2,\ldots). \tag{4} $$ The last equation states that $A_1$ is determined by the $A_i$s and $B_i$s alone and in fact is independent of $C_i$s, so $G_{ABC}$ in (2) must also be independent of them. Similarly, the equation (1) is independent of $C_i$s, which means if $(A_1^0,A_2^0,\ldots;B_1^0,B_2^0,\ldots;C_2^0,\ldots)$ is any solution of (1), then so is $(A_1^0,A_2^0,\ldots;B_1^0,B_2^0,\ldots;C_2,\ldots)$ for any admissible values of $C_2,\ldots$. Thus, the values of $C_i$s in (1) may be fixed arbitrarily, say with $C_i^0$s, and the obtained equation will be equivalent to the starting one: $$ \Phi_A(A_1,A_2,\ldots)\triangleq F_{AC}(A_1,A_2,\ldots;C_2^0,\ldots),\\ \Phi_B(B_1,B_2,\ldots)\triangleq F_{BC}(B_1,B_2,\ldots;C_2^0,\ldots), $$ $$ \Phi_A(A_1,A_2,\ldots)=\Phi_B(B_1,B_2,\ldots),\tag{5} $$ as if (1) is satisfied with some $C_i$s, it's also satisfied with $C_i^0$s as has been shown.

The above derivation has a substantial flaw. The condition (5) of equilibrium between A and B was derived from (1) which assumes there is an equilibrium with some system C in some state. Moreover, it should be assumed that all but $C_1$ variables of C can be fixed at given values $C_2^0,\ldots$ while reaching an equilibrium with A and B by adjusting $C_1$, and that the corresponding value of $C_1$ is determined uniquely. Otherwise it can be questioned whether a set of variables values with $C_i$s changed to $C_i^0$ remains in the domains of $F_{AC}$ and $F_{BC}$ functions. In fact, the system C with all but one variables being fixed serve as a thermometer here, and the assumption states the existence of thermometers. Then the zeroth law ensures the consistency of their readings.

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  • $\begingroup$ An alternative prove of temperature existence can be obtained utilizing properties of the energy of a system, namely its additivity and uniqueness of distribution in parts of a system. The last claim justifies the step of expressing the energy of a part of a system from the equilibrium condition. $\endgroup$
    – Seriy
    Jul 10, 2020 at 19:02
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I think I've actually got this after some work. According to (I.4)

$$F_{AC}(A_1, A_2, ..., C_1, C_2, ...)-F_{BC}(B_1, B_2, ..., C_1, C_2, ...)=0.$$

For this to be equal to (I.5) the above needs to be true no matter what the C coordinates are. Therefore we ought to be able to decompose the above functions as

$$F_{AC}(A_1, A_2, ..., C_1, C_2, ...)=\phi_A(A_1, A_2, ...) \chi(C_1, C_2, ...) + \epsilon(C_1, C_2, ...)$$

$$F_{BC}(B_1, B_2, ..., C_1, C_2, ...)=\phi_B(B_1, B_2, ...) \chi(C_1, C_2, ...) + \epsilon(C_1, C_2, ...)$$

and therefore the first equation I've written implies $\phi_A(A_1, A_2, ...)=\phi_B(B_1, B_2, ...)$. I'm still not entirely sure that my decomposition is the most general decomposition I can write that meets all of my requirements, however I think that must be the type of argument we're looking for.

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  • $\begingroup$ This is the conclusion I too arrived at when I first met this argument years ago. I think the comment by safkan (about multiple possible systems C) is also helpful. $\endgroup$ Sep 5, 2019 at 11:09

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