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:


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


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


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


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.

  • $\begingroup$ 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 '18 at 7:56

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.

| cite | improve this answer | |
  • $\begingroup$ 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. $\endgroup$ – Andrew Steane Sep 5 '19 at 11:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.