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I am studying the mathematical proof of existence of empirical temperature online. I have studied the same in the post Proving the existence of temperature from zeroth law in the MIT OCW notes .

Zeroth law states that if A and C are in joint (mutual) equilibrium and simultaneously B and C are in joint equilibrium then A and B are also in joint equilibrium.

I have tried to write the proof properly with the help of above references. But I have doubts in various steps of proof.

Consider three systems $A,\;B$ and $C$ described by macroscopic coordinates $(A_1,...,A_n),\;(B_1,...B_k),\;(C_1,...,C_m)$. If A and C are in joint equilibrium then $f_{AC}(A_1,...,A_n,C_1,...,C_m)=0\tag{1}$
Also if B and C are in joint equilibrium then $f_{BC}(B_1,...,B_k,C_1,...,C_m)=0\tag{2}$
$\text{From}\;(1)$, $\implies\;C_1=F_{AC}(A_1,...,A_n,C_2,...,C_m)\tag{3}$ $\text{Also from}\;(2),$ $\implies\;C_1=F_{BC}(B_1,...,B_k,C_2,...,C_m)\tag{4}$
As A and B are simultaneously in equilibrium with C, so from (3) and (4), $F_{AC}(A_1,...,A_n,C_2,...,C_m)=F_{BC}(B_1,...,B_k,C_2,...,C_m)\tag{5}$

As a result of zeroth law, A and B are also in joint equilibrium, so
$f_{AB}(A_1,...,A_n,B_1,...,B_k)=0\tag{6}$
So, equation (6) is independent of $C_i's$, so whatever is the value of $C_i's$, for some particular $A_i's$ and $B_i's$, equation (6) so also (5) continues to hold.
So by fixing $C_i's$ to $C_i^o$ in equation (5) $\forall\;i\in{\{2,...,m}\}$, define
$\Phi_A(A_1,...,A_n)\triangleq F_{AC}(A_1,...,A_n,C_2^o,...,C_m^o)\;\text{and}\\ \Phi_B(B_1,...,B_k)\triangleq F_{BC}(B_1,...,B_k,C_2^o,...,C_m^o)\\ C_1^o=\Phi_C(C_2^o,...,C_m^o)\\ \text{As system C is also in equilibrium with itself(reflexivity), so $C_1^o$ is determined uniquely}\tag{7}$

So, $\Phi_A(A_1,...,A_n)=\Phi_B(B_1,...,B_k)=\Phi_C(C_2^o,...,C_m^o)\tag{8}$

That empirical temperature is defined as $\Phi(A_1,...,A_n)$ or $\Phi_B(B_1,...,B_k)$ or $\Phi_C(C_2^o,...,C_m^o)$

I have the following doubts while writing the proof -

  1. In (3), we are considering $f_{AC}$ and $f_{BC}$ to be an explicit function in $C_1$. So, how we justify it for any general thermodynamic system, like the function $f_{AC}$ and $f_{BC}$ can be implicit in $C_1$ like we can't separate it like the way shown. So, in (3), are we considering only those macroscopic coordinates of C which are explicit or can be separated from $f_{AC}$ and $f_{BC}$? Or there is some different reasoning?

  2. The empirical temperature is basically $\Phi(A_1,...,A_n)$ which has the same unit as that of macroscopic coordinate $C_1$. If there are more than one separable macroscopic coordinates of C in $f_{AC}$ and $f_{BC}$ (like $C_2, C_3$ etc in addition to $C_1$) then can we take out any of these macroscopic coordinates from $f_{AC}$ and $f_{BC}$ and make the equation like (3) and (4) with $C_2$ (say) in their $L.H.S.$? And our unit or nature of empirical temperature then changes as $C_2, ...,C_k$ all have different units? But I think that doesn't happen. Please clarify this doubt

  3. In books, equation (8) is written as $\Phi_A(A_1,...,A_n)=\Phi_B(B_1,...,B_k)=\Phi_C(C_1,C_2,...,C_m) $. Why $C_1$ is included in $\Phi_C$ like it gets separated from $f_{AC}$ and $f_{BC}$. So shouldn't $\Phi_C$ be like $\Phi_C(C_1,C_2,...,C_m)$?

Please help me in clarifying the doubt, it takes somewhat long time for me to understand the proof of empirical temperature. But I am not able to clarify my above doubts. I am very very confused. Please help.
[Edit- Have fixed the equation (1) and (2) as per comment]

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  • $\begingroup$ Your equation (1) is already making an assumption beyond the argument in those lecture notes - namely that each system has a function $f$ such that $f_A = f_B$ when $A$ and $B$ are in equilibrium. I also don't know what you mean by saying that $f_{AB}$ could be implicitly a function of $C_1$, which is an independent thermodynamical variable which defines the state of system $C$. Through what other quantity could it implicitly enter? $\endgroup$
    – J. Murray
    Commented Mar 7, 2021 at 2:38
  • $\begingroup$ @J.Murray, in class we have been told that $f_A=f_B$ is equivalent to $f_{AB}=0$. May you please tell whether the above statement is true or not. By "$f_{AC}$ could be implicitly a function of $C_1$", I mean that it can be the case that $C_1$ can't be separated from $f_{AC}$ for any general thermodynamic system and we won't get equation (3). Ex- $x^2+y^2+x+y=0$, In this equation we can't separate out x, to forma a equation like f(y)=x? That s my question. Please help. $\endgroup$
    – Manu
    Commented Mar 7, 2021 at 4:13

1 Answer 1

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By "$f_{AC}$ could be implicitly a function of $C_1$", I mean that it can be the case that $C_1$ can't be separated from $f_{AC}$ for any general thermodynamic system and we won't get equation (3). Ex- $x^2+y^2+x+y=0$, In this equation we can't separate out $x$, to form an equation like $f(y)=x$?

In your example, $f(x,y)=x^2+y^2+x+y$ is an explicit function of both $x$ and $y$. You are concerned that we cannot rewrite this as $y(x)=\ldots$, which is the subject of the implicit function theorem. In this case, it says that given some point $(x_0,y_0)$, we can rewrite your expression as $y(x)=\ldots$ locally (that is, in a small neighborhood of $(x_0,y_0)$) if $\partial f/\partial y \neq 0$.

In your case, the singular points are those such that $\partial f/\partial y = 2y +1 = 0 \implies y=-1/2$. If we consider the point e.g. $(-1,0)$ (which lies on the curve), then it should be possible to find $y$ as an explicit function of $x$ in a neighborhood of this point. It is indeed possible; note that $$f(x,y)=x^2+y^2+x+y=\left(x+\frac{1}{2}\right)^2 + \left(y+\frac{1}{2}\right)^2 - \frac{1}{2}=0$$ $$\implies y=-\frac{1}{2} \pm \sqrt{\frac{1}{2}-\left(x+\frac{1}{2}\right)^2}$$ Because we are looking for a function in a neighborhood of $(-1,0)$, we should choose the positive branch (this yields $y(-1)=0$); therefore we have that

$$y(x) = -\frac{1}{2} + \sqrt{\frac{1}{2}-\left(x+\frac{1}{2}\right)^2}$$

This function is well-defined for $x\in [-\frac{1}{2}-\frac{1}{\sqrt{2}},-\frac{1}{2}+\frac{1}{\sqrt{2}}]$.

And our unit or nature of empirical temperature then changes as C2,...,Ck all have different units?

This argument isn't concerned with units. The claim is that there exists some functions $\Phi_A,\Phi_B,\Phi_C$ such that $\Phi_A=\Phi_B\iff A$ and $B$ are in thermal equilibrium. If you need to scale one of those functions to give them dimensions of temperature, then you can of course do so.

Why $C_1$ is included in $\Phi_C$ like it gets separated from $f_{AC}$ and $f_{BC}$?

I'm afraid I don't understand this question. You should look at the examples given in the lecture notes and apply them to an actual system; that should help clarify the argument.

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