# Doubts in the proof of existence of empirical temperature as a result of zeroth law of thermodynamics

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]

• 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? Commented Mar 7, 2021 at 2:38
• @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.
– Manu
Commented Mar 7, 2021 at 4:13

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.