I am studying Mehran Kardar's Statistical Physics of Particles. In chapter-1, they derive the existence of empirical temperature using Zeroth Law.
It is given as

Definition- If two systems, A and B, are separately in equilibrium with a third system, C, then they are also in equilibrium with one another.
Let the equilibrium state of systems $A,\;B,\;and\;C$ be described by the coordinates ${\{A1,A2,···,A_n}\},\;{\{B1,B2,··· ,B_n}\},\;and\;{\{C1,C2,··· ,C_n}\}$, respectively. The assumption that $A$ and $C$ are in equilibrium implies a constraint between the coordinates of A and C, that is, a change in $A_1$ must be accompanied by some changes in ${\{A2,··· ,C1,C2,···,C_n}\}$ to maintain equilibrium of A and C.
So due to equilibrium of $A$ and $C$,
and due to equilibrium of $B$ and $C$,
This is equivalent to solving each of the above equations for C1 to yield
Thus if C is separately in equilibrium with A and B, we must have
However, according to the zeroth law there is also equilibrium between A and B, implying the constraint
We can select any set of parameters ${\{A,B}\}$ that satisfy the above equation, and substitute them in $Eq. 5$. The resulting equality must hold quite independently of any set of variables ${\{C}\}$ in this equation. We can then change these parameters, moving along the manifold constrained by $Eq.6$, and $Eq.5$ will remain valid irrespective of the state of $C$. Therefore, it must be possible to simplify $Eq.5$ by canceling the coordinates of $C$. Alternatively, we can select any fixed set of parameters $C$, and ignore them henceforth, reducing the condition $Eq.6$ for equilibrium of $A$ and $B$ to

I am not able to understand the last paragraph of the proof.
How the equality in $Eq.5$ holds irrespective of $C's$. The line "We can then change these parameters, moving along the manifold constrained by Eq.6, and Eq.5 will remain valid irrespective of the state of C. Therefore, it must be possible to simplify Eq.5 by canceling the coordinates of C", is not clear to me.

In all books, like Adkins' Equilibrium Thermodynamics, similar argument that $Eq.5$ holds irrespective of $C's$, so there is a function $\Phi$, is used.

But I am not able to understand this mathematically, what is going on in this argument.
Please help me in understamding it.


1 Answer 1


Yes, the argument is a bit difficult to penetrate. I will try to reformulate it here.

We consider all three systems at equilibrium with each other, i.e. $A\sim B$, $A\sim C$ and $B\sim C$. Since $A\sim B$, any equilibrium state of $C$ in the $AC$ system ($C\sim A$), must also be an equilibrium state of $C$ in the $BC$ system due to the transitivity of equilibrium. This is equivalent to saying, that if $A\sim B$ and $C\sim A$, then, for any fixed $(C_2, C_3,..., C_n)$, equations (1) and (2) must hold for the same $C_1$, because if they didn't, you would have $A\sim B$ and $A\sim (C_1,...,C_N)$ but $B\nsim (C_1,...,C_N)$, which would break the transitivity. The last statement is equivalent to equating the right parts of equations (3) and (4), leading to equation (5).

Putting it all together, we get that if $A\sim B$, then for any choice $(C_2, C_3,..., C_n)$ equation (5) must hold.

I hope it is more clear this way.

  • 1
    $\begingroup$ Since the equation (5) holds for any $(C_2,.., C_N)$, we can put them all to zero and call $F(A_1,...A_N, 0,....,0)$ as $\Phi_A(A_1,...A_N)$ (and the same for the system $B$). $\endgroup$
    – Pavlo. B.
    Apr 26, 2021 at 18:06
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    $\begingroup$ Yes, the choice is arbitrary, but the argument doesn't prove that $\Phi$ is temperature. The argument essentially proves that you can introduce such "equivalence function" $\Phi(A)$, that all states $A$ satisfying the equilibrium equation $\Phi(A)=\text{const}$ are going to be in equilibrium with each other. If you choose different $C_i$, the function $\Phi$ is going to change $\Phi_1\rightarrow \Phi_x$, but the hypersurfaces of constant values of $\Phi$ are not going to change, since your system is the same. $\endgroup$
    – Pavlo. B.
    Apr 26, 2021 at 18:46
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    $\begingroup$ The only way this can happen, is if the new $\Phi_x$ is some function of the old $\Phi$, i.e. $\Phi_x(A)=f(\Phi_1(A))$. $\endgroup$
    – Pavlo. B.
    Apr 26, 2021 at 18:50
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    $\begingroup$ Yes, the "existence of temperature" usually means that for any systems $A$, $B$ and $C$ there exist such functions of state $T_A(A)$, $T_B(B)$ and $T_C(C)$, such that equality between any two such functions implies thermal equilibrium between corresponding systems. The functions are clearly not unique, since if you applied the same transformation to all of them (like, took a square of each of them), the equality of new temperatures would still imply thermal equilibrium of systems. $\endgroup$
    – Pavlo. B.
    Apr 27, 2021 at 5:08
  • 1
    $\begingroup$ Thank you so much for the discussion. The discussion helps me in understanding the proof clearly. $\endgroup$
    – Iti
    Apr 27, 2021 at 5:17

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