Entropy equation in Pathria

Question

In the first chapter of Pathria's statistical mechanics book, he shows that for two systems in thermal equilibrium, $$ \left( \frac{\partial \ln \Omega}{\partial E} \right)_{N,V} \equiv \beta$$ is the same for both. Also, from the first law of thermodynamics, their temperature which follows the equation $$\left( \frac{\partial S}{\partial E} \right)_{N,V} = \frac{1}{T}$$ is the same for both subsystems. He then says that from these two equations we can conclude that "for any physical system" $$\frac{\Delta S}{\Delta (\ln \Omega)} = \frac{1}{\beta T} = \textrm{constant}$$

I have two related questions here:

1. Why did he write it like that instead of $$\left( \frac{\partial S}{\partial (\ln \Omega)} \right)_{N,V} = \frac{1}{\beta T}$$
2. How does he know that it's a constant? At this stage, all we can tell is that $\beta$ can be a function of temperature because it's the same for both systems at thermal equilibrium, so $1/ \beta T$ should also be a function of temperature.

Answer 1

$\DeclareMathOperator{\D}{d\!}$

When two or more systems are in thermal equilibrium they necessarily share the same value of $\beta$ and $T$, yet they generally share no other state function (i.e., each system can have its own $E$, $\Omega$, $V$, $N$, composition, etc). Thus $\beta(T)$ is an invertible function of $T$ and only $T$. As you state, it follows that

\begin{align} k = \frac{1}{\beta(T) T} \end{align}

likewise can be a function of only $T$ or a constant.

From your last equation

\begin{align} \left( \frac{\partial S}{\partial \ln \Omega} \right)_{N,V} = k \end{align}

and the fact that only three state functions are necessary to completely define the state of a microcanonical system, we have, at constant $N$ and $V$,

\begin{align} \D S &= k \D \ln\Omega \end{align}

Note that both $\D S$ and $\D \ln\Omega$ are exact differentials when $N$ and $V$ are held constant, meaning that $S(\Omega)$ is only a function of $\Omega$. Thus $k$ must be either a function of $\Omega$ or a constant; $k$ cannot be a function of $T$. But $k$ cannot be a function of any state variable other than $T$, that is

\begin{align} k(T,\Omega, N, V)=k(T) \end{align}

and hence $k$ must be a constant, in fact a universal constant that is independent of state.

For question 1, both $S$ and $\ln \Omega$ are state functions so the equation with deltas is also valid (at least at constant $N$ and $V$ --- to show it is generally valid requires the third law).

Answer 2

By fundamental derivative definition, $$ \tag 1 \lim_{\Delta x \to 0} \frac {\Delta f(x)}{\Delta x} = \frac {df(x)}{d x} $$

So when $\Delta x \approx 0$, then $$ \tag 2 \frac {\Delta f(x)}{\Delta x} \approx \frac {df(x)}{d x} $$

i.e. then they are sort of interchangeable. But in general, the meaning of derivative and function change is different,- while $$\tag 3 \left( \frac{\partial S}{\partial E} \right)_{N,V} = \frac 1T$$ gives instantaneous temperature inverse of a system,

$$\tag 4 \left( \frac{\Delta S}{\Delta E} \right)_{N,V} = \left \langle \frac {1}{ T}\right \rangle$$ gives an average temperature inverse of a system.

Now if for a systems in thermal equilibrium $1/T = \text{const}$, it follows that average temperature inverse is a constant too, i.e. $\left \langle 1/ T\right \rangle = \text{const}$.

(By analogy, if car's instantaneous speed is constant, then it's average speed is also a constant.)