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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.
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Brain Stroke Patient
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  • Does he not prove $\beta=kT$ somewhere in there? – Jahan Claes Nov 20 '22 at 16:57
  • @JahanClaes No he doesn't – Brain Stroke Patient Nov 20 '22 at 17:10
  • Your second question answers the first question. If it is a constant, then the stronger $\Delta$ condition can be used. – Jbag1212 Nov 20 '22 at 18:45
  • Did you have a look at this related question? https://physics.stackexchange.com/questions/269538/proving-that-the-boltzmann-entropy-is-equal-to-the-thermodynamic-entropy?rq=1 Within the constraint on closed systems, the accepted answer is the correct one, and adresses your question n.2. About n.1, it is impossible to give a precise answer based on the book. If you have problems with that answer or if you wonder how to deal with the $N$ dependence, you may edit your question or ask another question. – GiorgioP-DoomsdayClockIsAt-90 Nov 20 '22 at 18:48
  • @GiorgioP Yes I saw that answer. I understand the accepted answer, but that approach is different from the one I have a question with. What I want to know is whether we can conclude $\beta T =$ constant based on what Pathria has told us, or do we need more analysis like in the accepted answer. The accepted answer also gives the equation in terms of total differentials which is an important step in concluding that $\beta T$ is constant, something that is also connected to my first question since we're not supposed to get a total derivative relation from the two partial derivative relations. – Brain Stroke Patient Nov 20 '22 at 19:48
  • @Jbag1212 I don't quite follow. I'm thinking of a counterargument like this: suppose $f = x + y + z$, then $(\partial f/ \partial x)_{y,z} =$ constant but $\Delta f / \Delta x \neq $ constant. So the partial derivative being constant shouldn't automatically imply the stronger $\Delta$ condition. – Brain Stroke Patient Nov 20 '22 at 19:53
  • Sure, but in the OP don’t you also have the assumption that the partial derivative is only a function of $T$? You say that $\beta$ is a function of temperature – Jbag1212 Nov 20 '22 at 21:18
  • Yes. It should be a function of temperature. – Brain Stroke Patient Nov 20 '22 at 21:35

1 Answers1

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).

ratsalad
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