I have been studying Statistical Mechanics from the Book "Statistical Mechanics by R.K. Pathria & Paul D. Beale".
So, in chapter 1, it discusses about Statistical basis of Thermodynamics, starting with defining number of microstates, $\Omega$ of a system with $(N,V,E)$ [where $N$ = number of particles, $V$ = Volume of the system, $E$ = Total Energy of the system] to get into a condition of equilibrium for the system :
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Now, from that we found a microscopic entity called $\beta$ (defined below) which must be same for the two systems A and B in order to be in equilibrium. $$\beta = \left( \frac {\partial \ln \Omega (N,V,E)}{\partial E} \right)_{N,V,E=\bar E}$$ Now, this must be related to Thermodynamic Temperature (since that also ensures Thermodynamic Equilibrium), $T$ defined by $$\frac 1 T = \left( \frac{\partial S} {\partial E} \right)_{N,V}$$

And, the relation was established by multiplying $\frac 1 \beta$ with $\frac 1 T$ and finding:
$$ \frac {\Delta S}{\Delta \ln \Omega} = \frac 1 {\beta T} = \text{constant} = k_B $$

Now, at this point, I had no clue as to how come this parameter can be a constant (which was named after Boltzmann, and we know that is a universal constant, but how is that defined here?). There was no convincing remarks about that in the book (or maybe, I am missing an obvious fact here)

  • $\begingroup$ What relationships (if any) had been established between macroscopic thermodynamics and statistical mechanics at this point? $\endgroup$ Aug 19, 2021 at 11:35
  • $\begingroup$ Maybe I am missing something here, but for a thermodynamical system in equilibrium, to my knowledge the definition of entropy is $S=k_b\log(\Omega)$, which gives the result immediately. $\endgroup$
    – Koschi
    Aug 19, 2021 at 14:22
  • $\begingroup$ @Koschi The aim is to show $T^{-1}=\left(\tfrac{\partial S}{\partial E}\right)_{N,\,V},\,S=k_B\ln\Omega$ are equivalent definitions of $S$. $\endgroup$
    – J.G.
    Aug 19, 2021 at 14:31
  • $\begingroup$ @J.G. or OP, In the expression $T^{-1}=\left(\frac{\partial S}{\partial E}\right)_{N, V}$ how is $S$ defined? $\endgroup$
    – Jagerber48
    Aug 19, 2021 at 16:55

2 Answers 2


The difficulty in answering this question lies in there being so many different ways of presenting the basics of statistical mechanics.

If you're prepared to accept that your first two equations are essentially microscopic and macroscopic versions of the same thing, then you can see that they will remain so if we multiply both sides of the first equation by a constant, $k_\text B$. We'll give $k_\text B$ the same units as those of $S$: $$k_\text B \beta =\left(\frac {\partial (k_\text B\ln \Omega)}{\partial E}\right)_{N, V, E=\tilde E}.$$ Why have we done this? Because $k_\text B \ln \Omega$ has the units of $S$, so if we were prepared to accept the original first equation as saying essentially the same as the second (macroscopic) equation, we can accept the modified equation, with the right numerical value for $k_\text B$, as exactly matching the second, with $\frac 1T=k_\text B \beta$.

  • 1
    $\begingroup$ I should have added that temperature could be measured in J according to $$\frac1T=\left(\frac {\partial (\ln \Omega)}{\partial E}\right)_{N, V, E=\tilde E},$$ eliminating $k_\text B$ (both its numerical value and its unit). You could blame our semi-arbitrary degree kelvin for having to have Boltzmann's constant! $\endgroup$ Aug 20, 2021 at 15:38

Wellcome, @Anirban Kopty. As you said, if two sub-systems (A and B) in contact are in thermal equilibrium, then $\beta$ is the same in these two parts ($\beta=\beta_A=\beta_B$). Well, this is exactly how the zeroth principle defines the temperature. Therefore, $\beta$ and T have to be somehow related.

How the energy appears in the definition of $\beta$ fixes that $\beta$ should be proportional to $\frac{1}{T}$. The proportional constant is universal (this reasoning didn't depend on particular details of these sub-systems) and turns out to be $k_B$.


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