Where does the ln come from in S = k ln(W) ? - Take 2

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Some years ago I wrote this post. Now I want to come at it from a different angle.

A closed system spontaneously goes towards the state with maximum multiplicity, $W$. For a system with $N$ molecules with energies $\varepsilon_1, \varepsilon_2, \ldots $ we therefore want to find the values of $N_i$ that maximises $W(N_1, N_2, \ldots)$.

This is easier to do for $\ln W$ than $W$, which is fine because $W$ is a maximum when $\ln W$ is a maximum, and this happens when
$$ \frac{N_i}{N}= p_i = \frac{e^{-\beta \varepsilon_i}}{q} $$
$\beta$ can be found by comparison to experiment.  For example, the energy of an ideal monatomic gas with $N_A$ particles is
$$ U^{\mathrm{Trans}} = N_A \langle \varepsilon^{\mathrm{Trans}} \rangle = N_A\sum_i p_i \varepsilon_i = \frac{3N_A}{2\beta} = \tfrac{3}{2}RT \implies \beta = \frac{1}{kT} $$
where $R$ is determined by measuring the temperature increase due to adding a known amount of energy to the system.

So far we have used $\ln W$ instead of $W$ for convenience, but is there something special about $\ln W$? Yes, the change in $\ln W$ has can be expressed quite simply
$$ d \ln W = \beta \sum_i \varepsilon_i dN_i = N \beta \sum_i \varepsilon_i dp_i =  \frac{dU}{kT} $$
So change in internal energy $U$ due to a redistribution of molecules among energy levels is equal to the change in $\ln W$ (as opposed to $W$) multiplied by $kT$
$$ dU = Td\left( k \ln W \right) = TdS$$
The final question is whether there is something special about $\ln = \log_e$ as opposed to say $\log_a$ where $a \ne e$? Well, $\log_a W$ is a maximum when
$$ \frac{N_i}{N}= p_i = \frac{a^{-\beta \varepsilon_i}}{q} $$
There is an extra term in the derivation but that cancels out in the end.  So no change there.

What about $\beta$?  There are two changes.  The previous derivation of $U^{\mathrm{Trans}}$ relied on this relation (I'll drop the "Trans" label for the moment)
$$  \varepsilon_i e^{-\beta \varepsilon_i}  = - \frac{d }{d \beta} e^{-\beta \varepsilon_i} \implies \langle \varepsilon \rangle = - \frac{1}{q} \frac{dq}{d\beta} $$
which now becomes
$$  \varepsilon_i a^{-\beta \varepsilon_i}  = - \frac{1}{\ln(a)} \frac{d }{d \beta} a^{-\beta \varepsilon_i} \implies \langle \varepsilon \rangle = - \frac{1}{q \ln(a)} \frac{dq}{d\beta}$$
While $q$ has an extra $\ln (a)$ term, the derivative wrt $\beta$ is still the same and
$$ U^{\mathrm{Trans}} = N_A \langle \varepsilon^{\mathrm{Trans}} \rangle = N_A\sum_i p_i \varepsilon_i = \frac{3N_A}{2 \ln (a) \beta} = \tfrac{3}{2}RT \implies \beta = \frac{1}{\ln (a) kT} $$
So, $S = k \log_e W$ is special in the sense that the proportionality constant $k$ is the experimentally measured ideal gas constant divided by Avogadro's number.  In any other base we have to write either $S = k \ln (a) \log_a W$ where $k = R/N_A$ or $S = k^\prime \log_a W$ where $k^\prime = \ln(a)R/N_A$

Clearly, $S = k \log_e W$ is the most natural choice, and this is because $e$ is the (only) value of $a$ for which
$$ \frac{d}{dx} a^x = a^x $$
In fact that is one way to define what $e$ actually is.



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