I've been trying to understand how we can equate the Boltzmann entropy $k_B \ln \Omega$ and the entropy from thermodynamics. I'm following the approach found in the first chapter in Pathria's statistical mechanics, and in many other texts. Many other questions on stackexchange come close to addressing this problem, but I don't think any of the answers get at my specific question.

So, we're considering two isolated systems 1 and 2, which are brought into thermal contact and allowed to exchange energy (let's assume for simplicity that they can only exchange energy). On the thermodynamic side of the problem, we have the necessary and sufficient condition for thermal equilibrium

$$T_1=\frac{\partial E_1}{\partial S_1}=T_2=\frac{\partial E_2}{\partial S_2},$$

where the temperatures $T_1$ and $T_2$, the internal energies $E_1$ and $E_2$, and the entropies $S_1$ and $S_2$ are all defined appropriately in operational, thermodynamic terms. On the other hand, we can show that the necessary and sufficient condition for equilibrium from the standpoint of statistical mechanics is given by

$$\beta_1 \equiv \frac{\partial \ln \Omega_1}{\partial E_1}= \beta_2 \equiv \frac{\partial \ln \Omega_2}{\partial E_2}.$$

Here, $\Omega_1$ and $\Omega_2$ are the number of microstates associated with the macrostate of each system. Now, since both of these relations are necessary and sufficient for equilibrium, one equality holds if and only if the other also holds. My question is: How can we proceed from here to show that $S=k_B \ln \Omega$, without limiting our scope to specific examples (like an ideal gas)? In Pathria's text and in other treatments, I don't see much explanation for how this step is justified.

My possibly wrong thoughts are: It seems like we first need to show that $\beta$ is a function of $T$ alone (and indeed the same function of $T$ for both systems), and then show that the form of this function is in fact $\beta \propto T^{-1}$. But I'm not sure how to prove either of those claims.

  • $\begingroup$ I think the reasoning is: if you define the temperature $T = \frac 1{k_B\beta}$, then the entropy can be $k_B\ln \Omega$. $\endgroup$ – user115350 Jul 24 '16 at 18:12
  • $\begingroup$ @user115350 i agree that that would be enough. so put another way, my question is: how can we know that? how can we prove the relation $\beta = 1/k_B T$, when $T$ is defined in thermodynamic terms and $\beta$ is defined in terms of microstate numbers? $\endgroup$ – Wade Hodson Jul 24 '16 at 20:52
  • $\begingroup$ Based on statistical theory, entropy is defined first $S=k_B \ln \Omega$. Then the thermodynanic temperature is defined as $\frac1{T}=\frac{\partial S}{\partial E}$, which leads to $T=\frac1{k_B\beta}$. The path is very clear. I guess your question might be: how can one prove that the statistical entropy is the classical entropy? I think there is a lot discussion on this in text book. $\endgroup$ – user115350 Jul 25 '16 at 19:38
  • $\begingroup$ @user115350 right, i'm trying to show that the statistical (Boltzmann) entropy and thermodynamic entropy are the same (and I'm most interested in a very general proof of this equality, one which isn't restricted to specific systems like ideal gases or something). we can certainly define statistical mechanical entropy as $S= k_B \ln \Omega$ if we so chose. but this definition will be worthless if we can't show how it is linked to the thermodynamic entropy, a quantity that we can actually measure in terms of energy flows and temperature changes. $\endgroup$ – Wade Hodson Jul 25 '16 at 20:01
  • $\begingroup$ @user115350 and to add a little more motivation - i've seen several proofs of the equality of the statistical mechanical and thermodynamic entropies in various works, but so far i've never been fully convinced by any of them. all of them seem to only apply in limited cases, and to not be fully general. the reason i posted this is because the line of argument outlined in my post looks like a very airtight proof of the equality in the general case, except for that last step. so i wanted to try to clarify the logic that goes into making the final step of the proof. $\endgroup$ – Wade Hodson Jul 25 '16 at 20:12

As discussed in the comments, your proof needs to show that: $$ \beta = \frac{1}{kT} $$ Following Gaskell's "Introduction to Thermodynamics", I think that this is a definition. The rationale comes from looking at $\beta$ as a parameter which controls the shape of the distribution of the Boltzmann distribution of energy among particles: $$ n_{i}=\frac{ne^{-\beta E_i}}{P} $$ where $n$ is the total number of particles, $E_i$ is the $i^{th}$ energy level, and $n_i$ is the occupation of the $i^{th}$ energy level, and $P$ is the partition function.

Having $\beta$ and $T$ inversely proportional makes sense because, as the plot of occupation vs energy below shows, you would expect the higher energy states to become more occupied when the temperature is raised. This happens when beta is lowered.

enter image description here

  • $\begingroup$ Good answer, but I really would like someone to say that we're not "proving" anything here but simply giving a motivation: we can make Boltzmann's / Shannon's conception of entropy behave in important ways very like the macroscopic, Clausius definition by postulating that the Lagrange multiplier $\beta$ is the reciprocal temperature (modulo a scale factor): this is what you are showing here. We then find that statistical mechanics foretells experimental results if we assume this. So it's really a theoretical hunch backed by experiment .... $\endgroup$ – Selene Routley Dec 12 '16 at 9:25
  • $\begingroup$ ...See my answer here for more information. $\endgroup$ – Selene Routley Dec 12 '16 at 9:26
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    $\begingroup$ @WetSavannaAnimalakaRodVance hey thanks for the response. i did some more research after asking this, and i'm now of the tentative opinion that it is possible to prove the connection. it's impossible to lay out my whole argument in this comment, but basically the idea is: since we can prove that beta = 1/T (T here is ideal gas temperature) for the ideal gas case, and we can show empirically that the ideal gas T is proportional to thermodynamic T, we can use the ideal gas as a "bridge" between thermo and stat mech to show that the thermodynamic T and 1/beta must be proportional for any system $\endgroup$ – Wade Hodson Dec 12 '16 at 21:12

$\newcommand{\mean}[1] {\left< #1 \right>}$ $\DeclareMathOperator{\D}{d\!}$

Proof that $\beta = \frac{1}{k T}$

This proof assumes only classical thermodynamics and the Boltzmann distribution of microstates. It does not assume anything about statistical entropy.

First recall the statistical mechanics expressions for the pressure $P$ and internal energy $E$ of a system

\begin{align} \label{eq:sm_pressure} P = \mean{P} = \frac{1}{\beta} \left( \frac{\partial \ln Z}{\partial V} \right)_{\beta, N} \end{align}

\begin{align} \label{eq:sm_energy} E = \mean{E} = -\left( \frac{\partial \ln Z}{\partial \beta} \right)_{V, N} \end{align}

where $Z$ is the partition function. These can both be derived from the Boltzmann distribution of microstates.

Lets take the partial derivative of the pressure with respect to $\beta$, holding $V$ and $N$ constant.

\begin{align} \left( \frac{\partial P}{\partial \beta} \right)_{N} &= \frac{1}{\beta} \left( \frac{\partial^2 \ln Z}{\partial \beta ~\partial V } \right)_{N} -\frac{1}{\beta^2} \left( \frac{\partial \ln Z}{\partial V} \right)_{\beta, N} \\ &= \frac{1}{\beta} \left( \frac{\partial^2 \ln Z}{\partial \beta ~\partial V } \right)_{N} -\frac{1}{\beta} \mean{P} \end{align}

Now we take the partial derivative of the energy with respect to volume $V$, holding $\beta$ and $N$ constant.

\begin{align} \left( \frac{\partial E}{\partial V} \right)_{\beta, N} &= -\left( \frac{\partial^2 \ln Z}{\partial \beta ~\partial V} \right)_{N} \end{align}

Combining these two partial derivatives gives

\begin{align} \label{eq:sm_pressure_eq} -\mean{P} &= \left( \frac{\partial E}{\partial V} \right)_{\beta, N} + \beta \left( \frac{\partial P}{\partial \beta} \right)_{N,V} \end{align}

This equation, which was derived completely from statistical mechanics assumptions, can be compared with a famous analogous equation from classical thermodynamics "the thermodynamic equation of state":

\begin{align} \label{eq:ct_pressure_eq} -P &= \left( \frac{\partial E}{\partial V} \right)_{T, N} - T \left( \frac{\partial P}{\partial T} \right)_{N,V} \end{align}

Because $\mean{P} = P$, we can combine these two equations:

\begin{align} \left( \frac{\partial E}{\partial V} \right)_{\beta, N} + \beta \left( \frac{\partial P}{\partial \beta} \right)_{N,V} &= \left( \frac{\partial E}{\partial V} \right)_{T, N} - T \left( \frac{\partial P}{\partial T} \right)_{N,V} \\ \label{eq:eq1} \left( \frac{\partial E}{\partial V} \right)_{\beta, N} + \left( \frac{\partial P}{\partial \ln \beta} \right)_{N,V} &= \left( \frac{\partial E}{\partial V} \right)_{T, N} - \left( \frac{\partial P}{\partial \ln T} \right)_{N,V} \qquad \textrm{because $\D \ln x = \frac{\D x}{x}$} \end{align}

Therefore, the left and right-hand sides are equal, and they are equal only when

\begin{align} \D \ln \beta &= - \D \ln T \end{align}

Integrating both sides:

\begin{align} \ln \beta &= - \ln T -\ln k \\ \beta &= \frac{1}{k T} \end{align}

Because $\beta$ is independent of composition, the proportionality constant $k$ is a universal constant which is also independent of composition.


To go from statistical mechanics to thermodynamics we assume that the quantity $\frac{\partial E}{\partial S}$ is equal to Inverse of temperature. Talking about Boltzmann's relation it can be verified by considering case of coins.

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    $\begingroup$ Please refrain from using short forms and use MathJax to format equations. $\endgroup$ – wavion Apr 18 '20 at 6:46

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