Proving that the Boltzmann entropy is equal to the thermodynamic entropy

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.

• I think the reasoning is: if you define the temperature $T = \frac 1{k_B\beta}$, then the entropy can be $k_B\ln \Omega$. Jul 24, 2016 at 18:12
• @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? Jul 24, 2016 at 20:52
• 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. Jul 25, 2016 at 19:38
• @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. Jul 25, 2016 at 20:01
• @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. Jul 25, 2016 at 20:12

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Proof that $$\beta = \frac{1}{k T}$$ and that $$S = k \ln \Omega$$

This proof follows from only classical thermodynamics and the microcanonical ensemble. It makes no assumptions about the analytic form of statistical entropy, and does not involve the ideal gas law.

First recall that the pressure of an individual microstate is given from mechanics as:

\begin{align} P_i &= -\frac{\D E_i}{\D V} \end{align}

When assuming only $$P$$-$$V$$ mechanical work, the energy of a microstate $$E_i(N,V)$$ is only dependent on two variables, $$N$$ and $$V$$. For example, consider a quantum mechanical system like particles confined in a box. Therefore, at constant composition $$N$$,

\begin{align} P_i &= -\left( \frac{\partial E_i}{\partial V} \right)_N \end{align}

In a system described by the microcanonical ensemble, there are $$\Omega$$ possible microstates of the system. The energy of an individual microstate $$E_i$$ is likewise trivially independent of the number microstates $$\Omega$$ in the ensemble. Therefore, the pressure of an individual microstate can also be expressed as

\begin{align} P_i &= -\left( \frac{\partial E_i}{\partial V} \right)_{\Omega,N} \end{align}

According to statistical mechanics, the macroscopic pressure of a system is given by the statistical average of the pressures of the individual microstates:

\begin{align} P = \mean{P} &= \sum_i^\Omega \pr_i P_i \end{align}

where $$\pr_i$$ is the equilibrium probability of microstate $$i$$. For a microcanonical ensemble, all microstates have the same energy $$E_i = E$$, where $$E$$ is the energy of the system. Therefore, from the fundamental assumption of statistical mechanics, all microcanonical microstates have the same probability at equilibrium

\begin{align} \pr_i = \frac{1}{\Omega} \end{align}

It follows that the pressure of a microcanonical system is given by

\begin{align} P = \mean{P} &= -\sum_i^\Omega \frac{1}{\Omega} \left( \frac{\partial E_i}{\partial V} \right)_{\Omega,N} \\ &= -\frac{1}{\Omega} \sum_i^\Omega \left( \frac{\partial E}{\partial V} \right)_{\Omega,N} \\ &= -\frac{\Omega}{\Omega} \left( \frac{\partial E}{\partial V} \right)_{\Omega,N} \\ P &= -\left( \frac{\partial E}{\partial V} \right)_{\Omega,N} \end{align}

This expression for the pressure of a microcanonical system can be compared to the classical expression

\begin{align} P &= -\left( \frac{\partial E}{\partial V} \right)_{S,N} \end{align}

which immediately suggests a functional relationship between entropy $$S$$ and $$\Omega$$.

Now we take the total differential of the energy of a microcanonical system:

\begin{align} \D E = \left(\frac{\partial E}{\partial \ln \Omega}\right)_{V, N} \D \ln\Omega + \left(\frac{\partial E}{\partial V}\right)_{\ln \Omega, N} \D V \end{align}

As stated in the OP, for the microcanonical ensemble, the condition for thermal equilibrium is:

\begin{align} \beta &= \left( \frac{\partial \ln \Omega}{\partial E} \right)_{V,N} \end{align}

Thus,

\begin{align} \D E = \frac{1}{\beta} \D \ln\Omega - P \D V \end{align}

Compare with the classical first law of thermodynamics:

\begin{align} \D E = T \D S - P \D V \end{align}

Because these equations are equal, we see that

\begin{align} T \D S &= \frac{1}{\beta} \D \ln\Omega \\ \D S &= \frac{1}{T \beta} \D \ln\Omega \end{align}

Note that both $$\D S$$ and $$\D \ln\Omega$$ are exact differentials, so $$\frac{1}{T \beta}$$ must be either a constant or a function of $$\Omega$$. Since $$\D S$$ and $$\D \ln\Omega$$ are both extensive quantities, $$\frac{1}{T \beta}$$ cannot depend on $$\Omega$$, and therefore

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

where $$k$$ is a universal constant that is independent of composition, since $$\beta$$ and $$T$$ are both independent of composition.

By integrating, we have

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

where $$C$$ is a constant that is independent of $$E$$ and $$V$$, but may depend on $$N$$. By invoking the third law we can set $$C=0$$ to arrive at the famous Boltzmann expression for the entropy of a microcanonical system:

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

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.

• 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 .... Dec 12, 2016 at 9:25
• @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 Dec 12, 2016 at 21:12

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Proof that $$\beta = \frac{1}{k T}$$ for the canonical ensemble

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.

• A late comment. Your "proof" goes in the right direction, but there is a problem. Your starting formulae for $E$ and $P$ are valid in the canonical ensemble, while Boltzmann's expression for the entropy, as in the question, is valid in the microcanonical ensemble. Mar 24, 2021 at 7:21
• My proof is more general, and applies to the OP's question, since Boltzmann's entropy can be derived from the canonical ensemble for the special case when all microstates have the same energy. Mar 25, 2021 at 12:14
• This is not correct. The microstates of the canonical ensemble do not have the same energy. The fact that a large majority of them has energy very close to the average energy is not a good reason for ignoring the deviations. As a matter of fact, the correct expression for the specific heat in the canonical ensemble directly exploits the non-zero variance of the energy distribution. Mar 25, 2021 at 14:59
• Imagine a quantum mechanical canonical ensemble, in which we lower the temperature until only the ground states are accessible. Then the energy variance is zero, as well as the heat capacity, as is well known experimentally. All microstates have the same energy, and the entropy is given by an expression equivalent to the Boltzmann expression, dependent only on the log of the redundancy of the ground state. Mar 26, 2021 at 15:23
• @GiorgioP You motivated me to produce a similar "proof" specifically for the microcanonical ensemble. It also addresses the OP's desire to derive the Boltzmann entropy equation. See my latest answer. Mar 27, 2021 at 21:25

There is no well-defined "thermodynamic entropy" outside of the Shannon or Von Neumann entropy because there is no well-defined concept of temperature at all without entropy. Entropy is foundational, and temperature is derived from it. And entropy is fundamentally statistical in nature.

In fact, it is entropy that should have its own base unit, with temperature having a derived unit. If entropy had a unit, $$B$$, then temperature would have the unit $$J/B$$.

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.

• Please refrain from using short forms and use MathJax to format equations. Apr 18, 2020 at 6:46