Why doesn't the Boltzmann entropy change in reversible adiabatic expansion?

Question

I tried to ask this question here, but I formulated it badly and as a result the answers, while correct, didn't address my real concern.

What I really want to know is, in reversible adiabatic expansion, why isn't there a change in the volume of phase space accessible to the system, i.e. why doesn't the Boltzmann entropy change? This post is probably the closest match, in which the first answer sums it up conceptually (the increase in entropy due to volume is offset by the decrease due to drop in temperature) and the last answer sets up the proof for the case of an ideal gas, but no general proof is offered.

In other words, I'm looking for a general proof, and NOT one that relies on Clausius' definition of entropy $dS = \frac{\delta q_{rev}}{T}$ and then asserts the equivalence of Boltzmann and Clausius entropies without justification.

Or to frame it a bit more mathematically, when a system at pressure $p$ and volume $V$ expands by $dV$ and loses $p dV$ worth of internal energy, why is its phase space volume $\Omega$ unaffected? Surely a proof must exist for such a fundamental process.

EDIT: I appreciate the answers and yet they have forced me to realize that I once again failed to articulate this properly, so let me try to expand. Let's go with a closed system; thus

$$\Omega = \Omega(E, V) \tag{1}\label{1}$$ $$d\Omega = \left.\frac{\partial \Omega}{\partial E}\right|_V dE + \left.\frac{\partial \Omega}{\partial V}\right|_E dV \tag{2}\label{2}$$

Also since it's adiabatic, and assuming only P-V work, we know $dE = -p dV$. Then, no change in $\Omega$ becomes equivalent to the condition

$$p\ \left.\frac{\partial\Omega}{\partial E}\right|_V = \left.\frac{\partial\Omega}{\partial V}\right|_E \tag{3}\label{3}$$

So this is what I want to prove. But I want to prove it in the following sense. In addition to $\eqref{1}$ we have

$$p = p(E, V) \tag{4}\label{4}$$

That is, both pressure and phase space volume are functions of energy and volume. And I want to understand why the forms of these two functions have to be so as to satisfy $\eqref{3}$.

In other words, I am looking for a justification based on statistical mechanics, not classical thermodynamics. In particular, pressure is defined mechanically as the flux of momentum through the bounding surface (although since we're in equilibrium, it's really the flux of momentum through any surface you like). So why do the dependence of momentum flux on energy and volume, and the dependence of phase space occupation on energy and volume, have to relate in such a way as to satisfy $\eqref{3}$?

Answer 1

in reversible adiabatic expansion, why isn't there a change in the volume of phase space accessible to the system, i.e. why doesn't the Boltzmann entropy change?

There are two phase volume functions important here, with possibly different values. The "Boltzmann" phase volume $B_t = B(E_t,V_t,N)$, which Boltzmann's entropy is based on, is volume of the phase region $\mathcal{R}(E_t,V_t,N)$ compatible with the macroscopic variables $E_t,V_t,N$ at any point $t$ of the process, which is a set with simple boundary defined by $E_t,V_t,N$, thus made of compact parts with no "holes" or "empty filaments"; and the phase volume $\Omega_t$ is volume of the phase region $\mathcal{O}_t$ made of points at time $t$ which evolved from the initial phase region $\mathcal{O}_0 = \mathcal{R}(E_0,V_0,N)$ defined by $E_0,V_0,N$, thus expected to be very "foamy" - mixed with holes and empty filaments that are not part of the set.

In mechanics, for models described by an Hamiltonian, the Liouville theorem holds, which says $\Omega_t$ is constant in time no matter what (including during an irreversible process), so $\Omega_t=\Omega_0$. No such thing holds in general for $B_t$; it can change in time.

However, in the special case the process is thermodynamically reversible, it is infinitely slow, and the system is described faithfully at all its stages by macroscopic variables $E_t,V_t,N$; and then, the phase region $\mathcal{R}$ implied by these is the same as the phase region $\mathcal{O}_t$ which evolves in phase space from the initial region $\mathcal{O}_0 = \mathcal{R}(E_0,V_0,N)$. So for reversible processes, there is no difference between the two regions, and $B_t=\Omega_t=const.$

In other words, if the macroscopic process is reversible (infinitely slow), the phase region implied by $E_t,V_t,N$ "follows" the phase region of all points evolved by Hamiltonian equations of motion from the initial condition $E_0,V_0,N$, and these regions are the same at all times.

If the process is irreversible, e.g. due to happening with non-zero speed, then the system does not pass only through exactly equilibrium macrostates. It gets into non-equilibrium macrostates, where description of macrostate by $E_t,V_t,N$, while possible, is inaccurate. $\mathcal{R}$ implied by $E_t,V_t,N$ (a set with no holes) becomes different from $\mathcal{O}_t$ (a set that is getting more foamy and complicated in time). Mathematically, in mechanics, $B_t$ can be evolve to be smaller or greater than $\Omega_t$; but in physically realizable processes, where observations and experiments imply 2nd law, this law restricts our models to the latter case. For ideal gas in adiabatic process, this implies $B_t$ can't decrease, thus Boltzmann's entropy implied by the macrostate can't decrease. In an irreversible processes, it typically increases.

For example, in irreversible adiabatic expansion, the gas undergoes turbulent motion, its pressure drops (Bernoulli theorem), and the gas pressure on the piston is less than the outside pressure. Thus work done for the same $dV$ is lower, and internal energy at any expanded value of $V$ will be higher; then it's obvious that $B_t$ becomes larger than $B_0$.

Answer 2

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Since you want the answer in terms of Boltzmann entropy, we will work with the microcanonical ensemble. Recall from the basic assumptions of the microcanonical that its energy $E$ is a function of three state variables, particle number $N$, volume $V$, and the number of microstates $\Omega$. From multivariate calculus we can then write the total differential for the energy $E(\Omega, V, N)$ of a microcanonical ensemble:

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

and at constant composition $N$ we have:

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

Now consider a reversible change of energy when we keep the number of microstates $\Omega$ constant. Under these two conditions:

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

This means that when $\Omega$ is constant, the total reversible change of energy is due only to a change in volume. But from classical thermodynamics we know that a reversible change of energy that is only due to a change in volume is one that involves only reversible work, and hence no heat transfer.

Thus we know three things about this microcanonical process, (1) it is adiabatic, (2) the factor relating $\D E$ and $\D V$ is the negative pressure of the system,

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

and (3)

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

Restated, for a reversible adiabatic change of state, $\Omega$ is constant.

Answer 3

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You derived this as the condition to satisfy:

\begin{align} P \left(\frac{\partial \Omega}{\partial E}\right)_{V} = \left(\frac{\partial \Omega}{\partial V}\right)_{E} \end{align}

From the cycle rule, we can substitute the RHS

\begin{align} P \left(\frac{\partial \Omega}{\partial E}\right)_{V} = -\left(\frac{\partial \Omega}{\partial E}\right)_{V} \left(\frac{\partial E}{\partial V}\right)_{\Omega} \end{align}

which gives

\begin{align} P = -\left(\frac{\partial E}{\partial V}\right)_{\Omega} \end{align}

which is known to be the pressure in a microcanonical ensemble, and hence $P=P$ is true, and your equation (3) is proven using only statistical mechanics, as requested.