Entropy increase from unstable representations

Key claim

The comfortable belief under pressure: “Irreversibility must come from a microscopic arrow of time.”

Theme

Entropy increase as a statement about unstable representations, not trajectories.

Guiding question

If microscopic dynamics is reversible, why does the same mathematical flow produce irreversible thermodynamics?

Where exactly does the arrow enter when moving between Hamiltonian mechanics, stochastic processes, and information theory?

Setup

Consider a classical Hamiltonian system with phase-space point $\mathrm{\Gamma }=(q,p)$, evolving by

Its probability density obeys the Liouville equation

or equivalently ${\partial }_t\rho =\{H,\rho \}$.

The fine-grained Gibbs entropy is

and is exactly conserved:

Now introduce a coarse-graining map $\mathcal{C}$, which replaces the microscopic distribution by a lower-resolution description:

The central question is not merely “why does entropy increase?” but:

Why does a loss of description behave like a physical law?

This is the minimal arena because it contains the entire tension: reversible microscopic flow, irreversible effective evolution, and information loss.

Analysis

Derivation — dynamical systems view

The clean derivation begins with Liouville’s theorem:

Hamiltonian flow preserves phase-space volume. Microscopic trajectories therefore never compress into equilibrium.

In a mixing chaotic system, however, an initially localized distribution stretches and folds:

The information is not destroyed; it moves into increasingly fine phase-space structure.

A coarse observer cannot resolve those filaments. Under suitable mixing and coarse-graining assumptions, the effective entropy grows because the reduced description projects away those correlations.

Scope and assumptions — statistical mechanics view

The hidden assumption is that coarse-graining is innocent.

It is not.

In the Boltzmann description, molecular chaos factorizes the distribution of incoming collision partners:

This closure discards correlations from the autonomous one-particle description. The $H$-theorem then gives

The exact microscopic two-particle distribution does not remain factorized after collisions. Correlations accumulate.

The time-asymmetric ingredient is the low-correlation boundary condition imposed on incoming states, schematically

not Newton’s equations themselves.

Physical interpretation — information theory and stochastic processes

The dictionary is:

Statistical mechanics	Information theory
microstate $\mathrm{\Gamma }$	full information state
coarse-graining	compression or marginalization
entropy increase	information unavailable to the reduced description
correlations	hidden memory

For a bipartite classical distribution,

where $I(A:B)$ is the mutual information.

A closed system can keep its fine-grained total entropy constant while subsystem entropies increase because information migrates into correlations.

This structurally parallels open-system quantum mechanics, where

But the statements are not identical. Thermodynamic coarse-graining is often a chosen family of macroscopic observables, while the quantum partial trace is defined by a selected tensor-product decomposition.

So we must distinguish:

• Shared mathematical structure: reduction by marginalization or partial trace.
• Controlled approximation: an autonomous Markovian master equation.
• Analogy: “the environment stores forgotten information.”

False claim to diagnose

Because microscopic equations conserve entropy, the second law is merely an illusion caused by human ignorance and has no objective content.

Why is this tempting?

The equation

looks decisive. If entropy is constant microscopically, thermodynamic entropy seems subjective.

The failure occurs at the identification

They are different functionals on different state spaces.

As a counterexample, take a gas initially confined to half a box. Under Hamiltonian evolution, its fine-grained distribution remains an incompressibly deformed region. Yet macroscopic measurements robustly predict expansion toward equilibrium.

The issue is not mere ignorance. Macroscopic observables are insensitive to the extremely fine correlations produced by mixing, and the low-entropy initial condition is a physical boundary condition.

The defensible statement is:

The second law is not a fundamental law of individual trajectories; it is a typical law for macroscopic descriptions conditioned on special low-correlation initial data.

What follows — and what does not

The disagreement is not about the microscopic equations of motion. All three lenses can accept reversible microscopic dynamics.

The conflict comes from the choice of observables, closure assumptions, and boundary conditions:

• Hamiltonian mechanics retains the full fine-grained distribution and all correlations.
• Boltzmann theory closes the dynamics of a one-particle distribution using molecular chaos.
• Information theory tracks where information excluded from the reduced description can reside.

The arrow is therefore not inserted into Hamiltonian dynamics. It appears when a reversible theory is mapped onto a restricted state space and supplied with asymmetric boundary data.

Exercise

Consider the symmetric two-state Markov process

with $p_1+p_2=1$ and Shannon entropy

1. Show explicitly that $S(t)$ increases until equilibrium, $p_1=p_2=1/2$.
2. Identify the mathematical assumption that makes this evolution different from Hamiltonian flow.

Hint 1

Write $p_2=1-p_1$ and compute $\mathrm{d}S/\mathrm{d}t$.

Hint 2

Use $x=p_1-p_2$ and solve its evolution first.

Oral check 1. If every microscopic velocity in a gas is reversed, why does entropy decrease only from an extraordinarily special correlated state?

Oral check 2. Does tracing out an environment by itself guarantee permanently irreversible decoherence?

Solution

From the master equations,

$$\frac{\mathrm{d}x}{\mathrm{d}t}=-2kx,$$Copy LaTeX

so

$$x(t)=x(0){\mathrm{e}}^{-2kt},p_{1,2}(t)=\frac{1}{2}(1\pm x).$$Copy LaTeX

The entropy is

$$S(x)=-\frac{1+x}{2}\ln \frac{1+x}{2}-\frac{1-x}{2}\ln \frac{1-x}{2}.$$Copy LaTeX

Its production rate is

$$\frac{\mathrm{d}S}{\mathrm{d}t}=kx\ln \frac{1+x}{1-x}\ge 0$$Copy LaTeX

for $|x|<1$, with equality only at $x=0$. Hence $S$ approaches

$$S_{max}=\ln 2.$$Copy LaTeX

The decisive step is the autonomous Markov rule

$$p(t+\mathrm{\Delta }t)=M_{\mathrm{\Delta }t}p(t).$$Copy LaTeX

Its inverse is generally not a stochastic map. In a coarse-grained derivation, the missing information resides in microscopic correlations and memory that are absent from the state vector $p$.

For this symmetric doubly stochastic process, $|x|$ contracts and the relative entropy to equilibrium decreases. This is the precise information-space contraction used here.

Check your understanding

What is the more fundamental source of the thermodynamic arrow: the dynamics, the initial condition, or the observer’s choice of variables?

Reply with your answer, your attempted solution, or simply say “deeper,” “too easy,” or “too hard.”

Connections and next step

• New pressure point: fine-grained conservation versus autonomous macroscopic irreversibility.
• Main domains: classical mechanics, statistical mechanics, and information theory.
• Toy models: a coarse-grained Hamiltonian gas and a symmetric two-state Markov process.
• Unresolved: when is entropy increase observer-relative, and when is it structurally robust across macroscopic descriptions?
• Revisit: equilibrium and typicality in three runs.