advanced27 min studyLesson 353 in Physics

Find the arrow of time in the closure assumption

Derive the sign of Boltzmann entropy production without confusing it with fine-grained entropy.

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What you’ll understand

  • State molecular chaos as a condition on incoming collision partners.
  • Track the opposite signs of H and kinetic entropy.
  • Identify boundary and reversibility assumptions behind the H theorem.

Microscopic elastic trajectories can be reversed, yet a dilute gas usually smooths away a deliberately prepared velocity imbalance. Boltzmann’s H theorem explains a precise version of this behavior for a closed kinetic equation. Its irreversibility is not obtained by secretly changing Newton’s laws. It enters through an assumption about incoming pairs, followed by a convexity inequality. Understanding that route clarifies what the theorem proves and why it does not force the fine-grained entropy of a Hamiltonian ensemble to increase.

Incoming independence is directional

The exact one-particle equation depends on a two-particle distribution, which in turn depends on higher correlations. Molecular chaos closes this hierarchy by approximating the distribution of incoming collision partners by a product of one-particle distributions at the encounter. The gas should be dilute enough that partners have generally lost relevant correlation with each other before meeting. The assumption concerns the incoming collision surface, not arbitrary independent positions and velocities at every instant.

A collision itself creates correlations between outgoing velocities through its shared conservation laws. Imposing independence before and after every encounter would discard that structure indiscriminately. Reversing all velocities in an evolved state turns outgoing correlations into carefully prepared incoming correlations; that reversed ensemble need not obey the original closure assumption. Thus the reduced equation’s time direction is consistent with reversible microscopic dynamics once its restricted statistical assumptions are made explicit.

fin(2)≈f(v)f(v1),H=∫fln⁡(f/f∗) d3x d3v,Skin=−kBH+constantf^{(2)}_{\rm in}\approx f(\mathbf v)f(\mathbf v_1),\qquad H=\int f\ln(f/f_*)\,d^3x\,d^3v,\qquad S_{\rm kin}=-k_BH+\text{constant}
f* is a fixed positive reference with the same units as f, making the logarithm dimensionless. Conserved particle number makes its choice irrelevant to dH/dt; Skin is kinetic entropy in J/K.↔ Scroll wide equations horizontally.

One inequality fixes the sign

Differentiate H in time and insert the kinetic equation. The extra derivative of f integrates to zero when particle number is conserved. For periodic space, an isolated specularly reflecting vessel, or sufficiently decaying whole-space data with no entropy flux at infinity, the streaming contribution produces no net change in the integrated H. The collision contribution can then be symmetrized under exchanging the two particles and reversing the elastic collision map.

Write A for the incoming product of distributions and B for the outgoing product. The symmetrized integrand is minus a positive kernel times (A−B) times the logarithm of A/B. Because logarithm increases monotonically, these two factors have the same sign. H therefore cannot increase and kinetic entropy cannot decrease. In an open vessel, entropy can enter or leave through the boundary, so the total time derivative includes an entropy flux as well as nonnegative production.

dHdt=−14∫d3x d3v d3v1 dΩ Bc(A−B)ln⁡AB≤0,(A−B)ln⁡(A/B)≥0\frac{dH}{dt}=-\frac14\int d^3x\,d^3v\,d^3v_1\,d\Omega\,B_c(A-B)\ln\frac AB\leq0,\qquad (A-B)\ln(A/B)\geq0
Bc=g dσ/dΩ≥0 is the collision kernel, A=ff₁ and B=f′f′₁. The sign statement assumes integrability, microreversibility, molecular chaos, and no net entropy flux through spatial boundaries.↔ Scroll wide equations horizontally.

Equality is local, and not every entropy is this entropy

Collision production vanishes when the paired products balance on all allowed scattering channels. Under ordinary nondegenerate elastic collision rules, this gives a local Maxwellian. Its density, flow velocity, and temperature may still vary in space. Streaming can disturb that local equilibrium, so vanishing collision production at one instant is not by itself proof of a uniform final equilibrium. A quantitative convergence rate also requires further hypotheses beyond the sign inequality.

The Markov lab offers an elementary entropy analogy. Reset and set p=q=0.25, then Step twice to inspect two transitions from state zero. Predict state-one probability 0.375. Reset and set p=q=0.5, then Step once; predict probability 0.5. The symmetric chain moves toward a uniform law, but it is a finite exact probability model, not a gas collision operator or a proof of the Boltzmann theorem. General asymmetric chains need not increase ordinary Shannon entropy.

A=B ⟹ ln⁡(f/f∗)=a+b⋅v+c v2,c<0A=B\ \Longrightarrow\ \ln(f/f_*)=a+\mathbf b\cdot\mathbf v+c\,v^2,\qquad c<0
a, b, c are local parameters chosen to make a normalizable Maxwellian; c has units s²/m². The implication uses the collision-invariant characterization for three-dimensional elastic encounters.↔ Scroll wide equations horizontally.

Evaluate the sign of one channel

A normalized illustrative collision channel has incoming product A=4 and outgoing product B=1. Its symmetrized H-production contribution is −(A−B)ln(A/B)/4 in reciprocal seconds. Compute that contribution and the associated entropy sign.

  1. The difference A−B is +3 and ln(A/B)=ln4 is positive. Their product is 3ln4≈4.159, so the positivity required by the theorem holds channel by channel in this simple example.

  2. Multiplication by −1/4 gives −3ln4/4≈−1.040 s⁻¹ for the stated normalized H contribution. Reversing the labels gives (−3)ln(1/4), the same positive product before the overall minus sign.

  3. Kinetic entropy is minus kB times H, so the corresponding entropy-production contribution is +1.040 kB per second. If A and B were equal, the contribution would vanish continuously; neither a logarithmic sign nor an arbitrary reference scale would reverse it.

The result

The specified channel contributes approximately −1.040 s⁻¹ to H and +1.040 kB/s to kinetic entropy.

Explore the model

Interactive lab

Interactive lab · markov

Compare two symmetric mixing laws

Reset and set p=0.25, q=0.25, then Step twice; predict P(state 1)=0.375. Reset and set p=0.5, q=0.5, then Step once; predict 0.5.

Compare two symmetric mixing lawsExact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. Current readings: P(state 1) at step 0: 0%; Stationary P(state 1): 60%; Chain behavior: Monotone convergence. All parameters are adjustable with the labeled controls below.Two-state Markov distribution after 0 steps: state zero probability 1, state one probability 0.Initial distribution (1, 0) · step 0p = 0.3q = 0.201100%0%1−p1−qP(0)P(1)Monotone convergence
P(state 0)P(state 1)Stationary distribution
P(state 1) at step 0
0%
Stationary P(state 1)
60%
Chain behavior
Monotone convergence
t = 0 s

Exact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forStep advances one chain transition because n=floor(2t). These exact probabilities illustrate symmetric mixing, not molecular chaos or gas entropy production.

Check your understanding

Which assumption distinguishes the kinetic arrow of time from exact reversible Hamiltonian evolution?

Try it yourself

Evaluate the same normalized channel formula for A=1 and B=3. Is the result consistent with the H theorem, and does it imply that exact fine-grained Gibbs entropy increases?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Both the difference and logarithm now change sign. Distinguish the reduced kinetic entropy from the full phase-space entropy.

Compare with the solution

The product is (1−3)ln(1/3)=2ln3, giving H contribution −ln3/2≈−0.5493 s⁻¹ and positive kinetic entropy production. This is consistent with the closed Boltzmann equation. It does not imply growth of fine-grained Gibbs entropy under Hamiltonian Liouville evolution, which preserves that entropy when the appropriate boundary assumptions hold.

Take it with you

  • Molecular chaos factorizes incoming partners and discards specific correlations.
  • Boltzmann H decreases while kinetic entropy increases.
  • Boundary entropy flux and fine-grained Hamiltonian entropy require separate accounting.

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