Questions: The H-Theorem and the Arrow of Time

Question 1

Loschmidt's reversibility paradox challenges the H-theorem by pointing out that:

• A. Boltzmann's collision integral violates conservation of energy for inelastic collisions
• B. If you take a gas that has just evolved from low to high entropy and reverse all molecular velocities, the time-reversed evolution must cause H to increase — apparently contradicting dH/dt ≤ 0
• C. The Maxwell-Boltzmann distribution cannot be reached from arbitrary initial conditions, contradicting the theorem's claim
• D. Molecular chaos is inconsistent with quantum mechanical uncertainty, invalidating the proof

Answer: (B)

Loschmidt's paradox: take a gas in an out-of-equilibrium state. It evolves toward equilibrium — H decreases. Now freeze the gas at some intermediate point and reverse all velocities. Newton's laws are time-symmetric, so the time-reversed evolution is equally valid. But this means H must now increase (running the same trajectory backward), apparently contradicting dH/dt ≤ 0. Boltzmann's response: the velocity-reversed state is extraordinarily special — it has precisely tuned inter-particle correlations designed to recreate the past. Such a state violates the molecular chaos assumption (Stosszahlansatz) that underlies the theorem. Naturally arising states do not have this property.

Question 2

Boltzmann's resolution of Loschmidt's paradox identifies which assumption as the key ingredient that breaks the time-symmetry in the H-theorem?

• A. Conservation of momentum in two-body elastic collisions
• B. Molecular chaos (Stosszahlansatz): the assumption that colliding molecules have uncorrelated velocities before collision, which holds for naturally arising states but fails for the artificially velocity-reversed state
• C. The classical (non-quantum) treatment of molecular velocities
• D. The assumption that the gas is spatially uniform

Answer: (B)

The H-theorem's proof requires that colliding molecules have uncorrelated velocities before they collide — molecular chaos. This is valid for typical out-of-equilibrium states, where molecules have not previously interacted in correlated ways. But the velocity-reversed state has been constructed so that each molecule's velocity is precisely correlated with its future collision partner — it was designed to undo the history of collisions. This finely tuned correlation violates molecular chaos from the outset, so the H-theorem's proof does not apply to it. This is how time-symmetry of dynamics is compatible with the directional decrease of H: the asymmetry enters through the assumption about initial correlations.

Question 3

The H-theorem demonstrates that entropy is expected to generally increase because the fundamental microscopic laws of physics are time-asymmetric — they distinguish past from future at the molecular level.

• T. True
• F. False

Answer: False

This is precisely backward. The microscopic laws (Newton's equations, quantum mechanics) are time-reversal symmetric — any valid trajectory run backward is also a valid trajectory. The H-theorem does NOT rely on time-asymmetric microscopic laws. Instead, entropy increase is statistical: there are overwhelmingly more high-entropy microstates than low-entropy ones, so almost all microscopic trajectories originating from a low-entropy state lead to higher entropy. The arrow of time is a consequence of the overwhelming probability of high-entropy configurations, not of any asymmetry in the fundamental laws.

Question 4

The Maxwell-Boltzmann velocity distribution corresponds to the minimum of Boltzmann's H function — the state of maximum entropy for an ideal gas in equilibrium.

• T. True
• F. False

Answer: True

The H-theorem shows that H = ∫ f ln(f) d³v is monotonically non-increasing (dH/dt ≤ 0). The evolution continues until H reaches its minimum, which occurs exactly at the Maxwell-Boltzmann distribution f(v) ∝ exp(−mv²/2kT). Since entropy S = −kH, this minimum of H is the maximum of entropy. The Maxwell-Boltzmann distribution is therefore the unique fixed point of the collision dynamics under molecular chaos — any other distribution will evolve toward it. This gives the H-theorem its physical content: it explains why the Maxwell-Boltzmann distribution is the universal equilibrium, not just a convenient assumption.

Question 5

If the fundamental laws of physics are time-reversal symmetric, why does entropy increase rather than decrease in practice? Use the insight from the H-theorem to explain.

Model answer: Time-symmetric laws mean that for every trajectory leading from low entropy to high entropy, there exists a time-reversed trajectory going from high to low. But these reversed trajectories require extremely special initial conditions — precisely correlated molecular velocities. Naturally arising low-entropy states have uncorrelated molecular velocities (molecular chaos), and under these typical conditions, the overwhelming majority of microscopic trajectories lead to higher-entropy states simply because there are vastly more high-entropy microstates than low-entropy ones. Entropy increases because it is statistically inevitable, not because the laws forbid decrease.

This is the statistical interpretation of the second law, which Boltzmann spent years defending against Loschmidt and Zermelo. The H-theorem makes it precise and quantitative: under molecular chaos, the collision dynamics drive any distribution toward the Maxwell-Boltzmann equilibrium because that distribution is the overwhelmingly probable one. The 'arrow of time' is not written into the microscopic laws but into the initial conditions: we observe entropy increase because we (and the universe around us) started in a low-entropy state, and almost all paths forward from a low-entropy state lead to higher entropy. This connects directly to the cosmological question of why the early universe had low entropy.