advanced27 min studyLesson 345 in Physics

Detailed balance and time reversal

Distinguish edge-by-edge detailed balance from global stationarity, use cycle products to diagnose driving, and relate reversible path statistics to the parity of observed variables.

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What you’ll understand

  • Compute stationary edge currents in a finite-state Markov model.
  • Test detailed balance rather than infer it from a time-independent distribution.
  • State the variable and protocol transformations required for physical time reversal.

A stationary population can hide ongoing cycles of transitions. Detailed balance is stronger: each transition channel is balanced by its reverse in the stationary ensemble. This distinction explains how a system can maintain unchanged state probabilities while dissipating energy. We will use small Markov networks to make the difference calculable, then state the limits of interpreting a simple reversal of recorded labels as a physical time-reversed experiment.

Balance each edge or only each node

Let wijw_{ij} be a continuous-time transition rate from state i to state j, with units of inverse seconds. For probabilities p, define the oriented edge current Jij=piwij−pjwjiJ_{ij}=p_iw_{ij}-p_jw_{ji}. The master equation adds incoming and subtracts outgoing currents at each node. A stationary distribution pi makes every node's total accumulation vanish. Detailed balance instead requires πiwij=πjwji\pi_iw_{ij}=\pi_jw_{ji} on every connected pair. Equal total inflow and outflow need not mean equal flow on each edge.

In a two-state network with both rates positive, stationarity forces the only edge current to zero, so the stationary chain obeys detailed balance. At least one cycle is needed to sustain an independent stationary circulation in such a finite-state model. For a three-state ring, clockwise currents can enter and leave each node at the same rate, preserving all populations. The absence of population change then says nothing about whether forward and reverse transitions are individually balanced. Counting transitions as well as occupations reveals the hidden motion.

p˙i=∑j≠i(pjwji−piwij),Jij=piwij−pjwji,detailed balance: πiwij=πjwji\dot p_i=\sum_{j\ne i}(p_jw_{ji}-p_iw_{ij}),\qquad J_{ij}=p_iw_{ij}-p_jw_{ji},\qquad \text{detailed balance: }\pi_iw_{ij}=\pi_jw_{ji}
The rates describe a time-homogeneous continuous-time chain and probabilities sum to one. Detailed balance is an edge condition; stationarity only requires the incoming current sum to vanish at each node.↔ Scroll wide equations horizontally.

A cycle ratio detects irreversibility

Multiplying detailed-balance relations around a closed cycle cancels the stationary probabilities. The product of forward rates must equal the product of reverse rates. Conversely, on an irreducible finite graph with bidirectional edges, satisfying this condition on all cycles allows a consistent reversible stationary distribution to be constructed. The logarithm of the product ratio is a dimensionless cycle affinity. A nonzero value rules out detailed balance even when the stationary probabilities are known exactly and appear symmetric.

For a thermally consistent transition model, local detailed balance relates log rate ratios to energy and reservoir entropy exchanges. A cycle affinity can then represent a chemical or mechanical driving force divided by thermal energy. Purely assigning a Markov matrix does not automatically specify that thermodynamic interpretation. Without energies, reservoirs, and coupling assumptions, the cycle test diagnoses stochastic irreversibility, while an actual heat or fuel consumption rate requires more physical information. This separation prevents a convenient mathematical chain from being mistaken for a fully calibrated thermodynamic machine.

A0120=ln⁡w01w12w20w10w21w02,detailed balance ⟹ A0120=0\mathcal A_{0120}=\ln\frac{w_{01}w_{12}w_{20}}{w_{10}w_{21}w_{02}},\qquad \text{detailed balance}\ \Longrightarrow\ \mathcal A_{0120}=0
All rates in the ratio must be positive. The numerator and denominator have identical units, making the affinity dimensionless. A nonzero cycle affinity implies stationary circulation for the irreducible three-state ring.↔ Scroll wide equations horizontally.

Reverse the right physical variables

For states representing positions or other time-even variables and time-independent reversible dynamics, detailed balance equates stationary probabilities of a path and its reversed sequence. If velocities are included, physical time reversal also changes their signs. Magnetic fields or moving boundary protocols may require corresponding reversals. Simply playing a movie backward without transforming the appropriate external controls is not the same experiment. The mathematical definition of a reverse path must match the physical parity and the chosen dynamics.

Reversibility is also distinct from convergence to stationarity. A discrete chain that alternates deterministically between two states obeys detailed balance with equal stationary weights, yet an ensemble started in one state never converges to those weights. Aperiodicity or appropriate continuous-time assumptions are needed for relaxation. These qualifications matter when a laboratory instrument uses discrete transitions while a physical model uses continuous rates. Both can teach balance, but their transition parameters, clock units, and convergence behavior must not be silently identified.

A two-state process with balanced traffic

A continuous-time system has rates w01=3 s−1w_{01}=3\,\mathrm{s^{-1}} and w10=1 s−1w_{10}=1\,\mathrm{s^{-1}}. Find its stationary probabilities, forward and reverse edge traffic, and the relaxation rate of a probability perturbation.

  1. Stationarity requires 3π0=π13\pi_0=\pi_1, and normalization requires π0+π1=1\pi_0+\pi_1=1. Thus π0=1/4\pi_0=1/4 and π1=3/4\pi_1=3/4.

  2. The two directed stationary traffic rates are π0w01=3/4 s−1\pi_0w_{01}=3/4\,\mathrm{s^{-1}} and π1w10=3/4 s−1\pi_1w_{10}=3/4\,\mathrm{s^{-1}}. Their difference is zero even though transitions continue occurring.

  3. Writing p˙1=3(1−p1)−p1\dot p_1=3(1-p_1)-p_1 gives p˙1=3−4p1\dot p_1=3-4p_1. A deviation from its stationary value therefore decays at rate 4 s−14\,\mathrm{s^{-1}}, with a relaxation time of one-quarter second.

The result

The stationary weights are one-quarter and three-quarters, edge traffic is balanced at 0.75 per second in each direction, and perturbations relax at 4 per second.

Explore the model

Interactive lab

Interactive lab · markov

Equal stationary weights can have different relaxation

Reset and set BOTH Transition p = 0.3 and Transition q = 0.1; predict stationary probabilities (0.25, 0.75). For a fresh comparison, Reset and set p = 0.6 and q = 0.2; predict the same stationary law but faster convergence. Step once and compare state-1 probabilities 0.3 and 0.6.

Equal stationary weights can have different relaxationExact 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 forThis instrument uses discrete transition probabilities and starts in state 0. It illustrates balance and relaxation but has only two states, so it cannot exhibit a driven three-state cycle or measure heat.

Check your understanding

Which observation rules out detailed balance in a stationary three-state ring?

Try it yourself

Three states form a ring. Every clockwise edge has rate 2 s−12\,\mathrm{s^{-1}} and every counterclockwise edge rate 1 s−11\,\mathrm{s^{-1}}. Show that the uniform distribution is stationary, then find each clockwise net edge current and the cycle affinity.

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

Give me a hint

At each node there is one incoming edge of each rate and one outgoing edge of each rate. Apply the product-ratio test separately from that node balance.

Compare with the solution

With πi=1/3\pi_i=1/3, each node's total inflow and outflow is (2+1)/3=1 s−1(2+1)/3=1\,\mathrm{s^{-1}}, so the distribution is stationary. Each clockwise oriented current is (1/3)(2)−(1/3)(1)=1/3 s−1(1/3)(2)-(1/3)(1)=1/3\,\mathrm{s^{-1}}. The cycle affinity is ln⁡(23/13)=3ln⁡2\ln(2^3/1^3)=3\ln2, which is nonzero. Thus stationary uniform occupations coexist with irreversible circulation, and detailed balance fails despite exact node-by-node conservation.

Take it with you

  • Stationarity balances nodes; detailed balance balances every reversible edge.
  • A nonzero cycle affinity detects driving hidden by stationary occupations.
  • Time reversal must transform odd variables and relevant external protocols consistently.

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