Non-equilibrium and stochastic thermodynamics

The histogram that hides a current

Imagine a small driven system with three possible positions around a ring. At a randomly chosen time, each position is equally likely. A histogram therefore has three equal bars. Does that mean the system is in equilibrium?

Not necessarily. Suppose clockwise jumps occur twice as readily as counterclockwise jumps. The bars can stay equal while a persistent circulation continues. A stationary probability distribution tells us where systems are found, but it need not tell us which transitions they make. Maintaining the directional bias requires a physical source of energy or another thermodynamic drive.

We will follow this assigned ring in two distinct experiments: relaxation from unequal probabilities, and fluctuations in a separately prepared stationary ensemble. It is a model with symbolic time and temperature scales, not a measurement of a particular molecular motor. It lets us connect individual jumps, ensemble averages, energy exchange and the second law without confusing a steady state with equilibrium.

Choose an ensemble and a reading path

A trajectory records one system’s random sequence of jumps and waiting times. An ensemble consists of many repetitions prepared under the same conditions. Its probability p_i(t) is the chance of finding state i at time t, with Σ_i p_i=1. A smooth probability curve is not the continuous orbit of one particle; each individual position trajectory jumps between the three states.

First read the rates, energy signs, entropy balance and worked ring example. The later return path explains path-probability ratios, work relations, cycle affinity, housekeeping production and precision bounds. Natural logarithms are used throughout. The Boltzmann constant k_B converts dimensionless entropy expressions into entropy units J K⁻¹.

Probability and statistical entropy provides preparation for averages and logarithms; the first-law and irreversible-transfer lessons connect the signs and macroscopic balances. Equilibrium response and kinetic transport provide useful comparisons, but neither their equilibrium assumptions nor a velocity-distribution closure will be silently transferred to this driven jump model.

Turn jump rates into probability flow

Let τ>0 be a time scale. On every link of the ring, the clockwise transition rate is 2/τ and the counterclockwise rate is 1/τ. A rate is probability per small time interval conditional on the present state: for a sufficiently short Δt, a specified clockwise jump has probability approximately 2Δt/τ. It is not a probability of 2, and rates do not predict a fixed waiting time.

Write s=t/τ and label states 0,1,2, with indices understood modulo 3. The dimensionless clockwise net current on edge i→i+1 is J_i=2p_i−p_(i+1). It is forward probability flow minus backward probability flow, not the total number of jumps. The physical net flow is J_i/τ.

Probability conservation gives dp_i/ds=J_(i−1)−J_i=2p_(i−1)+p_(i+1)−3p_i. The gain terms bring probability into state i and the loss term removes it. Summing all three equations gives zero because every departing contribution arrives elsewhere. The continuous-time Markov assumption means that these specified rates and the current state suffice to determine future jump probabilities; unresolved memory would require a different description.

Distinguish equilibrium from stationarity

The uniform distribution π_i=1/3 is stationary: incoming and outgoing flow balance at every state. Yet every oriented edge has J_i=2/3−1/3=1/3. Probability circulates without accumulating anywhere, much as steady flow can enter and leave each segment of a closed pipe at equal rates.

Detailed balance is the stronger condition π_i w_(i→j)=π_j w_(j→i) on every resolved reversible link. It eliminates each pair’s stationary net current. Under the thermal and state-resolution assumptions used here, it characterizes equilibrium. Stationarity requires only the total balance at each state and allows a nonequilibrium steady state.

For the ring, the product of clockwise-to-counterclockwise rate ratios around a complete cycle is 2³=8. An equilibrium potential would give ratios whose energy differences telescope around a closed cycle, leaving product 1. No reassignment of three equilibrium state energies can supply this persistent cycle bias. Equal bars cannot repair the failure of detailed balance.

State the first law and the energy source

Choose an explicit sign convention: W_on is work supplied to the system and Q_bath is heat released to the bath. Then ΔU=W_on−Q_bath. Positive heat here leaves the system. A different reference may define heat into the system; its signs must be translated before comparing formulas.

For our isothermal interpretation, let all three internal state energies be equal, with no differences in state-internal entropy, and let the bath have temperature T>0. Assign external nonconservative work +k_B T ln 2 to each clockwise jump and the opposite amount to each counterclockwise jump. Since a jump changes no internal energy, Q_bath=W_on for that jump. The local rate ratio is consistent with bath entropy change ±k_B ln 2 under this declared local-detailed-balance interpretation.

The word declared matters. A list of transition rates alone does not uniquely identify microscopic work, heat, reservoirs or state-internal entropy. Experimental thermodynamic inference needs that additional information and suitably resolved transition channels. The ring’s constant internal energy therefore does not mean zero work: the external drive supplies energy that is continually passed to the bath.

Give system entropy its probability meaning

For a trajectory occupying state X_t, define stochastic system entropy s_sys(t)=−k_B ln p_(X_t)(t). It depends on the ensemble probability of the actual state, not just the state’s name or internal energy. Averaging gives S_sys=−k_B Σ_i p_i ln p_i. This is the discrete Shannon entropy with the physical factor k_B.

A trajectory’s system-entropy change is Δs_sys=k_B ln[p_initial(X_0)/p_final(X_t)]. When an ensemble is changing, this quantity can change even during a waiting period with no jump. That change is not automatically heat. For the uniform stationary ensemble both endpoint probabilities equal 1/3, so Δs_sys=0 path by path.

System entropy is only one part of the balance. The bath receives entropy Q_bath/T in this one-bath interpretation; total trajectory production is Σ=Δs_sys+Q_bath/T. An open system’s own entropy need not always increase. The second-law restriction concerns the appropriate total average, with physical exchanges and the statistical description stated.

Follow the unequal initial probabilities

Now prepare the distinct transient experiment with p(0)=(3/5,3/10,1/10), keeping the same rates. The initial oriented currents are (0.9,0.5,−0.4), and dp/ds=(−1.3,0.4,0.9). The last current is initially counterclockwise even though its clockwise rate is larger: there is initially much more probability at the destination state. Direction depends on flow, which combines rates and populations.

The equations can be solved exactly. Let δ_i=p_i(0)−1/3, let Q act as (Qp)_i=2p_(i−1)+p_(i+1)−3p_i, and set b=(Q+(9/2)I)δ, where I is the identity. With ω=√3/2, the solution is p_i(s)=1/3+e^(−9s/2)[δ_i cos(ωs)+(b_i/ω)sin(ωs)]. The exponential decay and oscillatory factors come from the two nonstationary eigenmodes; no fitted decay time has been inferred from a material.

The supplied numerical samples cover s=0 to 4 in steps of 1/40. All probabilities stay strictly positive and remain normalized without clipping or rescaling. The selected values below are rounded display values. At s=4 the largest probability deviation from 1/3 is still about 4.21×10⁻⁹; a row whose displayed bars look equal is not exactly stationary.

For this doubly stochastic ring, S_sys increases toward k_B ln 3 and D(p||π)=Σ_i p_i ln(3p_i)=ln 3−S_sys/k_B decreases. D is a relative entropy to the stationary distribution. Here that distribution is driven, not equilibrium. Its disappearance says that the probabilities have relaxed, not that the ongoing circulation or its energy cost has disappeared.

Separate entropy flow from entropy production

At an ensemble level, the dimensionless total production rate is σ_tot τ/k_B=Σ_i(2p_i−p_(i+1))ln[2p_i/p_(i+1)]. Each of the three undirected edges is counted once using its clockwise orientation. There is no additional factor 1/2 in this once-counted convention. Reversing an edge changes both signs, leaving its contribution unchanged.

For positive a and b, (a−b)ln(a/b)≥0 because both factors have the same sign. Thus each edge contribution is nonnegative, including the edge whose net current is initially negative. This proves nonnegative average production for the assigned positive-rate Markov model; it does not say that every individual trajectory has positive entropy production.

The medium entropy flow obeys σ_med τ/k_B=ln 2 Σ_i J_i=ln 2, since Σ_i J_i=1. Meanwhile d(S_sys/k_B)/ds=Σ_i J_i ln[p_i/p_(i+1)]. Consequently σ_tot τ/k_B=ln 2+d(S_sys/k_B)/ds. During relaxation the system contribution is positive in this example; at stationarity it vanishes, leaving the persistent ln 2 bath flow.

Integrate production without losing the steady cost

Over an observation interval 0 to t=τs, the mean total entropy produced is ⟨Σ⟩/k_B=s ln 2+[S_sys(s)−S_sys(0)]/k_B. The first term is the accumulated bath contribution. The second records ensemble relaxation. Entropy production, its rate and an entropy change are different quantities with different units.

Initially S_sys(0)/k_B≈0.897945725, whereas ln 3≈1.098612289. The system’s eventual entropy increase is therefore only about 0.200666564 k_B. The bath term s k_B ln 2 continues to grow linearly for as long as the drive is maintained. A finite relaxation contribution does not cancel a continuing maintenance cost.

This distinction is useful when interpreting a molecular machine, a driven chemical network or a coarse position trace. A stationary histogram can coexist with continual work consumption and heat release. To quantify a real device, its transitions, reservoir transfers, force or chemical work, unresolved states and time calibration must be known; our rates alone do not specify them.

Further: compare a path with its time reverse

A Markov path probability includes its initial-state probability, all jump rates along the path, and exponential survival factors for its waiting intervals. Compare the forward path with the reversed sequence of states and waiting intervals under the appropriately reversed driving protocol. For these even position states and the constant imposed drive, that protocol has the same rates; reversing a realized jump uses the rate for the opposite jump.

The survival factors cancel in this ring, and each clockwise jump contributes ln 2 to the log probability ratio, while each reverse jump contributes −ln 2. With the reverse experiment initially sampled from the actual forward endpoint distribution, the full ratio is Σ/k_B=N ln 2+ln[p_initial(X_0)/p_s(X_s)], where N=N_+−N_− is the net number of directional jumps.

N is not necessarily a whole number of completed cycles. A single clockwise jump gives N=1; an unwrapped displacement in turns is N/3, with an endpoint convention needed for integer winding counts. Also, swapping the rates 2 and 1 constructs this ring’s stationary adjoint dynamics, which reverses its stationary currents. It is a distinct comparison, not the automatic definition of physical time reversal. Momentum, magnetic fields and other odd variables require their own parity rules.

Further: understand what a fluctuation theorem says

When Σ/k_B is the log ratio of the specified forward and normalized reversed path measures, with matching support, averaging exp(−Σ/k_B) over forward paths gives 1. The reason is direct: multiplication by the exponential converts each forward path weight into its reversed weight, and those weights sum to one. This is an integral fluctuation relation under its stated conditions, not a rule for an arbitrarily chosen entropy-like quantity.

Jensen’s inequality then gives 1=⟨e^(−Σ/k_B)⟩≥e^(−⟨Σ⟩/k_B), hence ⟨Σ⟩≥0. Individual negative values remain possible. A detailed relation comparing probabilities at opposite entropy values needs the corresponding forward/reverse distributions and extra symmetry or stationary conditions; it is not automatically the same assertion as the integral relation.

For a separately initialized stationary ring, directional counts are independent Poisson variables N_+ with mean 2s and N_− with mean s. Constant hazards make this exact. Here the endpoint system term is zero, so Σ/k_B=N ln 2. The complete generating function gives ⟨e^(−Σ/k_B)⟩=exp{2s(1/2−1)+s(2−1)}=1. No finite count cutoff or renormalized histogram was used. For the unequal transient preparation, the same N ln 2 exponential average also happens to be one, but it omits the system endpoint term and is not the full total entropy. Passing that exponential test alone would miss the omission.

Further: keep work identities and endpoint entropy distinct

A different class of experiment begins in canonical thermal equilibrium and changes a prescribed control parameter. For the classical setting used here, assume weak coupling to a common equilibrium bath at temperature T and use the work-on-system convention. Jarzynski’s equality is ⟨e^(−βW_on)⟩=e^(−βΔF), where β=1/(k_B T) and ΔF is the equilibrium Helmholtz free-energy difference of the bare system between the endpoint controls. Each repetition needs the specified canonical preparation; the driving may be fast, and the final actual state need not be equilibrated.

Crooks’s work relation compares that forward protocol with its appropriately reversed protocol, prepared canonically at the forward final control: P_F(W)/P_R(−W)=e^[β(W−ΔF)] where the compared probability densities have support. A work value in the tail of one ensemble is compared with the opposite work under a different preparation and protocol. Jarzynski’s equality itself does not require that a reverse experiment actually be performed.

These work relations do not license replacing the actual endpoint probability inside stochastic system entropy by an equilibrium probability. If the forward process starts canonically but ends with distribution p_t, its physical finite-time total entropy in the simple one-bath setting is Σ/k_B=β(W_on−ΔF)−ln[p_t(X_t)/p_eq,B(X_t)]. The distribution correction vanishes only with the appropriate endpoint equality; alternatively, include and analyse an additional relaxation stage. The supplied noncanonical transient ring cannot simply be inserted into the uncorrected canonical work formula.

For fixed equilibrium energies and a suitable thermal reference, a nonequilibrium free-energy functional can differ from equilibrium free energy by k_B T D(p||p_eq). That is a statement about a specified reference and allowed thermodynamics, not a universal work-extraction recipe. In our driven ring, D(p||π) tends to zero while external work is still consumed. A small relative entropy to a driven steady state is not evidence of cost-free operation.

Further: read cycles, affinities and power

The dimensionless affinity of the clockwise cycle is A=ln[(2·2·2)/(1·1·1)]=3 ln 2. It adds the three local bias contributions around a complete loop. At stationarity, conservation makes the common oriented edge current J_cycle=1/3 in units of 1/τ. Multiplying gives σ_tot=(k_B/τ)J_cycle A=(k_B/τ)ln 2.

The total net directional-jump rate is instead Σ_i J_i/τ=1/τ, three times the cycle current. Multiplying that rate by work k_B T ln 2 per net jump gives mean supplied power k_B T ln 2/τ. Both calculations agree once the observable and units are matched. Counting edge currents three times and also using the full-cycle affinity would overcount production.

Larger networks can have several independent cycles and multiple physical reservoirs. A cycle decomposition organizes the currents and affinities, but a single scalar histogram cannot determine all of them. Reservoir-resolved channels are essential when different transitions share the same coarse pair of state labels. Combining channels before computing a logarithmic rate ratio can discard thermodynamic information.

Further: separate housekeeping from relaxation

Housekeeping production is the cost associated with sustaining a driven stationary circulation. In this fixed-rate ring with uniform π, its trajectory value is k_B N ln 2 and its mean rate is k_B ln 2/τ, even while the initial probabilities relax. The extra ensemble relaxation rate is −k_B dD(p||π)/dt and is nonnegative for this model.

Here the stationary-potential change along a jump is zero, since −ln π_i is the same at every state. Thus the medium excess term defined from those stationary-potential changes is zero. The nonadiabatic or relaxation part of total production nevertheless contains the system-entropy change. Calling all of these quantities “excess heat” would erase a real distinction.

The term adiabatic is also used for the housekeeping part in one entropy-production decomposition. That usage does not mean that our driven ring is thermally insulated; it continually releases heat to its bath. Definitions, controls and ensemble must be stated when comparing decompositions, especially with time-dependent driving, odd variables or several reservoirs.

Further: connect current precision with dissipation

For a stationary, irreducible, time-homogeneous continuous-time Markov jump process with thermodynamically consistent bidirectional rates on resolved even-parity states, a finite-time thermodynamic uncertainty relation bounds an antisymmetric integrated current J by Var(J)/⟨J⟩²≥2/(⟨Σ⟩/k_B), when the mean is nonzero and the required moments are finite. Even-parity states, such as the ring’s positions, are unchanged under time reversal. The bound links relative current noise to total mean dissipation; its preparation, reversal and dynamics are part of the theorem.

For our stationary net count N, ⟨N⟩=s and Var(N)=3s because the independent Poisson variances add. For s>0, the relative variance is 3/s and ⟨Σ⟩/k_B=s ln 2. Their product is 3 ln 2≈2.07944154, just above the lower bound 2. The ratio at s=0 is undefined, not zero, because both the mean and variance vanish.

This is a bound, not a claim that precision determines a unique mechanism or that every process attains equality. It must not be transferred unchanged to arbitrary transient preparations, discrete-time dynamics, time-dependent protocols, active baths or quantum coherence. A noise measurement can constrain dissipation only after the applicable physical and statistical model is justified.

Ask what the observer has hidden

A camera may distinguish three visible positions while missing chemical states, rapid transitions or momentum. Two records with the same visible jumps may then carry different heat or work. Grouping states can introduce memory, and fitting constant rates to the grouped record does not automatically recover the thermodynamics of the hidden system.

For a genuine projection of forward and reverse path measures, their relative entropy cannot increase under information loss. That mathematical statement does not guarantee that every fitted Markov model of the projected data gives a certified entropy-production lower bound: the fitted measure and the actual projected measure can differ. Time resolution, missed events, reservoir channels and state-internal equilibration all matter.

An experimentally useful question is therefore not merely “Can I fit these probabilities?” but “Which paths and exchanges does this observation resolve?” A thermodynamic coarse-graining needs stated scale-separation and equilibration conditions. In particular, results derived when all transitions share one common reservoir cannot simply be applied to a model with several unresolved reservoirs.

Choose the model before extending the result

Jump models suit sufficiently resolved discrete configurations and reaction events. Continuous overdamped Langevin models can describe a position moving through a fluid under appropriate time-scale and bath assumptions. Underdamped models keep momentum and kinetic energy; heat bookkeeping, stochastic integration and reversal parity must be consistent. These descriptions are related tools, not interchangeable formulas.

Strong system–bath coupling, memory, multiple reservoirs, active environments and coherent quantum dynamics require separately justified energy, entropy and reversal constructions. A bare-system equilibrium free energy or a simple even-state rate ratio may no longer be the correct quantity. Mentioning these regimes identifies the boundary of the present lesson; it does not solve their dynamics.

The central lesson is experimentally concrete. A stationary histogram can conceal a sustained current. Follow the transitions, identify what supplies energy, distinguish trajectory quantities from averages, and retain actual endpoint probabilities. With those conditions in place, fluctuations and the second law become compatible parts of the same explanation.

Worked examples

Try it yourself

1. Do three time-independent equal probability bars establish equilibrium?

   Show the answer and reasoning

   No.

   Stationarity concerns net accumulation at each state. Detailed balance also requires each resolved forward flow to equal its reverse.

2. Does the rate 2/τ mean a transition has probability 2?

   Show the answer and reasoning

   No.

   For sufficiently small Δt its conditional probability is approximately 2Δt/τ. The exact waiting-time model keeps probabilities between 0 and 1.

3. Why are there three terms rather than six in the total-rate expression?

   Show the answer and reasoning

   Each undirected edge is counted once.

   Its forward-minus-reverse flow is already paired with the forward/reverse logarithm. A directed double sum uses a different prefactor convention.

4. If a<b, can (a−b)ln(a/b) be negative?

   Show the answer and reasoning

   No; it is positive.

   Both a−b and ln(a/b) are negative, so their product is positive; it vanishes only when a=b.

5. If the dimensionless edge current is 1/3, what is its physical rate?

   Show the answer and reasoning

   1/(3τ).

   The time coordinate is s=t/τ. Derivatives or currents per unit s acquire a factor 1/τ in physical time.

6. Does ΔU=0 imply W_on=0?

   Show the answer and reasoning

   No.

   With the chosen signs, ΔU=W_on−Q_bath. Equal work input and heat output leave internal energy unchanged.

7. Can stochastic system entropy change during a wait with no jump?

   Show the answer and reasoning

   Yes, if the ensemble probabilities change.

   s_sys=−k_B ln p_(X_t)(t) contains explicit time dependence. This statistical change is not automatically a heat transfer.

8. At s=4, may rounded equal bars be declared exactly stationary?

   Show the answer and reasoning

   No.

   The exact solution retains probability deviations of about 4.21×10⁻⁹. Display resolution is not a mathematical limit.

9. Is one net clockwise jump necessarily one completed cycle?

   Show the answer and reasoning

   No.

   The ring has three edges per turn. The net directional count and cycle current are different observables.

10. What must be added to N ln 2 for the transient total entropy divided by k_B?

    Show the answer and reasoning

    ln[p_initial(X₀)/p_s(X_s)].

    The stochastic system endpoint term vanishes for the separate uniform stationary ensemble, not the unequal transient preparation.

11. Does ⟨exp(−Σ/k_B)⟩=1 imply Σ≥0 for every path?

    Show the answer and reasoning

    No.

    Jensen gives a nonnegative mean under the stated relation. Individual negative fluctuations remain possible.

12. Would an exponential mean of 1 for N ln 2 alone prove that the transient endpoint term was included?

    Show the answer and reasoning

    No.

    For these constant directional clocks, that housekeeping functional also has exponential mean 1. Its identity alone cannot identify the full transient total entropy.

13. Can the noncanonical supplied transient preparation be inserted into the uncorrected standard Jarzynski equality?

    Show the answer and reasoning

    No.

    The standard equality requires its canonical initial preparation and stated work/bath assumptions; altered preparations need the appropriate correction.

14. Must the driven forward process already be equilibrated when its protocol ends for the work relation?

    Show the answer and reasoning

    No.

    The reverse experiment begins canonically at the forward final control. Endpoint reference free energy is distinct from the actual forward endpoint distribution.

15. What is Var(N)/⟨N⟩² at s=0?

    Show the answer and reasoning

    Undefined.

    Both numerator and denominator vanish. The expression 3/s and the stated nonzero-mean uncertainty comparison apply for s>0.

16. Does a good constant-rate fit to observed positions certify all heat and entropy production?

    Show the answer and reasoning

    No.

    Hidden states, reservoir channels and memory can change path probabilities and physical transfers. The observation and thermodynamic coarse-graining require separate justification.

What to carry forward

• Equal time-independent probabilities do not establish equilibrium. Resolved stationary currents can reveal a maintained nonequilibrium circulation.
• Rates specify stochastic dynamics; physical heat and work require a stated energetic and reservoir interpretation. Work can pass through a system with no internal-energy change.
• Total trajectory entropy includes the actual ensemble endpoint term. The uniform stationary and unequal transient preparations must not be interchanged.
• Average total production is nonnegative under the stated model, while individual negative fluctuations are possible. A fluctuation relation needs its specific forward/reverse path measures and support.
• Canonical work identities use equilibrium reference free energies and prescribed preparations; they do not silently replace actual nonequilibrium endpoint probabilities.
• A driven steady state can keep dissipating after its histogram relaxes. Precision bounds and hidden-state inferences are useful only within their physical and statistical conditions.