Technopedia

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Nonequilibrium Statistical Mechanics

Nonequilibrium statistical mechanics describes systems that are driven, relaxing, or transporting: heat flowing through a wire, particles diffusing in a concentration gradient, a fluid sheared by a moving wall, or a system far from equilibrium after a quench. It is relevant whenever the system is not in thermal equilibrium: biological cells, electronic devices, climate systems, active matter, and any system with sustained currents or time-dependent driving. It extends the equilibrium ensembles to time-dependent probability distributions. For the equilibrium building blocks, see Statistical Mechanics.

Authoritative sources

Master equation and Fokker–Planck equation

The probability distribution P(x,t)P(\mathbf{x}, t) of a system’s state evolves according to a master equation (discrete states) or a Fokker–Planck equation (continuous states):

∂P∂t=∑x′[W(x′→x) P(x′,t)−W(x→x′) P(x,t)]\frac{\partial P}{\partial t} = \sum_{\mathbf{x}'} \bigl[ W(\mathbf{x}' \to \mathbf{x})\, P(\mathbf{x}', t) - W(\mathbf{x} \to \mathbf{x}')\, P(\mathbf{x}, t) \bigr]

For a continuous variable xx with drift A(x)A(x) and diffusion D(x)D(x) (Itô convention):

∂P∂t=−∂∂x[A(x) P]+∂2∂x2[D(x) P]\frac{\partial P}{\partial t} = -\frac{\partial}{\partial x}\bigl[A(x)\,P\bigr] + \frac{\partial^2}{\partial x^2}\bigl[D(x)\,P\bigr]

Note. The form above is the Itô convention. In the Stratonovich convention the diffusion term reads ∂x[D(x) ∂xP]\partial_x[D(x)\,\partial_x P] instead. The two differ when D(x)D(x) is position-dependent; they coincide for constant DD.

State-transition schematic (two-state system):

State 0 State 1 W₀₁ W₁₀
Details

Step 1: The master equation as a probability flow.

The master equation says: the rate of change of probability in state xx equals the total probability flowing into xx minus the total flowing out. For a two-state system with states 0 and 1 and transition rates W01W_{01} (0→1) and W10W_{10} (1→0):

P˙0=−W01 P0+W10 P1,P˙1=W01 P0−W10 P1\dot{P}_0 = -W_{01}\,P_0 + W_{10}\,P_1, \qquad \dot{P}_1 = W_{01}\,P_0 - W_{10}\,P_1

Step 2: Solve the two-state example.

With P0+P1=1P_0 + P_1 = 1, the solution is

P0(t)=P0eq+(P0(0)−P0eq) e−(W01+W10) tP_0(t) = P_0^{\text{eq}} + \bigl(P_0(0) - P_0^{\text{eq}}\bigr)\, e^{-(W_{01}+W_{10})\,t}

where P0eq=W10/(W01+W10)P_0^{\text{eq}} = W_{10}/(W_{01}+W_{10}). The relaxation time is τ=1/(W01+W10)\tau = 1/(W_{01}+W_{10}). At equilibrium, detailed balance holds: W01 P0eq=W10 P1eqW_{01}\,P_0^{\text{eq}} = W_{10}\,P_1^{\text{eq}}.

Step 3: From discrete to continuous — the Fokker–Planck equation.

When the state space is continuous (e.g., position xx), the Kramers–Moyal expansion of the master equation truncates at second order if the steps are small. The result is the Fokker–Planck equation with drift A(x)A(x) (deterministic force) and diffusion D(x)D(x) (noise strength). For the Langevin equation mx¨=−γx˙+F+ξ(t)m\ddot{x} = -\gamma\dot{x} + F + \xi(t) with ⟨ξ(t)ξ(t′)⟩=2γkBT δ(t−t′)\langle\xi(t)\xi(t')\rangle = 2\gamma k_B T\,\delta(t-t'), the overdamped limit (m→0m\to 0) gives A(x)=F/γA(x) = F/\gamma and D=kBT/γD = k_B T/\gamma.

Step 4: Detailed balance and nonequilibrium currents.

The condition W(x→x′)Peq(x)=W(x′→x)Peq(x′)W(x\to x')P_{\text{eq}}(x) = W(x'\to x)P_{\text{eq}}(x') ensures zero net probability current at equilibrium. When violated (e.g., by a non-conservative driving force), a sustained current J(x)=A(x)P−∂x[D(x)P]≠0J(x) = A(x)P - \partial_x[D(x)P] \neq 0 persists, and the system produces entropy. See CanonicalEnsemble for the equilibrium fixed point Peq∝e−βEP_{\text{eq}} \propto e^{-\beta E}.

Step 5: Interactive simulation.

The simulation below shows a Brownian particle under the Langevin equation. You can adjust the friction γ\gamma and temperature TT to observe the transition from ballistic (inertial) to diffusive (overdamped) motion, and verify the Einstein relation D=kBT/γD = k_B T/\gamma from the mean-squared displacement.

Authoritative sources

Example problem

A two-state system has rates W01=2 s−1W_{01} = 2\,\text{s}^{-1} and W10=1 s−1W_{10} = 1\,\text{s}^{-1}. Starting from P0(0)=1P_0(0) = 1, find P0(t)P_0(t) and the equilibrium probability.

Solution

P0eq=W10/(W01+W10)=1/3P_0^{\text{eq}} = W_{10}/(W_{01}+W_{10}) = 1/3. The relaxation rate is W01+W10=3 s−1W_{01}+W_{10} = 3\,\text{s}^{-1}.

P0(t)=13+23 e−3tP_0(t) = \frac{1}{3} + \frac{2}{3}\,e^{-3t}

At t=1 st = 1\,\text{s}: P0(1)=13+23e−3≈0.34P_0(1) = \frac{1}{3} + \frac{2}{3}e^{-3} \approx 0.34.

Linear response and Kubo formula

For a small perturbation h(t)h(t) coupled to an observable BB, the response of observable AA is

⟨A(t)⟩−⟨A⟩eq=∫−∞tχAB(t−t′) h(t′) dt′\langle A(t) \rangle - \langle A \rangle_{\text{eq}} = \int_{-\infty}^{t} \chi_{AB}(t-t')\, h(t')\, dt'

where the response function (Kubo formula) is

χAB(t)=iℏ⟨[A(t),B(0)]⟩eq\chi_{AB}(t) = \frac{i}{\hbar} \langle [A(t), B(0)] \rangle_{\text{eq}}

The fluctuation–dissipation theorem (FDT) in its classical form connects the response to equilibrium correlations:

χAB(ω)=12kBT SAB(ω)\chi_{AB}(\omega) = \frac{1}{2k_B T}\, S_{AB}(\omega)

where SAB(ω)S_{AB}(\omega) is the symmetrized spectral density SAB(ω)=∫−∞∞dt eiωt 12⟨{A(t),B(0)}⟩eqS_{AB}(\omega) = \int_{-\infty}^{\infty} dt\, e^{i\omega t}\, \tfrac{1}{2}\langle\{A(t),B(0)\}\rangle_{\text{eq}}. The quantum generalization replaces the prefactor with tanh⁡(βℏω/2)/(2ℏ)\tanh(\beta\hbar\omega/2)/(2\hbar) (see Details).

FDT breakdown with driving strength:

Driving strength h Response / h Linear (FDT holds) Nonlinear (FDT breaks) h*
Details

Step 1: What the FDT says.

The FDT states that the linear response of a system at equilibrium to a small perturbation is completely determined by the equilibrium fluctuations of the same observable. In the classical limit, the susceptibility (response) is proportional to the symmetrized correlation function:

χAB(ω)=12kBT SAB(ω)\chi_{AB}(\omega) = \frac{1}{2k_B T}\, S_{AB}(\omega)

This is the nonequilibrium generalization of the equilibrium fluctuation relations: Var⁡(E)=kBT2CV\operatorname{Var}(E) = k_B T^2 C_V (canonical) and Var⁡(N)=kBT ∂μ⟨N⟩\operatorname{Var}(N) = k_B T\, \partial_\mu \langle N\rangle (grand canonical). See Statistical Foundations of Thermodynamics.

Step 2: Worked example — magnetization in a field.

Consider NN non-interacting spins in a magnetic field B(t)B(t). The observable is the magnetization M=∑iμσiM = \sum_i \mu \sigma_i. The perturbation is h(t)=B(t)h(t) = B(t).

Step 3: Quantum generalization.

In the quantum case, the FDT takes the form

χAB(ω)=12ℏ[1−e−βℏω] SAB(ω)=tanh⁡(βℏω/2)2ℏ SAB(ω)\chi_{AB}(\omega) = \frac{1}{2\hbar}\bigl[1 - e^{-\beta\hbar\omega}\bigr]\, S_{AB}(\omega) = \frac{\tanh(\beta\hbar\omega/2)}{2\hbar}\, S_{AB}(\omega)

where SAB(ω)S_{AB}(\omega) is the symmetrized spectral density. In the classical limit ℏω≪kBT\hbar\omega \ll k_B T, tanh⁡(βℏω/2)≈βℏω/2\tanh(\beta\hbar\omega/2) \approx \beta\hbar\omega/2, giving χAB(ω)≈ω/(2kBT) SAB(ω)\chi_{AB}(\omega) \approx \omega/(2k_B T)\, S_{AB}(\omega), which is consistent with the classical FDT up to the conventional factor of ω\omega relating the one-sided and two-sided spectra.

Step 4: When the FDT breaks down.

The FDT holds only for linear response (small hh). Far from equilibrium, the response is nonlinear: higher-order response functions χ(2),χ(3),…\chi^{(2)}, \chi^{(3)}, \ldots appear, and the simple proportionality between response and correlation no longer holds. The diagram above illustrates this: for h<h∗h < h^* the response is linear (FDT valid); for h>h∗h > h^* it saturates or becomes nonlinear (FDT invalid).

Authoritative sources

Example problem

A paramagnetic salt with N=1020N = 10^{20} spins of moment μ=10−23 J/T\mu = 10^{-23}\,\text{J/T} is in a field B0=0.1 TB_0 = 0.1\,\text{T} at T=300 KT = 300\,\text{K}. Compute the static susceptibility χMM\chi_{MM} and verify the FDT.

Solution

Since βμB0=(10−23×0.1)/(1.38×10−23×300)≈2.4×10−4≪1\beta\mu B_0 = (10^{-23} \times 0.1)/(1.38\times10^{-23}\times 300) \approx 2.4\times10^{-4} \ll 1, we are in the Curie-law regime:

χMM≈Nμ2kBT=1020×(10−23)21.38×10−23×300≈2.4×10−14 J/T\chi_{MM} \approx \frac{N\mu^2}{k_B T} = \frac{10^{20}\times(10^{-23})^2}{1.38\times10^{-23}\times 300} \approx 2.4\times10^{-14}\,\text{J/T}

The equilibrium correlation at zero frequency is SMM(0)=Nμ2=10−26 J2/T2S_{MM}(0) = N\mu^2 = 10^{-26}\,\text{J}^2/\text{T}^2.

FDT check: SMM(0)/(kBT)=10−26/(1.38×10−23×300)≈2.4×10−14S_{MM}(0)/(k_B T) = 10^{-26}/(1.38\times10^{-23}\times 300) \approx 2.4\times10^{-14} ✓.

Transport coefficients

In the linear-response regime, transport is described by Onsager reciprocal relations:

Ji=∑jLij XjJ_i = \sum_j L_{ij}\, X_j

where JiJ_i are fluxes (heat, particle, momentum) and XjX_j are thermodynamic forces (gradients of 1/T1/T, μ/T\mu/T, velocity). The Onsager matrix LijL_{ij} is symmetric: Lij=LjiL_{ij} = L_{ji}.

Examples:

Details

Step 1: Derivation from time-reversal symmetry.

The Onsager relations follow from the microscopic reversibility of the equations of motion. If the dynamics is invariant under t→−tt \to -t combined with p→−p\mathbf{p} \to -\mathbf{p}, then the cross-correlation of two fluxes satisfies Lij=LjiL_{ij} = L_{ji} (or Lij=−LjiL_{ij} = -L_{ji} if one flux is odd under time reversal, e.g., magnetization).

Step 2: Concrete example — thermoelectric coupling.

In a conductor with both a temperature gradient and an electric field, the coupled transport equations are

Je=σ E−σS ∇T,JQ=Π σ E−κ ∇T\mathbf{J}_e = \sigma\,\mathbf{E} - \sigma S\,\nabla T, \qquad \mathbf{J}_Q = \Pi\,\sigma\,\mathbf{E} - \kappa\,\nabla T

The Onsager symmetry requires S=Π/σS = \Pi/\sigma (Seebeck coefficient equals Peltier coefficient divided by conductivity, up to TT). This is the Thomson relation.

Step 3: Entropy production and the second law.

The local entropy production rate is σs=∑iJiXi≥0\sigma_s = \sum_i J_i X_i \geq 0. For a symmetric positive-semidefinite Onsager matrix, this is automatically satisfied. Far from equilibrium, the linear relation J=LXJ = LX breaks down and nonlinear transport (turbulence, pattern formation) emerges. See PhaseTransitions for how nonequilibrium driving can induce phase transitions.

Authoritative sources

Example problem

A metal wire of length L=1 cmL = 1\,\text{cm} has a temperature difference ΔT=1 K\Delta T = 1\,\text{K} across it. The thermal conductivity is κ=400 W/(m\cdotpK)\kappa = 400\,\text{W/(m·K)} and the Seebeck coefficient is S=2 μV/KS = 2\,\mu\text{V/K}. Find the open-circuit voltage and the entropy production rate per unit volume.

Solution

Open-circuit voltage (no current): V=S ΔT=2×10−6×1=2 μVV = S\,\Delta T = 2\times10^{-6}\times 1 = 2\,\mu\text{V}.

Entropy production per unit volume: σs=κ(∇T)2/T2=400×(100)2/3002≈4.4 W/(m3\cdotpK)\sigma_s = \kappa(\nabla T)^2/T^2 = 400\times(100)^2/300^2 \approx 4.4\,\text{W/(m}^3\text{·K)}.

Entropy production and the second law

The entropy production rate is the central organizing principle of nonequilibrium thermodynamics:

σs=∑iJi Xi≥0\sigma_s = \sum_i J_i\, X_i \geq 0

Near equilibrium, the minimum entropy production principle (Prigogine) states that a system with fixed boundary conditions relaxes to the steady state that minimizes σs\sigma_s. Far from equilibrium, the second law emerges as an inequality from the Jarzynski equality via Jensen’s inequality.

Details

Step 1: Local entropy balance.

For a continuum, the entropy density s(r,t)s(\mathbf{r},t) obeys

∂s∂t=−∇⋅Js+σs\frac{\partial s}{\partial t} = -\nabla \cdot \mathbf{J}_s + \sigma_s

where Js\mathbf{J}_s is the entropy flux and σs≥0\sigma_s \geq 0 is the local entropy production rate. In the linear regime, σs=∑iJiXi\sigma_s = \sum_i J_i X_i with fluxes and forces as in the Onsager section. The second law (σs≥0\sigma_s \geq 0) is a local statement: entropy is produced at every point, not just on average.

Step 2: Minimum entropy production (Prigogine, 1945).

For a system near equilibrium with fixed boundary conditions (e.g., fixed temperature difference across a slab), the steady state minimizes the total entropy production. Example: in a 1D slab with fixed T1,T2T_1, T_2, the steady-state temperature profile is linear, and any deviation increases ∫σs dV\int \sigma_s\, dV. This principle does not hold far from equilibrium.

Step 3: Second law from the Jarzynski equality.

The Jarzynski equality ⟨e−βWd⟩=e−βΔF\langle e^{-\beta W_d}\rangle = e^{-\beta\Delta F} is an exact identity. Applying Jensen’s inequality to the convex function e−xe^{-x}:

⟨Wd⟩≥−1βln⁡⟨e−βWd⟩=ΔF\langle W_d \rangle \geq -\frac{1}{\beta}\ln\langle e^{-\beta W_d}\rangle = \Delta F

This recovers the second law ⟨Wd⟩≥ΔF\langle W_d\rangle \geq \Delta F as a consequence of an exact equality. The equality holds only for reversible (quasistatic) processes; any irreversibility makes the inequality strict. This connects the macroscopic second law to the microscopic fluctuation theorems below.

Authoritative sources

Example problem

A system is driven from state A to state B with ΔF=5 kBT\Delta F = 5\,k_B T. In 10610^6 independent trials, the average work is ⟨Wd⟩=7 kBT\langle W_d\rangle = 7\,k_B T. Compute the average entropy production in units of kBk_B and verify it is consistent with the second law.

Solution

Average entropy production: ⟨Σ⟩=β(⟨Wd⟩−ΔF)=(7−5)=2 kB\langle \Sigma\rangle = \beta(\langle W_d\rangle - \Delta F) = (7-5) = 2\,k_B.

Since ⟨Σ⟩=2 kB>0\langle\Sigma\rangle = 2\,k_B > 0, the second law is satisfied. The irreversibility is moderate: the process is far from reversible (which would give ⟨Σ⟩=0\langle\Sigma\rangle = 0).

Fluctuation theorems

Far from equilibrium, the Jarzynski equality and Crooks fluctuation theorem generalize the second law:

⟨e−βWd⟩=e−βΔF\langle e^{-\beta W_d} \rangle = e^{-\beta \Delta F} PF(W)PR(−W)=eβ(W−ΔF)\frac{P_F(W)}{P_R(-W)} = e^{\beta(W - \Delta F)}

where WdW_d is the work done in a nonequilibrium process and ΔF\Delta F is the free-energy difference. These hold for any driving protocol, not just linear response.

Details

Step 1: The Jarzynski equality as an exact identity.

The Jarzynski equality is not an approximation: it holds for arbitrary driving protocols, arbitrary initial equilibrium states, and arbitrary system size. It reduces to the second law by Jensen’s inequality (see the Entropy production section above). The equality encodes the full statistics of work, not just the mean.

Step 2: The Crooks theorem and time-reversal.

The Crooks fluctuation theorem relates the probability PF(W)P_F(W) of observing work WW in a forward protocol to the probability PR(−W)P_R(-W) of observing work −W-W in the time-reversed protocol. The ratio is fixed by the free-energy difference. In the limit of large systems or long times, the ratio becomes sharply peaked and the second law emerges as a statement about typical trajectories.

Step 3: Experimental verification.

These theorems have been verified in single-molecule pulling experiments (RNA unfolding, DNA stretching) and in colloidal particles in optical traps. In the RNA experiments, the work distribution for unfolding and refolding shows the predicted exponential ratio, and the Jarzynski equality correctly extracts the free-energy difference even for fast, far-from-equilibrium pulling.

Step 4: Connection to equilibrium.

In the quasistatic limit (infinitely slow driving), Wd→ΔFW_d \to \Delta F with zero variance, and both the Jarzynski equality and Crooks theorem reduce to the equilibrium statement ΔF=Wrev\Delta F = W_{\text{rev}}. See MicrocanonicalEnsemble for the isolated-system limit where work is the only energy exchange.

Authoritative sources

Example problem

A two-level system with gap ε\varepsilon is driven by suddenly changing the gap from ε0\varepsilon_0 to ε1\varepsilon_1. The initial state is thermal at temperature TT. Compute the work distribution and verify the Jarzynski equality.

Solution

The Hamiltonian is H=ε∣1⟩⟨1∣H = \varepsilon|1\rangle\langle 1|. The work is W=⟨n∣H1−H0∣n⟩W = \langle n|H_1 - H_0|n\rangle. If the system is in state ∣0⟩|0\rangle: W=0W = 0. If in ∣1⟩|1\rangle: W=ε1−ε0W = \varepsilon_1 - \varepsilon_0.

The initial probability of the excited state is p=e−βε0/(1+e−βε0)p = e^{-\beta\varepsilon_0}/(1+e^{-\beta\varepsilon_0}).

⟨e−βW⟩=(1−p)⋅1+p⋅e−β(ε1−ε0)=1+e−βε11+e−βε0=e−βΔF\langle e^{-\beta W}\rangle = (1-p)\cdot 1 + p\cdot e^{-\beta(\varepsilon_1-\varepsilon_0)} = \frac{1 + e^{-\beta\varepsilon_1}}{1+e^{-\beta\varepsilon_0}} = e^{-\beta\Delta F}

where ΔF=−kBTln⁡[(1+e−βε1)/(1+e−βε0)]\Delta F = -k_B T\ln[(1+e^{-\beta\varepsilon_1})/(1+e^{-\beta\varepsilon_0})]. ✓

Quizz

A particle of mass mm in a viscous medium at temperature TT is subject to a constant force FF. Show that in the overdamped limit the mean velocity is F/γF/\gamma and the diffusion coefficient is kBT/γk_B T/\gamma.

Solution

Start from the Langevin equation:

mx¨=−γx˙+F+ξ(t),⟨ξ(t)ξ(t′)⟩=2γkBT δ(t−t′)m\ddot{x} = -\gamma\dot{x} + F + \xi(t), \qquad \langle\xi(t)\xi(t')\rangle = 2\gamma k_B T\,\delta(t-t')

Step 1: Overdamped limit.

In the limit m→0m \to 0 (or equivalently t≫m/γt \gg m/\gamma), the inertial term vanishes and the equation becomes

x˙=Fγ+η(t)\dot{x} = \frac{F}{\gamma} + \eta(t)

where η(t)\eta(t) is a white noise with ⟨η(t)η(t′)⟩=2D δ(t−t′)\langle\eta(t)\eta(t')\rangle = 2D\,\delta(t-t') and D=kBT/γD = k_B T/\gamma.

Step 2: Mean velocity.

Taking the ensemble average: ⟨x˙⟩=F/γ+⟨η⟩=F/γ\langle\dot{x}\rangle = F/\gamma + \langle\eta\rangle = F/\gamma. The particle drifts at the force divided by friction.

Step 3: Diffusion coefficient (Einstein relation).

The mean-squared displacement is

⟨[x(t)−x(0)]2⟩=F2γ2t2+2Dt\langle [x(t)-x(0)]^2 \rangle = \frac{F^2}{\gamma^2}t^2 + 2Dt

For F=0F = 0 (pure diffusion): ⟨Δx2⟩=2Dt\langle\Delta x^2\rangle = 2Dt, giving D=kBT/γD = k_B T/\gamma.

Step 4: Connection to the FDT.

The Einstein relation D=μmob kBTD = \mu_{\text{mob}}\, k_B T (where μmob=1/γ\mu_{\text{mob}} = 1/\gamma is the mobility) is a direct consequence of the fluctuation–dissipation theorem: the same friction γ\gamma that causes dissipation (the drift term) also sets the noise amplitude (the diffusion term). This is the simplest physical manifestation of the FDT.