advanced26 min studyLesson 342 in Physics

Ornstein–Uhlenbeck relaxation

Solve a linear noisy relaxation process, distinguish transient variance from stationary correlation, and connect its steady width to a thermal harmonic trap.

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What you’ll understand

  • Compute the conditional mean and variance of an Ornstein–Uhlenbeck process.
  • Distinguish a stationary ensemble from a deterministic initial condition.
  • Relate relaxation rate and equilibrium variance to trap stiffness and mobility.

A restoring force prevents Brownian spreading from growing indefinitely. The simplest example is the Ornstein–Uhlenbeck process, which combines linear relaxation with additive Gaussian noise. Its exact solution separates memory of the initial state from fluctuations accumulated afterward. That separation makes it a useful diagnostic model: an exponentially relaxing mean does not imply that the variance decays to zero, and a stationary variance does not imply that successive measurements are independent.

An integrating factor separates memory and noise

Write the Itô equation dx=−κ(x−x∗)dt+2D dWtdx=-\kappa(x-x_*)dt+\sqrt{2D}\,dW_t, with positive relaxation rate kappa and constant D. The fixed point is x star. Kappa has units of inverse seconds, unlike a spring stiffness, and D has units of length squared per second. Multiplying by the exponential integrating factor converts the deterministic part into a total differential. The stochastic integral then collects each past kick multiplied by the fraction of its effect that survives relaxation up to the observation time.

A fixed initial position gives a mean that approaches the fixed point exponentially. Itô isometry gives the variance by integrating the squared exponential weight. Early times recover free-diffusion variance because the restoring force has had little time to act. At late times the variance approaches D divided by kappa. The process therefore forgets its initial mean without losing all fluctuations. Because the noise is additive Gaussian and the equation is linear, the transition distribution is Gaussian; this conclusion does not extend automatically to nonlinear restoring forces or non-Gaussian kicks.

⟨xt∣x0⟩=x∗+(x0−x∗)e−κt,Var⁡(xt∣x0)=Dκ(1−e−2κt)\langle x_t\mid x_0\rangle=x_*+(x_0-x_*)e^{-\kappa t},\qquad \operatorname{Var}(x_t\mid x_0)=\frac{D}{\kappa}(1-e^{-2\kappa t})
The conditional variance assumes a deterministic initial x0 and independent future Wiener increments. The mean decay timescale is 1/kappa, while the approach of this variance has decay rate 2 kappa.↔ Scroll wide equations horizontally.

Stationarity retains time correlations

If the initial distribution already has mean x star and variance D divided by kappa, its one-time distribution remains unchanged. Its two-time fluctuations are nevertheless correlated. A fluctuation at an earlier time influences the conditional mean later by the same exponential factor that governs deterministic relaxation. In the stationary ensemble the covariance is therefore the stationary variance multiplied by an exponential of the time separation. The absolute value of the separation makes this equilibrium correlation symmetric under exchanging the two observation times.

For a nonstationary initial distribution, the variance includes a decaying contribution from the initial spread as well as newly accumulated noise. One must not insert a stationary covariance formula into an experiment prepared at an exact position without checking the waiting time. Likewise, recording many points at intervals short compared with the relaxation time does not create many independent samples. Correlations reduce the information gained from repeated measurements, even though a histogram collected over a sufficiently long stationary trajectory can reproduce the correct Gaussian distribution.

pst(x)=κ2πDexp⁡ ⁣[−κ(x−x∗)22D],C(t)=Dκe−κ∣t∣p_{\mathrm{st}}(x)=\sqrt{\frac{\kappa}{2\pi D}}\exp\!\left[-\frac{\kappa(x-x_*)^2}{2D}\right],\qquad C(t)=\frac{D}{\kappa}e^{-\kappa|t|}
C is the centered two-time covariance in the stationary ensemble, with units of length squared. The density integrates to one on the full real line and requires kappa and D to be positive.↔ Scroll wide equations horizontally.

A thermal trap fixes the ratio

For the overdamped harmonic potential U=k(x−x∗)2/2U=k(x-x_*)^2/2, the mobility gives κ=μk\kappa=\mu k. If the bath is equilibrium thermal noise, D=μkBTD=\mu k_BT, so the steady variance becomes kBT/kk_BT/k. Mobility controls how fast the particle relaxes, while stiffness and temperature control its equilibrium width. Increasing friction slows relaxation but leaves the width unchanged when the noise amplitude changes consistently with the same bath temperature. Changing friction while holding an incompatible noise strength fixed describes a different effective temperature or a nonthermal model.

The Ornstein–Uhlenbeck model can also describe velocity relaxation, but then x is replaced by velocity and the diffusion coefficient has velocity-squared per time units. Always declare the modeled variable before interpreting D. A stable linear drift guarantees a stationary Gaussian for this scalar additive-noise model. It does not prove that every stationary noisy system is thermal, nor that a measured Gaussian histogram alone establishes equilibrium. Later lessons will show stationary densities with nonzero probability circulation, which cannot be diagnosed from their shape alone.

A half-relaxed mean with newly created variance

A particle starts exactly at x0=2 μmx_0=2\,\mathrm{\mu m} with x∗=0x_*=0, κ=ln⁡2 s−1\kappa=\ln2\,\mathrm{s^{-1}}, and D=ln⁡2 μm2 s−1D=\ln2\,\mathrm{\mu m^2\,s^{-1}}. Find its mean, variance, and mean-square position at t = 1 s.

  1. The surviving mean factor is e−κt=1/2e^{-\kappa t}=1/2, so ⟨x⟩=1 μm\langle x\rangle=1\,\mathrm{\mu m}. The deterministic initial value has not yet been fully forgotten.

  2. The stationary variance is D/κ=1 μm2D/\kappa=1\,\mathrm{\mu m^2}. The conditional variance is only 1−(1/2)2=0.75 μm21-(1/2)^2=0.75\,\mathrm{\mu m^2} because the initial position had no ensemble spread.

  3. The uncentered second moment adds the squared mean, yielding ⟨x2⟩=0.75+12=1.75 μm2\langle x^2\rangle=0.75+1^2=1.75\,\mathrm{\mu m^2}. Its value is not the stationary variance, even though the noise and restoring parameters are already held constant.

The result

After one second the mean is 1 micrometre, variance is 0.75 square micrometres, and mean-square position is 1.75 square micrometres.

Explore the model

Interactive lab

Interactive lab · markov

Relaxation rate and stationary law are different data

Reset and set BOTH Transition p = 0.2 and Transition q = 0.3; predict stationary state-1 probability 0.4. For a fresh run, Reset and set p = 0.4 and q = 0.6; predict the same stationary probability but arrival there after one transition. Step once in each run; each Step advances one discrete transition.

Relaxation rate and stationary law are different dataExact 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 exact two-state chain starts in state 0. It illustrates relaxation toward a stationary law, not a continuous Ornstein–Uhlenbeck trajectory or Gaussian noise.

Check your understanding

What distinguishes stationary Ornstein–Uhlenbeck motion from an ensemble started at a fixed point?

Try it yourself

A centered process has κ=2 s−1\kappa=2\,\mathrm{s^{-1}}, D=8 μm2 s−1D=8\,\mathrm{\mu m^2\,s^{-1}}, and known initial position x0=3 μmx_0=3\,\mathrm{\mu m}. At t=(ln⁡2)/2 st=(\ln2)/2\,\mathrm s, find conditional mean and variance. Separately find the stationary covariance at that time separation.

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

Give me a hint

The mean factor is one-half. A conditional variance built from new noise and a covariance in an already stationary ensemble are different quantities.

Compare with the solution

The mean is 3/2=1.5 μm3/2=1.5\,\mathrm{\mu m}. The stationary variance is D/κ=4 μm2D/\kappa=4\,\mathrm{\mu m^2}, so the conditional variance is 4(1−1/4)=3 μm24(1-1/4)=3\,\mathrm{\mu m^2}. For a separately prepared stationary ensemble, the covariance across the same time separation is 4(1/2)=2 μm24(1/2)=2\,\mathrm{\mu m^2}. The last number measures memory between two fluctuating observations and should not replace the first ensemble's noise-generated variance.

Take it with you

  • Linear relaxation damps initial memory while additive noise builds a finite variance.
  • A stationary histogram can coexist with strongly correlated observations.
  • Thermal trap width is kBT/k, while mobility also controls the relaxation timescale.

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