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advanced27 min studyLesson 359 in Physics

Keep velocity memory when diffusion is premature

Derive a finite-time spreading law and locate the continuum regime.

Before you begin
On this page

The question for today

What you’ll understand

  • Relate displacement variance to a velocity autocorrelation.
  • Recover ballistic and diffusive limits of an exponential-memory model.
  • Use both length and time ratios to assess rarefied transport.

A tracer cannot forget its initial velocity instantaneously. Before collisions or drag randomize its motion, displacement grows approximately as velocity times time, producing a variance proportional to time squared. At later times many direction changes can produce ordinary diffusion. A continuum diffusion equation hides the early memory and can therefore fail on short times or in a channel shorter than a mean free path. The crossover is a physical limit, not merely a need for a finer numerical grid.

A velocity correlation remembers the past

In a stationary zero-mean process, displacement is the time integral of velocity. Squaring that integral and averaging gives a double integral of the velocity correlation. Time-translation invariance reduces it to one integral with a triangular weight. This relation does not assume that the final motion is diffusive; it is a kinematic identity whenever the correlation and integrals are well defined. A nearly constant early correlation produces quadratic displacement variance.

If the velocity correlation is integrable over time, its integral defines a long-time diffusion coefficient for the selected Cartesian component. The variance then grows asymptotically linearly, apart from possible slower corrections. Integrability is a substantive condition: persistent flow or slowly decaying correlations can change the asymptotic law. Subtract any mean drift first; otherwise a deterministic squared-mean term can imitate persistent ballistic spreading.

⟨Δx(t)2⟩=2∫0t(t−s)Cv(s) ds,Cv(s)=⟨vx(s)vx(0)⟩,D=∫0∞Cv(s) ds\langle\Delta x(t)^2\rangle=2\int_0^t(t-s)C_v(s)\,ds,\qquad C_v(s)=\langle v_x(s)v_x(0)\rangle,\qquad D=\int_0^\infty C_v(s)\,ds
Cv has units m²/s², D has units m²/s, and the mean displacement is zero here. The diffusion coefficient exists only if the correlation integral converges appropriately.↔ Scroll wide equations horizontally.

Exponential memory gives a useful exact bridge

For an equilibrium inertial Langevin model with constant linear drag, the one-component velocity correlation is an exponential with amplitude equal to its thermal velocity variance. Perform the triangular integral exactly to obtain a formula valid throughout that model’s crossover. Expanding the exponential at short times cancels the apparent linear term and leaves the expected quadratic variance. At late times only the linear diffusive term and a constant offset remain.

The crossover time is the velocity relaxation time, which is mass divided by drag for this bath-coupled model. A gas’s collision time plays a related but model-dependent role; not every collision fully randomizes velocity. The simple exponential is an analytically useful model, not an exact theorem for all fluids. Momentum-conserving hydrodynamic correlations, confinement, and non-Markovian friction can alter the crossover or the late-time approach to diffusion.

Cv(t)=C0e−t/τ,D=C0τ,⟨Δx2⟩=2D[t−τ(1−e−t/τ)]C_v(t)=C_0e^{-t/\tau},\qquad D=C_0\tau,\qquad \langle\Delta x^2\rangle=2D\left[t-\tau(1-e^{-t/\tau})\right]
C₀ is a one-component velocity variance in m²/s² and τ a positive relaxation time in s. This exponential-memory model assumes stationary initial velocities, no mean drift, and an unbounded homogeneous bath.↔ Scroll wide equations horizontally.

A small apparatus can remain kinetic

For transport through a channel, compare mean free path with the distance to boundaries as well as with bulk gradient lengths. In a ballistic regime particles cross much of the apparatus before randomizing, so wall reflection and the incoming velocity distribution directly shape the flux. In a hydrodynamic bulk, many collisions allow local constitutive laws, yet a thin kinetic boundary layer can still require slip, temperature-jump, or accommodation information. There is no universal sharp Knudsen threshold independent of observable and accuracy requirement.

The Markov lab isolates a simple loss-of-memory comparison. Reset and set p=q=0.1, then Step once; predict state-one probability 0.1 from the initial state zero. Reset and set p=q=0.5, then Step once; predict 0.5, complete mixing in one transition. These are exact distribution probabilities rather than trajectories. The chain contains no physical velocities or displacement coordinate and cannot display a ballistic path, compute a mean-square displacement, or establish a gas’s boundary regime.

t≪τ: ⟨Δx2⟩≃C0t2,t≫τ: ⟨Δx2⟩≃2Dt,Kn=ℓ/Lt\ll\tau:\ \langle\Delta x^2\rangle\simeq C_0t^2,\qquad t\gg\tau:\ \langle\Delta x^2\rangle\simeq2Dt,\qquad \mathrm{Kn}=\ell/L
The limiting forms belong to the stated exponential-memory model. The separate spatial ratio Kn compares a molecular path ℓ with a physical channel or gradient length L; its interpretation requires boundary assumptions.↔ Scroll wide equations horizontally.

Work it through

At one relaxation time

Use a stationary exponential-memory model with C₀=4 m²/s² and τ=0.50 s. Find its diffusion coefficient and exact displacement variance at t=0.50 s. Compare the early ballistic and late diffusive estimates at that same time.

  1. The diffusion coefficient is the area under the velocity correlation: D=C₀τ=4×0.50=2 m²/s. This is a long-time property even though the exact model also describes earlier motion.

  2. At t=τ, the bracket t−τ(1−exp(−t/τ)) equals τ/e. Thus the variance is 2×2×0.50/e=2/e≈0.7358 m², and the standard deviation is approximately 0.858 m.

  3. The ballistic estimate C₀t² gives 1 m², while the late diffusive estimate 2Dt gives 2 m². Neither limiting expression is accurate at the crossover itself, despite each having the correct dimensions and asymptotic trend.

The result

The model has D=2 m²/s and variance 0.7358 m² at 0.50 s; the ballistic and diffusive estimates are 1 and 2 m² respectively.

The living notebook

Explore the model

Interactive lab

Interactive lab · markov

Inspect one-step memory loss

Reset and set p=0.1, q=0.1, then Step once; predict P(state 1)=0.1. Reset and set p=0.5, q=0.5, then Step once; predict 0.5.

Inspect one-step memory lossExact 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 forThe exact two-state probabilities compare relaxation per step. They are not particle trajectories, velocity correlations, or ballistic-to-diffusive displacement data.

Pause & reason

Check your understanding

What produces the quadratic short-time displacement variance in the stationary exponential-memory model?

Your turn

Try it yourself

A gas has characteristic speed 500 m/s and randomization time 0.20 ns. Estimate mean free path and compare channels of width 10 μm and 20 nm using Kn and direct crossing time.

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

Give me a hint

Use ℓ=vτ and t_cross=L/v. Keep nanoseconds, nanometres, and micrometres consistent.

Compare with the solution

The path estimate is 100 nm. For a 10 μm channel, Kn=0.01 and crossing time is 20 ns, about 100 randomization times; a hydrodynamic bulk may be plausible with separate wall-layer corrections. For 20 nm, Kn=5 and crossing time is 0.040 ns, one fifth of a randomization time, indicating strong ballistic influence and the need for kinetic boundary data.

Keep the idea

Take it with you

  • Displacement variance is an integral of velocity memory.
  • Ballistic and diffusive formulas are limits with different time domains.
  • Bulk continuum behavior does not remove kinetic boundary layers.

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