---
title: "Two ways for a swimmer to lose its heading"
description: "Derive the free-swimmer displacement curve and compare continuous Brownian turning with sudden, random tumbles."
canonical_url: "https://sandboxphysics.com/stories/active-matter/swimming-and-tumbling/"
language: "en"
page_type: "story"
alternate_languages:
  zh-CN: "https://sandboxphysics.com/zh/stories/active-matter/swimming-and-tumbling/"
last_modified: "2026-10-28"
---

[Active matter · A course with experiments](https://sandboxphysics.com/stories/active-matter/)

L2 · Chapter 2

# Two ways for a swimmer to lose its heading

They give the same free-particle mean-squared displacement when their speed and heading-correlation time match. Individual tracks still look different. Open Section 2.1 of Cates and Tailleur’s arXiv:1406.3533v1 beside this derivation; the matching is a statement about a particular statistic, not an identity between all trajectories.

Sandbox Physics · 2026-10-28

In this chapter

[All eight chapters](https://sandboxphysics.com/stories/active-matter/)

Before you start

Read the preceding chapter. You will use a time integral, an exponential decay and a mean over repeated independent runs.

## Define the plane and the noise

Positions have length units and time has its own unit. The heading angle is measured counterclockwise from the positive horizontal direction, in radians. The active Brownian model moves at a fixed swim speed while the heading undergoes angular diffusion. Translational diffusion adds independent positional noise.

In the equations, the two-component position noise and the scalar angle noise are independent Wiener increments: their mean is zero and their variance is the time step. The rotational diffusion coefficient has units of inverse time; the translational coefficient has units of length squared per time. There are no walls, forces or neighbours in this reference problem.

$$
d\mathbf r=v_0(\cos\theta,\sin\theta)\,dt+\sqrt{2D_t}\,d\mathbf W,\qquad d\theta=\sqrt{2D_r}\,dW_\theta
$$

## Start with the heading correlation

Angular diffusion makes the angle difference Gaussian, with variance twice rotational diffusivity times elapsed time. Averaging the cosine of this angle difference gives an exponential. Its decay time is the inverse rotational diffusion coefficient.

A run-and-tumble swimmer instead holds its heading until a Poisson event resets it to an independent uniform direction. The old and new heading correlate only if no reset has occurred. The probability of no reset is also exponential. Set the tumble rate equal to the rotational diffusion coefficient in two dimensions, and the two heading correlations match.

A partial reorientation would change that relation. So would a finite tumble duration. State which kind of tumble is implemented before comparing numerical rates.

$$
C(t)=\langle\mathbf n(t)\cdot\mathbf n(0)\rangle=e^{-t/\tau},\qquad \tau_{\rm ABP}=D_r^{-1},\quad\tau_{\rm RTP}=\alpha^{-1}
$$

## Integrate the velocity twice

Displacement is the time integral of velocity. Squaring it produces a double integral over two observation times. A stationary swimmer’s velocity correlation depends only on their time difference, so the double integral reduces to twice the integral of the correlation multiplied by the remaining observation time.

The final expression separates positional Brownian spreading from propulsion. At short times, expand the exponential: the active term is swim speed squared times elapsed time squared. At long times, the slope corresponds to a two-dimensional effective diffusion coefficient equal to the translational diffusivity plus half the squared swim speed times the persistence time.

$$
\begin{aligned}\langle|\Delta\mathbf r(t)|^2\rangle&=4D_t t+2v_0^2\int_0^t(t-s)e^{-s/\tau}\,ds\\&=4D_t t+2v_0^2\tau\left[t-\tau(1-e^{-t/\tau})\right].\end{aligned}
$$

Worked example

## A number to compare with a simulation

Use length and time units in which swim speed is 2, persistence time is 0.5 and translational diffusivity is 0.1. At one time unit, the passive term is 0.4. The active term is four times the bracket one minus one half times one minus the exponential of minus two. It is approximately 2.271, so the full mean-squared displacement is approximately 2.671.

Compare an average over many independent seeds with this value. One trajectory is not expected to match it. Reduce the numerical time step and check that any systematic error shrinks; increase the number of runs and check that the sampling scatter shrinks. The passive limit gives a second, simpler check: turn propulsion off and recover the Brownian result.

Alignment is a separate interaction. A polar rule makes neighbouring headings point the same way; a nematic rule treats a heading and its reversal equally. Matching the free correlation times does not make these interactions equivalent. Keep alignment off when testing the free-swimmer formula.

$$
\langle|\Delta\mathbf r(1)|^2\rangle=0.4+4\left[1-\tfrac12(1-e^{-2})\right]\approx2.671
$$

An experiment to think through

## Predict, change, explain

1. Predict the short-time slope change when swim speed doubles.
2. Compare free Brownian-turning and run-and-tumble settings at matched persistence. Export repeated records with different seeds.
3. Explain which ensemble statistic should agree and which visible feature of an individual track can differ.

[Watch swimmers gather](https://sandboxphysics.com/labs/active-particle-clustering/)

These Labs generate synthetic particle or continuum records. Their displayed model notes state the implemented dynamics and units.

## Try explaining it yourself

### 1. Why is it unsafe to measure long-time displacement using wrapped positions in a periodic box?

Read an explained answer

A swimmer can cross the boundary many times and return near its starting coordinate. Keep an unwrapped trajectory for displacement; use wrapped coordinates only for rendering and local interactions.

### 2. Does equal long-time diffusion prove equal short-time motion?

Read an explained answer

No. Passive diffusion can be tuned to the same late-time slope. Its displacement begins linearly in time, while the propulsion contribution begins quadratically.

## Sources to read alongside

1. [Cates & Tailleur, Motility-Induced Phase Separation, arXiv:1406.3533v1 (2014); Annual Review of Condensed Matter Physics 6 (2015)](https://arxiv.org/abs/1406.3533v1)

   Sections 2–4: orientational persistence, accumulation in slow regions and the local density-instability criterion. The course derives its own worked examples from these stated assumptions.

2. [Howse et al., Self-Motile Colloidal Particles: From Directed Propulsion to Random Walk, Physical Review Letters 99, 048102 (2007)](https://arxiv.org/abs/0706.4406v1)

   Experimental context for the short directed segment and later diffusive motion. The worked numbers in this course are ideal examples, not measurements from this paper.

[Previous A swimming particle remembers its direction](https://sandboxphysics.com/stories/active-matter/persistent-random-walks/) [Next A crowd can grow without anyone pulling](https://sandboxphysics.com/stories/active-matter/crowding-without-attraction/)