L1 · Chapter 1

How far a trapped bead wanders at room temperature

Sandbox Physics · 2026-10-29

A plastic bead one micrometre across floats in water, held by a focused laser. It never sits at the centre of the spot; it jitters around it. How far it wanders and how quickly it changes direction depend on three things only: the stiffness of the trap, the viscosity of the water and the temperature. Once these are fixed, the work, heat and free energy of the following chapters have a scale.

A bead that never sits still

An optical trap acts like an invisible spring. The further the bead is from the centre of the light spot $\lambda$ , the harder it is pulled back. Add a constant load force $f$ , for example a molecule tugging gently from the other side, and the potential energy is the expression below.

We measure length in nanometres and force in piconewtons. Their product, $\mathrm{pN\,nm}$ , equals $10^{-21}\,\mathrm{J}$ , which suits this scale. At 300 K the thermal energy is $k_BT=4.14\,\mathrm{pN\,nm}$ , the size of the energy that the surrounding water exchanges with the bead.

V(x,\lambda)=\frac{k}{2}(x-\lambda)^2+fx

Here

x

is the bead centre and

k

the trap stiffness. The load pushes the bead toward negative

x

without changing the shape of the bowl.

Source: U. Seifert

Where the bead spends its time

With the trap held still for long enough, the chance of finding the bead near $x$ is proportional to $e^{-V/k_BT}$ . This is the Boltzmann distribution. Each extra $k_BT$ of potential energy makes a position less likely by a factor $e$ .

Completing the square leaves a single square in the exponent, so the distribution is Gaussian. Its centre is shifted from the spot by $f/k$ , and its width depends only on stiffness and temperature. The variance is the average squared distance from the mean position; its square root is the typical size of the wandering.

Multiply the variance by $k/2$ and the spring holds $k_BT/2$ of energy on average. This is equipartition: each degree of freedom that enters the energy as a square receives half of $k_BT$ on average.

p_{\rm eq}(x)\propto e^{-V(x,\lambda)/k_BT},\qquad \langle x\rangle=\lambda-\frac{f}{k},\qquad \langle(x-\langle x\rangle)^2\rangle=\frac{k_BT}{k}

The load only moves the centre. The constant left over from completing the square is absorbed by the normalization.

A bead in a harmonic trap. The bowl is the potential energy; the shaded curve is the Boltzmann distribution of positions, with width

\sqrt{k_BT/k}

. Original schematic, not measured data.

Source: U. Seifert

How quickly it forgets where it was

In water the bead is so light and slow that inertia does not matter. At every moment the viscous drag balances the trap force plus random kicks from the water. This balance is the overdamped Langevin equation. The drag coefficient $\gamma$ comes from Stokes’ law and grows with bead radius and water viscosity.

Averaging removes the random force, and the mean position returns to equilibrium as $e^{-t/\tau_r}$ with relaxation time $\tau_r=\gamma/k$ . Temperature does not enter this ratio: it sets how far the bead wanders, not how quickly it comes back. The strength of the random force, $2\gamma k_BT$ , is fixed by the requirement that the long-time distribution be exactly the Boltzmann distribution above.

\gamma\dot x=-\frac{\partial V}{\partial x}+\xi(t),\qquad \langle\xi(t)\xi(s)\rangle=2\gamma k_BT\,\delta(t-s),\qquad \gamma=6\pi\eta a

Here

\eta

is the viscosity of water and

a

the bead radius. The random force at different times is uncorrelated.

Source: U. Seifert · O. Mazonka and C. Jarzynski

Worked example: a one-micrometre bead at 12 pN/μm

Take $k=12\,\mathrm{pN/\mu m}=0.012\,\mathrm{pN/nm}$ at 300 K. The variance is $k_BT/k=4.14/0.012\approx345\,\mathrm{nm^2}$ , so the bead wanders by about $18.6\,\mathrm{nm}$ . That is more than fifty times smaller than the bead itself and easy to measure with a fast camera or a photodiode.

With water at $1\,\mathrm{mPa\,s}$ and radius $0.5\,\mu\mathrm{m}$ , the drag coefficient is $\gamma=9.42\times10^{-3}\,\mathrm{pN\,ms/nm}$ and the relaxation time $\tau_r\approx0.785\,\mathrm{ms}$ . In less than a millisecond the bead has forgotten where it was.

A load of $f=0.1\,\mathrm{pN}$ moves the centre by $f/k\approx8.3\,\mathrm{nm}$ and leaves the width alone. Stiffen the trap to $40\,\mathrm{pN/\mu m}$ and the wandering shrinks to $10.2\,\mathrm{nm}$ while the relaxation time drops to $0.236\,\mathrm{ms}$ . A stiffer trap holds the bead closer and pulls it back faster.

Try it in the Lab

Watch a cloud of beads in a still trap

Choose the Flat ground preset, which has no load, and watch the cloud of beads at the bottom of the bowl before the pull starts. Predict how much narrower the cloud becomes when the stiffness goes from 4 to 40 pN/μm.
Change the stiffness, play again and compare the widths. Ten times the stiffness should shrink the width by about $1/\sqrt{10}\approx0.32$ .
Add a load. Note which way the centre of the cloud moves and whether the width changes, and explain both with the completed square.

Turn it around: measure stiffness from the wandering

In the calibration experiment, predict how the position record changes when the trap is stiffer.
Change only the stiffness, read the position variance and recover the stiffness from $k=k_BT/\langle\delta x^2\rangle$ . Compare it with the set value.

Exercises with explained answers

At a stiffness of $4\,\mathrm{pN/\mu m}$ and 300 K, how far does the bead wander, and what is its relaxation time?
Show the answer
The variance is $4.14/0.004\approx1036\,\mathrm{nm^2}$ , so the bead wanders by about $32\,\mathrm{nm}$ . The relaxation time is $\gamma/k\approx2.36\,\mathrm{ms}$ . A soft trap lets the bead wander further and brings it back more slowly.
Suppose the viscosity stays fixed and the temperature doubles. How do the size of the wandering and the relaxation time change?
Show the answer
The variance is proportional to $k_BT$ and doubles, so the wandering grows by $\sqrt2$ . The relaxation time $\gamma/k$ contains no temperature and stays the same. Real water becomes less viscous when heated, which speeds up relaxation; that is a separate effect.
How much less likely is the bead to be found three standard deviations from its mean position than at the mean position itself?
Show the answer
With $\sigma^2=k_BT/k$ , the extra potential energy is $\tfrac{k}{2}(3\sigma)^2=\tfrac{9}{2}k_BT$ , so the ratio of probability densities is $e^{-4.5}\approx0.011$ . A record of a few thousand frames contains many such excursions.

Sources and what we recomputed

U. Seifert, Stochastic thermodynamics, fluctuation theorems and molecular machines, arXiv:1205.4176v1; Reports on Progress in Physics 75, 126001 (2012)
Version 1, 18 May 2012. Section 2: overdamped Langevin dynamics, Eq. (14) for the heat released into the medium and Eqs. (15)–(19) for the first law along one trajectory. Section 4: Eqs. (58) and (61), the Jarzynski and Crooks relations. Section 6.2: entropy production for master equations, Eqs. (120)–(129). Section 7.2: relative-entropy bounds on dissipation, Eqs. (146)–(150). Section 7.3: measurement and feedback, Eqs. (151)–(163).
O. Mazonka and C. Jarzynski, Exactly solvable model illustrating far-from-equilibrium predictions, arXiv:cond-mat/9912121v1 (1999)
Version 1, 7 December 1999. A bead dragged through a viscous fluid by a harmonic spring at constant speed. Eqs. (4)–(7) set up the model and the work rate; Eq. (26) defines the finite-time factor; Eqs. (43)–(53) add a constant force, give the free-energy change in Eq. (49) and the Gaussian mean and variance of the work in Eqs. (51)–(53). We recompute these results; the paper does not give our numerical examples.