The mean work and its spread when a trap moves at constant speed

The bead trails behind the trap

Move the trap at constant speed $v$ , so $\lambda=vt$ and the pull lasts $\tau=L/v$ . In the frame that moves with the trap, let $y=x-\lambda$ . To keep up, the bead must move through the water, so drag holds it back; in the Langevin equation this appears as an extra constant force $-\gamma v$ .

Averaging removes the random force. The mean lag moves exponentially, with relaxation time $\tau_r$ , from its resting value $-f/k$ toward a steady value. In the steady state the extra lag $v\tau_r$ stretches the spring just enough to balance the drag $\gamma v$ .

\gamma\dot y=-ky-f-\gamma v+\xi(t),\qquad \langle y(t)\rangle=-\frac{f}{k}-v\tau_r\left(1-e^{-t/\tau_r}\right)

At the start the bead is in equilibrium with the still trap, so the lag comes from the load alone.

Source: O. Mazonka and C. Jarzynski

The mean work: part stored, part dissipated

The power from Chapter 2 is $-ky\,v$ . Insert the mean lag. The first term, $fv$ , integrates to $fL$ , the free-energy change $\Delta F$ . The second comes from the speed-induced lag and integrates to the mean dissipated work $W_d$ .

Two limits check the result. For a slow pull, $\tau\gg\tau_r$ , the bracket is close to $\tau$ and $W_d\approx\gamma vL$ : drag force times distance, which vanishes as the speed goes to zero. For a very fast pull, $\tau\ll\tau_r$ , expanding the exponential to second order gives $W_d\to kL^2/2$ , the sudden jump of Chapter 2.

\langle W\rangle=fL+W_d,\qquad W_d=\gamma v^2\left[\tau-\tau_r\left(1-e^{-\tau/\tau_r}\right)\right]

Eq. (51) of Mazonka and Jarzynski gives the same result. The bracket is the time integral of the extra lag over the whole pull, in units of the steady lag.

Source: O. Mazonka and C. Jarzynski

The spread is fixed by the mean

The lag $y$ is a Gaussian random process and the work is a linear integral of it over time, so the work of each pull is Gaussian. More than that, its variance is exactly $2k_BT$ times the mean dissipated work, the same relation we found for the sudden jump in Chapter 2.

The reason is that the drag that makes the bead lag and the random kicks that make it jitter come from the same water molecules. That is how the noise strength $2\gamma k_BT$ of Chapter 1 was fixed. The relation is an instance of the fluctuation–dissipation theorem and holds in this model at any speed.

It gives a check that does not need $\Delta F$ in advance: measure the work of many pulls, subtract the variance divided by $2k_BT$ from the mean, and you should get the free-energy change. The next chapter shows that this is the Gaussian form of the Jarzynski equality.

\operatorname{Var}W=2k_BT\,W_d,\qquad \langle W\rangle-\frac{\operatorname{Var}W}{2k_BT}=\Delta F

Eqs. (52)–(53) of Mazonka and Jarzynski. Gore, Ritort and Bustamante use the same Gaussian relation for their Figure 1. The second equality is exact for Gaussian work and approximate otherwise.

Source: O. Mazonka and C. Jarzynski · J. Gore

Worked example: three pulling speeds

Keep the values of the previous chapters: $k=12\,\mathrm{pN/\mu m}$ , $\tau_r=0.785\,\mathrm{ms}$ , $L=100\,\mathrm{nm}$ , $\Delta F=2.41\,k_BT$ . At $10\,\mu\mathrm{m/s}$ the pull takes $10\,\mathrm{ms}$ and the steady extra lag is $v\tau_r=7.85\,\mathrm{nm}$ . The formulas give $W_d=2.10\,k_BT$ , a mean work of $4.51\,k_BT$ and a standard deviation of $\sqrt{2\times2.10}=2.05\,k_BT$ .

Slow down to $2\,\mu\mathrm{m/s}$ and the pull takes $50\,\mathrm{ms}$ , with $W_d=0.448\,k_BT$ and a standard deviation of $0.947\,k_BT$ . Speed up to $40\,\mu\mathrm{m/s}$ and it takes $2.5\,\mathrm{ms}$ , with $W_d=6.36\,k_BT$ and a standard deviation of $3.57\,k_BT$ . Twenty times the speed raises the dissipated work about fourteenfold but widens the distribution less than fourfold.

The fraction of pulls with work below $\Delta F$ is $\Phi\!\left(-\sqrt{W_d/2k_BT}\right)$ : about 32%, 15% and 3.7% at the three speeds. Such pulls are everywhere in a slow drag and become scarce in a fast one.

In the Lab the trap moves in small steps. Splitting the distance into $n$ steps, the difference from the continuous formula shrinks as $1/n^2$ ; the default setting uses 204 steps, with a relative difference of about $4\times10^{-4}$ .

Try it in the Lab

Compare a slow pull with a fast one

Predict first: going from the Nearly reversible preset to Rare pulls decide, which way do the centre and the width of the work histogram move? By roughly what factor does the width change?
Select the two presets in turn, compare the mean work and the histogram width, and use $\operatorname{Var}W=2k_BT\,W_d$ to check that the change in width matches the change in mean work.
Return to the Nearly reversible preset and change only the trap stiffness. From the slow-pull limit, predict whether the mean work changes, then test it. Repeat at the default speed and explain why a soft trap now dissipates slightly less.

Exercises with explained answers

In the slow-pull limit, halve the speed over the same distance. What happens to the dissipated work and to the standard deviation of the work?

Show the answer

For a slow pull $W_d\approx\gamma vL$ , so halving the speed halves the dissipated work. The standard deviation $\sqrt{2k_BTW_d}$ falls by $1/\sqrt2$ . Twice the time buys half the dissipation.

Keep the speed fixed and double the trap stiffness. In the slow-pull limit, does the mean dissipated work change? What about the lag?

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The mean dissipated work $\gamma vL$ contains no stiffness and stays the same. The lag $v\tau_r=\gamma v/k$ halves, but the spring is twice as stiff, so the force is still $\gamma v$ . Drag sets the dissipation; stiffness only sets how far the bead trails.

Two hundred pulls give a mean work of $4.41\,k_BT$ and a variance of $3.9\,k_BT^2$ . Estimate the free-energy change.

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In the Gaussian approximation $\Delta F\approx4.41-3.9/2\approx2.46\,k_BT$ . The sample variance of 200 pulls has a statistical error of about 10%, which alone moves the estimate by about $0.2\,k_BT$ ; adding the error of the mean gives an uncertainty of about $0.25\,k_BT$ .

Sources and what we recomputed

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.

J. Gore, F. Ritort and C. Bustamante, Bias and error in estimates of equilibrium free-energy differences from nonequilibrium measurements, PNAS 100, 12564–12569 (2003)

Open-access text on PubMed Central. Figure 1: Gaussian dissipated-work distributions with means of 0.5, 1, 2 and 4 thermal energy units. Figure 2: 100 000 estimates, each from 20 work values, at a mean dissipated work of 4 units; mean bias about 1.07 units. Figure 3: bias against sample size and the fall-off exponent for the constant C equal to 15. We reproduce the Figure 2 bias by our own sampling.