Work and heat for a single pull

Who does work and who receives heat

The potential energy $V(x,\lambda)$ can change in two ways. When the experimenter moves the trap, $\lambda$ changes before the bead has moved; that energy comes from the experimenter and counts as work. When the trap stays still and the bead moves, the water pushed it; that energy came from or went to the water and counts as heat.

Cut a pull into many small steps, sort each step this way and add them up. The result is the work $w$ done on the bead and the heat $q$ released into the water along that trajectory. Their difference is the change in potential energy from start to end: the first law for a single trajectory. Work done on the bead and heat released to the water are both counted positive, as in Eqs. (14)–(19) of the Seifert review.

The heat term must be evaluated with the force at the midpoint of each small step. Noisy trajectories need this care. With it, the first law holds exactly for every trajectory.

For our potential, $\partial V/\partial\lambda=-k(x-\lambda)$ . With the trap moving at speed $\dot\lambda$ , the power is $k(\lambda-x)\dot\lambda$ : the spring force on the trap times the trap velocity.

When the bead trails behind the trap, the spring is stretched and the experimenter pushes against it: the work is positive. If the water briefly kicks the bead ahead of the trap, the spring helps the trap along and that short stretch of work is negative. The total work of a pull depends on how far the bead lags on average.

w=\int_0^\tau\frac{\partial V}{\partial\lambda}\dot\lambda\,dt,\qquad q=-\int_0^\tau\frac{\partial V}{\partial x}\circ\dot x\,dt,\qquad w=q+\Delta V

The circle marks the midpoint rule.

\Delta V

is the potential energy at the end of this trajectory minus that at the start, and is itself random.

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

Worked example: the trap jumps by 100 nm

Take the bead of Chapter 1, with $k=0.012\,\mathrm{pN/nm}$ and load $f=0.1\,\mathrm{pN}$ . The bead is in equilibrium with the trap at $\lambda=0$ , then the trap jumps instantly to $L=100\,\mathrm{nm}$ . The bead has no time to move, so no heat flows and the work is the change in potential energy at the same position.

The starting point $x_0$ is Gaussian with mean $-f/k$ and variance $k_BT/k$ . The work is a linear function of $x_0$ , so it is Gaussian too. Its mean is $70\,\mathrm{pN\,nm}=16.9\,k_BT$ and its standard deviation $kL\sqrt{k_BT/k}=5.38\,k_BT$ .

Moving the trap from 0 to $L$ raises the equilibrium free energy by only $\Delta F=fL=10\,\mathrm{pN\,nm}=2.41\,k_BT$ . That is the work of an infinitely slow move. The extra $14.5\,k_BT$ spent by jumping is the dissipated work. After the jump the bead slides to the bottom of the new bowl and this energy ends up in the water as heat.

Chapter 3 shows that covering the same distance at constant speed in 10 ms costs only $4.51\,k_BT$ on average. The slower the move, the less is wasted.

W_{\rm jump}=\frac{k}{2}\big[(x_0-L)^2-x_0^2\big]=\frac{kL^2}{2}-kLx_0,\qquad \langle W_{\rm jump}\rangle=\frac{kL^2}{2}+fL

The first term,

kL^2/2

, is the dissipated work and the second the free-energy change. The variance of the work is

kL^2k_BT

, exactly

2k_BT

times the dissipated work. Chapter 3 explains the reason.

Some pulls cost less than the free-energy change

A single pull can do less work than $\Delta F$ . In the jump above, a bead that happens to start at the front of the bowl compresses the spring, and the jump costs less. The chance of such luck is $\Phi(-14.5/5.38)\approx0.004$ , where $\Phi$ is the cumulative standard normal distribution.

The second law is a statement about the average: $\langle W\rangle\ge\Delta F$ . For one bead, single trajectories below that line are common. In a 10 ms drag at constant speed, about 15% of pulls do less work than $\Delta F$ . Chapter 4 uses these lucky pulls to find the free energy directly.

Try it in the Lab

Follow the work and heat of one pull

Choose the Uphill drag preset. Predict which of the three curves, work, heat or potential-energy change, keeps changing after the pull ends.
In the chart of work, heat and potential energy for one typical pull, wait until the pull ends. Read the three values and check $W=Q+\Delta V$ .
Count the beads that turn green in the scene: so far they have done less work than the free-energy change at that moment. Do faster pulls give more green beads or fewer?

Exercises with explained answers

The trap stays still while the bead slides from the wall of the bowl to the bottom. What are the work and the heat?
Show the answer
The trap does not move, $\dot\lambda=0$ , so the work is zero. The first law gives $q=-\Delta V$ : whatever potential energy the bead loses goes into the water as heat.
Halve the jump in the worked example to 50 nm. What are the mean work, the dissipated work and the standard deviation of the work?
Show the answer
The dissipated work is $kL^2/2=15\,\mathrm{pN\,nm}\approx3.62\,k_BT$ , the free-energy change $fL=5\,\mathrm{pN\,nm}\approx1.21\,k_BT$ , so the mean work is about $4.83\,k_BT$ . The standard deviation is $\sqrt{2\times3.62}\approx2.69\,k_BT$ , only half as large as before. Halving the distance cuts the dissipated work to a quarter.
Why is the heat zero during an instantaneous jump? What happens afterwards?
Show the answer
During the jump only the trap position changes, so all of the energy change is work. Afterwards the trap is still and the bead slides toward its new equilibrium. Its extra potential energy goes gradually into the water as heat; on average the heat released equals the dissipated work.

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.