FR25 · Work along one path

Jarzynski Equality: Dragging a Trapped Bead

Can repeated irreversible pulls reveal the free-energy change of a reversible one? One pull of a trapped bead can cost more or less work than the free-energy change, because the surrounding water kicks it at random. On average a pull always costs more. Averaging the exponential of minus the work over many pulls, or finding where the forward and reversed work distributions cross, still recovers the free-energy change, as long as the rare low-work pulls are sampled.

Drag a trapped bead uphill hundreds of times. Each pull costs different work, yet the work distributions still reveal the free-energy change.

Mean work $4.41\,k_{\mathrm B}T$

Jarzynski estimate $2.08\,k_{\mathrm B}T$

Crooks crossing $2.3\,k_{\mathrm B}T$

Exact free-energy change $2.41\,k_{\mathrm B}T$

Pull, repeat, compare

Trap speed

10

v\;(\mu\mathrm{m/s})

Load force

0.1

f\;(\mathrm{pN})

Trap stiffness

12

k\;(\mathrm{pN}/\mu\mathrm{m})

Pulls each way

200

N

Drag the cyan trap to draw your own protocol.

Sampling seed

Seed of the forward pulls

2501

Model, units and sources

$k_{\mathrm B}T=4.142\,\mathrm{pN\,nm},\quad \gamma=6\pi\eta a=9.42\times10^{-3}\,\mathrm{pN\,ms/nm}$

Work on the bead is positive; heat released to the water is positive. Each step moves the trap at fixed bead position, then relaxes the bead with the exact Ornstein–Uhlenbeck propagator, so the discrete Jarzynski and Crooks relations hold exactly.

Mazonka & Jarzynski (1999), Eqs. 43–53 · Seifert (2012), Eqs. 14–19, 58, 61

Derive the Jarzynski and Crooks relations

Model assumptions

A one-dimensional overdamped bead of radius 0.5 µm in water at 300 K sits in a harmonic optical trap with a constant load force. Each time step first moves the trap with the bead held still, which is work, then lets the bead relax exactly in the fixed trap, which is heat released to the water. Inertia, wall corrections to the drag, trap nonlinearity, detector noise and laser heating are omitted. Every pull is a model simulation; no measured data are shown.

Forward and mirrored reverse work

Estimate as pulls accumulate

Work, heat and potential energy of one typical pull

More numbers and how they are computed

Exact mean work of this protocol: $0$
Exact dissipated work: $0$
Continuous-time mean work: $0$
Estimate from reverse pulls: $0$
Bennett acceptance ratio: $0$
Bootstrap error of the Jarzynski estimate: $0$
Expected mean bias at this pull count: $0$
Share of the largest single weight: $0$
Effective number of pulls: $0$
Largest mismatch of work minus heat minus potential change: $0$
Relaxation time: $0$
Time steps per pull: $0$

Exact values come from the mean and covariance recursions of the discrete protocol; their ratio checks the fluctuation–dissipation relation. The expected band uses 600 seeded Gaussian replicates at each pull count. Pulls are in simulation order, so jumps in the running curves are single rare pulls.

Start with the essentials

Focus question: Can repeated irreversible pulls reveal the free-energy change of a reversible one?
One-sentence intuition: The mean work overshoots the free-energy change, but the exponential average and the crossing of forward and reversed work distributions both land on it. The exponential average leans on rare low-work pulls, so too few pulls bias it upward.

Core mathematical model

Where the work of one pull goes

W=Q+\Delta V,\qquad V(x,\lambda)=\tfrac{k}{2}(x-\lambda)^2+fx

Work done by the moving trap either raises the bead potential energy or leaves as heat into the water. Work equals heat plus the potential-energy change for every single pull.

Jarzynski equality

\left\langle e^{-W/k_{\mathrm B}T}\right\rangle=e^{-\Delta F/k_{\mathrm B}T},\qquad \Delta F=fL

The average runs over pulls that start in equilibrium. Here the free-energy change equals the load force times the drag distance, so the estimate can be checked exactly.

Crooks fluctuation theorem

\frac{P_{\mathrm F}(W)}{P_{\mathrm R}(-W)}=e^{(W-\Delta F)/k_{\mathrm B}T}

Forward pulls and the same path run backwards have work distributions that cross exactly at the free-energy change.

Mean dissipated work at constant speed

\langle W\rangle-\Delta F=\gamma v^2\left[\tau-\tau_r\left(1-e^{-\tau/\tau_r}\right)\right]=\frac{\operatorname{Var}W}{2k_{\mathrm B}T}

Here $\gamma$ is the bead drag coefficient, $v$ the trap speed, $\tau$ the pull time and $\tau_r=\gamma/k$ the relaxation time. The work is Gaussian, and its spread is fixed by the extra work.

Common difficulties

A pull below the free-energy change

Typical misconceptionA pull that costs less than the free-energy change breaks the second law.

Better mental modelThe second law limits the average over many pulls. A single small system can borrow heat from the water now and then; such pulls are rare, and they are exactly what the exponential average needs.

Run the experiment

01
Watch the green beads
Play Uphill drag and watch the upper band.
What to observe: A few beads turn green: so far they have cost less work than the bowl has risen. The histogram shows them as the left tail.
02
Pull faster
Choose Rare pulls decide.
What to observe: Mean work rises well above the free-energy change, and the forward exponential average stays high while the crossing still finds it.
03
Drag the trap yourself
Drag the forward trap handle to any new position, then let go.
What to observe: Whatever path you draw, the estimate approaches the load force times your net displacement, and every pull still splits its work exactly into heat plus potential-energy change.

Jarzynski Equality: Dragging a Trapped Bead

Work fluctuates, free energy does not

Core mathematical model

Where the work of one pull goes

Jarzynski equality

Crooks fluctuation theorem

Mean dissipated work at constant speed

Common difficulties

A pull below the free-energy change

Run the experiment

Watch the green beads

Pull faster

Drag the trap yourself

Related experiments and articles

From a trapped bead to free energy from irreversible pulls