How much work can be extracted by measuring a trapped bead and moving the trap

Work and heat in one cycle

Let $z$ be the bead offset from the trap centre, in units of the thermal width $\sqrt{k_BT/k}$, with energies in units of $k_BT$. At the start of a cycle $z$ is Gaussian with variance $S$. A camera reads $y=z+\varepsilon$, where the error $\varepsilon$ has variance $R$. The trap then jumps by $gy$; $g$ is called the gain.

The bead has no time to move during the jump, so the work is just the change in potential energy, given below. Then the trap stays put while the bead relaxes for a time $t_w$. No work is done during the wait, and every drop in potential energy leaves as heat into the water. With $a=e^{-2t_w/\tau_r}$, the offset variance must return to $S$ after the wait, which fixes the steady variance.

Averaging uses $\langle zy\rangle=S$ and $\langle y^2\rangle=S+R$. A positive extracted work means the reading really did let the bead fall lower; the second term is the cost of the error, and it grows with the gain.

How much information one reading carries

How much the reading $y$ tells us about the bead is measured by the stochastic information $I=\ln[p(y\mid z)/p(y)]$, Eq. (151) of the Seifert review. It is large when the reading lands unexpectedly close to the true position, and it can be negative when the reading lands far off.

Its average is the mutual information. For a Gaussian reading $p(y)$ has variance $S+R$ and $p(y\mid z)$ has variance $R$. Averaging the difference of their logarithms gives the formula below. It depends only on the signal-to-noise ratio $S/R$ and is measured in nats.

Source: U. Seifert

The bound set by information

Sagawa and Ueda proved that with feedback $\langle W\rangle\ge\Delta F-k_BT\langle I\rangle$. After a full cycle the trap has the same shape, so $\Delta F=0$ and the mean extracted work cannot exceed $k_BT\langle I\rangle$. Our cycle starts from the steady state of the loop, and the matching exact equality is Eq. (156) of the Seifert review, where the total entropy production $\Delta s_{\mathrm{tot}}$ is the system entropy change plus $\beta Q$.

Averaging Eq. (156), and using the fact that in the steady state the mean system entropy change equals $-I_{\mathrm{left}}$, gives a tighter bound. $I_{\mathrm{left}}$ is the information about the bead that the reading still holds at the end of the cycle. Nobody used it, and the next measurement makes it obsolete. After a long wait it is almost zero.

The bound limits only the average. A single cycle can extract more than $k_BT\,I$, just as a single pull in Chapter 4 could cost less than the free-energy change. The sample average of the equality behaves like the Jarzynski average too: a few cycles whose reading points badly the wrong way dominate it, so small samples fall short of one.

Source: T. Sagawa and M. Ueda · U. Seifert

The best gain and the efficiency

After a long wait $a\to0$ and $S\to1$. Setting the derivative with respect to the gain to zero gives the best gain $g^\ast=1/(1+R)$, which extracts $\tfrac12\,\mathrm{SNR}/(1+\mathrm{SNR})$ thermal energies per cycle, with $\mathrm{SNR}=1/R$. The sharper the reading, the closer the best gain gets to one, which means moving the trap all the way to the reading.

The extracted work never exceeds half a thermal energy, while the information grows with the logarithm of the signal-to-noise ratio. Their ratio is the efficiency below. It approaches one as the signal-to-noise ratio goes to zero and is only about 0.21 at a ratio of 100. Section 7.3.5 of the Seifert review states that this bound cannot be reached by controlling the trap centre alone; using all the information also requires changing the trap stiffness.

Source: U. Seifert

Where the second law survives

The engine seems to draw work steadily from water at a single temperature. The missing piece is the controller’s memory: each reading must be recorded and then erased before the next one. Sagawa and Ueda showed in 2009 that measuring and erasing together cost at least $k_BT\,I$ of work. Include that cost, and the whole machine gains no net work per cycle; the second law holds.

Information therefore acts like a fuel. It can be turned into work, at most one thermal energy per nat, and making and clearing it costs at least as much.

Source: T. Sagawa and M. Ueda

Worked example: a blurry reading and fast cycles

Take a trap stiffness of $10\,\mathrm{pN/\mu m}$, so the thermal width is $20.4\,\mathrm{nm}$ and the relaxation time $\tau_r=0.942\,\mathrm{ms}$. A measurement error of $10\,\mathrm{nm}$ gives $R=(10/20.35)^2=0.241$, a signal-to-noise ratio of 4.14 and a best gain of 0.806. Use $g=0.8$ and wait $3\,\mathrm{ms}$, so $a=0.0017$ and $S=0.999$.

The mean work formula gives $-\beta\langle W\rangle=\tfrac12[(1.6-0.64)(0.999)-0.64(0.241)]=0.402$. The information is $\tfrac12\ln(1+0.999/0.241)=0.818\,\mathrm{nat}$, so the efficiency is 0.49. In physical units each cycle extracts $1.67\,\mathrm{pN\,nm}$, an average power of about $0.56\,\mathrm{aW}$.

Now cycle fast: error $2\,\mathrm{nm}$, $g=1$, and a wait of only $0.25\,\mathrm{ms}$. Then $a=0.588$, $R=0.0097$ and $S=0.417$: the bead has no time to spread back to its thermal width, so the feedback has cooled it. Each cycle extracts only $0.204\,k_BT$, less than the 0.494 of a full $3\,\mathrm{ms}$ wait, but cycles are twelve times shorter and the power rises from $0.165$ to $0.815\,k_BT/\mathrm{ms}$.