How much dissipation an observer can see when part of the motion is hidden

The dissipation an observer can prove

Call the stretch of trajectory an observer records $\gamma_t$, and its time reverse $\tilde\gamma_t$. For a process in equilibrium, the reversed film is as likely as the forward one; for a driven process the forward film is more likely. How fast the relative entropy between the two distributions grows with time is the dissipation visible in the record.

Eq. (1) of Martínez and co-workers (2019) states that this rate never exceeds the true entropy production rate, and equals it when the record holds every variable that is out of equilibrium. Section 7.2 of the Seifert review uses the same relative entropy to measure dissipation and notes that coarse graining can only lower it: the coarser the record, the harder it is to tell the two films apart.

Source: I. A. Martínez · U. Seifert

Snapshots at fixed intervals

A camera takes a frame every $\Delta t$. Each frame shows the state exactly, but nothing between frames. The probability that two consecutive frames read $i$ then $j$ is $p_iP_{ij}$, with transfer matrix $P=e^{W\Delta t}$ and $W$ the rate matrix of the previous chapter’s master equation. The snapshots still form a Markov chain, so comparing forward and reverse pairs of consecutive frames is enough, as in the first formula below.

For a short interval, $p_iP_{ij}\approx p_iw_{ij}\Delta t$ and the first formula reduces to Eq. (128) of the previous chapter: the snapshots see all of the dissipation. For a longer interval, several hops can happen between frames. On a three-state ring two hops forward look exactly like one hop back, and that confusion cancels part of the forward flow, so the snapshots prove less.

Doubling the interval is the same as deleting every other frame of the original film. Deleting data cannot make the forward and reverse distributions easier to tell apart, so the bound per unit time can only fall or stay. Once the interval is much longer than the relaxation time of the enzyme, the second frame barely remembers the first and the bound goes to zero.

Source: U. Seifert

Bound or empty

Now take an observer who cannot tell ET from ED and only knows whether the enzyme is empty (E) or holding something (call it B). The signal keeps switching, tens of times per second between E and B, yet the forward and reversed films are statistically identical.

Start with pairs of frames. With only two classes, the chance that the first frame shows E equals the chance that the second does. That gives the formula below at once: pairs from E to B are exactly as common as pairs from B to E, and the bound is zero.

The whole record behaves the same way. E is a single microstate, so each entry into E wipes out the past; such moments are called renewal points. Whether the enzyme leaves E forward into ET or backward into ED does not depend on how long it stayed in E, and how long it then stays in B depends only on which side it entered. The record is a sequence of alternating, independent dwell times. Reversed, it is again an alternating sequence with the same distributions and the same independence, so the relative entropy is exactly zero, even though the enzyme dissipates 140 $k_BT$ every second.

Any observer who splits a three-state ring into two classes meets the same wall, because one of the two classes always holds a single state.

Clues in waiting times

Seeing no net flow does not always mean nothing can be proved. Martínez and co-workers split the relative entropy of an observed record into two parts in their Eq. (2): one from asymmetry in the order of jumps and one from asymmetry in the distribution of dwell times.

In their partially hidden network, the observer sees states 1 and 2 plus a class H made of two hidden states, and labels each stay in H by the state before it. Going from 1 through H to 2 and from 2 through H to 1 can take differently distributed times; their Eq. (11) turns that difference into a bound on dissipation that works even when the observed net flow between 1 and 2 is zero.

For the bound-or-empty observer, every stay in B starts and ends in E. There is no direction to label, and both parts vanish. To see this enzyme dissipate, an observer must tell ET from ED, or count net turns.

Source: I. A. Martínez

Worked example: one enzyme, four observers

Keep the defaults of the previous chapter: $\sigma=139.6\,k_BT/\mathrm{s}$ and one hop every $16.1\,\mathrm{ms}$ on average. The two nonzero eigenvalues of the rate matrix are $(-149\pm52i)\,\mathrm{s^{-1}}$, so the relaxation time is about $6.7\,\mathrm{ms}$.

The camera takes a frame every $2\,\mathrm{ms}$. Out of every ten thousand pairs, 336.7 go from E to ET and 78.3 back; 330.1 go from ET to ED and 71.7 back; 275.2 go from ED to E and 16.7 back. Every edge has the same net excess, 258.4 pairs, so the snapshots see a turn rate of $0.02584/(2\,\mathrm{ms})=12.92\,\mathrm{s^{-1}}$, below the true 17.45. The three log ratios add to $\ln\frac{336.7}{78.3}+\ln\frac{330.1}{71.7}+\ln\frac{275.2}{16.7}=5.79$, also below the fuel of 8 per turn. Their product is $12.92\times5.79=74.8\,k_BT/\mathrm{s}$, 54% of the true value.

Slow the camera to one frame every $10\,\mathrm{ms}$, longer than the relaxation time, and the snapshots prove only $9.16\,k_BT/\mathrm{s}$, about 7%. Speed it up to one frame every $0.1\,\mathrm{ms}$ and they prove 97%. The observer who counts turns proves $59.8\,k_BT/\mathrm{s}$, or 43%. The bound-or-empty observer sees about 353.4 pairs each way between E and B in the $2\,\mathrm{ms}$ snapshots and proves nothing. All four numbers come from the same enzyme; only what is seen differs.