---
title: "How much dissipation an observer can see when part of the motion is hidden"
description: "Bound entropy production from timed snapshots, show why a bound-or-empty record looks perfectly reversible, and meet the clues in waiting times."
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language: "en"
page_type: "story"
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  zh-CN: "https://sandboxphysics.com/zh/stories/stochastic-thermodynamics/hidden-dissipation/"
last_modified: "2026-10-29"
---

[Stochastic thermodynamics](https://sandboxphysics.com/stories/stochastic-thermodynamics/)

L4 · Chapter 7

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

Sandbox Physics · 2026-10-29

The enzyme of the previous chapter produces about 140 $k_BT$ of entropy per second, but an experiment rarely sees every step. A signal might only say whether the enzyme is empty or holding something, and a camera takes one frame at a time. This chapter asks how much entropy production such a partial record can prove. The answer depends on what is seen. Timed snapshots prove part of it, less the slower they come. A bound-or-empty record runs the same forwards and backwards, statistically, and proves no dissipation at all.

## Before you begin

The master equation, cycle current and entropy production of the previous chapter. You also need the definition of relative entropy (the Kullback–Leibler divergence) and the fact that it is never negative. You only need to use a matrix exponential, not compute one.

- State the dissipation visible in a record as the relative entropy between forward and reversed trajectory probabilities, and explain why it never exceeds the true entropy production.
- Write the entropy production bound from timed snapshots and describe how it changes with the interval.
- Use renewal to prove that a bound-or-empty record is invariant under time reversal, and say when waiting-time distributions add information.

## 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.

$$
\sigma\ \ge\ \sigma_{\mathrm{obs}}=\lim_{t\to\infty}\frac{1}{t}\,D\big[P(\gamma_t)\,\big\|\,P(\tilde\gamma_t)\big]
$$

The right side uses only the observer’s own record, so any observer can use it to bound the entropy production.

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.

$$
\sigma_{\mathrm{snap}}(\Delta t)=\frac{1}{\Delta t}\sum_{i\ne j}p_iP_{ij}\ln\frac{p_iP_{ij}}{p_jP_{ji}},\qquad P=e^{W\Delta t}
$$

Entropy production per second that pairs of snapshots can prove.

$$
\sigma_{\mathrm{snap}}(2\Delta t)\le\sigma_{\mathrm{snap}}(\Delta t)\le\sigma,\qquad \lim_{\Delta t\to0}\sigma_{\mathrm{snap}}(\Delta t)=\sigma
$$

Denser frames prove more, up to the true value.

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.

$$
p(\mathrm{E},\mathrm{B})=p(\mathrm{E})-p(\mathrm{E},\mathrm{E})=p(\mathrm{B},\mathrm{E})
$$

The first equality sorts by the first frame, the second by the second frame; in the steady state both frames show $\mathrm{E}$ with the same probability.

## 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.

$$
\sigma_{\mathrm{obs}}=\sigma_{\mathrm{aff}}+\sigma_{\mathrm{WTD}}
$$

The first part comes from the order of jumps, the second from dwell-time distributions.

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.

## Try it in the Lab

### [Hide the dissipation](https://sandboxphysics.com/labs/hidden-entropy-network/)

1. Read the four readouts in the default state, then look at the record strip: the full record on the first row, snapshots on the second, bound or empty on the third. Which row makes it hardest to see that the enzyme turns forward?
2. Predict what fraction the snapshot bound keeps at a 10 ms interval, then check with “Slow camera misses the cycle”. Next drag the snapshot interval all the way left and watch it approach the entropy production.
3. Choose “Slow hydrolysis step”, where the enzyme spends most of its time in ET. Which comes closer to the entropy production, the counting bound or the snapshot bound?
4. Open the details and compare the bound-or-empty estimate from the simulated signal with the exact value $0$. E and B alternate in the record; why does that make the estimate negligibly small?

## Exercises with explained answers

1. Why can doubling the snapshot interval only lower the entropy production the snapshots prove, or leave it unchanged?

   Show the answer

   A record with interval $2\Delta t$ is obtained from one with interval $\Delta t$ by deleting every other frame. Deleting data cannot make the forward and reversed films easier to distinguish, so the relative entropy can only fall, and the entropy production proved over the same time cannot grow.

2. For the bound-or-empty observer, prove $p(\mathrm{E},\mathrm{B})=p(\mathrm{B},\mathrm{E})$. If the camera could tell three classes apart, would the pairs still have to be symmetric?

   Show the answer

   The chance that the first frame shows E, $p(\mathrm{E},\mathrm{E})+p(\mathrm{E},\mathrm{B})$, equals the chance that the second does, $p(\mathrm{E},\mathrm{E})+p(\mathrm{B},\mathrm{E})$. Subtract. With three classes, the steady state only requires each class to gain as much as it loses, so all three edges can share the same nonzero excess; in the worked example each edge of the 2 ms snapshots has 258.4 more pairs forward than back.

3. Suppose a camera recognises ET but lumps E and ED together. Is its record still reversible?

   Show the answer

   Yes. ET is a single state, so every entry into ET is a renewal point. The direction in which the enzyme leaves ET does not depend on how long it stayed, and the stay in the other class depends only on which side it entered. The record is again a sequence of alternating independent dwell times, unchanged by reversal. On a three-state ring, every two-class observer sees no dissipation.

[Previous: How much entropy a fuel-driven enzyme produces each second](https://sandboxphysics.com/stories/stochastic-thermodynamics/entropy-production-in-networks/) [Next: Reading the 2003 paper by Gore, Ritort and Bustamante on the bias of free energies from few pulls](https://sandboxphysics.com/stories/stochastic-thermodynamics/reading-the-jarzynski-bias-paper/)

## Sources and what we recomputed

1. [U. Seifert, Stochastic thermodynamics, fluctuation theorems and molecular machines, arXiv:1205.4176v1; Reports on Progress in Physics 75, 126001 (2012)](https://arxiv.org/abs/1205.4176v1)

   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).

2. [I. A. Martínez, G. Bisker, J. M. Horowitz and J. M. R. Parrondo, Inferring broken detailed balance in the absence of observable currents, arXiv:1809.02084v2 (2019)](https://arxiv.org/abs/1809.02084v2)

   Version 2, 23 July 2019. Eq. (1): the entropy production rate is at least the relative entropy rate between an observed trajectory and its time reverse. Eq. (2): that rate splits into an affinity part and a waiting-time part (Eqs. 3 to 6). The partially hidden network example observes states 1 and 2 and one hidden class. The journal version was not checked.