FR25 · Irreversibility

Hidden Entropy: How Much Dissipation Can You See?

How much of an enzyme’s dissipation can an observer prove from what it records? An enzyme that burns ATP dissipates at a steady rate, but an observer who sees only part of it can prove only a lower bound. Counting completed cycles proves almost all of it near equilibrium and less under strong drive; snapshots prove less as the interval grows; a signal that only shows bound or empty proves nothing.

An ATP-driven enzyme cycles through three states. Compare the entropy production with what cycle counts, snapshots or a bound/empty signal can prove.

Entropy production $140\,k_{\mathrm B}T/\mathrm{s}$

Proved by counting cycles $59.8\,k_{\mathrm B}T/\mathrm{s}$

Proved by snapshots $74.8\,k_{\mathrm B}T/\mathrm{s}$

Proved by bound or empty $0\,k_{\mathrm B}T/\mathrm{s}$

Hide the dissipation

Fuel per cycle

8

\Delta\mu\;(k_{\mathrm B}T)

Snapshot interval

2.0

\Delta t\;(\mathrm{ms})

Hydrolysis speed

1

\kappa_{\mathrm h}/\kappa_0

Drag the ↕ handles to raise or lower the bound states.

Energy levels and seed

ATP-bound level

-1

E_{\mathrm{ET}}\;(k_{\mathrm B}T)

ADP-bound level

1

E_{\mathrm{ED}}\;(k_{\mathrm B}T)

Seed of the 1024 records

2531

n

Model, units and sources

$w^{\pm}_i=\kappa_i\,e^{\pm(E_i-E_{i+1}+\Delta\mu/3)/2},\qquad \kappa_0=20\,\mathrm{s}^{-1}$

Rates obey local detailed balance; the attempt rates and drawn barrier heights are schematic. Each of the 1024 records lasts 200 mean dwell times. The bound-or-empty observer sees E apart from ET and ED together.

Barato & Seifert (2015), Eqs. 4, 10, 19–20 · Seifert (2012), Eqs. 128–129, 146–149 · Martínez et al. (2019), Eqs. 1–6

Derive the three bounds

Model assumptions

A single enzyme goes from empty (E) to ATP-bound (ET), releases phosphate to become ADP-bound (ED), then releases ADP and is empty again, consuming one ATP per cycle. Each step obeys local detailed balance on a tilted free-energy washboard; the attempt rates and barrier heights are schematic. 1024 independent enzymes are simulated exactly with the Gillespie algorithm; the bars and readouts are exact model values and the dots are estimates from the simulated records. No measured data are shown.

One enzyme, three records

What each record can prove

Slower snapshots prove less

Stronger fuel hides more from the counts

More numbers and how they are computed

Share proved by counting cycles: $0$
Share proved by snapshots: $0$
Entropy production from the simulated full records: $0$
Counting bound from the simulated cycle counts: $0$
Snapshot bound from the simulated snapshots: $0$
Bound-or-empty estimate from the simulated signal: $0$
Cycle rate: $0$
Spread of the cycle count: $0$
Uncertainty product and its minimum for three states: $0$
Time spent in E, ET and ED: $0$
Mean dwell time: $0$
Records times duration: $0$

Exact values come from the stationary distribution, the tilted generator and the matrix exponential of the three-state rate matrix. Simulated estimates use every jump of 1024 Gillespie records; the snapshot estimator adds one half to each pair count. The bound-or-empty record is exactly reversible because every exit from the bound states returns to the single empty state.

Start with the essentials

Focus question: How much of an enzyme’s dissipation can an observer prove from what it records?
One-sentence intuition: Every partial record gives a lower bound on the entropy production. Counting cycles proves almost all of it near equilibrium but only a fraction under strong drive. Snapshots lose it as the interval grows. A bound-or-empty signal proves nothing, because its record looks the same played backwards.

Core mathematical model

Entropy production of the cycle

\sigma=\sum_{i\ne j}p_i w_{ij}\ln\frac{p_i w_{ij}}{p_j w_{ji}}=J\,\Delta\mu

In the steady state the sum over all jumps reduces to the cycle rate $J$ times the fuel per cycle.

Bound from counting cycles

\sigma\ge\frac{J^2}{D}

Here $D$ measures how much the number of completed cycles spreads over time. The thermodynamic uncertainty relation turns a precise count into a minimum cost.

Bound from snapshots

\sigma\ge\frac{1}{\Delta t}\sum_{i\ne j}P_{ij}\ln\frac{P_{ij}}{P_{ji}}

$P_{ij}$ is the probability of seeing state $i$ and then state $j$ one interval later. A blurred record can only lose irreversibility, never gain it.

Common difficulties

No current, no dissipation?

Typical misconceptionIf the observed signal shows no net flow, the system is at equilibrium.

Better mental modelThe bound-or-empty signal has no net flow and is exactly time-symmetric, yet the enzyme keeps burning ATP. The dissipation is hidden inside the bound state.

Run the experiment

01
Watch the walker
Follow the large bead down the tilted staircase of wells.
What to observe: Most hops go downhill and flash red as heat enters the water; occasional uphill hops flash blue. The small dots drift and spread.
02
Slow the camera
Choose Slow camera misses the cycle.
What to observe: The snapshot bound drops to a few percent, while counting cycles still proves almost half.
03
Reshape the landscape
Drag the orange or pink wells up and down.
What to observe: The entropy production changes with the cycle rate, but the bound-or-empty signal stays at zero.

Hidden Entropy: How Much Dissipation Can You See?

Dissipation you cannot see

Core mathematical model

Entropy production of the cycle

Bound from counting cycles

Bound from snapshots

Common difficulties

No current, no dissipation?

Run the experiment

Watch the walker

Slow the camera

Reshape the landscape

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