The precision a clock pays for in heat

The second law, with a probability attached takes a small system, drags it from one equilibrium to another, and finds that the second law holds on average and fails in individual runs, with the failures counted exactly by a relation that has no free constant in it. That is a statement about a system pushed between two states of rest. Most of what small systems in nature actually do is different: they are held away from rest indefinitely, driven continuously by a supply of free energy, and they do something repeatedly — a motor protein takes step after step along a filament, an enzyme turns over molecule after molecule, a chemical oscillator keeps time.

Every one of these is a counter, and every counter has a problem that the counting of entropy as a count makes unavoidable. At the scale of molecules, every forward step has a reverse, and the reverse happens. A count that is not driven goes backwards exactly as often as forwards and gets nowhere. A count that is driven goes forwards on average, but it jitters, and the jitter is set by how hard it is driven. The driving is paid for in heat. The question this essay answers is what the heat buys, and whether there is a floor under the price — and there is, one of the cleanest results of the last decade of thermodynamics.

A clock that can tick backwards #

The simplest counter takes steps forwards at a rate $k_+$ and backwards at a rate $k_-$, at random, each step independent of the last. It could be a motor on a track, a ratchet, the hand of a molecular clock. How the two rates are related is not up to the designer. If each forward step releases a free energy $a$ — in units of $kT$ — then microscopic reversibility fixes their ratio at $k_+/k_- = e^{a}$, the Boltzmann factor of the exponential that decides everything applied to one step. This is local detailed balance, the same accounting that makes the engine a fluctuation cannot run stall at a single temperature: with nothing to spend, forwards and backwards are equally likely, and the only way to bias the count is to pay for the bias.

The two sets of traces have the same average slope, one net tick per unit of time, and very different textures. The weakly driven clock, with half a $kT$ per step, takes a forward step only about 62 per cent of the time, so it gets its one net tick by taking about four steps, three of which cancel; its count wanders widely. The strongly driven clock, with 8 $kT$ per step, almost never steps back, and its count is as regular as a sequence of independent random events can be. After sixty units of time, the first is spread by nearly sixteen ticks and the second by less than eight.

The difference is paid for. Each net forward step releases its free energy $a$ into the surroundings as heat, and the entropy of the surroundings rises by $a$ in units of $k$ per net step. The weakly driven clock, after sixty net ticks, has produced 30 units of entropy; the strongly driven clock has produced 480. Regularity, bought with heat.

The product that does not depend on time #

Put the three quantities side by side. The walker’s net count after time $t$ has mean $(k_+ - k_-)t$ and variance $(k_+ + k_-)t$, because each step, forward or back, adds one unit of independent randomness. The entropy produced is the net count times $a$. The squared relative uncertainty, $\varepsilon^2 = \mathrm{Var}/\mathrm{mean}^2$, falls as $1/t$; the entropy $\Sigma$ grows as $t$; and their product is independent of time altogether:

$$\varepsilon^2\,\frac{\Sigma}{k} = \frac{k_+ + k_-}{k_+ - k_-}\,a = a\coth\frac{a}{2}.$$

The time independence is the whole point. It says that waiting longer does not make precision cheaper: to halve the relative uncertainty of a count, a clock must run four times as long and produce four times as much entropy, and there is no way to average out of the price. The product is a property of the clock’s design, not of how long it has been running.

The curve has two regimes. When the driving is strong the product is simply $a$, the entropy per step, because the clock almost never steps back and its count is a Poisson process — the same memoryless randomness as a nucleus with no clock — whose relative uncertainty is $1/\sqrt{N}$ — no amount of extra driving can do better than never stepping back. When the driving is weak, the product approaches 2 and does not go below it. That floor is the content of the result. It was conjectured by Andre Barato and Udo Seifert in 2015 from exactly this kind of calculation, and proved the next year by Todd Gingrich, Jordan Horowitz, Nikolay Perunov and Jeremy England for any process described by random jumps between states and held in a steady state: for every current such a process carries, the squared relative uncertainty times the entropy produced is at least $2k$. It is called the thermodynamic uncertainty relation — an uncertainty in the sense of spread, not of quantum mechanics.

Where the two comes from #

The number 2 is not a fitted constant, and its origin can be seen in the walker at weak driving. Let both rates be close to a common value $r$, with the forward one larger by a factor $e^{a}$ for small $a$. The net current is the difference of the two rates, about $ra$. The noise is their sum, about $2r$, because a backward step adds exactly as much randomness to the count as a forward one. The entropy produced per unit time is the current times $a$, about $ra^2$, in units of Boltzmann’s constant. Squared relative uncertainty per unit time times entropy per unit time is then $2r/(ra)^2 \times ra^2 = 2$, and every $r$ and every $a$ cancels.

So the two is the ratio of noise to signal at equilibrium: both directions of stepping make noise, while only their difference carries the current and releases the heat. The same factor sits in Einstein’s relation between diffusion and mobility, and in the equation that only runs forwards, where a random walk with a small bias spreads at a rate set by the sum of its step rates and drifts at a rate set by their difference. Far from equilibrium the backward steps die away, the noise falls from the sum towards the forward rate alone, and the product grows; there is no arrangement of rates that makes the noise smaller than the equilibrium ratio implies, and that is what the theorem proves.

What each unit of heat buys, and where it stops buying #

Turn the question round. A clock is asked to count to a hundred on average; how much does it cost to do so with a given spread?

Going from half a $kT$ per step to 2 $kT$ quadruples the heat and nearly halves the spread, from twenty ticks to eleven and a half. Going from 2 to 8 quadruples it again and narrows the spread from 11.5 to 10.0. The weakly driven clock is close to the bound — it is getting nearly the most precision the thermodynamics allows for its heat — but it is imprecise, because it has spent very little. The strongly driven clock is precise and extravagant, spending four times what the bound demands for the precision it achieves.

The limit that stops the strong clock improving is not thermodynamic at all. A clock whose ticks come at independent random moments has a count spread by the square root of its mean, $\pm 10$ for a hundred ticks, whatever drives it. Beating that needs a clock whose ticks are not independent, and the way to get one is to build each tick out of several steps.

Splitting each tick into steps #

Suppose each tick is completed only after the clock passes through a ring of $N$ internal states, each step with an equal share $A/N$ of the tick’s free energy $A$. The ticks are now more regular than random events, because a tick requires $N$ steps to accumulate, and the sum of $N$ random waiting times is proportionally less variable than one. The product becomes $(A/N)\coth(A/2N)$.

At 20 $kT$ per tick — roughly the free energy a cell gets from hydrolysing one molecule of ATP — a single-step clock sits ten times above the bound. Split the same tick into three steps and it is a little over three times above; into thirty steps and it is within four per cent of the bound. The heat is the same in every case; what changes is how finely it is spent. The bound can be approached only by spending the free energy in small pieces, each one close to equilibrium — which is the same lesson as the reversible engine of what a system actually minimises, which reaches the Carnot efficiency only by running infinitely slowly through infinitely many small steps.

Read as a statement about precision per unit heat, the relation gives a hard number. After $n$ ticks with a free energy $A$ spent per tick, the relative uncertainty of the count cannot be less than $\sqrt{2/(nA)}$. A clock that spends 20 $kT$ per tick cannot, after a single tick, be surer than about 32 per cent that the tick has come; after a hundred ticks, not surer than 3 per cent. To time something to one part in a thousand, it must produce at least two million $k$ of entropy along the way. In a pendulum clock that is nothing — a weight-driven escapement dissipates something like $10^{15}\,kT$ per swing. For a chemical oscillator in a bacterium, keeping a 24-hour rhythm on a budget of ATP molecules, it is a real constraint, and it is one of the reasons biochemical clocks are studied this way: the relation says how many molecules of fuel a given regularity requires, whatever the chemistry.

No design lies below the line #

The walker and the uniform ring are special cases, chosen because their statistics can be written down. The claim is that the bound holds for any network of states with any rates, and it is worth checking against machines nobody designed.

The points are machines assembled by rolling dice for their rates: some with one step far slower than the others, some nearly symmetric, some strongly driven and some barely. Their statistics are computed exactly, from the largest eigenvalue of a “tilted” rate matrix whose first two derivatives give the mean and the variance of the count. Every one of them lies above 2. Every one lies above the curve of the uniform three-step ring as well, which makes the second observation of the drawing: for a given number of states and a given free energy per cycle, equal steps are the most precise design, because a slow step dominates the waiting time and restores the randomness the other steps had averaged away.

Reading the heat off the fluctuations #

The relation runs both ways, and the reverse direction has turned out to be the more useful. Suppose a motor protein is watched stepping along a filament under a microscope, and its steps are counted: the mean number in a given time, and their variance. Nothing is known about its internal states or its chemistry. The relation still says that the entropy it produced over that time was at least $2k\,N^2/\sigma^2$. Divided by the time, that is a lower bound on the rate at which it turns free energy into heat, read off a movie of its motion.

This is called thermodynamic inference, and it is the practical payoff. Heat at the scale of a single molecule is almost impossible to measure directly — a motor dissipating a hundred $kT$ per second produces about $4\times 10^{-19}$ watts — but positions are easy to measure, and their fluctuations carry a bound on the heat. The bound is strictly a lower one, in the same way that the bit that has to be paid for is a minimum rather than a measurement: the machine may be wasting far more, in cycles that consume fuel and produce no motion, and the relation says nothing about waste it cannot see. What it does do is rule out models. A proposed mechanism for a motor that dissipates less than the bound demands, given the motor’s measured regularity, is wrong.

The equilibrium end of the curve #

Near equilibrium the product approaches 2 for every clock, and that end of the curve is not new. It is the fluctuation–dissipation relation in disguise. When a system is pushed gently, the current it carries is proportional to the push, the entropy produced is proportional to the push squared, and the fluctuations of the current are the equilibrium fluctuations, which do not depend on the push at all. Multiply them and the product is fixed, as the size of an energy fluctuation is fixed by a heat capacity in the temperature a molecule does not have — the relation of half a kT in a piece of wire, where the noise voltage across a resistor is set by the same resistance that dissipates power when a current flows. Johnson noise and the thermodynamic uncertainty relation are the two ends of one statement: the resistance that costs heat when a current is driven is the resistance that makes the current fluctuate when it is not.

What is new is that the bound survives far from equilibrium, where fluctuation–dissipation relations break down. A strongly driven motor has no equilibrium fluctuations to speak of, and nothing in linear response constrains it. The relation holds there anyway, with the same number.

Where the bound holds, and where it does not #

Jumps between states, or overdamped motion. The proof covers systems whose dynamics is a set of random jumps between discrete states with memoryless rates, and systems moving through a viscous medium where inertia is negligible — which includes nearly all of molecular biology and much of chemistry.

Steady states. The system must be held at a fixed driving for long enough to settle. Versions exist for finite times and for systems driven periodically, and in the periodic case the bound is weaker and depends on the driving frequency; a clock pushed by an external oscillator is not bound in the same way, because some of its regularity is borrowed from the oscillator.

Currents odd under time reversal. The counted quantity must change sign if the film is run backwards — steps, reactions, charge passed. Quantities that are even under reversal, such as the time spent in a state, obey different bounds.

Classical dynamics. Systems with inertia and magnetic fields, and quantum systems carrying current coherently, can violate the classical bound. Coherent electron transport through a quantum dot has been shown theoretically to beat it, much as the noise pushed below the floor beats a classical noise limit, by an amount that coherence and not heat pays for, and how far quantum clocks can go below 2 is part of what is being worked out.

A product that hides its two factors #

The figures plot a product of two quantities, and a product hides its factors. A clock sitting at 2.04 is nearly optimal in the sense of the bound and nearly useless as a clock, because it achieves its efficiency by stepping back almost as often as forward; a clock at 8 is wasteful and much better at telling the time. The bound compares designs at equal precision or equal heat, not in general, and says nothing about speed at all — a clock can be arbitrarily slow and still sit on the bound. Real designs trade all three, and a biologist asking why a motor spends 20 $kT$ per step where the bound would accept much less is asking about speed and reliability together, which no curve here plots.

Still open: the price of a tick for the best clocks #

For a molecular clock the bound is a live constraint and for a pendulum clock it is irrelevant, and in between is a question nobody has settled: how much entropy does the most precise possible clock of a given size have to produce per tick? An experiment in 2021 on a clock made from a vibrating membrane, whose ticks were read by counting its oscillations in a laser, found the entropy it produced per tick growing in proportion to the precision it achieved, as the relation suggests. Whether quantum clocks, with coherence in place of many classical steps, can buy precision more cheaply, and by how much, remains open, as does the question of how tight the bound can be made when only part of a machine’s motion is visible.

The habit worth carrying away is to ask what a regular process spends. Anything that counts reliably at a temperature is being driven, and its regularity is bought with heat at a rate no design can go below: twice Boltzmann’s constant, divided by the square of the relative uncertainty. A clock that ticks precisely is not only a mechanism but a small furnace, and the second law sets the price of its precision as surely as it sets the price of the work it does.