advanced29 min studyLesson 347 in Physics

Stochastic work, heat, and entropy

Assign work and heat along overdamped trajectories with explicit signs, preserve the stochastic chain rule, and separate system entropy changes from bath entropy production.

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What you’ll understand

  • Apply a pathwise first law with work on the system and heat to the bath.
  • Use Stratonovich heat consistently with an Itô Langevin model.
  • Distinguish stochastic system entropy, bath entropy, and total entropy production.

Thermodynamic quantities fluctuate along microscopic trajectories. The first law still provides exact bookkeeping within a declared stochastic model, but work, heat, and internal energy must have compatible signs and integration conventions. Entropy requires another distinction: heat changes the bath entropy, while the system's stochastic entropy depends on the probability of its state in the evolving ensemble. We will keep both contributions and calculate a simple trap protocol without pretending that one observed trajectory determines the whole probability distribution.

Declare energy-transfer signs first

Consider an overdamped coordinate in a potential U(x,λ)U(x,\lambda), with constant mobility and an isothermal equilibrium bath satisfying D=μkBTD=\mu k_BT. Lambda is an externally prescribed control parameter. Define work W as energy supplied to the system by changing lambda, and heat QbathQ_{\mathrm{bath}} as energy delivered from the system to the bath. Then ΔU=W−Qbath\Delta U=W-Q_{\mathrm{bath}}. This heat sign is opposite to the convention in which heat into the system is positive. Either convention is valid, but their first laws cannot be interchanged.

For conservative force −∂xU-\partial_xU, the work increment is ∂λU dλ\partial_\lambda U\,d\lambda and the bath heat increment is −∂xU∘dx-\partial_xU\circ dx. The circle denotes a Stratonovich integral, ensuring the ordinary chain rule for U. The underlying Langevin equation can still be written in Itô form; with additive noise its drift is the same in both conventions. The distinction now concerns how the trajectory-dependent force is integrated against the noisy displacement, rather than a change of the physical dynamics.

dU=∂λU dλ+∂xU∘dx=δW−δQbath,δW=∂λU dλ,δQbath=−∂xU∘dxdU=\partial_\lambda U\,d\lambda+\partial_xU\circ dx=\delta W-\delta Q_{\mathrm{bath}},\qquad \delta W=\partial_\lambda U\,d\lambda,\quad \delta Q_{\mathrm{bath}}=-\partial_xU\circ dx
All three transfers have energy units. The potential and force are conservative, lambda is a prescribed finite-variation protocol, and heat is positive into the bath. Extra nonconservative forces require their own work contribution.↔ Scroll wide equations horizontally.

The Itô correction belongs in energy bookkeeping

If the displacement differential is interpreted in Itô form, the energy increment contains the additional term D∂x2U dtD\partial_x^2U\,dt. Dropping it would incorrectly violate the first law on average even at equilibrium. Equivalently, converting the Stratonovich heat to an Itô integral includes minus this same correction. A harmonic trap makes the issue visible: noise continually injects energy while drag removes it, and a constant equilibrium energy average results from their balance rather than from both effects individually vanishing.

At fixed lambda and with no additional force, heat to the bath equals the decrease in potential energy between the trajectory's endpoints. During an instantaneous parameter quench the coordinate has no time to change, so the potential-energy jump is work. During subsequent relaxation, endpoint energy differences alone do not specify the system's stochastic entropy, because they do not determine the probability densities. This separation is useful experimentally: positions and a calibrated potential may determine pathwise energy transfer, while entropy inference also needs ensemble or distribution information.

Δssys=−kBln⁡p(xf,tf)p(xi,ti),Δsbath=QbathT,Σ=Δssys+Δsbath\Delta s_{\mathrm{sys}}=-k_B\ln\frac{p(x_f,t_f)}{p(x_i,t_i)},\qquad \Delta s_{\mathrm{bath}}=\frac{Q_{\mathrm{bath}}}{T},\qquad \Sigma=\Delta s_{\mathrm{sys}}+\Delta s_{\mathrm{bath}}
The two continuous densities use the same coordinate and fixed reference measure, so their ratio is dimensionless. Entropies have units J/K. The bath temperature is fixed, and the ensemble density follows the stated stochastic dynamics.↔ Scroll wide equations horizontally.

Negative single-trajectory production is allowed

The stochastic system entropy can be represented by minus Boltzmann's constant times the logarithm of density relative to a fixed reference density. Its ensemble mean reproduces the corresponding Gibbs–Shannon entropy. A trajectory ending in a more probable state can have a negative system entropy change. Bath entropy can also be negative when a thermal fluctuation transfers heat into the system. Neither fact alone contradicts the second law, which constrains total entropy production on average under the complete model and preparation.

For regular thermally consistent Markov dynamics with mutually supported forward and reverse path ensembles, total production satisfies ⟨e−Σ/kB⟩=1\langle e^{-\Sigma/k_B}\rangle=1. Jensen's inequality then gives nonnegative mean total production, while individual negative values remain possible. The system boundary term is essential; bath heat alone does not generally satisfy this same finite-time identity. Hidden reservoirs, feedback selection, singular support, or omitted degrees of freedom can require modified statements. The theorem therefore tests a fully specified trajectory ensemble, not an arbitrary heat histogram divorced from its preparation.

Shift a trap, then follow a partial relaxation

A harmonic trap has U=k(x−λ)2/2U=k(x-\lambda)^2/2, k=0.008 pN μm−1k=0.008\,\mathrm{pN\,\mu m^{-1}}, and kBT=0.004 pN μmk_BT=0.004\,\mathrm{pN\,\mu m}. A selected trajectory starts at x = 0, lambda = 0. The centre jumps to 1 micrometre, then the particle reaches x = 0.5 micrometre at fixed centre. Find work, bath heat, and bath entropy change.

  1. Before the quench U is zero. During the instantaneous centre shift the coordinate stays zero, so the work is the energy jump W=k(1)2/2=0.004 pN μmW=k(1)^2/2=0.004\,\mathrm{pN\,\mu m}. There is no coordinate-displacement heat during that instantaneous shift.

  2. At the final position, Uf=k(0.5−1)2/2=0.001 pN μmU_f=k(0.5-1)^2/2=0.001\,\mathrm{pN\,\mu m}. With the centre fixed during relaxation, bath heat is 0.004−0.001=0.003 pN μm0.004-0.001=0.003\,\mathrm{pN\,\mu m}.

  3. The bath entropy in Boltzmann units is Qbath/(kBT)=0.003/0.004=0.75Q_{\mathrm{bath}}/(k_BT)=0.003/0.004=0.75. The full energy change satisfies 0.001=0.004−0.0030.001=0.004-0.003. Computing total entropy production would additionally require the initial and final ensemble densities at the observed endpoints.

The result

Work on the particle is 0.004 pN micrometres, heat to the bath is 0.003 pN micrometres, and bath entropy increases by 0.75 kB; total production is not determined by these energies alone.

Explore the model

Interactive lab

Interactive lab · function

Read potential differences without inventing a thermal path

For a fresh comparison, Reset and set BOTH Coefficient a = 1 and Position x = 2, then Reset and set a = 1 and x = 1. Predict quadratic values 4 and 1. Their difference illustrates the energy decrease at fixed potential between two endpoints. Inspect paused values or Step.

Read potential differences without inventing a thermal pathThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe graph is a deterministic quadratic with a tangent and signed area. It does not follow a stochastic path or compute heat, work, temperature, entropy, or a driving protocol.

Check your understanding

Why is the circle used in the pathwise heat integral for this overdamped potential model?

Try it yourself

A fixed trap has U=x2 pN μm−1U=x^2\,\mathrm{pN\,\mu m^{-1}}, with x in micrometres. A trajectory goes from x = 2 to x = 1 micrometre without protocol work. Let kBT=1 pN μmk_BT=1\,\mathrm{pN\,\mu m} and endpoint densities be 0.1 and 0.2 per micrometre. Find bath heat and total entropy production.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the potential decrease for heat to the bath and the logarithm of final-to-initial density for system entropy.

Compare with the solution

The potential drops from 4 to 1 pN μm1\,\mathrm{pN\,\mu m}, so Qbath=3 pN μmQ_{\mathrm{bath}}=3\,\mathrm{pN\,\mu m} and ΔU=−3=W−Qbath\Delta U=-3=W-Q_{\mathrm{bath}}. The bath entropy change is 3kB3k_B. The system entropy change is −kBln⁡(0.2/0.1)=−kBln⁡2-k_B\ln(0.2/0.1)=-k_B\ln2. Thus Σ=(3−ln⁡2)kB\Sigma=(3-\ln2)k_B. The system contribution is negative here because the final endpoint is more probable, while the combined production is positive for this selected trajectory.

Take it with you

  • Use one declared sign convention for work, heat, and the pathwise first law.
  • Stratonovich heat and Itô energy calculations agree only with the required correction.
  • Total entropy production includes both evolving system probability and bath heat.

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