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.
On this page
The question for today
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 , with constant mobility and an isothermal equilibrium bath satisfying . Lambda is an externally prescribed control parameter. Define work W as energy supplied to the system by changing lambda, and heat as energy delivered from the system to the bath. Then . 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 , the work increment is and the bath heat increment is . 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.
The Itô correction belongs in energy bookkeeping
If the displacement differential is interpreted in Itô form, the energy increment contains the additional term . 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.
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 . 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.
Work it through
Shift a trap, then follow a partial relaxation
A harmonic trap has , , and . 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.
Before the quench U is zero. During the instantaneous centre shift the coordinate stays zero, so the work is the energy jump . There is no coordinate-displacement heat during that instantaneous shift.
At the final position, . With the centre fixed during relaxation, bath heat is .
The bath entropy in Boltzmann units is . The full energy change satisfies . Computing total entropy production would additionally require the initial and final ensemble densities at the observed endpoints.
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.
The living notebook
Explore the model
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.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
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.
Pause & reason
Check your understanding
Your turn
Try it yourself
A fixed trap has , with x in micrometres. A trajectory goes from x = 2 to x = 1 micrometre without protocol work. Let 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 , so and . The bath entropy change is . The system entropy change is . Thus . The system contribution is negative here because the final endpoint is more probable, while the combined production is positive for this selected trajectory.
Keep the idea
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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