Physics theoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)

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Mean work is at least the finite free-energy difference

Physical scope

An inequality for the ensemble-mean work in the finite protocol under the Jarzynski model assumptions; it is not a per-path second-law inequality and is not an empirical statement.

Statement

Under the hypotheses of Jarzynski equality for a finite protocol with stationary kernels, the mean work obeys $E_{F} W \geq Δ F$ . The conclusion concerns the path ensemble only; individual paths may have $W < Δ F$ .

Facts & Assumptions

Given: A finite-valued work random variable $W$ under the forward path law and $β > 0$ .

[F1]

Jarzynski equality for a finite protocol with stationary kernels: $E_{F} e^{- β W} = e^{- β Δ F}$ .

[F2]

The two-point convexity inequality for the exponential function: $x \mapsto e^{x}$ is convex on $R$ .

[F3]

Finite Jensen inequality for a convex function and nonnegative weights summing to one: for a finite probability vector $(p_{j})$ and real values $(x_{j})$ , $e^{\sum p_{j} x_{j}} \leq \sum p_{j} e^{x_{j}}$ .

[F4]

The exponential function is strictly increasing: the exponential is strictly increasing on $R$ .

Proof

Given: [F1]--[F4].

1.1givenF1F2F3

Apply finite Jensen to the finitely many forward paths with $x_{ω} = - β W (ω)$ and weights $P_{F} (ω)$ . Then $e^{- β E_{F} W} \leq E_{F} e^{- β W} = e^{- β Δ F}$ .

2.1F4step 1.1algebra∎

Strict increase of the exponential yields $- β E_{F} W \leq - β Δ F$ . Since $β > 0$ , multiplying by $- 1 / β$ reverses the order and gives $E_{F} W \geq Δ F$ .

Mean work is at least the finite free-energy difference

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources