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⟨exp⁡(−ΔstotkB)⟩=1\left\langle \exp\left(-\frac{\Delta s_{\mathrm{tot}}}{k_{\mathrm B}}\right) \right\rangle = 1

Why this formula appears here

Exponentiating and averaging over forward paths produces the integral fluctuation theorem, ⟨exp⁡(−ΔstotkB)⟩=1\left\langle \exp\left(-\frac{\Delta s_{\mathrm{tot}}}{k_{\mathrm B}}\right) \right\rangle = 1. Jensen’s inequality then gives

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Δstot\Delta s_{\mathrm{tot}}

Symbol Δ s_tot

Δ sts_tot occurs above the fraction bar. The numerator is divided by the entire denominator below it.

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kBk_{\mathrm B}

Symbol k_mathrm B

kmk_mathrm B occurs below the fraction bar. The quantity above the bar is divided by this expression; zero is excluded as a denominator.

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How to interpret it

With a fixed numerator, increasing a nonzero denominator reduces the fraction. Read it with the definitions, units, and assumptions supplied by the article.

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Published contexts (1)

A symbol can carry a different meaning in another article. Each occurrence keeps its own guide and term definitions.

⟨exp⁡(−ΔstotkB)⟩=1.\left\langle \exp\left(-\frac{\Delta s_{\mathrm{tot}}}{k_{\mathrm B}}\right) \right\rangle = 1.

Equation 8 · Physics

The Statistical Mechanics of Irreversibility at Molecular Scale

This equation states an equality: the expressions on both sides have the same value under the article’s assumptions.

Exponentiating and averaging over forward paths produces the integral fluctuation theorem, ⟨exp⁡(−ΔstotkB)⟩=1\left\langle \exp\left(-\frac{\Delta s_{\mathrm{tot}}}{k_{\mathrm B}}\right) \right\rangle = 1. Jensen’s inequality then gives

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