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⟨e−W/kBT⟩=e−ΔF/kBT\left\langle e^{-W/k_B T} \right\rangle = e^{-\Delta F/k_B T}

Why this formula appears here

The next structural advance did not concern the demon at all; it concerned a much older and more general problem — how to extract equilibrium information from a process driven arbitrarily far from equilibrium. In 1997, Christopher Jarzynski published “Nonequilibrium Equality for Free Energy Differences” in Physical Review Letters [ 3 ] . The Jarzynski equality states that if a system is driven from one equilibrium state to another along some protocol, performing a fluctuating amount of work W on each repetition, the equilibrium free energy difference Δ\Delta F between the two states can be recovered exactly from an exponential average over the work distribution: ⟨e−W/kBT⟩=e−ΔF/kBT\left\langle e^{-W/k_B T} \right\rangle = e^{-\Delta F/k_B T}. This is a…

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e−W/kBTe^{-W/k_B T}

Symbol e^-W/k_B T

e−e^-W/kBk_B T is part of the quantity the equation computes from the expression on the right.

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e−ΔF/kBTe^{-\Delta F/k_B T}

Symbol e^-Δ F/k_B T

e−e^-Δ F/kBk_B T is one of the signed contributions combined to compute the quantity on the left.

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Published contexts (1)

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⟨e−W/kBT⟩=e−ΔF/kBT\left\langle e^{-W/k_B T} \right\rangle = e^{-\Delta F/k_B T}

Equation 10 · Physics

From Origins to Frontier: A History of Stochastic Thermodynamics and Complex Systems

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

The next structural advance did not concern the demon at all; it concerned a much older and more general problem — how to extract equilibrium information from a process driven arbitrarily far from equilibrium. In 1997, Christopher Jarzynski published “Nonequilibrium Equality for Free Energy Differences” in Physical Review Letters [ 3 ] . The Jarzynski equality states that if a system is driven from one equilibrium state to another along some protocol, performing a fluctuating amount of work W on each repetition, the equilibrium free energy difference Δ\Delta F between the two states can be recovered exactly from an exponential average over the work distribution: ⟨e−W/kBT⟩=e−ΔF/kBT\left\langle e^{-W/k_B T} \right\rangle = e^{-\Delta F/k_B T}. This is a…

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