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Kinetic coefficients: x˙=−λx=−γX\dot x=-\lambda x=-\gamma Xx˙=−λx=−γX

Proved
OnsagerReciprocal.kinetic_coefficients_relation

by Lucas · Oct 9, 2026 · Mathlib 0df444a (Lean v4.33.1)

mathematical-physicsonsagerthermodynamics

Let β\betaβ be a positive definite real symmetric n×nn\times nn×n matrix and λ\lambdaλ any real n×nn\times nn×n matrix. Put X=βxX=\beta xX=βx and γ=λβ−1\gamma=\lambda\beta^{-1}γ=λβ−1. Then for every x∈Rnx\in\mathbb R^nx∈Rn,

−λx=−γX.-\lambda x=-\gamma X .−λx=−γX.

That is, the quasi-stationary relaxation law x˙i=−λikxk\dot x_i=-\lambda_{ik}x_kx˙i​=−λik​xk​ can be rewritten in terms of the conjugate quantities as x˙i=−γikXk\dot x_i=-\gamma_{ik}X_kx˙i​=−γik​Xk​, which is how the kinetic coefficients γik\gamma_{ik}γik​ are introduced.

Preamble
import Mathlib
import Definitions.Def_OnsagerReciprocal_basic

open Matrix
Formal statement
namespace OnsagerReciprocal
theorem kinetic_coefficients_relation {n : ℕ} (β lam : Matrix (Fin n) (Fin n) ℝ)
    (hβ : β.PosDef) (x : Fin n → ℝ) :
    -(lam *ᵥ x) = -(kineticCoeff β lam *ᵥ conjugate β x) := by sorry
end OnsagerReciprocal
Source
Wikipedia, "Onsager reciprocal relations", https://en.wikipedia.org/w/index.php?title=Onsager_reciprocal_relations&oldid=1355014688, section "Abstract formulation" and its "Proof" subsection (pp. 5-6 of the PDF export), following L. D. Landau, E. M. Lifshitz, Statistical Physics, Part 1 (1975)
Read-back

What the Lean code literally says, in plain math · Aristotle (Harmonic)

Non-blind read-back — not independent testimony. This read-back was written by the same agent (Aristotle, by Harmonic) that drafted the Lean statement, with full knowledge of the source text and of the intended meaning. It was not produced by a blind, independent auditor, so reviewers must not treat it as independent testimony; compare the Lean code against the source yourself.

For every n∈Nn\in\mathbb Nn∈N, every real n×nn\times nn×n matrix β\betaβ that is positive definite (in Mathlib's sense for real matrices: β\betaβ is symmetric and vTβv>0v^{\mathsf T}\beta v>0vTβv>0 for every nonzero v∈Rnv\in\mathbb R^nv∈Rn), every real n×nn\times nn×n matrix λ\lambdaλ and every x∈Rnx\in\mathbb R^nx∈Rn:

−(λx)=−((λβ−1)(βx)),-(\lambda x)=-\big((\lambda\beta^{-1})(\beta x)\big),−(λx)=−((λβ−1)(βx)),

where β−1\beta^{-1}β−1 is the matrix inverse (genuine inverse here, since positive definite matrices are invertible). The statement is a single vector identity; no time variable or derivative appears in it. For n=0n=0n=0 both sides are the empty vector.

Human review
  • Endorsed by Shuze Chen · Oct 10, 2026

    Confirmed by the moderator at approval.

  • Endorsed by Lucas · Oct 10, 2026

    Confirmed by the mission captain (proposal self-audit).

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