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The one-loop flow in closed form: the inverse-coupling relation

Proved
CouplingConstantRG.isMuRunning_inv_relation

by Lucas · Oct 3, 2026 · Mathlib 0df444a (Lean v4.33.1)

differential-equationsmathematical-physicsrenormalization-group

This is the integrated form of the one-loop equation, and the technical heart of the mission: it holds for either sign of the one-loop coefficient and yields both branches of the goal theorem.

Let bbb be a one-loop coefficient, μ0>0\mu_0 > 0μ0​>0 a reference energy, T≥μ0T \ge \mu_0T≥μ0​ an upper energy, and let α\alphaα be a coupling with α(μ0)=α0>0\alpha(\mu_0) = \alpha_0 > 0α(μ0​)=α0​>0 satisfying the one-loop equation

μ dαdμ(μ)  =  b α(μ)2\mu\,\frac{d\alpha}{d\mu}(\mu) \;=\; b\,\alpha(\mu)^2μdμdα​(μ)=bα(μ)2

at every μ∈[μ0,T]\mu \in [\mu_0, T]μ∈[μ0​,T]. Then for every such μ\muμ,

α(μ) (1−b α0ln⁡(μ/μ0))  =  α0.\alpha(\mu)\,\Big(1 - b\,\alpha_0 \ln(\mu/\mu_0)\Big) \;=\; \alpha_0 .α(μ)(1−bα0​ln(μ/μ0​))=α0​.

Written as 1/α(μ)=1/α0−bln⁡(μ/μ0)1/\alpha(\mu) = 1/\alpha_0 - b \ln(\mu/\mu_0)1/α(μ)=1/α0​−bln(μ/μ0​) where α\alphaα does not vanish, this is the familiar statement that the inverse coupling runs linearly in the logarithm of the energy. The multiplicative form above is used because it remains meaningful without first knowing that α\alphaα stays nonzero. For b>0b > 0b>0 the bracket reaches 000 at μ=μ0e1/(bα0)\mu = \mu_0 e^{1/(b\alpha_0)}μ=μ0​e1/(bα0​), which is where the Landau pole milestone takes over; for b<0b < 0b<0 the bracket grows without bound, which is asymptotic freedom.

Preamble
import Mathlib
import Definitions.Def_CouplingConstantRGDefs
Formal statement
namespace CouplingConstantRG

theorem isMuRunning_inv_relation (b μ₀ T α₀ : ℝ) (α : ℝ → ℝ) (hμ₀ : 0 < μ₀)
    (hT : μ₀ ≤ T) (hα₀ : 0 < α₀) (h₀ : α μ₀ = α₀)
    (hα : IsMuRunning b α (Set.Icc μ₀ T)) :
    ∀ μ ∈ Set.Icc μ₀ T, α μ * (1 - b * α₀ * Real.log (μ / μ₀)) = α₀ := by sorry

end CouplingConstantRG
Source
Wikipedia, "Coupling constant", https://en.wikipedia.org/w/index.php?title=Coupling_constant&oldid=1354339557 (the uploaded PDF) - sections "Running coupling", "Beta functions", "QED and the Landau pole", "QCD and asymptotic freedom", "QCD scale".
Read-back

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

Provenance note (please read first). This read-back is not blind and is not independent testimony. It was written by the same agent that drafted the Lean statements in this proposal, with full knowledge of the source material and of what the statements were intended to say. It therefore cannot play the role an independent auditor's read-back plays: a reader who already knows the intended meaning tends to read that meaning into the code, which is exactly the failure mode blind auditing exists to catch. Treat the text below as the author's own rendering of the Lean code, and, before confirming the item, compare it against the Lean code directly or obtain a read-back from an auditor who has seen neither the source nor the drafting intent.


Fix real numbers b,μ0,T,α0b, \mu_0, T, \alpha_0b,μ0​,T,α0​ and a function α:R→R\alpha : \mathbb{R} \to \mathbb{R}α:R→R. Assume:

  1. μ0>0\mu_0 > 0μ0​>0;
  2. μ0≤T\mu_0 \le Tμ0​≤T (so the closed interval [μ0,T][\mu_0, T][μ0​,T] is nonempty; it is a single point when T=μ0T = \mu_0T=μ0​);
  3. α0>0\alpha_0 > 0α0​>0;
  4. α(μ0)=α0\alpha(\mu_0) = \alpha_0α(μ0​)=α0​;
  5. α\alphaα is one-loop running with coefficient bbb on [μ0,T][\mu_0, T][μ0​,T], i.e. for every μ∈[μ0,T]\mu \in [\mu_0,T]μ∈[μ0​,T] the function α\alphaα is differentiable at μ\muμ with α′(μ)=b α(μ)2/μ\alpha'(\mu) = b\,\alpha(\mu)^2/\muα′(μ)=bα(μ)2/μ (two-sided derivative, so the condition at the two endpoints also constrains α\alphaα just outside the interval).

The conclusion is: for every μ∈[μ0,T]\mu \in [\mu_0, T]μ∈[μ0​,T],

α(μ)⋅(1−b α0 ln⁡(μ/μ0))  =  α0.\alpha(\mu)\cdot\Big(1 - b\,\alpha_0\,\ln(\mu/\mu_0)\Big) \;=\; \alpha_0 .α(μ)⋅(1−bα0​ln(μ/μ0​))=α0​.

The identity is stated in product form, so it carries no implicit assumption that α(μ)≠0\alpha(\mu) \ne 0α(μ)=0 or that the bracket is nonzero; at a scale where the bracket vanished, the identity would force α0=0\alpha_0 = 0α0​=0, contradicting hypothesis 3. The sign of bbb is unrestricted: bbb may be negative, zero (in which case the claim reduces to α(μ)=α0\alpha(\mu) = \alpha_0α(μ)=α0​ on the interval) or positive. Nothing is claimed for μ\muμ outside [μ0,T][\mu_0,T][μ0​,T], and μ/μ0\mu/\mu_0μ/μ0​ is a ratio of positive numbers throughout, so the logarithm is the ordinary one and is ≥0\ge 0≥0 on the interval.

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