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First Friedmann equation in the dimensionless scale factor: H2=(a˙/a)2=8πG3[ρ+ρc−ρ0a2]H^2 = (\dot a/a)^2 = \frac{8\pi G}{3}\big[\rho + \frac{\rho_c-\rho_0}{a^2}\big]H2=(a˙/a)2=38πG​[ρ+a2ρc​−ρ0​​]

Proved
FriedmannEquations.first_friedmann_dimensionless

by Lucas · Sep 27, 2026 · Mathlib 0df444a (Lean v4.33.1)

cosmologyfriedmann-equationsgeneral-relativity

Let G>0G>0G>0, k∈Rk\in\mathbb Rk∈R, and let t0t_0t0​ ("now") and ttt be times with R(t0)≠0R(t_0)\ne0R(t0​)=0, R(t)≠0R(t)\neq 0R(t)=0. Suppose the first Friedmann equation with Λ=0\Lambda = 0Λ=0, H2=8πGρ3−kR2H^2 = \frac{8\pi G\rho}{3} - \frac{k}{R^2}H2=38πGρ​−R2k​, holds at both ttt and t0t_0t0​ (same kkk). Put a(s)=R(s)/R(t0)a(s) = R(s)/R(t_0)a(s)=R(s)/R(t0​), H0=H(t0)H_0 = H(t_0)H0​=H(t0​), ρ0=ρ(t0)\rho_0 = \rho(t_0)ρ0​=ρ(t0​) and ρc=3H02/(8πG)\rho_c = 3H_0^2/(8\pi G)ρc​=3H02​/(8πG). Then a˙(t)/a(t)=H(t)\dot a(t)/a(t) = H(t)a˙(t)/a(t)=H(t) and

(a˙(t)a(t))2=8πG3[ρ(t)+ρc−ρ0a(t)2].\Big(\frac{\dot a(t)}{a(t)}\Big)^2 = \frac{8\pi G}{3}\left[\rho(t) + \frac{\rho_c - \rho_0}{a(t)^2}\right].(a(t)a˙(t)​)2=38πG​[ρ(t)+a(t)2ρc​−ρ0​​].
Preamble
import Definitions.Def_FriedmannEquations_Defs
import Mathlib

open FriedmannEquations
Formal statement
theorem FriedmannEquations.first_friedmann_dimensionless
    (G k : ℝ) (hG : 0 < G) (R ρ : ℝ → ℝ) (t t₀ : ℝ) (hR : R t ≠ 0) (hR₀ : R t₀ ≠ 0)
    (h₁ : FirstFriedmannEq G 0 k R ρ t) (h₁₀ : FirstFriedmannEq G 0 k R ρ t₀) :
    hubble (fun s => R s / R t₀) t = hubble R t ∧
    hubble (fun s => R s / R t₀) t ^ 2 =
      8 * Real.pi * G / 3 *
        (ρ t + (criticalDensity G (hubble R t₀) - ρ t₀) / (R t / R t₀) ^ 2) := by sorry
Source
Wikipedia, "Friedmann equations", revision oldid=1374073387, https://en.wikipedia.org/w/index.php?title=Friedmann_equations&oldid=1374073387 (the PDF supplied with this proposal), section "Dimensionless scale factor", p. 3.
Read-back

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

NOTE — NON-BLIND READ-BACK. This read-back is not independent testimony. It was written by the same agent (Aristotle, by Harmonic) that drafted the Lean statements, with full knowledge of the source article and of the intended meaning. It was produced at the explicit request of the proposal owner. Reviewers must not treat it as a blind audit; an independent read-back is still recommended.

Fix reals G,kG,kG,k with G>0G>0G>0, functions R,ρ:R→RR,\rho:\mathbb R\to\mathbb RR,ρ:R→R and times t,t0t,t_0t,t0​ with R(t)≠0R(t)\neq0R(t)=0 and R(t0)≠0R(t_0)\ne0R(t0​)=0. Assume at time ttt and at time t0t_0t0​ (both with the same kkk and with Λ=0\Lambda=0Λ=0)

(R˙(s)R(s))2=8πGρ(s)3−kR(s)2,s∈{t,t0}.\Big(\frac{\dot R(s)}{R(s)}\Big)^2 = \frac{8\pi G\rho(s)}{3} - \frac{k}{R(s)^2},\qquad s\in\{t,t_0\}.(R(s)R˙(s)​)2=38πGρ(s)​−R(s)2k​,s∈{t,t0​}.

Let aaa be the function s↦R(s)/R(t0)s\mapsto R(s)/R(t_0)s↦R(s)/R(t0​) and H0=R˙(t0)/R(t0)H_0 = \dot R(t_0)/R(t_0)H0​=R˙(t0​)/R(t0​). Conclusion: (i) a˙(t)/a(t)=R˙(t)/R(t)\dot a(t)/a(t) = \dot R(t)/R(t)a˙(t)/a(t)=R˙(t)/R(t), and (ii)

(a˙(t)a(t))2=8πG3(ρ(t)+3H028πG−ρ(t0)(R(t)/R(t0))2).\Big(\frac{\dot a(t)}{a(t)}\Big)^2 = \frac{8\pi G}{3}\left(\rho(t) + \frac{\frac{3H_0^2}{8\pi G} - \rho(t_0)}{\big(R(t)/R(t_0)\big)^2}\right).(a(t)a˙(t)​)2=38πG​(ρ(t)+(R(t)/R(t0​))28πG3H02​​−ρ(t0​)​).

All derivatives are deriv (junk 000 where not differentiable); no differentiability is assumed.

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