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Eq. (3.6): the Friedmann equations imply ρ˙+3H(ρ+p)=0\dot\rho+3H(\rho+p)=0ρ˙​+3H(ρ+p)=0

Proved
HackCosmologicalAQFT.continuity_eq_of_friedmann

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

cosmologymathematical-physicsode

Let G>0G>0G>0, let I⊆RI\subseteq\mathbb RI⊆R be an open set of cosmological times, and let a,ρ,p:R→Ra,\rho,p:\mathbb R\to\mathbb Ra,ρ,p:R→R be the scale factor, energy density and pressure. Assume aaa is twice continuously differentiable on III with a>0a>0a>0 there, and that both Friedmann equations hold on III:

H2=8πG3ρ,a¨a=−4πG3(ρ+3p),H=a˙a.H^2=\frac{8\pi G}{3}\rho,\qquad \frac{\ddot a}{a}=-\frac{4\pi G}{3}(\rho+3p),\qquad H=\frac{\dot a}{a}.H2=38πG​ρ,aa¨​=−34πG​(ρ+3p),H=aa˙​.

Then for every t∈It\in It∈I the energy density obeys the conservation law

ρ˙(t)+3H(t)(ρ(t)+p(t))=0.\dot\rho(t)+3H(t)\big(\rho(t)+p(t)\big)=0.ρ˙​(t)+3H(t)(ρ(t)+p(t))=0.

This is the observation in Hack (2016), p. 80, that eq. (3.6), the FLRW form of ∇μTμν=0\nabla^\mu T_{\mu\nu}=0∇μTμν​=0, can be obtained directly from the Friedmann equations.

Formalization Note No differentiability of ρ\rhoρ is assumed. On III the first Friedmann equation determines ρ\rhoρ from aaa, so ρ˙\dot\rhoρ˙​ is meaningful.

Preamble
import Mathlib
import Definitions.Def_hack_flrw_friedmann
Formal statement
namespace HackCosmologicalAQFT

/-- Hack 2016, Sect. 3.1, eq. (3.6): the Friedmann equations imply `ρ̇ + 3H(ρ + p) = 0`. -/
theorem continuity_eq_of_friedmann (G : ℝ) (hG : 0 < G) (a ρ p : ℝ → ℝ) (I : Set ℝ)
    (hI : IsOpen I) (ha : ContDiffOn ℝ 2 a I) (hpos : ∀ t ∈ I, 0 < a t)
    (hF1 : FirstFriedmannEq G a ρ I) (hF2 : SecondFriedmannEq G a ρ p I) :
    ∀ t ∈ I, deriv ρ t + 3 * hubbleRate a t * (ρ t + p t) = 0 := by sorry

end HackCosmologicalAQFT
Source
T.-P. Hack, *Cosmological Applications of Algebraic Quantum Field Theory in Curved Spacetimes*, SpringerBriefs in Mathematical Physics 6, Springer, 2016, DOI 10.1007/978-3-319-21894-6 (https://doi.org/10.1007/978-3-319-21894-6), Sect. 3.1, p. 80, eqs. (3.5)–(3.6)
Read-back

What the Lean code literally says, in plain math · Aristotle (Harmonic) — non-blind, same agent as drafter

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 it may silently read the intended meaning into the code. Reviewers must not mistake it for independent testimony; compare the Lean code against the source directly.

Fix a real number GGG with G>0G>0G>0, three functions a,ρ,p:R→Ra,\rho,p:\mathbb R\to\mathbb Ra,ρ,p:R→R, and a set I⊆RI\subseteq\mathbb RI⊆R. Assume:

  1. III is open;
  2. aaa is twice continuously differentiable on III;
  3. a(t)>0a(t)>0a(t)>0 for every t∈It\in It∈I;
  4. for every t∈It\in It∈I: (a′(t)/a(t))2=8πG3ρ(t)\left(a'(t)/a(t)\right)^2=\frac{8\pi G}{3}\rho(t)(a′(t)/a(t))2=38πG​ρ(t);
  5. for every t∈It\in It∈I: a′′(t)/a(t)=−4πG3(ρ(t)+3p(t))a''(t)/a(t)=-\frac{4\pi G}{3}\big(\rho(t)+3p(t)\big)a′′(t)/a(t)=−34πG​(ρ(t)+3p(t)), where a′′a''a′′ is the derivative of the function s↦a′(s)s\mapsto a'(s)s↦a′(s).

Here a′(s)a'(s)a′(s) is the derivative of aaa at sss (taken as 000 where aaa is not differentiable). On the open set III, assumption 2 makes a′a'a′ and a′′a''a′′ the genuine derivatives.

Conclusion. For every t∈It\in It∈I,

ρ′(t)+3 a′(t)a(t) (ρ(t)+p(t))=0,\rho'(t)+3\,\frac{a'(t)}{a(t)}\,\big(\rho(t)+p(t)\big)=0,ρ′(t)+3a(t)a′(t)​(ρ(t)+p(t))=0,

where ρ′(t)\rho'(t)ρ′(t) is the derivative of ρ\rhoρ at ttt if ρ\rhoρ is differentiable there, and 000 otherwise. No hypothesis on differentiability of ρ\rhoρ or ppp, or on the signs of ρ,p\rho,pρ,p, is made. If III is empty the statement holds trivially.

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