Downstream Mach number of a shock (Ogilvie 2016, eq. 6.31)

Proved

AstrophysicalFluidDynamics.shock_downstream_mach_sq

by Lucas · Oct 4, 2026 · Mathlib 0df444a (Lean v4.33.1)

astrophysicsfluid-dynamicsshock-waves

Under the same hypotheses, the downstream Mach number $M_2 = u_2/(\gamma p_2/\rho_2)^{1/2}$ satisfies

M_2^2 = \frac{2+(\gamma-1)M_1^2}{2\gamma M_1^2-(\gamma-1)},

where $M_1 = u_1/(\gamma p_1/\rho_1)^{1/2}$ is the upstream Mach number.

This relation shows that the flow is supersonic on one side of the shock and subsonic on the other.

Preamble

import Definitions.Def_AstrophysicalFluidDynamics_ShockDefs
import Mathlib

open Filter Topology

Formal statement

namespace AstrophysicalFluidDynamics

theorem shock_downstream_mach_sq
    (γ ρ₁ u₁ p₁ ρ₂ u₂ p₂ : ℝ) (hγ : 1 < γ)
    (hρ₁ : 0 < ρ₁) (hu₁ : 0 < u₁) (hp₁ : 0 < p₁)
    (hρ₂ : 0 < ρ₂) (hu₂ : 0 < u₂) (hp₂ : 0 < p₂)
    (hRH : RankineHugoniot γ ρ₁ u₁ p₁ ρ₂ u₂ p₂)
    (hshock : (ρ₂, u₂, p₂) ≠ (ρ₁, u₁, p₁)) :
    (machNumber γ ρ₂ u₂ p₂) ^ 2 = shockDownstreamMachSq γ (machNumber γ ρ₁ u₁ p₁) := by sorry

end AstrophysicalFluidDynamics

Source

G. I. Ogilvie, *Astrophysical fluid dynamics* (lecture notes), J. Plasma Phys. 82 (2016) 205820301, https://doi.org/10.1017/S0022377816000489, §6.3.2, eq. (6.31), p. 39; Example A.13, eq. (A 27)

Read-back

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

Read-back disclosure — NOT blind, NOT independent. This read-back was written by the same agent (Aristotle, by Harmonic) that drafted the Lean statement below, with full knowledge of the source text and of the intended meaning. It is therefore not independent testimony and must not be mistaken for a blind audit. A reviewer should compare the Lean code against the source directly; an independent blind read-back is still recommended.

Fix real numbers $\gamma, \rho_1, u_1, p_1, \rho_2, u_2, p_2$ . Assume $\gamma > 1$ ; $\rho_1 > 0$ , $u_1 > 0$ , $p_1 > 0$ ; $\rho_2 > 0$ , $u_2 > 0$ , $p_2 > 0$ ; and that the three Rankine–Hugoniot equalities hold, namely

\rho_2 u_2 = \rho_1 u_1,\qquad \rho_2 u_2^2 + p_2 = \rho_1 u_1^2 + p_1,

Assume moreover that the triple $(\rho_2,u_2,p_2)$ is not equal to the triple $(\rho_1,u_1,p_1)$ (at least one component differs). Here $M_1 = u_1 / \sqrt{\gamma p_1/\rho_1}$ and $M_2 = u_2/\sqrt{\gamma p_2/\rho_2}$ (real square root; under the positivity assumptions the radicands are positive, so no junk value of the square root or of division by zero arises). Conclusion: $M_2^2 = \dfrac{2+(\gamma-1)M_1^2}{2\gamma M_1^2-(\gamma-1)}$ (the right-hand side being read as $0$ if its denominator vanishes).

Human review

Endorsed by Shuze Chen · Oct 8, 2026
Confirmed by the moderator at approval.
Endorsed by Lucas · Oct 8, 2026
Confirmed by the mission captain (proposal self-audit).

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