Physics theoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)

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Square-root action momentum equals massive mechanical momentum

Physical scope

Identity in the adopted massive point-particle model; the SR supplier is used only for the explicitly flat consistency check.

Statement

Let $I \subseteq R$ be an interval with nonempty interior, and let $γ : I \to M$ be a $C^{1}$ future-timelike curve on a prescribed smooth Lorentzian background, parameterized by $λ$ ; set $V = d γ / d λ$ . For the massive square-root Lagrangian $L = - m c \sqrt{- g (V, V)}$ with $m, c > 0$ , its fibre derivative is $p = m c V^{♭} / \sqrt{- g (V, V)}$ . If $τ$ is proper time along $γ$ , then $U = d γ / d τ = c V / \sqrt{- g (V, V)}$ , so the raised momentum is $p^{♯} = m U$ . In flat spacetime this agrees with the adopted special-relativistic mechanical momentum convention. This is a model identity, not a measurement.

Facts & Assumptions

Given: A future timelike $C^{1}$ worldline on a smooth Lorentzian background, $m > 0$ , $c > 0$ , and the square-root action model.

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): Assume countable choice ACω, meaning every sequence of nonempty sets has a choice function; no full Axiom of Choice is assumed.

[F1]

Timelike worldlines, proper time, length and four-velocity on a Lorentzian four-manifold supplies $d τ = \sqrt{- g (V, V)} d λ / c$ on a future-timelike curve, and $U = d x / d τ$ . The parameter and SI quantity conventions are those of Lorentz background, particle quantities and parameter units.

[F2]

Massive square-root Legendre map and its future shell: fibre derivative of $L = - m c \sqrt{- g (V, V)}$ is $p = m c V^{♭} / s$ .

[F3]

Adopted massive-particle momentum, observer measurement and isolated-collision conservation adopts the flat massive-particle mechanical momentum $P = m U$ , with its component identities supplied by Energy, momentum, invariant mass and four-force of relativistic particles, and necessary collision thresholds. The geometric four-velocity and proper-time relation are supplied by Norms of four-velocity and four-acceleration, and the observer rest-space decomposition.

[F4]

Free massive test-particle action: the massive square-root action is the adopted free-particle model.

[F5]

First-order shell action yields massive and null geodesics: for a smooth stationary first-order triple on the future massive branch with positive multiplier, its canonical covector equals the square-root fibre derivative. This comparison requires that supplier's stationary-triple hypotheses.

Proof

1.1givenA1F1F4algebra

Use the inherited smooth tangent bundle structure under [A1], adopt the square-root model [F4], and use the proper-time convention [F1]. Put $s = \sqrt{- g (V, V)} > 0$ . The proper-time definition gives $d τ / d λ = s / c$ , so $U = V / (d τ / d λ) = c V / s$ .

2.1givenF2F5step 1.1algebra

By [F2], $p^{♯} = m c V / s = m U$ . This derives the mechanical-momentum identity from the adopted action by differentiation; it is not inferred merely from matching units. This is pointwise on the given $C^{1}$ curve and requires no stationarity. If the curve is additionally part of a smooth stationary first-order triple with the hypotheses of [F5], then the covector of that triple equals this fibre derivative.

3.1F3step 2.1∎

In the flat specialization, [F3] uses the same mass-times-four-velocity convention, so the expression in step 2.1 agrees componentwise there. The curved statement itself used only the metric and proper-time definitions, not the flat theorem.

Depends on

Dependency tree · 2 levels

Sources

Sean M. Carroll, Lecture Notes on General Relativity, complete lecture notes (arXiv:gr-qc/9712019v1)