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

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System invariant mass and final-state thresholds

Physical scope

Necessary kinematic rest-mass thresholds for a finite, complete, isolated event in flat relativistic-particle mechanics; no interaction probability, coupling, selection rule, or claim that a channel occurs.

Statement

Let a mass-specified isolated collision configuration of Finite event collisions and observer quantities satisfy the balance postulate Isolated event four-momentum balance. Write its common total as $Q : = Q_{i n} = Q_{o u t}$ . It is nonzero future causal, and it is timelike if either side contains a massive leg. If $Q$ is future timelike, let $M : = \sqrt{- η (Q, Q)} / c$ , and let $μ_{i n}$ and $μ_{o u t}$ be the sums of the recorded rest masses on the two sides. Then

$M \geq μ_{i n}, M \geq μ_{o u t} .$

In particular, a proposed final collection with total rest mass $μ$ requires the invariant squared energy $σ : = - c^{2} η (Q, Q) = M^{2} c^{4}$ to satisfy $σ \geq μ^{2} c^{4}$ . If an actual final collection contains a nonzero null leg, then the inequality is strict. Equality is kinematically attained exactly when every leg on the relevant side is massive and all its four-momenta are proportional to the same future unit timelike vector. This is a necessary kinematic condition; it does not establish that the interaction realizes the channel.

If instead $Q$ is future null, each side consists only of null legs on one common future null ray; no center-of-momentum frame exists and no positive-mass final leg is compatible with that total. In SI units $m_{i}, M, μ$ are kilograms, $M c^{2}$ is joules, and $σ$ is joules squared.

Facts & Assumptions

Given: A finite, mass-specified collision datum in the flat model, no omitted legs or external four-momentum transfer, and the isolated-event balance postulate.

[A1]

Isolated event four-momentum balance: for the complete isolated event, $Q_{i n} = Q_{o u t}$ .

[F1]

Finite event collisions and observer quantities: each side is a finite nonempty list of nonzero future-causal momenta with a recorded mass for every massive or null leg; its side total is the corresponding finite sum.

[F2]

Finite causal sums and center-of-momentum rest observer: each side total is future causal and nonzero; if timelike its invariant mass is at least the sum of its leg masses, with equality exactly for a massive co-moving list; a null total consists only of co-directed null legs.

[F3]

Mass shell inversion, observer work and variable mass: for a massive leg of constant mass $m > 0$ , the adopted momentum $P = m U$ has invariant shell $η (P, P) = - m^{2} c^{2}$ , so the mathematical invariant mass of that leg is its physical rest-mass parameter.

[F4]

Total four-momentum, invariant mass and center-of-momentum frame of a finite system: for a future-timelike total $Q$ , $M = \sqrt{- η (Q, Q)} / c$ and its rest energy is $M c^{2}$ .

Proof

technique · apply the finite causal-sum result to both sides of the adopted balance

1.1A1F1F2F4

By the isolated-event assumption [A1], the incoming and outgoing total vectors are the same $Q$ . The collision data [F1] give finite nonempty lists of nonzero future-causal summands, so [F2] applies to each side and shows that $Q$ is future causal and nonzero. Thus the two sides have the same causal character; if $Q$ is timelike, it has one common invariant mass $M$ as defined in [F4].

1.2F2F3F4algebra

In the timelike case, apply the mass inequality of [F2] first to the incoming list and then to the outgoing list. By [F3], each massive leg's invariant mass equals its recorded rest mass, and each null leg has invariant mass zero; hence the sums of leg masses in [F2] are $μ_{i n}$ and $μ_{o u t}$ . Since both side totals equal $Q$ , each has invariant mass $M$ , so $M \geq μ_{i n}$ and $M \geq μ_{o u t}$ . For a proposed final list, set $μ = μ_{o u t}$ and multiply $M \geq μ$ by $c^{2} > 0$ before squaring nonnegative quantities; this gives $σ = M^{2} c^{4} \geq μ^{2} c^{4}$ .

2.1A1F2F4step 1.2∎

By [F2], equality for a timelike side holds exactly when all its legs are massive and co-moving; any nonzero null leg makes the mass inequality strict. If the common total is null, [F2] says both lists contain only null legs on the same ray, so no massive leg occurs and no rest frame is available. The inequalities constrain kinematics only; [A1] is the sole physical collision assumption and asserts no interaction mechanism or channel occurrence.

System invariant mass and final-state thresholds

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources