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Two null momenta and collision angle

Physical scope

Invariant energy of two nonzero future-null incoming momenta as resolved in one named inertial laboratory observer; no collision outcome or reaction channel is asserted.

Statement

Let $P_{1}, P_{2}$ be two nonzero future-null incoming momenta in a finite collision configuration, and let $n$ be the future unit observer of a named inertial laboratory frame. Define $E_{i} : = E_{n} (P_{i}) > 0$ , $p_{i} : = q_{n} (P_{i})$ , and equip $n^{⊥}$ with the induced Euclidean inner product $⟨ u, v ⟩_{n} : = η (u, v)$ and norm $∣ u ∣_{n} : = \sqrt{⟨ u, u ⟩_{n}}$ . Let $θ \in [0, π]$ be their spatial angle, $\cos θ = ⟨ p_{1}, p_{2} ⟩_{n} / (∣ p_{1} ∣_{n} ∣ p_{2} ∣_{n})$ . For $Q : = P_{1} + P_{2}$ and $σ : = - c^{2} η (Q, Q)$ ,

$σ = 2 E_{1} E_{2} (1 - \cos θ), M (Q) = \frac{\sqrt{σ}}{c^{2}} .$

The total $Q$ is null exactly when $θ = 0$ (the two incoming vectors lie on one future null ray); it is timelike for $θ > 0$ . In particular, for head-on momenta $θ = π$ , $M (Q) = 2 \sqrt{E_{1} E_{2}} / c^{2}$ ; if $E_{1} = E_{2} = E$ , this is $2 E / c^{2}$ . The angle and the two energies depend on the named observer, while $σ$ and $M (Q)$ are invariant.

In SI units $E_{1}, E_{2}$ are joules, $σ$ is joules squared, and $M (Q)$ is kilograms; $θ$ is dimensionless.

Facts & Assumptions

Given: Two nonzero future-null incoming momenta, a future unit laboratory observer, and the finite collision-data conventions.

[F1]

Finite event collisions and observer quantities: each null leg is a nonzero future-null vector; its momentum scale is specified independently of proper time.

[F2]

Mechanical momentum and observer energy: $E_{n} (P) = - c η (P, n)$ , $q_{n} (P) = P + η (P, n) n$ , and the observer rest space has its Euclidean inner product.

[F3]

Total four-momentum, invariant mass and center-of-momentum frame of a finite system: the total four-momentum is the finite vector sum; its invariant mass is $M (Q) = \sqrt{- η (Q, Q)} / c$ when $Q$ is future timelike, and $M (Q) = 0$ when $Q$ is future null, with no rest frame in the null case.

[F4]

Positivity, mass shell and kinetic energy of the observer quantities of a four-momentum: for a nonzero future-null vector, the observer energy is positive and equals $c$ times the norm of its spatial momentum.

[F5]

The closed future cone in inertial coordinates: description, finite sums, and the coordinate criterion for causal reachability: the closed future cone is closed under finite sums; in an inertial chart it is given by $Q^{0} \geq ∣ Q_{s} ∣$ , its timelike part by $Q^{0} > ∣ Q_{s} ∣$ , and its nonzero null part by $Q^{0} = ∣ Q_{s} ∣ > 0$ .

Proof

technique · expand the Lorentzian inner product using the observer decomposition

1.1F1F2F4

By [F2], $P_{i} = (E_{i} / c) n + p_{i}$ , with $p_{i} \in n^{⊥}$ . The null mass-shell identity of [F4] gives $∣ p_{i} ∣ = E_{i} / c$ , and [F4] also gives $E_{i} > 0$ , so the spatial angle is well-defined.

2.1F2F3F5step 1.1algebra

Orthogonality to $n$ gives $η (P_{1}, P_{2}) = - E_{1} E_{2} / c^{2} + ⟨ p_{1}, p_{2} ⟩_{n}$ . Since $P_{i}$ are null, $σ = - c^{2} η (P_{1} + P_{2}, P_{1} + P_{2}) = - 2 c^{2} η (P_{1}, P_{2}) = 2 E_{1} E_{2} (1 - \cos θ)$ . By [F5], $Q$ lies in the closed future cone, and its laboratory time component is $Q^{0} = (E_{1} + E_{2}) / c > 0$ by step 1.1, so it is nonzero. In that chart $η (Q, Q) = - (Q^{0})^{2} + ∣ Q_{s} ∣^{2} = - σ / c^{2}$ . If $σ > 0$ , this identity and $Q^{0} > 0$ give $Q^{0} > ∣ Q_{s} ∣$ , so [F5] makes $Q$ future timelike and [F3] gives $M (Q) = \sqrt{σ} / c^{2}$ ; if $σ = 0$ , then $Q$ is future null and [F3] gives $M (Q) = 0 = \sqrt{σ} / c^{2}$ .

3.1F1F2F3F4F5step 1.1step 2.1algebra∎

Because $E_{1} E_{2} > 0$ and $1 - \cos θ \geq 0$ , $σ = 0$ exactly when $\cos θ = 1$ , or $θ = 0$ ; then the spatial vectors have the same direction and the null four-vectors are positive multiples of one future-null vector. For $θ > 0$ , $σ > 0$ , so the coordinate argument in step 2.1 gives $Q$ future timelike. At $θ = π$ , substitution gives $M = 2 \sqrt{E_{1} E_{2}} / c^{2}$ , and equal energies give $2 E / c^{2}$ .

Two null momenta and collision angle

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources