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Finite causal sums and center-of-momentum rest observer

Statement

Let $(A, V, η, C^{+})$ be the oriented time-oriented affine Minkowski model with $c > 0$ , and fix an inertial frame. For a finite nonempty list $P_{1}, \dots, P_{N}$ of nonzero future-causal vectors in $V$ , let $Q : = \sum_{i = 1}^{N} P_{i}$ , let $m_{i} : = \sqrt{- η (P_{i}, P_{i})} / c$ (so $m_{i} = 0$ exactly for a null leg), and let $M (Q)$ be the invariant mass of the total when $Q$ is timelike, as defined in Total four-momentum, invariant mass and center-of-momentum frame of a finite system.

Then $Q$ is nonzero and future causal. If at least one $P_{i}$ is timelike, then $Q$ is future timelike. If all $P_{i}$ are null, then $Q$ is null exactly when all $P_{i}$ lie on the same future null ray; otherwise $Q$ is future timelike. Whenever $Q$ is timelike, the center-of-momentum frame exists, its time-axis observer $n_{Q} = Q / (M (Q) c)$ has zero total spatial momentum and total observer energy $M (Q) c^{2}$ , and

$M (Q) \geq \sum_{i = 1}^{N} m_{i} .$

Equality holds exactly when every $P_{i}$ is massive and there is one future unit timelike vector $n$ such that $P_{i} = m_{i} c n$ for every $i$ . A nonzero future-null total has no center-of-momentum frame.

Facts & Assumptions

Given: A positive scale $c$ , an inertial frame, and a finite nonempty list of nonzero future-causal vectors $P_{i}$ in its Lorentz vector space.

[F1]

Total four-momentum, invariant mass and center-of-momentum frame of a finite system defines the finite total $Q = \sum_{i} P_{i}$ , its invariant mass $M (Q) = \sqrt{- η (Q, Q)} / c$ when $Q$ is future timelike, and the meaning of a center-of-momentum frame.

[F2]

Observer energy, spatial momentum, invariant mass, rest energy and kinetic energy of a four-momentum defines $E_{n} (P) = - c η (P, n)$ , $q_{n} (P) = P + η (P, n) n$ , and the nonnegative invariant mass $m (P) = \sqrt{- η (P, P)} / c$ for a nonzero future-causal vector.

[F3]

Inertial frames and inertial charts of the affine Minkowski model supplies a future orthonormal basis; The closed future cone in inertial coordinates: description, finite sums, and the coordinate criterion for causal reachability says the future closed cone has coordinate form $v^{0} \geq ∣ v_{s} ∣$ and is closed under finite sums, with a strictly future-timelike sum when at least one summand is timelike.

[F4]

Center-of-momentum frames of a future timelike four-momentum and their unique time vector: for future-timelike $Q$ , $n_{Q} = Q / (M (Q) c)$ is a future unit observer, a frame with time vector $n_{Q}$ exists, and $q_{n_{Q}} (Q) = 0$ , $E_{n_{Q}} (Q) = M (Q) c^{2}$ .

[F5]

Positivity, mass shell and kinetic energy of the observer quantities of a four-momentum: for every nonzero future-causal $P$ and future unit observer $n$ , $E_{n} (P) > 0$ and $q_{n} (P) \in n^{⊥}$ ; the mass shell is $E_{n} (P)^{2} = m (P)^{2} c^{4} + c^{2} ∣ q_{n} (P) ∣^{2}$ for a timelike $P$ , while $E_{n} (P) = c ∣ q_{n} (P) ∣$ for a null $P$ .

[F6]

The Cauchy-Schwarz inequality for finite sums: for finite Euclidean vectors, $∣ \sum_{i} u_{i} \cdot v_{i} ∣ \leq (\sum_{i} ∣ u_{i} ∣^{2})^{1 / 2} (\sum_{i} ∣ v_{i} ∣^{2})^{1 / 2}$ , with equality only for proportional coordinate lists.

Proof

technique · use the future-cone sum rule, then decompose in the total rest frame

1.1F3given

By [F3], applying the finite-sum cone rule to the nonzero future-causal $P_{i}$ shows that $Q$ is future causal. In the fixed inertial frame each nonzero future-causal summand has positive time component, so the time component of $Q$ is a finite sum of positive numbers and $Q \neq 0$ . If one summand is timelike, [F3] gives that $Q$ is future timelike.

1.2F3F6algebra

Suppose every $P_{i}$ is null. Write its coordinates as $P_{i} = (a_{i}, u_{i})$ in the fixed frame. By [F3], $a_{i} = ∣ u_{i} ∣ > 0$ . Then $Q$ is null exactly when $∣ \sum_{i} u_{i} ∣ = \sum_{i} ∣ u_{i} ∣$ . Squaring both sides, this equality is equivalent to $\sum_{i < j} (∣ u_{i} ∣ ∣ u_{j} ∣ - u_{i} \cdot u_{j}) = 0$ . For each pair, apply [F6] to singleton coordinate lists to get $u_{i} \cdot u_{j} \leq ∣ u_{i} ∣ ∣ u_{j} ∣$ ; therefore every summand is nonnegative. Equality of the sum forces equality for every pair, and equality with positive dot product forces each pair to be positively proportional. Thus all $P_{i}$ lie on one future null ray. Conversely, positive multiples of one future-null vector sum to a future-null vector. If the directions are not all the same, the inequality is strict, so the already future-causal $Q$ is timelike.

1.3F1F4F5algebra

Let $Q$ be future timelike and use the center-of-momentum frame of [F4]. The total spatial momentum is zero and $E_{n_{Q}} (Q) = M (Q) c^{2}$ . By linearity of $E_{n_{Q}}$ , $M (Q) c^{2} = \sum_{i} E_{n_{Q}} (P_{i})$ . Each term satisfies $E_{n_{Q}} (P_{i}) \geq m_{i} c^{2}$ by the mass-shell identities in [F5], proving $M (Q) \geq \sum_{i} m_{i}$ .

2.1F2F4F5step 1.3∎

Equality in step 1.3 requires equality in every nonnegative term $E_{n_{Q}} (P_{i}) - m_{i} c^{2}$ . A nonzero null $P_{i}$ has $m_{i} = 0$ but $E_{n_{Q}} (P_{i}) > 0$ by [F5], so equality excludes null legs. For a timelike leg, [F5] gives $E_{n_{Q}} (P_{i}) = m_{i} c^{2}$ exactly when $q_{n_{Q}} (P_{i}) = 0$ ; then its observer decomposition is $P_{i} = (E_{n_{Q}} (P_{i}) / c) n_{Q} = m_{i} c n_{Q}$ . Conversely, if every leg has the form $P_{i} = m_{i} c n$ with one common future unit timelike $n$ , then $Q = c (\sum_{i} m_{i}) n$ , so $M (Q) = \sum_{i} m_{i}$ directly.

Finite causal sums and center-of-momentum rest observer

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources