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Complete two-body rest-frame kinematics

Physical scope

Exact two-product momentum-balance kinematics in the center-of-momentum frame of a timelike total in flat relativistic-particle mechanics; no interaction probability, cross section, or claim that a channel occurs.

Statement

Let a finite isolated collision have future-timelike total $Q$ of invariant mass $M > 0$ , and set $W : = M c^{2}$ . Suppose the complete outgoing list consists of exactly two legs, with recorded rest masses $a, b \geq 0$ ; by the adopted balance postulate, their four-momenta must sum to $Q$ . In a center-of-momentum frame with time-axis observer $n_{Q}$ , a nonzero two-product state with $p_{a} + p_{b} = 0$ exists exactly when

$W \geq (a + b) c^{2} if a, b > 0, W > (a + b) c^{2} if a = 0 or b = 0.$

When it exists, there are $k \geq 0$ and a unit vector $e \in n_{Q}^{⊥}$ such that $p_{a} = k e$ and $p_{b} = - k e$ . The energies and momentum magnitude are

$E_{a} = \frac{W^{2} + (a^{2} - b^{2}) c^{4}}{2 W}, E_{b} = \frac{W^{2} + (b^{2} - a^{2}) c^{4}}{2 W}, k^{2} = \frac{[W^{2} - (a + b)^{2} c^{4}] [W^{2} - (a - b)^{2} c^{4}]}{4 W^{2} c^{2}} .$

For $a, b > 0$ at threshold, $k = 0$ and the direction is immaterial. If a massless product is present, the strict threshold is required because a zero momentum would be the forbidden zero four-vector. These equations describe kinematic balance only. In SI units $a, b, M$ are kilograms, $W, E_{a}, E_{b}$ are joules, and $k$ is kilogram metres per second; the norm and dot product on $n_{Q}^{⊥}$ are induced by $η$ .

Facts & Assumptions

Given: A timelike total $Q$ , a center-of-momentum observer $n_{Q}$ , two specified nonnegative rest masses $a, b$ , and the adopted finite collision-balance model.

[A1]

Isolated event four-momentum balance: a complete isolated event balances the finite incoming and outgoing four-momentum sums.

[F1]

Finite event collisions and observer quantities: a massive final leg is future timelike with its recorded positive mass; a massless leg is a nonzero future-null vector with recorded mass zero.

[F2]

System invariant mass and final-state thresholds: the total mass is $M > 0$ , the center-of-momentum energy is $W = M c^{2}$ , and any final list obeys the rest-mass threshold.

[F3]

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

[F4]

Mechanical momentum and observer energy: relative to a future unit observer, the energy and spatial momentum are the observer decomposition of the same four-vector.

[F5]

For a massive leg of recorded mass $a > 0$ , Finite event collisions and observer quantities and Mechanical momentum and observer energy give $P = a U$ with $η (U, U) = - c^{2}$ , hence $η (P, P) = - a^{2} c^{2}$ by bilinearity; The closed future cone in inertial coordinates: description, finite sums, and the coordinate criterion for causal reachability gives the future-cone coordinate criterion in the center-of-momentum frame.

Proof

technique · solve the two on-shell equations in the total rest frame and verify the converse construction

1.1A1F1F2F3F4F5

In the center-of-momentum frame from [F3], $Q = (W / c) n_{Q}$ has zero total spatial momentum, so balance [A1] forces $p_{b} = - p_{a}$ . Put $k : = ∣ p_{a} ∣ \geq 0$ ; if $k > 0$ , write $p_{a} = k e$ with $∣ e ∣ = 1$ in the Euclidean rest space $n_{Q}^{⊥}$ . The massive and null mass shells from [F1] and [F5] give $E_{a}^{2} = a^{2} c^{4} + c^{2} k^{2}$ and $E_{b}^{2} = b^{2} c^{4} + c^{2} k^{2}$ , while the time component of balance is $E_{a} + E_{b} = W$ by [F2] and [F4].

2.1step 1.1algebra

Subtracting the shell equations and using $E_{a} + E_{b} = W > 0$ gives $(E_{a} - E_{b}) W = (a^{2} - b^{2}) c^{4}$ . Solving this and $E_{a} + E_{b} = W$ yields the stated $E_{a}, E_{b}$ formulas. Substituting the formula for $E_{a}$ into $c^{2} k^{2} = E_{a}^{2} - a^{2} c^{4}$ and factoring gives the stated expression for $k^{2}$ .

3.1F1F2step 2.1algebra

Necessity of $W \geq (a + b) c^{2}$ follows from [F2]. Conversely, under this bound the second factor in $k^{2}$ is nonnegative because $W \geq (a + b) c^{2} \geq ∣ a - b ∣ c^{2}$ . Also $E_{a} - a c^{2} = ((W - a c^{2})^{2} - b^{2} c^{4}) / (2 W) \geq 0$ and symmetrically $E_{b} - b c^{2} \geq 0$ , so both energies are future-directed. If $a, b > 0$ , equality at $W = (a + b) c^{2}$ gives $k = 0$ and two valid massive legs. If either mass is zero, that equality gives zero momentum and zero energy for that leg, which is not an admissible nonzero null vector; hence the strict bound is necessary and sufficient in that case.

4.1F1F2F3F4F5step 2.1step 3.1algebra∎

Under the stated admissible bound, choose any unit vector $e \in n_{Q}^{⊥}$ (this rest space is three-dimensional by [F3]) and define $P_{a} = (E_{a} / c) n_{Q} + k e$ , $P_{b} = (E_{b} / c) n_{Q} - k e$ . Their spatial sum is zero and their energies sum to $W$ , so $P_{a} + P_{b} = (W / c) n_{Q} = Q$ . Their orthogonal decompositions and the formulas for $E_{a}, E_{b}, k$ give $η (P_{a}, P_{a}) = - a^{2} c^{2}$ and $η (P_{b}, P_{b}) = - b^{2} c^{2}$ ; the energies are positive, and any zero-mass leg has $k > 0$ , so both vectors are nonzero future causal by the cone criterion in [F5]. Thus the constructed pair satisfies the balance and both mass shells.

Complete two-body rest-frame kinematics

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources