Two-body decay
B and C have equal and opposite momenta. Therefore,
E²ᵦ - M²ᵦ = p² = E²c - M²c. (12.99)
Also, conservation of energy gives
Eᵦ + Ec = Mₐ. (12.100)
Solving the two previous equations for Eᵦ and Ec gives (using the shorthand
a ≡ Mₐ, etc.)
Eᵦ = (a² + b² - c²)/(2a), and Ec = (a² + c² - b²)/(2a). (12.101)
Equation (12.99) then gives the momentum of the particles as
p = (1/2a)√(a⁴ + b⁴ + c⁴ - 2a²b² - 2a²c² - 2b²c²). (12.102)
Remark: It turns out that the quantity under the radical can be factored into
(a + b + c)(a + b - c)(a - b + c)(a - b - c). (12.103)
This makes it clear that if a = b + c, then p = 0, because there is no leftover energy for the particles to be able to move. Interestingly, the form of p in Eq. (12.102) looks very much like the area of a triangle with sides a, b, c, which is given by Heron's formula as
A = (1/4)√(2a²b² + 2a²c² + 2b²c² - a⁴ - b⁴ - c⁴). ♣ (12.104)