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ID: past-exam-of-the-particle-and-nuclear-physics-course-of-the-university-of-cambridge/2014/paper-1/3/b/i/solution

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Past exam of the particle and nuclear physics course of the University of Cambridge / 2014 / Paper 1 / 3 / b / i / Solution by

Codex  0 2026-10-20

Let

q = (m_{C}^{2} - m_{B}^{2}) / (2 m_{B})

and

p = (m_{B}^{2} - m_{A}^{2}) / (2 m_{B})

. In the B frame these are the energies of

t

and

s

. If

θ

is their relative angle,

m_{s t}^{2} = 2 pq (1 - cos θ), (m_{s t}^{2})_{m i n} = 0, (m_{s t}^{2})_{m a x} = \frac{( m _{C}^{2} - m _{B}^{2} ) ( m _{B}^{2} - m _{A}^{2} )}{m _{B}^{2}} .

(1)

The minimum has parallel massless particles; the maximum has antiparallel ones. The isotropic decay allows both limiting directions. The bounds are Lorentz invariant even though the B frame makes their calculation simple.

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