 Solution

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

c = 1

and use metric

(+ - - -)

. In the rest frame of B,

p_{B} = (m_{B}, 0)

and

p_{C} = p_{B} + p_{t}

. Since

p_{t}^{2} = 0

m_{C}^{2} = m_{B}^{2} + 2 m_{B} E_{t}, ∣ p_{C} ∣ = ∣ p_{t} ∣ = \frac{m _{C}^{2} - m _{B}^{2}}{2 m _{B}} .

m_{B}

because this is the B rest frame, not the C rest frame. In this frame C and

t

have equal, parallel three-momenta.

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