Past exam of the particle and nuclear physics course of the University of Cambridge / 2014 / Paper 1 / 3 / a

Table of contents
- i a
  - Solution i
- ii a
  - Solution ii

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}} .

The denominator is

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.

ii

Solution

The second decay has

p_{A} = - p_{s}

in the B rest frame. Using

p_{A}^{2} = m_{A}^{2}

and

p_{s}^{2} = 0

gives

∣ p_{A} ∣ = ∣ p_{s} ∣ = \frac{m _{B}^{2} - m _{A}^{2}}{2 m _{B}}, E_{A} = \frac{m _{B}^{2} + m _{A}^{2}}{2 m _{B}} .

This is an ordinary two-body decay in its parent's rest frame.

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