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Past exam of the mathematics course of the University of Cambridge

2024

Paper 4

The <four-momentum> of a massive particle is
$$
P^\mu=(E/c,\mathbf p)=m\gamma(c,\mathbf v),
\qquad P^2=m^2c^2,
$$
while for a massless particle
$$
P^\mu=(E/c,\mathbf p),\qquad E=c|\mathbf p|,\qquad P^2=0.
$$
For $P_1=P_2+P_3$, invariance gives
$$
(P_1-P_2)^2=P_3^2.
$$
In the rest frame of particle 1, $P_1=(m_1c,0)$, so
$$
\boxed{E_2=\frac{m_1^2+m_2^2-m_3^2}{2m_1}c^2},
\qquad
\boxed{E_3=\frac{m_1^2+m_3^2-m_2^2}{2m_1}c^2},
$$
with $E_1=m_1c^2$. Equal and opposite final <momenta> are real exactly when
$$
\boxed{m_1\geq m_2+m_3},
$$
the threshold condition for a <relativistic two-body decay>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 4 / 12C / a / Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 4 / 12C / a / Solution

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