Past exam of the mathematics course of the University of Cambridge / 2013 / ia / Paper 4 / 12B / b

Solution

The total relativistic energy and momentum in the original frame are

E = c^{2} (m_{1} γ_{1} + m_{2} γ_{2}), p = m_{1} γ_{1} u_{1} + m_{2} γ_{2} u_{2} .

V

makes the total spatial momentum

p^{'} = γ_{V} (p - V E / c^{2})

V = \frac{c ^{2} p}{E} = \frac{m _{1} γ _{1} u _{1} + m _{2} γ _{2} u _{2}}{m _{1} γ _{1} + m _{2} γ _{2}} .

It is a positive-weighted average of subluminal velocities, so

∣ V ∣ < c

P_{3} = P_{1} + P_{2}

provided no energy or momentum escapes in additional radiation. For the single outgoing particle,

u_{3} = c^{2} p / E = V

, so its speed in the original frame is $∣ V ∣$ . Its rest mass is the invariant norm of the total four-momentum:

m_{3}^{2} = \frac{E ^{2}}{c ^{4}} - \frac{p ^{2}}{c ^{2}} .

Expanding and using

γ_{i}^{2} (1 - u_{i}^{2} / c^{2}) = 1

gives

m_{3}^{2} = m_{1}^{2} + m_{2}^{2} + 2 m_{1} m_{2} γ_{1} γ_{2} (1 - \frac{u _{1} u _{2}}{c ^{2}}) .

The positive square root is taken for

m_{3}

. The factor multiplying

2 m_{1} m_{2}

is the relative Lorentz factor, at least one. Therefore

m_{3} \geq m_{1} + m_{2}

, with equality when

u_{1} = u_{2}

; otherwise the initial relative kinetic energy becomes internal rest energy.

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