 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2013/3/a/solution

c = 1

as stipulated, so masses are measured in energy units. The incoming momentum is

p = E^{2} - M^{2}

and the invariant squared total four-momentum is

s = (E + m)^{2} - p^{2} = M^{2} + m^{2} + 2 m E .

β_{*} = p / (E + m)

, with

γ_{*} = (E + m) / s

. In this frame the incident particle has

E_{M}^{*} = \frac{M ^{2} + m E}{s}, p_{M}^{*} = γ_{*} (p - β_{*} E) = \frac{m p}{s} .

A nontrivial head-on elastic collision reverses

p_{M}^{*}

and preserves

E_{M}^{*}

. Boosting back to the laboratory frame gives

E^{'} = γ_{*} (E_{M}^{*} - β_{*} p_{M}^{*}) = \frac{( E + m ) ( M ^{2} + m E ) - m ( E ^{2} - M ^{2} )}{s},

E^{'} = \frac{2 m M ^{2} + E ( m ^{2} + M ^{2} )}{2 E m + m ^{2} + M ^{2}} .

The no-interaction forward branch instead leaves

E^{'} = E

; the requested formula describes the actual one-dimensional encounter and recoil.

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