 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1/section-a/a6/solution

The threshold must conserve both energy and momentum. For head-on photons, the total energy is

E_{1} + E_{2}

, while the total momentum has magnitude

∣ E_{2} - E_{1} ∣/ c

. Their combined invariant mass

M

satisfies

M^{2} c^{4} = (E_{1} + E_{2})^{2} - c^{2} (\frac{E _{2} - E _{1}}{c})^{2} = 4 E_{1} E_{2} .

M = 2 m_{e}

. Therefore

E_{2, m i n} = \frac{( m _{e} c ^{2} ) ^{2}}{E _{1}} = \frac{( 5.110 \times 1 0 ^{5} eV ) ^{2}}{1 0 ^{- 3} eV} ≃ 2.61 \times 1 0^{14} eV = 261 TeV .

Simply equating the laboratory energy to

2 m_{e} c^{2}

would miss the very large laboratory momentum: the newly produced pair must carry it away.

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