When a high energy photon passes near a heavy nucleus, a process known as pair p

Question

When a high energy photon passes near a heavy nucleus, a process known as pair production can occur. As a result, an electron and a positron (the electron's antiparticle) are produced. In one such occurence, a researcher notes that the electron and positron fly off in opposite directions after being produced, each traveling at speed 0.677c. The researcher records the time that it takes for the electron to travel from one position to another within his detector as 22.1 ns.

How much time would it take for the electron to move between the same two positions as measured by an observer moving along with the positron?

Explanation / Answer

v =0.677 c, t =22.1 ns

In the researcher's frame the distance traveled by the electron is
L = (0.677c)t

Velocity of electron relative to positron (relativistic):
v' = (0.677c + 0.677c) / [1 + (0.677c)(0.677c)/c²]
v' = 0.928 c

The distance "between the same two positions" contracts in the positron frame.
L' = L sqrt[1 - (0.677c)²/c²]
L' = L sqrt[1 - 0.677²]

So the observed time in the positron frame for the electron to travel this distance would be
t' = L' / v'
t' = L sqrt[1 - 0.677²] / v'
t' = (0.677c)t sqrt[1 - 0.677²] / v'
t' = (0.677c)(22.1 ns) sqrt[1 - 0.677²] / (0.928c)
t' = 11.866 ns

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