 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2015/6/solution

Align both frames'

x

axes with the motion and make their origins coincide at

t = t^{'} = 0

. With

S^{'}

moving along

+ x

v_{2}

relative to

S

, the Lorentz transformation and its inverse are

x^{'} = γ_{2} (x - v_{2} t), t^{'} = γ_{2} (t - \frac{v _{2} x}{c ^{2}}), x = γ_{2} (x^{'} + v_{2} t^{'}), t = γ_{2} (t^{'} + \frac{v _{2} x ^{'}}{c ^{2}}),

where

γ_{2} = (1 - v_{2}^{2} / c^{2})^{- 1/2}

; the transverse coordinates are unchanged. Along the object's path,

d x^{'} / d t^{'} = v_{1}

. Taking differentials of the inverse transformation gives

\frac{d x}{d t} = \frac{γ _{2} ( d x ^{'} + v _{2} d t ^{'} )}{γ _{2} ( d t ^{'} + v _{2} d x ^{'} / c ^{2} )}, v = \frac{v _{1} + v _{2}}{1 + v _{1} v _{2} / c ^{2}} .

v_{1}, v_{2} ≪ c

it reduces to the Galilean sum; if

v_{1} = c

, the result is still

c

. Two subluminal positive speeds give a subluminal result.

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