 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2018/5/solution

S^{'}

moving at

+ v

along

x

relative to

S

x^{'} = γ (x - v t), y^{'} = y, t^{'} = γ (t - \frac{vx}{c ^{2}}), γ = (1 - v^{2} / c^{2})^{- 1/2} .

Differentiate along the object's trajectory and divide the spatial differentials by

d t^{'}

. Using

u_{x} = d x / d t

and

u_{y} = d y / d t

u_{x}^{'} = \frac{u _{x} - v}{1 - u _{x} v / c ^{2}}, u_{y}^{'} = \frac{u _{y}}{γ ( 1 - u _{x} v / c ^{2} )} .

In the rest frame of ship A, take

v = c /2

and the Earth-frame velocity of B as

(0, - c /2)

. Here

γ = 2/ 3

and the common denominator is one, so

u_{B x}^{'} = - \frac{c}{2}, u_{B y}^{'} = - \frac{3}{4} c .

The required relative speed is the magnitude of this two-component velocity, not either component separately:

∣ u_{B}^{'} ∣ = c \frac{1}{4} + \frac{3}{16} = \frac{7}{4} c ≃ 0.6614 c .

The inverse Lorentz factor check is

γ_{A B} = γ_{A} γ_{B} (1 - u_{A} \cdot u_{B} / c^{2}) = 4/3

, which gives the same relative speed.

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