Discussion
An inertial observer sees two spacecrafts S and T flying away from each other along \(x\)-axis with individual speed \(0.5c\), where \(c\) is the speed of light. The speed of T with respect to S is
Answer: A
The speed of T relative to S is \(\dfrac{4}{5}c\).
Velocities do not simply add in special relativity. In the observer's frame S moves at -0.5c and T at +0.5c. Using the relativistic subtraction formula:
\[ \begin{aligned} v_{TS} &= \frac{u - v}{1 - uv/c^{2}} = \frac{0.5c + 0.5c}{1 + 0.25} \\ &= \frac{c}{1.25} = 0.8c \end{aligned} \]
c is the Galilean sum 0.5c + 0.5c, and 4/3 c exceeds the speed of light, which no relative speed of material bodies can. 2/3 c comes from dividing by 1.5 instead of 1.25.
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