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Exercise 6.9.6

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Note that if $S_{2}$ is moving past $S_{1}$ at a velocity $v > 0$ as measured on $S$ , the $T_{v}$ is the transformation from space-time coordinates of $S_{1}$ to that of $S_{2}$ . So now we have

{\begin{matrix} T_{v_{1}} : S \to S^{'}, \\ T_{v_{2}} : S^{'} \to S^{″}, \\ T_{v_{3}} : S \to S^{″} . \end{matrix}

The given condition says that $T_{v_{3}} = T_{v_{2}} T_{v_{3}}$ . This means that

B_{v_{2}} B_{v_{1}} = (\begin{matrix} \frac{1}{\sqrt{1 - v_{2}^{2}}} & 0 & 0 & \frac{- v_{2}}{\sqrt{1 - v_{2}^{2}}} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \frac{- v_{2}}{\sqrt{1 - v_{2}^{2}}} & 0 & 0 & \frac{1}{\sqrt{1 - v_{2}^{2}}} \end{matrix}) (\begin{matrix} \frac{1}{\sqrt{1 - v_{1}^{2}}} & 0 & 0 & \frac{- v_{1}}{\sqrt{1 - v_{1}^{2}}} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \frac{- v_{1}}{\sqrt{1 - v_{1}^{2}}} & 0 & 0 & \frac{1}{\sqrt{1 - v_{1}^{2}}} \end{matrix})

= (\begin{matrix} \frac{1 + v_{2} v_{1}}{\sqrt{(1 - v_{2}^{2}) (1 - v_{1}^{2})}} & 0 & 0 & \frac{- v_{2} - v_{1}}{\sqrt{(1 - v_{2}^{2}) (1 - v_{1}^{2})}} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \frac{v_{2} - v_{1}}{\sqrt{(1 - v_{2}^{2}) (1 - v_{1}^{2})}} & 0 & 0 & \frac{1 + v_{2} v_{1}}{\sqrt{(1 - v_{2}^{2}) (1 - v_{1}^{2})}} \end{matrix}) = B_{v_{3}} .

Hence we know that

{\begin{matrix} \frac{1 + v_{2} v_{1}}{\sqrt{(1 - v_{2}^{2}) (1 - v_{1}^{2})}} = \frac{1}{\sqrt{1 - v_{3}^{2}}}, \\ \frac{- v_{2} - v_{1}}{\sqrt{(1 - v_{2}^{2}) (1 - v_{1}^{2})}} = \frac{- v_{3}}{\sqrt{1 - v_{3}^{2}}} . \end{matrix}

By dividing the second equality by the first equality, we get the result

v_{3} = \frac{v_{1} + v_{2}}{1 + v_{1} v_{2}} .

jephianlin

2011-06-27 00:00

Exercise 6.9.6

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