Exercise 6.9.5

Answers

We only know that

T_{v} (\begin{matrix} 0 \\ 0 \\ 0 \\ 1 \end{matrix}) = (\begin{matrix} - v t^{″} \\ 0 \\ 0 \\ t^{″} \end{matrix})

for some $t^{″} > 0$ . Compute that

⟨ T_{v}^{∗} L_{A} T_{v} (\begin{matrix} 0 \\ 0 \\ 0 \\ 1 \end{matrix}), (\begin{matrix} 0 \\ 0 \\ 0 \\ 1 \end{matrix}) ⟩ = ⟨ L_{A} (\begin{matrix} - v t^{″} \\ 0 \\ 0 \\ t^{″} \end{matrix}), (\begin{matrix} - v t^{″} \\ 0 \\ 0 \\ t^{″} \end{matrix}) ⟩ = {(t^{″})}^{2} (v^{2} - 1)

and

⟨ T_{v}^{∗} L_{A} T_{v} (\begin{matrix} 0 \\ 0 \\ 0 \\ 1 \end{matrix}), (\begin{matrix} 0 \\ 0 \\ 0 \\ 1 \end{matrix}) ⟩ = ⟨ L_{A} (\begin{matrix} 0 \\ 0 \\ 0 \\ 1 \end{matrix}), (\begin{matrix} 0 \\ 0 \\ 0 \\ 1 \end{matrix}) ⟩ = - 1

by Theorem 6.41. Hence we have ${(t^{″})}^{2} (v^{2} - 1) = - 1$ and so $t^{″} = \frac{1}{\sqrt{1 - v^{2}}}$ .

jephianlin

2011-06-27 00:00