- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
2.21 Miscellaneous review exercises on matrices
6. $\quad$ The special theory of relativity makes use of a set of equations of the form $$x' = a(x - vt), \quad y' = y, \quad z' = z, \quad t' = a(t - vx/c^2).$$ Here $v$ represents the velocity of a moving object, $c$ the speed of light, and $a = c/\sqrt{c^2 - v^2},$ where $|v| \lt c.$ The linear transformation which maps the two-dimensional vector $(x, t)$ onto $(x', t')$ is called a Lorentz transformation. Its matrix relative to the usual bases is denoted by $L(v)$ and is given by $$L(v) = a \begin{bmatrix} 1 & -v \\ -vc^{-2} & 1 \end{bmatrix}.$$ Note that $L(v)$ is nonsingular and that $L(0) = I.$ Prove that $L(v)L(u) = L(w),$ where $w = (u + v)c^2/(uv + c^2).$ In other words, the product of two Lorentz transformations is another Lorentz transformation.
$\quad$ Proof. $\quad$ If \begin{align*} L(v) &= \frac{c}{\sqrt{c^2 - v^2}} \begin{bmatrix} 1 & -v \\ -vc^{-2} & 1 \end{bmatrix} \end{align*} we have \begin{align*} L(v)L(u) &= \frac{c^2}{\sqrt{(c^2 - v^2)(c^2 - u^2)}}\begin{bmatrix} 1 & -v \\ -vc^{-2} & 1 \end{bmatrix} \begin{bmatrix} 1 & -u \\ -uc^{-2} & 1 \end{bmatrix} \\ &= \frac{c^2}{\sqrt{(c^2 - v^2)(c^2 - u^2)}} \begin{bmatrix} 1 + uvc^{-2} & -u - v \\ -vc^{-2} - uc^{-2} & uvc^{-2} + 1 \end{bmatrix}. \end{align*} To prove that $L(v)L(c) = L(w),$ we must show that \begin{align*} \frac{c^2(1 + uvc^{-2})}{\sqrt{(c^2 - v^2)(c^2 - u^2)}} &= \frac{c}{\sqrt{c^2 - w^2}} \\ \\ \frac{c^2(u + v)}{\sqrt{(c^2 - v^2)(c^2 - u^2)}} &= \frac{cw}{\sqrt{c^2 - w^2}}. \end{align*}
$\quad$ First, we take \begin{align*} \frac{c^2(1 + uvc^{-2})}{\sqrt{(c^2 - v^2)(c^2 - u^2)}} &= \frac{c^2 + uv}{\sqrt{(c^2 - v^2)(c^2 - u^2)}} \\ &= \frac{1}{\sqrt{\frac{(c^2 - v^2)(c^2 - u^2)}{(c^2 + uv)^2}}} \\ &= \frac{1}{\sqrt{\frac{c^4 - c^2u^2 - c^2v^2 + u^2v^2}{(c^2 + uv)^2}}} \\ &= \frac{1}{\sqrt{1 - \frac{c^2(u + v)^2}{(c^2 + uv)^2}}}. \end{align*} Then, if we multiply the terms inside the square root by $c^2/c^2,$ noting that $w^2 = \frac{c^4(u + v)^2}{(c^2 + uv)^2},$ we get \begin{align*} \\ \frac{c^2(1 + uvc^{-2})}{\sqrt{(c^2 - v^2)(c^2 - u^2)}} &= \frac{1}{\sqrt{1 - \frac{c^2(u + v)^2}{(c^2 + uv)^2}}} \\ &= \frac{1}{\sqrt{\frac{c^2}{c^2} - \frac{c^4(u + v)^2}{c^2(c^2 + uv)^2}}} \\ &= \frac{1}{\frac{1}{c}\sqrt{c^2 - \frac{c^4(u + v)^2}{(c^2 + uv)^2}}} \\ &= \frac{c}{\sqrt{c^2 - w^2}}. \end{align*} Multiplying this result by $w = \frac{c^2(u + v)}{c^2 + uv}$ gives us \begin{align*} \\ \frac{c^2(u + v)}{\sqrt{(c^2 - v^2)(c^2 - u^2)}} &= \frac{c^2(1 + uvc^{-2})}{\sqrt{(c^2 - v^2)(c^2 - u^2)}} \frac{c^2(u + v)}{c^2 + uv} \\ &= \frac{cw}{\sqrt{c^2 - w^2}} \end{align*} Thus, we can rewrite the product $L(v)L(u)$ as: \begin{align*} L(v)L(u) &= \frac{c^2}{\sqrt{(c^2 - v^2)(c^2 - u^2)}} \begin{bmatrix} 1 + uvc^{-2} & -u - v \\ -vc^{-2} - uc^{-2} & uvc^{-2} + 1 \end{bmatrix} \\ \\ &= \frac{c}{\sqrt{c^2 - w^2}} \begin{bmatrix} 1 & -w \\ -wc^{-2} & 1 \end{bmatrix} \\ \\ &= L(w). \quad \blacksquare \end{align*}