47102 愛因斯坦對勞侖茲變換的簡單推導

### 一、引言

\begin{aligned} x'=\,&\frac{x-vt}{\sqrt{1-\frac{v^2}{c^2}}},\\ t'=\,&\frac{t-\frac v{c^2} x}{\sqrt{1-\frac{v^2}{c^2}}},\\ & y'=y,\\ & z'=z. \end{aligned}\label{1}

### 二、推導過程的第一步

\begin{aligned} &y'=y,\\ &z'=z,\\ \left(\begin{array}{c} x'\\ t' \end{array}\right)=&\left(\begin{array}{ccc} a&~~&b\\ c&&d \end{array}\right)\left(\begin{array}{c} x\\ t \end{array}\right). \end{aligned}\label{2}

\begin{aligned} x'-ct'=\,&\lambda (x-ct),\\ x'+ct'=\,&\mu (x+ct). \end{aligned}\label{3}

\begin{align*} -ct'=\,&\lambda (vt-ct),\\ ct'=\,&\mu (vt+ct). \end{align*}

### 三、推導過程的第二步---勞侖茲收縮

\begin{aligned} \Delta x'-c\Delta t'=\,&\lambda \Delta x,\\ \Delta x'+c\Delta t'=\,&\mu \Delta x. \end{aligned}\label{5}

\begin{aligned} 2\Delta x'=\,&(\lambda +\mu )\Delta x,\\ &\hskip -200pt {\hbox{或}}\\ \frac{\Delta x}{\Delta x'}=\,&\frac 2{\lambda +\mu}, \end{aligned} \label{6}

\begin{align*} &\frac 1\lambda (x'-ct' )=x-ct,\\ &\frac 1\mu (x'+ct' )=x+ct. \end{align*}

\begin{align*} &\frac 1\lambda \delta x'=\delta x-c\delta t,\\ &\frac 1\mu \delta x'=\delta x+c\delta t. \end{align*}

\begin{align} \frac{\delta x'}{\delta x}=\frac 2{\frac 1\lambda +\frac 1\mu}.\label{7} \end{align}

\eqref{6} \eqref{7} 兩式都是快照 $\Delta x,\delta x'$ 與靜止長度之比, 因此, 基於對稱的考量

\begin{align*} \frac{\Delta x}{\Delta x'}=\,&\frac{\delta x'}{\delta x},\\ {\hbox{或 }} \lambda +\mu =\,&\frac 1\lambda +\frac 1\mu =\frac{\lambda +\mu} {\lambda\mu}, \end{align*}

\begin{align} \lambda \mu =1.\label{8} \end{align}

\eqref{8} 與 \eqref{4} 合併

$$\frac \lambda \mu =\frac{c+v}{c-v},\qquad \lambda \mu =1,$$

\begin{align} \lambda =\sqrt{\frac{c+v}{c-v}},\qquad \mu =\sqrt{\frac{c-v}{c+v}}.\label{9} \end{align}

\begin{align*} x'-ct'=\,&\lambda (x-ct),\\ x'+ct'=\,&\mu (x+ct). \end{align*}

\begin{align*} x'=\,&\frac{x-vt}{\sqrt{1-\dfrac{v^2}{c^2}}},\\ t'=\,&\frac{t-\dfrac v{c^2} x}{\sqrt{1-\dfrac {v^2}{c^2}}}. \end{align*}

$$\lambda +\mu =\sqrt{\frac{c+v}{c-v}}+\sqrt{\frac{c-v}{c+v}}=\frac{2c}{\sqrt{c^2-v^2}}=\frac 2{\sqrt{1-\frac{v^2}{c^2}}}.$$