双生子佯谬3
坐标系O’相对于O以速度v沿x轴正方向匀速运动,当坐标重合时: \( (x=0,x’=0,t=0,t’=0) \), 当O’坐标系原点走到O坐标系的\(x_0\)位置,然后停止, 此时O坐标系看来: \( x=x_0,t=x_0/v,x’=0\), \(t’=\gamma(t-xv/c^2)=\frac{x_0}{v\gamma} \) 即: \( (x_1,x’_1,t_1,t_1’)=(x_0,0,\frac{x_0}{v},\frac{1}{\gamma}\frac{x_0}{v}) \) 要折返,初始坐标为: \( x=x_0, x’=0, t=0, t’=0 \) 回来时(相对速度变成-v): \( \Delta x’=\gamma(\Delta x+x_0+v\Delta t) \) \( \Delta t’=\gamma(\Delta t+(\Delta x+x_0) v/c^2) \) \( \Delta x+x_0=\gamma(\Delta x’-v\Delta t’) \) 则会得到: 当认为\((x=0)\)时, \( \Delta x=-x_0, \Delta t=x_0/v \), \( \Delta x’ =\gamma(v\Delta t)=\gamma x_0 \) \( \Delta t’ =\gamma t =\gamma x_0/v \) 即:\( (x_2,x’_2,t_2,t_2’)=(0,\gamma x_0, \frac{x_0}{v},\frac{\gamma x_0}{v}) \) ...