User:AbiLtoCen/Sandbox

$$ \begin{bmatrix} c\,t' \\ x' \\ y' \\ z' \end{bmatrix} = \begin{bmatrix} \gamma&-\gamma\,\beta_x&-\gamma\,\beta_y&-\gamma\,\beta_z\\ -\gamma\,\beta_x&1+(\gamma-1)\dfrac{\beta_x^2}{\beta^2}&(\gamma-1)\dfrac{\beta_x \beta_y}{\beta^2}&(\gamma-1)\dfrac{\beta_x \beta_z}{\beta^2}\\ -\gamma\,\beta_y&(\gamma-1)\dfrac{\beta_y \beta_x}{\beta^2}&1+(\gamma-1)\dfrac{\beta_y^2}{\beta^2}&(\gamma-1)\dfrac{\beta_y \beta_z}{\beta^2}\\ -\gamma\,\beta_z&(\gamma-1)\dfrac{\beta_z \beta_x}{\beta^2}&(\gamma-1)\dfrac{\beta_z \beta_y}{\beta^2}&1+(\gamma-1)\dfrac{\beta_z^2}{\beta^2}\\ \end{bmatrix} \begin{bmatrix} c\,t \\ x \\ y \\ z \end{bmatrix}\, $$

Hence:

$$ \begin{bmatrix} c\,t_1' \\ x_1' \\ y_1' \\ z_1' \end{bmatrix} = \begin{bmatrix} \gamma&-\gamma\,\beta_x&-\gamma\,\beta_y&-\gamma\,\beta_z\\ -\gamma\,\beta_x&1+(\gamma-1)\dfrac{\beta_x^2}{\beta^2}&(\gamma-1)\dfrac{\beta_x \beta_y}{\beta^2}&(\gamma-1)\dfrac{\beta_x \beta_z}{\beta^2}\\ -\gamma\,\beta_y&(\gamma-1)\dfrac{\beta_y \beta_x}{\beta^2}&1+(\gamma-1)\dfrac{\beta_y^2}{\beta^2}&(\gamma-1)\dfrac{\beta_y \beta_z}{\beta^2}\\ -\gamma\,\beta_z&(\gamma-1)\dfrac{\beta_z \beta_x}{\beta^2}&(\gamma-1)\dfrac{\beta_z \beta_y}{\beta^2}&1+(\gamma-1)\dfrac{\beta_z^2}{\beta^2}\\ \end{bmatrix} \begin{bmatrix} c\,t_1 \\ x_1 \\ 0 \\ 0 \end{bmatrix}\,=$$ $$ \begin{bmatrix} \gamma\cdot c\,t_1-x_1\cdot\gamma\cdot\beta_x \\-\gamma\cdot\beta_x\cdot c\,t_1+x_1\cdot(1+(\gamma-1)\cdot\dfrac{\beta_x^2}{\beta^2}) \\ -\gamma\cdot\beta_y\cdot c\,t_1+x_1\cdot(\gamma-1)\cdot\dfrac{\beta_x\cdot\beta_y}{\beta^2}  \\ -\gamma\cdot\beta_z\cdot c\,t_1+x_1\cdot(\gamma-1)\cdot\dfrac{\beta_x\cdot\beta_z}{\beta^2} \end{bmatrix} $$

As in this case $$x_1=-c\,t_1$$ one gets:

$$c\,t_1'=\gamma\cdot c\,t_1+c\,t_1\cdot\gamma\cdot\beta_x = \gamma\, c\,t_1(1+\beta_x)$$

$$x_1'=-\gamma\cdot\beta_x\cdot c\,t_1-c\,t_1\cdot(1+(\gamma-1)\cdot\dfrac{\beta_x^2}{\beta^2}) =-c\,t_1\left(1+\beta_x\gamma\left(1+\dfrac{\gamma-1}{\gamma \beta^2}\cdot\beta_x\right)\right) =-c\,t_1\left(1+\beta_x\gamma\left(1+\dfrac{1-\sqrt{1-\beta^2}}{\beta^2}\cdot\beta_x\right)\right)$$

$$y_1'=-\gamma\cdot\beta_y\cdot c\,t_1-c\,t_1\cdot(\gamma-1)\cdot\dfrac{\beta_x\cdot\beta_y}{\beta^2} =-c\,t_1\beta_y\gamma\left(1+\dfrac{\gamma-1}{\gamma\beta^2}\cdot\beta_x\right) =-c\,t_1\beta_y\gamma\left(1+\dfrac{1-\sqrt{1-\beta^2}}{\beta^2}\cdot\beta_x\right)$$

$$z_1'=-\gamma\cdot\beta_z\cdot c\,t_1-c\,t_1\cdot(\gamma-1)\cdot\dfrac{\beta_x\cdot\beta_z}{\beta^2} =-c\,t_1\beta_z\gamma\left(1+\dfrac{\gamma-1}{\gamma\beta^2}\cdot\beta_x\right) =-c\,t_1\beta_z\gamma\left(1+\dfrac{1-\sqrt{1-\beta^2}}{\beta^2}\cdot\beta_x\right)$$

whereby $$0.5+0.125 \beta^2 \le\ \dfrac{1-\sqrt{1-\beta^2}}{\beta^2} \le 0.5 + r \beta^2$$ with r = 0.5 for  |&beta;| ≤ 1 and r = 0.144 for |&beta;| ≤ 0.5 and r = 0.126 for |&beta;| ≤ 0.125