相对论速度变换

$$\begin{array}{r}\begin{array}{ll}{v}_x^{\prime }& =\frac{v_x-u}{1-\frac{u{v}_x}{c^2}}\\ v_y^{\prime }& =\frac{v_y}{\gamma (1-\frac{u{v}_x}{c^2})}\\ v_z^{\prime }& =\frac{v_z}{\gamma (1-\frac{u{v}_x}{c^2})}\end{array}\}\leftrightarrows \{\begin{array}{ll}{v}_x& =\frac{v_x^{\prime }+u}{1+\frac{u{v}_x^{\prime }}{c^2}}\\ v_y& =\frac{v_y^{\prime }}{\gamma (1+\frac{u{v}_x^{\prime }}{c^2})}\\ v_z& =\frac{v_z^{\prime }}{\gamma (1+\frac{u{v}_x^{\prime }}{c^2})}\end{array}\end{array}$$