相对论速度变换

52字

$$\begin{aligned} \left. \begin{array}{ll} v_x’ &= \frac{v_x - u}{1 - \frac{uv_x}{c^2}} \newline v_y’ &= \frac{v_y}{\gamma(1 - \frac{uv_x}{c^2})} \newline v_z’ &= \frac{v_z}{\gamma(1 - \frac{uv_x}{c^2})} \end{array} \right \rbrace \quad \leftrightarrows \quad \left\lbrace \begin{array}{ll} v_x &= \frac{v_x’ + u}{1 + \frac{uv_x’}{c^2}} \newline v_y &= \frac{v_y’}{\gamma(1 + \frac{uv_x’}{c^2})} \newline v_z &= \frac{v_z’}{\gamma(1 + \frac{uv_x’}{c^2})} \end{array} \right. \end{aligned}$$

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