速度の変換則 \[ \vec{v}_0=(v_{0x}\ ,\ \ v_{0y})\ \rightarrow\ \vec{v} =(v_x\ ,\ \ v_y)=\left(\frac{v_{0x}+V}{1+\frac{Vv_{0x}}{c^2}}\ ,\ \ \frac{v_{0y}\sqrt{1-\frac{V^2}{c^2}}}{1+\frac{Vv_{0x}}{c^2}}\right) \] から \[ 1-\frac{v^2}{c^2}=\frac{\left(1-\frac{v_0^2}{c^2}\right)\left(1-\frac{V^2}{c^2}\right)}{(1+\frac{Vv_{0x}}{c^2})^2} \] よって運動量の変換は \[ \frac{m\vec{v}_0}{\sqrt{1-\frac{v_0^2}{c^2}}}\ \rightarrow\ \frac{m\vec{v}}{\sqrt{1-\frac{v^2}{c^2}}} =\left(\frac{m(v_{0x}+V)}{\sqrt{1-\frac{v_0^2}{c^2}}\sqrt{1-\frac{V^2}{c^2}}}\ , \ \ \frac{mv_{0y}}{\sqrt{1-\frac{v_0^2}{c^2}}}\right) \] 従って \[ \vec{p}_0\ \rightarrow\ \vec{p} =\left(\frac{p_{0x}}{\sqrt{1-\frac{V^2}{c^2}}}+ \frac{mV}{\sqrt{1-\frac{v_0^2}{c^2}}\sqrt{1-\frac{V^2}{c^2}}}\ ,\ \ p_{0y}\right) \]