運動量の座標変換

速度の変換則

$${\overrightarrow{v}}_0=(v_{0x} , v_{0y}) \to \overrightarrow{v}=(v_x , v_y)=\left(\frac{v_{0x}+V}{1+\frac{V{v}_{0x}}{c^2}} , \frac{v_{0y}\sqrt{1-\frac{V^2}{c^2}}}{1+\frac{V{v}_{0x}}{c^2}}\right)$$

から

$$1-\frac{v^2}{c^2}=\frac{\left(1-\frac{v_0^2}{c^2}\right)\left(1-\frac{V^2}{c^2}\right)}{{(1+\frac{V{v}_{0x}}{c^2})}^2}$$

よって運動量の変換は

$$\frac{m{\overrightarrow{v}}_0}{\sqrt{1-\frac{v_0^2}{c^2}}} \to \frac{m\overrightarrow{v}}{\sqrt{1-\frac{v^2}{c^2}}}=\left(\frac{m(v_{0x}+V)}{\sqrt{1-\frac{v_0^2}{c^2}}\sqrt{1-\frac{V^2}{c^2}}} , \frac{m{v}_{0y}}{\sqrt{1-\frac{v_0^2}{c^2}}}\right)$$

従って

$${\overrightarrow{p}}_0 \to \overrightarrow{p}=\left(\frac{p_{0x}}{\sqrt{1-\frac{V^2}{c^2}}}+\frac{mV}{\sqrt{1-\frac{v_0^2}{c^2}}\sqrt{1-\frac{V^2}{c^2}}} , p_{0y}\right)$$