ラグランジアンとガリレイ変換

ラグランジアンとガリレイ変換

$$\begin{eqnarray} L&=&\frac{m}{2}\dot{q}^2\\ \end{eqnarray}$$ $$\begin{eqnarray} q'&=&q + Vt\;\dots\;系がV(一定値)で移動している\\ \dot{q'}&=&\dot{q}+V\;\dots\;\frac{\mathrm{d}}{\mathrm{d}t}(q + Vt)=\frac{\mathrm{d}}{\mathrm{d}t}q + \frac{\mathrm{d}}{\mathrm{d}t}Vt\\ L'&=&\frac{m}{2}(\dot{q’})^2\\ &=&\frac{m}{2}(\dot{q}+V)^2\\ &=&\frac{m}{2}(\dot{q}^2+2\dot{q}V+V^2)\\ &=&\frac{m}{2}\dot{q}^2+\frac{m}{2}(2\dot{q}V+V^2)\\ &=&L+\frac{m}{2}\left(2\dot{q}V+V^2\right)\\ \frac{\partial L'}{\partial q}-\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L'}{\partial \dot{q}} &=&\left\{\frac{\partial L}{\partial q} + \frac{\partial}{\partial q}\left(2\dot{q}V+V^2\right)\right\} - \left[ \frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot{q}} + \frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial}{\partial \dot{q}} \left\{ \frac{m}{2} (2\dot{q}V+V^2) \right\} \right]\\ &=&\frac{\partial L}{\partial q} - \left( \frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot{q}} + \frac{\mathrm{d}}{\mathrm{d}t}mV \right)\;\dots\;\frac{\mathrm{d}}{\mathrm{d}q}\left(2\dot{q}V+V^2\right) = 0\;\dots\;qと\dot{q}は独立\\ &=&\frac{\partial L}{\partial q} - \frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot{q}} \;\dots\;\frac{\mathrm{d}}{\mathrm{d}t} mV = 0\;\dots\;ここではVは定数である\\ \end{eqnarray}$$