ランダウ力学 §40問題3 解説

\begin{align} L &= \frac{m}{2} \sum_a \bm{v}_a^2 - \frac{m^2}{2 \left( M + nm \right)} \left( \sum_a \bm{v}_a \right)^2 - U \end{align}

\begin{align} E &= \frac{m}{2} \sum_a \bm{v}_a^2 - \frac{m^2}{2 \left( M + nm \right)} \left( \sum_a \bm{v}_a \right)^2 + U \label{eq_40-3e1} \end{align}

\begin{align} \bm{p}_a &= \frac{\partial L}{\partial \bm{v}_a} \nonumber \\ &= m \bm{v}_a - \frac{m^2}{M + nm} \sum_b \bm{v}_b \end{align}

\begin{align} \sum_a \bm{p}_a &= m \sum_a \bm{v}_a - \frac{nm^2}{M + nm} \sum_b \bm{v}_b \nonumber \\ &= \left( \frac{m \left( M + nm \right)}{M + nm} - \frac{nm^2}{M + nm} \right) \sum_a \bm{v}_a \nonumber \\ &= \frac{mM}{M + nm} \sum_a \bm{v}_a \end{align}

\begin{align} \bm{p}_a &= m \bm{v}_a - \frac{m^2}{M + nm} \frac{M + nm}{mM} \sum_b \bm{p}_b \nonumber \\ &= m \bm{v}_a - \frac{m}{M} \sum_b \bm{p}_b \end{align}

\begin{align} \bm{v}_a &= \frac{\bm{p}_a}{m} + \frac{1}{M} \sum_b \bm{p}_b \label{eq_40-3e2} \end{align}

\begin{align} H &= \frac{m}{2} \sum_a \left( \frac{\bm{p}_a}{m} + \frac{1}{M} \sum_b \bm{p}_b \right)^2 \nonumber \\ &\quad - \frac{m^2}{2 \left( M + nm \right)} \frac{\left( M + nm \right)^2}{m^2M^2} \left( \sum_a \bm{p}_a \right)^2 + U \nonumber \\ &= \frac{1}{2m} \sum_a \bm{p}_a^2 + \frac{1}{M} \left( \sum_a \bm{p}_a \right)^2 \nonumber \\ &\quad + \frac{nm}{2M^2} \left( \sum_a \bm{p}_a \right)^2 - \frac{M + nm}{2M^2} \left( \sum_a \bm{p}_a \right)^2 + U \nonumber \\ &= \frac{1}{2m} \sum_a \bm{p}_a^2 + \frac{1}{2M} \left( \sum_a \bm{p}_a \right)^2 + U \end{align}