Discussion
For a particle the kinetic energy and the potential energy is given by
\[ \begin{aligned} T &= \frac{1}{2}m\dot{r}^{2}, \\ U &= \frac{1}{r}\left(1 + \frac{\dot{r}^{2}}{c^{2}}\right), \end{aligned} \]
where \(c\) is a constant. Then the Hamiltonian H of the particle is :
Answer: D
The Hamiltonian is \(H = \dfrac{1}{2}\dfrac{p_{r}^{2}\, r c^{2}}{(mrc^{2} - 2)} + \dfrac{1}{r}\).
The Lagrangian is \(L = T - U = \frac{1}{2}m\dot{r}^{2} - \frac{1}{r} - \frac{\dot{r}^{2}}{rc^{2}}\). The momentum is
\[ \begin{aligned} p_{r} &= \frac{\partial L}{\partial \dot{r}} = \dot{r}\left(m - \frac{2}{rc^{2}}\right) \\ &= \dot{r}\, \frac{mrc^{2} - 2}{rc^{2}} \end{aligned} \]
Then
\[ \begin{aligned} H &= p_{r}\dot{r} - L \\ &= \frac{1}{2}\dot{r}^{2}\left(m - \frac{2}{rc^{2}}\right) + \frac{1}{r} \end{aligned} \]
and substituting \(\dot{r} = p_{r} rc^{2}/(mrc^{2} - 2)\):
\[ \begin{aligned} H &= \frac{1}{2}\, \frac{p_{r}^{2}\, rc^{2}}{mrc^{2} - 2} \\ &\quad + \frac{1}{r} \end{aligned} \]
\(H = T + U\) fails because the potential depends on velocity. Option B writes \(p_{r}\dot{r}^{2}\) instead of \(p_{r}\dot{r}\) and is not expressed in \(p_{r}\), and option C has the wrong sign on \(1/r\).
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