Discussion
A Lagrangian of a particle of mass \(m\) is given by
\[ \begin{aligned} L &= \frac{1}{2}m\dot{r}^{2} \\ &\quad - \frac{1}{r}\left(1 + \frac{\dot{r}^{2}}{c^{2}}\right), \end{aligned} \]
\(c\) is a constant. Then :
Answer: B
The correct statement is H is conserved.
The Lagrangian does not depend explicitly on time, so the Hamiltonian (the Jacobi integral) is a constant of the motion: \(dH/dt = -\partial L/\partial t = 0\).
H is not the total energy here, because the potential depends on the velocity. From the previous computation,
\[ \begin{aligned} H &= \frac{1}{2}m\dot{r}^{2} - \frac{\dot{r}^{2}}{rc^{2}} + \frac{1}{r} \\ E &= T + U \\ &= \frac{1}{2}m\dot{r}^{2} + \frac{\dot{r}^{2}}{rc^{2}} + \frac{1}{r} \end{aligned} \]
They differ by \(2\dot{r}^{2}/(rc^{2})\), which changes along the motion. So 'H represents the total energy' is false, E is not conserved, and 'both H and E are conserved' fails with it. \(H = T + U\) needs a potential independent of velocity and a time-independent constraint.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development