Discussion
The Lagrangian
\[ L_{0}=\frac{1}{2}m\dot{q}^{2}-\frac{1}{2}m\omega^{2}q^{2} \]
with the generalized coordinate \(q\) is transformed to \(L = L_{0} + \alpha\frac{df(q)}{dt}\). Consider the following statements:
(i) Expression for the canonical momentum does not change.
(ii) The equation of the motion does not change.
Which of the following options is correct for the above statements?
Answer: D
(i) is not correct and (ii) is correct.
Adding a total time derivative \(\alpha\,df(q)/dt = \alpha f'(q)\,\dot{q}\) changes the canonical momentum:
\[ \begin{aligned} p &= \frac{\partial L}{\partial \dot{q}} \\ &= m\dot{q} + \alpha f'(q) \end{aligned} \]
so statement (i) is false. But the extra term contributes nothing to the Euler-Lagrange equation:
\[ \tfrac{d}{dt}(\alpha f'(q)) - \alpha f''(q)\dot{q} = 0 \]
The equation of motion stays \(m\ddot{q} = -m\omega^{2}q\), so (ii) is true. Lagrangians differing by a total time derivative give the same action up to end-point terms and hence the same dynamics, while the canonical momentum is shifted by a gauge-like term.
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