Question

The Lagrangian \(L_0 = \frac{1}{2}m\dot{q}^2 - \frac{1}{2}m\omega^2q^2\) 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 motion does not change.
Which of the following options is correct for the above statements?

• A. Both (i) and (ii) are correct.
• B. Both (i) and (ii) are not correct.
• C. (i) is correct and (ii) is not correct.
• D. (i) is not correct and (ii) is correct.

Hint:

Adding \(\alpha \frac{df(q)}{dt} = \alpha f'(q)\dot{q}\) to a Lagrangian only adds a boundary term to the action, so the Euler-Lagrange equation is unchanged; but since \(p=\partial L/\partial \dot q\) is evaluated pointwise, the momentum does shift by \(\alpha f'(q)\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The new Lagrangian is built by adding \(\alpha \frac{df(q)}{dt}\) to \(L_0\). Since \(f\) depends only on \(q\), the chain rule gives \(\frac{df(q)}{dt} = f'(q)\dot{q}\), so \(L = L_0 + \alpha f'(q)\dot{q}\). This is a term linear in \(\dot{q}\) whose coefficient depends on \(q\), which is exactly the form of a total time derivative added to a Lagrangian. We need to check what such a term does to the canonical momentum \(p = \frac{\partial L}{\partial \dot{q}}\) and to the Euler-Lagrange equation of motion.

Step 2: Key Formula or Approach:
1. Canonical momentum: \(p = \frac{\partial L}{\partial \dot{q}}\).
2. Euler-Lagrange equation: \(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right) - \frac{\partial L}{\partial q} = 0\).
We work out both quantities for the given \(L\) and compare them with the ones from \(L_0\) alone, which give simple harmonic motion.

Step 3: Detailed Explanation:
Canonical momentum from the new \(L\):
\[ p = \frac{\partial L}{\partial \dot{q}} = m\dot{q} + \alpha f'(q) \]
The momentum from \(L_0\) alone is \(m\dot{q}\), so the extra term \(\alpha f'(q)\) makes the new momentum different from the old one whenever \(f'(q)\) is not zero. So statement (i) is not correct.
Now the equation of motion. First,
\[ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right) = \frac{d}{dt}\left(m\dot{q} + \alpha f'(q)\right) = m\ddot{q} + \alpha f''(q)\dot{q} \]
since \(f'(q)\) depends on time only through \(q(t)\). Next,
\[ \frac{\partial L}{\partial q} = -m\omega^2 q + \alpha f''(q)\dot{q} \]
because differentiating the extra term \(\alpha f'(q)\dot{q}\) with respect to \(q\) (treating \(\dot{q}\) as independent) gives \(\alpha f''(q)\dot{q}\). Putting these into the Euler-Lagrange equation:
\[ m\ddot{q} + \alpha f''(q)\dot{q} - \left(-m\omega^2 q + \alpha f''(q)\dot{q}\right) = 0 \]
\[ m\ddot{q} + \alpha f''(q)\dot{q} + m\omega^2 q - \alpha f''(q)\dot{q} = 0 \]
The \(\alpha f''(q)\dot{q}\) term cancels exactly, leaving
\[ m\ddot{q} + m\omega^2 q = 0 \]
which is the same simple harmonic oscillator equation you get from \(L_0\) by itself. So statement (ii) is correct.

Step 4: Final Answer:
Adding \(\alpha \frac{df(q)}{dt}\) changes the canonical momentum but leaves the equation of motion unchanged, so (i) is not correct and (ii) is correct.\[ \boxed{\text{(D)}} \]