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: Recall the general rule about adding a total time derivative.
There is a standard result in Lagrangian mechanics: if you add \(\frac{d}{dt}F(q,t)\) to any Lagrangian, the equations of motion never change, because the extra piece contributes only a boundary term to the action \(\int L\,dt\), and the variation of a boundary term vanishes when the endpoints of \(q(t)\) are held fixed. Here \(F(q) = \alpha f(q)\), so this rule applies directly and predicts that the dynamics cannot change. What the rule does not protect is the canonical momentum, which is defined pointwise as $p = \partial L/\partial \dot q$ and is free to shift by whatever extra $\dot q$ dependence the added term carries.

Step 2: Check statement (i), the momentum.
Write out the added piece explicitly: $\alpha \frac{df(q)}{dt} = \alpha f'(q)\dot q$, a term proportional to $\dot q$. Differentiating the full $L$ with respect to $\dot q$ picks up this whole term as a plain additive shift:
\[ p_{\text{new}} = m\dot q + \alpha f'(q), \qquad p_{\text{old}} = m\dot q \]
The two momenta differ by $\alpha f'(q)$, a nonzero quantity for a generic $f$. Since the question does not restrict $f$ to have $f'=0$, the momentum genuinely changes, so (i) is FALSE.

Step 3: Check statement (ii), the equation of motion, using the boundary-term argument directly on the action.
The action changes as $S o S + \alpha \int \frac{df(q)}{dt}\,dt = S + \alpha\big[f(q(t_2)) - f(q(t_1))\big]$. Varying $q(t)$ with the endpoints $q(t_1)$ and $q(t_2)$ held fixed (as is always assumed in deriving the Euler-Lagrange equation) makes this extra piece a constant under variation, so $\delta\left(\alpha[f(q(t_2)) - f(q(t_1))]\right) = 0$. The stationary-action condition on the new $L$ therefore reduces to exactly the stationary-action condition on $L_0$, giving the same Euler-Lagrange equation $m\ddot q + m\omega^2 q = 0$. So (ii) is TRUE.

Step 4: Cross-check with the direct algebra.
As a consistency check, note that the extra momentum term $\alpha f'(q)$ contributes $\frac{d}{dt}[\alpha f'(q)] = \alpha f''(q)\dot q$ to $\frac{d}{dt}(\partial L/\partial \dot q)$, and the same $\alpha f''(q)\dot q$ appears with the opposite sign in $\partial L/\partial q$ (from differentiating $\alpha f'(q)\dot q$ with respect to $q$). These identical terms cancel in the Euler-Lagrange equation, matching the boundary-term argument above.

Final Answer:
The momentum picks up an extra term and is not preserved, while the dynamics stay exactly the same as for $L_0$: (i) is not correct, (ii) is correct.\[ \boxed{\text{(D)}} \]