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GATE 2026 PH – Question 46

Classical Mechanics · D'Alembert's principle, Euler-Lagrange equation, Hamilton's principle, calculus of variations · 2 marks · Multiple choice

The Lagrangian $$L_0=\frac12m\dot q^2-\frac12m\omega^2q^2$$ 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 the motion does not change. Which of the following options is correct for the above statements?

  1. Both (i) and (ii) are correct.
  2. Both (i) and (ii) are not correct.
  3. (i) is correct and (ii) is not correct.
  4. (i) is not correct and (ii) is correct.

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Correct answer: (D) (i) is not correct and (ii) is correct.

Explanation

Adding a **total time derivative** to a Lagrangian, $L=L_0+\alpha\dfrac{df(q)}{dt}$, does not change the equation of motion but does change the canonical momentum.

**Rewrite the extra term.** Since $f$ depends on $q$ only,
$$\frac{df}{dt}=f^\prime(q)\,\dot q .$$

**(i) Canonical momentum.**
$$p=\frac{\partial L}{\partial\dot q}=m\dot q+\alpha f^\prime(q).$$
This differs from the old $p=m\dot q$. So statement (i) is **not correct**.

**(ii) Equation of motion.** A total derivative contributes the same amount to both sides of the Euler-Lagrange equation:
$$\frac{d}{dt}\frac{\partial}{\partial\dot q}\left(\alpha f^\prime\dot q\right)=\alpha f^{\prime\prime}\dot q=\frac{\partial}{\partial q}\left(\alpha f^\prime\dot q\right),$$
so they cancel. The equation of motion is unchanged: statement (ii) is **correct**.

Answer: (i) is not correct and (ii) is correct (option D).