The GATE Grind

GATE 2023 PH – Question 40

Classical Mechanics · canonical transformations: Poisson bracket. Special theory of relativity: Lorentz transformations, relativistic kinematics, mass-energy equivalence. · 2 marks · Multiple choice

A rod PQ of proper length L lies along the x-axis and moves towards the positive x direction with speed v= 3c 5 with respect to the ground (see figure), where c is the speed of light in vacuum. An observer on the ground measures the positions of P and Q at different times tP and tQ respectively in the ground frame, and finds the difference betwen them to be 9L 10 . What is the value of tQ−tP?

Moving rod with endpoints P and Q.
  1. $L/(3c)$
  2. $L/(5c)$
  3. $L/(6c)$
  4. $2L/(3c)$

Practise this question in The GATE Grind →

Show answer and explanation

Correct answer: (C) $L/(6c)$

Explanation

**Setting.** The rod has proper length $L$ and moves along $+x$ at speed $v=3c/5$, so $\gamma=\dfrac{1}{\sqrt{1-9/25}}=\dfrac54$.

**Length of the rod in the ground frame** (both ends measured at the same time): the contracted length is
$$L^\prime=\frac{L}{\gamma}=\frac45L .$$

The observer, however, records the positions of P and Q at **different times** $t_P$ and $t_Q$, and finds the difference $x_Q-x_P=\dfrac{9L}{10}$. Each end moves during the time between the two recordings, so
$$x_Q(t_Q)-x_P(t_P)=\underbrace{\tfrac45L}_{\text{simultaneous separation}}+v\,(t_Q-t_P).$$

**Solve:**
$$\frac{9L}{10}=\frac{4L}{5}+v\,\Delta t\;\Rightarrow\;v\,\Delta t=\frac{L}{10}\;\Rightarrow\;\Delta t=\frac{L}{10}\cdot\frac{5}{3c}=\frac{L}{6c}\quad(\text{option C}).$$