Question

Three events, \( E_1(ct = 0, x = 0) \), \( E_2(ct = 0, x = L) \) and \( E_3(ct = 0, x = -L) \) occur, as observed in an inertial frame \( S \). Frame \( S' \) is moving with a speed \( v \) along the positive x-direction with respect to \( S \). In \( S' \), let \( t'_1, t'_2, t'_3 \) be the respective times at which \( E_1, E_2, \) and \( E_3 \) occurred. Then,

• A. \( t'_2<t'_1<t'_3 \)
• B. \( t'_1 = t'_2 = t'_3 \)
• C. \( t'_3<t'_1<t'_2 \)
• D. \( t'_3<t'_2<t'_1 \)

Hint:

In Lorentz transformations, time depends on position when \( t = 0 \) in the original frame. The event farther along the direction of motion occurs earlier in the moving frame.

Answer 1

The Correct Option is A

Step 1: Apply the Lorentz transformation for time.
\[ t' = \gamma \left( t - \frac{v x}{c^2} \right). \] Given \( t = 0 \) for all three events, this simplifies to: \[ t' = -\gamma \frac{v x}{c^2}. \] Step 2: Substitute for each event.
For \( E_1: x = 0 \Rightarrow t'_1 = 0. \)
For \( E_2: x = +L \Rightarrow t'_2 = -\gamma \frac{vL}{c^2}. \)
For \( E_3: x = -L \Rightarrow t'_3 = +\gamma \frac{vL}{c^2}. \)
Step 3: Compare the times.
\[ t'_2<t'_1<t'_3. \] Step 4: Final Answer.
Hence, the correct order is \( t'_3<t'_1<t'_2 \) after interpreting the negative direction.