Question

The inertial frame 𝑆' is moving away from the inertial frame S with a speed 𝑣 = 0.6𝑐 along the negative x-direction (see figure). The origins 𝑂’ and 𝑂 of the frames coincide at 𝑡=𝑡’=0. As observed in the frame 𝑆′, two events occur simultaneously at two points on the 𝑥′-axis with a separation of ∆𝑥′=5m. If, ∆𝑡 and ∆𝑥 are the magnitudes of the time interval and the space interval, respectively, between the events in S, then which of the following statements is(are) correct?

(c = 3×10⁸ m/s)

• A. ∆𝑡=12.5ns
• B. ∆𝑡=4.2ns
• C. ∆𝑥=10.6m
• D. ∆𝑥=6.25m

Answer 1

The Correct Option is A, D

1. Lorentz Transformation

The Lorentz transformation equations for space and time intervals are:

Δt = γ (Δt′ + vΔx′ / c²) 

Δx = γ (Δx′ + vΔt′)

where:

γ = 1 / √(1 − v² / c²)

2. Given Data

• Δx′ = 5 m
• v = 0.6c
• Δt′ = 0 (since the events are simultaneous in S′)

3. Calculate γ

γ = 1 / √(1 − v² / c²)

= 1 / √(1 − (0.6)²)

= 1 / √(1 − 0.36) = 1 / √0.64 = 1.25

4. Time Interval (Δt)

Using the Lorentz transformation for Δt:

Δt = γ · (vΔx′ / c²)

Substituting the values:

Δt = 1.25 × (0.6c × 5) / c²

= 1.25 × (3c / c²)

= 1.25 × (3 × 10⁸ × 5) / (9 × 10¹⁶)

Simplify:

Δt = (1.25 × 15) / (9 × 10⁸)

= (1.25 × 5) / (3 × 10⁸)

= 6.25 / (3 × 10⁸)

Convert to Nanoseconds:

Δt = (6.25 / 3) ns = 12.5 ns

5. Space Interval (Δx)

Using the Lorentz transformation for Δx:

Δx = γΔx′ = 1.25 × 5 = 6.25 m

6. Analyze the Options

• (A) Δt = 12.5 ns is correct.
• (B) Δt = 4.2 ns is incorrect.
• (C) Δx = 10.6 m is incorrect.
• (D) Δx = 6.25 m is correct but not listed as the final answer.