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 𝑑=𝑑’=0. As observed in the frame 𝑆′, two events occur simultaneously at two points on the π‘₯β€²-axis with a separation of βˆ†π‘₯β€²=5m. If, βˆ†π‘‘ and βˆ†π‘₯ 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?
The inertial frame
(c = 3Γ—108 m/s)

Updated On: Feb 6, 2025
  • βˆ†π‘‘=12.5ns
  • βˆ†π‘‘=4.2ns
  • βˆ†π‘₯=10.6m
  • βˆ†π‘₯=6.25m
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The Correct Option is A, D

Solution and Explanation

1. Lorentz Transformation

The Lorentz transformation equations for space and time intervals are:

Ξ”t = Ξ³ (Ξ”tβ€² + vΞ”xβ€² / cΒ²) 

Ξ”x = Ξ³ (Ξ”xβ€² + vΞ”tβ€²)

where:

Ξ³ = 1 / √(1 βˆ’ vΒ² / cΒ²)

2. Given Data

  • Ξ”xβ€² = 5 m
  • v = 0.6c
  • Ξ”tβ€² = 0 (since the events are simultaneous in Sβ€²)

3. Calculate Ξ³

Ξ³ = 1 / √(1 βˆ’ vΒ² / cΒ²)

= 1 / √(1 βˆ’ (0.6)Β²)

= 1 / √(1 βˆ’ 0.36) = 1 / √0.64 = 1.25

4. Time Interval (Ξ”t)

Using the Lorentz transformation for Ξ”t:

Ξ”t = Ξ³ Β· (vΞ”xβ€² / cΒ²)

Substituting the values:

Ξ”t = 1.25 Γ— (0.6c Γ— 5) / cΒ²

= 1.25 Γ— (3c / cΒ²)

= 1.25 Γ— (3 Γ— 10⁸ Γ— 5) / (9 Γ— 10¹⁢)

Simplify:

Ξ”t = (1.25 Γ— 15) / (9 Γ— 10⁸)

= (1.25 Γ— 5) / (3 Γ— 10⁸)

= 6.25 / (3 Γ— 10⁸)

Convert to Nanoseconds:

Ξ”t = (6.25 / 3) ns = 12.5 ns

5. Space Interval (Ξ”x)

Using the Lorentz transformation for Ξ”x:

Ξ”x = Ξ³Ξ”xβ€² = 1.25 Γ— 5 = 6.25 m

6. Analyze the Options

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