Question

A rocket (S') moves at a speed \(\frac{c}{2}\)m/s along the positive x-axis, where c is the speed of light. When it crosses the origin, the clocks attached to the rocket and the one with a stationary observer (S) located at x = 0 are both set to zero. If S observes an event at (x, t), the same event occurs in the S' frame at

• A. \(x'=\frac{2}{\sqrt3}(x-\frac{ct}{2})\ \text{and}\ t'=\frac{2}{\sqrt3}(t-\frac{x}{2c})\)
• B. \(x'=\frac{2}{\sqrt3}(x+\frac{ct}{2})\ \text{and}\ t'=\frac{2}{\sqrt3}(t-\frac{x}{2c})\)
• C. \(x'=\frac{2}{\sqrt3}(x-\frac{ct}{2})\ \text{and}\ t'=\frac{2}{\sqrt3}(t+\frac{x}{2c})\)
• D. \(x'=\frac{2}{\sqrt3}(x+\frac{ct}{2})\ \text{and}\ t'=\frac{2}{\sqrt3}(t+\frac{x}{2c})\)

Answer 1

The Correct Option is A

To solve this problem, we will utilize the Lorentz transformation equations which are used to relate the space and time coordinates of the same event in different inertial frames of reference moving relative to each other at a constant velocity.

The general form of the Lorentz transformation equations when moving with velocity \(v\) along the x-axis is:

1. \(x' = \frac{x - vt}{\sqrt{1 - \frac{v^2}{c^2}}}\)
2. \(t' = \frac{t - \frac{vx}{c^2}}{\sqrt{1 - \frac{v^2}{c^2}}}\)

In the problem, the rocket moves with a speed \(v = \frac{c}{2}\). We will substitute this value into the Lorentz transformation equations.

Substituting \(v = \frac{c}{2}\):

1. The Lorentz factor, \(\gamma\), is computed as follows:

\(\gamma = \frac{1}{\sqrt{1 - \left(\frac{1}{2}\right)^2}} = \frac{1}{\sqrt{1 - \frac{1}{4}}} = \frac{1}{\sqrt{\frac{3}{4}}} = \frac{2}{\sqrt{3}}\)

1. Now, substitute \(\gamma\) and \(v\) into the Lorentz transformation equations:

\(x' = \frac{2}{\sqrt{3}} \left(x - \frac{ct}{2}\right)\)

\(t' = \frac{2}{\sqrt{3}} \left(t - \frac{x}{2c}\right)\)

This correctly matches the first option provided in the question:

\(x' = \frac{2}{\sqrt{3}}(x-\frac{ct}{2}) \ \text{and} \ t' = \frac{2}{\sqrt{3}}(t-\frac{x}{2c})\)