Question:

A rod \(PQ\) of proper length \(L\) lies along the \(X\)-axis and moves towards the positive \(X\)-direction with speed \(v = \frac{3c}{5}\) 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 \(t_P\) and \(t_Q\) respectively in the ground frame, and finds the difference between them to be \(\frac{9L}{10}\). What is the value of \(t_Q - t_P\)? 

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The difference in position in a moving frame is affected by both length contraction and the relativity of simultaneity. Apply the appropriate formula for time difference.
Updated On: Aug 30, 2025
  • \(\frac{L}{3c}\)
  • \(\frac{L}{5c}\)
  • \(\frac{L}{6c}\)
  • \(\frac{2L}{3c}\)
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The Correct Option is C

Solution and Explanation

- The proper length of the rod is \(L\), and it moves with a velocity \(v = \frac{3c}{5}\). The length contraction formula gives the contracted length in the ground frame: \[ L' = L \sqrt{1 - \frac{v^2}{c^2}} = L \sqrt{1 - \frac{9}{25}} = L \times \frac{4}{5}. \] - The difference in positions between \(P\) and \(Q\) observed in the ground frame is \(L' = \frac{4L}{5}\). However, the observer measures \(\frac{9L}{10}\) as the difference. This discrepancy arises from the relativity of simultaneity, given by: \[ \Delta t = \frac{\Delta x}{v} = \frac{\frac{9L}{10} - \frac{4L}{5}}{v} = \frac{\frac{9L}{10} - \frac{8L}{10}}{\frac{3c}{5}} = \frac{L}{6c}. \] Hence, the time difference is \(\boxed{\frac{L}{6c}}\).
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