Question:

Two observers \( O \) and \( O' \) observe two events \( P \) and \( Q \). The observers have a constant relative speed of 0.5c. In the units, where the speed of light, \( c \), is taken as unity, the observer \( O \) obtained the following coordinates:
\text{Event P: } x = 5, y = 3, z = 5, t = 3
\text{Event Q: } x = 5, y = 1, z = 3, t = 5
\text{The length of the space-time interval between these two events, as measured by \( O' \), is \( L \). The value of \( |L| \) (in integer) is \(\underline{\hspace{2cm}}\).}

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The space-time interval is invariant in special relativity. Use the formula \( L^2 = - (c^2 \Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2) \) to calculate it.
Updated On: Dec 29, 2025
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Correct Answer: 2

Solution and Explanation

The space-time interval \( L \) is given by: \[ L^2 = - (c^2 \Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2), \] where \( \Delta t = t_Q - t_P = 5 - 3 = 2 \), \( \Delta x = x_Q - x_P = 5 - 5 = 0 \), \( \Delta y = y_Q - y_P = 1 - 3 = -2 \), and \( \Delta z = z_Q - z_P = 3 - 5 = -2 \). Substitute into the equation: \[ L^2 = - (2^2 - 0^2 - (-2)^2 - (-2)^2) = - (4 - 4 - 4) = 4. \] Thus, the space-time interval is: \[ |L| = \sqrt{4} = 2. \] The value of \( |L| \) is 2.
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