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Two events occur at the same place in a certain inertial frame and are separated by a time interval of \SI{4}{s}. What is the spatial separation between these two events in an inertial frame in which the events are separated by a time interval of \SI{6}{s}?
Lorentz transformation gives for time and space: \begin{align} \Delta x' &= \gamma (\Delta x  v\cdot \Delta t)\\ \Delta t' &= \gamma \left(\Delta t  \frac{v \cdot \Delta x}{c^2} \right) \end{align} The second equation tells us, that \begin{align} \gamma &= \frac{\Delta t'}{\Delta t}\\ &= 1.5 \end{align} because \emph{same place} event means $\Delta x =0$. With $\gamma$ known, it the mutual velocity is straightforward: \begin{align} v &= \sqrt{1\frac{1}{1.5^2}} \cdot c\\ &= 0.745c \end{align} The spatial difference between the two events is in the second frame of reference is hence: \begin{align} \Delta x' &= \gamma (\Delta x  v\cdot \Delta t)\\ &= \gamma v \Delta t\\ &= \SI{6.7e8}{m} \end{align}
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