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Two events occur at the same time in an inertial frame $\mathcal{S}$ and are separated by a distance of \SI{1}{km} along the $x$ axis. What is the time difference between these two events as measured in a frame $\mathcal{S'}$ moving with constant velocity along $x$ and in which their spatial separation is measured as \SI{2}{km}?
Lorentz transformation between the two systems gives: \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} Because \emph{same time} event means $\Delta t=0$, we get from the first equation: \begin{align} \gamma &= \frac{\Delta x'}{\Delta x}\\ &= 2\\ \end{align} And with $\gamma$ known, we can easily calculate the relative velocity of the two systems to each other: \begin{align} v &= \sqrt{1\frac{1}{2^2}} \cdot c\\ &= 0.866c\\ &= \SI{2.596e8}{\meter\per\second} \end{align} The time difference between the two events in $\mathcal{S'}$ follows then directly from the second equation: \begin{align} \Delta t' &= \gamma \frac{v\cdot\Delta x}{c^2}\\ &= \SI{5.777e6}{s} \end{align} That means, in the second frame of reference, the second event took place first, \SI{5.777}{\micro s} before it.
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