Answer:
To the right relative to the original frame.
Explanation:
In first reference frame S,
Spatial interval of the event, [tex]\rm \Delta x=1000\ m-0\ m=1000\ m.[/tex]
Temporal interval of the event, [tex]\rm \Delta t = 1\ \mu s=10^{-6}\ s.[/tex]
In the second reference frame S', the two flashes are simultaneous, which means that the temporal interval of the event in this frame is [tex]\rm \Delta t'=0\ s.[/tex]
The speed of the frame S' with respect to frame S = v.
According to the Lorentz transformation,
[tex]\rm \Delta t'=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\left( \Delta t-\dfrac{v\Delta x}{c^2}\right ).\\\\Since,\ \Delta t'=0,\\\therefore \dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\left( \Delta t-\dfrac{v\Delta x}{c^2}\right )=0\\\Rightarrow \Delta t-\dfrac{v\Delta x}{c^2}=0\\\dfrac{v\Delta x}{c^2}=\Delta t\\v=\dfrac{c^2\Delta t}{\Delta x }.\\\\Also, \ \Delta t,\ \Delta x>0\ \Rightarrow v>0.[/tex]
And positive v means the velocity of the second frame S' is along the positive x-axis direction, i.e., to the right direction relative to the original frame S.
