Question

A light flashes at position x=0m. One microsecond later, a light flashes at position x=1000m. In a second reference frame, moving along the x-axis at speed v, the two flashes are simultaneous. Is this second frame moving to the right or to the left relative to the original frame?

Answer 1

To the right relative to the original frame.

Step-by-step explanation:

In first reference frame S,

Spatial interval of the event,
`\rm \Delta x=1000\ m-0\ m=1000\ m.`

Temporal interval of the event,
`\rm \Delta t = 1\ \mu s=10^(-6)\ s.`

In the second reference frame S', the two flashes are simultaneous, which means that the temporal interval of the event in this frame is
`\rm \Delta t'=0\ s.`

The speed of the frame S' with respect to frame S = v.

According to the Lorentz transformation,

`\rm \Delta t'=\frac{1}{\sqrt{1-(v^2)/(c^2)}}\left( \Delta t-(v\Delta x)/(c^2)\right ).\\\\Since,\ \Delta t'=0,\\\therefore \frac{1}{\sqrt{1-(v^2)/(c^2)}}\left( \Delta t-(v\Delta x)/(c^2)\right )=0\\\Rightarrow \Delta t-(v\Delta x)/(c^2)=0\\(v\Delta x)/(c^2)=\Delta t\\v=(c^2\Delta t)/(\Delta x ).\\\\Also, \ \Delta t,\ \Delta x>0\ \Rightarrow v>0.`

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.