Question

Two events occur at locations separated by a distance of 49.5m and by a time interval of 0.528 ?s, according to observerO.ObserverO′ isinmotionawayfromOwitha speedof0.685cinthexdirection.AccordingtoO′,what are the spatial and time separations of the events?

Answer 1

Given : t = 0.528

u = 0.685 c

  x = 49.5 m

Time separation:

t^' = (t -ux/c²) (1-u²/c²)

= (0.528 - 0.685 x 49.5 / 300 ) ( (1- (0.685)² ) )  

= 0.5695  

Spatial separation:

x' = (x - ut) (1-u²/c²)

= (49.5 - (0.685 x 300 x 0.528) (1 - (0.685)² ) m

= -111.17 m

The negative sign of Δx′ indicates that O′ finds the two events in inverted locations compared with O; for example, if O finds that event 1 occurs at a smaller x coordinate than event 2, then O′ finds that event 1 occurs at a larger x′ coordinate than event 2. That is, O sees event 1 to the left of event 2, while O′ sees event 1 to the right of event 2. Note than both observers find the time interval to be positive – event 2 occurs after event 1 to both observers.