Consider ∆t = γ(∆t’ + v∆x’/c2 ). If two events occur at different places in reference frame S’, then ∆x’ in this equation is not zero. It follows that even if the events are simultaneous in S’, they will not be simultaneous in frame S. The time interval between events in S will be
∆t= γ v∆x’/c2
Suppose now that two events occur at the same place in S’ (so ∆x’ = 0) but at different times (∆t’ ≠ 0). Equation ∆t = γ(∆t’ + v∆x’/c2 ) then reduces to ∆t = γ∆t’.
This confirms time dilation. Because the two events occur at the same place S’, the time interval ∆t’ between them can be measured with a single clock, located at that place. Under these conditions, the measured interval is a proper time interval, and we can label it ∆tp.
Hence we get ∆t = γ∆tp. (Time dilation)
Consider ∆x’ = γ(∆x – v∆t). If a rod lies parallel to the x and x’ axes and is at rest in reference frame S’, an observer in S’ can measure its length at leisure. One way to do so is by subtracting the coordinates of the end points of the rod. The value of ∆x’ that is obtained will be the proper length Lp of the rod.
Suppose the rod is moving in frame S. This means that ∆x can be identified as the length L of the rod in frame S only if the coordinates of the rod’s end points are measured simultaneously – that is, if ∆t = 0. If we put ∆x’ = Lp , ∆x = L, and ∆t = 0 in ∆x’ = γ(∆x – v∆t), we get
Lp= γL (Length contraction)