# Simultaneity in relativity

Let there are two inertial frames of references S and S’. S is the stationary frame of reference and S’ is the moving frame of reference. At time t=t’=0 that is in the start, they are at the same position that is Observers O and O’ coincides. After that S’ frame starts moving with a uniform velocity v along x axis.Let two events in frame S occur simultaneously at positions P1 and P2. The coordinates of the P1 will be (x1,y1,z1,t1) and of P2 will be (x2,y2,z2,t2). The events will be simultaneous (occur at the same time) according to the observer in frame S. Therefore

t1 = t2 (1)

The question arises here will the event be simultaneous from frame S’? Let us discuss it?

From Lorentz transformation equations of time:

t’1 = (t1 – x1v/c^{2})/(√1 – v^{2}/c^{2}) (2)

and t’2 = (t2 – x2v/c^{2})/(√1 – v^{2}/c^{2}) (3)

Subtracting equation (2) from equation (3), we get

t’2 – t’1 = (t2 – x2v/c^{2})/(√1 – v^{2}/c^{2}) – (t1 – x1v/c^{2})/(√1 – v^{2}/c^{2})

or t’2 – t’1 = (t2 – t1)/ )/(√1 – v^{2}/c^{2}) – v/c^{2}(x2 – x1)/(√1 – v^{2}/c^{2})

Put equation (1) in above equation, we get

t’2 – t’1 = – v/c^{2}(x2 – x1)/(√1 – v^{2}/c^{2}) (4)

As the two events occur at different positions, that is

x2 ≠x1

Therefore L.H.S of equation (4) will not be zero, thus

t’2 – t’1 ≠ 0

or t’1 ≠ t’2

It proves that the same events will not be simultaneous from frame S’, that is it will not appear to occur at the same time from S’.