Simultaneity

1. Definition

In galilean relativity, the time is considered absolute, 
that is the same for all the observers in different frames
. 
If two events occured at the same time, that is simultaneously, 
in one reference frame, they are also simultaneous in all 
reference frames. 

This absolute simultaneity is not valid. 

According to Lorentz transformastionx , we have:
 x'= γ(x - vt ),
 t' = γ (t- vx/c2),
 
 If, in a frame, two simultaneous events (x1,t1) and (x2,t2) 
 occur , we have t1 = t2, that is Δt = t2 - t1
 
In another frame O' moving at a speed v with respect to the 
 first frame O, we have:
 
Δt' = t'2 - t'1 = γ (t2- vx2/c2) - γ (t1- vx1/c2) =
γ [(t2 - t1) - v(x2 - x1)/c2)] =  - γ vΔ/c2)]

Since Δt' is not zero, these two events are not 
simultaneous in the new frame O'. Two observers in different 
frames do never agree on events being simultaneous.
 

2. Example Vaulter's pole through a barn

v = 0.888 c
β = 0.888

(1 - β2)1/2 = 
(1 - (0.888)2)1/2 = 0.46

γ = 1/(1 - β2)1/2 = 2.17

To make things simple let's set γ = 2

2.1. Points of view of observers

Barn's frame:

The barn's length is 10 m
The pole is moving at v = 0.888c toward me (the barn), 
its length is contracted by γ, that is 16m/γ = 
16/2 = 8 m.
Therefore the pole's length (8 m) is enough to get in 
the barn and then close the two doors of the barn.


Pole's frame:

The pole's length is 16 m
The barn is moving at v = - 0.888c toward me (the pole), 
its length is contracted by γ, that is 10m/γ 
= 10/2 = 5 m.
Therefore the barn's length (5 m) is insufficient 
to contain the pole (16 m). If the two doors close
just as the front tip of the pole reaches the back 
door, the pole will break off the last 
16 - 5 = 11 m of the pole:

2.2 Simultaneity

Let's set the following events:

For the barn:

We close the two doors at t = 0, therefore:
Event1 (closing the left door) = (x1 = 0 , t1 = 0)
Event2 (closing the right door) = (x2 = 10m , t2 = 0)
These two events are simultaneous in the frame 
of the barn.


For the pole:

Right now at t'1 = 0 and x'1 = 0, the right door is 
already closed. This right  door was closed at t'2 = - 0.058µs  
(already 0.058µs ), the right door closed  0.058 
before the left closes. This time is taken by the barn 
to move toward the pole, then the related displacement 
is 0.058µs  x 0.888c = 15m.

Event'2 (closing the right door) = (x'2 = 20, t'2 = - 0.058 µs )
Event'1 (closing the left door) = (x'1 = 0, t'1 = 0)

The left door closes at t'1 = 0 and the right dooe 
was already closed 0.058 µs before. These two 
events Event'1 and Event'2 are not simultaneous.




Another event:

Let's consider the event3 at x'2 = 16m:
The right door of the barn closed at x2 = 20 m and at 
the time t = - 0.058 µs. By moving left toward 
the pole, the right door reached the front tip of the 
pole at x'2 = 16 m after traveling 20 - 16 = 4 m, that 
corresponds to the time = - 0.058 µs + 4 m/0.888c µs = 
- 0.058 µs  + 0.015 µs = - 0.0423 µs.

We can write for the event 3 = event'2 at x' = 16m:
Event3 = (x'2, t'2) = (16 m, -0.043µs)

Next, verify the value of the time t'2 by using Lorentz's 
transformations:
 x'3 = γ(x2 - vt2) 
 x2 and t2 are the space time coordinates of the tip of the 
 pole in the barn frame. That is x2 = 10 - 8 = 2 and t2 = 2/0.888c 
 = 0.0077 µs 

x'3= γ(10 - 0.888c x 0.0075 µs ) = 16 m,
t'2 = γ (t2 - vx2/c2) = 2(0.007 - 0.888 x 10/c) 
= - 0.043
Therefore, the results agree.