Snip3r said:
When 2 objects A,B are moving wrt to each other (lets say @0.86c)then from frame A if local time is 10 years then time at B is 5 years. What does this mean?
You do understand that when you say 2 objects are moving with respect to each other at 0.86c, you don't mean they are both moving at the same time at that rate, do you? Don't you in your mind think about one frame at a time--from A's frame, B is moving, and from B's frame, A is moving? You don't think that both are moving at the same 0.86c which would add up to a total speed of 1.72c, do you? So when you are thinking about A's frame, B is the one that is experiencing speed which means B is the one that is experiencing time dilation. Conversely, when you are thinking about B's frame, A is the one that is experiencing speed which means A is the one that is experiencing time dilation.
Time dilation simply means that moving clocks are ticking at a slower rate than stationary clocks according to whichever frame you are considering using the Einstein's formula:
τ = t√(1-β
2) = t/γ
where τ is the time on a moving clock and t is the time on a stationary clock.
So, yes, in frame A, when 10 years ticks by, 5 years will tick by for B since he is moving at 0.86c in A's frame.
Snip3r said:
is it any event happening now in A(10 years) is simultaneous with the events happened in B when its clock ticked 5 years?
When you're talking about comparing the time coordinate of an event in two frames of reference, you need to use the Lorentz Transform which is:
t' = γ(t-βx)
So clearly, these are not the same formulas because the LT involves the location of the event as well as the time. But if you are careful to always calculate the location of B as being x = β*t, then:
t' = γ(t-β
2t) = γt(1-β
2)
And since γ = 1/√(1-β
2) then
t' = t(1-β
2)/√(1-β
2) = t(√(1-β
2)) = t/γ