Use the relativistic coordinate transformation
(x, y, z, t) − (x′, y′, z′, t′) shown and given
below where the latter frame S′, (x′, y′, z′, t′),
has a velocity 2.69813 × 108 m/s in the
positive direction relative to the frame S,
(x, y, z, t).
For convenience, assume that the origins of the two reference frames overlap at time
t = t′ = 0.
what is the observed time interval T = t2 − t1 in the stationary reference frame given T′ = t′2 − t′1 in the moving reference frame?
The speed of light is 2.99792 × 108 m/s and the time measured in the primed frame is 52 s
there were some other details for this problem but the above is the question..what is the time interval...
so far i've got the speed of light, velocity, i found B= 0.900000667, Y=2.294164587, and t'=52 seconds
and now i'm stuck as to how to plug them in to find the answer. please help thanks
so i figured it out....its simply the y* t' = answer
To find the observed time interval T = t2 - t1 in the stationary reference frame, given T' = t'2 - t'1 in the moving reference frame, we can use the relativistic coordinate transformation equations.
The equation for time dilation in special relativity is given by:
T = Y * T'
where T is the observed time interval in the stationary frame, Y is the Lorentz factor, and T' is the time interval in the moving frame.
You mentioned that the Lorentz factor Y is already calculated as 2.294164587, and T' is given as 52 seconds. Now we can plug in these values into the equation to find T.
T = 2.294164587 * 52
T ≈ 119.4105453 seconds
Therefore, the observed time interval T in the stationary reference frame is approximately 119.4105453 seconds.