Derive expression of time dilation in relativity?

Answer 1

In order to derive this, we have to consider two systems S & S'.Let S' be moving with a velocity v w.r.t S in the positive direction of X-axis.

Clock too!! has to be included as the word dilation means to lengthen at time.

So,suppose a clock is situated in the system S at position x and gives signs of interval ∇t, i.e., ∇t = t2-t1 ___________________(i)

If this interval is observed by an observer in system S',then interval ∇t' recorded by an observer is given by ∇t' = t1'-t2' _____________(ii)

From Lorentz transformation, t1' = t1-(vx/c2)/(1-v2/c2)1/2_____________(iii)

and t2' = t2-(vx/c2)/(1-v2/c2)1/2 ______________(iv)

Substituting the values of t1' and t2' from equations (iii) & (iv) in (ii), we get

∇t' =t2-(vx/c2)/(1-v2/c2)1/2 - t1-(vx/c2)/(1-v2/c2)1/2

= t2-vx/c2 - t1+ vx/c2/(1-v2/c2)1/2 = t2-t1/(1-v2/c2)1/2 = ∇t/ (1-v2/c2)1/2

So, ∇t' = ∇t/ (1-v2/c2)1/2, where ∇t'>∇t.

Hey buddies,just imagine the graph yourself..!!

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IZ

Answer 2

Time dilation in special relativity is given by the equation ( t' = \gamma \cdot t ), where ( t' ) is the dilated time interval, ( t ) is the proper time interval, and ( \gamma ) is the Lorentz factor ( \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} ). This equation shows that as an object's velocity approaches the speed of light, time dilation becomes more significant, meaning that time passes more slowly for the moving object relative to a stationary observer.