Last time we have discussed and derive the length contraction in relativity, let us today discuss the concept of time dilation and derive the relation for time dilation.

Let us today discuss and derive an interesting concept in relativity called the time dilation. You must be astonished to know that when an object moves with a speed comparative to the speed of light, its time dilated or more appropriate term is that it is appeared to move slow.

Let there are two inertial frames of references S and S1. 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 a clock is placed in the frame S’. The time coordinate of the initial time of the clock will be t1 according to the observer in S and the time coordinate of the final tick (time ) will be will be t2 according to same observer.

The time coordinate of the initial time of the clock will be t’1 according to the observer in S’ and the time coordinate of the final tick (time ) will be will be t’2 according to same observer.Therefore the time of the object as seen by observer O’ in S’ at the position x’ will be

t’ = t’2 – t’1 (1)

The time t’ is called the proper time of the event. **Proper time** is defined as the time of an event measured by the observer which is in the same frame in which the event is occurred.

The apparent or dilated time of the same event from frame S at the same position x will be

t = t2 – t1 (2)

Now use Lorentz inverse transformation equations for time (that I discussed during the derivation of Lorentz transformation equations), that is

t1 = (t’1 + x’v/c^{2)}/(√1 – v^{2}/c^{2}) (3)

t2 = (t’2 + x’v/c^{2)}/(√1 – v^{2}/c^{2}) (4)

By putting equations (3) and (4) in equation (2) and solving, we get

t = (t’2 – t’1)/ (√1 – v^{2}/c^{2})

Substitute equation (1) in above equation,

t = t’/(√1 – v^{2}/c^{2})

This is the relation of the time dilation.

Why it known as time dilation, if we solve (√1 – v^{2}/c^{2}), then the value will always be in decimal (except when v = c).

If we divide t’ with a decimal value, then t will be more than t’. This is the reason that it is called **time dilation** as dilation means to get increase. Thus the clock will appear to be slower now.

**For example let us discuss a numerical:** If an object is moving with speed 0.8c and its keeping the time 1 second, then what will be its apparent time?

Solution:

Given v = 0.8c

Proper length t’ = 1 sec

To calculate t

As t = t’/(√1 – v^{2}/c^{2})

If we substitute the given values, we get

t = 1.67 sec

Thus t > t’

**Special Case:**

If v <<< c, then v^{2}/c^{2 } will be negligible in t’/(√1 – v^{2}/c^{2}) and it can be neglected

Then t = t’

You only get time dilation if you assume c to be constant; otherwise t is invariant. It cannot be independently proved that c is constant. If c is constant in all frames of reference, why do we have to measure it in a vacuum? Shouldn’t it be constant in all media?

In the derivation, the c is assumed to be constant, that is is free space or vacuum.