 In this video lecture I have discussed about the time dilation, which comes under the Special theory of Relativity Unit. To understand this concept the basics like frame of reference, Observer and events idea should be clearer. Here both S and S’ frame are considered as inertial and relative speed of S’ frame with respect to the S is v, this is constant, so there is no acceleration and obviously frame will be inertial.

Suppose an event occurs in to the S’ frame which O’ observer of S’ frame notes to that in terms of (x’, y’,z’, and t’). at the same time the stationary observer O, observe the events in terms of (x,y,z and t). Here interesting thing is that both see the events in their frame as real but on the other hand when they came to know how it seems from his frame of reference then it becomes interesting for both of them. How one can see an event in the other frame this is solved by the special theory of relativity perfectly with the help of Lorentz Transformation.

Now, to explain the time dilation in special theory of relativity, suppose a clock is fixed at x’ point in the S’ frame, the time interval between two ticks of the clock according to the observer O’ are t1 prime and t2 prime. So time interval is t0=t2prime – t1 prime; this is proper time which is observed by the observer O’ in S’ frame. Now this time interval between two ticks of the clock according to the observer O in S frame will be t1 and t2. The time interval t= t2-t1 according to the observer O in S frame.

Further, we have to find the relationship between the t and t0 by using the Lorentz Transformation. The stationary observer O (the person who is in rest) see that moving frame clock run slowly. It means that one hour time interval between any of the two events will be larger than the one hour at the rest frame (it might be 2, 3 or more than this depends upon the relative speed that 0.6c, 0.8c or 0.9c or more than this.)

Here some example is chosen to make this concept more clearly to students.

Ex.3 A wrist watch keeping correct time on the earth is worn by the pilot of a spaceship. How much will it appear to lose per day with respect to an observer on earth when spaceship leaves the earth with a constant velocity of 10^7 m/sec.
In this video example number 3 is tricky, where a pilot who corrects the wrist watch on earth before taking flight into the space. Pilot is interested to know that how much he will lose the time to travel in the space per day or per hour in his frame with respect to the stationary observer, he wants to find it with the help of a day on the earth relative to by which he is moving in to the space. So obviously he knows about the time at the earth (i.e. 24 hours), which is t, because this is time for a day for the person who is living on the earth with respect to the space ship or any other moving frame (on the other hand you can understand it time for the rest observer). So problem is to find the time in space with the help of it and that will be t0. So after finding the time with the relative speed one can find the difference between the day at earth and on the space, so by that difference of time the pilot is slow as compared to the stationary observer. Convert this time into hour, minutes and ultimately into seconds.

Ex.4.How fast must a spacecraft travel relative to the earth for each day on the space craft to correspond to 2 days on the earth?
In the 4th example, you can see that one have to find the speed of spacecraft which is moving relative to the earth. Now what the pilot says! he says that each day on the spacecraft (it means one day that is 24 hours, obviously telling the time and it will be t0) will correspond to the 2 days on the earth (so again told the time on the earth that is a stationary frame, rest frame and time is observed here by the observer O, so it will be t). Now if one knows about the t and t0, he/she can find the relative speed in terms of the c i.e. speed of light. Or directly can put the value of speed of light i.e. 3×10^8 m/sec.