Irodov Problem 1.353

As seen from AB's reference frame: A'B' is moving from left to right with a speed v and so has a length of . AB is at rest has a length of l0.

The distance traveled by A'B' when A and A' align is and so the time taken is


The distance traveled by A'B' when B and B' align is l0 and so the time taken is l0/v.

As seen from A'B''s reference frame: AB is moving from right to left at a speed v and so has a length of . A'B' is at rest and has a length of l0.

The distance traveled by AB when B and B' align is and hence the time required is .

The distance traveled by AB when A and A' align is lo and so the time required is lo/v.

Isn't it interesting that according to AB, A and A' align first and then B and B' while according to A'B', B and B' align first and then A and A'. This demonstrates that in relativistic mechanics sequence of events as seen by different observers may in fact be different!

Irodov Problem 1.352

a) Let the proper length of the rod be lo. Since the rod is moving at a speed v with respect to frame K, the length of the rod as seen from Frame K would be lesser than its proper length and equal to . At time instant tA, the front end of the rod was at xA. This implies that the rear end must have been at . At time tB, the rear end of the rod was at xB, this implies that the rear end of the rod moved a distance of in a time interval of tB - tA. Since the rod was moving at a speed v, we have,








b) In order to determine the time interval required to measure the proper length of the rod in equation (1) we can set xB - xA = lo, and we have,