A person in a train moving at a speed ` 3 xx 10 ^(7 ) m s ^(-1)` sleeps at 10 .00 p.m. by his watch and gets up at 4.00 a.m. How long did he sleep according to the clocks at the stations?

Explanation

The time interval measured by the watch is the proper time interval because the events , 'sleeping ' and 'getting up' , are recorded by the single clock (the watch). The clocks at the stations represent the ground frame and in this frame he sleeps at one place and gets up at another place. Thus, the time interval measured by the station clocks is improper time interval and is more than the proper time interval. The duration of his sleep in the ground frame is ` delta t' = gamma Delta t = ( Delta t / (sqrt 1 - v^(2) / c^(2))) = (6 h/ (sqrt 1 - (3 xx 10 ^(7 m s ^(-1 ) / 3 xx 10 ^(8 m s ^(-1 ) )^(2))))`
`= 6 h (sqrt 100 / 99 ) = 6 hours 1.8 minutes`.
The speed of the train in this example is hypothetical. A typical fast train today runs at about` 300 km h ^(-1) .`
Repeat the exercise with such a train .