problem as shown in the picture below A person walks very slowly at speed u from
ID: 3893338 • Letter: P
Question
problem as shown in the picture below
A person walks very slowly at speed u from the back of a train of proper length L to the front. The time-dilation effect in the train frame can be made arbitrarily small by picking u to be sufficiently small (because the effect is second order in u). Therefore, if the person's watch agrees with a clock at the back of the train when he starts, then it also (essentially) agrees with a clock at the front when he finishes. Now consider this setup in the ground frame, where the train moves at speed v. The rear clock reads Lv /c2 more than the front, so in view of the preceding paragraph, the time gained by the person's watch during the process must be Lv/c2 less than the time gained by the front clock. By working in the ground frame, explain why this is the case. Assume u v.Explanation / Answer
The phenomena which is explained by using time dilation concept in the train's frame ( or the person's frame) can be explained in the ground's frame using the concept of length contraction. This always happen and so the concepts are complementary to each other and the 2 sides of the same coin.
This happens as follows:
From the ground frame of reference, the length of the train is reduced to L * sqrt( 1 - v^2/c^2). so the contraction in L is given by = L - L * sqrt( 1 - v^2/c^2).
which can be approximated by binomial expansion and retaining lower order terms as:
delta L = L - ( L - (1/2) v^2/c^2 ) = (1/2)* L * v^2/c^2.
Effective velocity with which person covers the distance can be shown to be v/2
so, effective time gain = delta L / (v/2) = Lv / c^2
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