Consider two points A and B (Fig.1) at rest in S frame having a distance L0 between them (a long scale may be imagined as placed between A and B). ------ (1)(adsbygoogle = window.adsbygoogle || []).push({});

The ‘traveler’ twin (S-frame) starts from A at time t = 0 and suddenly accelerates to attain a constant velocity v in the view point of the S frame. After reaching B it suddenly decelerates, comes to rest, and returns back. Neglecting the acceleration and deceleration times, the ‘stay-at-home’ twin will check that the ‘traveler’ twin has reached the point B in a time (two synchronized clocks can be placed respectively at points A & B to check this time in S frame; the stay-at home's own frame)

t = L0 /γ (1)

Further in the view point of the ‘traveler’ twin (S1 frame), a ‘Lorentz contracted’ length L0 / (see Figure 2) passes by him (which he can check by comparing with a scale longitudinally placed in his own frame) taking a time shown on his (S1) clock as

t1= L0 /γ v (2)

where γ = (1- v2/c2)-1/2

From equations (1) and (2), we get

t1= t (1- v2/c2)1/2 (3)

Eq. (3) shows that t1 is less than t.

After completing the to and fro journey, the ‘traveler’ twin has spent a total time

T1= 2 t1 = 2 t (1- v2/c2)1/2 = T (1- v2/c2)1/2 (4)

or T1 = T (1- v2/c2)1/2 .

This shows that the ‘traveler twin’ has aged less than the ‘stay-at-home’.

Traveler twin (S1 frame) also checks that he has aged less

In the above discussion the time readings on the clocks positioned at A and B of the relatively moving S frame can also be checked from the view point of the S1 frame. Due to the time dilation effect in the view point of S1 frame, the time t1 on the S clock at A ought to be

t2= t1/γ = L0 /γ2v (5)

Now according to S1 criteria of simultaneity, the S clocks will show a progressive difference of readings (see Fig.2) so that in the view point of S1 frame, the time at the point B1 of the relatively moving S frame comes out to be

t2 = t1 + L0 v/c2 = L0 /γ2v + L0 v/c2 = L0/v = t (6)

which is the same value of time as is read in S frame. So the traveler twin also finds that the time passed in S frame is more than that passed in his own frame.

Hence Proved.

(FOR FIGURES SeeIAPT Physics Bulletin (Kanpur India) Feb 2014; p.32.

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# Too Simple & Short Solution of Twin Paradox

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