Two twins, one stays at home, while the other travels to the stars at speeds close to the speed of light. Upon her return, the traveling twin will look much younger than her stay-at-home twin, why?

The twin who leaves and returns has the trajectory depicted in red-blue in fig 1. While for the twin staying on earth, the trajectory is in green. The diagram is deceiving as it gives the appearance that the traveling twin has covered more grounds. But remember that the “distance” on a space-time graph is really t

^{2}– x

^{2}, which is the proper time. Moving clocks are slower, and therefore the traveling twin’s clock will register a smaller time than the stay-at-home twin.

We will assume that the traveling twin moves from a

_{1}to a

_{2}at a velocity v, and when she returns back, she moves with the same velocity, but in the opposite direction from a

_{2}to a

_{3}. Therefore it takes an equal amount of time T for each leg of the trip. According to the stay-at-home twin, the round trip amounts to 2T. For the traveling twin, on the first leg of the trip, from a

_{1}to a

_{2},

Ta

_{1}to Ta

_{2}= T(1 – v

^{2}/c

^{2})

^{½}

On the return trip,

Ta

_{2}to Ta

_{3}= T (1 – (–v)

^{2}/c

^{2})

^{½}= T(1 – v

^{2}/c

^{2})

^{½}

Therefore the total trip = 2T(1 – v

^{2}/c

^{2})

^{½}, and this is less than 2T.

How can we tell which twin moved and which twin was at rest? In this case, the traveling twin measures both events, departure and arrival, with the same clock on the first leg of the trip. The twin on earth would need two clocks to measure these two events. Ditto for the return trip. Hence the traveling twin's clock measures the proper time, and is designated as the moving clock.

*Moving clocks slow down.*