Problem : A spacecraft travels at 0.99c to a star 3.787×1013 kilometers away. How long will a roundtrip to this star take from the point of view of someone on the earth?If we calculate the number of seconds in a year it turns out that 3.787×1016 meters is about 4 light-years (the distance light travels in one year at c ). The spacecraft is traveling virtually at c , so the trip to the star takes 4 years of earth time. The roundtrip takes 8 years.
Problem : With reference to the previous problem, how long will the roundtrip take for someone in the spaceship, according to someone measuring from the earth?According to an observer on the earth, since the spacecraft is moving, its passengers' time is dilated. The factor by which this occurs is γ = = 7.09 . The passengers measure less time so, the roundtrip time is (1/7.09)×8 = 0.14×8 = 1.1 years.
Problem : Now in the reference frame of someone in the spaceship, what is the time taken for the roundtrip as observed by a passenger, and by someone on earth (ignoring the times when the spaceship is accelerating or decelerating).The whole point of the twin paradox is that a passenger on the spaceship apparently measures the opposite: that is, that the trip takes 8 years for them, but only 1.1 years for those standing back on the earth. It turns out that this reasoning is incorrect and in fact the passengers measure the same times as an observer on the earth when the (General Relativistic) effects of acceleration and deceleration are taken into account.
Problem : If one person stays on earth and one person travels to the distant star, who will age more during the trip and by what amount?As we have seen, the reasoning of the passenger on the spaceship is erroneous because the spaceship is not in an inertial reference frame. The reasoning of the person on earth is correct (the earth is approximately inertial). They measure the passenger as aging less than themselves by an amount 8 - 1.1 = 6.9 years.
Problem : Twin A floats freely in outer space. Twin B flies past in a spaceship at speed v 0 . Just as they pass each other they both start timers at t = 0 . At the instant of passing B also turns on his engines so as to decelerate at g . This causes B to slow down and eventually to stop and accelerate back towards A so that when the twins pass each other again B is traveling at speed v 0 again. If they compare their clocks, who is younger?This is just a variation of the same problem (that is, the twin paradox as stated in Section 2). Twin A is in an inertial reference frame so she can successfully apply the logic of Special Relativity to find that B's time is dilated and hence that B is younger. B is not in an inertial reference frame so the opposite reasoning does not apply, and we conclude that when all the effects of the acceleration are accounted for he must agree with his twin that he is younger.
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