One can resolve the apparent paradox without having to appeal to General Relativity, by realizing that we’re not using two (in which case there would be equivalence), but actually at least three different inertial frames (i.e. frames moving at a constant speed). One, the rest frame of the Earth. Two, the rest frame of the spaceship travelling away from earth. Three, the rest frame of the space ship travelling towards earth.
So, then, let’s take a look at what happens to our two twins: One, Trippy, embarks on a journey to Planet X ten light years away, while the other, Homey, stays on earth. Let’s say at the same time (in the Earth’s rest frame), on a planet ten light years further away than Planet X (so twenty light years away from Earth in the direction of Planet X) another spaceship embarks on a journey in the opposite direction at the same speed as Trippy’s ship (0.866c for convenience, because that gives a Lorentz factor of 2) (let’s call Trippy’s ship Ship 1, and the other one Ship B).
From Earth, the whole thing is simple enough: Trippy hops onto the ship going at 0.866c, goes to Planet X – which takes him 11.55 years --, then he hops onto the ship going the opposite direction, rides that for another 11.55 years, and is back home after 23.1 years – i.e., when Trippy comes back, Homey is 23.1 years older.
Now, to Trippy, things look slightly different – in the rest frame of Ship 1, the distance between Earth and Planet X is length contracted, half as long in fact, since we have a Lorentz factor of 2. Cruising at 0.866c, it thus takes him a cool 5.78 years to get there. And once he’s there, he decides – in no time at all! – that Planet X is really a bit of a shithole, hops onto Ship B, and makes back – again, covering five light years in 5.78 years. Thus, to him, the whole trek took a mere 11.55 years, half of what it seemed like to his brother Homey.
Of course, that’s exactly what Homey expected – Trippy moved at 0.866c for 23.1 years, meaning his clock ticked half as fast as the one on Homey’s wall, registering, in toto, 11.55 years.
However, depending on how well versed Trippy is in Special Relativity, he might be in for a surprise – it sounds entirely reasonable for him to expect that, since he was in an inertial frame of reference all the time and thus can view the rest frame of the Earth as moving, it ought to be Homey that has only experienced half the time passing – to wit, since Trippy’s trip took him 11.55 years, he might expect that Homey ought only have aged 5.78 years!
And the thing is, for each of the two legs of his journey, Trippy is entirely correct in his reasoning – Homey ages 2.89 years during the 5.78 years it takes Trippy to get to Planet X, and again 2.89 years during the journey back. Nevertheless, once Trippy arrives back home, he will find his brother having aged 23.1 years since his departure.
So when does he suddenly accumulate those missing 17.32 years?
The answer to that lies in the relativistic concept of simultaneity. Two events (x[sub]1[/sub], y[sub]1[/sub], z[sub]1[/sub], t[sub]1[/sub]) and (x[sub]2[/sub], y[sub]2[/sub], z[sub]2[/sub], t[sub]2[/sub]) are simultaneous, almost trivially, when t[sub]1[/sub] = t[sub]2[/sub]. Now, the rub is that if two events are simultaneous in one inertial system, they need not be in another that is moving relatively to the first one. That’s because the math you use to ‘translate’ coordinates in one inertial system into coordinates in another inertial system is that of Lorentz transformations, which mix up time and space – crucially, the time coordinate in the system you translate into depends on both time and location in the system you translate from; if we denote with a ’ the coordinates in an inertial frame moving at speed v in x-direction, the relation for the transformed time coordinate is t’ = γ(t - vx/c[sup]2[/sup]), with γ being the Lorentz factor. Hence, even though t[sub]1[/sub] = t[sub]2[/sub], it’s generally not the case that t’[sub]1[/sub] = t’[sub]2[/sub] (because x[sub]1[/sub] ≠ x[sub]2[/sub]), and two events that are simultaneous in one frame of reference may occur at different times in another, relatively moving, one.
Thus, the events on Earth a traveller on Ship 1 would consider simultaneous with himself generally will differ from the events on Earth a passenger on Ship B considers simultaneous with himself; and, when changing over from one ship to the other, the notion of simultaneity changes accordingly.
All that’s left to do, therefore, is to calculate the difference in time between a event on Earth that Trippy on Ship 1 considers simultaneous with himself (t[sub]S[sub]1[/sub][/sub]), and a point in time that Trippy on Ship B considers simultaneous with himself (t[sub]S[sub]B[/sub][/sub]). With the transformation given above, that’s:
t[sub]S[sub]B[/sub][/sub] - t[sub]S[sub]1[/sub][/sub] = γ(t + vx/c[sup]2[/sup]) - γ(t - vx/c[sup]2[/sup]) (with the ‘+’ sign in front of the first v coming from the fact that the inertial frame in which Ship B is at rest moves at v in negative x direction in Earth’s rest frame)
= 220.866*5 (because c is 1 in lightyears per year, and x is equal to five light years – the distance between Earth and Planet X from the rest frame of either ship)
So, when Trippy makes the jump from one ship to the other, the age of Homey, as calculated by Trippy, ‘jumps’ 17.32 years – there’s no paradox at all!
Of course, this isn’t actually physically possible – generally, you’d have to decelerate the ship to a relative full stop, then re-accelerate it, and decelerate it again at arriving on Earth, making the whole thing quite a bit more complicated; but, since acceleration can be modelled as a series of infinitesimal jumps from inertial systems of infinitesimally different speeds, the reasoning is essentially the same, although in that case, the ageing will occur smoothly, not in a sudden jump, which is much more in line with what one might expect (though I sometimes feel like having taken a great leap in age, particularly on weekend mornings when I’ve been out the night before).