Quote by ghwellsjr
Like I said, you don't have to assume that one twin is stationary in the first frame and the other twin is stationary in the second frame, but if you do, then they can never reunite and there will never be any way to compare their ages.
But you can pick the frame in which just one twin is at rest and the other twin is traveling, not in his own frame, but in the frame of the atrest twin. The you can use the formula that Einstein worked out in section 4 of his 1905 paper introducing Special Relativity which looks like this:
τ=t√(1v ^{2}/c ^{2})
where τ, tau, represents the time dilation of the traveling twin and t is the normal time for the atrest twin.

I am not assuming that one of the twins is stationary, but that one is moving with a different speed than the other. But this situation is symmetric, when A is moving away from B, then also B is moving away from A. Both can make the calculation of their ages and A comes to the conclusion that B will be younger, then B will make the calculation and to him A should be younger.
Quote by ghwellsjr
So if you consider the traveling twin to always be traveling at speed v in the rest frame of the other twin, in other words, from the time he leaves until he gets back, he is always traveling at "v", although his direction can be changing, the ratio of their accumulated ages is simply √(1v^{2}/c^{2}).

When the twin wants to return then his speed changes to v. Then we cannot make a new calculation because their origins of their reference frames are not in the same place.
Einstein is assuming that the reference frame can change its speed or direction and the Lorentz transformation still can be used. Which is true it can or cannot?