Can someone explain how this situation works in special relativity?

OK, so let's say there's twin brothers, Al and Bert. Let's say Al goes off in a rocket at .9c away from Bert for two years of Bert's time. When Al comes back, he'll be younger than Bert is. I'm fine with that. But now let's say that Al stays put and Bert shoots off in the opposite direction at .9c for two years of his time. Wouldn't Bert, when he comes back from his trip, be younger than Al? But aren't the two situations exactly the same (since velocity is relative)? Obviously, they can't both be younger than the other brother.Can someone explain how this situation works in special relativity?
ADDED: sorry, I misread your question. The answer is that whoever is travelling will get younger. So Al will be younger than Bert when he gets back, and then Bert goes away and ages less than Al and comes back. their difference in ages will depend on who has travelled further and quicker. If they travelled at the same speed and for the same distance, they'll be the same age again :)





... but are you sure that's what you mean to ask? because there is a very famnous paradox called 'the twin paradox', composed by Einstein, which is based on what you say. imagine Al goes off for his trip and comes back younger. But can't Bert claim, well velocity is relative, so shouldn't Bert also be younger? (There is no second trip involved in the twin paradox.)





If this is what you mean to ask (I think maybe), then the answer is below.


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The resolution of the paradox is that Al is undergoing brief but significant periods of acceleration (speeding up away from earth to reach the constant speed 0.9c, then decelerating down to low speed when he arrives; and doing this once more on the way back), during which (due to Lorenz contraction) Al will reckon that the distance he travelled is only a fraction of the distance that Bert sees him travel; since space shrinks up close to the speed of light.





An accelerating frame is not inertial, and special relativity does not say that all observers are equivalent, it says that all -inertial- observers are equivalent. hence Al and Bert are not subject to the same 'any observer is relevant' postulate (the equivalence postulate) of relativity in this case, since Al has transferred to several different inertial frames during his travel. Hence we can definitely say it is the accelerated frame (Al's) which has experienced less time than the stationary one (Bert's).





(btw, the fact that the 'paradox' has a resolution means it's not really a paradox, it just seems that way. The name just stuck.)Can someone explain how this situation works in special relativity?
';I'm fine with that?';


Brilliant physicists and scientists have been wrestling with this concept for a hundred years - and you're ';fine with it?';


I envy you.......





In answer to your question, which ever twin is in a situation that slows time relative to the other twin, he will be the younger one when they get back together in the same time reference - regardless of whether it's Bert or Al and regardless of the direction.
The important distinction is which twin actually turns his rocket around to make the return trip. He's the one that will come back younger. The difference between the twins is that one simply sits and waits the whole time but the other actually fires rocket engines to turn around and come back.





It may seem that the situation can be viewed as if the ';waiting'; twin is in motion while the rocketeer is not, so it makes for an identical scenario. It isn't identical though, because the 'waiting' twin wouldn't feel any acceleration at the turn around time the way the way the rocket-riding twin would.





Having said this, I read your question as saying that the other twin rides the rocket the second time. If that's the case then the other twin would come back younger. Whoever rides the rocket and fires engines to turn around and come back will be younger when they meet again.