I know this one may be hard to state clearly, but may be worth trying to write down anyway, with the hopes that maybe Aaron or Collector had the idea down more clearly than I did...Jack and Jill are exactly the same age and start from the exact same point at the exact same time going at the exact same constant speed in exact opposite directions for some fixed amount of time/distance A. Now they play a game, where the object is to choose a direction and speed at which to move such that whichever of the two is younger when they meet wins. To avoid the infinite race to the speed of light, assume that neither of them can travel faster than .9375c, but can travel in either direction along the one dimensional line that they started travelling along before meeting each other. Is there a strategy that either Jack or Jill can follow to try to be the younger when they meet?

1. They start at the the same point at the same constant speed travelling in opposite directions. This implies they were both already travelling toward each other at this constant speed. Correct?2. You say that they can travel no faster than 0.9375c. Why this (funny) number, and what was there initial constant speed?

Last edited by TraderJoe on May 7th, 2005, 10:00 pm, edited 1 time in total.

QuoteOriginally posted by: TraderJoe1. They start at the the same point at the same constant speed travelling in opposite directions. This implies they were both already travelling toward each other at this constant speed. Correct?I only detail this to try and remove any doubt that they start the game at the same age. Assume they were born at the exact same time and leave earth at the exact same time in opposite directions exactly mirroring each others' acceleration and speed for a fixed amount of time to get to their starting points. Once we know they are a fixed distance x apart and still the same age, the game begins. In the original twin paradox, the twins are a distance apart, but the one who has (assymetrically) accelerated becomes the younger.Quote2. You say that they can travel no faster than 0.9375c. Why this (funny) number, and what was there initial constant speed?I only picked this number arbitrarily to make the math easy for Lorentz transformations (1 - 1/16), and you can pick any other number < 1 to use a the limit. Similarly, you can put a limit on their acceleration (say 1g). It doesn't matter what their initial constant speed was, the given is that they start distance X apart at the same age.

Thanks Exotiq, This is quite different from your original post? Anyway, think I got it now... Is it the one who decelarates/accelerates the quickest to travel in the opposite direction?

Last edited by TraderJoe on May 8th, 2005, 10:00 pm, edited 1 time in total.

I think you'll need some more constraints. The problem is that whenever they are about the meet, the older one can run away to avoid defeat, making the game last forever. We need to limit the game somehow. Also, we have to define meet to mean sharing the same location and speed, so we have to determine what happens if they approach the same place at different speeds.I think there's an interesting brainteaser in here, but it will take some refinement.

Because of the symmetry of the problem and no radomizing variables participating in the "optimal strategy" ... there may not be a winning strategy at all.Shouldn't we look at this as a game theory problem. Whatever is optimal for Jack, will be optimal for Jill too.So their initial action at time t=0, will be identical. Assuming that the information of initial action of one will reach the other at the same time, their subsequent actions will also be the same (since, game theory will ask them to changeover to same next step).So, if both are always doing the same thing all the times, their ages cannot differ at all !Or is it that there are a range of "initial actions" (with more than one best) ... the "best" depending upon what the other person chose among the same set of "best" ? (I'll need to go thru relativistic theory to list the possible first steps .. like (i) accelerate towards the competitor at maximum available acceleration, until top speed is reached, (ii) or do the same thing in the reverse directions, or (iii) --(probably not optimal) stay put where you are) )

- alexandreC
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The limiting constraint Aaron is refering to can be solved if we assume the universe is finite (if you start moving in one point and don't change your direction, you will end up in the same point - just like if you start moving in the earth and never change direction, you should go arround the earth and come back to the same spot, without ever changing direction.)Assuming this, my strategy to win the game is to never change my direction nor the initial speed.Alex

QuoteAssuming this, my strategy to win the game is to never change my direction nor the initial speed.AlexAssuming no other theoretical twist available to slow your aging process ... this strategy guarantees, you won't lose.But does not guarantee that you will win, either (in the case your competitor also does the same)

Last edited by bhutes on May 27th, 2005, 10:00 pm, edited 1 time in total.

- alexandreC
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Fair enough, you are right.Rephrasing it:My strategy in order to maximise my chances of wining the game is ......

I don't think this works. If both twins do the same thing, they pass at the other end of the universe. Each looks out of the spaceship and sees the other one as older. No one loses, but the game is never over because the twins don't end up at the same place at the same time and speed.But suppose Jack does this strategy. Jill stops and waits for Jack to approach her from the other direction. She accelerates away from him, but at less than maximum velocity. She gradually speeds up until Jack catches up and they are going the same speed. Jill will be younger, and will win the game.

- alexandreC
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I don't think this works. If both twins do the same thing, they pass at the other end of the universe. Each looks out of the spaceship and sees the other one as older. No one loses, but the game is never over because the twins don't end up at the same place at the same time and speed.yes, bhutes has pointed out that my strategy does not guarantee you to win.But it guarantees you not to loose- and I think its the only strategy that gives you that guarantee.But suppose Jack does this strategy. Jill stops and waits for Jack to approach her from the other direction. She accelerates away from him, but at less than maximum velocity. She gradually speeds up until Jack catches up and they are going the same speed. Jill will be younger, and will win the game.I am not sure if this is correct.In the inertial frame of Jack, Jill acelarates (several times), so if we compare their the ages in this reference frame (which is common for both of them in the final state) Jill will be older.

Last edited by alexandreC on June 1st, 2005, 10:00 pm, edited 1 time in total.

If Jack is holding a pendulum, it will show acceleration starting and stopping, but never reversing. Jill will meet him at the same location, speed and time. Her pendulum will show both acceleration and deceleration, in fact two of each. So Jill will be younger.

- alexandreC
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If Jack is holding a pendulum, it will show acceleration starting and stopping, but never reversing.Jack's strategy is "to never change direction nor the initial speed".Therefore he will not notice any acceleration.

Last edited by alexandreC on June 2nd, 2005, 10:00 pm, edited 1 time in total.

You're right. I was thinking of both starting at the same place and speed, and accelerating to the constant speed. But the problem says they start at that speed.But my argument is still the same. Without a pendulum reversal, Jack will be older.

I'm not familiar with relativity, but I thought the age only depends on how fast they move. What does it have to do with acceleration and deceleration?QuoteOriginally posted by: AaronIf Jack is holding a pendulum, it will show acceleration starting and stopping, but never reversing. Jill will meet him at the same location, speed and time. Her pendulum will show both acceleration and deceleration, in fact two of each. So Jill will be younger.

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