1
$\begingroup$

We have two twins, one on earth and one on a rocket traveling at very very high speed, near the speed of light, away from earth

After some time, the rocket decelerates so that the distance between him and his earth twin is nearly constant.

Suppose a third individual exists always exactly in the middle between the two twins(or belonging to the perpendicular bisector of the fictional line between the two twins), the observer uses a telescope to observe each twin.

So which one of the two twins aged more, the one on the rocket or the one on earth?

From the earth twin point of view he is stationary while from that of the rocket twin he is one stationary so who aged according to the third individual? Or does this not fall in the case of the twin paradox because they never meet?

$\endgroup$
  • 1
    $\begingroup$ possible duplicate physics.stackexchange.com/q/87602 $\endgroup$ – onurcanbektas May 28 '17 at 14:33
  • $\begingroup$ no the questions are different $\endgroup$ – Bilal Fares May 28 '17 at 14:35
  • 1
    $\begingroup$ If Bob is moving away from Alice at speed $v$, and Carl stays halfway between them, then Carl is moving away from Alice at speed $v/2$ in one direction and from Bob at speed $v/2$ in the opposite direction. You can use this to transform Alice's coordinates into Carl's, or Carl's into Bob's, or vice versa. What is confusing you about this? $\endgroup$ – WillO May 28 '17 at 14:51
  • 1
    $\begingroup$ @BilalFares: And what is stopping you from figuring that out? $\endgroup$ – WillO May 28 '17 at 15:33
  • 3
    $\begingroup$ I believe WillO is seeing a symmetry in the problem which makes it trivial to find the answer. What he's poking at is trying to understand your thought processes so that he can understand why the symmetry is not obvious to you. Often on stack exchange we ask questions in comments to understand where the OP got lost so that we can focus on that part and help you learn. Otherwise, we'd just throw a bunch of equations at you and walk away confident that we were helpful! $\endgroup$ – Cort Ammon May 28 '17 at 16:23
1
$\begingroup$

After some time, the rocket decelerates so that the distance between him and his earth twin is nearly constant.

In this case (even though the twins never meet!) it would be possible for them to come to agreement about who is older. F.e. they could start sending messages to each other. As soon as one of them receives message he appends his own age to the end of the message and sends it back. The resulting list can look like:

  • I am first twin. I am 1 year old now.
  • I am second twin. I am 2 year's 1 month old now.
  • I am first twin. I am 3 year's old now.
  • I am second twin. I am 4 year's 1 month old now. ...

Having analysed the message they can both agree that the second twin is 1 month older than the first one.

Same result would get the third observer who stays in the middle of the (now staying still) twins.

The older twin would be the one who stayed on Earth.

The paradox is that from the point of view of the travelling twin he was staying still and the other twin was actually moving! The situation seems symmetric! Both twins describe the situation like this: I was staying still, the other twin moved away from me very fast, than stopped and now we compare our ages.

Solution of this paradox is that situation IS NOT symmetric. Because the frame of reference the second twin is using is not inertial.

$\endgroup$
0
$\begingroup$

The neat thing about special relativity is that you can do the calculations in any reference frame and you will arrive at an answer which is consistent. In this case, the key is that the information is traveling to this third observer, and they are making all of the decisions. Thus, it makes sense to solve the problem in the reference frame of this third observer.

As far as the third observer is concerned, the rocket is shooting away from them at 1/2 the velocity that ground observers saw. Why? Because he's staying half way between the planet and the rocket, so he's always going to see the rocket having 1/2 the velocity ground observer see. (Edit: as Lesnik points out, at relatavistic speeds it wont be exactly 1/2 the velocity. However, it is guaranteed to be exactly the same velocity in each direction, and that's what matters)

He also is going to see a planet rocketing away in the other direction at the same velocity! The fact that planets are big doesn't matter. From his frame of reference, the planet is falling away. This is just like how a skydiver perceives the plane shooting up away from them when a ground observer would say that the plane was flying level and the skydiver was falling.

Thus, we have a very simple situation. Two twins, traveling away from the third observer at the same velocities, with the same acceleration profiles. End result: the third observer will see them at exactly the same age.

$\endgroup$
  • $\begingroup$ Hmmm. 1/2 of velocity. I don't think so. Suppose the observer sees the Earth and the rocket both having speed 0.9c. What was the original speed of the rocket as seen from Earth? $\endgroup$ – lesnik May 28 '17 at 16:40
  • $\begingroup$ @lesnik Point. I added an edit to that effect. It doesn't change the logic anyway, because its the symmetry that matters, but you're right that I ignored the relatavistic effects in a relativity problem! $\endgroup$ – Cort Ammon May 28 '17 at 16:42
  • 1
    $\begingroup$ Second objection. The middle-point observer would see both twins having same age as long as rocket flies: with this I agree. But the question is about the situation after the rocket decelerated and have has zero speed (according to Earth). The twins would not be of the same age after that. $\endgroup$ – lesnik May 28 '17 at 16:43
  • $\begingroup$ @lesnik From the third person's perspective, held at they mythical half-way-point, the planet would also decelerate at exactly the same relative rate, so you would see the same effects for the planet-side twin as well. This is why I find the symmetry so useful. Without it, we'd have to solve this in a more complicated manner by treating the observer as an accelerating and decelerating frame of their own and basically solve two twin-paradox problems. $\endgroup$ – Cort Ammon May 28 '17 at 16:53
  • $\begingroup$ Imagine that after that the second twin flies back to Earth. According to you, for the man-in-the-middle the situation is still absolutely symmetric. And when all three of them meet at Earth both twins would be of the same age! There is no paradox at all! Or is there? $\endgroup$ – lesnik May 28 '17 at 17:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.