1. Jan 7, 2006

### Anders Lundberg

There has been a lot of discussion about the twin paradox in this forum before, but I have another angle of this problem which I have problems figuring out.
What if we have these triplets, one takes off in a rocket in one direction, one takes of in another rocket in a perpendicular direction to the first one and the last one stays put at mother earth. Now the two space travelers continues their travel, near lightspeed, for one year and then turns around and returns to earth, again in one year.
According to the twin paradox, the two space travelers should be younger than the one who stayed on earth, so far so good. The travelers where the ones that "worked their way around", so I have no problem accepting this. Now to the problem.
What about the relation between the two space travelers? They also have been moving near lightspeed in relation to each other (not only to the one who stayed), but they have both contributed equally to this fact. No one has accelerated more than the other, so they are equal in this matter. But if traveler A looks at traveler B, he will see B moving att near lightspeed and therefore he will also see the time moving slower at B, and vice versa for B looking at A. Now, when A and B meets on earth again after their journey, how can we avoid a contradiction where it is a fact for A that B's clock has moved slower than A's and that it simultainously is a fact for B that A's clock has moved slower than B's?
Of course there must be a perfectly sound explanation for this, so can somebody please give that explanation???

2. Jan 7, 2006

### Hurkyl

Staff Emeritus
You're making the same omission that many people make when analyzing the ordinary twin paradox.

"The spacebound twin sees the Earthbound twin moving at near lightspeed, and therefore he will also see time moving slower for the Earthbound twin. Therefore, when they meet, the Earthbound twin should be younger."

3. Jan 8, 2006

### Anders Lundberg

Well, as far as I have understood these matters, the spacebound twin is the only one changing velocity (synonymous to "changing reference system") and this fact is the reson (really not knowing why, I have to confess) that the earthbound twin gets older and not younger. But in my example I am not comparing with the earthbound guy, I'm comparing between the two spacebound guys, who both have changed their reference system.
Well, when thinking about it, it could of course be a question of both travelers changing their speed to an equal amount, where it does not really matter that they do not share the same reference system. What matters is to what extent they change reference system, which in this case happens to be to an equal extent. Thats why they are of equal age when returning to earth.
Am I right?
When spacebound triplet A returns to earth, he will meet his earthbound triplet C, who now is older than A. He will also meet his spacebound triplet B, who has "avoided" this agening because he also has been traveling in space, let be in another direction, but nevertheless in the same manner (same acceleration pattern) as spacebound triplet A.

4. Jan 8, 2006

### Hurkyl

Staff Emeritus
Then let me explain!

The facts about time dilation are generally presented for inertial reference frames. In any given inertial reference frame, faster clocks tick slower.

The mistake that leads to the twin paradox is the following analysis:

"From the spacebound twin's reference frame, he is always stationary, and the Earthbound twin is the one moving, and thus his clock is ticking slower."

The mistake is that the spacebound twin's reference frame is not (always) an inertial reference frame. The theorem "In any given inertial reference frame, faster clocks tick slower." cannot be applied to the spacebound twin's reference frame when he's accelerating!

You can work out how time dilation works in accelerated frames too. You start with the ordinary amount of time dilation. Then, you say:
(1) If I'm accelerating towards a clock, you add some amount K to its rate. (So, it speeds up)
(2) If I'm accelerating away from a clock, you subtract some amount K from its rate. (So, it slows down, or even runs in reverse)

The amount, K, is proportional to distance (so things really far away speed up a lot, or run really fast in reverse, while nearby things aren't affected much), and also depends on how much acceleration.

This is certainly a valid analysis -- one of the points of relativity is that you're permitted to do the analysis in whichever frame you choose, and it's usually best to pick a frame where the problem is simplest!

(And the point of the twin paradox is to try and prove that you cannot do the analysis in whichever frame you choose... thankfully the twin paradox argument fails!)

5. Jan 8, 2006

### Staff: Mentor

In this forum, the "Classical Physics" forum? I think not. Perhaps you're thinking of the "Special & General Relativity" forum, a bit further down the list.

6. Jan 11, 2006

### yogi

Well - the author of this thread has posed a problem similar to the issue i wished to discuss in the thread on the reality of relativity. Here, in Lundbergs formulation, we have all three triplets in the same frame on earth and all three clocks are brought to sync. The two traveling siblings must return to find their clocks read the same - but the stay at home clock will have logged more time. The apparent paradox arises because during their journey, each traveler can use any number of techniques (e.g., reflected radar pulses) to make measurments of the clock in the other traveler's spaceship - and because of symmetry, each of the travelers will measure the other moving clock to be running slow - just as they will meaure distances in the other spaceship to be contracted - but these are apparent observations - these reciprocal findings of tardy clocks made during the time when they were in relative motion don't survive when the two travelers are returned to earth and all 3 clocks are compared while at rest in the earth frame - the traveling siblings clocks will have both logged less time than the stay at home clock by the same amount as predicted by Einstein in Part 4 of his 1905 paper. Most relativists explain the twin and triplet paradoxes by treating as real the apparent time differences that accumulate during the outward, inward and turn around times. However, even though the apparent observations of the other twin during these phases of the trip cannot be justified by logic or experiment, one nonetheless arrives at the same result - so I suppose if one is comfortable with a particular method that works - so be it

7. Jan 11, 2006

### pervect

Staff Emeritus
The thread has been moved to the SR forum by the moderators, that's where I'm reading it now (and it's been here for some time, now).

8. Jan 11, 2006

### Staff: Mentor

I just noticed today that this thread had been moved. Unfortunately, I can't edit my posting above any more. The moderators are welcome to delete it!

9. Jan 11, 2006

### Hurkyl

Staff Emeritus
You've fallen into the twin paradox trap -- this statement is wrong.

The correct statement is:

Each of the travellers will, while travelling inertially, observe the other clocks to be running slow. However, while accelerating towards Earth, each of the travellers will observe the other clocks to be running extraordinarily fast.

10. Jan 12, 2006

### yogi

hurkyl - if you are using reflected light (radar methods) yes - the clocks look like they are running fast if they are approaching at constant relative velocity- this is an apparent speeding up - just as the clocks appear to be running slow when they are separating if you use transmitted and reflected em waves to draw conclusions about what is going on with time. Some authors (such as Resnick) resolve the entire twin scenerio using the doppler shifts - curiously it works - but it is illusory ...the clocks are not running faster or slower on the inbound and outbound trips simply because of the rate at which pulses are received relative to the rate of transmission.

11. Jan 12, 2006

### JesseM

I don't think Hurkyl was talking about doppler shifts and how each twin would appear to the other visually, I think he was talking about analyzing the situation from the point of view of a non-inertial coordinate system where the travelling twin is at rest throughout the journey. See the 'general relativity' explanation of the twin paradox.

12. Jan 12, 2006

### Hurkyl

Staff Emeritus
Right -- when I say "each spacebound triplet, while accelerating, sees the other clocks running extremely fast", I mean that in the exact same sense as "each spacebound triplet, when moving inertially, sees the other clocks running slow".

To put it differently, without any reference to how the accelerated frame appears, the following is true. Let Joe and Susan be the spacebound triplets.

(1) Just before beginning his acceleration, Joe makes a note of what Susan's clock says right now, according to his current inertial reference frame.
(2) Joe begins his acceleration back towards Earth.
(3) Joe completes his acceleration back towards Earth.
(4) Joe writes down what Susan's clock says right now, according to his current inertial reference frame.

Joe will find that the times he recorded in (4) and (1) are vastly different -- in fact, when he does (4), he will find that Susan's clock actually reads much more than his own clock.

(Technical point -- of course, Joe is unable to actually observe what Susan's clock says)

13. Jan 12, 2006

### yogi

Ok jesse and hurkyl - i stand corrected.