Another twin paradox variation


by Ookke
Tags: paradox, twin, variation
Ookke
Ookke is offline
#1
May15-08, 04:39 PM
P: 95
Sorry about twin paradox again, I'll try to keep this simple as simple as I can.

Initial situation:
All participants are at rest, A1 and A2 are together at midpoint between B1 and B2. All clocks are in sync.

_____________A1, A2
_____B1____________________B2

----

A1 and A2 move, say, with speed 0,8c to opposite directions, measured in B's frame (the relative speed between A's being somewhat higher of course).


____________<-A1 A2->
_____B1____________________B2

---

When A2 and B2 meet, A2 makes instantaneous turnaround and begins to move to A1's direction with the same speed, thus entering A1's reference frame.

___<-A1__________________<-A2
_____B1____________________B2

When A's compare their clocks, they note that A2's clock is behind since A2 is the one that has made the acceleration. (1)

A's are inertial observers from now on, so this clock difference should be permanent.

---

B1 and B2 catch up the respective A's with identical acceleration, B1 and B2 never moving relative to each other. All four are now in the same reference frame. This is the end situation.

___<-A1__________________<-A2
___<-B1__________________<-B2

---


Different observers cannot disagree of local events, so the clock readings at meeting points should be absolute. Let
tA1 = clock time of A1 when A1 and B1 meet
tA2 = clock time of A2 when A2 and B2 meet
tB1 = clock time of B1 when A1 and B1 meet
tB2 = clock time of B2 when A2 and B2 meet


The events "A1 and B1 meet" and "A2 and B2 meet" are simultaneous in the B's frame.

Just before the B's accelerate, B's agree that
tA1 = tA2, tB1 = tB2 and tA1 < tB1. (2)

B's do not move relative to each other, so after acceleration is still tB1 = tB2. Because A1 and B1 are at the same place during the acceleration, their clocks go at the same rate; as with A2 and B2. Therefore, after the acceleration, B's still agree of (2).

(1) and (2) seem to contradict, all four observers nevertheless being in the same reference frame.

Does this have something to do with the "identical" acceleration of B's, or is there a more embarassing flaw somewhere? I'm not actually believing this is a contradiction ;)
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yuiop
yuiop is offline
#2
May15-08, 09:39 PM
P: 3,966
Hi Ookkie,
nice thought experiment and a little bit tricky. The solution is that when B1 and B2 accelerate to join A1 and A2 they no longer see their spatial separation the same as when they were stationary. If we introduce a fith observer D who never moves it is easy to see. If B1 and B2 are initially a distance of x apart when at rest then they should be a distance of 0.6x apart in D's frame when they reach 0.8c relative to D if they are to measure their separation the same as when they were at rest with D. Since their separation is x in D's frame when co-moving with A1 and A2 they now see their own separation as 1.666x so one has moved relative to other and their clocks will be out of sync. To observer D the clocks of B1 and B2 will look like they are syncronised so it follows the clocks will not be synchronised from the point of view of b1 and B2 in thier new reference frame.

So these statements:

(a) "The events "A1 and B1 meet" and "A2 and B2 meet" are simultaneous in the B's frame."
(b) "B's do not move relative to each other, so after acceleration is still tB1 = tB2"
(c) "Because A1 and B1 are at the same place during the acceleration, their clocks go at the same rate; as with A2 and B2."
(d) "Therefore, after the acceleration, B's still agree of (2).

are not true for the following reasons:

(a) The events are simultaneous in D's frame but not in B's frame.
(b) B's do move relative to each other in their own frame.
(c) When A1 arrives at B1 and A2 arrives at B2 there is the same offset between the a clocks and the B clocks at that point. When A1 passes the stationary B1, it has a relative velocity of 0.8c and continues with constant velocity relative to D while B1 is accelerating during the same period so their clocks do not advance the same amount. A1 and B2 accelerate at the same rate at the same time from the same place so their clock advance at the same rate during that phase. If the A clocks and the B clocks had the same offset before the acceleration phase then they will have different offsets after the accleration phase.
(d) Is not true for reasons a,b and c.
Ookke
Ookke is offline
#3
May16-08, 01:43 PM
P: 95
Hi, thanks for the reply.

Quote Quote by kev View Post
(b) B's do move relative to each other in their own frame.
Ok. Actually, this thought experiment is somewhat similar to Bell's spaceship paradox, which I didn't realize.

I find it somewhat counter-intuitive that identical acceleration may cause stretching of distance, but I'm happy with the explanation.

By the way: anyone knows if there have been experiments regarding Bell's spaceship paradox?


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