Re: Variation of Twins Problem

seppala51_at_sbcglobal.net
Date: 08/19/04 

Date: Thu, 19 Aug 2004 02:32:19 GMT

On Thu, 19 Aug 2004 01:05:02 GMT, Tom Roberts <tjroberts@lucent.com>
wrote:

>seppala51@sbcglobal.net wrote:
>> Two identical twins (clocks) are traveling toward each other in
>> identical spacecrafts along the x-axis. Their relative velocity is
>> V=0.866c. When they meet at x=0, they verify they are indeed
>> identical and the same age. They set both of their clocks to zero at
>> this meeting point. They also press a deceleration button on each of
>> their respecitve spacecrafts which causes their crafts to change
>> velocity. When the deceleration ends, their relative velocity is V=0.
>
>OK. You have implicitly described this in a single frame, the one the
>twins end up at rest in when they ultimately have zero relative
>velocity. But you seem not to realize this. Moreover, you never stated
>it is an inertial frame -- I'll assume that, because without it nothing
>makes sense (i.e. if that frame isn't inertial then SR does not apply
>here). I call this the "final inertial frame".
>
>I will also assume that the deceleration of the two twins is identical
>in that they have identical proper accelerations as a function of their
>proper times. That just means they use identical rockets to slow down with.
>
>
>> Here's what I don't understand. When they meet, both twins are the
>> same age and the clocks are set to zero, and when they have zero
>> relative velocity both twins are the same age and both of their clocks
>> show the same time (and are now running at the same rate). If the
>> clocks start at zero, and then show the same time when they have zero
>> relative velocity, doesn't that imply that the clocks ran at the same
>> rate when they had relative motion?
>
>Yes it does, WHEN "RATE" IS MEASURED IN THE FINAL INERTIAL FRAME (and
>simultaneously in that frame).
>
>
>> If spacecraft A observers say
>> that craft B's clock ran at a slower rate during some part of the
>> deceleration, then during some part of the deceleration craft's B
>> clock must have run a faster rate also otherwise they would not show
>> the same time when they have zero relative velocity. Observers in
>> craft A cannot say that craft B stopped decelerating before craft A
>> stopped and they cannot say that craft B stopped decelerating as each
>> of these can be experimentally shown to be false. Please explain how
>> this is resolved with Einstein's notion of time.
>
>You got yourself all confused by attempting to reference "rate" to the
>non-inertial frames of the decelerating spacecrafts. This takes much
>care, and in practice is so error-prone that it is advisable to not
>attempt it. Use the final inertial frame and it's quite clear that at
>every instant in that frame the two twins' velocities are equal in
>magnitude and opposite in direction, and their clock tick-rates wrt that
>frame are at all times equal.
As viewed in the final rest frame, the problem is symmetrical, and yes
their clock tick-rates wrt that frame are at all times equal. I know
that. And also that frame says both clocks ended up with fewer ticks
elapsing during the deceleration than clocks in the final rest frame
ticked. That is not what I'm asking. What I'm looking for is how the
situation is viewed from say spacecraft A. If observers in this frame
record their measurement of B's clock wrt their clock's reading, what
do they get? I can do the point at the start of the deceleration
(both clocks read 0) and the points once the deceleration stops (both
clocks read t=T. But during the deceleration observers on craft A do
not measure craft's B clock to run at the same rate as clock B. So
at some point during the deceleration the clocks must lag each other
and at some other point they must catch up to each other because when
they are in the final rest frame they must have identical tick-rates,
and from the symmetry of the problem they must also show the same
time. Or of course, they could all show the equal times through out
the deceleration as measured in the A frame, and also in the B frame -
but that's Newton's view, not Einstein's.
    You suggest that this "catch up" happens at the instant the
deceleration stops, or at least that what I'm inferring from your
posts. But I see no reason why one part of the acceleration is any
different from any other part of the acceleration.
David
<snip?