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Feenburg’s Twin Paradox Still Unsolved?
Author: Erich Schoedl
Added: 07/19/2003
Type: Review
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2

coordinate axes could be visible to a third observer, the corresponding y-axis of both twins would remain parallel.  No big deal, right?

But if there is a true frame-rotational symmetry, then why is twin B aging slower than the variable star?  It could just as easily be stationed so that the star, the Earth, and twin A are accelerated and move relative to twin B.  Of course, if they must in turn decelerate and reverse velocity to return to a stationary B, then the three entities would have aged less than B since they would have covered a longer world line relative to any reference frame observer.  In the first segment of the journey, what should distinguish between the star aging slower, and twin B aging slower?

So let’s consider case-2, and assume it an equivalent example to case-1 as follows.  The experiment should remain identical to the original scenario if once the Earth, star, and twin A’ accelerated to velocity v (simultaneous for A’), twin B’ waits motionless for a period of proper time ½*tb (just as when B were moving away in the original experiment), then twin B’ would have to accelerate to a new velocity toward A’ of u = 2v/(1 + v2/c2) for an additional proper time of ½*tb.

With simple coordinate transform by rotation of axes, the two cases should be symmetric, as we assume they would be from the perspective of twin A maintaining rotation symmetry (the y-axis, perpendicular to motion, doesn’t change direction between the two frames).  Obviously the cycles observed from the star by twin B’ in case-2, should at first appear slow due to the time dilation associated with the star’s movement, then faster due to B’’s time dilation from motion toward A’.  Yet if the perpendicular of the y-axis of the frames is to remain unchanged between the two cases, the symmetry of the experiment is contradictory, and pure “relativity” is kaput on these grounds without invoking the smack of the ether or frame dragging of the surrounding stars.  So the paradox contends, how can case-2 be equivalent to the original experiment?  The cases “appear” visually symmetrical, but the observed cycle time of the variable star is contradictory to that assumption between the two cases.

True relative symmetry can be completely restored for a simple price.  Perpendicularity of axes transverse to motion between relative frames must be capitulated to an extent.  Please refer to the paper Distortions of Relativistic Measure (also posted at PF) for a more visually protracted description of the idea I’m going to use to explain.  For the “Feenburg’s Twin Paradox”, here’s how the correct symmetry of relativity in this peculiar case can be logically established.



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  Article Comments   Add Comment | View All (4)


Poster: Erich
Added: -1/1-/2003

It's probably too late to help. But if you have any specific questions, just email me.

Poster: stefanie
Added: -1/0-/2003


Doing a 1 page research paper on the twin paradox. Dont understand anything about this stuff. This is for an English 1A class, if you can believe it. Can anybody help me out with some info?

Poster: es
Added: -0/8-/2003

Oops, thanks Neil.
Poster: Neil
Added: -0/7-/2003

off-shoot or offshoot, NOT off-chute.
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