Re: A Symmetric Twins Conundrum
 From: Mike Fontenot <mlfasf@xxxxxxxxxxx>
 Date: Wed, 20 Sep 2006 13:46:02 0600
I finally figured out how to determine twin 2's age, according
to twin 1, as a function of twin 1's age, for the case where
both twins are accelerating arbitrarily and independently.
I arrived at the solution by staring for quite a while at
a Minkowski diagram...the same way I inferred the CADO
equation quite a few years ago. (The CADO equation
applies to the case where twin 2 never accelerates (the
stayathome twin)).
Using that procedure for the special case of the symmetrical
"twin paradox" problem, the result turns out to be what I had
initially guessed would be the answer (which was also the
solution given by jem). My incorrect attempt to adapt my
CADO equation to that problem produced my second guess as
to the solution, which differed from my first guess, and which
I now definitely know to be incorrect.
The procedure I've found for determining the age correspondences
of the two twins, when they both accelerate, is not an adaptation
of the CADO equation (i.e., there is no one equation which replaces
the CADO equation, in the general case). But the new procedure
has a graphical interpretation on a Minkowski diagram which is
a generalization of the graphical interpretation on a Minkowski
diagram which originally led me to the CADO equation. The
new procedure is conceptually straightforward, but it is more
involved computationally. It does allow idealized problems
with instantaneous speed changes ("twin paradox"type problems)
to be solved by hand. But more complicated problems, like
those with piecewiseconstant (e.g., 1 g) accelerations, or
those with arbitrary acceleration profiles, require more
extensive calculations on a computer.
I'll post an outline of the procedure as soon as time permits.
For readers who don't remember what my first prospective solution
to the symmetrical "twin paradox" problem was, here's a repeat
of that description:
A plot of twin 2's age, versus
twin 1's age, according to twin 1, starts from the origin as
a straight line (of some positive slope 1/gamma < 1). Working
from the other end (when the twins are reunited, say, at ages
40 and 40), the plot is another straight line with the same
slope 1/gamma, ending at the point (40,40). At the midpoint
(age 20 years old for twin 1), the plot jumps discontinuously
from the lower straight line to the upper straight line.
This plot is just like the plot for the standard (asymmetrical)
twin paradox (plotting the home twin's age versus the traveler's
age, according to the traveler), except that the magnitude of the
discontinuity is smaller.
But I had wanted to adapt my CADO equation to handle the symmetrical
problem so that I could obtain the plot without having to work
from both ends. (In a more complicated scenario (with multiple
speed changes), the twins would not necessarily be reunited, so
that working backward from the reunion wouldn't be possible).
That's one of the advantages of my CADO equation (for problems
in which one of the twins is perpetually inertial): it allows
the magnitude of the discontinuity to be computed directly, without
using information about the future parts of the trip.
Mike Fontenot
.
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