# 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

stay-at-home 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 piecewise-constant (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

.

**Follow-Ups**:**Re: A Symmetric Twins Conundrum***From:*shuba

- Prev by Date:
**Re: Classical and modern physics** - Next by Date:
**Lagrange in GR ?** - Previous by thread:
**Re: A Symmetric Twins Conundrum** - Next by thread:
**Re: A Symmetric Twins Conundrum** - Index(es):