pccx@ihlpg.UUCP (p cunetto) (06/17/85)
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Several responses to the twin paradox have been posted that correctly
note the situation is not symmetrical with respect to the two twins,
and therefore there is no paradox.
Q *
| \
| \
| \
| \
| * R
| /
| /
| /
| /
P *
Twin A who follows PQ is in an inertial frame the entire "journey",
while twin B who follows PRQ is not. He accelerates at R
(he need not accelerate at P or Q, they can compare ages in passing).
However, the fundamental cause of the age difference when they reunite at Q
is NOT the acceleration at R. Any age difference caused by the acceleration at
R can be made an arbitrarily small fraction of the total age difference
merely by making the PR and RQ legs sufficiently long.
So the real cause of the age difference must be sought elsewhere.
Your don't have to look to General Relativity for that cause, it can be
found in Special Relativity.
The "cause" is the wrong way triangle inequality of Minkowski geometry.
Namely, if R is in the future of P, and Q is in the future of R, then
d(P,Q) >= d(P,R) + d(R,Q)
where d(x,y) is the interval (not the spatial distance) between x and y.
(Outline of the proof follows below)
For an observer at rest in an inertial frame (viz. twin A between P and Q,
and twin B between P and R and between R and Q), the interval between
two events on his world line is just the elapsed time between them.
Therefore, the wrong way triangle inequality says:
The elapsed time from P to Q for A is greater than the total
elapsed time from P to R to Q for B, i.e, A is older than B at Q.
As stated above, the age difference caused by the acceleration at
R can be made arbitrarily small compared to the age difference "caused"
by the wrong way triangle inequality.
So it is the geometry of spacetime and the physical significance of
the interval that causes the age difference, not the acceleration.
------------------
Wrong way triangle inequality. (2 dimensions)
* Q (x3,t3)
| \
| \
| \
| \
| * R (x2,t2)
| /
| /
| /
| /
P * (x1,t1)
Given that R is in the future of P, i.e., x2-x1 <= c(t2-t1)
and that Q is in the future of R, i.e., x3-x2 <= c(t3-t2)
show d(P,Q) >= d(P,R) + d(R,Q)
where d(P,Q) is the interval between P and Q: i.e.
d(P,Q) = (c**2)(t3-t1)**2 - (x3-x1)**2
Outline of the proof:
Expand d(P,Q) >= d(P,R) + d(R,Q) and cancel like terms on each side.
The resulting inequality (call it [1]) is what we are going to prove.
Multiply corresponding sides of the two inequalities given in the
assumptions to get a single inequality. Multiply it by -1, rearrange
terms, and you get [1]. QED
Phil Cunetto