Re: The Twin Paradox explained from the moving twin ?
- From: alanmc95210@xxxxxxxxx
- Date: 17 Sep 2006 13:54:08 -0700
Nicolaas Vroom wrote:
Question:
Is it possible to explain the Twin Experiment from the moving point of view
?
IMO this is not possible.
(cut)
Assume Alpha Centari is exactly 4 light years away, and one twin is
traveling
there at 4/5 speed of light. (Using a 3,4,5 triangle I avoid
irrational numbers in my computations.
Traveling at 4/5 the speed of light, from the point of view of the
stay at home twin, the trip will take 10 years, 5 years there, 5 years
back.
Time for the traveller T' = T( sqrt( 1- (v^2/c^2))) = 3/5 T
Likewise, the distance for the traveler, D' = 3/5 D
The traveler on the spaceship sees himself traveling a distance of
4*3/5 = 2 2/5 light years in a time of 3 years, and likewise the 2 2/5
light years back
in a time of 3 years, so the traveler will see the trip as lasting 6
years.
Say the twins have super telescopes and can see each other throughout
the trip. As long as
they are receeding from each other, they will SEE each other as moving
at ((1 - 4/5)/(1+ 4/5))^(1/2) = 1/3 normal speed. When they
approach each other, they will
see each other as moving at 3 times normal speed.
((1 + 4/5)/(1 - 4/5)^1/2
As long as they are traveling apart, the twins will see each other as
aging at 1/3 speed. As long as they are traveling towards each other,
the twins will see each other as aging at triple speed. As long as they
are travelling
apart, they will compute each other's time as slow, they will SEE each
other age even slower than the relativistic time, because of the extra
time it takes light to reach them.
When they move towards each other, they will compute each other's
time as running slow but they will SEE each other as aging fast because
of the continually shrinking amount of time it takes light to reach
them.
The difference is, the traveling twin will see the stay at home twin
as aging at 1/3 speed for the 3 years to Alpha Centauri, for a total
of 1 year,
and at triple speed for the 3 year trip back to earth = 3*3=9.
The traveling twin will see the stay at home age 1 year during the
trip out, and 9 years during the trip back, for a total of 10 years.
The stay at home twin will see the travel age at 1/3 speed for 9
years, the 5 years it takes the traveler to get to Alpha Centauri,
plus the 4 years it takes the light to get back to earth. Since the
total trip will take 10 years, the stay at home twin will see the
traveler age at triple speed during the one year he observes the
traveler coming back to earth. The earth observer sees the traveler
age at 1/3 speed for 9 years, for a total of 3 years, and at triple
speed for 1 year, for another 3 years, giving 6 years for the round
trip.
Both observers see each other aging at the same slow rate while moving
apart, they see each other aging at the same fast rate while moving
together. The difference lies in one observer deliberately changes
the relative motion of his rocket from moving away from earth to
moving towards earth, and the other observer remaining passive, and
not seeing the change until the light from the
traveler reaches earth. If the earth could be accelerated like a
rocket ship, and the earthbound observer decided to change his frame
so the rocket appeared to be moving towards him at 4/5 lightspeed
rather that away at 4/5 lightspeed, while the rocket remained in
motion past Alpha Centauri, then it would have been the earth twin who
appeared to age less.
Of course you could have some intermeditate situation where BOTH
observers decide to change their relative motion before they see the
other observer change his motion.
.
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