Time Dillation Loophole
Knappoleon
I have a question about time dilation, but more specifically the twin
paradox. Keep in mind that I have a very basic understanding of relativity.
Now, in the twin paradox a one twin gets into a spaceship and zooms away
from earth at near the speed of light. Years later, the twin returns only to
find that everyone else has grown old and the world he once knew is gone. This
has all been proven according to Einstein's Theory of Special Relativity.
There is one problem here, though. All motion is taken to be relative to the
twin on earth.
If motion is taken to be relative to the twin in the spaceship there is a
whole other side to this paradox. To the twin in the spaceship he is not
moving at all; it is the twin back on earth who is zooming away at near the
speed of light. Then, when the twin returns to earth (or, if we are assuming
that motion is relative to the twin in the spaceship, when the earth returns to
him) the twin on earth will have stayed the same while the twin in the ship
will have probably been long dead.
Can somebody please explain this to me? If you can, send your reply to
knapp...@aol.com.
Thanks,
Julius (Jules) Knapp
Edward Schaefer
It's time to post this baby again:
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To solve this "paradox", let's first set the stage with two twins.
One goes off the Alpha Centauri @ .866c, gets there after
5 years (Earth coordinates), and then turns around and comes back
@ .866c. The travelling twin will return after 10 years (Earth
time) but will have aged only 5 years. [sqrt (1 - [.866c]^2/c^2) =
1/2 is why I chose v = .866c to begin with.]
I have just described things from the standpoint of the stay-at-home
twin. Note that the space and time covered by the spaceship twin is
the same in both the outbound and inbound legs. It is commonly
assummed that the distance covered by the stay-at-home twin wrt the
spaceship twin is therefore also the same in both the outbound and
inbound legs of the trip. This is not the case. In order to come
back, the spaceship twin had to accelerate and move himself from one
inertial frame-of-reference to another. In the process, the
position of the stay-at-home twin got pushed away in space and back
in time. As a result, the spaceship twin observed the stay-at-home
twin to cover 20 years of time, with his clock ticking at half-speed
all the while.
So let's do some math.
At turnaround, the spaceship twin has been outbound for 2.5
years by his clock. He looks back, and sees his stay-at-home
cousin as he was at time t_o, where
t_o = t_n / (1 + v/c)
and t_n is the current time for the spaceship twin (2.5 yr.)
Therefore t_o = 2.5 / (1 + 0.866) = 1.34yr; and at that time the
stay-at-home twin was 1.16 light-years away in the frame-of-reference
of the spaceship twin.
Now the spaceship twin fires his engines to change his relative
motion from v = 0.866c to v = -0.866c in the frame-of-reference
of the stay-at-home twin. The relativistic addition of velocities
states that his total change-of-velocity in the frame-of-reference
of the outbound leg is -0.990c. So we will now use the Lorentz
Transformations to determine the spaceship twin's view of spacetime
at turnaround in the frame of reference of the inbound leg of the
trip. In this case, the origin of coordinates is placed at
turnaround.
x' = (x - vt) / sqrt (1 - v^2/c^2)
= {-1.16 ly - [(-.990 ly/y) * (-1.16 y)]} / sqrt (1 - 0.99^2)
= -16.16 ly
t' = -16.16 y
This is merely a coordinate shift caused by the spaceship twin
changing frames of reference. Physically, the stay-at-home twin
is in the same place and the same time as before turnaround.
To get together again, the stay-at-home twin is observed to come
to the spaceship twin @ .866c from (x' = -16.16, t' = -16.16).
This takes 18.66 years, placing the event at t = 2.5 y in the
frame-of-reference of the travelling twin on the inbound leg,
which is EXACTLY what we want, and results in the spaceship
twin locally observing the passage of a total of 5 years of time.
(Remember: t=0 is at turnaround in this frame-of-reference.)
At the same time, the stay-at-home twin was observed to cover
1.34 + 18.66 = 20 years of coordinate time in the frames of reference
of the spaceship twin, with his clock ticking at half-time all the
while. Therefore the spaceship twin observed the stay-at-home twin
to age 10 years.
End of "paradox".
EMS
P.S. Dill-ation? It would seem that you are more into spices than
Relativity. :-)
John Wasser
the "To," "Cc," and "Newsgroups" headers for details. ]]
In article <19990420211647...@ng-fv1.aol.com>, Knappoleon
<knapp...@aol.com> wrote:
> I have a question about time dilation, but more specifically the twin
> paradox.
> If motion is taken to be relative to the twin in the spaceship there is a
> whole other side to this paradox.
That threw me, too, until I remembered the contraction.
For simplicity, assume a relative speed that dilates time (and
contracts space) by half.
From the point of view of the twin on earth, the twin on the ship
travels for 10 years to a distant star and turns around and takes
another 10 years to get back. The ship looks contracted while
moving (only half its usual length) but that is to be expected.
The twin on earth is now 20 years older.
From the point of view of the twin on the ship the earth moves
away at a great speed. The universe contracts so the distant
star is only half as far as it appeared from earth. The star,
approaching from only half as far, arrives in 5 years. The star
(and the rest of the universe) reverse direction and head back the
other way. The earth aproaches the ship again in 5 years and stops.
The twin on the ship has aged 10 years.
Frank Wappler
Knappoleon wrote:
> Now, in the twin paradox a one twin gets into a spaceship
> and zooms away from earth at near the speed of light.
And, as the twin in the spaceship might describe it:
the other twin zooms away from the spaceship, together with earth.
> Years later, the twin returns only to find that everyone else
> has grown old and the world he once knew is gone.
Yes, that's one possible outcome of such an experiment.
Another outcome might be that spaceship and earth meet again,
and earth finds the spaceship older ("than expected").
> This has all been proven according to Einstein's
> Theory of Special Relativity.
There's no way to "prove" trial by trial experimental outcomes.
What Einstein prescribed were procedure how to _measure_
which of the two is older and which is younger, trial by trial.
Using their measurement and the "principle of extremal action"
they can express "the most probable potential" in the region
through which they travelled separately.
They can determine that one of them (who's measured "younger") was
reflected by a repulsive potential (such as "a wall", for instance),
while the other (who's measured "older") was not, or at least less so.
> There is one problem here, though.
> All motion is taken to be relative to the twin on earth.
> To the twin in the spaceship he is not moving at all [...]
Distance, velocity, acceleration are measured by the twins only wrt.
each other. That's right, but that's not a problem. They'll measure
their one "age difference" that is specific to this trial anyways;
and derive _from it_ which twin was reflected, and which twin was not.
Best regards, Frank W ~@) R
p.s.
> Can somebody please explain this to me?
> If you can, send your reply to knapp...@aol.com.
Copy sent by e_mail.