Re: Controversy over Twin Paradox

From: Jon Bell (jtbellj3p_at_presby.edu)
Date: 07/24/04 

Date: Sat, 24 Jul 2004 01:06:49 +0000 (UTC)

In article <c22a881.0407230153.15303e83@posting.google.com>,
Amol <amol12882@yahoo.co.in> wrote:
>
>Therfore the explanation can be given from both point of view, hence
>the age of both the persons should be same, because time is relative.

Yes. Here's a numerically worked-out example, using the Lorentz
transformation equations, that shows that both the traveller and the
stay-at-home must agree on the elapsed time shown by each person's clock,
even though in each person's reference frame, the other clock appears to
"run slower."

The scenario: You stay behind on Earth while your twin goes on a
round-trip space journey to Star Base Alpha, which is 4 light-years away.
His ship can travel at a speed of 0.8c, so he needs 5 years for the
outbound trip and another 5 years for the return trip. The total trip
duration is 10 years by your reckoning.

Relativity predicts that your twin experiences less elapsed time
because of time dilation:

   (5 years) * sqrt (1 - 0.8^2) = 3 years

for each leg of the trip, and from his point of view the round trip lasts
only 6 years.

The relativistic time dilation equation predicts that each twin's clocks
"run slower" in the other twin's reference frame. So why can't your twin
conclude that the trip must be shorter for you, than it is for him? The
answer lies in the fact that your experiences are not symmetrical. Your
twin is at rest in two different inertial reference frames, one during the
outbound trip and another one during the inbound trip. You remain at rest
in a single inertial reference frame during the entire journey. Your twin
has to fire his spaceship's engines at the turnaround point. You do
nothing.

To convince ourselves that both you and your twin do in fact agree that
the elapsed time for the trip is longer on your clock than on his, we need
to look carefully at how the two clocks behave in your twin's reference
frames (note plural). In order to do this properly, we need to use the
full Lorentz transformation equations, and not just the length-contraction
and time-dilation equations, which don't contain the "full story" of
relativistic kinematics.

Let's call your reference frame S, in which an event has coordinates (x,
t); let's call your twin's reference frame during the outbound trip S', in
which an event has coordinates (x', t'); and finally, let's call your
twin's reference frame during the return trip S", in which an event has
coordinates (x", t").

To simplify the analysis, let's assume that your twin's ship can
accelerate so rapidly that it can effectively change velocity
instantaneously.

If S and S' are set up so that at the instant when your twin departs
(moving in the +x direction with speed v), both you and he are at x = x' =
0, and your clock and your twin's clock are synchronized so that they both
read zero, we can use the usual textbook version of the Lorentz
transformation equations

     x' = g*(x - v*t)
     t' = g*(t - v*x/c^2)

or inversely

     x = g*(x' + v*t')
     t = g*(t' + v*x'/c^2)
     
where g = 1 / sqrt (1 - (v/c)^2), i.e. what most books call "gamma." In
our example, v = 0.8 light-year per year, and c = 1 light-year per year,
so g = 1.6667 = 5/3.

Your twin travels at a speed of 0.8c, so in your frame, S, he takes 5
years to cover the 4 light-years to Star Base Alpha, where he turns
around (instantaneously). In frame S, his turnaround takes place at
x = 4 light-years and t = 5 years.

We calculate the position and time of the turnaround in *his* frame as
follows:

    x' = (5/3)(4 - 0.8 * 5) = 0 (he's stationary in S')
    t' = (5/3)(5 - 0.8 * 4) = 3 years

So, during the twin's outbound trip (5 years according to you), 3 years
elapse on his clock. That is, in your frame, S, his clock runs slower
than yours, by a factor of g = 0.6.

Now, how does *your* clock behave in *his* frame, S'?

First, let's calculate your position in his frame at t' = 3 years. In S',
you are moving "backwards" (-x' direction) at speed 0.8 light-years per
year. So at t' = 3 years, you are located at x' = -(0.8)(3) = -2.4
light-years.

We can now calculate what your clock (which shows "S time") reads at this
point in time in your twin's frame, S', using the inverse Lorentz
transformation:

    t = (5/3)(3 + (0.8)(-2.4)) = 1.8 years

So, during your twin's outbound trip (3 years according to him), 1.8 years
elapse on your clock, in his frame. That is, in his frame, S', your clock
runs slower than his, by a factor of g = 0.6.

Now, your twin fires his rocket engines to turn around very quickly so
that he is now moving towards you with speed 0.8 light-years per year.
He is now in a new reference frame, call it S".

We need a Lorentz transformation between S and S". We can't simply
change primes to double primes in the equations that we used before,
because S and S" do not coincide at x = x" = 0 and t = t" = 0. We have to
use a more general version of the Lorentz transformation:

     x" - x_0" = g*[(x - x_0) - v*(t - t_0)]
     t" - t_0" = g*[(t - t_0) - v*(x - x_0)/c^2]

Inverse:

     x - x_0 = g*[(x" - x_0") + v*(t" - t_0")]
     t - t_0 = g*[(t" - t_0") + v*(x" - x_0")/c^2]

where some event (call it a "reference event") has coordinates x_0 and
t_0 in frame S, and coordinates x_0" and t_0" in frame S".

I don't think I've seen these written down anywhere before, but I'm
99.999% sure that they must be correct, because they simply incorporate a
translation or shift of the origin of coordinates for both frames. Note
that if we set x_0 = x_0" = 0 and t_0 = t_0" = 0, we get back the original
textbook version of the Lorentz transformation.

For this case, let the turnaround of your twin's ship be the reference
event. Then

     x_0 = 4 light-years
     t_0 = 5 years
     x_0" = 0
     t_0" = 3 years
     v = -0.8c (because now the ship is moving in the -x direction in S)
     
and the transformation equations between S and S" are

     x" = (5/3)[(x - 4) - (-0.8)(t - 5)]
     t" - 3 = (5/3)[(t - 5) - (-0.8)(x - 4)]

     x - 4 = (5/3)[ x" + (-0.8)(t" - 3)]
     t - 5 = (5/3)[(t" - 3) + (-0.8)x"]

In your twin's new frame S", just after the turnaround, you are at the
same position as just before the turnaround, because we're assuming that
the twin can turn instantaneously, so your coordinates in S" are x" = -2.4
light-years and t" = 3 years. In your frame, S:

     x - 4 = (5/3)[-2.4 + (-0.8)(3 - 3)]
     t - 5 = (5/3)[(3 - 3) + (-0.8)(-2.4)]

which gives x = 0 (you're still at your own origin) and t = 8.2 years.

Therefore, in your twin's first reference frame, just before the
turnaround, your clock reads 1.8 years; and in his second reference frame,
just after the turnaround, your clock reads 8.2 years. Apparently your
clock has gained 6.4 years during the turnaround! But this has no
physical significance as far as your clock is concerned. It's simply
because your twin is measuring space-time using a different coordinate
system. It's analogous to taking the axes on a two-dimensional graph on
paper and rotating them quickly. A point's coordinates change suddenly,
but nothing actually happens to the point itself.

The return trip is similar to the outgoing trip as far as elapsed time is
concerned. In the classic words of textbook writers :-), "the details are
left as an exercise for the reader." 5 years elapse in your frame, S, and
the total elapsed time on your clock, according to you, is 5 years (time
at turnaround) + 5 years = 10 years. Meanwhile, in your twin's frame, S",
1.8 years elapse on your clock, so according to your twin, the total
elapsed time on your clock is 8.2 years (time immediately after
turnaround) + 1.8 years = 10 years.

The two of you agree that 10 years have elapsed on your clock. Likewise,
you both agree that 6 years have elapsed on his clock. The fact that each
of your clocks "runs slow" compared to the other one, during each leg of
the trip, does not lead to an actual physical contradiction when the
clocks are re-united at the end of the trip.

I have tried to avoid making any reference to what you and your twin
actually *see* the other clock doing during the trip, because then we have
to take into account the fact that light (by which we see) travels at a
finite speed. For an analysis of the "paradox" from this perspective, see

<http://www.google.com/groups?as_umsgid=bnq8m5%24i2i%241@jtbell.presby.edu> 

-- 
Jon Bell <jtbellm4h@presby.edu>                     Presbyterian College
Dept. of Physics and Computer Science        Clinton, South Carolina USA

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