The twin paradox

[pvalev <pvalev@BAS.BG>]

I am going to draw conclusions from the simple setup below.

--- pvalev <pvalev@bas.bg> wrote:

>      The new problem. Let the traditional beam be replaced with a
> perfectly elastic ball bouncing up and down on the train, along the
> y'-axis. On the train, its speed is w' and the time of a one-way
> movement is t'. Special relativity assumes that distances
> perpendicular to the axis of relative motion are equal in the two
> frames so in the track frame the vertical, y-component of the path
of
> the ball is h=w't'. Now the Pythagorean theorem gives
>
> (w't')^2 + (vt)^2 = (st)^2                 /1/
>
> where s is the speed of the ball in the track frame (along a path
> oblique to the axis of relative motion).
>      Eq. /1/ can be transformed into
>
> t'/t = [(s/w')^2 - (v/w')^2]^(1/2)          /2/
>
> This is a general result giving time dilation for any speed
> perpendicular to the axis of relative motion in the primed frame.
> When this speed is c (the ball is replaced with a beam again and
> w'=s=c), /2/ gives the familiar
>
> t'/t = 1/gamma                               /3/


As someone really interested in SR might have found out, there is no
way to develop the result /2/ in the direction of a "non-light" clock
so we are forced to deal with light clocks only. Let us assume that
light clocks in the two reference frames do indeed show time dilation
in accordance with /3/. The result /2/ can be used to show that,
despite the assumption, nothing at all can be said about the rate of
processes in the two frames - whether the twin on the spaceship will
be younger or older cannot be resolved within SR.
     Let the train be at rest initially, and two identical trees
grow - one on the train, the other on the ground. Then at t'=t=0
(Event 1) the train starts moving with a speed v along the x-axis. On
the train, the tree grows with a speed w'. As it reaches the ceiling
(Event 2) the times are respectively t' and t. Eq. /2/ can be
transformed into

t'/t = w/w'                                     /4/

where w is the vertical component of the speed of growth of the tree
ON THE TRAIN, in the track frame. (In the track frame, the tree on
the train grows obliquely to the x-axis).
     On the other hand, the height of the tree on the ground is
w_o*t, where w_o is oviously the speed of growth of that tree in the
track frame. So we have:

wt = w't' : height of the tree on the train

w_o*t  :  height of the tree on the ground

Now there is absolutely no information in SR about the relation
between w_o and w. So one cannot say which tree is taller. Of course
we can advance more or less reasonable assumptions about w_o and w,
but these would not be assumptions advanced by SR.
     The most reasonable assumption is, of course, w=w_o. If so, this
would mean that the growths in the two frames are identical. True,
light clocks in the train have made a smaller number of
ticks, but this is compensated by the fact that w' is larger than
w_o, according to /4/.

Pentcho Valev