sci.physics.research

Jacques,

On Feb 17, 10:54am, jacques <...@> wrote:
> > Notice also that this solution does not describe the situation when
> > the 2 rockets are accelerating, but the result of such situation when
> > freezed..

Freezing the situation is not really necessary.

> But I guess that if from this status, the two rocket now deccelerate
> at the same rate for the same time this stretch would be cancelled
> which shows some antisymetry in the process depending on relative
> directions of motion and acceleration, which looks quite odd in SR
> (motion is usually considered as not absolute).

Look at /2007/10/,
maybe it will make the whole thing a little more intuitive.
If you consider the diagram at the bottom of the blog post (direct
link for the diagram:
/_UnfX9_V6JCc/RwxwoODXZJI/AAAAAAAAAAk/SkOsruyghSs/s1600-h/Bells+),
it shows the equivalent "paradox" in the Euclidean case. Hopefully
that makes things clearer.

In the diagram, the equivalent of deceleration is when the green and
red curve both become horizontal again, meaning that the distance
between them returns to its original value.

> I find this disymmetry
> between distance and time intriguing and I wonder how physical is a
> distance (therefore this stretch) in Relativity.

The stretch is as physical as the loss of tension you would get if you
put a piece of string with one end on the red curve and one end on the
green curve along one of the arrows in the Euclidean case. In reality,
the string will follow a slightly more complicated curve than the
straight lines I drew, since it will be perpendicular to the local
time line at every point, but the difference in length is really minor
and can be ignored in first approximation.