On Mon, May 23, 2016 at 02:25:09AM -0700,  (via tml list) wrote:
> On 22 May 2016 at 18:03, Tim wrote:
> > With tau 0.5, the relative velocity is sqrt(3/4) c.  If the velocity
> > is parallel with the jump direction, the Lorentz transformation gives
> >
> >  x' = (x - v t) / sqrt(1 - (v/c)^2)
> >
> > so that the distance is 1.99 parsecs.
>
> No, *by definition* it's 2 parsecs. Because that's what a tau of .5
> means. If it helps, that's a gamma of 2.

I know exactly what it means.  If the jump endpoints had been
simultaneous in the ship's frame, the distance would indeed have been
exactly 2 parsecs.  However, there is also a time component of a week
in the ship's frame, and that changes things slightly.  It's only
slight since a parsec is about 170 light-weeks, and the 1-week time
component in the ship's frame is small compared with the
170-light-week spatial component.

> 1.99 is due to rounding errors in your calcs.

No, there is no rounding error there.  A more precise figure is
1.98982 parsecs.

Look at the numerator in the Lorentz transformation: it includes time
as well as distance.  The 1-week time difference is rotated by the
boost transformation partially into the space axis, which accounts for
the 0.01 parsec difference from what you may have been expecting if
you ignored the time component.

The same applies to the Lorentz transformation for time, which is why
fairly small relative speeds with FTL travel can produce large enough
time variations to affect causality.  Even 1/170 c is enough to turn a
1-parsec jump into a time-reversed one; this comes from the fact that
a parsec is about 170 light-weeks.

> Just out of curiosity, what do you get for distance if the time is
> two weeks?

For a jump that is exactly 1 parsec and 1 week in the ship's reference
frame?  1.000052 parsecs in the observer's frame.

- Tim