On 22 May 2016 at 18:03, Tim wrote:

> On Fri, May 20, 2016 at 06:03:52PM -0700,  (via tml list) wrote:
> > But only the "it took two weeks" observers would say that the ship
> > had a tau of 0.5   :-)
>
> I did make an error in my previous post, but not that great of one.
>
> 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.

1.99 is due to rounding errors in your calcs.

> However, the time interval they
> observe between the ship going into jump and coming out is
>
>  t' = (t - vx/c^2) / sqrt(1 - (v/c)^2),
>
> which is -293 weeks.  If the relative velocity is in the opposite
> direction, the distance observed is 2.01 parsecs and time +297 weeks.

Ooo-kay. thanks for the calc.

> A very simple sanity check can be made without involving the Lorentz
> transformations: A relative velocity corresponds to a rotation of
> spacetime axes, preserving the x^2 - (c t)^2 invariant interval. This
> is similar to Euclidean rotations preserving Euclidean distance x^2 +
> y^2, except with a conversion factor of c between space and time, and
> a different sign.
>
> So if in one frame, it emerges at a point 1 parsec away and 1 week
> later, that's an invariant interval of extremely close to 1 parsec
> (it's about 0.99998 parsecs).  If some other observer sees the
> distance as being 2 parsecs, then they must see the time as being
> +/-295 weeks to preserve the invariant interval.  2 parsecs and 2
> weeks is not possible.

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

--
Leonard Erickson (aka shadow)
shadow at shadowgard dot com