"Paul B. Andersen" <paul.b.andersen@hia.no> wrote in message
news:bksuop$70v$1@dolly.uninett.no...
>
> "Androcles" <jp006f9750@blurbblueyonder.co.uk> skrev i melding
news:FShcb.289$O25.268@news-binary.blueyonder.co.uk...
> >
> > "Paul B. Andersen" <paul.b.andersen@hia.no> wrote in message
> > news:bks41d$sov$1@dolly.uninett.no...
> > [snip as read]
> > > 2.  Androcles thinks that the transform:
> > >      tau = (t+vx/c^2)/sqrt(1-v^2/c^2)
> > >      implies some kind of "time contraction" as opposed to "time dilation".
>
> > Yep. Care to show why it isn't?
>
> Happy to oblige.
> We consider a stationary clock in the "Latin frame",  x = constant.
> This clock is obviously moving in the "Greek frame",
> and we can answer the question:
>
>   At which rate  dt/dtau is the _moving_ clock observed to run
>   at in the "Greek frame"?
>
> by differentiating:
> dtau/dt = d/dt ((t+vx/c^2)/sqrt(1-v^2/c^2)) = 1/sqrt(1-v^2/c^2))
> thus:
> dt/dtau = sqrt(1-v^2/c^2)
>
> This is good ol' "time dilation".
>
> We might also consider a stationary clock in the "Greek frame",
> and ask the question:
>
>   At which rate  dtau/dt is the _moving_ clock observed to run
>   at in the "Latin frame"?
>
> (The simplest would of course be to use the inverse transform,
>  but let's use the one above.)
> Let us assume the clock is at xi = 0.
> This clock will be moving in the "Latin frame", x = -v*t.
> tau = (t+v*(-v*t)/c^2)/sqrt(1-v^2/c^2) = t*sqrt(1-v^2/c^2)
> thus:
> dtau/dt = sqrt(1-v^2/c^2)
>
> This is good ol' "time dilation".
>
> Maybe you can explain where you see your "time contraction"?
>
> Paul

Glad to oblige.
First, we set w = -v, so that the "Greek frame" moves in the opposite
direction.
 We consider a stationary clock in the "Latin frame",  x = constant.
 This clock is obviously moving in the "Greek frame",
 and we can answer the question:

   At which rate  dt/dtau is the _moving_ clock observed to run
   at in the "Greek frame"?

 by differentiating:
 dtau/dt = d/dt ((t+wx/c^2)/sqrt(1-w^2/c^2)) = 1/sqrt(1-w^2/c^2))

 then we substitute for w its value. (w^2 is of course v^2),
 and we note that v has a negative slope.
 Hence
 dtau/dt = d/dt ((t-vx/c^2)/sqrt(1-v^2/c^2)) = -1/sqrt(1-v^2/c^2))

 thus:
 dt/dtau = -sqrt(1-v^2/c^2)

This is the  infamously incomprehensible and not understood "time
contraction".
Need I go on?
Androcles