# Re: starship-design: FTL idea

```Isaac,

>- - - - - - - 8< - - - - - - - cute here - - - - - - - 8< - - - - - - -
>For instance, let's say a particular gizmo, "The Foo Radio", can
>transmit (and receive) a morse code message at 1.8c (relative to
>the frame of reference of the gizmo).
>
>What you need to send a message backwards in time is two
>space ships travelling very fast apart from each other (for
>instance .9c).  Both have Foo Radios pointed at each
>other.  Space ship A has his Foo Radio set up to simply

I guess that should be ship B.

>For simplicity, assume I am talking about ship A's frame of
>reference (and a clock on ship A) unless specified otherwise.
>
>Let's say ship A decides to send a message to ship B when
>ship B is .9 light hours away; the time is 10:00.  Ship
>B will receive the message at 11:00 when ship B is 1.8
>light hours away.  Ship B echoes the message back towards
>ship A at 1.8c in _its_ frame of reference.  You need to
>do a Lorentz transformation to figure out where/when this
>message goes in A's frame of reference.  The result is
>that in A's frame of reference, ship B's Foo Radio beam
>moves at 1.42 c _backwards_ in time!

How nice it would have been if you'd not left out the calculations for this
transform... Now I still don't see why it goes backwards in time.

BTW.
Ship B echoes the message back towards ship A at 1.8c
in _its_ frame of reference.

That means it is moving at only 1.03053c in the frame where B moves with 0.9c
-1.8=(0.9-1.03053)/(1+(0.9*-1.03053))

So for A the beam has approaches with:
(0.9+1.03053)/(1+(0.9*1.03053))=1.001584c

(I haven't the faintest idea how you get 1.42)

Timothy

```