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Re: starship-design: FTL idea



Timothy van der Linden wrote:
>Isaac,

BTW, I wrote this FTL example once a while ago, but I didn't keep records
of my calculations.

>>- - - - - - - 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
>>echo whatever he receives from his Foo Radio receiver.

>I guess that should be ship B.

It's worded correctly, but confusingly.  It is ship _A_'s
receiver, but of course this receiver is set up to receive
Foo messages from 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.

It's very difficult to explain why it moves backwards in time without
space-time graphs.  ASCII art wouldn't suffice for this example.

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

Of course.  As I said.

>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))

1.8=(.9-1.45)/(1+(.9*(-1.45)), whatever the significance of this
equation.  You have to be very careful about whether you're looking
at a beam sent in the same direction or opposite to the direction
ship B is moving in.
-- 
    _____     Isaac Kuo kuo@bit.csc.lsu.edu http://www.csc.lsu.edu/~kuo
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