From: hinson@maxwell.physics.purdue.edu (Jason W. Hinson)
Newsgroups: rec.arts.startrek.tech
Subject: Re: FTL travel/communication
Date: 14 Aug 92 14:52:25 GMT
Organization: Purdue University Physics Department

In article <1992Aug13.140545.28418@jyu.fi> tm@jyu.fi (Timo Mannikko) writes:
>
>>I wrote the following message to someone who asked me to explain why FTL
>>travel/communication impies a causality problem, and I thought it might
>>be of some interest to readers of r.a.s.tech:
>   [Message deleted]
>This was a very good explanation, thanks!
>
>>Hope this was helpful... feel free to ask other questions
>>
>>-Jay
>
>I will! Namely, I have always wondered one thing in this so called twin
>paradox. You know, twin A stays home, twin B makes a trip with velocity
>c minus epsilon, and returns home younger than A. Everything OK so far.
>But: From B's point of view it is A who has moved, so shouldn't A be the
>younger brother?
>
>First I thought that it is the acceleration which makes the difference.
>But then again: Let us suppose that B accelerates with constant one g,
>which is the same acceleration as A is experiencing. So, where *is* the
>difference?
>- Timo

Good question.  First, to say they are experiencing the same thing because
they are both undergoing the same acceleration (g) is not quite right. One
of them is undergoing g due to gravity which doesn't make them move any
faster or slower with respect to anything.  The other is being accelerated
so that he moves at a different velocity with respect to the earth...
see how that makes a difference.

To answer the question, let me introduce you to space-time diagrams 
(this is my first try at drawing this with ascii characters)
To make it easier, I will show the trip out and the trip back in two 
seperate diagrams.  (Note: v = 3/5 c)
                               !
(the trip out)                 ! (the trip back)
time                           ! time
   ^                           !    ^         
   |              time'        !  10|8
   |            +              !    |*          
   |           +    space'     !    | *          
   |          +   '            !    |  * 
   |         +  '              ! 6.8|,  *
   |        + '                !    |  , * 
   |       +'                  !    |    ,* 
  5|      *4                   !   5|      *4
   |    '*                     !    |       +,
   |  ' *                      !    |        + ,
3.2|'  *                       !    |         +  ,space'
   |  *                        !    |          +
   | *                         !    |           +time'
   |*                          !    |
---+--------------->space      ! ---+--------------->space

OK, and now the explanation... Person A sits on earth and follows the
straight up and down time line with the horizontal space line.
Person B follows the time' line with the space' line as shown.
    "A" sees B traveling away from him at first (the diagram on the left).
If this isn't clear, look where A's clock reads 5. Follow A's space line
at that point (you will be looking directly to the right of the 5). At that
point, you have to move through A's space line a bit before you hit 
the place where B is at that point. So, after 5 years, B has moved a 
particular distance from A.    
    Now, When A's clock reads 5, he thinks B's clock reads 4 (If you follow
A's SPACE line horrizontally over to B's time line, you will see I have
written 4).  HOWEVER, B will look back through HIS space line (space' on the
diagram to the left) and will say that A's clock reads 3.2 as I have shown.
(Each think the other' clock is running slower).
    BUT, look what happens when B turns to head back home (the diagram
on the right).  After he turns, his space frame is shifted. SO, now he 
looks back through his new space' line and sees that A's clock reads
6.8.  So, in the relativly small amount of time it takes B to turn back
for home (here we have assumed it is an instantanious turn) B will see
A's clock speed up to go from 3.2 to 6.8 in a FLASH.  HERE is where
the difference occurs, during that acceleration.
    So when B finally gets home, his clock reads 8 years and A's clock
reads 10 years.

By the way, the effect of such an acceleration will become greater the
farther B travels away from A. (examine the diagram and see if that 
becomes clear).  SO when he accelerates to leave earth, the effect
he sees on his brother's watch won't be that great.

ANYWAY, hope this helps to clear things up a little.

-Jay