From: Xela@yabbs
To: JasonLee@yabbs
Subject: okay...
Date: Mon Mar  7 23:18:51 1994


So... T(v)[v, 0, 0, 1] = [0, 0, 0, (1-v^2)^.5]

Measuring S and C in terms of S' and C' 1 second after the origins 
coincide, there exists a time t'' > 0 on C' such that

T(v)[0, 0, 0, 1] =  [-vt'', 0, 0, t'']  and using the same techniques as 
above, 

t'' = 1/ (1-v^2)^.5

so T(v)[0, 0, 0, 1] = [ -v/(1 - v^2)^.5, 0, 0, 1/(1-v^2)^.5]

TIME CONTRACTION: this is the basis for my point about the twins...

Say the origin of S' corresponds to the twin on the plane, and the origin 
of S corresponds to the stationary twin.  Start them off at the same 
point, so that their origins and their clocks both read zero at point of 
departure.

As viewed from S, the space-time coord. of the plane at t > 0 measured by 
C are [vt, 0, 0, t].  From S' the s-t coord. of the plane at t > 0 is 
measured by [0, 0, 0, t'].  The two sets of points describe the same 
event, so T(v)[vt, 0, 0, t] = [0, 0, 0, t']

T(v) = { 1/(1-v^2)^.5  0  0  -v/(1-v^2)^.5 }
       {        0      1  0        0       }
       {        0      0  1        0       }
       {-v/(1-v^2)^.5  0  0   1/(1-v^2)^.5 }

After matrix multiplication... t' = t(1-v^2)^.5

So the time on the plane whic h the twin experiences, t', is some fraction 
of the time, t, which the stationary twin experiences.  So the plane twni 
leaves the plane younger than the twin who stayed still.

That's about it! Phew!

-Alex