Re: How is the relativistic equation L_r=L_0/gamma derived?

The contraction of a moving rod is most easily shown from the Lorentz 
transformation relating x',t' in the system S' in which the rod is at rest 
and x,t, the space-time coordinates in the system S in which the rod is 
moving with velocity v.  The Lorentz transformation can be written as 
x'=g(x-vt), t'=g(t-vx) in units where c=1.   I have used g to represent 
what is usually called gamma=1/sqrt{1-v^2}>1.  Note that, if x'=const., 
corresponding to something at rest in S', then dx/dt=v.  (If you don't 
know calculus, just disregard the previous sentence.)   To measure the 
length of a rod at rest in L', you measure one end (x1') and then the 
other end (x2').  The length of the rod is then L'=x2'-x1'.  Since the rod 
is at rest in S', you can casually walk from one end to the other in 
making your measurement.  They don't have to be made at the same time for 
a rod at rest.  The key thing that distinguishes the system in which the 
rod is moving from that in which it is at rest is that you MUST measure 
each end at the SAME time if the rod is moving.  You would get nonsense if 
you measured x1 and x2 at different times for the moving rod.
This means that you must have t1=t2, while t1' and t2' needn't (and won't) 
be equal.  So you can write L'=x1'-x2' =g(x1-vt)-g(x2-vt), where t must be 
the same in each term.
Subraction gives L'=g(x1-x2)=gL.  This is usually written as L=L'/g 
showing that a correct measurement of the moving rod will give a smaller 
answer than a measurement of the rod at rest.