Re: How does the Lorentz transformation for time hold up to relativity?

The Lorentz transformation for time IS the relativistic transformation for 
time.
The Galilean principle of relativity says that it is not possible to 
determine an 'absolute' velocity. That is, if you are in a space-craft 
with no windows moving at a constant speed, you would have no idea what 
speed you were moving at. Newton's laws obey this principle.
The problem came with the development of Maxwell's laws of 
electromagnetism (in 1865). These showed a finite speed of light, c (= 
300,000,000 metres per second). This, together with Newton's laws, would 
imply that a car moving at speed u would see light to go at a speed c-u, 
and thus, by the speed of light in your window-less space rocket, you 
could determine your speed. However, experiments showed that this was not 
the case.
Consider the following thought experiment, but when describing it, let us 
assume that the principle of relativity holds. That is, the speed of light 
is the same for all observers.

MIRROR _____ C___________________
            /\              |
           /  \             |  - Height L
          /    \            |
LIGHT    /      \           |_
SOURCE: A       B: DETECTOR

Length AB=d, Length AC=h. Speed of light=c

If I was stationary (with respect to the apparatus) and watching this 
experiment, you would see the light to move a distance of 2h between A and 
B. Therefore, the time taken 9from my point of view) must be t=2h/c.
But what if you were moving in the direction AB at speed v, such that you 
are at A when the light leaves A, and at B when the light reaches B. To 
you, the experiment would look like this:
_____C_____
     |
     |
     |
     |
    A,B
So, you see the light to move a distance of 2L. So your time must be 
t'=2L/c.
But from the Pythagorean theorem, (d/2)^2+(L)^2=(h)^2.
BUT, t=2h/c => h=ct/2; t'=2L/c => L=ct'/2. And (from my point of view), 
you were moving at speed v, and covered the distance d in time t, 
therefore d=vt.
So, (d/2)^2+(L)^2=(h)^2 => (vt/2)^2+(ct'/2)^2=(ct/2)^2
			=>     (c^2-v^2)(t^2)=(ct')^2
                        =>                  t=1/[[1-(v/c)^2]^.5]
This is the Lorentz transformation for time. It was developed so that 
Maxwell's laws remained unchanged when worked out for a moving observer, 
i.e. so that the speed of light appears the same to all observers. It was 
discovered by Lorentz, but Einstein is the one who decided to transform 
Newton's Laws in this way. Mass, and distance, are also transformed in the 
same way, yielding results that quite a few people have heard of, e.g. E=m
(c^2).

Let's try out this formula on a problem:
Imagine that an astronaut goes off at 0.8c, relative to the Earth, and 
returns after 30 years have elapsed on the rocket. We have v=0.8c, and 
t'=30 years. Therefore,   
                       t= 30/[[1-(0.8)^2]^.5]
                        = 30/[1-0.64]^.5
                        =30/0.6
                        = 50 years have gone by on Earth!
Special relativity is quit difficult to get your head round (it made 
Einstein's hair stand on end). Well done for trying. I reccommend the 
following books:
'Introducing Einstein' (formerly 'Einstein for Beginners') by Joseph 
Scwarz and Michael McGuiness.
'Six Not-So-Easy Pieces,' a collection of Feynman's brilliant lectures.
'E=mc^2, an Equation That Changed the World' (I can't remember who it's 
by, but it is very good).