PARENTS:
 LORENTZ TRANSFORMATIONS 
 RELATIVITY WORKSHOP 
 WORKSHOPS 
 ARCHIVE                                                                                                         
IMPLICATIONS OF LORENTZ TRANSFORMATIONS
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Introduction:
^^^^^^^^^^^^^
My latest contribution to the workshop consisted of a derivation of 
the Lorentz transformation:

y1=gamma*(x1+beta*x2) 		(1)
y2=gamma*(beta*x1+x2)
y3=x3
y4=x4

with:
x1=c*t 			(2)
beta=V/C
gamma=1/sqrt(1-beta^2)

These transformations shows us the connexion between a 'stationary' referential 
'x', to a moving referential 'y' which moves at speed V in the direction of x2 
(relatively to the 'stationary' referential 'x').

So where is SR today in this workshop? We've made a strange, counter intuitive 
assumption: That light moves at the same speed in all referentials. We've 
played with math, and found that transforming coordinated follows a wierd
transformation. However we have absolutely no idea what this actually means. 
All of our discourse has been in MS, and non of it pertained to PS predictions.

Let us therefor try to see what we can deduce from the Lorentz transformation 
about observable phenomena.

Index:
^^^^^^
1) Time dilation
2) Lorentz contraction
3) Speed cummulation

1) Time dilation
^^^^^^^^^^^^^^^^
Consider a stationary clock. It's 4D Mikowski space (MinSp) position is 
Clock_X=(ct,0,0,0) as it shows the time t. Now suppose we view the clock from 
a moving referential. Using the Lorentz transformation, we get that it's 
position (in the new referential Y) will now be: 

Clock_Y=(gamma*ct,gamma*beta*ct,0,0). 

Where:  beta=V/c ; gamma=1/sqrt(1-beta^2)

What can we say about it?

1) The ratio: x_Y/t_Y=gamma*beta*c*t/gamma*t=beta, which indeed show that the 
clock is moving at a speed V in the current referential.

2) This position at time gamma*t will be attained when the clock shows the 
time t. 

Since (V < c) (the speed is smaller then the speed of light), we know that:
(beta < 1) ;  (1-beta^2 < 1) ; and therefor (gamma > 1).
We can therefor conclude that the clock is retarded!
The clock shows time t when the time we proceve is gamma*t. 

If we have a clock moving with our referential, it shows the time t when the 
'stationary' clock shows the time t/gamma *in the moving referential*. 
Similarily and symmetrically, in the stationary referential the moving clock 
is retarded as it shows t/gamma while the stationary clock shows t. This may 
seem paradoxical to our intuition (each clock is slower then the other) but
it is infact paradox free. It's a direct result of the fact that in different 
referentials, different events are considered to be simultanious.

2) Lorentz contraction
^^^^^^^^^^^^^^^^^^^^^^
Consider a stationary ruler of length L. It has two ends, which occupy the 
positions:
Ruler1_X=(ct,0,0,0)
Ruler2_X=(ct,L,0,0)

Using the Lorentz transformations we get:

Ruler1_Y=gamma*(ct,beta*ct,0,0)
Ruler2_Y=gamma*(ct+L*beta,beta*ct+L,0,0)

A naive analiser (let's call him Dave) of these equations might say: The length 
of the ruler in the moving coordinates is the difference between the spatial 
distance of these two coordinates, i.e.

L_Y=Ruler2_y2-Ruler1_y2=gamma*(beta*ct+L)-gamma*beta*ct=gamma*L

Therefor naive Dave could say that the ruler is longer by a factor of gamma 
(gamma>1).

However, since I'm now way past the age of naivity, I have no choice but to 
examine the equations more closely, and see that naive Dave substracted the 
spatial coordinates of events taking place at different times. The length of 
the ruler can only be calculated correctely by substracting the spatial 
positions of the rulers' ends at the same time.  i.e. we must change the time 
t->t' in:

Ruler2_Y=gamma*(ct'+L*beta,beta*ct'+L,0,0)

Such that it's time gamma(t'+L*beta/c) would equal the time of the first end 
gamma*t. Therefor t'=t-L*beta/c, and we get:

Ruler1_Y=gamma*(ct,beta*ct,0,0)
Ruler2_Y=gamma*(c(t-L*beta/c)+L*beta,beta*c(t-L*beta/c)+L,0,0)

Or, with some agebric manipulations:

Ruler2_Y=gamma*(ct,beta*ct-L*beta^2+L,0,0)
Ruler2_Y=gamma*(ct,beta*ct+L/gamma^2,0,0)

Now that both positions are at the same time, we can calculate the ruler's 
length by substracting the positions of it's edges:

L_Y=Ruler2_y2-Ruler1_y2=gamma*(beta*ct+L/gamma^2)- gamma*beta*ct=L/gamma

And so the ruler is shorter by a factor of gamma (gamma>1), and most of us have 
heard of the Lorentz contraction rather then a Lorentz expansion.

3) Speed cummulation
^^^^^^^^^^^^^^^^^^^^
Consider a point object moving at a constant speed W with respect to the 
stationary system X. It's positions are the worldline:

Point_X=(ct,Wt,0,0) for all real values of t.

How does this object look like from the Y referential moving at a constant 
speed V with respect to the stationary referential? Using the Lorentz 
transformation we get:

Point_Y=gamma*(ct+Wt*beta,beta*ct+Wt,0,0)

So at what velocity is our Point object move at the Y referential? One simply 
needs to observe the ratio c*Point_y2/Point_y1:

V_Y=c*[gamma*(beta*ct+Wt)]/[gamma(ct+Wt*beta)]=c*(c*beta+W)/(c+W*beta)

Remembering beta=V/c

V_Y=(V+W)/(1+(W/c)*(V/c))

This is the called the formula of speed cumulation. It tells us how to 
calculate the speed of a moving object in a new coordinate frame. It replaces 
the intuitive formula that would simply say V_Y=V+W. Let us notice the 
following points:

1) If all speeds are much smaller then the speed of light: W<  LORENTZ TRANSFORMATIONS 
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