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Derivation of Lorentz Transformation (Mishehu)

The homogeneity of space and time leads to the demand that the
Lorentz transformations are linear (I've no time to explain further
right now). They can be written as a composition of spacial rotations
and Lorentz boosts.
All possible transformations - called the poincare group are Lorentz
transformations plus trasnlations: x'=x+k, or t'=t+k.
The Lorentz boosts:
^^^^^^^^^^^^^^^^^^^
Lorentz boosts are transformations of the form:
t'=L00*t+L01*x
x'=L10*t+L11*x
with the parameters L00,L10,L01,L11 as constants, which we'll now
find.
Now let us make three demands:
1) A stationary object in the system K, will move at speed V in
system K'.
2) A light ray moving at velocity C in system K, will move at speed C
in K'.
3) A light ray moving at velocity -C in system K, will move at speed -
C in K'.
Demand 1 means that the worldline (t,0) will become worldline
(t',Vt'). This form garauntees that the ratio x'/t' will be V.
(t',Vt'). This form garauntees that the ratio x'/t' will be V.
L00*t+L01*0=t' ����� (1)
L10*t+L11*0=Vt'
Dividing these equation one by the other leads to:
L00*V=L10. (2)
Demand 2 means the worldline (t,Ct) will become (t',Ct').
L00*t+L01*C*t=t' (3)
L10*t+L11*C*t=C*t'
so: (L00+L01*C)*C=L10+L11*C (4)
If we observe a light ray moving to the opposite direction (demand 4)
we get (t,-Ct) ==> (t',-Ct')
L00*t-L01*C*t=t'� (5)
L10*t-L11*C*t=-C*t'
so: (-L00+L01*C)*C=L10-L11*C (6)
Equations (4) + (6) lead to:
L01*C^2=L10 (7)
Equations (4) - (6) lead to:
L00*C=L11*C (8)
To sum up, equations (2) (7) and (8)
t'=L00*t+L00*V/C^2*x (9)
x'=L00*V*t+L00*x
x'=L00*V*t+L00*x
Now we still have one parameter, which we need to determine. This
parameter is called (by most physicists) gamma=L00. We'll denote it
by g for brevity.
What this parameter does, is expand all of space time. Clearly such a
transformation perserves the velocity of light (space and time are
multiplied by the same constant). If we put V=0, we see that our
Lorentz transformation reduces to:
t'=g*t  (10)
x'=g*x
This makes it tempting to say we should pick g=1, so that we do not
stretch space-time. gamma however can be velocity dependent. In order
to find g(V), what we can do, is suggest the following route:
1) Transform from K to K' using the Lorentz transformation L(V).
2) Transform from K' back to K using the Lorentz transformation L(-V).
The result must be in the form of a space-strech, and the gammas can
be choosen such that it would be the identity. Here I add the claim
that g(V)=g(-V). There is a simple argument explaining it. If no one
else would suggest it, I'd reveal it in a future post.
t'=g*t+g*V/C^2*x (11)
x'=g*V*t+g*x
t=g*t'-g*V/C^2*x' (12)
x=-g*V*t'+g*x'
t=g*(g*t+g*V/C^2*x)-g*V/C^2*(g*V*t+g*x) (13)
x=-g*V*(g*t+g*V/C^2*x)+g*(g*V*t+g*x)
1=g^2*V^2/C^2+g^2   (14)
g=sqrt(1-V^2/C^2)
We now have the final form of the Lorentz equation:
t'=g*(t+V/C^2*x) (15)
x'=g*(V*t+x)
It is customary to define x0=ct, and then get:
x0'=gamma*(x0+beta*x) (16)
x'=gamma*(beta*t+x)
with:
beta=V/C      (17)
gamma=sqrt(1-beta^2)

PARENTS:
 LORENTZ TRANSFORMATIONS 
 RELATIVITY WORKSHOP 
 WORKSHOPS 
 ARCHIVE