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Invariant Galilean Transformations On All Laws
Section - 13. But Isn't (x'-x.c')=(x-x.c) Almost the Definition of a Linear Transform?

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Previous Document: 12. But Isn't (x'-x.c')=(x-x.c) a Tautology?
Next Document: 15. But The Transform Won't Work On Wave Equations?
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Now, how on earth can we relate a tautology to a basic
definition in math?

From the top, bottom, middle, and other books in the stack
we get this definition:
--------------------------------------------------------------


A linear transformation, A, on the space is a method of corr-
esponding to each vector of the space another vector of the
space such that for any vectors U and V, and any scalars
a and b,

	A(aU+bV) =  aAU + bAV.
-------------------------------------------------------------

Let points on the sphere satisfy the vector X={x,y,z,1}, 
and the circle center satisfy C={x.c,y.c,z.c,1}. Let a=1, 
and b=-1.

Let A= ( 1   0   0  -ut )
       ( 0   1   0  -vt )
       ( 0   0   1  -wt )
       ( 0   0   0   1  )

A(aX+bC) = aAX + bAC.

      aX+bC  =  (x-x.c, y-y.c, z-z.c,  0  ).


The left hand side:

     A( x - x.c ,  y - y.c,  z - z.c,  0  ) 
     
     = ( x-x.c ,  y-y.c,  z-z.c,  0  ). 
     
The right hand side:

       aAX= ( x-ut, y-vt, z-wt, 1 ).
       bAC= (-x.c+ut, -y.c+vt, -z.c+wt, -1 ).
and
  
  aAX+bAC = ( x-x.c, y-y.c, z-z.c,  0  ).

Need it be said?  

Sure:  QED.  On the galilean transform the 
definition of a linear transform, 

       A(aU+bV)=aAU + bAV, 
       
is completely satisfied.

The generalized form transforms exactly and 
non-redundantly - with ONE TRANSFORM, not a
transform and reverse transform - and non-
tautologically, just as the very definition
of a linear transform says it should.

And does so with absolute invariance, with this
galilean transformation.


------------------------------

       
Subject: 14. But The Transform Won't Work On Time Dependent Equations?
       
The main crackpot that has asserted such a thing was referring
to equations such as in Subject 4, above. The Light Sphere
equation; for which we have shown repeatedly elsewhere that the
numerical calculations are identical for any primed values as
for the unprimed values.

The presence - before transformation - of a velocity term
seems to confuse the crackpots. It turns out there is ex-
treme historical reason for this, as you will see in the
subject on Maxwell's equations.

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Top Document: Invariant Galilean Transformations On All Laws
Previous Document: 12. But Isn't (x'-x.c')=(x-x.c) a Tautology?
Next Document: 15. But The Transform Won't Work On Wave Equations?

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Last Update March 27 2014 @ 02:12 PM