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? See reader questions & answers on this topic! - Help others by sharing your knowledge 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. User Contributions: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? Single Page [ Usenet FAQs | Web FAQs | Documents | RFC Index ] Send corrections/additions to the FAQ Maintainer: Thnktank@concentric.net (Eleaticus)
Last Update March 27 2014 @ 02:12 PM
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