Top Document: Invariant Galilean Transformations On All Laws Previous Document: 10. But Isn't (x'-x.c')=(x-x.c) Actually Two Transformations? Next Document: 12. But Isn't (x'-x.c')=(x-x.c) a Tautology? See reader questions & answers on this topic! - Help others by sharing your knowledge That particular crackpottery is perhaps more corrupt than moronic, since it includes deliberately hiding a vt term from view, and pretending it isn't there. [However, we have seen above that it is a familiar incompetency, and not likely an original.] "Look," the crackpots say, "there is a time term in the transformed (x' - x.c+vt). The transform isn't invariant! It's time dependent!" Just put x' in its original axis form, also, which reveals the other time term, the one they hide: (x'-x.c+vt) = (x-vt - x.c+vt) = (x-x.c). So, at any and all times, the transform reduces to the original expression, with no time term on which to be dependent. Then there is the fact that if you leave the equation in any of the various notation forms - with or without reducing them algebraicly - the arithmetic always comes down to the same as (x-x.c). That means nothing to crack- pots, but may mean something to you. User Contributions:Top Document: Invariant Galilean Transformations On All Laws Previous Document: 10. But Isn't (x'-x.c')=(x-x.c) Actually Two Transformations? Next Document: 12. But Isn't (x'-x.c')=(x-x.c) a Tautology? 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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