Top Document: Invariant Galilean Transformations On All Laws Previous Document: 9. But Doesn't x.c'=x.c? Next Document: 11. But Doesn't (x'-x.c+vt) Prove The Transformation Time Dependent? See reader questions & answers on this topic! - Help others by sharing your knowledge One crackpot puts the (x'-x.c')=(x-vt - x.c+vt) relationship like this: (x-vt+vt - x.c). See, he says, that is transforming x (with x-vt - x.c) and then reversing the transform (x-vt+vt - x.c). That's just another crackpot form of the idiocy that x.c' <> x.c-vt. You'll have noticed the implication is that there is no transform vt term relating to x.c. User Contributions:Top Document: Invariant Galilean Transformations On All Laws Previous Document: 9. But Doesn't x.c'=x.c? Next Document: 11. But Doesn't (x'-x.c+vt) Prove The Transformation Time Dependent? 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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