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Invariant Galilean Transformations On All Laws
Section - 10. But Isn't (x'-x.c')=(x-x.c) Actually Two Transformations?

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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.

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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?

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Last Update March 27 2014 @ 02:12 PM