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# Invariant Galilean Transformations On All LawsSection - 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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Last Update March 27 2014 @ 02:12 PM