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Top Document: Invariant Galilean Transformations (FAQ) On All Laws
Previous Document: 8. What does sci.math have to say about x0'=x0-vt?
Next Document: 10. But Isn't (x'-x.c')=(x-x.c) Actually Two Transformations?


9. But Doesn't x.c'=x.c?


That idea is one of the most idiotic to come up, and it does
so frequently. And in a number of guises.

The idea being that x.c' <> x.c-vt, with x.c being what
we have called x0 above; the notation makes no difference.

Some crackpots have managed to maintain that position even
after graphs have illustrated that such an idea means that
after a while a circle center represented by x.c' could be
outside the circle.

The leading crackpot just make that explicit, as far as
one can tell from his befuddled post in response to a line
about "active" transforms, which are actually moving body
situations, not coordinate transformations:
--------------------------------------------------------------------

e>An active transform is not a coordinate transform, ...

 Right, it is a transform of the center (in the opposite direction)
 done to effect the change of coordinates without a coordinate
 transform.  ...

E: Transform of the center?  Center of a circle?
    He really is saying a circle center moves in
    the opposite direction of the circle! Right?
--------------------------------------------------------------------

If r=10 and x.c was at x.c=0, then the points on the circle
(10,0), (-10,0), (0,10) and (0,-10) could at some time become
(-10,0), (-30,0), (-20,10), and (-20,-10), but with x.c'=x.c,
the circle center would be at (0,0) still!  The circle is here
but its center is way, way over there! Indeed, although a change
of coordinate systems is not movement of any object described in
the coordinates, the x.c'=x.c crackpottery is tantamount to the
circle staying put but the center moving away. Or vice versa.



Top Document: Invariant Galilean Transformations (FAQ) On All Laws
Previous Document: 8. What does sci.math have to say about x0'=x0-vt?
Next Document: 10. But Isn't (x'-x.c')=(x-x.c) Actually Two Transformations?

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