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Invariant Galilean Transformations On All Laws
Section - 11. But Doesn't (x'-x.c+vt) Prove The Transformation Time Dependent?

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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.

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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?

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Last Update March 27 2014 @ 02:12 PM