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Invariant Galilean Transformations On All Laws
Section - 7. The Crackpots' Version of the Transforms.

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It has become apparent - whether misleading or not -
that the crackpot responses to the obvious derive from
a common source, whether it be bandwagoning or their
SR instructors.

Below, in the sci.math subject, we see that all sci.math
respondents agree with the basic "controversial" position
of this faq: every coordinate is transformed, whether a
supposed "constant" or not.

Think about it, the generalized coordinate of a circle
center, x0, applies to infinities upon infinities of
circle locations (given y and z, too); it is a constant 
only for a given circle, and even then only on a given 
coordinate axis.

And even "variables" are often held 'constant' during 
either integration or differentiation. 

The utility of a "variable" is that you can discuss all 
possible particular values without having to single out 
just one.  That utility does not make particular - singled 
out - values on the variable's axis not values of the 
variable just because they have become named values.

In any case, all that is preamble to the incompetent idea
they have proposed for a transform of coordinates. It is
based on the idea that the circle center, point of emission,
whatever, has coordinates that cannot be transformed.

Let there be an equation, say (x)^2 - (ict)^2 = 0.

What is the transformed version of that equation?

Answer: (x')^2 - (ict')^2 = 0.   That's the one thing the
Brittanica got right. Note that the leading crackpot just
criticized this faq for presuming to correct the Britt-
anica, but it then and before poses the incompetent pseudo-
transform we analyze here in this section.

x to x' and t to t' are obviously coordinate transforms; 
the x and t coordinates have been replaced by the coord-
inates in the primed system.  

A tranform of an equation from one coordinate system to 
another is NOT a substitution of the/a definition of x 
for itself; that is not a coordinate transformation.
The most that can said for such a substitution is that
it is a change of variable.

But the crackpots are calling this a coordinate trans-
form of the original equation:

    (x'+vt)^2 - (ict')^2 = 0.

It is not a coordinate transform, of course, except 
accidentally. (x'+vt) is not the primed system
coordinate, it is another form/expression of x. They 
get that substitution by solving x'=x-vt for x; x=x'+vt.

So, by incompetent misnomer, they accomplish what they
have been railing against all along.

It has been the generalized coordinate form in question all
this time:

    (x-x0)^2 - (ict)^2 = 0.

Here they substitute for x instead of transforming to the
primed frame:

	   (x'+vt-x0)^2 - (ict')^2.
	    -----
	      ^
	      |
	      ^
	      |
It is still x ^ but see what they have accomplished  
by their mis/malfeasance:

	   [x'+vt-x0]=[x'+(vt-x0)]=[x'-(x0-vt)]. 
				  =[x'-x0']

The crackpots have been bragging about how you don't
have to transform the circle center's coordinate to
transform the circle center's coordinate.  Bragging 
that what they were doing was not what they said 
they were doing.

This does give us insight as to some of the crackpot
variations on their x0'<>x0-vt theme, which in all the
variations will be discussed in later sections..

They are used to seeing the mixed coordinate form,
(x'+vt-x0) without realizing what it respresented,
so - accompanied with a lack of understanding of
the term 'dependent' - they are used to seeing just
the one vt term, and not the one hidden in the defi-
nition of x' and are used to imagining it makes the
whole expression time dependent and thus not invariant.

About which, let x=10, let, x0=20, v=10, and t 
variously 10 and 23:

(x-x0)=-10.  Using their (x'+vt-x0):

For t=10, we have (x'+vt-x0) = [ (10-10*10) + (10*10) - (20) ]
			     =      -90     +   100   -  20 
			     = -10 
			     = (x-x0)

For t=23, we have (x'+vt-x0) = [ (10-10*23) + (10*23) - (20) ]
			     =      -220    +   230   -  20 
			     = -10 
			     = (x-x0)

The result depends in no way on the value of time;
we showed the obvious for a couple of instances of t
just so you can see that the crackpots not only do
not understand the obvious logic of the algebra
{ (x'-x0')=[ (-vt)-(x0-vt) ]=(x-x0) } - which shows
that the transform has no possible time term effect -
but they don't understand even a simple arithmetic
demonstration of the facts.

Oh. Their (x'+vt-x0) or (x'+vt'-x0) reduces the same 
way since t'=t:

    (x-vt+vt-x0)=(x-x0).

Their process, which says (x'+vt') is the transform
of x, says that (x'+vt') is the moving system location 
of x, but it can't be because x is moving further in
the negative direction from the moving viewpoint. 

That formula will only work out with v<0 which is indeed
the velocity the primed system sees the other moving at.
However, that formula cannot be derived from x'=x-vt,
the formula for transformation of the coordinates from 
the unprimed to the primed,

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Top Document: Invariant Galilean Transformations On All Laws
Previous Document: 6. The data scale degradation absurdity.
Next Document: 8. What does sci.math have to say about x0'=x0-vt?

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