Search the FAQ Archives

3 - A - B - C - D - E - F - G - H - I - J - K - L - M
N - O - P - Q - R - S - T - U - V - W - X - Y - Z
faqs.org - Internet FAQ Archives

Invariant Galilean Transformations On All Laws
Section - 5. Transformations on Generalized Coordinate Laws

( Single Page )
[ Usenet FAQs | Web FAQs | Documents | RFC Index | Business Photos and Profiles ]


Top Document: Invariant Galilean Transformations On All Laws
Previous Document: 4. The Encyclopedia Brittanica Incompetency.
Next Document: 6. The data scale degradation absurdity.
See reader questions & answers on this topic! - Help others by sharing your knowledge
The traditional Gallilean transform is correct:

     t'   = t  

     x'   = x - vt.

But remember this: a transform of x doesn't effect 
just some values of x, but all of them, whether they 
are in the formula or not.  This is important if you 
want to do things right. The crackpot position is
strongly against this sci.math verified position, and
the apparently standard coordinate pseudo-transformation
they suggest is perhaps the result. {See Table of
Contents.]

Let's use a simple equation: x^2 + y^2 = r^2, which is 
the formula for a circle with radius r, centered at a
location where x=0. 

But what if the circle center isn't at x=0?  Well, we'd 
want to use the form analytic geometry, vector algebra, 
and elementary measurement theory tells us to use, a form 
where we make explicit just where the circle center is, 
even if it is at x=x0=0:

   (x-x0)^2 + (y-y0)^2 = r^2.

The circle center coordinate, x0, is an x-axis coordinate, 
just like all the x-values of points on the circle.

So, in proper generalized cartesian coordinate forms 
of laws/equations we want to transform every occurence 
of x and x0 - by whatever name we call it: x.c, x_e,
whatever.

So, what is the transformed version of (x-x0)?  Why,
(x'-x0'); both x and x0 are x-coordinates, and every
x-coordinate has a new value on the new axis. 

So, what is the value of (x'-x0') in terms of the original
x data?

From the transform equations we see that x'=x-vt, which
is also true for x0'=x0-vt:

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

In other words, when we use the generalized coordinate form
specified by analytic geometry, we find that the value of 
(x'-x0') does not depend on either time or velocity in any
way, shape, form, or fashion.

Similarly for (y-y0).

We can treat time the same way if necessary: (t-t0).

The above is a proof that any equation in x,y,z,t is
invariant under the galilean transforms. Just use the
generalized coordinate form, with (x-x0)/etc, in the 
transformation process, not the incompetently selected 
privileged form, with just x/etc. 

[The form is "privileged" because it assumes the circle 
center, point of emission, whatever, is at the origin of 
the axes instead at some less convenient point. After 
transform the coordinate(s) of the circle center/origin 
are also changed but the privileged form doesn't make 
this explicit and screws up the calculations, which 
should be based on (x'-x0') but are calculated as (x'-0).]

The value of (x'-x0') is the same as (x-x0).  That makes
sense. 

Draw a circle on a piece of paper, maybe to the right 
side of the paper. On a transparent sheet, draw x and y 
coordinate axes, plus x to the right, plus y at the top. 

Place this axis sheet so the y-axis is at the left side 
of the circle sheet. 

Now answer two questions after noting the x-coordinate of 
the circle center and then moving the axis sheet to the right:

(a) did the circle change in any way because you moved
the axis sheet (ie because you transformed the coordin-
nate axis)?

(b) did the coordinate of the circle center change?

The circle didn't change [although SR will say it did];
that means that (x'-x0') does indeed equal (x-x0).

The coordinate of the circle center did change, and it
changed at the same rate (-vt) as did every point on
the circle.   That means that x0'<>x0, and the fact the
circle center didn't change wrt the circle, means that 
the relationship of x0' with x0 is the same as that of 
any x' on the circle with the corresponding x: x'=x-vt; 
x0'=x0-vt.

This is to prepare you for the True Believer crackpots that 
say 'constant' coordinates can't be transformed; some even 
say they aren't coordinates. These crackpots include some 
that brag about how they were childhood geniuses, btw.

QED: The galilean transformation for any law on
generalized Cartesian coordinates is invariant under 
the Galilean transform.

The use of the privileged form explains HOW the transformed 
equation can be messed up, the next Subject explains what 
the screwed up effect of the transform is, and how use
of the generalized form corrects the screwup.

User Contributions:

Comment about this article, ask questions, or add new information about this topic:

CAPTCHA




Top Document: Invariant Galilean Transformations On All Laws
Previous Document: 4. The Encyclopedia Brittanica Incompetency.
Next Document: 6. The data scale degradation absurdity.

Single Page

[ Usenet FAQs | Web FAQs | Documents | RFC Index ]

Send corrections/additions to the FAQ Maintainer:
Thnktank@concentric.net (Eleaticus)





Last Update March 27 2014 @ 02:12 PM