# net answers Invariant Galilean Transf

## Found at: ftp.icm.edu.pl:70/packages/usenet/sci.physics/Invariant_Galilean_Transformations_On_All_Laws

```From: Thnktank@concentric.net (Eleaticus)
```
```Newsgroups: sci.physics,sci.physics.relativity,alt.physics,sci.math,sci.answers,alt.answers,news.answers
```
```Subject: Invariant Galilean Transformations On All Laws
```
```Supersedes: <physics-faq/criticism/galilean-invariance_931434544@rtfm.mit.edu>
```
```Followup-To: poster
```
```Date: 23 Jul 1999 12:37:14 GMT
```
```Organization: none
```
```Lines: 1311
```
```Approved: news-answers-request@MIT.EDU
```
```Expires: 21 Aug 1999 12:36:23 GMT
```
```Message-ID: <physics-faq/criticism/galilean-invariance_932733383@rtfm.mit.edu>
```
```NNTP-Posting-Host: penguin-lust.mit.edu
```
```Summary: All laws/equations are Galilean invariant when expressed
```
```	 in the generalized cartesian coordinates demanded by basic
```
```	 analytic geometry, vector algebra, and measurement theory.
```
```Originator: faqserv@penguin-lust.MIT.EDU
```

```Disclaimer: approval for *.answers is based on form, not content.
```
```    Opponents of the content should first actually find out what
```
```    it is, then think, then request/submit-to arbitration by the
```
```    appropriate neutral mathematics authorities. Flaming the hard-
```
```    working, selfless, *.answers moderators evidences ignorance
```
```    and despicable netiquette.
```
```Archive-Name: physics-faq/criticism/galilean-invariance
```
```Version: 0.04.03
```

```	     Invariant Galilean Transformations On All Laws
```
```		   (c) Eleaticus/Oren C. Webster
```
```		      Thnktank@concentric.net
```

```An obvious typo or two corrected.
```
```The Brittanica section revised to less
```
```'pussy-footing' and to more directly
```
```anticipate the elementary measurement
```
```theory and basic analytic geometry
```
```that is applied to the transformation
```
```concept.
```

```------------------------------
```

```Subject: 1. Purpose
```

```The purpose of this document is to provide the student of Physics,
```
```especially Relativity and Electromagnetism, the most basic princ-
```
`ples and logic with which to evaluate the historic justification`
```of Relativity Theory as a necessary alternative to the classical
```

```We will prove that all laws are invariant under the Galilean
```
```transformation, rather than some being non-invariant, after
```

```We shall also show that another primal requirement that SR
```
```exist is nonsense: Michelson-Morley and Kennedy-Thorndike do
```
`ndeed fit Galilean (c+v) physics.`

```------------------------------
```

```Subject: 2. Table of Contents
```

```	1. Foreword and Intent
```
```	2. Table of Contents
```
```	3. The Principle of Relativity
```
```	4. The Encyclopedia Brittanica Incompetency.
```
```	5. Transformations on Generalized Coordinate Laws
```
```	6. The data scale degradation absurdity.
```
```	7. The Crackpots' Version of the Transforms.
```
```	8. What does sci.math have to say about x0'=x0-vt?
```
```	9. But Doesn't x.c'=x.c?
```
```       10. But Isn't (x'-x.c')=(x-x.c) Actually Two Transformations?
```
```       11. But Doesn't (x'-x.c+vt) Prove The Transformation Time
```
```	   Dependent?
```
```       12. But Isn't (x'-x.c')=(x-x.c) a Tautology?
```
```       13. But Isn't (x'-x.c')=(x-x.c) Almost the Definition of
```
```	   a Linear Transform?
```
```       14. But The Transform Won't Work On Time Dependent Equations?
```
```       15. But The Transform Won't Work On Wave Equations?
```
```       16. But Maxwell's Equations Aren't Galilean Invariant?
```
```       17. First and Second Derivative differential equations.
```

```------------------------------
```

```Subject: 3. The Principle of Relativity and Transformation
```

`t is not one law, but at least two, and leaves us`
`n doubt about any third location. This is the`
```may be perceived or measured, and whether or not the
```
```observer is moving.
```

```The idea of location translates to a coordinate
```
```be considered as having a coordinate system origin
```
```moving with it. If you perceive me moving relative
```
```to you - who have your own coordinate system - will
```
```your measurements of my position and velocity fit
```
```the same laws my own, different measurements fit?
```

```called covariant. If it is identical in form, var-
```
```ables, and output values, it is called invariant.
```

```What we're asking is that if the x-coordinate, x,
```
```on one coordinate axis works in an equation, does
```
```the coordinate, x', on some other, parallel axis
```
```the coordinate, x' is the 'transformed' coordinate.
```

```The situation is complicated because we're talking
```
```about coordinates - locations -  but in most mean-
```
`ngful laws/equations, it is lengths/distances (and`
```time intervals) the equations are about, and x coord-
```
`nates that represent good, ratio scale measures of`
```axis. [See Table of Contents for discussion of scales.]
```

```So, if we have an x-coordinate in one system, then
```

`n the form of the Galilean transformation, which`
```the transform equations x'=x-vt, y'=y, z'=z, t'=t in
```
```the simplified case where attention is focused only
```
```on transforming the x-axis, and not y and z. In the
```
```case of Special Relativity, the x' transform is the
```
```and t'=(t-xv/cc)/sqrt(1-(v/c)^2). In either case, v
```
`s the relative velocity of the coordinate systems;`
`f there is already a v in the equations being trans-`
```formed use u or some other variable name.
```

```------------------------------
```

```Subject: 4. The Encyclopedia Brittanica Incompetency.
```

```One example of the traditional fallacious idea
```
```that an equation is not invariant under the galilean
```
```transformation comes from the Encyclopedia Brittanica:
```

```"Before Einstein's special theory of relativity
```
```that the time coordinates measured in all inertial
```
```frames were identical and equal to an 'absolute
```
```time'.  Thus,
```

```      t = t'.              (97)
```

```"The position coordinates x and x' were then
```
```assumed to be related by
```

```       x' = x - vt.         (98)
```

```"The two formulas (97) and (98) are called a
```
```Galilean transformation. The laws of nonrelativ-
```
`stic mechanics take the same form in all frames`

```"The position of a light wave front speeding from
```
```the origin at time zero should satisfy
```

```       x^2 - (ct)^2 = 0          (99)
```

`n the frame (t,x) and`

```      (x')^2 - (ct')^2 = 0       (100)
```

`n the frame (t',x'). Formula (100) does not`
```transform into formula (99) using the transform-
```
```ations (97) and (98),  however."
```
```.................................................
```

```Besides the trivially correct statement of what the
```
```Galilean 'transform' equations are, there is exactly
```
```one thing they got right.
```

```     the question of invariance, given that eq-99 is
```
```     the correct 'stationary' (observer S) equation.
```
```     [Let observer M be the 'moving'system observer.]
```

```     In particular, eq-100 is of exactly the same
```
```     form [the square of argument one minus the square
```
```     of argument two equals zero (argument three).]
```

```     eq-100; for one thing, the transforms are TO x' and
```
```     t' from x and t, not the other way around, and the
```
```     idea that either observer's equation should contain
```
```     within itself the terms to simplify or rearrange to
```
```     get to the other is ridiculous. As the transform
```
```     equations say, the relationship of t', x' to t, x
```
```     is based on the relative velocity between the two
```
```     systems, but neither the original (eq-99) equation
```
```     nor the M observer equation is about a relationship
```
```     between coordinate systems or observers. One might
```
```     as well expect the two equations to contain banana
```
```     export/import data; there is no relevancy. The
```
```     'transform' equations are the relationships between
```
```     x' and x, t' and t and have nothing to do with what
```
```     one equation or the other ought to 'say'.  The
```
```     equations' content is the rate at which light emitted
```
```     along the x-axes moves.
```

```     most despise the consequences of measurement theory
```
```     (demonstrable fact) contained in this document are
```
```     those who want to argue against our saying the Britt-
```
```     anica got eq-100 right;
```

```     They insist that the correct equation is derived
```
```     directly from x'=x-vt and t'=t. Solve for x=x'+vt
```
```     and replace t with t', then substitute the result
```
```     in eq-99: (x'+vt')^2 - (ct')^2 = 0.
```

```     Besides the fact that this results in an equation
```
```     with arguments exactly equal to eq-99, they will
```
```     insist the transform is not invariant.
```

```     the correct M system equation on which to base the
```
```     the discussion of invariance, is that the variables
```
```     are M system variables, never mind the fact that
```
```     the arguments are S system values.
```

```     That argument of theirs is arrant nonsense. The
```
```     velocity v that S sees for the M system relative
```
```     to herself is the negative of what the M system
```
```     sees for the S system relative to himself.
```

```     In other words, x'+vt' is a mixed frame expression
```
```     and it is x'+(-v)t' that would be strictly M frame
```
```     notation, and that equation is far off base. [Work
```
```     it out for yourself, but make sure you try out an
```
```     S frame negative v so as not to mislead yourself.]
```

```V.   In I. we said: "given that eq-99 is the correct
```
```     'stationary' equation. Let's look at it closely:
```

```       x^2 - (ct)^2 = 0          (99)
```

```     This whole matter is supposed to be about coordinate
```
```     transforms. Is that what t is, just a coordinate?
```

```     No. It isn't, in general.  Suppose you and I are both modelling
```
```     the same light event and you are using EST and I'm using PST.
```
```     'Just a time coordinate' is just a clock reading amd your t clock
```
```     reading says the light has been moving three hours longer
```
```     than my clock reading says. Well, that's what the idea that
```
```     t is a coordinate means.
```

```     Eq-99 works if and only if t is a time interval, and in
```
```     particular the elapsed time since the light was emitted.
```
```     Thus, that equation works only if we understand just
```
```     what t is, an elapsed time, with emissioon at t=0.
```

```     However, we don't have to 'understand' anything if we use
```
```     a more intelligent and insightful form of the equation:
```

```     (x)^2 - [ c(t-t.e) ]^2 = 0,
```

```     where t.e is anyone's clock reading at the time of light
```
```     emission, and t is any subsequent time on the same clock.
```

```     Similarly, x is not just a coordinate, but a distance
```
```     since emission.
```

```     (x-x.e)^2 - [ c(t-t.e) ]^2 = 0        (99a)
```

```VI.  In the spirit of 'there is exactly one thing
```
```     they got right', the correct M system version
```
```     of eq-99a is eq-100a:
```

```     (x'-x.e')^2 - [ c(t'-t.e') ]^2 = 0   (100a)
```

```     Every observer in the universe can derive their
```
```     eq-100a from eq-99a and vice versa, not to mention to and
```
```     from every other observer's eq-99a.
```

```     Now, THAT's invariance. [You do realize that every
```
```     eq-100a reduces to eq-99a, when you back substitute
```
```     from the transforms, right? t.e'=t.e, x.e'=x.e-vt.]
```

```------------------------------
```

```Subject: 5. Transformations on Generalized Coordinate Laws
```

```The traditional Gallilean transform is correct:
```

```     t'   = t
```

```     x'   = x - vt.
```

```But remember this: a transform of x doesn't effect
```
```are in the formula or not.  This is important if you
```
```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
```
```and elementary measurement theory tells us to use, a form
```
```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,
```

```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,
```

```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
```
`s also true for x0'=x0-vt:`

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

```(x'-x0') does not depend on either time or velocity in any
```

```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
```
`nvariant under the galilean transforms. Just use the`
```transformation process, not the incompetently selected
```

```[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
```

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

```Draw a circle on a piece of paper, maybe to the right
```
```coordinate axes, plus x to the right, plus y at the top.
```

```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
```
```that brag about how they were childhood geniuses, btw.
```

```QED: The galilean transformation for any law on
```
```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.
```

```------------------------------
```
```
```
```Subject: 6. The data scale degradation absurdity.
```

```The SR transforms and the Galilean transforms both
```
```convert good, ratio scale data to inferior interval
```
```coordinate forms specified by analytic geometry and
```
```vector algebra.
```

```Both sets of transforms are 'translations' - lateral
```
```movements of an axis, increasing over time in these
```
```cases - but with the SR transform also involving a
```
```transform to x', and -xv/cc in the t transform to t',
```
```that degrades the ratio scale data to interval scale
```
```quality in the size-of-units sense we have here.
```

```SR likes to consider its transforms just rotations,
```
```they were 'translations' (movements) - and in the case
```
```of 'good' rotations, ratio scale data quality is indeed
```
```tations; they are not rigid rotations and they don't
```
```appropriately rescale all the axes that must be rescaled
```
```to preserve compatibility.
```

```The proof is in the pudding, and the pudding is the
```
```combination of simple tests of the transformations.
```
```We can tell if the transformed data are ratio scale
```
```or interval.
```

```Ratio scale data are like absolute Kelvin. A measure-
```
```ment of zero means there is zero quantity of the
```
`tion, subtraction, multiplication, and division.`

```The test of a ratio scale is that if one measure
```
```looks like twice as much as another, the stuff
```
```being measured is actually twice as much. With
```
```absolute Kelvin, 100 degrees really is twice the
```
```as much as 100.
```

`s why your science teacher wouldn't let you use it`
`n gas law problems.  There is only one mathematical`
```operation interval scales support, and that has to
```
```be between two measures on the same scale: subtraction.
```

```as much as 50; we have to convert the data to absolute
```
```Kelvin to tell us what the real ratio of temperatures
```
`s.`

```However, whether we use absolute Kelvin or relative
```
```Celsius, the difference in the two temperature readings
```
`s the same: 50 degrees.`

```Thus, if we know the real quantities of the 'stuff'
```
```being measured, we can tell if two measures are on
```
```a ratio scale by seeing if the ratio of the two
```
```measures is the same as the ratio of the known quant-
```
`ties.`

`s automatically a pass.`

```test becomes the next in line.
```

`n two interval scale measures are ratio scale, so`
`t is ratios of two differences that tell the tale.`

```Let's do some testing, and remember as we do that our
```
```concern is for whether or not the data are messed up,
```
```not with 'reasons', excuses, or avoidance.
```
```------------------------------------------------------
```

```Are we going to take a transformed length (difference)
```
```and see whether that length fits ratio or interval scale
```

```Of course, not. Interval scale data are ratio after
```
```one measure is subtracted from another. That is the
```
```major reason the SR transforms can be used in science.
```

```Let there be three rods, A, B, C, of length 10, 20, 40,
```
```our original x-axis, with one end of each rod at the
```
```origin, where x=0, and the other end at the coordinate
```
```that tells us the correct lengths.
```

```Note that these x-values are ratio scale only because
```
```one end of each rod is at x=0. That may remind you of
```
```the correct way to use a ruler or yard/meter-stick:
```
```measuring. Put the 1.00 mark there instead of the zero,
```
```and you have interval scale measures.
```

```Let A,B,C,   be 10, 20, 40.
```
```Let a,b,c    be x' at v=.5, t=10.
```

```x'=x-vt.
```

```A   B   C         a      b      c
```
```----------------  --------------------
```
```----------------  --------------------
```
```B/A = 2           b/a = 3
```
```C/A = 4           c/a = 7
```
```C/B = 2           c/b = 2.333
```

```			       Obviously, the transformed
```
```			       values are no longer ratio
```
```			       scale. The effect is less on
```
```			       the greater values.
```

```C-A = 10          b-a = 10
```
```C-A = 30          c-a = 30
```
```C-B = 20          c-b = 20
```

```			       Obviously, the transformed
```
```			       values are now interval scale.
```
```			       This will hold true for any
```
```			       value of time or velocity.
```

```(C-A)/(B-A) = 3   (c-a)/(b-a) = 3
```
```(C-B)/(B-A) = 2   (c-b)/(b-a) = 2
```

```			       Obviously, the ratios of the
```
```			       differences are ratio scale,
```
```			       being identical to the ratios
```
```			       of the corresponding original
```
```			       - ratio scale - differences.
```

```The main difference between these results and the SR
```
```neatly to the original, ratio scale, differences.
```

```This is due only to the rescaling by 1/sqrt(1-(v/c)^2).
```
```The ratios of the differences on the transformed values
```

```Using the generalized coordinate form, such as (x-x0),
```
```the transform produces an interval scale x' and an
```
`nterval scale x0'. That gives us a ratio scale (x'-x0'),`

```------------------------------
```

```Subject: 7. The Crackpots' Version of the Transforms.
```

```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
```
```of this faq: every coordinate is transformed, whether a
```

```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
```
```out - values on the variable's axis not values of the
```
```variable just because they have become named values.
```

```they have proposed for a transform of coordinates. It is
```
```based on the idea that the circle center, point of emission,
```

```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-
```
`nates 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
```
`t 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.
```

```accidentally. (x'+vt) is not the primed system
```
```coordinate, it is another form/expression of x. They
```

```So, by incompetent misnomer, they accomplish what they
```

```this time:
```

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

```Here they substitute for x instead of transforming to the
```

```	   (x'+vt-x0)^2 - (ict')^2.
```
```	    -----
```
```	      ^
```
```	      |
```
```	      ^
```
```	      |
```
```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
```
```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,
```
```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
```

```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;
```
```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
```

```Oh. Their (x'+vt-x0) or (x'+vt'-x0) reduces the same
```

```    (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,
```

```------------------------------
```

```Subject: 8. What does sci.math have to say about x0'=x0-vt?
```

```The crackpots' positions/arguments were put to sci.math
```
`n such a way that at least two or three who posted re-`

```Their responses:
```

```----------------------------------------------------------
```

```     values on the x-axis are not subject to the transform".
```

```AA: ====================================================================
```

```  No.  x0' = x0 - vt.
```

```  Well, if you want, you could define "constant values on the x-axis", but
```
`n the context of the question that is not relevant.  The relevant fact is`
```that if the unprimed observer holds an object at point x0, then the
```
```numerically related to x0 by x0'= x0 -vt.
```

```AA: ====================================================================
```
```EE: ====================================================================
```

```What does this mean? The line x=x0 will give x'=x-v*t=x0-vt', so if x0'
```
`s to give the coordinate in the (x',t',)-system, it will be given by`
```x0'=x0-v*t': ie., it is not given by a constant. Thus, being at rest
```
```(constant x-coordinate) is a coordinate-dependent concept.
```

```EE: ====================================================================
```
```GG: ====================================================================
```

```Sounds very false. We can say that the representation of the point X0 is
```
```the number x0 in the unprimed system, and x0' in the primed system.
```
```Clearly x0 and x0' are different, if vt is not zero. However one may say
```
```that (though it sounds/is stupid) the point X0 itself "is the same
```
```throughout the transformation". However that expression sounds
```
```meaningless, since a transform (ok, maybe we should call it a change of
```
```basis) is only a function that takes the point's representation in one
```
```x0' for the points' representations in some coordinate systems.
```

```GG: ====================================================================
```

```------------------------------
```
```
```
```Subject: 9. But Doesn't x.c'=x.c?
```

```That idea is one of the most idiotic to come up, and it does
```

```The idea being that x.c' <> x.c-vt, with x.c being what
```

```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
```
```--------------------------------------------------------------------
```

```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?
```
```--------------------------------------------------------------------
```

```(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.
```

```------------------------------
```
```
```
```Subject: 10. But Isn't (x'-x.c')=(x-x.c) Actually Two Transformations?
```

```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
```

```That's just another crackpot form of the idiocy that
```
```x.c' <> x.c-vt. You'll have noticed the implication
```
`s that there is no transform vt term relating to x.c.`

```------------------------------
```
```
```
```Subject: 11. But Doesn't (x'-x.c+vt) Prove The Transformation
```
```	     Time Dependent?
```
```
```
```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!
```

```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
```

```Then there is the fact that if you leave the equation
```
`n any of the various notation forms - with or without`

```------------------------------
```
```
```
```Subject: 12. But Isn't (x'-x.c')=(x-x.c) a Tautology?
```

```My dictionary relates 'tautology' to needless repetition.
```

```That's another form of the x.c' <> x.c-vt idiocy.
```

```The repetition involved is the vt transformation term.
```
```Apply the -vt term to the x term, and it is needless
```
```to the x.c term.  The x.c' = x.c crackpot idiocy.
```

```The repetition of the vt terms is required by the presence
```
```of two x values to be transformed.
```

```Be sure to note the next section.
```

```------------------------------
```

```Subject: 13. But Isn't (x'-x.c')=(x-x.c) Almost the Definition of
```
```	     a Linear Transform?
```

```Now, how on earth can we relate a tautology to a basic
```

```From the top, bottom, middle, and other books in the stack
```
```--------------------------------------------------------------
```

```A linear transformation, A, on the space is a method of corr-
```
```esponding to each vector of the space another vector of the
```
```a and b,
```

```	A(aU+bV) =  aAU + bAV.
```
```-------------------------------------------------------------
```

```Let points on the sphere satisfy the vector X={x,y,z,1},
```
```and the circle center satisfy C={x.c,y.c,z.c,1}. Let a=1,
```
```and b=-1.
```

```Let A= ( 1   0   0  -ut )
```
```       ( 0   1   0  -vt )
```
```       ( 0   0   1  -wt )
```
```       ( 0   0   0   1  )
```

```A(aX+bC) = aAX + bAC.
```

```      aX+bC  =  (x-x.c, y-y.c, z-z.c,  0  ).
```

```The left hand side:
```

```     A( x - x.c ,  y - y.c,  z - z.c,  0  )
```
```
```
```     = ( x-x.c ,  y-y.c,  z-z.c,  0  ).
```
```
```
```The right hand side:
```

```       aAX= ( x-ut, y-vt, z-wt, 1 ).
```
```       bAC= (-x.c+ut, -y.c+vt, -z.c+wt, -1 ).
```
```and
```
```
```
```  aAX+bAC = ( x-x.c, y-y.c, z-z.c,  0  ).
```

```Need it be said?
```

```Sure:  QED.  On the galilean transform the
```

```       A(aU+bV)=aAU + bAV,
```
```
```
`s completely satisfied.`

```The generalized form transforms exactly and
```
```non-redundantly - with ONE TRANSFORM, not a
```
```transform and reverse transform - and non-
```
```tautologically, just as the very definition
```
```of a linear transform says it should.
```

```And does so with absolute invariance, with this
```

```------------------------------
```

```
```
```Subject: 14. But The Transform Won't Work On Time Dependent Equations?
```
```
```
```The main crackpot that has asserted such a thing was referring
```
```to equations such as in Subject 4, above. The Light Sphere
```
```equation; for which we have shown repeatedly elsewhere that the
```
```numerical calculations are identical for any primed values as
```
```for the unprimed values.
```

```The presence - before transformation - of a velocity term
```
```treme historical reason for this, as you will see in the
```

```------------------------------
```
```
```
```Subject: 15. But The Transform Won't Work On Wave Equations?
```

```See Subject 17, below, for a discussion of Second Derivative
```
```forms and the galilean transforms.
```

```------------------------------
```
```
```
```Subject: 16. But Maxwell's Equations Aren't Galilean Invariant?
```

```Oh? Just what is the magical term in them that prevents
```
```(x'-x.c')=(x-vt - x.c+vt)=(x-x.c) from holding true?
```

```eralized coordinate form(s) of Maxwell: there are no coordi-
```
```nates to transform!
```

```When True Believer crackpots are shown the simple
```
```their first defense is usually an incredibly stupid
```
```"x0'=x0, because the coordinate of a circle center,
```
```or point of emission, etc, is a constant and can't
```
```be transformed."
```

```The last defense is "but Maxwell's equations are not
```
`nvariant under that coordinate transform."  When`
```asked just what magic occurs in Maxwell that would
```

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

```from working, and when asked them for a demonstration,
```
```they will never do so, however many hundreds of
```
```times their defense is asserted.
```

```The reason may help you understand part of Einstein's
```
```Relativity derivation:
```

```THERE ARE NO COORDINATES IN THE EQUATIONS TO BE TRANSFORMED.
```

```Einstein gave the electric force vector as E=(X,Y,Z)
```
```and the magnetic force vector as B=(L,M,N), where the
```
```force components in the direction of the x axis are
```
```the z direction.
```

```Those values are not, however, coordinates, but values
```
```very much like acceleration values.
```

```BTW, the current fad is that E and B are 'fields', having
```
```been 'force fields' for a while, after being 'forces'.
```

```So, when Einstein says he is applying his coordinate
```
```transforms to the Maxwell form he presented, he is
```
```either delusive or lying.
```

```(a) there are no coordinates in the transform equations
```
```    he gives us for the Maxwell transforms, where
```
```    B=beta=1/sqrt(1-(v/c)^2):
```
```
```
```    X'=X.                L'=L.
```
```    Y'=B(Y-(v/c)N).      M'=B(M+(v/c)Z).
```
```    Z'=B(Z+(v/c)M).      N'=B(N-(v/c)Y).
```

```    X is in the same direction as x, but is not a coordinate.
```
```    Ditto for L. They are not locations, coordinates on the
```
```    x-axis, but force magnitudes in that direction.
```

```    Similarly for Y and M and y, Z and N and z.
```

```(b) the v of the "coordinate transforms" is in Maxwell
```
```    before any transform is imposed; Einstein's transform
```
```    v is the velocity of a coordinate axis, not the velocity
```
```    of a particle, which is what was in the equation before
```
```    he touched it.
```

```(c) if they were honest Einsteinian transforms, they'd be
```
```    incompetent. The direction of the particle's movement is
```
```    x, which means it is X and L that are supposed to be
```
```    transformed, not Y and M, and Z and N. And when SR does
```
```    transform more than one axis, each axis has its own
```
```    velocity term;  using the v along the x-axis as the v
```
```    for a y-axis and z-axis transform is thus trebly absurd:
```
```    the axes perpendicular to the motion are not changed
```
```    according to SR, the v used is not their v, and the v
```
```    is not a transform velocity anyway.
```

```(d) as everyone knows, the effect of E and B are on the
```
```    particle's velocity, which is a speed in a particular
```
```    direction.  Both the speed and direction are changed
```
```    by E and B, but v - the speed - is a constant in SR.
```

```As absurd as are the previously demonstrated Einsteinian
```
```blunders, this one transcends error and is an incredible
```
```example of True Believer delusion propagating over decades.
```

```The components of E and B do differ from point to point,
```
```and in the variations that are not coordinate free,
```
```they are subject to the usual invariant galilean trans-
```
```formation when put in the generalized coordinate form.
```

```-------------------------------------------------------------
```

```The SR crackpots don't know what coordinates are. The
```
```various things they call coordinates include coordin-
```
```nates, but also include a variety of other quantities.
```

```------------------------------------------------------
```

```     manner [like x^2+y^2=r^2] but it is the use of vector
```
```     notation that shows us what is going on. In vector
```
```     notation the triplet x,y,z [or x1,x2,x3, whatever]
```
```     represents the three spatial coordinates, but there
```
```     are so-called basis vectors that underlie them. Those
```
```     may be called i,j,k. Thus, what we normally treat as
```
```     x,y,z is a set of three numbers TIMES a basis vector
```
```     each.
```

```
```
```     If e, f, j are distances from the origin of i,j,k then
```
```     e*i, f*j, g*k are coordinates: distances in the directions
```
```     of i,j,k respectively, from their origin. That makes the
```
```     triplet a coordinate vector that we describe as being an
```
```     x,y,z triplet; perhaps X=(x,y,z).
```

```     The e*i, f*j, g*k products could be directions; take any
```
```     of the other vectors described above or below and divide the
```
```     e,f,g numbers by the length of the vector [sqrt(e^2+f^2+g^2)].
```
```     That gives us a vector of length=1.0, the e,f,g values of
```
```     which show us the direction of the original vector. That
```
```     makes the triplet a direction vector that we describe as
```
```     being an x,y,z triplet; perhaps D=(x,y,z).
```

```     The e*i, f*j, g*k products could be velocities; take any
```
```     of the unit direction vectors described above and multiply
```
```     by a given speed, perhaps v. That gives a vector of length
```
```     v in the direction specified. That makes the triplet a
```
```     velocity vector that we describe as being an x,y,z triplet;
```
```     perhaps V=(x,y,z). Each of the three values, e,f,g, is the
```
```     velocity in the direction of i,j,k respectively.
```

```     The e*i, f*j, g*k products could be accelerations; take any
```
```     of the unit direction vectors described above and multiply
```
```     by a given acceleration, perhaps a. That gives a vector of
```
```     length a in the direction specified. That makes the triplet
```
```     an acceleration vector that we describe as being an x,y,z
```
```     triplet; perhaps A=(x,y,z). Each of the three values, e,f,g,
```
```     is the acceleration in the direction of i,j,k respectively.
```
```
```
```     The e*i, f*j, g*k products could be forces (much like accel-
```
```     erations); take any of the unit direction vectors described
```
```     above and multiply by a given force, perhaps E or B. That
```
```     gives a vector of length E or B in the direction specified.
```
```     That makes the triplet a force vector that we describe as
```
```     being an x,y,z triplet; perhaps E=(x,y,z) or B=(x,y,z). Each
```
```     of the three values, e,f,g, is the force in the direction of
```
```     i,j,k respectively.
```

```Einstein's - and Maxwell's - E and B are
```
```not coordinate vectors.
```

```============================================================
```

```There is another variety of intellectual befuddlement that
```
```misinforms the idea that Maxwell isn't invariant under the
```

```Velocities With Respect to Coordinate Systems.
```
```-----------------------------------------------
```
```Aaron Bergman supplied the background in a post to a sci.physics.*
```
```newsgroup:
```
```===============================================================
```

```Now, according to simple E&M, each current generates a magnetic
```
```field and this causes either a repulsion or attraction between
```
```the wires due to the interaction of the magnetic field and the
```
```current. Let's just use the case where the currents are parallel.
```
```Now, suppose you are running at the speed of the current between
```
```the wires. If you simply use a galilean transform, each wire,
```
`n this frame, there is no force between the wires. But this is a`
```contradiction.
```

```================================================================
```

```First of all, the invariance of the galilean transform (x'-x.c')
```
```=(x-x.c),  insures that it is an error to imagine there is any
```
```the usual, convenient rest frame is the best frame and only frame
```
```(x'-x.c')=(x-x.c).]
```

```Second, given that you decide unnecessarily to adapt a law to
```
```a moving frame, don't confuse coordinate systems with meaningful
```

```coordinate system have to do with physics?
```

```Nothing. Certainly not anything in the example Bergman gave.
```

```What is relevant is not current velocity with respect to a
```
```coordinate system, but current velocity with respect to wires
```
```and/or a medium.  The velocity of an imaginary coordinate sys-
```
```tem has absolutely nothing to do with meaningful physical vel-
```
```ocity. You can - if you are insightful enough and don't violate
```
`tem (e) - identify a coordinate system and a relevant physical`
```object, but where some v term in the pre-transformed law is
```
`n use, don't confuse it with the velocity of the coordinate`
```transform.
```

```Velocities With Respect to ... What?
```
```-----------------------------------------------
```
```Albert Einstein opened his 1905 paper on Special Relativity
```
```===============================================================
```

```The equations of the day had a velocity term that was taken
```
```as meaning that moving a magnet near a conductor would create
```
```a current in the conductor, but moving a conductor near a
```

```The important velocity quantity is the velocity of the
```
```magnet and conductor with respect to each other, not to
```
```not to an arbitrary coordinate system.
```

```One possible cause was the idea: "but the equation says the magnet
```
```must be moving wrt the coordinate system" or "... the absolute
```

```There not being anything in the equation(s) to say either of
```
```those, it is amazing that folk will still insist the velocity
```
```term has nothing to do with velocity of the two bodies wrt
```
```each other.
```
```-----------------------------------------------------------
```

```------------------------------
```
```
```
```Subject: 17. First and Second Derivative differential equations.
```

```One of the intellectually corrupt ways of
```
`n the generalized coordinate form demanded`
```by analytic geometry, vector analysis, and
```
```measurement theory
```

```    [ (x'-x.c')=[ (x-vt)-(x.c-vt) ]=(x-x.c) ]
```

`s the assertion that those equations 'over there'`
```(usually Maxwell or wave) are somehow immune to
```
```the elementary laws of algebra used to demon-
```
```assertions are never accompanied by reference
```
```to the magical math that makes elementary al-
```

```based on the incompetent transformation of
```
```the privileged form of an equation instead
```
```of the correct form. [Evidence of this is
```
```any reference to an effect due to the velocity
```
```of the transform; it falls out algebraicly
```
```- as you see above - and cancels out arith-
```
```metically - as you can see above.]
```

```But usually it is just whistling in the dark,
```
```the mean old vampire.
```

```The most general equation that could be conjured
```
```up is a differential with either First or Second
```
```Derivatives.
```

```Let's examine the plausibility of such magical
```
```magical, non-invariance assertions.
```

```(a) to get a Second Derivative you must have
```
```    a First Derivative.
```
```(b) to get a First Derivative you must have
```
```    a function to differentiate.
```
```(c) to get a Second Derivative you must have
```
```    a function in the second degree.
```

```So, let us examine the question as to whether
```
```any such common Maxwell/wave equation will
```

```(a) the common, privileged form, represented
```
```    as ax^2, with a being an unknown constant
```
```    function.
```

```(b) the generalized cartesian form, represented
```
```    as a(x-x.c)^2 = ax^2 -2ax(x.c) + ax.c^2,
```
```    with a being an unknown constant function.
```

```(c) the transformed generalized cartesian form,
```
```    represented as a(x-vt -x.c+vt)^2, same as for
```
```    (b), = ax^2 -2ax(x.c) + ax.c^2, of course,
```
```    with a being an unknown constant function.
```

```    and that this version is only correct because
```
```    x.c=0, otherwise (b) is the correct form:
```
```
```
```     d/dx    ax^2  = 2ax
```
```    (d/dx)^2 ax^2  = 2a
```

```     d/dx    (ax^2 -2ax(x.c) + ax.c^2) = 2ax - 2ax.c
```
```    (d/dx)^2 (ax^2 -2ax(x.c) + ax.c^2) = 2a
```

```So, what we have seen so far is
```

```(1)  differential equations in the second degree
```
```- the wave equations - must clearly be the same for
```
```all forms: the privileged form in x, the generalized
```
```cartesian form in x and the centroid, x.c, or the
```
```transformed generalized cartesian form.
```

```That is, anyone who imagines that correct usage
```
```frames is at first showing his ignorance, and in
```
```the end showing his intellectual corruption.
```

```(2) As far as the First Derivatives are concerned, the
```
```only cases in which there really is a difference between
```
```the two forms is where x.c <> 0, and in that case, the
```
```use of the privileged form is obviously incompetent.
```

```So, how do you correctly use the differential equations?
```

```at x=0, etc, you can't go wrong without trying to
```

```not at x=0, you must use (x-x.c) anyplace x appears
```
`n the equation.`

```moving frame centroid as well as the light front
```
```(or whatever) moving frame data itself, perhaps first
```
```calculating (x'-x.c'), which equals (x-x.c) which is
```
```obviously correct, and which is obviously the plain old
```
```correct x of the privileged form.
```

```Unless, of course, there really is some magical term
```
```or expression that invalidates the obvious and elemen-
```
```tary algebra of the invariance demonstration.
```

```Or maybe you just whistle when you don't want basic
```
```algebra to hold true.
```

```Eleaticus
```

```!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?
```
```! Eleaticus        Oren C. Webster         ThnkTank@concentric.net  ?
```
```! "Anything and everything that requires or encourages systematic   ?
```
```!  examination of premises, logic, and conclusions"                 ?
```
```!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?
```

```.
```