net answers Invariant Galilean Transf

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From: Thnktank@concentric.net (Eleaticus)
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Subject: Invariant Galilean Transformations On All Laws
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Summary: All laws/equations are Galilean invariant when expressed
	 in the generalized cartesian coordinates demanded by basic
	 analytic geometry, vector algebra, and measurement theory.
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Archive-Name: physics-faq/criticism/galilean-invariance
Version: 0.04.03

	     Invariant Galilean Transformations On All Laws
		   (c) Eleaticus/Oren C. Webster

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


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
       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

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; 

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

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-

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.


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:


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 


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

"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 ).
  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:


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
    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

     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.* 

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


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 

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

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

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

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