Bill Hobba wrote:
I believe that the symmetry properties of an inertial reference frame implies
it must be linear. [...]

Rather than pounding on "linear or not?", let me suggest that the
definition of "inertial coordinates" is:

Inertial coordinates are such that _ANY_ object that moves
inertially traces a uniform straight line wrt the coordinates.

Linearity of transforms between inertial coordinates follows
immediately. Remarkably, one does not need to define what "moving
inertially" actually means....

This trivially dismisses Eugene Shubert's objection.


Tom Roberts

"Bill Hobba" writes:

In another thread and in some private correspondence the issue of the
linearity of inertial system transformations was called into question. I
believe that the symmetry properties of an inertial reference frame implies
it must be linear. By symmetry properties I mean homogeneity and isotropy
in space and homogeneity in time.

I have seen a number of proofs this must be the case an outline of one I
will give below.

let x' = f(x,t); delta x'/delta x is the ratio of a rod of length delta x to
the same rod in the moving system.

I hate to be picky, but if you are going to hold t constant, then
delta x'/delta x is actually the ratio of a rod of length x' which
is stationary in the primed frame with respect to the length of
the same rod in the unprimed frame. This matches the fact that
the coefficient of x in the formula for x' is greater than or
equal to 1, since that corresponds to length contraction (delta x
is less than delta x').

Homogeneity of space and time implies
this must be independent of x or t. In taking the limit we have dx'/dx = c1
where c1 is not dependant on either x or t. Integrating we have x' = c1x +
g(t). Taking the derivative wrt to t and noting again from homogeny in
space and time it must be independent of x or t we have x' = c1x + c2t + c3.
A similar proof follows for the time coordinate.

Does anyone see anything wrong with the proof, in particular does anyone see
how is it possible to have an extra term v/c f(x) added to the Lorentz
transform?

No, I don't see anything wrong with the proof, and if you have the
extra term, then the metric after the transformation is not the
Minkowski metric, so I don't see how anybody could claim that there
is such an extra term.

David McAnally

--------------

Tom Roberts wrote in message ...

Rather than pounding on "linear or not?", let me suggest that the
definition of "inertial coordinates" is:

Inertial coordinates are such that _ANY_ object that moves
inertially traces a uniform straight line wrt the coordinates.

The problem with your definition is that it's entirely a definition.
It's completely empty of physical content. Be aware that my NON-LINEAR
TRANSFORMATION in exercise 1 and 2 of
http://www.everythingimportant.org/r...eneralized.htm has the
property that every "inertial" observer traces out equal distances in
equal *PROPER* time intervals.

Linearity of transforms between inertial coordinates follows
immediately. Remarkably, one does not need to define what "moving
inertially" actually means....

Unremarkably, if you don't want to grapple with physics, then you've
left in the ignorance of not understanding physics.

This trivially dismisses Eugene Shubert's objection.

Unremarkably, trivial definitions often dismiss substantive ideas.

For example:

Inertial frames in VSL (variable speed of light) special relativity
don't move linearly but you're quite content with throwing out this
interesting generalization of SR with a definition. Why? For the sole
reason of winning an empty argument or, possibly, to keep the debate
out of territory unfamiliar to you but accessible to me?

Eugene Shubert
http://www.everythingimportant.org

(David McAnally) wrote in message ...
"Bill Hobba" writes:

Does anyone see anything wrong with the proof, in particular does anyone see
how is it possible to have an extra term v/c f(x) added to the Lorentz
transform?

No, I don't see anything wrong with the proof, and if you have the
extra term, then the metric after the transformation is not the
Minkowski metric, so I don't see how anybody could claim that there
is such an extra term.

David McAnally

--------------

David,

The point about the extra term is a simple test in SR understanding.
See exercise 1 and 2 of
http://www.everythingimportant.org/r...eneralized.htm

(Perfectly Innocent) writes:

(David McAnally) wrote in message ...
"Bill Hobba" writes:

Does anyone see anything wrong with the proof, in particular does anyone see
how is it possible to have an extra term v/c f(x) added to the Lorentz
transform?

No, I don't see anything wrong with the proof, and if you have the
extra term, then the metric after the transformation is not the
Minkowski metric, so I don't see how anybody could claim that there
is such an extra term.

David McAnally

--------------

David,

The point about the extra term is a simple test in SR understanding.
See exercise 1 and 2 of
http://www.everythingimportant.org/r...eneralized.htm

Your questions at the end of the webpage completely ignore the second
postulate. You have transformation equations

x' = x_0 + ((x-x_0) - v (t-t_0) + v/c \zeta(x))/sqrt(1-v^2/c^2),

t' = t_0 + ((t-t_0) - v/c^2 (x-x_0) - 1/c \zeta(x))/sqrt(1-v^2/c^2)

+ 1/c \zeta(x').

In order to be consistent with the second postulate, ct'-x' must
be functionally dependent on ct-x and v only, and ct'+x' must
be functionally dependent on ct+x and v only. Now,

ct'+x' = ct_0 + x_0 + {(c-v) (t-t_0) + (1-v/c) (x-x_0)

- (1-v/c) \zeta(x)}/sqrt(1-v^2/c^2) + \zeta(x')

= ct_0 + x_0 + sqrt((c-v)/(c+v)) [c (t-t_0) + (x-x_0)]

- sqrt((c-v)/(c+v)) \zeta(x) + \zeta(x'),

so, since ct'+x' and sqrt((c-v)/(c+v)) [c (t-t_0) + (x-x_0)]
are functionally dependent on ct+x and v only, it is required
that -sqrt((c-v)/(c+v)) \zeta(x) + \zeta(x') be functionally
dependent on ct+x and v only. Also,

ct'-x' = ct_0 - x_0 + {(c+v) (t-t_0) - (1+v/c) (x-x_0)

- (1+v/c) \zeta(x)}/sqrt(1-v^2/c^2) + \zeta(x')

= ct_0 + x_0 + sqrt((c+v)/(c-v)) [c (t-t_0) - (x-x_0)]

- sqrt((c+v)/(c-v)) \zeta(x) + \zeta(x'),

so, since ct'-x' and sqrt((c+v)/(c-v)) [c (t-t_0) - (x-x_0)]
are functionally dependent on ct-x and v only, it is required
that -sqrt((c+v)/(c-v)) \zeta(x) + \zeta(x') be functionally
dependent on ct-x and v only.

Since -sqrt((c-v)/(c+v)) \zeta(x) + \zeta(x') must be
functionally dependent on ct+x and v only, then its
derivative with respect to v, evaluated at v=0, must
be functionally dependent on ct+x only, and so

\zeta(x)/c - (t-t_0) \zeta'(x) + \zeta(x) \zeta'(x)/c

must be functionally dependent on ct+x only.

Since -sqrt((c+v)/(c-v)) \zeta(x) + \zeta(x') must be
functionally dependent on ct-x and v only, then its
derivative with respect to v, evaluated at v=0, must
be functionally dependent on ct-x only, and so

- \zeta(x)/c - (t-t_0) \zeta'(x) + \zeta(x) \zeta'(x)/c

must be functionally dependent on ct-x only.

So

\zeta(x)/c - (t-t_0) \zeta'(x) + \zeta(x) \zeta'(x)/c = F(ct+x),

- \zeta(x)/c - (t-t_0) \zeta'(x) + \zeta(x) \zeta'(x)/c = G(ct-x),

(*)

for some functions F and G. Upon subtraction, this yields that

2 \zeta(x)/c = F(ct+x) - G(ct-x).

Differentiation with respect to t yields 0 = c F'(ct+x) - c G'(ct-x),
so that F'(ct+x) = G'(ct-x) = K for some constant K, and so
F(ct+x) = K (ct+x) + L, G(ct-x) = K (ct-x) + M for some constants
L and M. So 2 \zeta(x)/c = 2 K x + L - M, so that \zeta(x) = K c x + B
for B constant. Substituting into (*), then

K x + B/c - K c (t-t_0) + K (K c x + B) = K (ct+x) + L,

- K x - B/c - K c (t-t_0) + K (K c x + B) = K (ct-x) + M.

Equating the coefficients of t on both sides of either equation,
we get -Kc = Kc, and so K = 0. It follows that the only function
\zeta(x) consistent with the second postulate is constant, and
so the transformation is an element of the Poincare group.

On final note, the transformations from the earlier part of the
page form a one-dimensional subgroup of the conformal group
for 1+1 dimensional space. The conformal group is well-known,
and is infinite-dimensional. The generators for the Lie algebra
of the conformal group are (ct+x)^(n+1) d/d(ct+x) for all integers
n and (ct-x)^(n+1) d/d(ct-x) for all integers n, where d/d(ct+x)
is partial differentiation with respect to ct+x where ct-x is held
constant, and d/d(ct-x) is partial differentiation with respect
to ct-x where ct+x is held constant. The Lie algebra of the
conformal group is isomorphic to the direct sum of a pair of
centerless real Virasoro algebras.

David McAnally

---------------

"Bill Hobba" writes:

Does anyone see anything wrong with the proof, in particular does anyone see
how is it possible to have an extra term v/c f(x) added to the Lorentz
transform?

By ignoring the second postulate. If the coordinates are to reflect
the fact that the speed of light is c in both frames of reference,
then the transformations as on the relevant webpage (as supplied by
Perfectly Innocent) require that f(x) be constant.

David McAnally

--------------

On 8/10/2003 6:41 AM, Perfectly Innocent wrote:
Tom Roberts wrote in message ...
Rather than pounding on "linear or not?", let me suggest that the
definition of "inertial coordinates" is:
Inertial coordinates are such that _ANY_ object that moves
inertially traces a uniform straight line wrt the coordinates.
The problem with your definition is that it's entirely a definition.
It's completely empty of physical content.

Coordinates are geometrical, not physical. So it makes good sense to
describe them in geometrical terms, as I did.

One can use physically-based coordinates (e.g. laid out with
a meter stick), and that can give them physical content.
But that is not essential.... Traditionally physicists
always did this (including Einstein's 1905 paper), but
since the advent of GR it is known to be too restrictive.


Be aware that my NON-LINEAR
TRANSFORMATION in exercise 1 and 2 of
http://www.everythingimportant.org/r...eneralized.htm has the
property that every "inertial" observer traces out equal distances in
equal *PROPER* time intervals.

OF COURSE an inertially-moving object traces out equal distances in
equal proper time intervals -- that's a direct consequence of timelike
geodesics in a flat manifold. But that's not the issue, the issue is
whether it traces a uniform straight line WRT THE COORDINATES (i.e. the
coordinates are linear functions of proper time or some other affine
parameter). After all, the goal is to define inertial coordinates. Your
coordinates do not do that except for some special choices of zeta(.).

Don't forget to include y and z, and consider paths not
along the axes....


Linearity of transforms between inertial coordinates follows
immediately. Remarkably, one does not need to define what "moving
inertially" actually means....
Unremarkably, if you don't want to grapple with physics, then you've
left in the ignorance of not understanding physics.

But this is "grappling" with geometry, not physics.


Inertial frames in VSL (variable speed of light) special relativity
don't move linearly but you're quite content with throwing out this
interesting generalization of SR with a definition. Why?

Because if inertially-moving objects don't move in uniform straight
lines wrt the coordinates, there's no point in calling the coordinates
"inertial" -- that is far too enormous a break with Newtonian usage.


Tom Roberts

Tom Roberts wrote in message ...

Coordinates are geometrical, not physical. So it makes good sense to
describe them in geometrical terms, as I did.

Why are my coordinates any less geometrical than yours?

Be aware that my NON-LINEAR
TRANSFORMATION in exercise 1 and 2 of
http://www.everythingimportant.org/r...eneralized.htm has the
property that every "inertial" observer traces out equal distances in
equal *PROPER* time intervals.

OF COURSE an inertially-moving object traces out equal distances in
equal proper time intervals. ... After all, the goal is to define
inertial coordinates.

I just did.

Eugene Shubert
http://www.everythingimportant.org/relativity/

(Perfectly Innocent) writes:

(David McAnally) wrote in message ...
"Bill Hobba" writes:

Does anyone see anything wrong with the proof, in particular does anyone see
how is it possible to have an extra term v/c f(x) added to the Lorentz
transform?

By ignoring the second postulate. If the coordinates are to reflect
the fact that the speed of light is c in both frames of reference,
then the transformations as on the relevant webpage (as supplied by
Perfectly Innocent) require that f(x) be constant.

David McAnally

--------------

David,

I am fairly confident that our disagreement is based on different
presuppositions. The speed of light is c is my equations.

Not if you define the velocity in the primed frame of reference
by the traditional definition of dx'/dt'. If you define velocity
by the traditional definition of dx'/dt', then that requires that
f(x) is a constant.

We simply
define speed differently.

Yes. I define the velocity as the derivate of the spatial
coordiantes with respect to the temporal coordiante.

Furthermore, I suspect that you're using
simultaneity

You are welcome to suspect what you want, even if you are completely
wrong.

and imposing this obsolete and delusive prejudice onto
your equations whereas I am certain that I interpret my equations
without this ancient and subtle prejudice.

No. You are using a completely unfamiliar and unknown definition
for velocity.

I asked before why the first half of the webpage was so completely
devoted to determining a one-dimensional subgroup of the conformal
equation. Your complaints about the fact that I define the
velocity in the primed frame to equal dx'/dt' make it all the
more astounding that you even bothered trying to come up with
the subgroup.

David McAnally

--------------

(David McAnally) wrote in message ...
(Perfectly Innocent) writes:

David,

I am fairly confident that our disagreement is based on different
presuppositions. The speed of light is c is my equations.

Not if you define the velocity in the primed frame of reference
by the traditional definition of dx'/dt'. If you define velocity
by the traditional definition of dx'/dt', then that requires that
f(x) is a constant.

The traditional definition of velocity being dx'/dt' does indeed evoke
a notion of simultaneity, which I refuse to depend upon.

We simply
define speed differently.

Yes. I define the velocity as the derivate of the spatial
coordiantes with respect to the temporal coordiante.

The traditional definition of velocity, as the time rate of change of
position, is based on the spooky notion of tracking distant motion
according to an absolute, stationary clock. And that's a source of
trouble that you don't even realize.

Furthermore, I suspect that you're using
simultaneity

You are welcome to suspect what you want, even if you are completely
wrong.

If you're certain, then let's discuss concrete counterexamples to your
beliefs. Where's the error in my derivation of the Lorentz
transformation based as it is on a Galilean definition of time?

http://www.everythingimportant.org/relativity/

Where's the physical content of velocity as dx'/dt' if we were to
invoke a change of coordinates from the Einsteinian synchronization to
a Galilean synchronization?

http://www.everythingimportant.org/relativity/

and imposing this obsolete and delusive prejudice onto
your equations whereas I am certain that I interpret my equations
without this ancient and subtle prejudice.

No. You are using a completely unfamiliar and unknown definition
for velocity.

In my studies of generalized SR, I use proper velocity to understand
the physics. Proper velocity is very well known. Because there is no
frame that moves at light speed, for that special case, I used a
reasonable, operational definition to define the speed of light:

http://groups.google.com/groups?hl=e...g.goog le.com

I asked before why the first half of the webpage was so completely
devoted to determining a one-dimensional subgroup of the conformal
equation. Your complaints about the fact that I define the
velocity in the primed frame to equal dx'/dt' make it all the
more astounding that you even bothered trying to come up with
the subgroup.

Regarding: http://www.everythingimportant.org/r...eneralized.htm
I have an interest in generalizations of SR that preserve the
equivalence of all frames of reference (the principle of relativity).
To that end it occurred to me that examining groups that generalize
the Lorentz transformation might be reasonable objects to investigate.

Why is a group with one space dimension, one time dimension and one
parameter describing a continuum of different states of motion
one-dimensional?

Eugene Shubert
http://www.everythingimportant.org/r...multaneity.htm