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. 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?

Thanks
Bill

"Bill Hobba" wrote in message
...
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. 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?

The Lorentz transformations relate to actual space and time measurements.

Special Relativity abolishes the idea of absolute space and absolute time...
prefering to use inertial frames of reference.

Read Mach for further understanding, a good primar for Einsteint... as
Einstein tried to incorporate his ideas into General Relativity.

You're on the right track!

"Bill Hobba" wrote in message ...

I believe that the symmetry properties of an inertial reference frame implies
it must be linear.

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. Homogeneity of space and time implies
this must be independent of x or t.

Thanks
Bill

If delta x'/delta x varied like Exp(kt), I'd call that homogeneous.

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

"Bill Hobba" wrote in message ...

Bill Hobba wrote:
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. 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?

I quite agree, although with a caveat below....

Mark replied:

The Lorentz transformations relate to actual space and time
measurements.

Special Relativity abolishes the idea of absolute space and
absolute time...
prefering to use inertial frames of reference.

Read Mach for further understanding, a good primar for Einsteint...
as
Einstein tried to incorporate his ideas into General Relativity.

You're on the right track!

My proof related to inertial reference frames only so GR is not
required;
SR is all that is needed.

Bill, you're using finite relations in the transformation, (quote from
above),

x' = c1x + c2t + c3, (Hobba1)

something that is difficult to do when relating
any CS at a point, (General Covariance). The quoted equation relates
CS's separated in location. It becomes unreasonably difficult to
compare
distance origins without the benefit of GR even if no g-fields are
present,

You go on to say,
"A similar proof follows for the time coordinate."

presuming a solution like,

t = (x' - c1x - c3)/c2, which is getting complicated.

The reason for the complication is due to non-coincident origins.
But a transformation of location will first simplify the relation of
CS's K and K' to have a coincident origin. When this is done,
your question,

" 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?"

That's right, you can then generally substitute

g_i4 = - g_ij dx^j/dx^4

(which is your v/c f(x) term ).

The lorentz transformations are an integral part of
SR and in no way incorporate the idea of absolute space and time -
they
describe how space time measurements in one inertial coordinate
system
relate to space time measurements in another coordinate system. Note
that
Landau in Mechanics defines inertial reference frames as those that
posses
the symmetry properties I stated and is equivalent to the usual
definition
via the POR.

BTW I am not a supporter of Mach's principle

Me too, an inertial force is sensed by a free-falling object
if it deviates from a geodesic solution, and this solution is
the solution of Newton's 1st law of motion.

as I have never seen an exact
statement of it that is experimentally verifiable or refutable. If
you know
of one I would like to hear it. However I have seen quite a few and
non
really look that good to me.

To me the fundamental idea of GR is that the metric must be a
dynamical
variable

Well Bill, the metric defined by,

g_i4 = - g_ij dx^j/dx^4

is termed the *dynamic non-orthogonal spacetime metric*

and have its own lagrangian - once you assume that the EFE more or
less follow.

Thanks
Bill

Your welcome, and thanks for the good question,
Ken S. Tucker

Ken S. Tucker wrote;
Bill, you're using finite relations in the transformation, (quote from
above),

I have no idea what your trying to say.

Ken S. Tucker wrote;

x' = c1x + c2t + c3, (Hobba1)

something that is difficult to do when relating
any CS at a point, (General Covariance).

The principle of general covariance (really the principle of general
invariance but I will not be that picky) is a GR concept - I am dealing with
SR here so it is a concept that is not applicable.

Ken S. Tucker wrote;
'The quoted equation relates CS's separated in location. It becomes
unreasonably difficult to compare distance origins without the benefit of GR
even if no g-fields are present,'

Again this has nothing to do with GR - it is an inertial frame so gravity is
precluded.

Ken S. Tucker wrote;
You go on to say,
"A similar proof follows for the time coordinate."

presuming a solution like,

t = (x' - c1x - c3)/c2, which is getting complicated.

No it is the time coordinate in the ' system ie

t' = c4x + c5t + c6

Ken S. Tucker wrote;
The reason for the complication is due to non-coincident origins.

Coincidence of origins will remove the constant at the end but is linear in
either case ie if the origins are coincident then the transformation is:

x' = c1x + c2t and t' = c3x + c4t.

Ken S. Tucker wrote;
But a transformation of location will first simplify the relation of
CS's K and K' to have a coincident origin. When this is done,
your question,

" 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?"

That's right, you can then generally substitute

g_i4 = - g_ij dx^j/dx^4

(which is your v/c f(x) term ).

Again I do not follow. If a term v/cf(x) was added then the transformation
would be:

x' = c1x + c2t + c3 + v/cf(x). Taking the ratio of a rod of length delta x
in the x system to the rod measured in the x' system we have delta x'/delta
x = c1deltax + v/cf(xstart) - v/cf(xend)/deltax which is not independent of
position as the homogeneity property of an inertial reference frame implied
it should be. Remember we are dealing with SR here and inertial reference
frames.

Thanks
Bill

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

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. Homogeneity of space and time
implies
this must be independent of x or t.

Perfectly Innocent wrote:
If delta x'/delta x varied like Exp(kt), I'd call that homogeneous.

Remember in this sense homogeneous in time means that the same experiment
done at a different time will give the same results. It is a physical
requirement - not a mathematical one. Thus if you added your term to the
equation you would have the ratio of a rod of length delta x stationary in
the non ' system to the same rod measured in the ' system as exp(kt) which
is dependant on when the experiment is carried out. The homogeneity
property is broken - you can tell when the experiment was carried out and is
not allowed by the properties of an inertial reference frame.

Thanks
Bill

"Bill Hobba" wrote in message ...
Ok, Bill, pardon the top post, I may be confused but
I'm not doing GR when I use,

ds^2 = g_uv dx^u dx^v (kst1)

since this is an abbreviated way of writing

ds^2 = c^t^2 - dx^2 -dy^2 -dz^2 (kst2)

when g11...g33 =-1, g00=1.

So (kst1) is SR when the metrics are constants, nothing
precludes the use of tensor analysis merely because one
is doing SR.

Would you agree that Eq. (kst1) essentially contains
everything necessary about the Lorentz transform?

For example in SR, it is possible to integrate (kst1) to find,
(after specifyng metrics that are constant in integration)

s^2 = g_uv x^u x^v + constant of integration. (kst3)

Obviously Eqs (kst1,3) are expressed in the prime system,

ds^2 = g'_uv dx'^u dx'^v (kst1')

s^2 = g'_uv x'^u x'^v + constant' of integration. (kst3')

Any SR or Lorentz transform must comply with kst1,3
and kst'1,3 and resolve kst2. So if you begin with
these equations and derive what you want, that should
be ok.

That said, you can substitute the following metrics,
g00.....g33 =1 and g14 = -v/c in Eq. (kst1) to get
(I'll just use t and x),

ds^2 = g00 dt^2 + 2 g01 dt dx + g11 dx^2

= dt^2 -2 v/c dt dx + dx^2

v = dx/dt

ds^2 = dt^2 - 2 (dx/dt) dt dx + dx^2 = dt^2 - dx^2

and is Eq. (kst2).

Regards
Ken S. Tucker

Ken S. Tucker wrote;
Bill, you're using finite relations in the transformation, (quote from
above),

I have no idea what your trying to say.

Ken S. Tucker wrote;

x' = c1x + c2t + c3, (Hobba1)

something that is difficult to do when relating
any CS at a point, (General Covariance).

The principle of general covariance (really the principle of general
invariance but I will not be that picky) is a GR concept - I am dealing with
SR here so it is a concept that is not applicable.

Ken S. Tucker wrote;
'The quoted equation relates CS's separated in location. It becomes
unreasonably difficult to compare distance origins without the benefit of GR
even if no g-fields are present,'

Again this has nothing to do with GR - it is an inertial frame so gravity is
precluded.

Ken S. Tucker wrote;
You go on to say,
"A similar proof follows for the time coordinate."

presuming a solution like,

t = (x' - c1x - c3)/c2, which is getting complicated.

No it is the time coordinate in the ' system ie

t' = c4x + c5t + c6

Ken S. Tucker wrote;
The reason for the complication is due to non-coincident origins.

Coincidence of origins will remove the constant at the end but is linear in
either case ie if the origins are coincident then the transformation is:

x' = c1x + c2t and t' = c3x + c4t.

Ken S. Tucker wrote;
But a transformation of location will first simplify the relation of
CS's K and K' to have a coincident origin. When this is done,
your question,

" 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?"

That's right, you can then generally substitute

g_i4 = - g_ij dx^j/dx^4

(which is your v/c f(x) term ).

Again I do not follow. If a term v/cf(x) was added then the transformation
would be:

x' = c1x + c2t + c3 + v/cf(x). Taking the ratio of a rod of length delta x
in the x system to the rod measured in the x' system we have delta x'/delta
x = c1deltax + v/cf(xstart) - v/cf(xend)/deltax which is not independent of
position as the homogeneity property of an inertial reference frame implied
it should be. Remember we are dealing with SR here and inertial reference
frames.

Thanks
Bill

Bilge said
'Your "proof" just states that the length of the rod is unchanged under an
infinitessimal displacement. Since the infinitessimal displacement depends
only on the first derivatives, and a finite displacement is built from a
series of infinitessimal ones, the transform is linear.'


Dill Hobba replied (misspelling intended)
I am a little confused here. I said delta x'/delta x is independent of x
or
t where delta x' and delta x are finite. I then take the limit to give
the
derivative and integrate back up to obtain the original function f(x,t).
I
probably am missing something. Could you clarify?

God I am stupid. I see Bilge's point. The fact that detla x'/delta x = c
where c is impendent of x or t implies more or less by the definition of
linearity it must be linear (same increase in x always leads to same
increase in x') without the calculus. That comes from having a math
background where you try to prove things without understanding what they
mean. Thus taking the origin as the start point in both systems you have
from the fact above ax = c1x for a fixed t say t = 0. Same for the origin
(ie ax=0) delta x/delta t = c. Thus the origin transforms as ax = c2t. To
obtain the full function you add the distance the rod moved from the origin
to the length of the rod to get x' = c1x + c2t.

Thanks
Bill

Bilge said
'Your "proof" just states that the length of the rod is unchanged under
an
infinitessimal displacement. Since the infinitessimal displacement
depends
only on the first derivatives, and a finite displacement is built from a
series of infinitessimal ones, the transform is linear.'


Dill Hobba replied (misspelling intended)
I am a little confused here. I said delta x'/delta x is independent of
x
or
t where delta x' and delta x are finite. I then take the limit to give
the
derivative and integrate back up to obtain the original function f(x,t).
I
probably am missing something. Could you clarify?


Bill Hobba said
God I am stupid. I see Bilge's point. The fact that detla x'/delta x = c
where c is impendent of x or t implies more or less by the definition of
linearity it must be linear (same increase in x always leads to same
increase in x') without the calculus. That comes from having a math
background where you try to prove things without understanding what they
mean. Thus taking the origin as the start point in both systems you have
from the fact above ax = c1x for a fixed t say t = 0. Same for the origin
(ie ax=0) delta x/delta t = c. Thus the origin transforms as ax = c2t.
To
obtain the full function you add the distance the rod moved from the
origin
to the length of the rod to get x' = c1x + c2t.

After reading what I wrote a few typo's and the way I said things may not
convey exactly what I was trying to say. As Bilge said finite displacements
are built from infinitesimal ones so the ratio of an infinitesimal rod in
one frame to the rod viewed in the other is independent of where the rod
is - this leads to the ratio of finite rods being the same ie linear.

However me being the way I am I kind of like the calculus argument ie dx'/dx
= c1 then integrating up to get x' = c1x + c(t) etc. Of course since there
are more than one variable I really should have used partial derivatives.

Thanks
Bill

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


Eugine Shubert replied.
Homogeneity and isotropy is properly understood as applying to space,
not coordinates. Furthermore, explicit examples exist proving that
non-linear functions are allowed to be the change of coordinate
transformations between physically equivalent inertial frames of
reference in a relativistic, isotropic and homogeneous space.

See http://www.everythingimportant.org/r...eneralized.htm

I take the principle of relativity to mean the physical equivalence
of all inertial frames of reference.

It applies to the points in a reasonable coordinate such as a Cartesian one
(basically a stationary one) As points have no objective existence apart
from a coordinate system your statement does not even make much sense. I
have seen your counter example and does not obey the space homogeneity
property - the v/cf(x) term forbids it - you can tell the difference between
points.

If you believe in the POR then you believe that an inertial system is
homogeneous and isotropic. Simply shift all the clocks a certain amount and
mover and /or rotate you coordinate system and you again have an inertial
coordinate system. Indeed that's is how Landau in his book Mechanics
defines an inertial coordinate system.

After thinking about what Bilge wrote I have now come up with the most
elegant way I know to prove it.

Consider the transformation as a vector relation L(X) where X = (x1, x2, x3,
t). For small delta x from the calculus we know that L(X + delta x) = L(X)
+ A(delta x) where A is a 4x4 matrix. Now from homgeinity in space and time
A must not depend on X - if it did you would be able to tell the difference
between differnt points and times by the value of A. Thus L(X) = L(0 +
delta x1 + delta x2 +++ delta xn) where each delta xi is small and adds up
to X. Hence L(X) = L(0) + A(delta x1) + A(delta x2) ++++ A(delta xn) = L(0)
+ A(delta x1 + delta x2 ++++ delta xn) = L(0) + A(X) = C + A(X) where C is
the vector L(0). Thus the transformation is linear.

For you to still allow your v/cf(x) term you must show the fault in my
reasoning.

Thanks
Bill