Tom Roberts wrote in message ...
On 6/25/2003 7:25 AM, Perfectly Innocent wrote:
I believe that this supposed necessity of linearity is a very common
misconception.

Not really. It seems you don't really understand what "homogeneous and
isotropic" mean, and what the specific coordinates of SR require (see
below).

In the context of this discussion, "homogeneous and isotropic" is a
property of spacetime. The only initial requirement for each inertial
frame of reference is a suitable metric space and a clock at each
point to specify the time of local events.

Yes. Look below and see that you misunderstand.

I have challenged the claim wherever I have encountered
it and find it strange that, on this one point, I have never received
so much as a single word of doubt. Doesn't anyone understand what's
being said?

Yes.


Jackson is wrong. The isotropy and homogeneity of space-time has
nothing to do with linearity. See exercise 1 and 2 of the following
link for an explicit counterexample.
http://www.everythingimportant.org/r...eneralized.htm

Your transform (your ex. 1) is manifestly nonlinear. And
non-homogeneous.

I couldn't possibly claim a counterexample if my transform was linear.

And violates the PoR (the frame with v=0 is special).

Where's your proof that my transform violates the PoR?
What do you mean "the frame with v=0 is special"?

When v=0, my transform reduces to x'=x, t'=t, just like in the Lorentz
transformation! Why can't we apply your logic to Einstein's SR and
dismiss it immediately by concluding that the frame with v=0 is
special?

You seem to be equating a single frame with a transformation between
frames. I know that you like to accuse others of being confused. Do
you ever point a finger to yourself?

It is not a counterexample to the claim, it is merely a more general
coordinate transform than the Lorentz transform. shrug

shrug My transformation was built from the ground-up using the
Lorentz transformation. I only added or subtracted some constant to
every Einsteinian clock in the standard setting in every inertial
frame of reference. How does this insolent and disrespectful act
against the demigods of spacetime violate the isotropy and homogeneity
of spacetime?

While it may well be experimentally indistinguishable from SR
(I haven't checked),

What would you check?

it is clear that standard synchronized
clocks and rulers at rest in the moving frame do not correspond
to the coordinate values x' and t'.

The standard synchronization isn't sacrosanct. I think you're speaking
gibberish.

This is just one more
example of using a "funny" coordinate system, and becoming
confused by not dealing with INVARIANT QUANTITIES

Correct. You are confused because you are not dealing with INVARIANT
QUANTITIES.

(the funny
coordinate system implies funny metric components, and you
forgot to look at them).

shrug I don't need to look at the components of a pseudo metric. The
Shubertian clock technique is the obvious way to understand SR.
http://www.everythingimportant.org/relativity/

Correction: I shouldn't say, "Jackson is wrong." It would be more
meaningful to say Jackson probably never followed the reasoning to the
statement that "the isotropy and homogeneity of space-time requires
the linearity of the Lorentz transformation."

Remember, please, that "homogeneous and isotropic" apply to the metric
of spacetime, and here we're really discussing the projection of the
metric onto Cartesian coordinate systems constructed using standard
clocks and rulers at rest in the inertial frames. It is an elementary
exercise to show linearity is required among coordinates meeting those
requirements. Elementary, at least, if one understands both real
analysis and the basics of diffrential geometry.

You're not qualified to debate this until you understand the meaning
of arbitrary coordinate systems. Consult all your textbooks on real
analysis and differential geometry if you think they are superior to
Shubertian clock technology. Up to this point you've firmly
demonstrated total bewilderment on the meaning of coordinates.

Here is my counterexample.

x' = b + Y(v)[(x-b) -v(t-a) + (v/c)f(x)]

t' = a + Y(v){(t - a) - v(x-b)/c^2 -f(x)/c} + (1/c)f{b + Y(v)[(x-b)
-v(t-a) + (v/c)f(x)]}

f(x) is any function of x.
a = t_0, b = x_0, are constants.
Y(v) = [1-(v/c)^2]^(-1/2).
Note that the transformations I specified form a group.

What does it mean physically?

Yes, this goes beyond the mathematical basis of most presentations of
SR. That's why they merely state linearity, rather than proving it is
required. Many/most physicists are cavalier about mathematical rigor,
including textbook authors....

It's easy to see that linearity in SR is an arbitrary choice favoring
convenient and easily managed clock synchronizations.

Hmmm. I think that's correct. At least in some sense. But SR also
includes the idea of using coordinates based on standard clocks and
rulers at rest in the inertial frame(s) of interest. This last is really
what requires linearity of the coordinate transforms, not just clock synch.

Tom Roberts

What is non-standard about my clocks and rulers? My definition of
distance is based on the very best certified rulers and my definition
of time is built from genuine Shubertian clock components. I think
you're reasoning in a circle.

There is nothing in the assumption of spatial homogeneity and isotropy
that disallows arbitrary coordinate systems. If arbitrary coordinate
systems can be conceived, then nonlinear transformations between
different inertial frames of reference are expected.

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

Perfectly Innocent wrote:
In the context of this discussion, "homogeneous and isotropic" is a
property of spacetime.

OK. But the standard approach to SR also applies homogeneous and
isotropic to the coordinates, because it demands that the coordinates be
laid out in every frame using standard clocks and rulers. I suspect this
is the part you missed.


Tom Roberts wrote in message ...
[your transform] violates the PoR (the frame with v=0 is special).

Where's your proof that my transform violates the PoR?
What do you mean "the frame with v=0 is special"?

Just look at it. In the frame with v=0 the coordinate speed of light is
isotropically c; in general that holds in no other frame. That is a
quite special property and it singles out that specific frame, violating
the PoR.


When v=0, my transform reduces to x'=x, t'=t, just like in the Lorentz
transformation! Why can't we apply your logic to Einstein's SR and
dismiss it immediately by concluding that the frame with v=0 is
special?

Because unlike your transformation, using Lorentz transforms does not
single out the v=0 frame -- the coordinate speed of light is
isotropically c in every frame. Other properties are similar....

Note also that only in the v=0 frame can you use standard clocks and
rulers as coordinate clocks and rulers -- in every other frame you need
to manufacture special ones, and must do so differently in each frame.


It is not a counterexample to the claim, it is merely a more general
coordinate transform than the Lorentz transform. shrug
shrug My transformation was built from the ground-up using the
Lorentz transformation. I only added or subtracted some constant to
every Einsteinian clock in the standard setting in every inertial
frame of reference. How does this insolent and disrespectful act
against the demigods of spacetime violate the isotropy and homogeneity
of spacetime?

It doesn't affect spacetime of course, but it affects the coordinates.


While it may well be experimentally indistinguishable from SR
(I haven't checked),
What would you check?

Whether the round-trip coordinate speed of light is isotropically c in
every frame. If that is true then these transforms belong the the
equivalence class of theories that are experimentally indistinguishable
from SR.


(the funny
coordinate system implies funny metric components, and you
forgot to look at them).

Remember, please, that "homogeneous and isotropic" apply to the metric
of spacetime, and here we're really discussing the projection of the
metric onto Cartesian coordinate systems constructed using standard
clocks and rulers at rest in the inertial frames. It is an elementary
exercise to show linearity is required among coordinates meeting those
requirements. Elementary, at least, if one understands both real
analysis and the basics of diffrential geometry.
[...]
What is non-standard about my clocks and rulers?

First remember that in the standard SR one demands that the coordinates
be determined by standard clocks and rulers. When I said "clocks and
rulers" I meant COORDINATE clocks and rulers. Yours are clearly
nonstandard, because in your coordinates the metric components are not
diag(-1,1,1,1).


There is nothing in the assumption of spatial homogeneity and isotropy
that disallows arbitrary coordinate systems.

Of course not! But there is also an expectation in standard SR that the
coordinates be determined by standard clocks and rulers at rest in the
frame. It is this later condition that implies homogeneity and isotropy
of the coordinates in any inertial frame, and also the linearity of
transforms between such coordinate systems.


If arbitrary coordinate
systems can be conceived, then nonlinear transformations between
different inertial frames of reference are expected.

Sure. But they don't meet the expectations of standard SR.


Tom Roberts

Tom Roberts wrote:
Just look at it. In the frame with v=0 the coordinate speed of light is
isotropically c; in general that holds in no other frame. That is a
quite special property and it singles out that specific frame, violating
the PoR. [...]

Hmmm. In thinking about this, I think there is another "hidden"
postulate of SR: the coordinates in each inertial frame are to be laid
out using standard clocks and rulers. I suspect Einstein would have
called that a definition rather than a postulate.... But it's definitely
required to obtain the Lorentz transforms, as Pmb points out (indirectly).


Tom Roberts

Bilge wrote:
Tom Roberts:
Tom Roberts wrote:
Just look at it. In the frame with v=0 the coordinate speed of light is
isotropically c; in general that holds in no other frame. That is a
quite special property and it singles out that specific frame, violating
the PoR. [...]

Hmmm. In thinking about this, I think there is another "hidden"
postulate of SR: the coordinates in each inertial frame are to be laid
out using standard clocks and rulers. I suspect Einstein would have
called that a definition rather than a postulate.... But it's definitely
required to obtain the Lorentz transforms, as Pmb points out (indirectly).

I think you are trying to make more out of special relativity than it is
or could possibly be and/or trying to remedy a defect which einstein could
only remedy by inventing general relativity. First of all, special relativity
cannot specify anything about laying out rulers and clocks in inertial
frames because special relativity doesn't define an inertial frame as
anything but a frame in which rulers and clocks are laid out such that
the first postulate is satisfied [...]

I disagree, and will be posting a more complete article shortly. It will
be replying to my own message, to which you replied.


Tom Roberts

Tom Roberts wrote:
Hmmm. In thinking about this, I think there is another "hidden"
postulate of SR: the coordinates in each inertial frame are to be laid
out using standard clocks and rulers. I suspect Einstein would have
called that a definition rather than a postulate.... But it's definitely
required to obtain the Lorentz transforms, as Pmb points out (indirectly).

After further thought, I no longer think this is needed. Certainly not
to merely establish the linearity of the transforms.

And I did not discuss a much simpler and straightforward way to
demonstrate the necessity of linearity in the transforms of SR:

Einstein's 1905 article on SR restricted consideration to systems of
coordinates "in which the equations of Newtonian mechanics hold good"
[A.Einstein, "Zur Elektrodynamik bewegter Koerper", Ann. d. Physik, 17,
1905, trans. Perrett and Jeffery; start of section I].

So ask: Do the equations of Newtonian mechanics hold good in Pmb's
coordinates? Clearly not, except for very special choices of the
arbitrary function f(.).

In fact, Newton's first law is sufficient to apply stringent kinematic
constraints on the transforms:

In the v=0 frame, consider a pointlike object with trajectory
x = a + b t [a,b,d,e,g,h are arbitrary
y = d + e t real constants]
z = g + h t
Then given the coordinate transforms in question we must have
d^2 x' / dt'^2 = 0
d^2 y' / dt'^2 = 0
d^2 z' / dt'^2 = 0
This merely says that uniform linear motion of the particle in
the unprimed coordinates must also be uniform linear motion in
the primed coordinates. [#]

Applying those requirements to all possible values of {a,b,d,e,g,h,v}
will impose constraints on the arbitrary function f(.). This is far more
algebra than I am interested in performing, but from other approaches I
know that one of the constraints will be:

f''(.) = 0 [here "'" means DIFFERENTIATION, and is NOT
a frame designator]

This of course means that f(.) is not really an arbitrary function, and
Pmb's transforms are constrained to be linear. In order to be used in SR
anyway.


Note also that Pmb's transforms are not the most general possible. It is
possible to select a different f(.) for each frame, and still have a
group. This amounts to making f(.) either labeled by v (i.e. f_v(.)), or
better, making it a function of two variables f(v,x) [%]. Yes, the
structure of the group will be different, but it will still be a group.
Linearity in its second argument will still be required (but not in its
first argument, v, whch is essentially a label). With this
generalization, I'm pretty sure the transforms will be equivalent to
those discussed in my ancient article "Theories Equivalent to SR" posted
to sci.physics.relativity on 11/21/1999:
http://www.google.com/groups?q=group...ent.com&rnum=1

The wider transform group of that article does not incorporate
Einstein's clock synchronization convention; impose that as well and one
is led uniquely to the Lorentz transforms.

[%] One must also impose some constraint on the functional form
of f(v,x). I believe continuity in v is sufficient (i.e. one can
select a different f(.) in each frame, but it must be done in
a continuous manner as a function of v). Newton's first law
will impose constraints on the functional form of f(v,x) for
both arguments (see the references in that ancient article)....

[#] This leaves open the question of whether the equations of
Newtonian mechanics hold good in the v=0 frame. That's not an
issue for the question of whether or not linearity is required
of the transforms, but it is relevant to the larger question I
opened above. Einstein discussed this in his 1907 Jahrbuch
article [thanks to Stephen Speicher for pointing this out],
and I am not interested enough to pursue it further.


Tom Roberts

Tom Roberts wrote

So ask: Do the equations of Newtonian mechanics hold good in Pmb's
coordinates?

And what exactly are my coordinates??

pmb

In article ,
Gauge wrote:
Tom Roberts wrote

So ask: Do the equations of Newtonian mechanics hold good in Pmb's
coordinates?

And what exactly are my coordinates??

pmb

(0,0,0,0)?


--
"Is that plutonium on your gums?"
"Shut up and kiss me!"
-- Marge and Homer Simpson

Tom Roberts:
Bilge wrote:


I think you are trying to make more out of special relativity than it is
or could possibly be and/or trying to remedy a defect which einstein could
only remedy by inventing general relativity. First of all, special
relativity cannot specify anything about laying out rulers and clocks in
inertial frames because special relativity doesn't define an inertial
frame as anything but a frame in which rulers and clocks are laid out such
that the first postulate is satisfied [...]


I disagree, and will be posting a more complete article shortly. It will
be replying to my own message, to which you replied.

I read your post and I don't see how you disagree. In fact, since I
read that post before this one, I had thought the opposite until
reading this post.

(Gregory L. Hansen) wrote in message ...
In article ,
Gauge wrote:
Tom Roberts wrote

So ask: Do the equations of Newtonian mechanics hold good in Pmb's
coordinates?

And what exactly are my coordinates??

pmb

(0,0,0,0)?

A highly original remark.

(Gregory L. Hansen) wrote in message ...
In article ,
Gauge wrote:
Tom Roberts wrote

So ask: Do the equations of Newtonian mechanics hold good in Pmb's
coordinates?

And what exactly are my coordinates??

pmb

(0,0,0,0)?

What does that mean?