On Fri, 17 Jun 2022 18:51:08 -0700 (PDT), Stan Fultoni
<fultonistan@gmail.com> wrote:

> If you go back and actually state all of your premises, correctly and clearly, you will reproduce the standard axiomatic derivation.

What are some good references for the "standard axiomatic derivation".
TIA

On Wednesday, June 15, 2022 at 2:31:22 PM UTC-7, Ricardo Jimenez wrote:
> To derive the Lorentz transforation (2D)
>
> 1. It is a 2x2 real matrix with non 0 determinent.
> 2. It transforms x=vt into x'=0; its inverse transforms x'=-vt to x=0.
> 3. It transforms x=ct, x=-ct to x'=ct', x'=-ct' respectively.
> 4. v, c are real with c>0 and |v|<c.
>
> Nothing more, nothing less is needed. Correct?

It's just "is it what it is", ....

All those you'd agree are "continuous" or "real-valued" variables.

Then there's leaving out coordinates above light, how that affects
the ratios, when in fact classically it all has to be computed
instantaneously from the classical, "inside" and "outside" the
natural units of "light's speed: maximum in space terms".
I.e. in Special Relativity v is always < 1.

Or rather, it results that it is always < 1, after computing inputs,
then computing those back down under the ratios of the local invariance,
about the arithmetic only [0,1] and the properties of their products
accordingly, all in the same range [0,1], where modeling an infinite
speed would be [0,oo), in terms of what is computed the instantaneous
impulse, classically, according to the current distance(s).

Or, for example, there is the arithmetic where [1,2] is the "post-classical"
bound and models an invariant, twice light's speed, an invariant,
under what transforms the products results according to invariant what
are bounded.

Here though the idea of gravity's speed being infinite, or a theory
where it is so, then mathematically it's the space of still
what result the classical velocities, that gravity's force is so small,
here what it means for light, mass-less, and mass-energy equivalence, and
bodies, massy, and mass-energy equivalence, and why linear accelerators,
and cyclotron accelerators, are fundamentally different.

In terms of potential, travel seems results this according to SR, GR, ....

Under these so many more terms, ....

The Lorentz is a linear impulse I think will agree.

Bounded down under the quadratic is for inverse square, ....

There are of course many or various ways to write a Lorentz transform.
I.e. there are several ways writing it and saying "Lorentz, Hamiltonian,
LaGrangians, Galileo, ...".

Invariant theory these days, unified field theory, invariant theory
and measure theory, singularity theory, ....

On Saturday, June 18, 2022 at 5:37:12 PM UTC-7, Ricardo Jimenez wrote:
> > If you go back and actually state all of your premises, correctly and clearly,
> > you will reproduce the standard axiomatic derivation.
>
> What are some good references for the "standard axiomatic derivation".

Every good book on the foundations of relativity theory (since about 1910) covers this extensively. It was already "old hat" when Pauli wrote his famous encyclopedia article on relativity in 1920. Even good introductory texts usually cover it, such as Rindler's "Essential Relativity" (1969). There are whole books devoted to presenting literally dozens of derivations based on a variety of premises. I urge you to acquaint yourself with the vast literature on this subject.

El sábado, 18 de junio de 2022 a las 20:37:12 UTC-4, Ricardo Jimenez escribió:

>
> > If you go back and actually state all of your premises, correctly and clearly, you will reproduce the standard axiomatic derivation.
> What are some good references for the "standard axiomatic derivation".
> TIA

Good and very clear derivations of the Lorentz Transformations is available in two very good sources:

1) Landau, Liftshitz, "The classical Theory of fields", chapter 1, section 4. This first chapter has only 21 pages but covers the whole algebra for Special Relativity.
2) Lecture 1 of the course Special Relativity of Professor Susskind (available at https://www.youtube.com/watch?v=toGH5BdgRZ4&t=4368s). There he derives the geometrical version, showing both the transformations and the relativity of simultaneity. Note, with c=1, how the LTEs become quite symmetrical as x' = (x - vt)/√(1-v²) and t' = (t - vx)/√(1-v²)

On Sat, 18 Jun 2022 21:09:00 -0700 (PDT), Stan Fultoni
<fultonistan@gmail.com> wrote:

>On Saturday, June 18, 2022 at 5:37:12 PM UTC-7, Ricardo Jimenez wrote:
>> > If you go back and actually state all of your premises, correctly and clearly,
>> > you will reproduce the standard axiomatic derivation.
>>
>> What are some good references for the "standard axiomatic derivation".
>
>Every good book on the foundations of relativity theory (since about 1910) covers this extensively. It was already "old hat" when Pauli wrote his famous encyclopedia article on relativity in 1920. Even good introductory texts usually cover it, such as Rindler's "Essential Relativity" (1969). There are whole books devoted to presenting literally dozens of derivations based on a variety of premises. I urge you to acquaint yourself with the vast literature on this subject.

I am consulting all the books mentioned in this thread. Thanks for
the suggestions. None seem to have a list of the mathematical
assumptions needed to make a formal derivation of the Lorentz
transformation by the 4 coefficient method. One more I found
necessary to add is:
6. If (x,t1) and (x,t2) are events with t2>t1, then t'2>t'1.

This seems to be necessary in order to choose the positive square root
when solving for gamma at the end and it doesn't seem to be a
consequence of the principle of relativity, homogeneity or isotropy. I
still haven't decided on the best axioms for handling linearity and
y'=y, z'=z. I'll keep reading.

On Monday, June 20, 2022 at 6:43:02 AM UTC-7, Ricardo Jimenez wrote:
> I am consulting all the books mentioned in this thread. Thanks for
> the suggestions. None seem to have a list of the mathematical
> assumptions needed to make a formal derivation of the Lorentz
> transformation by the 4 coefficient method.

It isn't clear what you think is missing. See, for example, the simple derivation in Appendix 1 of Einstein's popular book on relativity (written in 1916), where he shows for the "4 coefficient method" (which he reaches honestly, not by fiat) that it suffices to stipulate that x=+-ct maps to x'=+-ct', and that the transformation and its inverse (specifically length relations) are reciprocal (which they must be by relativity), meaning the inverse has the same form with v replaced by -v.

But this differs from what more sophisticated approaches are trying to do, which is to see how far we can get before even invoking light speed invariance, purely proceeding from relativity and isotropy. This alone suffices to yield the Lorentz transformation with one degree of freedom, which we can call the parameter k, so it can yield either Euclidean, Galilean, or Lorentzian relativity. If k=0 the transformation is Galilean, whereas if k=1/c^2 the transformation is Lorentzian. It can also be shown that the inertia of a quantity E of energy is kE, so the inertia of energy suffices to deduce special relativity.

> One more I found necessary to add is:
> 6. If (x,t1) and (x,t2) are events with t2>t1, then t'2>t'1.

That's is usually expressed as a result, not a premise, i.e., Lorentz transformations map timelike to timelike intervals, and also spacelike to spacelike intervals.

> I still haven't decided on the best axioms for handling linearity

That just comes from the definition of an inertial coordinate system as one for which the laws of inertia hold good, combined with continuity to rule out linear fractional.

> and y'=y, z'=z.

That isn't an axiom, it's an immediate deduction from isotropy.

On 6/17/22 6:26 PM, Stan Fultoni wrote:
> Tom Roberts wrote:
>> Note that paper uses the group property (B), but the author does
>> not call it that.
>
> There is no meaningful group theory used in the derivation.

Apparently you did not bother to actually read the paper cited. It
computes the composition of two transforms and requires the result to
have the same form as the original (single) transform. That is applying
group theory, even through the author didn't call it that.

> Again, it is a transformation, not a transform (sheesh),

Picayune triviality. The two words are used interchangeably, as in this
context they are synonyms. (IOW: this is just nouning a verb.)

> and as every competent intellect knows, it suffices to consider the
> 1+1 dimensional case, since the composition of boosts with ordinary
> rotations is trivial and unambiguous.

Apparently you have not actually studied Lorentz transforms in 3+1
dimensions -- the composition of boosts and rotations is CERTAINLY not
trivial [#]. For instance, the composition of two boosts in different
directions generates a spatial rotation (as well as a boost) -- that
would not be expected on the basis of the transforms in 1+1 dimensions,
or on the basis of Galilean transforms. Moreover, the 1+1-D subgroup is
abelian, but the full 3+1-D Lorentz group is not.

[#] If you really think this is "trivial", show us all
how incredibly complicated the composition of velocities
actually is, for two arbitrary velocities in 3+1-D.
No sensible person would consider that "trivial".
Hint: it's too complicated even for Wikipedia, where
they restricted the first to be coordinate aligned.

Tom Roberts

On Tuesday, 21 June 2022 at 17:27:37 UTC+2, tjrob137 wrote:

> Apparently you have not actually studied Lorentz transforms in 3+1
> dimensions -- the composition of boosts and rotations is CERTAINLY not
> trivial [#]. For instance, the composition of two boosts in different
> directions generates a spatial rotation (as well as a boost) -- that
> would not be expected on the basis of the transforms in 1+1 dimensions,
> or on the basis of Galilean transforms. Moreover, the 1+1-D subgroup is
> abelian, but the full 3+1-D Lorentz group is not.

In the meantime in the real world, of course, forbidden
by your insane Shit GPS and TAI keep measuring t'=t,
just like all serious clocks always did.

On Tuesday, June 21, 2022 at 8:27:37 AM UTC-7, tjrob137 wrote:
> > There is no meaningful group theory used in the derivation.
>
> the composition of two transforms and requires the result to
> have the same form as the original (single) transform. That is applying
> group theory...

No it's not. You clearly don't even know what "applying group theory" would look like. The fact that the composition of two linear transformations is a linear transformation is not "applying group theory", and the proposition that the transformation between inertial-based coordinates (as defined) is necessarily reciprocal follows physically from the relativity principle and isotropy, not "from group theory". Sheesh.

> even through the author didn't call it that.

He didn't call it that because it is not that. Sheesh.

> > Again, it is a transformation, not a transform (sheesh),
>
> Picayune triviality. The two words are used interchangeably...

Nope they are not. A transform is not the same things as a transformation. Please try to educate yourself at least a little bit and stop making a fool of yourself.

> > and as every competent intellect knows, it suffices to consider the
> > 1+1 dimensional case, since the composition of boosts with ordinary
> > rotations is trivial and unambiguous.
>
> Apparently you have not actually studied Lorentz transforms in 3+1
> dimensions...

You're so funny. As I said, all derivations proceed from the standard configuration, which reduces to the 1+1 analysis, and then we can trivially apply conjugated rotations to the boosts in the obvious way to give arbitrary 3+1 transformations. That fact that this entails Thomas precession, etc., doesn't negate this fact.

> No sensible person would consider that "trivial".

Well, trivial is certainly a relative concept. I don't doubt that many things that are trivial for most people are not trivial for you, but that's a separate discussion...

On Mon, 20 Jun 2022 11:33:25 -0700 (PDT), Stan Fultoni
<fultonistan@gmail.com> wrote:

>On Monday, June 20, 2022 at 6:43:02 AM UTC-7, Ricardo Jimenez wrote:
>> I am consulting all the books mentioned in this thread. Thanks for
>> the suggestions. None seem to have a list of the mathematical
>> assumptions needed to make a formal derivation of the Lorentz
>> transformation by the 4 coefficient method.
>
>It isn't clear what you think is missing. See, for example, the simple derivation in Appendix 1 of Einstein's popular book on relativity (written in 1916), where he shows for the "4 coefficient method" (which he reaches honestly, not by fiat) that it suffices to stipulate that x=+-ct maps to x'=+-ct', and that the transformation and its inverse (specifically length relations) are reciprocal (which they must be by relativity), meaning the inverse has the same form with v replaced by -v.
>
Einstein uses a variant of the 4 coefficient method but essentially
assumes the same axioms. He rotates the coordinate axes to x-ct=0 and
x+ct=0 and similarly for the (x',ct') frame. With these axes, the
light cone assumption together with the linearity assumption implies
that the transformation is given by a diagonal matrix with real
entries he denotes by gamma and mu. This is conjugate to the matrix
for the (x,ct) frame which turns out only to have 2 unknown
parameters. Determining these parameters is done similarly to what is
done in the usual 4 coefficient proof. Also he just assumes y=y' and
z=z' without explanation, at least in the body of Appendix 1.
>
>> One more I found necessary to add is:
>> 6. If (x,t1) and (x,t2) are events with t2>t1, then t'2>t'1.
>
>That's is usually expressed as a result, not a premise, i.e., Lorentz transformations map timelike to timelike intervals, and also spacelike to spacelike intervals.
>
I take back 6. because it is unnecessary. The assumption in 1. that
the determinant of the Lorentz transformation is positive forces the
choice of the positive square root for gamma.

>> I still haven't decided on the best axioms for handling linearity
>
>That just comes from the definition of an inertial coordinate system as one for which the laws of inertia hold good, combined with continuity to rule out linear fractional.
>
Where can I find the details? Can lines of the form t=constant be
considered world lines and so must map into other world lines by the
law of inertia?

>> and y'=y, z'=z.
>
>That isn't an axiom, it's an immediate deduction from isotropy.

I notice some writers just assume it while others give a physical
argument involving a cylinder moving along the x axis at velocity v. I
would like something more precise than just using the word "isotropy"
to wave it away.

On Tue, 21 Jun 2022 17:25:07 -0400, Ricardo Jimenez
<rickyjim@earthlink.net> wrote:

>>> and y'=y, z'=z.
>>
>>That isn't an axiom, it's an immediate deduction from isotropy.
>
>I notice some writers just assume it while others give a physical
>argument involving a cylinder moving along the x axis at velocity v. I
>would like something more precise than just using the word "isotropy"
>to wave it away.

I think I see it now. Isotropy means invariance under rotation. Here
y' = Y(v, x, y, z, ct), z' = Z(v, x, y, z, ct) should not change under
a rotation of the y-z plane while v, x and t are held fixed. The same
is true for y = Y(-v, x', y', z'), z = Z(-v, x', y', z'). Thus there
is a contradiction unless y'= y and z'= z. Am I approximately right?

On Tuesday, June 21, 2022 at 2:25:17 PM UTC-7, Ricardo Jimenez wrote:
> >See, for example, the simple derivation in Appendix 1 of Einstein's popular book on relativity (written in 1916), where he shows for the "4 coefficient method" (which he reaches honestly, not by fiat) that it suffices to stipulate that x=+-ct maps to x'=+-ct', and that the transformation and its inverse (specifically length relations) are reciprocal (which they must be by relativity), meaning the inverse has the same form with v replaced by -v.
> >
> Einstein uses a variant of the 4 coefficient method but essentially assumes
> the same axioms.

He just stipulates that x=+-ct maps to x'=+-ct', and that the transformation and its inverse (specifically length relations) are reciprocal (which they must be by relativity), meaning the inverse has the same form with v replaced by -v. Is that what you mean by "the same axioms"?

> ...he just assumes y=y' and z=z' without explanation, at least in the body of Appendix 1.

Not so. After using the "4 coefficient method" to derive the 1+1 transformation, he explain that in 3+1 dimensions we stipulate that r=+-ct implies r'=+-ct', so there is a constant "s" such that (x^2 + y^2 + z^2 - c^2t^2) = s(x'^2 + y'^2 + z'^2 - c^2t'^2), and from the result on the x axis we know s=1, so the quadratic expression is invariant, and by substituting the transformed expressions for x' and t' into this equality, we get y^2 + z^2 = y'^2 + z'^2. So, the only degree of freedom would be to spatially rotate about the x axis, but by convention we agree to keep the axes aligned for our standard configuration (pure boost), and hence y=y' and z=z'. (You could write the transformation with a rotation about the x axis, but then you would be combining a boost with a spatial rotation.)

> >> I still haven't decided on the best axioms for handling linearity
> >
> >That just comes from the definition of an inertial coordinate system as one for which the laws of inertia hold good, combined with continuity to rule out linear fractional.
> >
> Where can I find the details? Can lines of the form t=constant be
> considered world lines and so must map into other world lines by the
> law of inertia?

That comes from homogeneity and isotropy, e.g., laying out identical rulers with equally-spaced clocks can't map to non-uniformly distributed rulers and clocks. The non-linearity would violate homogeneity and isotropy.

> I think I see it now. Isotropy means invariance under rotation. Here
> y' = Y(v, x, y, z, ct), z' = Z(v, x, y, z, ct) should not change under
> a rotation of the y-z plane while v, x and t are held fixed. The same
> is true for y = Y(-v, x', y', z'), z = Z(-v, x', y', z'). Thus there
> is a contradiction unless y'= y and z'= z. Am I approximately right?

That's on the right track, although Einstein applies it more simply. In his 1905 he derives y'=phi(v)y for v in the x direction, and then notes that by isotropy (same in both directions) we must have phi(v)=phi(-v), and we already have phi(v)*phi(-v)=1, so we have phi(v)=1, meaning y'=y, and same for z'=z. It's actually even simpler: If you imagine a rod moving perpendicular to its axis, it's length can't depend on the direction in that perpendicular plane.