Unstated Special Relativity Axiom

[Ricardo Jimenez]

As far as I know, the following does not follow from Einstein's 2
postulates:
Suppose two inertial systems are in uniform relative motion along the
same direction. If system 1 measures the speed of system 2 as v (as a
fraction of the speed of light, say), then system 2 will measure the
speed of system 1 as -v.

This axiom seems to be necessary to establish a fundamental fact of
relativity theory:

Suppose in an inertial system two events have coordinates (0,t'(1)),
(0,t'(2)) with t'(2)>t'(1). Then in an inertial system moving at
positive speed v, rectilinearly to the first system, the coordinates
of the same events in system 1, (x(1),t(1)) (x(2),t(2)), satisfy t(2)
- t(1) = gamma*(t'(2) - t'(1)), |x(2) - x(1)| = gamma*v*(t'(2) -
t'(1)).

What make relativity seem strange to many is that the derived concept
of speed measures the same in two moving coordinate systems while the
more basic concepts of length and time do not.

[Ricardo Jimenez]

On Sat, 04 Sep 2021 13:13:15 -0400, Ricardo Jimenez
<rick...@earthlink.net> wrote:

Sorry please fix my previous post by:
>
>Then in an inertial system 2 moving at

>positive speed v, rectilinearly to the first system, the coordinates

>those events in system 2, (x(1),t(1)) (x(2),t(2)), satisfy t(2)

[Tom Roberts]

On 9/4/21 12:13 PM, Ricardo Jimenez wrote:
> As far as I know, the following does not follow from Einstein's 2
> postulates: Suppose two inertial systems are in uniform relative
> motion along the same direction. If system 1 measures the speed of
> system 2 as v (as a fraction of the speed of light, say), then
> system 2 will measure the speed of system 1 as -v.
>
> This axiom seems to be necessary to establish a fundamental fact of
> relativity theory:

No additional axiom is needed. One derivation of the equations of SR
uses Einstein's two postulates, plus his "hidden postulates" (discussed
in 1907: a. clocks and rulers have no memory, b. space is homogeneous
and isotropic, c. time is homogeneous). These yield the Lorentz
transform from one inertial frame to a second frame moving with velocity
v relative to the first. Invert it to obtain the transform from the
second frame to the first: a Lorentz transform with v replaced by -v.

Tom Roberts

[Ricardo Jimenez]

I don't understand your statement that no additional axiom is needed
above Einstein's 2 postulates but then you bring up a, b and c in
order to prove the Lorentz transform. So from what does it follow
that if observer 1 measures observer 2 moving at speed v, observer 2
will measure observer 1 moving at speed -v?

[rotchm]

On Saturday, September 4, 2021 at 6:08:47 PM UTC-4, Ricardo Jimenez wrote:

> So from what does it follow
> that if observer 1 measures observer 2 moving at speed v, observer 2
> will measure observer 1 moving at speed -v?

By the relativity principle, both observers will measure the same magnitude v.
[Or, you can require linearity of the eqs...].

The minus sign, is there because YOU put it there, YOU chose the arbitrary orientation of the coordinate systems.
Note that when you use the vector form, 3D say, of the LT's, this minus is not there; its simply a velocity vector. And as you must know, vectors don't have a "sign". They have an orientation. The LT's are independent of the "origins", "orientations" of the frames.
I.e. solutions to ds² = dx² + dy² + dz² - dt²

https://en.wikipedia.org/wiki/Lorentz_transformation#Homogeneous_Lorentz_group

Once YOU chose your coordinate system(s) (& orientations), then you may "simplify" and rewrite your eqs, generating your minus sign(s).
This is the same thing as in Galilean relativity, or plain ol' kinematics; its a choice of direction/orientation/sign.

[rotchm]

On Saturday, September 4, 2021 at 9:46:16 PM UTC-4, rotchm wrote:

e.g.: let's say we are both side-by-side one meter apart. I can say that you're at plus one meter from me or I can say you're at -1 meter from me; It depends on my choice orientation of my coordinate system. It has nothing to do with the physics And is not part of physics. The minus sign, I put it there for convenience.

[Ricardo Jimenez]

On Sat, 4 Sep 2021 18:46:15 -0700 (PDT), rotchm <rot...@gmail.com>
wrote:

Velocity vectors are in tangent spaces which are vector spaces and so
negation makes sense. It even says so in your link. But no matter.
To make things simple I should have said that both systems move with
respect to each other along the x axis. But still you are begging the
question. How does the relativity principle (whatever that is)
account for the same measured absolute value of velocity of observers
while it doesn't imply that distances and time are measured the same
by both systems? Did you read the analysis in my original post? I
still think that my suggested axiom, or something equivalent, is
needed to derive the basics of relativity.

[Maciej Wozniak]

On Saturday, 4 September 2021 at 22:33:39 UTC+2, tjrob137 wrote:
> On 9/4/21 12:13 PM, Ricardo Jimenez wrote:
> > As far as I know, the following does not follow from Einstein's 2
> > postulates: Suppose two inertial systems are in uniform relative
> > motion along the same direction. If system 1 measures the speed of
> > system 2 as v (as a fraction of the speed of light, say), then
> > system 2 will measure the speed of system 1 as -v.
> >
> > This axiom seems to be necessary to establish a fundamental fact of
> > relativity theory:
> No additional axiom is needed.

We just have to realize we're FORCED!! Forced to THE BEST WAY!
(the way of your idiot guru and his idiot minions).

[Tom Roberts]

On 9/4/21 5:08 PM, Ricardo Jimenez wrote:
> On Sat, 4 Sep 2021 15:33:31 -0500, Tom Roberts
> <tjrobe...@sbcglobal.net> wrote:
>> On 9/4/21 12:13 PM, Ricardo Jimenez wrote:
>>> As far as I know, the following does not follow from Einstein's 2
>>> postulates: Suppose two inertial systems are in uniform relative
>>> motion along the same direction. If system 1 measures the speed
>>> of system 2 as v (as a fraction of the speed of light, say), then
>>> system 2 will measure the speed of system 1 as -v.
>>>
>>> This axiom seems to be necessary to establish a fundamental fact
>>> of relativity theory:
>>
>> No additional axiom is needed. One derivation of the equations of
>> SR uses Einstein's two postulates, plus his "hidden postulates"
>> (discussed in 1907: a. clocks and rulers have no memory, b. space
>> is homogeneous and isotropic, c. time is homogeneous). These yield
>> the Lorentz transform from one inertial frame to a second frame
>> moving with velocity v relative to the first. Invert it to obtain
>> the transform from the second frame to the first: a Lorentz
>> transform with v replaced by -v.
>>
>> Tom Roberts
>
> I don't understand your statement that no additional axiom is needed
> above Einstein's 2 postulates but then you bring up a, b and c in
> order to prove the Lorentz transform.

The axioms I labeled a, b, and c were implicitly assumed in Einstein;'s
1905 derivation, but he did not describe them until 1907. They are not
"additional", they are part and parcel of his derivation. They are
easily missed, because they are simply stating rather obvious things.

> So from what does it follow that if observer 1 measures observer 2
> moving at speed v, observer 2 will measure observer 1 moving at
> speed -v?

As I said, it follows from the Lorentz transform. When O1 and O2 are
both inertial coordinates, and the transform O1->O2 is a Lorentz
transform with velocity v, then O1 will measure the velocity of O2
relative to O1 as v. Ditto for the inverse transform O2->O1 implying
that O2 measures the velocity of O1 as -v.

[If one does this properly in 3+1 dimensions, then v is
a 3-vector, and there is no requirement that the O1
and O2 spatial coordinate axes align with each other.]

So no "additional axiom" is needed. IOW: your supposed "axiom" is easily
derived from the Lorentz transform.

Tom Roberts

[Ricardo Jimenez]

On Mon, 6 Sep 2021 09:35:50 -0500, Tom Roberts

You need the axiom in order to rigorously prove the Lorentz transform.
Your reasoning is circular.

[Townes Olson]

On Saturday, September 4, 2021 at 8:12:38 PM UTC-7, Ricardo Jimenez wrote:
> How does the relativity principle (whatever that is) account for the same m

> measured absolute value of velocity of observers ...

The principle of relativity asserts that for every material particle there is a system of coordinates (called inertia-based coordinates) in terms of which the particle is instantaneously stationary and all the equations of physics take the same homogeneous and isotropic form. This is a very strong principle. This principle (noting the isotropy) entails reciprocity, as can be seen by considering two identical charged particles, initially adjacent and at mutual rest; after they are released they repel each other in opposite directions. Each particle is instantaneously at rest in terms of some system of inertia-based coordinates, and the equations of physics take the same isotropic form in terms of both, so in terms of those coordinates the other is moving away at some speed v. Since the same (isotropic) laws of physics apply to both, and the situation is perfectly symmetrical, the spatial origin of the inertia-based coordinate system in which each particle is at rest is receding from the other at the same speed. We can construct this for any two systems of inertia based coordinates, so they all exhibit reciprocity.

> while it doesn't imply that distances and time are measured the same
> by both systems?

No, the principle of relativity doesn't imply that there is only a single system of inertia-based coordinates. It implies that there is some constant k such that, for any two systems of inertia-based coordinates, we have x'=(x-vt)g and t'=(t-kvx)g where g=1/sqrt(1-kv^2). If k=0 this is the Galilean transformation, and if k=1/c^2 this is the Lorentz transformation.

[Tom Roberts]

On 9/6/21 11:50 AM, Townes Olson wrote:
> the principle of relativity doesn't imply that there is only a single
> system of inertia-based coordinates. It implies that there is some
> constant k such that, for any two systems of inertia-based
> coordinates, we have x'=(x-vt)g and t'=(t-kvx)g where
> g=1/sqrt(1-kv^2). If k=0 this is the Galilean transformation, and if
> k=1/c^2 this is the Lorentz transformation.

Yes. And if k=-1/c^2 this is a Euclidean rotation in the x-t plane.

Of course only k=1/c^2 agrees with all the experiments.

Tom Roberts

[Ross A. Finlayson]

Huh, thanks, that's a neat way to help organize the concern.

I don't see how b and curved space-time fit but I suppose that's RoS.

(Or rather my aside is that's my comment.)

Then, I might write about v and -v as the "light speed rest frame theory",
as what puts that up front.

What as just a theory is flexible in that respect thank you....

Thus then I plan to use boost and pull-backs what result Fitzgerald
or space-contraction, where otherwise that might be written instead
in the usual, space terms, or where it's GR first then SR so derived,
as instead of vice-versa which is the only free observable component
in optical terms, where SR is as so.

[Maciej Wozniak]

And we're FORCED!!! To THNE BEST WAY!!!

[Thomas 'PointedEars' Lahn]

True, but (you know that) since it must be rotation-invariant, it suffices
to consider it in one spatial dimension (rotate the spatial coordinate
system such that the motion is in the positive x-direction, do the
transformation for x, then do the inverse rotation).

>>So no "additional axiom" is needed. IOW: your supposed "axiom" is easily
>>derived from the Lorentz transform.
>>
>>Tom Roberts
>
> You need the axiom in order to rigorously prove the Lorentz transform.

No, because from the two postulates of OEDoMB¹, which have their foundation
in observations/experiments, *follows* the Lorentz transform(ation). [It
was Einstein’s great achievement to find independently the same
transformation as Lorentz and Poincaré, but *without* requiring an aether.]

¹ Einstein, A.(1905). “On the Electrodynamics of Moving Bodies”.
Official Translation from German into English:
<https://einsteinpapers.press.princeton.edu/vol2-trans/154> pp.

If a frame S' with coordinates t' and x' is moving relative to a frame
S, with coordinates t and x, parallel to the x-axis in the positive x-
direction with speed v (i.e. velocity v in this one-dimensional case), then
the coordinate transformation from S to S' is the Lorentz transformation:

t' = γ (t − v/c² x)
x' = γ (x − v t)
y' = y
y' = z,

where the Lorentz factor is

γ ≔ 1/√(1 − v²/c²).

Cf. <https://einsteinpapers.press.princeton.edu/vol2-trans/159> pp.

[The only assumption that Einstein makes there is basically that for the
measured length of the rod it would not matter whether, in *same*
(stationary) reference frame, the rod would be moving to the right or
the left.]

Given the above, the coordinate transformation from S' to S can be obtained
(by rather simple algebra) as follows:

t' = γ (t − v/c² x)
⇔ t'/γ = t − v/c² x
⇔ t = t'/γ + v/c² x.

x' = γ (x − v t)
⇔ x'/γ = x − v t
⇔ x = x'/γ + v t
= x'/γ + v (t'/γ + v/c² x)
= x'/γ + v t'/γ + v²/c² x
⇔ x − v²/c² x = x'/γ + v t'/γ
⇔ (1 − v²/c²) x = x'/γ + v t'/γ
⇔ 1/γ² x = x'/γ + v t'/γ
⇔ x = γ²/γ x' + γ²/γ v t'
= γ x' + γ v t'
= γ (x' + v t')

⇔ x = γ (x' − (−v) t').

[Similarly you can see that t = γ (t' + v/c² x') = γ (t' − (−v)/c² x').]

Note that the special principle of relativity (postulate no. 1; paraphrased:
“the laws of physics are the same in all inertial reference frames”²)
implies then that the velocity of S relative to/measured in S' is v' = −v. ∎

[If this symmetry would not follow from the Lorentz transformation,
then the transformation would be erroneous as we observe that symmetry
in nature. Note that if the Lorentz transformation is correct, it MUST
be approximable in the low-speed limit by the well-confirmed Galilean
transformation in *both*/*all* directions; it is.³]

² cf. <https://einsteinpapers.press.princeton.edu/vol2-trans/157>
³ I think I gave the proof here several times before.

> Your reasoning is circular.

Your translation, reading comprehension, or logic is flawed instead.


PointedEars
--
«Nec fasces, nec opes, sola artis sceptra perennant.»
(“Neither high office nor power, only the scepters of science survive.”)

—Tycho Brahe, astronomer (1546-1601): inscription at Hven

[Thomas 'PointedEars' Lahn]

True, but (you know that) since it must be rotation-invariant, it suffices
to consider it in one spatial dimension (rotate the spatial coordinate
system such that the motion is in the positive x-direction, do the
transformation for x, then do the inverse rotation).

>>So no "additional axiom" is needed. IOW: your supposed "axiom" is easily
>>derived from the Lorentz transform.
>>
>>Tom Roberts
>
> You need the axiom in order to rigorously prove the Lorentz transform.

No, because from the two postulates of OEDoMB¹, which have their foundation
in observations/experiments, *follows* the Lorentz transform(ation). [It
was Einstein’s great achievement to find independently the same
transformation as Lorentz and Poincaré, but *without* requiring an aether.]

¹ Einstein, A.(1905). “On the Electrodynamics of Moving Bodies”.
Official Translation from German into English:
<https://einsteinpapers.press.princeton.edu/vol2-trans/154> pp.

If a frame S' with coordinates t' and x' is moving relative to a frame
S, with coordinates t and x, parallel to the x-axis in the positive x-
direction with speed v (i.e. velocity v in this one-dimensional case), then
the coordinate transformation from S to S' is the Lorentz transformation:

t' = γ (t − v/c² x)
x' = γ (x − v t)
y' = y

z' = z,

[Tom Roberts]

On 9/6/21 8:25 PM, Ross A. Finlayson wrote:
> On Saturday, September 4, 2021 at 1:33:39 PM UTC-7, tjrob137 wrote:
>> One derivation of the equations of SR uses Einstein's two
>> postulates, plus his "hidden postulates" (discussed in 1907: a.
>> clocks and rulers have no memory, b. space is homogeneous and
>> isotropic, c. time is homogeneous). These yield the Lorentz
>> transform from one inertial frame to a second frame moving with
>> velocity v relative to the first. Invert it to obtain the
>> transform from the second frame to the first: a Lorentz transform
>> with v replaced by -v.
>

> I don't see how b and curved space-time fit but I suppose that's
> RoS.

Relativity of simultaneity has nothing to do with it.

Both b and c are inconsistent with curved spacetime. But no matter, as
this is Special Relativity, and it deals only with flat spacetime.
Curved spacetime is needed to accurately model gravitation, and that
requires General Relativity. GR does not conform to these postulates.

> [... further nonsense ignored]

Tom Roberts

[Julio Di Egidio]

On Tuesday, 7 September 2021 at 16:36:13 UTC+2, tjrob137 wrote:

> Both b and c are inconsistent with curved spacetime. But no matter, as
> this is Special Relativity, and it deals only with flat spacetime.
> Curved spacetime is needed to accurately model gravitation, and that
> requires General Relativity.

Not true: there are formulations of gravity as a gauge theory (i.e. gravity as a force). In particular, since it's a novel and very promising approach, look up so called "Geometric Algebra": <https://youtu.be/x7eLEtmq6PY>

Julio

[NaCl]

Thomas 'PointedEars' Lahn wrote:

> If a frame S' with coordinates t' and x' is moving relative to a frame
> S,
> with coordinates t and x, parallel to the x-axis in the positive x-
> direction with speed v (i.e. velocity v in this one-dimensional case),
> then the coordinate transformation from S to S' is the Lorentz
> transformation:
> t' = γ (t − v/c² x) x' = γ (x − v t) y' = y z' = z,
> where the Lorentz factor is γ ≔ 1/√(1 − v²/c²).

prove it. Don't come to me with definitions, like Kenseto.

[Ross A. Finlayson]

Thanks for your reply.

So, Relativity of Simultaneity isn't part of SR at all but SR + GR?

I.e., "RoS is a 'hidden' assumption in SR and GR together".

What assumptions are basically shared or erased between "SR" and
"GR" in terms of "GR, then SR", and, "SR, then GR"?

The optical and photon sector is really quite interesting, ...,
it's satisfying to think that the gravitational vector also
points or is strongest pointing to the source not the image,
as that the geodesy is everywhere always instantaneous.

There's much to be said for a usual clock principle, as for
example discussed a bit in the "Light Speed Rest Frame Theory",
which makes for modeling light as flux the wave-packet,
for writing out in some algebraic terms the instantaneous
state of image.

I hope it's clear that I appreciate the usual adherence to
what is mostly the _standard_ model while of course there's
that physics has an entire conundrum among SR, GR, gravity,
and quantum mechanics, which of course is a continuum mechanics.

[Ricardo Jimenez]

On Tue, 07 Sep 2021 15:31:41 +0200, Thomas 'PointedEars' Lahn
<Point...@web.de> wrote:

>> Your reasoning is circular.
>
>Your translation, reading comprehension, or logic is flawed instead.

Are you trying to communicate information or beat your chest trying to
convince others of your superiority? The point of physics is to build
up mathematical models that will predict the results of observations.
The unlying principle used to build up the model for special
relativity is the symmetry of observations made by two observers in
relative motion with each other. While at rest with respect of each
other, the two observers have synchronized their respective clocks and
chosen equal length rulers. After the relative motion starts along a
straight line:

1. Each makes the same observation of the speed of light.
2. Each makes the same observation of the speed of the other observer
relative to him/her self but with a sign reversal.
3. Each makes the same observation of the ratio of the number of the
other observer's clock ticks to his own and it is less than 1 and
depends only on their relative speed.
4. Each makes the same observation of the ratio of the length of the
other observer's rulers to his own and it is the same as the ratio in
3.

Notice that the observations in 1 and 2 are different than those of 3
and 4 in that that in the former the ratios of what the 2 observers
measure is one. I am interested if any of these modeling principles
are redundant or a shorter set of principles can be used to set up the
theory. Obviously 4 follows from the first three, but can one do
better? You can't just say that the mathematical definition of the
Lorentz group is all that is needed. You have to give all the
observations needed to ensure that the Lorentz group describes nature.

[Ricardo Jimenez]

Slight clarification. The first sentence of the last paragraph should
be: Notice that the observations in 1 and 2 are different than those
of 3 and 4 in that in 3 and 4 we are talking about an observer
measuring two things, his own stationary clock/ruler versus the moving
item of the same kind.

[Tom Roberts]

On 9/7/21 6:09 PM, Ross A. Finlayson wrote:
> On Tuesday, September 7, 2021 at 7:36:13 AM UTC-7, tjrob137 wrote:
>> On 9/6/21 8:25 PM, Ross A. Finlayson wrote:
>>> I don't see how b and curved space-time fit but I suppose that's
>>> RoS.
>> Relativity of simultaneity has nothing to do with it.
>>
>> Both b and c are inconsistent with curved spacetime. But no matter, as
>> this is Special Relativity, and it deals only with flat spacetime.
>> Curved spacetime is needed to accurately model gravitation, and that
>> requires General Relativity. GR does not conform to these postulates.
>

> So, Relativity of Simultaneity isn't part of SR at all but SR + GR?

You need to learn how to read. RoS is part and parcel of SR, and
therefore part of GR. But AS I SAID, RoS has nothing to do with "how b
and curved space-time fit".

> I.e., "RoS is a 'hidden' assumption in SR and GR together".

Nope. RoS is a conclusion, not any sort of assumption.

Tom Roberts

[Maciej Wozniak]

On Wednesday, 8 September 2021 at 19:10:01 UTC+2, tjrob137 wrote:
> On 9/7/21 6:09 PM, Ross A. Finlayson wrote:
> > On Tuesday, September 7, 2021 at 7:36:13 AM UTC-7, tjrob137 wrote:
> >> On 9/6/21 8:25 PM, Ross A. Finlayson wrote:
> >>> I don't see how b and curved space-time fit but I suppose that's
> >>> RoS.
> >> Relativity of simultaneity has nothing to do with it.
> >>
> >> Both b and c are inconsistent with curved spacetime. But no matter, as
> >> this is Special Relativity, and it deals only with flat spacetime.
> >> Curved spacetime is needed to accurately model gravitation, and that
> >> requires General Relativity. GR does not conform to these postulates.
> >
> > So, Relativity of Simultaneity isn't part of SR at all but SR + GR?
> You need to learn

that you're FORCED!!! To THE BEST WAY!!!!
That's what THE COMMUNITY OF PHYSICISTS
has secured!!!

[mitchr...@gmail.com]

On Saturday, September 4, 2021 at 10:13:24 AM UTC-7, Ricardo Jimenez wrote:
> As far as I know, the following does not follow from Einstein's 2
> postulates:
> Suppose two inertial systems are in uniform relative motion

If they have the same mutual relative speed comparison their clocks
ought to be the same.

[dean rector]

nonsense, you both imply a third central reference observer. There is no
need for that.

[Thomas 'PointedEars' Lahn]

Ricardo Jimenez wrote:

> […] Thomas 'PointedEars' Lahn […] wrote:
>>> Your reasoning is circular.
>> Your translation, reading comprehension, or logic is flawed instead.
>
> Are you trying to communicate information or beat your chest trying to
> convince others of your superiority?

How disappointing, though not entirely unexpected, that you would ignore the
whole argument that I gave, including the mathematical proof.

> The point of physics is to build up mathematical models that will predict
> the results of observations.

And you know neither the physics nor the mathematics very well. Otherwise
you would have been able to come up with the same reasoning and proof as
I did.

> The unlying principle

The *what*?

> used to build up the model for special relativity is the symmetry of
> observations made by two observers in relative motion with each other.

As you could have seen with more careful reading and thinking, the symmetry
also *follows* from the two postulates of OEDoMB through the Lorentz
transformation itself which is *derived* (by Einstein) based *only* on them.
Whereas the formulation of the transformation does NOT *require* it.

So the symmetry does NOT have to be an axiom, contrary to your repeated
claims.

However, consider this: I am looking at you moving at v to my right, which I
call the positive x-direction:

1. you
I

2. you
I

3. you
I

In 3 units of time, lets call it a “blubb” (unit symbol “b”), in my rest
frame you have moved 6 units of length, lets call that a “snurk” (unit
symbol “k”). So your velocity in my frame is v = 3/6 b/k = 1/2 b/k.

Now from your perspective, but you are looking in the same direction as I:

1. you
I

2. you
I

3. you
I

In 3 blubbs, I have moved 6 snurks to your left. We have defined “left” as
the negative direction. So my velocity in your rest frame is v' = −3/6 b/k
= −1/2 b/k = −v.

Isn’t it blatantly *obvious* that, if the principle of relativity applies,
my velocity in your rest frame *has* to be the additive inverse of your
velocity in my rest frame?

> [tl;dr]

I have studied this at university (and passed the respective exam). Tell me
something that I do not know yet.

> […] You can't just say that the mathematical definition of the

> Lorentz group is all that is needed.

Again, the Lorentz transformation follows from the two postulates which are
based on observation. In NONE of the two postulates is it stated that there
would be a symmetry of velocity.

That the transformation also makes for a group operation can be considered a
happy coincidence, although not an unexpected one given the *previously*
experimentally confirmed principle of relativity (Galileo’s ship etc.).

> You have to give all the observations needed to ensure that the Lorentz
> group describes nature.

Einstein already did in 1905. You need to read more carefully, and stop
applying wishful thinking to what you read. Without that change in
attitude, you will be yet another hopeless case.


PointedEars
--
Two neutrinos go through a bar ...

(from: WolframAlpha)

[Ross A. Finlayson]

It just seems like it could be assumed the other way.

I.e. in SR by itself it's not necessary and it's so in SR + GTR
because otherwise there wouldn't exist the geodesy having
a state at any one time.

I don't know that RoS is falsifiable in SR.

(And rather that's a polite way of saying I think not.)



Here's some reading from the Wikipedia, of course,
for following your advice and checking my head before
writing on relativity, https://en.wikipedia.org/wiki/Relativity_of_simultaneity ,
it basically agrees with "gravity points to the source not the image".

When I hear "in the frame" it's of course "in all the nested and containing frames".

From the Wiki article:

"In general the second observer traces out a worldline
in the spacetime of the first observer described by t = x/v,
and the set of simultaneous events for the second observer
(at the origin) is described by the line t = vx. Note the multiplicative
inverse relation of the slopes of the worldline and simultaneous events,
in accord with the principle of hyperbolic orthogonality. "

See it looks they've attached the language of GTR.

[Ross A. Finlayson]

I think you mean "there is a need for not that".

[Tom Roberts]

On 9/9/21 11:46 PM, Ross A. Finlayson wrote:
> On Wednesday, September 8, 2021 at 10:10:01 AM UTC-7, tjrob137 wrote:
>> RoS is a conclusion, not any sort of assumption.
>

> It just seems like it could be assumed the other way.

Hmmm. As usual, there are many alternative sets of axioms from which the
equations of a theory can be derived. While I have not investigated, I
suspect that these two axioms could be used to derive SR:
1. the Principle of Relativity
2. relativity of simultaneity
But nobody seriously does that, as RoS does not seem like a sensible
concept on which to base a theory.

In all other derivations, RoS is a conclusion (theorem), not an assumption.

> I.e. in SR by itself it's not necessary and it's so in SR + GTR
> because otherwise there wouldn't exist the geodesy having
> a state at any one time.

RoS is part and parcel of SR -- you cannot have SR without it. I have no
idea what you are thinking in the rest of that statement, but it looks
like complete nonsense to me.

> I don't know that RoS is falsifiable in SR.

In principle it certainly is. It's just that no experiment has observed
it, yet, due to technology difficulties.

[Full disclosure: I have proposed an experiment to
directly measure RoS, but no funding agency has
funded it. It is based on high-accuracy lasers and
requires about $30k in equipment.]

Tom Roberts

[Townes Olson]

On Sunday, September 12, 2021 at 9:02:13 AM UTC-7, tjrob137 wrote:
> It's just that no experiment has observed [RoS], yet, due to technology difficulties.

Not true. The relativity of inertial simultaneity has been demonstrated countless times. In order to understand these demonstrations, you must first understand the dynamical meaning of relativity of inertial simultaneity... which you do not.

> I have proposed an experiment to directly measure RoS...

No, you haven't. Your proposal is just to measure the first-order Sagnac effect, which (1) is not a measurement of the relativity of inertial simultaneity and (2) has been measured countless times with off-the-shelf equipment.

> no funding agency has funded it.

It's obvious to any competent reviewer that you misunderstand the subject, and all you are proposing to measure is the mundane Sagnac effect, which you confuse with the relativity of inertial simultaneity. Hopefully the funding agencies will continue to assign your appeals to competent reviewers (since you obviously are never going to stop).

[Ross A. Finlayson]

(Basically that the geodesy is always current: and that the force
vector of the gravitic field always points between two sources
not images, has that nothing ever "curves away" or diverges,
because, before it does the geodesy is immediately re-evaluated,
in the _continuum_, of the geodesy in GR the contents according
to the occupation numbers the frame of space-time, at any given time.)

(Thanks again for your reply and as usual I'll hope to make clear
when items mentioned are "in SR, only", or, "in GR, only", or,
in "SR +GR", or, "in light-speed rest frame theory with real space
contraction after fall gravity into GR and then SR". It seems neatest
that supergravity results this way from fall gravity, then also there's
for true polar unipotential as any gradient of anisotropy, that at
any given moment is in its instantaneous state.)

(Quantum mechanics: continuum mechanics.)

[Maciej Wozniak]

On Sunday, 12 September 2021 at 18:02:13 UTC+2, tjrob137 wrote:
> On 9/9/21 11:46 PM, Ross A. Finlayson wrote:
> > On Wednesday, September 8, 2021 at 10:10:01 AM UTC-7, tjrob137 wrote:
> >> RoS is a conclusion, not any sort of assumption.
> >
> > It just seems like it could be assumed the other way.
> Hmmm. As usual, there are many alternative sets of axioms from which the
> equations of a theory can be derived. While I have not investigated, I
> suspect that these two axioms could be used to derive SR:
> 1. the Principle of Relativity
> 2. relativity of simultaneity

3. That we're FORCED!!! To THE BEST WAY!!!

[Tom Roberts]

On 9/12/21 12:57 PM, Townes Olson wrote:
> On Sunday, September 12, 2021 at 9:02:13 AM UTC-7, tjrob137 wrote:
>> It's just that no experiment has observed [RoS], yet, due to technology difficulties.
>
> Not true. The relativity of inertial simultaneity has been demonstrated countless times.

Reference, please.

> [... further nonsense omitted]

Tom Roberts

[Townes Olson]

On Sunday, September 12, 2021 at 9:02:13 AM UTC-7, tjrob137 wrote:

> no funding agency has funded it.

It is obvious to any competent reviewer that you misunderstand the subject, and all you are proposing to measure is the mundane Sagnac effect, which you confuse with the relativity of inertial simultaneity. Hopefully the funding agencies will continue to assign your appeals to competent reviewers (since you obviously are never going to stop).

> I have proposed an experiment to directly measure RoS...

No, you have not. Your proposal is just to measure the first-order Sagnac effect, which (1) is not a measurement of the relativity of inertial simultaneity and (2) has been measured countless times with off-the-shelf equipment.

> It's just that no experiment has observed [RoS], yet, due to technology difficulties.

Not true. The relativity of inertial simultaneity has been demonstrated countless times. In order to understand these demonstrations, you must first understand the dynamical meaning of relativity of inertial simultaneity... which you do not.

> Reference, please.

A quick review of the google archive shows that you've been given the references to the demonstrations of the relativity of inertia (and hence the relativity of inertial simultaneity) multiple times before. They have not changed. Your problem is not lack of references, your problem is lack of understanding. In order to understand these demonstrations, you must first understand the dynamical meaning of relativity of inertial simultaneity... which you do not.

[Emmet Kahl]

Townes Olson wrote:

>> Reference, please.
>
> A quick review of the google archive shows that you've been given the

just give the reference, you fucking moron, the best you have. Stop
wasting my time with "google".

[Tom Roberts]

On 9/12/21 4:12 PM, Townes Olson wrote:
> [...] The relativity of inertial simultaneity has been demonstrated countless times.

> > On Sunday, September 12, 2021 at 9:02:13 AM UTC-7, tjrob137 wrote:

>> Reference, please.
>
> [...]

So you don't have any reference for actual measurements of relativity of
simultaneity. As I thought (because I have looked for such experiments).
Instead, you choose to bullshit, which is not very helpful to anyone.

Tom Roberts

[Townes Olson]

On Sunday, September 12, 2021 at 9:02:13 AM UTC-7, tjrob137 wrote:

> no funding agency has funded it.

It's obvious to any competent reviewer that you misunderstand the subject, and all you are proposing to measure is the mundane Sagnac effect, which you confuse with the relativity of inertial simultaneity. Hopefully the funding agencies will continue to assign your appeals to competent reviewers (since you obviously are never going to stop).

> I have proposed an experiment to directly measure RoS...

No, you haven't. Your proposal is just to measure the first-order Sagnac effect, which (1) is not a measurement of the relativity of inertial simultaneity and (2) has been measured countless times with off-the-shelf equipment.

> It's just that no experiment has observed [RoS], yet, due to technology difficulties.

Not true. The relativity of inertial simultaneity has been demonstrated countless times. To understand these demonstrations, you must first understand the dynamical meaning of relativity of inertial simultaneity... which you do not.

> Reference, please.

A quick review of the previous threads on this topic in this very forum shows that you've been given the references to the demonstrations of the relativity of inertia (and hence the relativity of inertial simultaneity) multiple times before. They have not changed. Your problem is not lack of references, your problem is lack of understanding. To understand the referenced demonstrations, you must first understand the dynamical meaning of relativity of inertial simultaneity... which you do not.

> So you don't have any reference...

You mis-read. Again, I directed you to the same references that have been given to you multiple times before. They have not changed, nor will they. Your problem is not lack of references, your problem is lack of understanding. To understand the referenced demonstrations, you must first understand the dynamical meaning of relativity of inertial simultaneity... which you do not. I urge you to re-read (or, perhaps, read) the previous threads in which this was all explained to you in detail, along with the referenced demonstrations. The referenced demonstrations have not changed, and the explanations have not changed, nor will they.

[Tom Roberts]

On 9/12/21 5:53 PM, Townes Olson wrote:
> [to me] your problem is lack of understanding.

No, YOUR problem is lack of understanding.

I am asking for references to EXPERIMENTS that measure the relativity of
simultaneity. That is, an experiment that unequivocally demonstrates
that two events that are simultaneous in one frame are not simultaneous
in a frame moving relative to the first, and measures the time
difference between them in the second frame, as a function of a) the
relative velocity between the two frames, and b) the distance between
the events in the first frame.

NONE of the references you purport to identify actually do that, but
that _IS_ what my proposed experiment will do. Sure, RoS is required for
internal consistency of SR, but that is NOT an experiment.

Note: it simply does not matter that the experiment can also be modeled
using the (generalized) Sagnac effect -- OF COURSE the result can be
calculated in either frame (hint: Lorentz invariance); in the second
frame it is due to relativity of simultaneity, and in the first frame it
is due to variations in the speed of light in moving fibers (aka the
(generalized) Sagnac effect).

Tom Roberts

[Townes Olson]

On Sunday, September 12, 2021 at 7:31:19 PM UTC-7, tjrob137 wrote:
> It's just that no experiment has observed [RoS], yet, due to technology difficulties...
> It simply does not matter that the experiment can also be modeled using the
> (generalized) Sagnac effect...

You contradict yourself. First you claim to be proposing an "experiment" to observe something (which you erroneously identify as RoS) never observed before, and then you concede that your "experiment" shows nothing but the mundane Sagnac effect, which you admit has been demonstrated trillions of times. That's the first problem. The second problem is erroneously equating an observation of the Sagnac effect with a demonstration of the relativity of inertial simultaneity. (A third problem is that you just refer to simultaneity rather than inertial simultaneity.)

> I am asking for references to EXPERIMENTS that measure the relativity of
> simultaneity.

First, you need to specify inertial simultaneity, not just simultaneity. Second, you aren't really asking about that, you are asking about experiments to measure the open-loop Sagnac effect, which is actually demonstrated trillions of times each day, but which is not a demonstration of the relativity of inertial simultaneity (because it would also exist in a stationary ether, for example). Third, if you were really interested in the empirical demonstration of the relativity of inertial simultaneity (which you aren't, because you don't even understand what that is), you would recognize that you already know of the actual demonstrations of the relativity of inertia and hence inertial coordinate systems. The reason you don't associate the experiments with the relativity of inertial simultaneity is because you don't understand how inertial-based coordinate systems are defined. Fourth, you naively fail to realize that all measurements involve rational inference. Fifth, you fail to distinguish between the definitional and the empirical content of Lorentz invariance. Sixth... well, the list goes on...

> That is, an experiment that unequivocally demonstrates that two events that
> are simultaneous in one frame are not simultaneous in a frame moving relative

> to the first...

You need to replace "simultaneous" with "inertially simultaneous", and yes, that is exactly what the relevant experiments unequivocally demonstrate, and no, the first-order Sagnac effect (such as your proposal and trillions of other examples) does not demonstrate this.

> NONE of the references you purport to identify actually do that...

You are mistaken. The experiments showing the relativity of inertia automatically show the relativity of inertial simultaneity.

> but that _IS_ what my proposed experiment will do.

No, you proposal is to demonstrate the Sagnac effect, which is demonstrated trillions of times each day, and which does not demonstrate the relativity of inertial simultaneity.

[Ross A. Finlayson]

The sampling, measurement, observer effects, as they were:
what exactly's a "rational inference measurement"?

[RichD]

On September 4, rotchm wrote:
>> So from what does it follow
>> that if observer 1 measures observer 2 moving at speed v, observer 2
>> will measure observer 1 moving at speed -v?
>
> By the relativity principle, both observers will measure the same magnitude v.

So if an observer 1 in frame K measures a bullet's velocity as 2000 feet/sec,
then observer 2 in frame L will measure the same. By the relativity principle -


--
Rich

[rotchm]

No. Your logic is flawed. You misapplied your "correspondence".
I said a ~b => b ~a. You replied that: thus a ~c => b ~c
See your error?

[Ricardo Jimenez]

On Wed, 15 Sep 2021 05:27:40 -0700 (PDT), rotchm <rot...@gmail.com>
wrote:

Just saying "It follows by the reativity principle" is unsatisfactory.
The relativity principle is not a mathematical statement since if
refers to all laws of physics. Not as bad as saying it follows from
the Lorentz transform which of course assumes the axiom in all
derivations I've seen.

[Maciej Wozniak]

On Wednesday, 15 September 2021 at 14:27:41 UTC+2, rotchm wrote:
> On Tuesday, September 14, 2021 at 10:46:44 PM UTC-4, RichD wrote:
> > On September 4, rotchm wrote:
> > >> So from what does it follow
> > >> that if observer 1 measures observer 2 moving at speed v, observer 2
> > >> will measure observer 1 moving at speed -v?
> > >
> > > By the relativity principle, both observers will measure the same magnitude v.
> >
> > So if an observer 1 in frame K measures a bullet's velocity as 2000 feet/sec,
> > then observer 2 in frame L will measure the same. By the relativity principle -
> No. Your logic is flawed.

And your "facts" are gedanken, not empirical, poor halfbrain.

[Tom Roberts]

On 9/15/21 9:22 AM, Ricardo Jimenez wrote:
> Not as bad as saying it follows from the Lorentz transform which of
> course assumes the axiom in all derivations I've seen.

Then you CLEARLY have not looked.

The "axiom" in question:
When inertial frame A measures inertial frame B to travel
with relative velocity v, then B will measure A to have
relative velocity -v. (Here v is a 3-vector; for derivations
in 1+1 dimensions, v is a (signed) real number.)

Indeed I know of no common derivation of the Lorentz transform that
requires that as an axiom; it is a direct consequence of the derivation.

IOW: every common derivation obtains the transform A -> B, given the
velocity of B relative to A. So there's no need to consider how B might
measure the relative speed of A. Once you have the transform A -> B,
obtaining the transform B -> A is simply an algebraic inverse, and
v => -v follows directly from that.

Tom Roberts

[Townes Olson]

On Wednesday, September 15, 2021 at 6:35:20 PM UTC-7, tjrob137 wrote:
> > Not as bad as saying it follows from the Lorentz transform which of
> > course assumes the axiom in all derivations I've seen.
> Then you CLEARLY have not looked. The "axiom" in question:
> When inertial frame A measures inertial frame B to travel
> with relative velocity v, then B will measure A to have
> relative velocity -v. (Here v is a 3-vector; for derivations
> in 1+1 dimensions, v is a (signed) real number.)
> Indeed I know of no common derivation of the Lorentz transform that
> requires that as an axiom; it is a direct consequence of the derivation.

No, every derivation makes use of reciprocity, but (as explained early in this thread) reciprocity follows from the principle of relativity and isotropy, so it doesn't need to be taken as a *separate* premise. See the explanation in the prior message, or see any of the derivations in the vast literature on this subject.

[RichD]

On September 15, rotchm wrote:
>>>> So from what does it follow
>>>> that if observer 1 measures observer 2 moving at speed v, observer 2
>>>> will measure observer 1 moving at speed -v?
>
>>> By the relativity principle, both observers will measure the same magnitude v.
>
>> So if an observer 1 in frame K measures a bullet's velocity as 2000 feet/sec,
>> then observer 2 in frame L will measure the same. By the relativity principle -
>
> No. Your logic is flawed. You misapplied your "correspondence".
> I said a ~b => b ~a. You replied that: thus a ~c => b ~c
> See your error?

No idea what you're on about.

But no matter. YOU are the one who claimed that any two
observers will measure the same v, "by the relativity principle".
This is preposterous. Is it actually necessary to spell
out the transform formula?

--
Rich

[rotchm]

On Wednesday, September 15, 2021 at 10:38:42 PM UTC-4, RichD wrote:
> On September 15, rotchm wrote:

> > No. Your logic is flawed. You misapplied your "correspondence".
> > I said a ~b => b ~a. You replied that: thus a ~c => b ~c
> > See your error?
> No idea what you're on about.

Basic symbolic logic, language.
I said I said a ~b => b ~a.
You gave an example to this as: a ~c => b ~c.
You do see that these are different?

> But no matter. YOU are the one who claimed that any two
> observers will measure the same v, "by the relativity principle".

Yes, by the "relativity principle", in quotes; RP.
Stronger, by symmetry principle: observer 1 performs a given exp, then observer 2, doing the *same* exp, will get the same result (this 'version' of the RP is sometimes used instead, as Poincare used).

IOW, observer 2 (who has set up his measurements procedures identically to observer 1, and measures the speed of observer 1) will get the same results as observer 1. Here, observers 1 & 2 are merely labels; they are mutually measuring each other's properties (here, their speeds).
However if they were measuring the speed of something else (like your bullet example), then you have three objects and you're breaking your symmetry, you are misapplying this symmetry principle between the two observers.

Again, a ~b => b ~a does not imply a ~c => b ~c.
Hence, the second postulate to "force" the SoL to be C for all observers, " a ~C <=> b ~C "

[Maciej Wozniak]

On Thursday, 16 September 2021 at 06:23:41 UTC+2, rotchm wrote:
> On Wednesday, September 15, 2021 at 10:38:42 PM UTC-4, RichD wrote:
> > On September 15, rotchm wrote:
>
> > > No. Your logic is flawed. You misapplied your "correspondence".
> > > I said a ~b => b ~a. You replied that: thus a ~c => b ~c
> > > See your error?
> > No idea what you're on about.
> Basic symbolic logic, language.
> I said I said a ~b => b ~a.
> You gave an example to this as: a ~c => b ~c.
> You do see that these are different?
> > But no matter. YOU are the one who claimed that any two
> > observers will measure the same v, "by the relativity principle".
> Yes, by the "relativity principle", in quotes; RP.
> Stronger, by symmetry principle: observer 1 performs a given exp, then observer 2, doing the *same* exp, will get the same result (this 'version' of the RP is sometimes used instead, as Poincare used).

In the meantime in the real world, however, the clocks of GPS
keep indicating t'=t, fucking your precious symmetry and its
precious principle.

[Ross A. Finlayson]

Ah, but that's after _correcting_ the clocks, the Parametric Post-Newtonian,
keeping in mind then that the resulting computations each start local.

Why and how various kinds of clocks operate in various regimes of gravity,
measuring time forwardly, makes for that a clock hypothesis has never been
falsified.

In General Relativity with mass-energy equivalence and constant speed of light,
according to the speed of light 0 = 0 and c = 1.

That being c_light, in General Relativity c_gravity is also 1, while,
in Newtonian physics, c_gravity is infinity. (Which in general relativity
according to mass-energy equivalence under velocity in the kinetic,
is where 0 = 0 and infinity = 1. )

[Tom Roberts]

On 9/15/21 8:52 PM, Townes Olson wrote:
> On Wednesday, September 15, 2021 at 6:35:20 PM UTC-7, tjrob137
> wrote:
>>> Not as bad as saying it follows from the Lorentz transform which
>>> of course assumes the axiom in all derivations I've seen.
>> Then you CLEARLY have not looked. The "axiom" in question: When
>> inertial frame A measures inertial frame B to travel with relative
>> velocity v, then B will measure A to have relative velocity -v.

Note: this "axiom" is called "reciprocity" below.

>> (Here v is a 3-vector; for derivations in 1+1 dimensions, v is a
>> (signed) real number.) Indeed I know of no common derivation of
>> the Lorentz transform that requires that as an axiom; it is a
>> direct consequence of the derivation.
>
> No, every derivation makes use of reciprocity, but (as explained
> early in this thread) reciprocity follows from the principle of
> relativity and isotropy, so it doesn't need to be taken as a
> *separate* premise.

I see statements of yours that reciprocity is implied by the PoR and
isotropy, with a demonstration for a very specific situation, but no
general case. I see no mention of how reciprocity is used in any
derivation of the Lorentz transform.

As I said, reciprocity is not needed in any of the common derivations.

If you think "every derivation makes use of reciprocity", please show
HOW AND WHERE IT IS USED in any common derivation. It would be best to
discuss Einstein's original derivation in 1905, but any common
derivation will serve.

> See the explanation in the prior message, or see any of the
> derivations in the vast literature on this subject.

AFAICT none of them describe how reciprocity is used in deriving the
Lorentz transform. You seem to be making stuff up like any crank.

Tom Roberts

[Ross A. Finlayson]

Which, uh, agrees with observation.

As an object approaches c things receding also approach c respectively.

Or, as the space-time wheel would have it, 0.5c, moving frames.
(That the frame moves to keep everything moving.)

I think that the more one learns about theory, the more
that variations of theories of Fitzgerald and Fatio, make sense.

(In their place after the modern and 21st century, for their
place what was modern and 18-19th century.)

Then for Fritz London, as, having the clearest models of solving
for unipotential or the super, the polar or highest non-linear,
of course is for constants, because the usual systems are
only either linear or asymptotically orthogonal, is why for
symmetry-invariance and phase transition, the quasi-invariant
or symmetry-flex, is for charge, parity, and time invariance,
what is conserved.

[Tom Roberts]

On 9/12/21 10:35 PM, Townes Olson wrote:
> On Sunday, September 12, 2021 at 7:31:19 PM UTC-7, tjrob137 wrote:
>> It's just that no experiment has observed [RoS], yet, due to
>> technology difficulties... It simply does not matter that the
>> experiment can also be modeled using the (generalized) Sagnac
>> effect...
>
> You contradict yourself.

No. You just did not read what I wrote.

> First you claim to be proposing an "experiment" to observe something
> (which you erroneously identify as RoS) never observed before, and
> then you concede that your "experiment" shows nothing but the
> mundane Sagnac effect, which you admit has been demonstrated
> trillions of times.

You are supposed to read what I write. In the lab frame the effect is
the (generalized) Sagnac effect, and in the moving frame it is
relativity of simultaneity. This has never been demonstrated in the
moving frame.

> (A third problem is that you just refer to simultaneity rather than
> inertial simultaneity.)

I have been discussing inertial frames throughout. I see no reason to
use the phrase "inertial simultaneity", as it is not common [#], and is
redundant.

[#] A Google search for "inertial simultaneity" yields
just 62 results, many of which are irrelevant.
scholar.google.com yields just 2 results, one of which
is bogus. This is the first I have seen that particular
phrase, and I am pretty sure it does not appear in any
of the many relativity textbooks I own and have used.

>> I am asking for references to EXPERIMENTS that measure the
>> relativity of simultaneity.
>

> you aren't really asking about that, you are asking about experiments
> to measure the open-loop Sagnac effect,

No, I am asking for experiments that measure the relativity of
simultaneity. That is, experiments which arrange for two events to be
simultaneous in one inertial frame and measure the time difference
between them in an inertial frame moving relative to the first. Why is
it so hard for you to understand this?

I phrase it that way, because that is what so many students find
difficult to accept about SR. No demonstration of the Sagnac effect will
convince them. but a direct measurement of the relativity of
simultaneity will help a lot.

> [... further obfuscation]

>> That is, an experiment that unequivocally demonstrates that two
>> events that are simultaneous in one frame are not simultaneous in
>> a frame moving relative to the first...
>
> You need to replace "simultaneous" with "inertially simultaneous",

No, I don't. Hint: read what I wrote, remembering that here I used
"frame" to mean "inertial frame". And see above.

IOW: "simultaneous in an inertial frame" has exactly the same meaning.

> and yes, that is exactly what the relevant experiments unequivocally
> demonstrate,

So give a reference to an experiment that arranged for two events to be
simultaneous in one inertial frame and measured the time difference
between them in an inertial frame moving relative to the first.

> and no, the first-order Sagnac effect (such as your proposal and
> trillions of other examples) does not demonstrate this.

In the right circumstances, such as my proposed experiment, the
first-order Sagnac effect is merely a description in another frame of
the first-order relativity of simultaneity. (Here "first order" refers
to v/c, the relative velocity between the two frames.)

>> NONE of the references you purport to identify actually do that...
>
> You are mistaken. The experiments showing the relativity of inertia
> automatically show the relativity of inertial simultaneity.

So give a reference to an experiment that arranged for two events to be
simultaneous in one inertial frame and measured the time difference
between them in an inertial frame moving relative to the first.

>> but that _IS_ what my proposed experiment will do.
>
> No, you proposal is to demonstrate the Sagnac effect,

(Sigh. You repeatedly fail to understand that there are TWO FRAMES
involved, and the descriptions are different in the two frames.)

In the moving frame it demonstrates relativity of simultaneity; in the
lab frame it demonstrates the (generalized) Sagnac effect. It will
measure the moving-frame time difference between a pair of events that
are simultaneous in the lab.

Tom Roberts

[Townes Olson]

On Thursday, September 16, 2021 at 6:23:10 PM UTC-7, tjrob137 wrote:
> I see statements of yours that reciprocity is implied by the PoR and
> isotropy, with a demonstration for a very specific situation, but no
> general case.

The proof is fully general, as explained in the post.

> As I said, reciprocity is not needed in any of the common derivations.

As always, you are mistaken... see below.

> If you think "every derivation makes use of reciprocity", please show
> HOW AND WHERE IT IS USED in any common derivation. It would be best to
> discuss Einstein's original derivation in 1905, but any common
> derivation will serve.

In that derivation it is invoked when, after defining system k moving at speed +v in terms of K, a third system K' is defined moving at speed -v in terms of k, and then the paper asserts that K' must be the same as K, which is to say that if the two-fold application of the transformation, which at this stage is determined only up to a factor phi(v), is applied first with +v and then with -v the result is the identity transformation, which enables the paper to assert phi(v)phi(-v)=1. We are entitled to assert this reciprocity because of the principle of relativity along with spatial isotropy, as explained in the previous message. We could make other assumptions about the relation between K and K', leading to other transformations, which of course would not be applicable to the inertia-based coordinates which satisfy relativity and isotropy (and hence reciprocity).

Be aware that the widely-published English translation of Perrett & Jeffrey contains an error, saying that k is moving at speed -v in terms of K', which of course is backwards. This was corrected in the English translation in Einstein's Collected Papers.

> AFAICT none of them describe how reciprocity is used in deriving the
> Lorentz transform.

That just confirms that you've never understood any of the derivations... which is obvious from all your posts. In more modern derivations the invocation of reciprocity is just as explicit.

> You seem to be making stuff up like any crank.

You're so funny.

[Townes Olson]

On Thursday, September 16, 2021 at 7:07:51 PM UTC-7, tjrob137 wrote:
> > You contradict yourself. First you claim to be proposing an "experiment" to

> > observe something (which you erroneously identify as RoS) never observed
> > before, and then you concede that your "experiment" shows nothing but the
> > mundane Sagnac effect, which you admit has been demonstrated trillions
> > of times.
>

> In the lab frame the effect is the (generalized) Sagnac effect, and in the moving
> frame it is relativity of simultaneity.

The Sagnac effect is the Sagnac effect. It is not one thing in terms of one system of coordinates and another thing in terms of another system of coordinates. Sheesh. A demonstration of the relativity of inertial simultaneity is quite different, and obviously relies on the properties of inertia, which you are not considering at all.

> This has never been demonstrated in the moving frame.

As always, you are mistaken. The Sagnac effect has been described in terms of many different systems of coordinates, not just in terms of one particular system of coordinates that you consider to be "at rest".

> I see no reason to use the phrase "inertial simultaneity", as it is not common [#],
> and is redundant.

The reason for using it is because it would remind you that what you are proposing has nothing to do with demonstrating inertial simultaneity, because you aren't showing anything about the relativity of inertia.

> [#] A Google search for "inertial simultaneity" yields ...

You are sliding into the Ed Lake school of research and "analysis".

> > you aren't really asking about that, you are asking about experiments
> > to measure the open-loop Sagnac effect,
>
> No, I am asking for experiments that measure the relativity of
> simultaneity.

Well, what you call "relativity of simultaneity" is just the Sagnac effect. So you think you are asking for experiments that demonstrate the relativity of simultaneity, but you are really just asking for experiments that demonstrate the Sagnac effect.

> Why is it so hard for you to understand this?

Ed Lake asks me the same about oscillating photons and radar guns that measure the speed of a truck from inside a truck.

> > and yes, that is exactly what the relevant experiments unequivocally
> > demonstrate,
>
> So give a reference to an experiment that arranged for two events to be
> simultaneous in one inertial frame and measured the time difference
> between them in an inertial frame moving relative to the first.

What you are describing is just the Sagnac effect, which has been demonstrated trillions of times. You can't even begin to comprehend the experiments that demonstrate the relativity of inertial simultaneity because you don't even know what that is (to the point that you're actually googling it... sheesh). You need to first learn the basics. What you are describing is nothing but the Sagnac effect, which has been demonstrated trillions of times every day, and it does not demonstrate the relativity of inertial simultaneity. In contrast, actual scientists have long ago unequivocally demonstrated the relativity of inertial simultaneity, in experiments that you are fully aware of, but you don't understand.

[Maciej Wozniak]

On Friday, 17 September 2021 at 03:04:30 UTC+2, Ross A. Finlayson wrote:
> On Wednesday, September 15, 2021 at 9:30:26 PM UTC-7, maluw...@gmail.com wrote:
> > On Thursday, 16 September 2021 at 06:23:41 UTC+2, rotchm wrote:
> > > On Wednesday, September 15, 2021 at 10:38:42 PM UTC-4, RichD wrote:
> > > > On September 15, rotchm wrote:
> > >
> > > > > No. Your logic is flawed. You misapplied your "correspondence".
> > > > > I said a ~b => b ~a. You replied that: thus a ~c => b ~c
> > > > > See your error?
> > > > No idea what you're on about.
> > > Basic symbolic logic, language.
> > > I said I said a ~b => b ~a.
> > > You gave an example to this as: a ~c => b ~c.
> > > You do see that these are different?
> > > > But no matter. YOU are the one who claimed that any two
> > > > observers will measure the same v, "by the relativity principle".
> > > Yes, by the "relativity principle", in quotes; RP.
> > > Stronger, by symmetry principle: observer 1 performs a given exp, then observer 2, doing the *same* exp, will get the same result (this 'version' of the RP is sometimes used instead, as Poincare used).
> > In the meantime in the real world, however, the clocks of GPS
> > keep indicating t'=t, fucking your precious symmetry and its
> > precious principle.
> Ah, but that's after _correcting_ the clocks

And? Since when your physics has banned calibrated measurement
devices and disqualified the result obtained with them?

[Maciej Wozniak]

On Friday, 17 September 2021 at 03:23:10 UTC+2, tjrob137 wrote:
> On 9/15/21 8:52 PM, Townes Olson wrote:
> > On Wednesday, September 15, 2021 at 6:35:20 PM UTC-7, tjrob137
> > wrote:
> >>> Not as bad as saying it follows from the Lorentz transform which
> >>> of course assumes the axiom in all derivations I've seen.
> >> Then you CLEARLY have not looked. The "axiom" in question: When
> >> inertial frame A measures inertial frame B to travel with relative
> >> velocity v, then B will measure A to have relative velocity -v.
> Note: this "axiom" is called "reciprocity" below.

And note: we're FORCED!!! To THE BEST WAY!!! The way of
our beloved guru and his obedient minion Tom!!

[Ross A. Finlayson]

I know nothing but local time and network time - that it's the same for intent.

Solar noon and clock time: right once a day.

I.e. whether the year defines the day or second defines the day,
here is where my basic understanding ends at the crystal clock
and the atomic clock, that, the atomic clock beats faster, outside
gravity.

[Ross A. Finlayson]

"The time transformation t = t′ in Eqs. (3) is deceivingly simple.
It means that in the rotating frame the time variable t′ is really
determined in the underlying inertial frame. It is an example of
coordinate time. A similar concept is used in the GPS."

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5253894/

[Maciej Wozniak]

You know very little; but that was obvious.
So, since when has your physics banned calibrated measurement

devices and disqualified the result obtained with them?

> "The time transformation t = t′ in Eqs. (3) is deceivingly simple.
> It means that in the rotating frame the time variable t′ is really
> determined in the underlying inertial frame. It is an example of
> coordinate time. A similar concept is used in the GPS."

"Coordinate time", you say? Time as a coordinate? What an
alien, how non-physical concept! Of course, physics has
to ignore time being a mere coordinate; but isn't it accidentally
a part of the very core of your SR shit? And a part of galilean
transformations as well?

[RichD]

On September 15, tjrob137 wrote:
>> Not as bad as saying it follows from the Lorentz transform which of
>> course assumes the axiom in all derivations I've seen.
>
> Then you CLEARLY have not looked.

What's clear is that YOUR thinking is superficial.

> The "axiom" in question:
> When inertial frame A measures inertial frame B to travel
> with relative velocity v, then B will measure A to have
> relative velocity -v.

> Indeed I know of no common derivation of the Lorentz transform that
> requires that as an axiom; it is a direct consequence of the derivation.

You have it it exactly backwards, as you commit petitio principii, blatantly.
The usual Lorentz derivation ASSUMES that v looks the same in both
directions. One cannot therefore claim it as a consequence.

More precisely, the Lorentz derivation does NOT depend on assuming such
a property.

> IOW: every common derivation obtains the transform A -> B, given the
> velocity of B relative to A. So there's no need to consider how B might
> measure the relative speed of A. Once you have the transform A -> B,
> obtaining the transform B -> A is simply an algebraic inverse, and
> v => -v follows directly from that.

All wrong.
Al sees Bob receding with velocity v.
Bob sees Al receding with velocity w.

As the theory currently stands, v may differ from w. And everything works.
You want to claim this cannot happen?
Show a contradiction.

Señor Jimenez tiene razon. Logically, "v = w" must be added as a third postulate,
the symmetric velocity postulate. Physically, it must be tested; either it's true, or not.

--
Rich

[RichD]

On September 15, rotchm wrote:
>> YOU are the one who claimed that any two
>> observers will measure the same v, "by the relativity principle".
>
> Yes, by the "relativity principle", in quotes; RP.

You misunderstand the principle.

> Stronger, by symmetry principle: observer 1 performs a given exp, then observer 2,

> doing the *same* exp, will get the same result.

Partial credit.
Relativity states that observers, WITHIN SEPARATE FRAMES, doing identical
experiments, must see identical results.

But there is no symmetry principle, that's your invention (perhaps absorbed from this group).

> IOW, observer 2 (who has set up his measurements procedures identically to observer 1,
> and measures the speed of observer 1) will get the same results as observer 1.

> Here, observers 1 & 2 are merely labels; they are mutually measuring each other's properties.

You're confused, regarding the definition of relativity.
"measuring each other's properties" doesn't constitute performing
identical experiments, within a frame.  It doesn't support the claim that
the laws of mechanics are uniform within every inertial frame.
REVIEW THE DEFINITION.

Al and Bob inhabit separate spacecraft, separating. Al attaches a
1/2" diameter hose to a pump, at 15 psi. He measures the water flow.
Bob does likewise, with identical equipment. They ought to see the
same result. THAT is relativity.

However, when they measure each other's receding velocity, the context
is very different.
Al watches Bob fly away. Bob does not watch Bob fly away.
Bob watches Al fly away. Al does not watch Al fly away.
These are DISTINCT experiments.

It isn't comparable to the water hoses. And it doesn't tautologically follow
from the theory, as presented by Einstein.

--
Rich

[RichD]

On September 6, Townes Olson wrote:
>> How does the relativity principle account for the same m
>> measured absolute value of velocity of observers ...
>
> The principle of relativity asserts that for every material particle there is a system of coordinates
> in terms of which the particle is instantaneously stationary and all the equations of physics take
> the same homogeneous and isotropic form.

Yes, WITHIN any system of coordinates, etc.
Which doesn't speak to a PAIR of such systems, in relative motion.

> This principle (noting the isotropy) entails reciprocity, as can be seen by considering two
> identical charged particles, initially adjacent and at mutual rest; after they are released they
> repel each other in opposite directions. Each particle is instantaneously at rest in terms of
> some system of inertia-based coordinates, and the equations of physics take the same isotropic
> form in terms of both, so in terms of those coordinates the other is moving away at some
> speed v.

Your example is artificial, a straw man. There's no need for dynamical interactions
or mutual history.

Simplify down to essentials: two inertial systems are in relative motion. That's all.
Strip all other baggage. Then ask: do they measure identical receding velocities?
The theory, as it stands, doesn't address this question. It needs be augmented by
a third postulate.

> Since the same laws of physics apply to both, and the situation is perfectly symmetrical,
> the spatial origin of the inertia-based coordinate system in which each particle is at rest
> is receding from the other at the same speed.

The same laws, WITHIN each.

Running my word processor over the 1905 paper, I discovered only one
reference to 'reciprocity', involving a magnet and conductor. It offers no
insight into this particular issue.

I discovered several instances of 'symmetry', of which only one pertains to this
question, in section 6. There, Einstein ASSUMES that v and -v identically apply
to the transforms, inversely. i.e. he commits the same fallacy of petitio principii
as everyone else here (apparently, the entire physics community).

>> while it doesn't imply that distances and time are measured the same
>> by both systems?
>
> It implies that there is some constant k such that, for any two systems of inertia-based coordinates,
> we have x'=(x-vt)g and t'=(t-kvx)g where g=1/sqrt(1-kv^2). if k=1/c^2 this is the Lorentz transformation.

No violation of relativity occurs, if they measure differing velocities, then use
those different values in their respective Lorentz applications.

--
Rich

[rotchm]

On Saturday, September 18, 2021 at 8:24:25 PM UTC-4, RichD wrote:
> On September 15, rotchm wrote:

> Al and Bob inhabit separate spacecraft, separating. Al attaches a
> 1/2" diameter hose to a pump, at 15 psi. He measures the water flow.
> Bob does likewise, with identical equipment. They ought to see the
> same result. THAT is relativity.

Yes, that's what I said. If they do the same measurement procedures, they will get the same results, no mater the exp.

> However, when they measure each other's receding velocity, the context
> is very different.

Nope, *no matter the exp*.

> Al watches Bob fly away. Bob does not watch Bob fly away.
> Bob watches Al fly away. Al does not watch Al fly away.
> These are DISTINCT experiments.

They are equivalent, the "same".
For instance, let Al & Bob each have an identical (light)source.
Al measures Bob's speed via doppler. Al will get a certain result (v).
Bob doing the same, will get the same result.

They both performed the same measurement procedure and therefore they will get the same result.
They have the same "v".

[Ross A. Finlayson]

Ah, no, Malu: I think a clock hypothesis is correct that there is basically
a "universal time" as it were. (That it's never been falsified, ....)

It's interesting what clock implemented in crystal and implemented in
the atomic clocks, is, why the atomic clocks are important, is, they
measure variations in the field of gravity, so, maybe you should look at
the clocks and their corrections on the satellites, that, they both are
advised for dead reckoning and absolute positioning, and, they
advise for what are field variations, courtesy the _atomic_ clock according
to its inertial acceleration like "gravity" observing relativity. (While
crystal and pendulum clocks for example after Allan discrepancy,
don't so much.)


If you read that article I found it helps explain some of these things
and why they are the way they are.

Speaking of pendulums and the Allais effect, it's pretty interesting
that alignment is a natural laboratory.

[Mike Gale]

Careful now. Relative velocity is not an invariant quantity, Travellers perceive one other with equal and opposite velocities, but you have to invoke the velocity addition formula to ascertain their relative velocity in other contexts. The missing axiom (or rather the neglected axiom) is, the reference frame in which the time axis is parallel to the centre of momentum has the fastest (proper time) clock. It's kind of a corollary to the isomorphism axiom. Consider for example the case where both twins go walkabout with equal and opposite velocities. They age at the same rate despite their relative velocity, but not as fast as the launchpad. Or try this. Throw a clock at a wall and then throw the wall at the clock from the same distance. The elapsed time on the clock on impact is different in each case even if the relative velocity is the same. In general, time slows down when you deviate from the centre of momentum (spacetime) heading and speeds up again when you return to that heading.

[Maciej Wozniak]

As for universal time, there are at least 3 of them in the world we inhabit.
GPS, UTC, TAI. The differences are slight, but they are obviously
3 different entities; all are coordinates, all are timelike, all are
observer_independent/universal/absolute.
UTC has even spelled directly in the name - "universal time coordinate".
Not enough, however, for a relativistic idiot to notice that there is
an universal time coordinate.

> It's interesting what clock implemented in crystal and implemented in
> the atomic clocks, is, why the atomic clocks are important, is

Is that some insane self appointed gurus have announced they
are. While what you say may be interesting a bit, the clocks keep
indicating t'=t, making time (as defined by your idiot guru himself)
galilean.
Common sense was warning you.

[Ross A. Finlayson]

https://ui.adsabs.harvard.edu/link_gateway/1992CeMDA..53...81K/ADS_PDF

https://www.itu.int/dms_pub/itu-r/opb/hdb/R-HDB-55-2010-OAS-PDF-E.pdf

(I just think that any time 'dilation' is explained by space contraction.)

[Maciej Wozniak]

You can explain any wonder of your gedankenwelt any way you
want to; in the meantime in the real world, however, GPS clocks
keep indicating t'=t, just like all serious clocks always did.

Have you, BTW, ever considered a possibility that
claiming "this is important!" you can be mistaken?

[Ross A. Finlayson]

(It only seemed important that the timekeeping according to
the various technologies varies, that only for the atomic clocks
does space contraction appear to result the clocks faster time,
and as where "time only slows" as according to some space contraction.)

(Basically the atomic clocks are for measuring changes in the field.)

(And no I don't know why g Earth's gravity constant ranges after
the first digit or two quite variously over the Earth according to
the geodetic survey.)

(It looks like that besides that the satellites once at equilibrium in
orbit in the inertial frame would be one rotating frame, the atomic
clocks then are basically accelerometers, though given the entire
constellation, that providing source-and-source synchronization.)

(I.e. where they wouldn't be useful for setting the time, they would
be useful for detecting anomalies.)

[Thomas 'PointedEars' Lahn]

Thomas 'PointedEars' Lahn wrote:

> That the [Lorentz] transformation also makes for a group operation can be
> considered a happy coincidence, although not an unexpected one given the
> *previously* experimentally confirmed principle of relativity (Galileo’s
> ship etc.).

To elaborate: A defining property of a group G is that for every element t
of the group there is a so-called *inverse element* t⁻¹ such that under the
group operation the composition of both is the identity operation
(id: G → G, x ↦ x):

t⁻¹ ∘ t = id.

Obviously if the special principle of relativity holds, then the Lorentz
transformation (LT) from S to S' must be reversible: An observer at rest in
S' must be able to obtain the corresponding coordinates in S using the LT
(it must not matter for them that an observer at rest in S considers them to
be moving; they consider themselves to be at rest and the observer at rest
in S to be moving instead).

That means that the LT from S' to S is the inverse element of the group: If
you perform the LT from S to S', and then the LT from S' to S, you must
obtain the coordinates in S again.¹

And indeed:

x'' = γ (x' − (−v) t')
= γ ([γ (x − v t)] − (−v) [γ (t − v/c² x)])
= γ (γ (x − v t) + γ (v t − v²/c² x))
= γ² (x − v t + v t − v²/c² x)
= γ² (x − v²/c² x)
= γ² (1 − v²/c²) x
= γ²/γ² x
= x. ∎

[Proving this for the other coordinates, and proving closure and
associativity of the Lorentz group, is left as an exercise to the reader.]


PointedEars
___________
¹ And one can easily prove that the inverse element of an element has to be
unique: Let t⁻¹ ∘ t = id, i.e. t⁻¹ is the inverse of t. Assume that there
is another inverse element t⁻¹' of t, i.e. t⁻¹' ∘ t = id, too. Then
t⁻¹ = id ∘ t⁻¹ = (t⁻¹' ∘ t) ∘ t⁻¹ = t⁻¹' ∘ (t ∘ t⁻¹) = t⁻¹' ∘ id = t⁻¹'.²
² t⁻¹ ∘ t = id ⇒ t = t ∘ id = t ∘ (t⁻¹ ∘ t) = (t ∘ t⁻¹) ∘ t ⇒ t ∘ t⁻¹ = id.
--
Q: How many theoretical physicists specializing in general relativity
does it take to change a light bulb?
A: Two: one to hold the bulb and one to rotate the universe.
(from: WolframAlpha)

[Tom Roberts]

On 9/8/21 9:12 PM, Thomas 'PointedEars' Lahn wrote:
> That the transformation also makes for a group operation can be considered a
> happy coincidence,

Not at all! It is ABSOLUTELY REQUIRED that the set of coordinate
transforms form a group (with composition as the group operation). That
is part and parcel of what it means to be a coordinate transformation.

The basic group for coordinate transformations is the diffeomorphism
group. But one can select subgroups of it by imposing various criteria
on the coordinates of interest. For inertial frames, in physics, the
relevant subgroup is SO(3,1), the Lorentz group.

Tom Roberts

[Thomas 'PointedEars' Lahn]

OK.


PointedEars
--
Two neutrinos go through a bar ...

(from: WolframAlpha)

[Julio Di Egidio]

On Sunday, 26 September 2021 at 12:12:26 UTC+2, Thomas 'PointedEars' Lahn wrote:
> Tom Roberts wrote:
>
> > On 9/8/21 9:12 PM, Thomas 'PointedEars' Lahn wrote:
> >> That the transformation also makes for a group operation can be
> >> considered a happy coincidence,
> >
> > Not at all! It is ABSOLUTELY REQUIRED that the set of coordinate
> > transforms form a group (with composition as the group operation). That
> > is part and parcel of what it means to be a coordinate transformation.
> >
> > The basic group for coordinate transformations is the diffeomorphism
> > group. But one can select subgroups of it by imposing various criteria
> > on the coordinates of interest. For inertial frames, in physics, the
> > relevant subgroup is SO(3,1), the Lorentz group.
>
> OK.

For flat space, not just for inertial frames...

Julio

[Julio Di Egidio]

For flat space-time...

Julio

[Tom Roberts]

No. In flat spacetime, no element of the Lorentz group can connect any
non-inertial coordinates to an inertial frame -- they only connect
inertial frames to other inertial frames.

You certainly can apply a Lorentz transform to one set of non-inertial
coordinates to obtain a second set of non-inertial coordinates, but that
won't form a group -- you cannot cover the set of all non-inertial
coordinates that way. For instance, no Lorentz transform will generate
spherical or cylindrical coordinates from Cartesian coordinates (etc.).

Tom Roberts

[Maciej Wozniak]

On Sunday, 26 September 2021 at 14:53:17 UTC+2, tjrob137 wrote:

> No. In flat spacetime, no element of the Lorentz group can connect any
> non-inertial coordinates to an inertial frame -- they only connect
> inertial frames to other inertial frames.

While, on the other hand, in the real world - the clocks of GPS
keep indicaating t'=t, just like all seerious clocks always did.

[Ross A. Finlayson]

Both inertial and momental....

It is usual continuous deformations.

The flat space time that according to Einstein is curving.

Here though anyone would note "what's then the momental
frame is not the inertial frame, those inertial frames", is of
course under charge and inertia what basically in the terms
of the one and the other the charge static and the thermo
for inertia or the kinetic, usually also the Einstein's mass-energy
equivalence, classically defining classical inertia.

Then, this "charge static" or for "torque is static", is that the
potential moment of waves is crests. (Though overpressure.)

I.e. the both the "Lorentz group frames".

(Or as many there are, multipole moment.)

Frames is the most usual establishment of geometry,
particularly Euclidean.

[Maciej Wozniak]

In the meantime in the real world, however, the clocks of
GPS keep indicating t'=t, just like all serious clocks always
did.

[carl eto]

GPS cannot be used to detect a time difference since the electrons of the GPS circuit are propagating at 10^6 m/s yet the velocity of light is 10^8 m/s.

[Ross A. Finlayson]

On Sunday, September 26, 2021 at 2:15:24 PM UTC-7, carl eto wrote:
> GPS cannot be used to detect a time difference since the electrons of the GPS circuit are propagating at 10^6 m/s yet the velocity of light is 10^8 m/s.

The "holes", the electron "holes" in the circuit.

Which move at c if not "propagate".

Electron holes and the NPN junction combine to make gates, it's so.

https://en.wikipedia.org/wiki/Electron_mobility

[Julio Di Egidio]

On Sunday, 26 September 2021 at 14:53:17 UTC+2, tjrob137 wrote:

I am still not sure if you are retarded, a fucking agent of the enemy, or both...

<https://en.wikipedia.org/wiki/Lorentz_group>

*Plonk*

Julio

[Kye Fox]

Julio Di Egidio wrote:

>> > For flat space-time...
>>
>> No. In flat spacetime, no element of the Lorentz group can connect any
>> non-inertial coordinates to an inertial frame -- they only connect
>> inertial frames to other inertial frames.
>> You certainly can apply a Lorentz transform to one set of non-inertial
>> coordinates to obtain a second set of non-inertial coordinates, but
>> that won't form a group -- you cannot cover the set of all non-inertial
>> coordinates that way. For instance, no Lorentz transform will generate
>> spherical or cylindrical coordinates from Cartesian coordinates (etc.).
>
> I am still not sure if you are retarded, a fucking agent of the enemy,
> or both...

you dont undrestand mathematics, physics and everything, you fucking
stoopid.

[Kye Fox]

Ross A. Finlayson wrote:

>> > For flat space, not just for inertial frames...
>> For flat space-time... Julio
>
> Both inertial and momental....

you both stupid like hell, what the fuck is "momental"??

[Ross A. Finlayson]

Hmm..., try "momentary", maybe Google will help you.

How is it that I am getting the last word in these things?

[Kye Fox]

Ross A. Finlayson wrote:

> On Monday, September 27, 2021 at 2:37:15 AM UTC-7, Kye Fox wrote:
>> Ross A. Finlayson wrote:
>>
>> >> > For flat space, not just for inertial frames...
>> >> For flat space-time... Julio
>> >
>> > Both inertial and momental....
>> you both stupid like hell, what the fuck is "momental"??
>
> Hmm..., try "momentary", maybe Google will help you.

not physics. Next.

[Ross A. Finlayson]

On Monday, September 27, 2021 at 11:01:38 AM UTC-7, Kye Fox wrote:
> Ross A. Finlayson wrote:
>
> > On Monday, September 27, 2021 at 2:37:15 AM UTC-7, Kye Fox wrote:
> >> Ross A. Finlayson wrote:
> >>
> >> >> > For flat space, not just for inertial frames...
> >> >> For flat space-time... Julio
> >> >
> >> > Both inertial and momental....
> >> you both stupid like hell, what the fuck is "momental"??
> >
> > Hmm..., try "momentary", maybe Google will help you.
> not physics. Next.

It's "momental" in case you asked.