On the Definition of an Inertial Frame of Reference

[Eugene Shubert]

Apparently, you all misunderstand my thesis

The paper you linked hasn't been cited in the five years since it was written. What does that mean?

It means that the conclusions are so obvious, you'd have to be a knucklehead to disagree with them. And certainly, if there was a significant error, someone would have published a quick rebuttal.

The implication for physics however is that my key argument in my quintessence paper is flawless. And that means that the arm-waving that allegedly connects transformational linearity to the homogeneity and isotropy of spacetime is sheer pretense.

It also means that all my raving detractors misunderstand The Quintessence of Axiomatized Special Relativity Theory.

 

[Tach]

It also means that all my raving detractors misunderstand .

No, it means that your paper violates the principle of relativity making it worthless.

 

[AlexG]

“ Originally Posted by brucep
The paper you linked hasn't been cited in the five years since it was written. What does that mean? ”


It means that the conclusions are so obvious, you'd have to be a knucklehead to disagree with them.

It means that everyone who has read the paper considers it to be the worthless product of a crank.

 

[Tach]

I reject the claims made by backward thinking physicists that mathematicians must not define inertial frames of reference in a coordinate-free way.

Are you for or against the sacred oracles of the ancient opinions?

Show how your transforms leave Maxwell's laws covariant. You can't. Hence, your "theory" is false.

 

[rpenner]

What linearity means:

$$ \forall A \, \left( { t'_A \\ x'_A \\ y'_A \\ z'_A } \right) = \Lambda \, \left( { t_A \\ x_A \\ y_A \\ z_A } \right) \\ \Rightarrow \quad \forall A, B \, \left( { \Delta t'_{AB} \\ \Delta x'_{AB} \\ \Delta y'_{AB} \\ \Delta z'_{AB} } \right) = \Lambda \, \left( { \Delta t_{AB} \\ \Delta x_{AB} \\ \Delta y_{AB} \\ \Delta z_{AB} } \right) \\ \Rightarrow \quad \left( \vec{a}' = \Lambda \, \vec{a} \, \wedge \, \vec{b}' = \Lambda \, \vec{b} \quad \rightarrow \quad \vec{a}' + \vec{b}' = \Lambda \, ( \vec{a} + \vec{b} ) \, \wedge \, n \vec{a}' = \Lambda \, ( n \vec{a} ) \right) \\ \Rightarrow \quad \vec{0} = \Lambda \, \vec{0}$$

so that not only do all observers agree things that happen at the same time and same place do in fact happen at the same time and same place, but that the halfway point between ruler marking or clock times or any two events in four-space is also agreed upon, and that the concept of uniform motion is agreed upon, and that ruler markings or clock times being uniform is agreed upon, and that finally, there is no special place or time in the universe since the position of the chosen origin does not factor into the transform of vectors. Importantly, vector spaces map to vector spaces, so a physics based on geometric reasoning is valid for all observers.

Hypothetical nonlinear transforms between observers destroys the ability to describe a physical law common to all observers which isn't tied to a place and time and state of motion. Indeed, both SR and GR call out special observers where the physical law is simplest -- the inertial observer for SR and the free-falling observer for GR.

When you don't argue to this point, you aren't doing physics. As a matter of fact, the elements of your paper used to talk about uniform motion, rulers with uniform spacing and clocks that tick regularly. Without a background model that contains linear transforms of coordinates between observers, such definitions make no absolute sense, and so you have used the assumption of linearity throughout. Consequently, when you try to build to nonlinearity from such a framework, you are no longer doing mathematics.

 

[Guest254]

Shubee's theory fails one of the basic tests: Maxwell's equations aren't covariant in his formalism.

To avoid confusion - Maxwell's equation's are not "Eugene Shubert covariant", i.e. the standard Maxwell equations do not take on the same form under his coordinate transformation. Whereas, they are Lorentz covariant, i.e. they do take on the same form under Lorentz transformations.

However, the generally covariant form of Maxwell's equations is obviously invariant under his (and anyone else's) coordinate transformation.

 

[AlphaNumeric]

The solution to the problem requires solving equations (18)-(20) but I confess--I don't know how to do that. So I assumed that $$g$$ is odd and found a solution. Then I realized that $$\epsilon + g$$ is also a solution.

Except that (28) doesn't satisfy (26) either if $$\epsilon \neq 0$$. As I said, you are solving them individually, not simultaneously. Altering a solution of equation X to become a solution to equation Y doesn't mean it is still a solution to equation X, X puts constraints on how you can solve Y. This is a pretty basic concept yet within 3 minutes of opening your 'work' I came across it twice. I stopped reading then.

If you can find a more general solution, then I will rewrite my paper and add a note of acknowledgement that you were the one that solved equations (18)-(20) with great ingenuity and utmost generality. Or if you can prove that my solution is already the most general one, then I will acknowledge that you have devised a mathematical proof of uniqueness.

Your 'solution' isn't a solution and even if it were your entire premise is based on your misunderstanding. Why should I help you when you don't help yourself?

It means that the conclusions are so obvious, you'd have to be a knucklehead to disagree with them.

That isn't how science research works. Papers with trivial results are not published and if a result is sufficient to be deemed worth publishing by a reputable journal then it means its worth citing. After all reviewers for journals are people in the community and thus they can decide if a paper is worth the attention of the community as a whole.

And certainly, if there was a significant error, someone would have published a quick rebuttal.

Like we've done with you you mean?

The implication for physics however is that my key argument in my quintessence paper is flawless.

No good scientist would ever make such a carte blanche statement. In minutes I've seen two mistakes on basic algebra, stuff school children can do, in your 'flawless' work. And that's just me, other people have other comments against your 'flawless' argument.

And that means that the arm-waving that allegedly connects transformational linearity to the homogeneity and isotropy of spacetime is sheer pretense.

And no one likes people making pretension claims, like the kind you make about your work.

It also means that all my raving detractors misunderstand The Quintessence of Axiomatized Special Relativity Theory.

Is this a 'no true Scotsman' argument you're using? Anyone who disagrees doesn't understand and you embrace anyone who agrees as someone who understands. Its a shifting of the goal posts, like religious people who say "He was never a true christian/jew/muslims" when someone becomes an apostate.

Like Tach, you obviously never published anything or earned a degree in math or physics or even derived an equation in your life.

I have as well. Specifically in space-time structures.

You're just clutching at straws. You ignore anything anyone here says because you can just say "I don't believe you have any education or research experience in this" and for those people whose credentials you can't ignore (ie reputable journal reviewers) you cry conspiracy of 'Einstein worshippers'. Its Catch 22, anyone who isn't part of the mainstream community you can dismiss for lack of experience and anyone in the mainstream community you can claim is sticking to dogma. That way you can live in your own little detached world of self delusion.

 

[przyk]

When you don't argue to this point, you aren't doing physics.

A year ago Shubert made it clear he was using his own home-made concept of a "scientific theory" which he's stretched to the point it includes invisible pink unicorns:

Off the top of my head: a model of a population of invisible pink unicorns that are governed by Newton's laws but don't interact with ordinary matter.

I happily applaud your axiomatization because it delights me to no end. Why should anyone be alarmed by me calling that a scientific theory?

[POST=2350865]link[/POST]. This isn't the first thread Shubert has started about the definition of a term whose implications he wants to leverage, only to descend to the usual disingenuous rhetoric and name-calling when he doesn't get the external validation he wanted.

 

[Tach]

To avoid confusion - Maxwell's equation's are not "Eugene Shubert covariant", i.e. the standard Maxwell equations do not take on the same form under his coordinate transformation. Whereas, they are Lorentz covariant, i.e. they do take on the same form under Lorentz transformations.

That was the meaning all along.

However, the generally covariant form of Maxwell's equations is obviously invariant under his (and anyone else's) coordinate transformation.

Not so sure, can you do the math to prove that?

 

[Guest254]

I should hope so!

 

[Eugene Shubert]

Except that (28) doesn't satisfy (26) either if $$\epsilon \neq 0$$.

I did not claim that equation (28) satisfies equation (26). I claimed that equation (27) satisfies equations (18)-(20) and therefore equation (28) satisfies equations (18)-(20).

 

[Tach]

I should hope so!

I would really like to see that. Would turn the thread away from the crackpot theory to something really interesting. How will you deal with the fact that t'=F(x,t, S(x,t))?

 

[James R]

20 posts deleted due to personal insults. Please stay on topic.

 

[Eugene Shubert]

To avoid confusion - Maxwell's equation's are not "Eugene Shubert covariant", i.e. the standard Maxwell equations do not take on the same form under his coordinate transformation. Whereas, they are Lorentz covariant, i.e. they do take on the same form under Lorentz transformations.

However, the generally covariant form of Maxwell's equations is obviously invariant under his (and anyone else's) coordinate transformation.

Clearly, The Quintessence of Axiomatized Special Relativity Theory is a derivation of special relativity from extraordinarily general principles and that spirit of generality has the very quintessence of general covariance built into it from the very start.

 

[Tach]

Clearly, The Quintessence of Axiomatized Special Relativity Theory is a derivation of special relativity from extraordinarily general principles and that spirit of generality has the very quintessence of general covariance built into it from the very start.

Your tall claim is contradicted by your inability to demonstrate even the simplest form of covariance, the one of the Maxwell laws. You have not done your homework and you don't even know where to start.

 

[Eugene Shubert]

Hypothetical nonlinear transforms between observers destroys the ability to describe a physical law common to all observers…

Without a background model that contains linear transforms of coordinates between observers, such definitions make no absolute sense … Consequently, when you try to build to nonlinearity from such a framework, you are no longer doing mathematics.

You are arguing against general covariance. I accept it. Like the early Einstein and perhaps most authors of elementary textbooks on special relativity, you believe in the overriding importance of clock synchronization. I do not.

 

[Eugene Shubert]

That would be a total waste of time since your transforms fail the covariance condition.

That's a contradiction in terms. By definition, a generally covariant theory is invariant under all possible coordinate transformations, even mine.

 

[Eugene Shubert]

demonstrate even the simplest form of covariance

Regardless of your proper time and location, there is a mathematical form of the classical theory of electrodynamics that is already generally covariant, and you know nothing about it.

 

[Guest254]

Aha - you made an edit!

That was the meaning all along.

I should hope so!

Not so sure, can you do the math to prove that?

The maths is hardly deep!

$$ \eta_{ab} \mapsto g_{ab}, \qquad \partial_a \mapsto \nabla_a $$

So Maxwell's equations can be written

$$ \nabla^a F_{ab} =0 , \qquad \nabla_{[a} F_{bc]} =0 $$

which are clearly invariant under coordinate transformations. Alternatively, use the exterior calculus version of Maxwell's equations and use the fact that pull-backs commute with exterior derivatives.

 

[Tach]

That's a contradiction in terms. By definition, a generally covariant theory is invariant under all possible coordinate transformations, even mine.

Nope, your ignorance is showing. If it were true (it isn't) you would have done the simple homework to show that Maxwell's equations are invariant wrt your cockamamie "transforms". You fail. Again.