# On the Definition of an Inertial Frame of Reference

Discussion in 'Pseudoscience Archive' started by Eugene Shubert, Oct 15, 2010.

1. ### Eugene ShubertRegistered Senior Member

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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?

You only have two choices:

The definition of an inertial frame of reference should be modernized, corrected and improved.

Inertial frames of reference must be dependent on a linear coordinate system.

Last edited: Oct 15, 2010
2. ### przyksquishyValued Senior Member

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Er, they *are* special coordinate systems. Inertial reference frames are the coordinate systems in which the laws of physics take their canonical, textbook form. Not all coordinate systems possess this property. Special relativity is a theory which postulates that the laws of physics are symmetrical under the group of Poincaré transformations. The Poincaré group is a greatly restricted subset of the most general possible group of coordinate transformations.

3. ### Eugene ShubertRegistered Senior Member

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450

The definition of an inertial frame of reference must be independent of the choice of coordinates. Sure, if the general case is too difficult to construct, then we have to settle for a linear theory.

4. ### przyksquishyValued Senior Member

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Nope. Inertial reference frames *are* coordinate systems. You can certainly express physical laws in ways that are independent of any choice of coordinates, but that means specifically not referring to coordinate systems, inertial or otherwise. "Inertial frames of reference in a coordinate-free way" is a contradiction in terms.

EDIT: actually, to split hairs, the term "reference frame" is a bit vague. Sometimes it's used interchangeably with "coordinate system", and sometimes it's less specific. You might even be able to come up with a definition of one of the less specific notions of a reference frame without referring to a coordinate system. For example, on a flat Lorentzian manifold, you could take a geodesic parameterised by its proper time (representing an observer's concept of time), partition space-time into hyperplanes orthogonal to that geodesic (representing simultaneity), and take the induced metric on the hyperplanes (spatial distance). There's really not much point in this though. Nobody bothers to give a precise definition of a reference frame because no-one ever needs one.

Last edited: Oct 15, 2010
5. ### Eugene ShubertRegistered Senior Member

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A precise definition of a reference frame is needed and given in The Quintessence of Axiomatized Special Relativity Theory, p. 2 and it's essentially the idea that you have outlined:

Obviously, the "mathematical clocks" in the quintessence paper are understood to be geometric points moving on geodesics parameterized by proper time.

Last edited: Oct 16, 2010
6. ### przyksquishyValued Senior Member

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Then what's your point? You're not saying anything new or even particularly interesting about relativity.

7. ### Eugene ShubertRegistered Senior Member

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It's just a personal victory for me that goes back to my sophomore year in high school. I was taking second year algebra with Mr. Carson. At the end of the course, Mr. Carson offered extra credit to any student willing to give a presentation to the class by taking the class through a challenging problem and solving it on the blackboard. I had read a book on special relativity that year and was amazed at the remarkable simplicity of the Lorentz transformation derivation. I was very eager to share that derivation and the beauty of such an amazing idea in physics that only requires elementary algebra. The only part of the derivation in the book that seemed completely unsupported to me was the claim that the transformation had to be linear. I don't recall anything about the rest of the book, certainly not the title or the author. That was many years ago. Because I couldn't see past the nonsense claim in that derivation, I felt that I couldn't make that elementary algebra presentation. I felt cheated by the obscurity of the nonsensical arm-waving but felt vindicated many years later by discovering that the transformation equations need not be linear and by deriving all the obvious nonlinear Lorentz-equivalent transformation equations.

The key idea is in recognizing the essential meaning of an inertial frame of reference and then doing the required math, which I would have loved to share in my high school algebra class.

8. ### przyksquishyValued Senior Member

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This just follows from the principle of homogeneity, which is the idea - consistent with experience and experimental evidence - that physics should be the same everywhere in space and time. The fact that the Lorentz transformation is linear means that you can write it in terms of intervals:
\begin{align} \Delta ct^{\prime} &\,=\, \gamma ( \Delta ct \,-\, \beta \Delta x ) \\ \Delta x^{\prime} &\,=\, \gamma ( \Delta x \,-\, \beta \Delta ct ) \end{align}​
without referring to a particular coordinate origin.

What do you mean by "need not be linear"? It's part of their definition.

If you're living in an imaginary universe where the only condition is the coordinate invariance of the speed of light then yes, you're not limited to linear transformations. There's a larger group of transformations (the conformal group) which preserves the speed of light and are generally non-linear. But in physics the point of Lorentz transformations isn't just that they're coordinate transformations that leave the speed of light invariant (or whatever other principle you used to derive them from). The point of them is that they're a symmetry of physical laws. For example, take any solution to Maxwell's equations of electrodynamics, and applying a Lorentz transformation to it will get you another solution to Maxwell's equations. This isn't true of more general transformations.

9. ### Eugene ShubertRegistered Senior Member

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There is no experimental evidence that prevents the universe from being described by wildly nonlinear transformation equations.

When physicists refer to the homogeneity and isotropy of space and time in the context of simplifying their derivation of the Lorentz transformation by arguing that everything is linear, they are just arm-waving to justify their pretense that their argument is geometrical. They have no definition of spacetime geometry at that point.

Exactly. It's purely definitional. I recognize that. But "The aim of The Quintessence of Axiomatized Special Relativity is to remove from Einstein's relativity theory everything that is confused, unnecessary and not amenable to experimental verification."

10. ### przyksquishyValued Senior Member

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What are you even talking about? In principle, you can parameterise space and time using any coordinates you want. The only restriction is that coordinates should faithfully represent the topology of space-time.

If you understand that the Lorentz transformation is supposed to represent a symmetry of existing physical laws, then you're plain wrong. The Standard Model of particle physics successfully describes all non-gravitational interactions seen in nature, and is well verified experimentally. General relativity is also quite well verified. Between them, both theories cover the currently accessible experimental domain. Both have clear mathematical definitions and there is no ambiguity concerning the symmetries they possess. You are not entitled to just make up any claims you want about physics and present them as fact.

No they're not. They're requiring that the laws of physics remain the same at every point in space and time - ie. they should be invariant under coordinate translations. Mathematically, we require that the Lorentz and translation groups commute. In other words if we choose to apply a translation to a system followed by a Lorentz boost, there should be a way of first applying a boost followed by a translation which produces the same result. In mathematical terms, if $x^{\prime} \,:\, \mathbb{R}^{4} \,\rightarrow\, \mathbb{R}^{4} \,:\, x \,\mapsto\, x^{\prime}(x)$ is your boost, then for all $a \,\in\, \mathbb{R}^{4}$ there must exist $b \,\in\, \mathbb{R}^{4}$ such that
$x^{\prime}(x) \,+\, b \,=\, x^{\prime}(x \,+\, a)$
for all $x \,\in\, \mathbb{R}$. Taking the partial derivative of both sides produces
$\frac{\partial x^{\prime}}{\partial x}(x) \,=\, \frac{\partial x^{\prime}}{\partial x}(x \,+\, a) \;.$​
This has to hold for all $x$ and $a$, so $\frac{\partial x^{\prime}}{\partial x}$ must be a constant, and the boost $x \,\mapsto\, x^{\prime}(x)$ must be a linear or affine transformation. Anything else is inevitably going to rely on the existence of a privileged coordinate origin.

I don't call that "arm-waving".

Then you're trying to fix what isn't broken. You may be confused about relativity. That doesn't mean the rest of us are.

Last edited: Oct 16, 2010
11. ### Eugene ShubertRegistered Senior Member

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As David Hilbert has said, "The art of doing mathematics consists in finding that special case which contains all the germs of generality." For special relativity, the special example of spacetime just happens to be the spacetime as it is defined in The Quintessence of Axiomatized Special Relativity. It is slightly more general than the group structure of the Poincaré group. Nonlinearity is built in from the very start. The linear representation of it is just a special case.

12. ### przyksquishyValued Senior Member

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Well then you're going wrong right there. Special relativity isn't mathematics. It's physics. It has a mathematical expression, but its goal is correctness as a statement about reality, not generality.

13. ### Eugene ShubertRegistered Senior Member

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Believe the man that taught Einstein how to derive the field equations for general relativity. Physics is mathematics.

14. ### przyksquishyValued Senior Member

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1) Citation needed. Where and when did Hilbert claim physics is mathematics? There's a big difference between saying that physics aims to formulate mathematical models of the universe, and saying that physics is mathematics. Do you seriously not grasp this?
2) In the offhand chance you do find such a quote, argument from authority. And a rather disingenuous one at that. Physicists define what physics is. Not David Hilbert.

15. ### Eugene ShubertRegistered Senior Member

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I don't recall the where and when. The exact wording in the Hilbert reference I had in mind reveals Hilbert's perfect comparison between geometry and physics according to his mathematical philosophy. See David Hilbert's Philosophy of Physics. For a perfect example of Hilbert identifying physics with mathematics, see the book, David Hilbert and the Axiomatization of Physics (1898-1918): From Grundlagen der Geometrie to Grundlagen der Physik. There, author Leo Corry mentions why Hilbert's published papers in physics greatly annoyed physicists. This is hilarious. Physicists were angered by reading Hilbert's scientific papers because Hilbert took the time to identify all his postulates and would then prove that his postulates were logically independent.

That argument is false. The authority Hilbert said, "Physics is much too hard for physicists."

16. ### DywyddyrPenguinaciously duckalicious.Valued Senior Member

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Wrong.

Oh, right. And Hilbert was "the authority" on what?
Certainly not physics.

17. ### brucepValued Senior Member

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Shubee has been spewing this nonsense for the last 20 years at Google groups. The chance that he's going to realize the error of his way now is nil.

18. ### Eugene ShubertRegistered Senior Member

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450

Consider the facts:

It took Einstein 8 years to come up with a solution to his “grand problem” and he couldn’t have done it without the assistance of mathematicians. It only took Hilbert several weeks to solve the problem. Hilbert’s derivation was based on an intuitive, formally unproved, invariant theoretical fact. Einstein’s solution, most probably a hint from Hilbert, contained no derivation at all. [1] Then Einstein spend another 8 years trying to unify gravity and electromagnetism, which culminated in Einstein’s 1923 papers on affine unified field theory but this unification, as physicists admit, is identical to Hilbert’s theory achieved in 1915. [2]

I have no reason to doubt Kip Thorne's fascinating historical account of the development of general relativity, where Hilbert is the hero, not Einstein.

See David E. Rowe's journal article, Einstein Meets Hilbert: At the Crossroads of Physics and Mathematics.

The abstract to that article says:

In that article, Kip Thorne is quoted from his 1994 book saying:

Also see Corry, Renn and Stachel Fingered as Likely Suspects in the Mutilation of David Hilbert's Page Proofs: [4] [5] [6] [7].

Last edited: Oct 17, 2010
19. ### brucepValued Senior Member

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2,816
Hilbert didn't think general relativity was his theory.

"Kip Thorne concludes, based on Hilbert's 1924 paper, that Hilbert regarded the General Theory of relativity as Einstein's: "Quite naturally, and in accord with Hilbert's view of things, the resulting law of warpage was quickly given the name the Einstein field equation rather than being named after Hilbert. Hilbert had carried out the last few mathematical steps to its discovery independently and almost simultaneously with Einstein, but Einstein was responsible for essentially everything that preceded those steps...".[29] However, Kip Thorne also stated, "Remarkably, Einstein was not the first to discover the correct form of the law of warpage[. . . .] Recognition for the first discovery must go to Hilbert."[30]"

Spot where I can't cite a source because I've made < 20 posts?

None of this supports your goofy unpublished paper.

20. ### przyksquishyValued Senior Member

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Then you've got nothing. It's not my job to start scouring references looking for quotations supporting your claims.

Hilbert is not an authority on what constitutes physics. If an argument from authority is all you've got, then, again, you've got nothing.