# Thread: On the Definition of an Inertial Frame of Reference

1. Originally Posted by przyk
No, Alphanumeric and Guest know perfectly well what they're talking about.

If you think about it, you'll realise that calling a theory or equation "covariant" requires that you accept that certain quantities are frame-dependent. For example, the usual textbook expression of Maxwell's equations ($\bar{\nabla} \cdot \bar{E} \,=\, \frac{\rho}{\varepsilon_{0}}$ and so on) isn't covariant, because it implicitly assumes that there are separate $\bar{E}$ and $\bar{B}$ fields.
Let's stop right here. The wave equation, expressed either in terms of E or B is covariant wrt the Lorentz transforms. This is precisely what got Lorentz and Einstein started, the search for a set of transforms that preserve the form of the wave equation.

In the Lorentz covariant formulation we accept that the electric and magnetic fields are frame-dependent.
Yes, B and E transform a certain way. The wave equation , on the other hand, is frame invariant (see attachment).

2. Originally Posted by przyk
$g_{\mu\nu} \,\mapsto\, \frac{\partial x^{\mu}}{\partial \tilde{x}^{\rho}} \, \frac{\partial x^{\nu}}{\partial \tilde{x}^{\lambda}} \, g_{\mu\nu}$
Would you please calculate $g_{\mu\nu}$ in the context of Shubert's formalism? This is the same problem I have given Guest254 and AlphaNumeric. Once you do that, please calculate the Christoffel symbols. Thank you.

3. Originally Posted by AlphaNumeric
You ignore when I (or Guest) say something which demonstrates beyond high school understanding and try to pretend we don't know what we're talking about.

I never said, hinted or implied that Guest ever made a mathematical mistake in his entire life.

Originally Posted by AlphaNumeric
YouTube link is to bring things down to my level) because I didn't pay much attention to you.

Is there no end to your misrepresentations?

Originally Posted by AlphaNumeric
You mistake apathy and disinterest for ignorance.

That's false. I equate obvious righteous indignation that is clearly rooted in proven ignorance as the very definition of bigotry.

Originally Posted by AlphaNumeric
If I gave every crank I've ever come across my full attention I'd never do anything else, there's so many of them.

Shameless slurs confirm words. You have condemned yourself.

Originally Posted by AlphaNumeric
Such a description would be classed as using 'weasel words' on Wikipedia, as you insert your opinion as to the worth of something you came up with, regardless of what anyone else has to say.

Originally Posted by AlphaNumeric
Yes yes, you're over everyone's head

The Black Knight understood what you did not. And his mathematical abilities are at the average high school level. I know. I've seen one of his self-published papers. He goes by the name Dono at google groups.

Originally Posted by AlphaNumeric
Like I said previously, a classic sign of a crank is someone who presents their work on forums, not journals. If you're so good and your work so 'valid' and 'flawless' (your words) then you should have been able to get it published.

Journals don't publish obvious results that high school students can understand even though professional physicists cannot.

Originally Posted by AlphaNumeric
Then you asked about Newton's laws and I gave a lengthy answer only for you to ask a question which suggested you didn't understand the response.

I tire of programmed responses without discernment like your cut-and-paste stuff out of the same material I learned in graduate school. You obviously failed to understand my questions and comments.

Originally Posted by AlphaNumeric
Like I said, if you don't like responses on forums then don't post your work.

I'm certain that I have said that I only trust mathematicians. That's not an invitation for physicists to intrude where they're not welcome and to find entertainment value in behaving like judgmental trolls.

Originally Posted by AlphaNumeric
Go to a journal instead so you can get feedback from people whose job it is to know relativity inside out, if you don't think I'm up to it.

I don't know of any physics journal that caters to bright high school students.

Originally Posted by AlphaNumeric
You want professionalism go to a reputable journal or university.

Been there, done that. Personal conversations with Professors Frankel (UCSD), Rindler (UTD) and others confirm my position.

Originally Posted by AlphaNumeric
I've done reviewing in the past and I did it in a professional manner, as there's rule of etiquette for journals but this isn't a journal and I'm not here in a professional capacity. Forums are entertainment for me

Yes. I can see how much you must enjoy pontificating and acting like a willfully close-minded troll.

Originally Posted by AlphaNumeric
Even the best physicists in the mainstream wouldn't regard work as 'flawless' before peer review (or even after), everyone makes mistakes.

I am certain about high school level algebra. My paper was obviously too difficult for you.

Originally Posted by AlphaNumeric

It's funny that you condemn me for what you are guilty of.

4. Originally Posted by Tach
One look at your idiotic thread on the Sagnac effect is enough to tell that you don't know the very basic SR.

5. Originally Posted by Eugene Shubert
Been there, done that. Personal conversations with Professors Frankel (UCSD), Rindler (UTD) and others confirm my position.
Er, they said your work was trivial. This supports what various people in this thread have been telling you: there may be a way of interpreting your non-linear transformation as being "correct", but only as a special case of the generally covariant formulation of relativity. Which many physicists are already familiar with. So your work is of no interest to either physicists or mathematicians, because we already have the generally covariant formulation and pseudo-Riemannian geometry.

6. As I have already pointed out, it is very easy to prove that a nonlinear model of the Lorentz transformations is an extraordinarily simple idea that physicists are confused about and it is precisely this determined confusion and ignorance that justifies Sir Knight in his holy crusade against me. He has been hounding me for years.

7. Originally Posted by Tach
Listen,

What you have been asked to do (and evaded repeatedly) is to show whether the equation:

$\Bigg(\nabla^2 - { \mu\epsilon } {\partial^2 \over \partial t^2} \bigg) \mathbf{E}\ \ = \ \ 0$

is invariant wrt the Shubert crackpot transform.
The fact you ask this demonstrates, once again, that you completely fail to understand the concept of general covariance. An equation obeys the principle of general covariance if its form is invariant under arbitrary coordinate changes. The equation you have given, in its current form, clearly does not obey the principle of general covariance. This is obvious!

However, if you write the wave equation in the form

$\nabla^a \nabla_a \phi =0$

with $g_{ij}=\eta_{ij}$, then this is obviously the same equation, but the form it is written in is preserved under general coordinate transformations. So we have expressed the equation in a generally covariant form - i.e. a form that is preserved under arbitrary coordinate changes.

The fact you're finding it so difficult to understand is ridiculous. You continue to beat your chest and insist everyone else is wrong! Let's look at some of the clangers you've dropped whilst insisting you know what you're on about:

Go you!

8. Tach, you appear to have either completely ignored almost all of my post or you simply didn't understand it. I'm leaning towards the former because even you should have managed to understand it when I repeatedly said I wasn't defending Eugene's work but making a broader point, yet you continue in your reply to say "Show me how Eugene's equations are right!!!", both failing to grasp the points I'm making about covariance and the very clear statement I made about how Eugene's work is irrelevant to the point I'm making.

Originally Posted by Tach
This is in the category "not even wrong" since it was the lack of covariance of the Maxwell laws under Galilei transforms that sparked the quest for the discovery of the Lorentz transforms. You have all your basics thoroughly screwed up. So, no, the Maxwell equations are not covariant in the Galilei formalism. They are in the Einstein-Lorentz formalism.
I explained the difference between a general coordinate transformation, which need not be a linear operator and whose properties are examined in a particular way, and a linear operator which leaves an inner product invariant. You failed to respond to any of it and you appear to have failed in understanding it.

I'm well aware of how Maxwell's equations are Lorentz invariant. This is not the same as saying "The only coordinate transformations valid are Lorentz ones". You can write Maxwell's equations in polar coordinates (and many physicists do for things like magnetohydrodynamics), which aren't a Lorentz transformation away from Cartesians, and still have valid expressions.

Maxwell's equations are Lorentz invariant in the sense of the inner product invariance. It's a little over the top but the most general way of writing down a Yang Mills gauge theory Lagrangian is along the lines of $\mathcal{L} = \textrm{Tr}(F_{\mu\nu}F^{\mu\nu}) = F^{a}_{\mu\nu}F^{a}^{\mu\nu}$ where 'a' is the Lie algebra generator index. The Lie algebra structure has a particular kind of symmetry, to do with the Killing form, and the space-time indices have a different kind. I am willing to go into the Killing form kind if you wish but lets stick to the space-time indices for the time being. The Lagrangian density can be rewritten as $F_{\rho\xi}\eta^{\rho \lambda} \eta^{\xi\phi} F_{\lambda \phi}$. Suppose now we act on the space-time indices with a linear operator in the manner of $v_{\mu} \to M_{\mu}^{\nu}v_{\nu}$ and $w^{\nu} \to M^{\nu}_{\mu}w^{\mu}$. Doing this for the F indices and factorising in a particular way it follows (the algebra is too lengthy to type here, I dislike typing excessive LaTeX on forums) that the Lagrangian density is invariant if $M\cdot \eta \cdot M^{\top} = \eta$ (if you don't see why, ask).

Thus the form of Maxwell's equations picks out a set of linear operators as 'special', which due to the form of the equations can be reexpressed as those linear operators which leave the metric (ie the associated inner product) invariant. Galilean transforms, in general (though the rotational subgroup does), don't do this specific thing.

However, if you wish to do a coordinate transformation via a Galilean transform, you're welcome to do so on Maxwell's equations. The specific form of the individual equations will change, this is not a surprise. For instance, the form of the Laplacian $\Delta = \sum_{j} \frac{\partial^{2}}{\partial x_{j}^{2}}$ changes completely when you go into polars (too long to type out, see here). The fact the form has changed isn't a problem, it doesn't mean writing the parameters of space in terms of spherical polars is wrong, just as using a Galilean transform to construct new space parameters will lead to a new expression which isn't 'wrong' either.

In the case of the inner product preserving Lorentz transformation if you have a set of equations involving say the electric and magnetic fields E and B in terms of parameters t,x,y,z then afterwards you'll have a set of equations which are exactly the same but with E changed to E', B to B' and (t,x,y,z) to (t',x',y',z'). That's 'special' because if you changed to polar coordinates you obviously don't just change x to r, y to $\theta$ etc but never the less the coordinate transformation is a valid one and one which doesn't change the tensor structure of Maxwell's equations.

Despite repeated explanation from myself, Guest and przyk you haven't demonstrated you even see the distinction, never mind understand it.

Originally Posted by Tach
Let's cut the crap and prove that the Jacobian for the Shubert formalism is valid . Roll up your sleeves for once and do some calculations. (hint: You won't be able to prove the Jacobian is valid. It is obvious why).
Where did I say I thought it was? I said assuming the Jacobian is valid then Eugene's transformation doesn't alter the tensor structure of Maxwell's equations. I repeatedly explained that the point about Jacobians is not dependent on the specific form of Eugene's expressions.

Seriously, how many times do I have to say something as simple as "Eugene's work is irrelevant to the point I'm making" before you get that its irrelevant to the point I'm making?

Originally Posted by Tach
Why don't you cut the crap and calculate the operator $\nabla^{a}$? You have the Shubert formalism at your disposal. Start with the Christoffel symbols. Enough blathering , let's see some calculations. You have two ways to show that you really know what you are talking about.
Actually I have three ways, one of which is explaining the mistake you've made and then elaborating on the specifics of that mistake and how you could correct your understanding if you bothered. And that's the route I've just gone down.

When you come back from your ban please don't come out with this "Show that Eugene's equations work!!" strawman again. Repeatedly I've said your mistake is nothing to do with Eugene's equations and I've explained it in the much wider context of tensor calculus and inner product spaces. This isn't a matter of "Opps, I forget to carry the two, turns out that transformation doesn't leave the metric invariant", its a matter of "The validity of a coordinate transformation is independent of the metric".

Originally Posted by Eugene
Journals don't publish obvious results that high school students can understand even though professional physicists cannot.
The cry of the internet crank.

9. Originally Posted by Tach
Would you please calculate $g_{\mu\nu}$ in the context of Shubert's formalism?
In "Shubert coordinates" the metric would be
$\text{d}s^{2} \,=\, -\, \text{d}t^{2} \,-\, 2 \, (\partial_{i}S) \, \text{d}t \, \text{d}x^{i} \,+\, \bigl(\delta_{ij} \,-\, ( \partial_{i}S)(\partial_{j}S) \bigr) \, \text{d}x^{i} \, \text{d}x^{j} \;.$
Shubert's transformation, assuming it's well defined (I don't care enough to check) leave the components of this metric invariant. In terms of the decomposition $\theta \,\circ\, \Lambda \,\circ\, \theta^{-1}$ of his transformation, $\theta^{-1}$ brings you back to the Minkowski metric, the Lorentz transformation $\Lambda$ leaves the Minkowski metric invariant, and $\theta$ takes you back to the metric above. That's the least trivial interpretation of Shubert's transformation that could be considered in any way "correct".

Now that you've got the metric, working out the Christoffel symbols is tedious but straightforward. If you really need them (as opposed to demanding people do calculations for you in the hope of stalling the thread while you try to get your act together), they're given in terms of the metric components by
$\Gamma^{\rho}_{\mu\nu} \,=\, \frac{1}{2} \, g^{\rho\kappa} \, \bigl[ \partial_{\mu} g_{\kappa \nu} \,+\, \partial_{\nu} g_{\mu \kappa} \,-\, \partial_{\kappa} g_{\mu \nu} \bigr] \;.$

10. Originally Posted by AlphaNumeric
The cry of the internet crank.

Go to your resting place ghost. It is obvious to anyone that has read my paper, your responses, post #189 or has seen the gruesome video reenactment, that Sir Knight put his sword through your head and that you have already died in shame and defeat and that you will only be remembered, if at all, as a bumbling jouster that received his just reward.

11. Originally Posted by Eugene Shubert
Go to your resting place ghost. It is obvious to anyone that has read my paper, your responses, post #189 or has seen the gruesome video reenactment, that Sir Knight put his sword through your head and that you have already died in shame and defeat and that you will only be remembered, if at all, as a bumbling jouster.
I think anyone who reads that will conclude you're completely insane.

12. I thought it would be fun to explain this thread to a general audience that might not grasp the intricacies of high school algebra and general covariance.

13. Originally Posted by przyk
If you really need them (as opposed to demanding people do calculations for you in the hope of stalling the thread while you try to get your act together), they're given in terms of the metric components by
$\Gamma^{\rho}_{\mu\nu} \,=\, \frac{1}{2} \, g^{\rho\kappa} \, \bigl[ \partial_{\mu} g_{\kappa \nu} \,+\, \partial_{\nu} g_{\mu \kappa} \,-\, \partial_{\kappa} g_{\mu \nu} \bigr] \;.$
Of course its a minor technicality but, since its a technicality I'm sure Tach won't be aware of, I should point out that that formula only works if you're neglecting torsion.

Originally Posted by Eugene Shubert
Go to your resting place ghost. It is obvious to anyone that has read my paper, your responses, post #189 or has seen the gruesome video reenactment, that Sir Knight put his sword through your head and that you have already died in shame and defeat and that you will only be remembered, if at all, as a bumbling jouster that received his just reward.
I take it from your inability to actually have a back and fore discussion when I reply to your questions that you're reduced to just attempting to insult people.

Tach's claims about covariance have been completely refuted and explained again and again by myself and others. Your claims have not gained any hold, you've convinced no one. Where's the defeat? So my comment about that specific part of your work wasn't correct, does that mean I was wrong about covariance and coordinate transforms? No. Does it mean your work is 'flawless'? No. Does it mean you're taken any more seriously? No.

When I've tried to engage you in conversation, giving detailed replies to your questions, you ignore me or reply with some cryptic response which suggests you didn't understand what I'd just explained to you.

My mistake was not giving you enough attention, it was not because I got something fundamentally wrong. In regards to getting something fundamentally wrong Tach has had his mistakes explained again and again. I have nothing to be 'shamed' about in that regard.

Tach doesn't think your work is valid either, so you're hardly putting yourself in a better position siding with him against me. No one here thinks your claims of 'flawless' work have merit. You couldn't get published in a reputable journal, you can't even convince people on the internet. If you want to talk about shaming I'd say that's pretty shameful. How many years have you been pushing your work now? I asked before but you didn't answer. I guess you're ashamed of the answer so you don't want to say.

I have no reason to hide or slink away. I stand by everything I've said to Tach about covariance and transformations and the moderators seem to have a similar view, along with Guest and przyk. As for wading through your work, like przyk says, I have no real wish to waste my time. The sorts of questions you ask, the responses you give and your clearly bitter and twisted view of how mainstream physics is done and the people who do it tells me enough to know that your work is very unlikely to be worth the time and effort of going through with a fine tooth-comb.

If you're so confident in your work and you think I'm in such a shameful position why aren't you responding to direct questions? Whenever I try to raise the level of discussion by asking direct questions of a technical nature both you and Tach ignore them and reply with something either irrelevant or which demonstrates you haven't actually understood (or perhaps even read) what I have said. It would seem the ones who are ashamed of the questions they are asked are you two.

So are you willing to answer direct questions or are you going to forever just make references to Monty Python? Its no skin off my nose either way at the end of the day, unlike yourself the contributions I've made to science amount to more than pushing pdfs I've written on forums, so it doesn't bother me if I didn't read your pdf with enough attention, I can at least say I've contributed to science.

14. Originally Posted by AlphaNumeric
Of course its a minor technicality but, since its a technicality I'm sure Tach won't be aware of, I should point out that that formula only works if you're neglecting torsion.
Thanks. I've only ever been faced with torsion free metrics in a couple of GR courses so I didn't have torsion in mind, though of course the metric I gave Tach is torsion free since it's just Minkowski in disguise.

15. Guest254,

I'd like to hear your opinion. In the mathematical model called Lorentzian spacetime, is it possible to define inertial frames of reference in a coordinate-free way? Is the existence of inertial frames of reference in that mathematical model a law of physics? If so, is that law of physics generally covariant?

16. My own answers, based on my understanding of relativity:
In the mathematical model called Lorentzian spacetime, is it possible to define inertial frames of reference in a coordinate-free way?
Defining inertial coordinate systems in a coordinate-free way is a contradiction in terms. If you want to express physical laws in a coordinate free way, it's possible, but then you're specifically avoiding the use of coordinate systems, inertial or otherwise.

If you mean "inertial frame" in some weaker sense than "inertial coordinate system" then it may be possible to define an "inertial frame" in a partially coordinate free way. I gave an example of how you might do this in an early post in this thread (not that I consider the result terribly useful). You only need a time-like geodesic parameterised by its proper time, which is a weaker requirement than a complete coordinate system.

Is the existence of inertial frames of reference in that mathematical model a law of physics?
Yes, though it doesn't have to be required directly. We can require that the laws of physics are defined on a flat Lorentzian manifold, where the metric has signature (-, +, +, +). This implies the existence of inertial coordinate systems without explicitly requiring them.

If so, is that law of physics generally covariant?
Yes, since the metric signature is an invariant property of the metric. To be clear on a subtle point, the definition of an inertial coordinate system is not covariant (ie. not all coordinate systems defined on a Lorentzian manifold are inertial), but the requirement that they exist is covariant.

Why do you need Guest's approval for all of this? If you don't feel competent enough to evaluate the answers I'm giving you and you'd rather hear them from a mathematician, then fine, and a good mathematician may well give you more elegant responses than I could. But then why are you filling this thread with inflammatory remarks like "physics is too hard for physicists" when you're obviously unable to evaluate what physicists are capable of?

17. Originally Posted by przyk
In "Shubert coordinates" the metric would be
$\text{d}s^{2} \,=\, -\, \text{d}t^{2} \,-\, 2 \, (\partial_{i}S) \, \text{d}t \, \text{d}x^{i} \,+\, \bigl(\delta_{ij} \,-\, ( \partial_{i}S)(\partial_{j}S) \bigr) \, \text{d}x^{i} \, \text{d}x^{j} \;.$

First off, thank you for taking the time and attempting to do the calculations. As you will see, my request had a purpose.

1. I suspect that you started from the Minkowski metric $ds^2=(cdt)^2-dx^2$ and you passed it through the Shubert transforms, right?
If this is the case then the calculations are incorrect. A quick examination shows that you missed the fact that $S_i=S_i(x,t,S_j(x,t)$. But this is the least of the problems.

Shubert's transformation, assuming it's well defined (I don't care enough to check)
2. Well, you should. The Shubert crcakpot tranform suffers from some serious drawbacks that make it invalid. Firstly, you cannot prove that the jacobian is non-degenerate. The dependence on the arbitrary synchronization functions $S_i$ and $S_j$ make it possible for the cancellation of determinant of the jacobian over an infinity of points (x,t). Secondly, in order for the covariance of the metric to be preserved under transformation, the jacobian needs to be a unit matrix[1] , a condition that is clearly not satisfied.

leave the components of this metric invariant.
3. The problem is that it doesn't (see above). You should have checked.

In terms of the decomposition $\theta \,\circ\, \Lambda \,\circ\, \theta^{-1}$ of his transformation, $\theta^{-1}$ brings you back to the Minkowski metric,
4. Not if $\theta^{-1}$ does not exist since the jacobian cannot be proven non-degenerate.

the Lorentz transformation $\Lambda$ leaves the Minkowski metric invariant, and $\theta$ takes you back to the metric above. That's the least trivial interpretation of Shubert's transformation that could be considered in any way "correct".
5. Well, they don't. My point all along.

Now that you've got the metric, working out the Christoffel symbols is tedious but straightforward.
6. Here you made another error. The spacetime being flat, the Riemann tensor is null so the Christoffel symbols are all null. [2]. The covariant derivatives reduce to standard derivatives.

If you really need them (as opposed to demanding people do calculations for you in the hope of stalling the thread while you try to get your act together),
7. I had a purpose: to get you to do your homework. Rather than talking down to me, I would hope that you would show some respect and realize not only your errors but also (more importantly) the deficiencies in the Shubert crackpot formalism. For a more detailed description of the shortcomings of the Shubert crackpot formalism , see here.

[1] C. Moller, Theory of relativity (pp.94-96)
[2] C. Moller, Theory of relativity (pp.377)

18. Originally Posted by Tach
I would hope that you would show some respect and realize not only your errors but also (more importantly) the deficiencies in the Shubert crackpot formalism.

It is clearly obvious Sir Knight that you are asking for something that is impossible. The persons seemingly most capable of judging my paper are actually most incapable of doing so fairly. They don't want to be ostracized by the community that greatly reveres Albert Einstein.

19. Originally Posted by Eugene Shubert
It is clearly obvious Sir Knight that you are asking for something that is impossible. The persons seemingly most capable of judging my paper are actually most incapable of doing so fairly. They don't want to be ostracized by the community that greatly reveres Albert Einstein.
They revere Einstein for good reason: his theory is correct. Yours is not.

20. Originally Posted by Tach
They revere Einstein for good reason: his theory is correct. Yours is not.

Keep believing that Sir Knight. You just might prod the pompous to reveal a little backbone.