On the Definition of an Inertial Frame of Reference

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

  1. Tach Banned Banned

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    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.


    Yes, B and E transform a certain way. The wave equation , on the other hand, is frame invariant (see attachment).
     
    Last edited: Oct 30, 2010
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  3. Tach Banned Banned

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    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.
     
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  5. Eugene Shubert Valued Senior Member

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


    Is there no end to your misrepresentations?


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


    Shameless slurs confirm words. You have condemned yourself.


    I'm not soliciting comments from superficial readers and non-thinkers.


    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.


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


    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.


    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.


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


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


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


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


    It's funny that you condemn me for what you are guilty of.
     
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  7. Neddy Bate Valued Senior Member

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    Just out of curiosity, what thread are you talking about?
     
  8. przyk squishy Valued Senior Member

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    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.
     
  9. Eugene Shubert Valued Senior Member

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    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.
     
  10. Guest254 Valued Senior Member

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    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!
     
  11. AlphaNumeric Fully ionized Registered Senior Member

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    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.

    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.

    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?

    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".

    The cry of the internet crank.
     
  12. przyk squishy Valued Senior Member

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    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
    \(\Gamma^{\rho}_{\mu\nu} \,=\, \frac{1}{2} \, g^{\rho\kappa} \, \bigl[ \partial_{\mu} g_{\kappa \nu} \,+\, \partial_{\nu} g_{\mu \kappa} \,-\, \partial_{\kappa} g_{\mu \nu} \bigr] \;.\)​
     
    Last edited: Oct 30, 2010
  13. Eugene Shubert Valued Senior Member

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    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.
     
    Last edited: Oct 30, 2010
  14. przyk squishy Valued Senior Member

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    I think anyone who reads that will conclude you're completely insane.
     
  15. Eugene Shubert Valued Senior Member

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    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.
     
  16. AlphaNumeric Fully ionized Registered Senior Member

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    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.

    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.

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  17. przyk squishy Valued Senior Member

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    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.
     
  18. Eugene Shubert Valued Senior Member

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    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?
     
    Last edited: Oct 31, 2010
  19. przyk squishy Valued Senior Member

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    My own answers, based on my understanding of relativity:
    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.

    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.

    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?
     
  20. Tach Banned Banned

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    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.

    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.



    3. The problem is that it doesn't (see above). You should have checked.


    4. Not if \(\theta^{-1}\) does not exist since the jacobian cannot be proven non-degenerate.

    5. Well, they don't. My point all along.


    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.

    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)
     
    Last edited: Nov 2, 2010
  21. Eugene Shubert Valued Senior Member

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    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.
     
  22. Tach Banned Banned

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    They revere Einstein for good reason: his theory is correct. Yours is not.
     
  23. Eugene Shubert Valued Senior Member

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

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