On the Definition of an Inertial Frame of Reference

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

  1. Neddy Bate Valued Senior Member

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    I had similar suspicions about Tach after devoting about five pages trying to correct the many errors he had made, (in a different thread). He even managed to get that thread locked, apparently because the moderator found it too painful to watch anymore. I'm better off, because I probably would have wasted a lot more time there.

    Eugene Shubert said it best:

     
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  3. Eugene Shubert Valued Senior Member

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    Taken from the Star Wars trilogy, the last line in Monty Python Meets Star Wars.
     
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  5. Eugene Shubert Valued Senior Member

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    You'll have to do a lot better than that.
     
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  7. AlphaNumeric Fully ionized Registered Senior Member

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    You and Tach both seem to be doing the same thing in that 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.

    You asked about Newton's laws in a covariant manner and I went into details on geodesics to answer you. Rather than acknowledge that you decide to try to insult me by calling me thick (ie saying that YouTube link is to bring things down to my level) because I didn't pay much attention to you.

    You mistake apathy and disinterest for ignorance. 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.

    That's your view. 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.

    Yes yes, you're over everyone's head

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    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. Instead you're here and like all cranks whining there's some kind of conspiracy or defence of orthodoxy holding you back, rather than any issues your claims have.

    Also I've gauged your questions and responses somewhat. I replied to you right at the beginning of the thread to answer what reason there is that Lorentz transforms are linear, which suggested you're unfamiliar with Lorentz transforms' formalisation. They are linear in the sense I just explained to Tach in regards to inner product invariance. Vectors exist in a vector space, space-time is equipped with a metric (Newtonian or Lorentzian, doesn't matter which you prefer) and has a complete basis. Thus we're working in a Hilbert space and operators on elements work by linear actions, ie \(L(a\mathbf{v}+b\mathbf{w}) = aL(\mathbf{v}) + bL(\mathbf{w})\). Lorentz transformations are linear in that sense, acting on elements of the tangent space of space-time, so the rotation of the sum of two vectors is equal to the sum of the rotations of the vectors, that sort of thing.

    Composing Lorentz transforms or adding non-transformed vectors to transformed ones, say something like \(\mathbf{v} \to L(\mathbf{v} + L(L(\mathbf{v} + L(\mathbf{v})))\) is still a linear operation, the expression is still linear in \(\mathbf{v}\). The transformations themselves, L, might be non-linear in terms of their parameters. For instance, a Lorentz boost by velocity V, \(L_{V}\) does not satisfy \(L_{V}(L_{V'}(\mathbf{v})) = L_{V+V'}(\mathbf{v})(\mathbf{v}))\), the relativistic velocity formula applies, \(L_{V}(L_{V'}) = L_{\frac{V+V'}{\sqrt{1+VV'}}}\). Rotations can be linear in this regard, if you work only in an abelian subgroup, like an SO(2) subgroup which satisfies \(L_{\theta}L_{\theta'} = L_{\theta+\theta'}\).

    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.

    Oh no, I was a little bit rude to someone, how shameful.

    Like I said, if you don't like responses on forums then don't post your work. 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. But I imagine you already have and you didn't like the answer, so you ended up pushing it on forums and if Brucep's right you've been doing this for years. Clearly even years of you harping on about about your 'flawless' work has gotten you nowhere, another 10 minutes of my time wouldn't have helped you.

    You want professionalism go to a reputable journal or university. 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, not work and if someone is pushing their 'flawless work' here then it means that they've already been turned down by people whose job it is to review papers or that they haven't bothered to submit it. Either way it raises serious questions as to the worth of their work and their attitude to viable science.

    Just as I don't come here to be a professional journal reviewer I don't expect anyone else to either, hence why none of my work (and likewise for the other people on these forums with research published) I post here for 'review'.

    Ah, because I didn't give you enough attention I'm completely uncredible.

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    Perhaps if you had not represented yourself as correcting the mistakes of all of theoretical physics with your 'flawless' work, ie acted a little more professionally and displaying some humility, I might have been more accommodating but when someone opens with claims of flawless work and how all of physics is some kind of group think orthodoxy I immediately think "This person has no clue as to how to present their work or how the research community works". Even the best physicists in the mainstream wouldn't regard work as 'flawless' before peer review (or even after), everyone makes mistakes. Peer review is about exposing flaws so people can repair them, presenting your work as 'flawless' demonstrates you've already made up your mind about your own work (and its always easy to convince yourself of your own preconceptions....) and yet you complain about mainstream orthodoxy!

    Tell me, how many years have you been pushing your work? Which journals have you submitted it to and what did they say?
     
    Last edited: Oct 29, 2010
  8. przyk squishy Valued Senior Member

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    What's wrong with that?
     
  9. AlphaNumeric Fully ionized Registered Senior Member

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    Why are you being so short in your answers? You asked about Newton's laws and when I give a lengthy explanation you reply with some curt response about curvature and now you're doing the same. I'm not going to give you the benefit of the doubt, as I don't think you understood my response or przyk's. You ask what you think is going to be a difficult question, someone answers it with ease and with detail and (it seems to me) you realise you don't understand the response so you just come up with some cryptic response in the hopes the matter stops there and you aren't pushed into admitting you didn't understand.

    What precisely do you want przyk to elaborate on?
     
  10. Tach Banned Banned

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    Just because you are grossly ignorant on basic SR doesn't mean that you "corrected" me. One look at your idiotic thread on the Sagnac effect is enough to tell that you don't know the very basic SR.

    I cannot speak for the moderator but I can tell you that it is your thread, with your repeated idiocies that got locked. I simply tried (and failed) to teach you some basic SR.
     
    Last edited: Oct 29, 2010
  11. Tach Banned Banned

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    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.

    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).


    Ok,

    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.
     
    Last edited: Oct 29, 2010
  12. arfa brane call me arf Valued Senior Member

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    Tach:

    See if you can figure out what this diagram represents. Note that \( \epsilon^2 \) represents a small parallelogram, or "area" of spacetime, which is curved.

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  13. Tach Banned Banned

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    Maybe that you cannot follow the discussion and you compensate by posting random jpegs picked from John Baez' site?

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    Last edited: Oct 29, 2010
  14. arfa brane call me arf Valued Senior Member

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    Or maybe that the random jpeg is a diagram of something that an expert like you should be able to explain to a dumbass like me?

    Or are you only an expert on general histrionics?
     
  15. Tach Banned Banned

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    Open a different thread and I'll explain, don't troll this thread.
     
  16. arfa brane call me arf Valued Senior Member

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    No, explain first why the diagram has nothing to do with this thread, which has digressed into a discussion about choosing coordinates and covariance.

    Or find another excuse, more likely. I expect you'll have something to say about curvature and how it has nothing to do with covariance.
     
  17. Tach Banned Banned

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    I remember you now, you are the guy who did not understand arc length and thought that it has something to do with curvature. So, you are now retaliating by trolling this thread. I posted a quick tutorial for you in your thread, so maybe you will give me the courtesy and stop trolling this thread? Thank you.
     
    Last edited: Oct 29, 2010
  18. arfa brane call me arf Valued Senior Member

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    That's an amazing memory you have.
    So you still think arc length doesn't have anything to do with curvature.

    The first question that comes to mind is: is an arc curved? Do you know the answer?
    Next up: Is the corresponding chord curved?

    Think carefully...
     
  19. Tach Banned Banned

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    I answered your question in your thread. It is not my fault that you are ignorant, you can only blame yourself for flunking differential geometry. I even posted a remedial class for you, so why don't go troll elsewhere.
     
    Last edited: Oct 30, 2010
  20. arfa brane call me arf Valued Senior Member

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    7,832
    So you are just a troll then?
    You've got exactly nothing to contribute of any value whatsoever.
    Thanks for the heads-up. I take it you don't understand John Baez' dissertation either, like how you don't understand the notion of arc and chord lengths as metric spaces?
     
  21. Tach Banned Banned

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    Barf, plonk.
     
  22. arfa brane call me arf Valued Senior Member

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    That's very witty and, dare I say, rather enlightening.
     
  23. przyk squishy Valued Senior Member

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    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. In the Lorentz covariant formulation we accept that the electric and magnetic fields are frame-dependent. Then the principle of Lorentz covariance claims that
    \( \begin{gather} x^{\mu} \,\mapsto\, \Lambda^{\rho}_{\mu} \, x^{\mu} \\ \\ F^{\mu\nu} \,\mapsto\, \Lambda^{\rho}_{\mu} \, \Lambda^{\lambda}_{\nu} \, F^{\mu\nu} \\ \\ j^{\mu} \,\mapsto\, \Lambda^{\rho}_{\mu} \, j^{\mu} \end{gather} \)​
    is a symmetry of Maxwell's equations, where \(\Lambda\) is a Lorentz transformation. Maxwell's equations possess no analogous symmetry with respect to Galilean transformations.

    In Lorentz-covariant expressions appears the symbol \(\eta_{\mu\nu}\) in tensor contractions, which plays the same sort of role the Kronecker delta does. The generally covariant formulation promotes this symbol to a tensor \(g\) and allows its components to be frame dependent. The principle of general covariance then claims that
    \( \begin{gather} x^{\mu} \,\mapsto\, \tilde{x}^{\rho}(x^{\mu}) \\ \\ F^{\mu\nu} \,\mapsto\, \frac{\partial \tilde{x}^{\rho}}{\partial x^{\mu}} \, \frac{\partial \tilde{x}^{\lambda}}{\partial x^{\nu}} \, F^{\mu\nu} \\ \\ j^{\mu} \,\mapsto\, \frac{\partial \tilde{x}^{\rho}}{\partial x^{\mu}} \, j^{\mu} \\ \\ g_{\mu\nu} \,\mapsto\, \frac{\partial x^{\mu}}{\partial \tilde{x}^{\rho}} \, \frac{\partial x^{\nu}}{\partial \tilde{x}^{\lambda}} \, g_{\mu\nu} \end{gather} \)​
    is a symmetry of the generally covariant formulation of Maxwell's equations, for arbitrary coordinate transformations. Proof: see just about any introduction to general relativity.

    General covariance is possible but avoidable if you're just doing SR. But it's absolutely vital to GR due to the non-existence of globally defined inertial coordinate systems. For this reason, the first thing GR texts teach you is how to do SR in arbitrary coordinate systems.
     

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