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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    I cited two books that contradict your crackpot claims. I can cite more.

    I think that you should start co-authoring papers with Shubert. After all, you are as much a crackpot as he is. You could contribute the electromagnetic part that is currently missing from his paper.
     
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  3. Guest254 Valued Senior Member

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    Come on now - you've just got the hump because you've been publicly schooled in basic tensor calculus. Go you!
     
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  5. Tach Banned Banned

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    By a crackpot who thinks that a change in notation turns non-covariant into covariant? Get real.
     
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  7. Guest254 Valued Senior Member

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    Isn't it all the more embarrassing, if I'm a crackpot?
     
  8. Tach Banned Banned

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    No, because your "schooling" unveiled your deep misunderstandings about the underlying physics. The only person that you embarrassed is yourself. You and Shubert should start co-authoring papers, you two make a fine pair, you both manipulate equations without understanding what they mean. You should contribute the currently missing electromagnetic section to his paper, it is going to be a smash.
     
    Last edited: Oct 27, 2010
  9. temur man of no words Registered Senior Member

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    What's Shubert transform?
     
  10. Tach Banned Banned

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    A non-linear, implicit, transform invented by the crackpot Eugene Shubert, http://www.everythingimportant.org/relativity/special.pdf.

    , the starter of this thread and peddled by him on the www. Rejected by mainstream physics with the exception of his new found convert Guest254.
     
  11. Tach Banned Banned

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    Check you basics, in order for the above to hold true, the coordinate transformation is assumed to be of the form:

    \(x'_u=x'_u(x_0,x_1,x_2,x_3)\)


    The Shubert crackpot transformations are of the form:


    \(x'_u=x'_u(x_0,x_1,x_2,x_3,f(x_0,x_1,x_2,x_3))\)

    where f is an arbitrary function. So, you have been talking nonsense all along.
     
  12. Guest254 Valued Senior Member

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    Please, please stop making a fool of yourself - the thing you wrote is still a function of the coordinates (x_0, ... , x_3). News flash: you can define a function x'' so that:

    \( x'' (x_0, x_1, x_2, x_3) = x' (x_0, x_1, x_2, x_3, f(x_0,x_1,x_2,x_3)) \)

    as long as everything is sufficiently smooth and invertible (i.e. they're coordinates), then it's fine. But top marks for really cementing down your level of understanding!
     
  13. temur man of no words Registered Senior Member

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    So it has 5 variables? What is f? Why f is in the last place?
     
  14. Tach Banned Banned

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    It needs to be an explicit function of the coordinates, check your books idiot.


    Sure you can, the point is that the partial derivatives no longer follow the tensor transformation rules. This is why I kept asking you to calculate the partial derivative wrt t, to convince yourself of your folly.

    Since you are that stupid to think that a change in notation changes the underlying physics, you are a really good candidate to contributing the missing electromagnetic section to Shubert's paper.
     
    Last edited: Oct 27, 2010
  15. Tach Banned Banned

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    It is all hopeless crackpottery. Yet, is has mesmerized the pretender Guest254 into aspiring to become a co-author.
     
  16. Guest254 Valued Senior Member

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    It is a function of (x_0, ... , x_3). How can you not get this? If I write the function G as

    G(x,y,z) = F(f(x,y,z),g(x,y,z),h(x,y,z))

    for invertible functions f,g,h and an appropriate function F, then G is still just a function of (x,y,z). Just because the derivatives of G will be complicated if I choose to write them in terms of F, doesn't change anything.

    Erm, they do, as I've tried to explain to you above.

    Are you still confused by this whole general covariance thing? I refuse to believe that all my explanations have gone to waste!
     
  17. AlphaNumeric Fully ionized Registered Senior Member

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    Tach, what Guest has been saying is valid. I think you're confusing the linear action of a Lorentz transform, where you can express the coordinate transform as a matrix, \(\tilde{x}^{\mu} = \Lambda^{\mu}_{\nu}x^{\nu}\). This is a particular kind of coordinate transform, where you can easily use the Jacobian matrix \(\frac{\partial \tilde{x}^{\mu}}{x^{\nu}} = \Lambda^{\mu}_{\nu}\) and you manifestly get contractions being Lorentz invariant.

    More general transformations need not be linear. The polar coordinate transformation, ie \((x,y) \to (\rho,\theta)\) is not linear, ie the polar coordinates are not linear functions of the Cartesians but a tensor expression remains in the same form. For instance if \(T_{ab} = 0\) in Cartesians then you have \(T_{xx} = T_{xy} = T_{yx} = T_{yy} = 0\). Do a change to polars and you get \(T_{ab} = 0\) again except a,b are now \(\rho,\theta\), not x and y, ie \(T_{\rho\rho} = T_{\rho\theta} = T_{\theta\rho} = T_{\theta\theta} = 0\). To express \(T_{\rho\theta}\) etc in terms of \(T_{xy}\) etc you use the transformations Guest has been saying.

    A tensor expression is either covariant or not. What coordinates you consider are irrelevant if the expression is covariant, that's the definition!! Provided Eugene's coordinates have a nowhere zero Jacobian when expressed in terms of standard Cartesians then they are a valid set of coordinates everywhere. They can even have zeros at finitely many places and still be 'good', polars have a problem at the origin and they are still useful.
     
  18. Tach Banned Banned

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    How come that you are unable to see that in trying to calculate the partial derivatives wrt the coordinates you end up, in the case of Shubert crackpot transforms, with extraneous terms due to having the partial derivative wrt to f? It is precisely this defect that destroyed the covariance of the classical Maxwell formalism (at least you got that part).
    How can you be so blind and think that a change in notation changes the underlying physics? If you agree that the classical formulation of Maxwell's equations is not covariant wrt to the Shubert transforms (you agreed to this long ago), what makes you think that a change in notation (to tensor notation) makes them become covariant?


    You are a good match for Shubert, both mathematicians, both with no understanding of physics. If anything, Shubert is a better mathematician than you. You make a fine pair, anyways.
     
  19. Eugene Shubert Valued Senior Member

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    I'm certain that you have wasted your time with Tach. However, I would greatly appreciate receiving your unquestionably intelligent opinion in a brief review of The Quintessence of Axiomatized Special Relativity Theory and I'm especially interested if you can find any conceptual or mathematical errors. Please keep in mind that it is a work in progress.
     
  20. Tach Banned Banned

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    Agreed, I am very familiar with the transformation between cartesian and polar coordinates.. Now please try the same exercise with the non-linear, implicit transforms as in the crackpot Shubert transforms.
     
  21. Tach Banned Banned

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    Shubert,

    You got yourself a fan (Guest254) that is cracked as you are. He is a potential co-author for your paper. Ask him nicely and he'll add the missing chapter on electromagnetism.
    While at it, he might extend the transforms from 1+1 spacetime to 3+1 and get the botched Doppler effect equations right.
    For good measure, he might straighten all your misconceptions in the "Twins paradox" chapter. Good luck!
     
  22. Guest254 Valued Senior Member

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    This is now quite unbelievable. You still seem to have no clue with regards the difference between Lorentz covariance and General covariance. I've been through this with you in detail. I've explained that under any coordinate transformation, tensor equations remain the same. How can you not get this? I assume you have absolutely no familiarity with general relativity?

    Are you trying to pair us up? I'm married!
     
  23. Tach Banned Banned

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    Not for IMPLICIT functions of coordinates, you dork. Check your books since you are too lazy to calculate a chain of of derivatives.

    But you two are a fit , like twins. You have the dubious honor of being his biggest supporter on the www. You two belong together.
     

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