On the Definition of an Inertial Frame of Reference

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

  1. Guest254 Valued Senior Member

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    If only you would apply the same passion in learning tensor calculus!
     
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  3. Guest254 Valued Senior Member

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    Ah - you keep editing!
    You have trouble with the chain rule?
     
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  5. Tach Banned Banned

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    No, but you obviously do. Because you fail to account for the extra terms that show up when you use the Schubert functions. Really, you two make the perfect crackpot pair, help him out to finish his paper, you have all the information that he needs.
     
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  7. Eugene Shubert Valued Senior Member

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    I have clearly demonstrated that my nonlinear transformation is invertible, see equation (46), (59) and (60), therefore, if it's differentiable, then the Jacobian exists and is never zero. Please keep in mind however that, in general, my transformation is so general that it's easy to come up with non-differentiable examples.
     
  8. Guest254 Valued Senior Member

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    Erm, the coordinates are just functions... the derivatives of coordinates are just functions... You don't seriously need to be told how to use the chain rule do you?
     
  9. Tach Banned Banned

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    have you even looked at his functions, idiot?
    did you notice the S(ax+bt+S(S(x,t)) ?
     
  10. Guest254 Valued Senior Member

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    You can't differentiate that?
     
  11. AlphaNumeric Fully ionized Registered Senior Member

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    Yes, I'm sure you are but I mention them because they are examples of non-linear transformations where tensor covariance is demonstrated. I doubt Guest thinks Eugene is onto anything, I certainly don't, he's instead talking about the general concept of coordinate transformations and tensors.

    If Eugene's coordinates have non-singular \(J^{a}_{b} = \frac{\partial \tilde{x}^{a}}{\partial x^{b}}\) then they are 'valid coordinates' and expressing Maxwell's equations in terms of them would still give the form \(\nabla^{a}F_{ab} = 0\), just F would be different from the F in the Cartesian form.

    You said that you think Guest doesn't know what he's talking about. I can assure you that Guest's knowledge in the likes of tensor calculus is formidable, I doubt that many of even the PhDs in physics here would even come close, myself included. I would have thought you might get the hint when he talked about exterior derivatives of things in cotangent bundles.

    I wouldn't trust your mathematical abilities as far as I can throw a 2 ton rhino so forgive me if I don't take your word for it and I'm not going to waste my time checking. The point Guest is talking about is entirely independent from the specifics of your nonsense.
     
  12. Tach Banned Banned

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    That's not the point, anything can be differentiated. What do you get when you do it?
     
  13. Guest254 Valued Senior Member

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    Hey, you were worried about derivatives - not me! I know the chain rule.
     
  14. Eugene Shubert Valued Senior Member

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    Hopefully Guest, after he's finished playing, will take the time and educate you also.
     
  15. Tach Banned Banned

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    Nah, he's a crackpot at par with Shubert.


    This is why I asked Guest254 to calculate something simple , like the partial derivative wrt to a coordinate of his choice.


    That might be but:

    -he doesn't understand the underlying physics, this became clear when he maintained that a change of notation changes the equations from being non-covariant to becoming covariant

    -he has no proof that the tensor transformation rules apply to implicit functions other than his unsubstantiated claims'
     
  16. Tach Banned Banned

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    But you get a garbled mess when you try getting the tensor transforms. You claim to be a mathematician (you are definitely not a physicist) but you do not understand the conditions under which the formulas you are citing have been derived. The nonsense that Shubert is doing (S(ax+bt+S(x,t)) is definitely not mainstream , you you seem to believe that it will retain the form of the tensor, while all along refusing to perform any calculations.
     
  17. Guest254 Valued Senior Member

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    I'm afraid I probably think you're a little bit whacko (what I'm having to explain to Tach is largely independent of whatever you've written - I'm talking about any coordinate transformation). On the up side, Tach thinks we have a future together. Call me?
     
  18. Guest254 Valued Senior Member

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    News flash: some coordinate transformations might be messy!
     
  19. Tach Banned Banned

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    It is not ONLY that they are "messy" , it is that their application does not result into the standard forms of tensor transformation. Alpha says that you are an expert, why are you so reluctant to do a simple calculation to prove it to yourself?
     
  20. Eugene Shubert Valued Senior Member

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    I understand that. I took Ted Frankel's graduate level geometry of physics course when I was a math student at UCSD.
     
  21. Guest254 Valued Senior Member

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    What bit can't you do? You say you're ok with the differentiation, so which bit is too tough?

    A coordinate system is a coordinate system is a coordinate system. Any set of diffeomorphisms defines a coordinate change. If you're terribly worried about implicitly defined functions, I'll let you into a secret... the implicit (and inverse) function theorems. Now go yonder and enjoy your new found knowledge!
     
  22. Guest254 Valued Senior Member

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    Good for you.

    Please Register or Log in to view the hidden image!

     
  23. Tach Banned Banned

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    Good, so you two can write the missing chapter on electrodynamics. While at it, you might collaborate on getting the Doppler effect right and removing the embarrassing chapter on the Twins "paradox".
     

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