Chinese Scholar Yang Jian liang Putting Wrongs to Rights in Astrophysics

Discussion in 'Pseudoscience' started by heyuhua, Apr 22, 2018.

  1. Michael 345 Looking for Bali in Nov Valued Senior Member

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    6,182
    I don't know

    and

    and

    and

    and you have been sharply told to be humble and learn

    I'm humble enough to say I has no idea what you both are talking about

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    I'll leave you big boys to get on with your whatever it is. As long as I can safely go outside I'm happy

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  3. NotEinstein Valued Senior Member

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    1,495
    You are missing the point: the contribution of the stress-energy tensor is being inverted. Sure, that probably doesn't express itself as a sign-flip when written out as Newtonian gravity, but it sure as heck completely changes the behavior of the gravity.

    Erm, you didn't even post a single calculation yourself, so it's just as much you that's not doing any of the calculations here.

    I do know that. And that metric determines how gravity behaves. I just skipped that step for simplicity.

    Not directly, no. But you are effectively putting a minus sign in front of all rest-masses (among others), so that is where my statement came from.

    Cool, let's see it! Please post it here.

    Evidence please.

    I'm learning lots of things from it; but perhaps not the things you are referring to.
     
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  5. heyuhua Registered Senior Member

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    If you can check Yang's calculation step by step your confusions will disappear and you will keep up with Yang's ideas
     
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  7. NotEinstein Valued Senior Member

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    1,495
    Why don't you post the clarifications here? Or is this entire thread just an advertisement for Yang's work?
     
  8. exchemist Valued Senior Member

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    I think HuHee = Yang.

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  9. NotEinstein Valued Senior Member

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    1,495
    If that's true, heyuhua should have no difficulties answering my question.

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  10. heyuhua Registered Senior Member

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    Obviously, here mathematical equations are not supported and thus you can only read Yang's article recommended in the head post, not reading mathematical equation step by step you cann't get into Yang's work. The head post is only a brief introduction to Yang's work
     
    Last edited: Apr 27, 2018
  11. heyuhua Registered Senior Member

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    your questions don't require special answer because they had already been answered in Yang's articles, if you once grasp Yang's work your questions will automatically be removed, you don't read Yang's articles hard, I'm sorry to say
     
    Last edited: Apr 27, 2018
  12. NotEinstein Valued Senior Member

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    1,495
    Obviously, you haven't done your homework: \(y=x^2\) Use the tex-tag. So please post the clarifications here.

    Then please point me to the specific part of Yang's article that answers my question. Better yet, post that part here.
     
  13. heyuhua Registered Senior Member

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    432
    your questions had been answered in Yang's articles via mathematical calculation, I can not post them here because mathematical equations cann't be writed on the forum
     
  14. NotEinstein Valued Senior Member

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    1,495
    "Then please point me to the specific part of Yang's article that answers my question."

    "Obviously, you haven't done your homework: \(y=x^2\) Use the tex-tag."
     
  15. heyuhua Registered Senior Member

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    432
    in Yang's article, for example, "Correction of Standard Model in View of Improved Gravity Equation" , the equation (1.15) treats your euqations
     
  16. NotEinstein Valued Senior Member

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    1,495
    Are you sure that's the right equation number? That one is about Hubble constant, not about the sign of the force of gravity.
     
  17. heyuhua Registered Senior Member

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    432
    Yes, it indeed is about Hubble law, but it matches well the observational data (reshifts and distances) with the conclusion of decelerated expansion, see the data in Fig. 1 recommended by you in the article :"
    In Defense of an Accelerating Universe: Model Insensitivity of the Hubble Diagram".
    as for to the sign of the force of gravity, see the equation (1.5) in Yang's article, which reflects attraction or repulsive
     
    Last edited: Apr 27, 2018
  18. heyuhua Registered Senior Member

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    432
    1111
     
    Last edited: Apr 27, 2018
  19. NotEinstein Valued Senior Member

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    1,495
    Where is the value of \(q_0\) coming from? It's being set without justification, which pretty much looks like it was set after the fact in order to fit the data.

    Except not really. That equation is derived for a static universe, where all forces cancel. So a sign-flip wouldn't show up in a equation of motion, because there literally is no motion.
     
  20. NotEinstein Valued Senior Member

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    1,495
    Let's focus on one step that's missing from the article. How does:
    \(h_{00}=-\frac{\gamma}{4\pi}\int\frac{\rho+3p}{\xi}dx'dy'dz'\)
    where \(\xi=\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}\)

    Get you to:
    \(h_{00}=\frac{2GM}{r}\)
    with \(\gamma=4\pi G\)?

    Edit: (The given equation for \(g_{jj}\) is obviously wrong in this article; a typo, I suspect.)
     
    Last edited: Apr 27, 2018
  21. NotEinstein Valued Senior Member

    Messages:
    1,495
    Or how to go from:
    \(\nabla^2 h_{\lambda\lambda}-\frac{\partial^2 h_{\lambda\lambda}}{\partial t^2}=\gamma[(\rho+p)U_{\lambda}^2+(\rho-p)\eta_{\lambda\lambda}]\)

    to:
    \(h_{00}=-\frac{\gamma}{4\pi}\int\frac{\rho+3p}{\xi}dx'dy'dz'\)
    where \(\xi=\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}\)
     
  22. NotEinstein Valued Senior Member

    Messages:
    1,495
    But more interesting is the missing minus sign in the partially contracted Ricci tensor. Right under "Omitting the terms \(o(h^2)\) we can infer that:" there's a formula that seems to be a minus-sign off from what it should be. Look at: https://en.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry#Ricci_and_scalar_curvatures The term with the derivative on the RHS where the indices match those on the LHS doesn't carry a minus-sign, yet in Yang's article it does. Could this be where Yang makes a mistake, that carries through to mess with the determination of \(\gamma\)? (Also, the last two terms are missing too.)
     
  23. heyuhua Registered Senior Member

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    432
    q_0 is obtaind from calculation according to observational cosmic density and Hubble parameter value of today. (1.15) is geodesic equation which is equivalent to the standard geodesic equation, don't means to be applied to static universe, about its derivation you may see the following Yang's article page 142,or arbitrary standard textbook of post Newton's mechanics
    Yang's crticle:
    http://www.journalrepository.org/media/journals/BJMCS_6/2011/May/1306478...
     

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