Correlating Newtonian Model with Einstein's GR

Discussion in 'Physics & Math' started by hansda, May 8, 2017.

  1. exchemist Valued Senior Member

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    This is a day to break out the champagne!!

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    P.S. Just received a laconic response to my reporting of this morning's efforts by The God, saying "resolved". Indeed so.
     
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  3. Kittamaru Now nearly 40 pounds lighter. Staff Member

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    In my defense... I had six reports to close all relating to or resulting from this individual

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  5. exchemist Valued Senior Member

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    Well I guessed I would not be the only one. But I am proud to have added my brick to the wall.
     
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  7. DaveC426913 Valued Senior Member

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    No way!!

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  8. James R Just this guy, you know? Staff Member

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    The God brought a permanent ban on himself due to accumulated warning points. The permanent ban was automatic, not instigated by moderators. The God can't say he wasn't warned (39 times).
     
  9. NotEinstein Registered Senior Member

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    Please elaborate; what do I not understand?

    Please elaborate; where have I misquoted you?

    And Newtonian physics is made up to corroborate to reality. How is that a bad thing? Of course they have to match reality; they would be useless as models otherwise!

    From a fundamental perspective.

    Absolutely not. Please stop putting words in my mouth.

    Since we both agree that Lagrangian mechanics doesn’t introduce any new physics when compared to Newtonian mechanics, there was no need to do so.

    I haven’t seen any strong argument from you either. “made up”, “conceptually different”. How is that not sloppy or hand wavy?

    I did not withdraw it. I only admitted it doesn't provide a working version Newtonian physics with a limited speed of gravity. It does however show that the limited speed of gravity was something being pondered about long before Einstein. In other words, people were already trying to modify/reject Newtonian physics' infinite speed of gravity.

    What? I do not understand what you are referring to?

    “Hand waving” doesn’t mean “wrong”. Please look up the meaning of "hand waving" as used in science.

    I did not. Please stop misrepresenting my position.

    Shall we also count the number of times you backed out of pointing out the mistake(s) in the mathematical derivation of Newtonian physics from GR?

    Yes, indeed it is. If I have done so, I apologize about that and will withdraw those statements. I’ve already pointed out so many instances of you doing the same (shall we count those?), I hope you’ll do the same?
     
  10. NotEinstein Registered Senior Member

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    (I wrote the above post before I knew about the permanent banning of The God.)

    I guess that leaves just one thing... For hansda to respond to my inquiry about his text not mentioning anything about the correlation between GR and Newtonian physics.
     
    Last edited: Oct 10, 2017
  11. QuarkHead Remedial Math Student Valued Senior Member

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    OK, here's one thing I don't understand.

    Newton's playground was flat space. Assume we can extend this to spacetime - that's what the weak field limit seems to mean.

    Given the field equations of gravitation, it is easy to see that in the absence of a gravitational source, spacetime is flat. This is because the Ricci curvature tensor everywhere vanishes, which can only happen when the metric field is constant (the Ricci tensor field is a second order differential of the metric field with respect to the coordinates - simple calculus). This is especially pretty because it provides an analogue of the famous Laplace equation - the divergence of the gradient of a scalar field is identically zero in the absence of a source. This is generally written as \(\nabla^2\phi =0\) where \(\phi\) is the scalar field.

    With this rambling, I get to my question: if GR reduces to the Newtonian law in the weak field limit, and if there exists a source, how can it be that, even so, the curvature tensor vanishes everywhere?
     
  12. NotEinstein Registered Senior Member

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    In the weak field limit the metric takes the form of a flat metric plus some small perturbations: \(G_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}\), where the contributions of \(h_{\mu\nu}\) are small. For more details, see this link I posted somewhere in this thread before: https://www.quora.com/How-can-we-de...avitation-from-Einsteins-theory-of-relativity
     
  13. QuarkHead Remedial Math Student Valued Senior Member

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    I am unconvinced that the term "flat metric" has any mathematical meaning. Moreover, you are just reiterating the assertion for which I was seeking clarification.

    Nonetheless, I thank you for your response. However, I have figured it out, so if anyone is interested, here goes........

    In general, an \(n\)-manifold is defined as a "jazzed-up" point set that is, at some arbitrary scale, indistinguishable from the plane \(R^n\). Clearly \(R^n\) is itself a manifold.

    A manifold with a metric is similarly defined, but with \(R^n=E^n\), by which I mean that the Euclidean metric applies locally. Assume that spacetime is a 4-manifold with a metric. Now because of the fact that one of the coordinates is signed differently from the other 3, at least in the definition of the metric, the Euclidean metric cannot apply, so the Minkowski 4-metric \(\eta_{jk}\) is used instead. This is, for example, is the metric used in the Special Theory, a theory that ignores forces an acceleration, where this metric is the same at each and every spacetime point for any choice of coordinates. I assume this implies that our 4-manifold is trivial i.e. that the "arbitrary scale" I referred to above is the whole banana.

    So if you want the Special Theory to be compatible with the General Theory, it follows that in the absence of a gravitational source (\(T_{jk}=0\)), and for the algebraic reasons I can earlier that the curvature field vanishes (\(R_{jk}=0\)), implying the metric field is constant globally. Again implying (if not down-right proving) that in the absence of a source, spacetime is locally and globally indistinguishable from itself (surprise!!) or flat.

    But since the inverse is true - a constant metric field implies the absence of a gravitational source - from the definition of the spacetime 4-manifold, we can take "local" to be a region of weak gravitational source.
     
  14. arfa brane call me arf Valued Senior Member

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    The special theory implies Minkowski spacetime (I think Quarkhead has just shown this), i.e. the absence of a gravitational source. But isn't Minkowski spacetime just one solution for "the" vacuum?

    And isn't Minkowski spacetime always a local solution (light doesn't bend much over relatively small distances)? Unless of course, you have a universe with no matter in it?
     
  15. Schmelzer Valued Senior Member

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    The weak field limit would have been better named weak field approximation. It is only an approximation, not really a limit. (The usual sloppy language of physics.)
    In the weak field approximation, the clock time is approximately the same as Newtonian absolute time, and the curvature is approximately zero.

    If there would be a real limit, it would have to be exactly zero. But this would be only in the case when the gravitational potential is exactly constant.
     
  16. NotEinstein Registered Senior Member

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    First you say that you are unconvinced that "flat metric" means anything, but then you prove that \(\eta_{\mu\nu}\) results in flat spacetime... Should I instead have said that "in the weak field limit the metric takes the form of the Minkowski metric..." etc., and that that results in a (nearly) flat spacetime?

    And I'm not sure the curvature tensor disappears: I don't think it does in the link I posted. The perturbation term still gives a contribution, and it's that which gives rise to Newtonian gravity.
     
  17. arfa brane call me arf Valued Senior Member

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    According to Wikipedia:
    "
    It is a mathematical fact that the Einstein tensor vanishes if and only if the Ricci tensor vanishes. This follows from the fact that these two second rank tensors stand in a kind of dual relationship; they are the trace reverse of each other:

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    where the traces are:

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    .

    A third equivalent condition follows from the Ricci decomposition of the Riemann curvature tensor as a sum of the Weyl curvature tensor plus terms built out of the Ricci tensor: the Weyl and Riemann tensors agree,

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    , in some region if and only if it is a vacuum region."
    --https://en.wikipedia.org/wiki/Vacuum_solution_(general_relativity)
     
  18. QuarkHead Remedial Math Student Valued Senior Member

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    Which is OK when physicists talk among themselves, I suppose, but confusing on a forum such as this.

    In the present context, I cite 3 more examples.....

    The General Theory is a FIELD Theory. This means that every term in the Einstein field equations are themselves fields. So it only makes sense to talk about "the" metric tensor or "the" curvature tensor (say) when you are referring to a specified point in spacetime. And if you are, it makes very little sense geometrically.

    Second, the so-called tensor, say, \(A_{jk}\) is not a tensor at all, neither is it a tensor field - in spacetime coordinates (where it is assumed that all tensor fields are symmetric), it refers rather to the 10 scalar components of a tensor (or outer) product of vector fields.(If you really have no life, read this. I did it here many years ago http://www.sciforums.com/threads/sr-so-whats-a-tensor-for-chrissake.74301/

    Third, and annoyingly, the set of all tensor fields on a given manifold is itself a vector space by the usual axioms. Therefore any element in this set i.e. tensor field is by definition a vector.

    Confused? you should be!![/quote]
     
    Last edited: Oct 12, 2017 at 9:44 PM
  19. Schmelzer Valued Senior Member

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    That's not the point. The language of physicists is sloppy if considered from point of view of mathematicians. The language in such a forum, full of laymen, is even more sloppy.
     
  20. hansda Valued Senior Member

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    This thread is about co-relation of various models of Physics. NM and GR are two models of physics. So they can be correlated. I have explained this in the OP.

    This thread is not about derivation.
     
  21. origin Trump is the best argument against a democracy. Valued Senior Member

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    I think that NotEinstein is pointing out that the only TOE you have are those 10 things in your shoes, but that is just my opinion.
     
  22. hansda Valued Senior Member

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    Good joke.
     
  23. hansda Valued Senior Member

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    Thanks for reading my paper. Also thanks for your views.
     

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