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Mathematica notebook for conversion to newtonian metric ?

  1. Apr 30, 2007 #1
    Basic postulate of GR is that locally spacetime looks flat i.e. any metric can be reduced to a local inertial metric at a point. The local inertial metric is exactly the flat Minkowski metric AT the point, and has zero derivatives AT the point but those are not true away from the point.

    Then Newtonian metric is the metric felt by a free fall observer. In it, the deviations from the Minkowski metric, to first order in 1/c^2, are described by a 'Newtonian gravitational potential'. I think this is the same thing as writing a general metric in 'Newtonian approximation', 'Newtonian gauge' or 'Weak field approximation'.

    I have a general metric and need a Mathematica notebook that will convert it to the Newtonian metric felt by a specified free fall observer. Does anyone know of something like that ?
     
    Last edited: May 1, 2007
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  3. May 1, 2007 #2

    Chris Hillman

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    What would Gauss say?

    Be careful, the physics literature unfortunately uses some badly inconsistent terminology here: "locally" should mean "in a local neighborhood", not "at the level of jet spaces" or even "eventwise", but older books/papers confuse these completely different notions.

    You should say, "i.e., we assume that spacetime can be modeled as a Lorentzian manifold". Gtr involves much more than this assumption, of course, which is shared with a large class of gravitation theories.

    You are probably thinking of Riemann normal coordinates, in which the Christoffel symbols vanish at some event, but this is a mathematical construction valid on any Lorentzian manifold and has nothing to do with physics.

    Careful, there is no "Newtonian metric" which is identifiable with any Lorentzian metric. (There is a geometric approach sometimes called "Newtonian spacetime", but that uses degenerate metrics, which are certainly not Lorentzian metrics, which are nondegenerate indeterminate.)

    Sounds like you are thinking of constructing a Riemann normal chart in a Lorentzian spacetime which is a weak-field approximation to a static vacuum solution in general relativity?

    Well, I don't think such a notebook could exist (unless it were cranky!*), and in any case, you can't use software tools if you don't understand what you are computing!

    If you can type in your metric as pseudolatex, I can probably help you

    1. find a "weak-field approximation",

    2. find a normal chart,

    3. find a frame suitable for studying the experience of observers with specified properties, unless you are doing something absurdly complicated or mistaken,

    or some combination.

    *FYI: I have been told that a certain crank with a "cult following" has directed his acolytes (who are not physicists) to seek out software on the net for purposes of computing nonsense. Sad.
     
  4. May 1, 2007 #3
    Forget the weak field, the exact math question I have in mind is:

    given are:
    a general metric in some coordinates
    a point of spacetime specified by its coordinates
    4 orthonormal vectors at that point specified by their components

    construct:
    the Riemann normal coordinates around that point i.e. I need the metric expressed in those coordinates up to terms of order 1/c^2


    I've seen the general procedure in Hartle's GR textbook but it's hard to implement in practise when the solutions of the geodesic equations are not obvious. Since I need the Riemann coordinates only up to terms of order 1/c^2, I hope there is a simplified way to find those. I am interested in article/book that shows how that approximate Riemman coordinate metric can be found or Mathematica notebook that can do it for me.
     
    Last edited: May 1, 2007
  5. May 2, 2007 #4

    Stingray

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    See section 7.2 of http://arxiv.org/abs/gr-qc/0306052. There's an expansion for the metric about the origin of a Riemann normal coordinate system in powers of distance. I don't know what you mean by expanding up to order 1/c^2, but to lowest nontrivial order, all you need is the Riemann tensor at the origin.
     
  6. May 2, 2007 #5
    Thanks a lot Stingray, your answer hit right on target.
     
    Last edited: May 2, 2007
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