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General Relativity and Minkowski space question

  1. Aug 19, 2009 #1
    Is the curved spacetime of General Relativity possible to be described by Minkowskian space which is flat, or do one need some other geometry?
     
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  3. Aug 19, 2009 #2

    Fredrik

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    If you look at a small enough region of spacetime, it will look approximately flat, and in the limit where the size of the region goes to zero, it will look exactly like Minkowski space, assuming of course that there are no singularities in the region.
     
  4. Aug 19, 2009 #3

    atyy

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    And provided you don't measure second derivatives (which can be argued are non-local in common sense, if not the maths sense).
     
  5. Aug 19, 2009 #4

    atyy

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    Also there is an interesting sentence in Kip Thorne's book about black holes, time warps etc in which he says general relativity is equivalent to a theory in flat spacetime, but where clocks run fast, and rulers shrink. However the closest mathematical statement I have been able to find is Eqn 62 in section 4.3 of http://relativity.livingreviews.org/Articles/lrr-2006-3/ [Broken] which requires that harmonic coordinates can be used.

    I'd be interested to know if this is really what Thorne was referring to, or whether he meant something else.
     
    Last edited by a moderator: May 4, 2017
  6. Aug 19, 2009 #5

    tiny-tim

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    Welcome to PF!

    Hi stolbein! Welcome to PF! :smile:
    It's like trying to map the Earth onto flat paper without losing the geometry … however you do it, the map is going to be distorted. :wink:
     
  7. Aug 19, 2009 #6
    As has been pointed out, it is possible to make the metric [tex] g_{\mu \nu} [/tex] be equal to the Minkowski metric [tex] \eta_{\mu \nu} [/tex] at any given (nonsingular) point, and to make all of its first derivatives vanish there, by using Riemann normal coordinates. However, the tensor [tex] R \indices{^{\mu}_{\nu}_{\sigma}_{\rho}} [/tex] (the Riemann curvature tensor) is an isometric invariant of the manifold, which is to say that if you keep the metric [tex] g_{\mu \nu} [/tex], then the curvature stays, too.

    However, if we are only dealing with weak gravitational fields, then we may regard general relativity as the theory of a symmetric tensor [tex] h_{\mu \nu} [/tex] propagating against a flat, Minkowskian background; this is called linearized gravity, and is used to study gravitational waves. Specifically, we can write [tex] g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu} [/tex], where the perturbation [tex] h_{\mu \nu} [/tex] is assumed to contribute significantly to measurable quantities only to first order (this is the "weak-field" assumption). We then have [tex] g^{\mu \nu} = \eta^{\mu \nu} - h^{\mu \nu} [/tex] (again, to first order), and we can raise and lower indices using [tex] \eta [/tex] (in fact, [tex] h^{\mu \nu} [/tex] is defined here as [tex] \eta^{\mu \sigma} \eta^{\nu \rho} h_{\sigma \rho} [/tex]). We can then go on to derive the Riemann, Ricci, and Einstein tensors, and the Ricci scalar, to get the field equations. Alternatively, and in keeping with the viewpoint of a field theory on a flat background, we can define the Lagrangian
    [tex]\displaystyle \mathcal{L} = \frac{1}{4} [2h \indices{^{\mu}^{\nu}_{,\mu}} h_{,\nu} - 2h \indices{^{\rho}^{\sigma}_{,\mu}} h \indices{^{\mu}_{\sigma}} + \eta^{\mu \nu} h \indices{^{\rho}^{\sigma}_{,\mu}} h_{\rho \sigma, \nu} - \eta^{\mu \nu} h_{,\mu} h_{,\nu} ] \textrm{,} [/tex]
    which, when varied with respect to [tex] h_{\mu \nu} [/tex], gives the linearized Einstein equations.
     
    Last edited: Aug 20, 2009
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