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Nordstrom Gravity

  1. May 1, 2013 #1
    1. The problem statement, all variables and given/known data

    Question: "A theory of gravity devised by physicist G. Nordstrom, relates [itex]g_{μ\nu}[/itex] to [itex]T^{μ\nu}[/itex] by the equation:

    [itex]R=κg_{μ\nu}T^{μ\nu}[/itex]

    where the metric has the form [itex]g_{μ\nu}=e^{2\Phi}[/itex] with [itex]\Phi=\Phi(x^{μ})[/itex] a function of the spacetime coordinates (the special form of the metric follows from requiring the vanishing of the Weyl curvature tensor [itex]_{αβγδ}=0[/itex]

    a) Show that in the Newtonian limit [itex]\Phi^{2}<<1[/itex] the geodesic equation for a test body moving slowly in this spacetime reproduces the kinematics of Newtonian gravity.

    b) Calculate the ricci scalar R in the Newtonian limit showing that it is just a second order differential operator acting on [itex]\Phi[/itex].



    3. The attempt at a solution

    My first question is really just about the metric i need to use. from a couple of things i've read on the internet, it seems i should just use the simple flat-space minkowski metric (ct, x, y, z), but i am not really sure how to calculate the christoffel coefficients in this metric as all [itex]g_{μ\nu}[/itex] must be a function of r (from [itex]\Phi=2GM/r[/itex]). wouldn't that mean that the christoffel coefficients all vanish (ie. [itex]∂_{t}g_{xx}=0[/itex])??? does this mean i need to use the shwarzschild metric??
     
  2. jcsd
  3. May 1, 2013 #2

    fzero

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    You're missing some things, [itex]g_{μ\nu}=e^{2\Phi}[/itex] doesn't make sense, since the left-hand side is a tensor and the right-hand side is a scalar. What you want is (referring to http://en.wikipedia.org/wiki/Nordström's_theory_of_gravitation#Features_of_Nordstr.C3.B6m.27s_theory)

    $$ g_{\mu\nu} = e^{2\Phi} \eta_{\mu\nu},$$

    where ##\eta_{\mu\nu}## is indeed the flat Minkowski metric. Since ##\Phi## is a function of the ##x^\mu##, the Christoffel symbols for ##g_{\mu\nu}## will not vanish, but because of the high amount of symmetry, they will have somewhat simple expressions.

    It is not the case that [itex]∂_{t}g_{xx}=0[/itex], rather [itex]∂_{t}g_{xx}=2e^{2\Phi}(∂_{t}\Phi) [/itex].
     
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