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Newtonian perturbations to Minkowski

  1. Dec 20, 2007 #1
    The Newtonian gravity limit of GR is represented by a perturbation to Minkowski metric:

    [tex] ds^2 = - (1+ 2\Phi(x,y,z,t)) dt^2 + (1-2 \Phi(x,y,z,t))(dx^2 + dy^2 + dz^2) [/tex]

    The geodesic equations of this metric correctly reproduce the Newtonian equations for acceleration of a test particle in the gravitational potential Phi:

    [tex] d^2 x / dt^2 = - \nabla \Phi [/tex]

    The Poisson equation that describes how matter generates the gravitational potential :

    [tex] \nabla^2 \Phi(x,y,z,t) = 4 \pi \rho(x,y,z,t) [/tex]

    should be derivable from Einstein equations. In Newtonian limit of Einstein equations, we describe the gravitating matter with energy momentum tensor of dust

    [tex] T_{\mu \nu} = \rho U_\mu U_\nu [/tex],

    where rho is the matter density.

    We have to assume the gravitating matter is moving much slower than light in the perturbed Minkowski coordinates above because only under such conditions the above Newtonian equations apply - they don't show GR or SR effects and behave as if signals propagate at infinite speed. If the matter is moving slowly, the U_0 components of the 4-velocity of gravitating matter will be ~1 and the other components suppressed by v/c, v = coordinate speed of gravitating matter. Correspondingly, the only significant component of the energy momentum tensor will be the T_00 ~ rho and the other components will be suppressed by a factor of (v/c)^2. The same applies to the components of Ricci and Einstein tensors which the Einstein equations set proportional to the energy momentum tensor.

    So we expect the above Minkowski perturbed metric to produce Ricci and Einstein tensors with components supressed by (v/c)^2 compared to the 00 component. Let's see if that is really the case. I plugged the above perturbed metric in Mathematica and neglecting factors of second and higher order in phi<<1 I got:

    [tex] R_{00} = 3 \partial_t \partial_t \Phi + \nabla^2 \Phi [/tex]
    [tex] R_{11} = R_{22} = R_{33} = - \partial_t \partial_t \Phi + \nabla^2 \Phi [/tex]
    [tex] R_{0i} = 2 \partial_t \partial_i \Phi [/tex]

    and the other components were zero or of second order in phi. The usual saying, neglect the time derivatives because gravitating matter is moving slowly, doesn't work because then R_11= R_22 = R_33 components won't be 'suppressed' but equal to the R_00 component. Requiring R_11= R_22 = R_33 = 0 imposes a wave equation condition on the gravitational potential - gravitational influence propagates at the speed of light. That is OK cause that is what we expect anyways. On the other hand, if we allow for the nonzero double time derivative, we get

    [tex] R_{00} = 4 \nabla^2 \Phi [/tex]

    instead of

    [tex] R_{00} = \nabla^2 \Phi [/tex]

    That means the coefficient in Einstein equation which is fixed by the Newtonian limit must be 4 times 8piG instead of 8piG.

    I am also not sure how the condition R_0i = 0 should be interpreted ??

    The above shows there is something improper in the Newtonian metric above - it either doesn't describe slowly moving dust (time derivatives neglected but R_11 not suppressed compared to R_00) or the coefficient in Einstein equation has to be adjusted to 4 times 8piG (time derivatives not neglected). Anyone knows what the metric must be in more proper treatment of Newtonian gravity, one that takes into consideration finite speed of gravitational influences? Is that 'linearized gravity' ?
    Last edited: Dec 20, 2007
  2. jcsd
  3. Dec 20, 2007 #2
    I got it. The Einstein tensor components for the above metric are

    [tex] G_{00} = 2 \nabla^2 \Phi [/tex]

    [tex] G_{11} = G_{22} = G_{33} = 2 \partial_t \partial_t \Phi [/tex]

    [tex] G_{0i} = 2 \partial_t \partial_i \Phi [/tex]

    the other components are either zero or of second order in Phi. So dropping the time derivatives really shows the 00 component dominates and the others are zero. That was expected because the Einstein tensor is proportional to the energy momentum tensor via the Einstein equations.

    The Ricci tensor 00 component is the same as the Einstein 00 after dropping the time derivative. On the other hand Ricci 11, 22, and 33 are not suppressed but equal to Ricci 00, and that is OK because Ricci doesn't have to follow the structure of the energy momentum tensor, only the Einstein tensor has to. That was the origin of the whole confusion.
    Last edited: Dec 20, 2007
  4. Dec 20, 2007 #3


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    My calculations with GRtensor agree. The exact expression I get for the time component of the Einstein tensor, [itex]G_{tt}[/itex] is:

    \frac{1}{1 -2\Phi^3} \left[ (2-8 \Phi^2)(\frac{\partial^2 \Phi}{\partial x^2}+\frac{\partial^2 \Phi}{\partial y^2}+\frac{\partial^2 \Phi}{\partial z^2} ) + (3 + 6\Phi) \left( (\frac{\partial \Phi}{\partial x})^2 + (\frac{\partial \Phi}{\partial y})^2 + (\frac{\partial \Phi}{\partial z})^2 \right) + (3 - 6 \Phi) (\frac{\partial \Phi}{\partial t})^2 \right]

    (I didn't look at the other components very closely).
    Last edited: Dec 20, 2007
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