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EFE in empty space

  1. Nov 20, 2007 #1
    Hi I'm a newbie for GR so please answer my basic question

    in empty space why we say for 2 or 3 Dimensions field equations in that region must have a curvature tensor equal to 0

    but when we say about 4-D or higher , the curvature tensor which non-vanish can satisfy EFE?

    I think when we say "empty space" it should mean that no energy or momentum so curvature must be zero

  2. jcsd
  3. Nov 20, 2007 #2
    Empty space means zero energy-momentum tensor which according to EFE sets to zero the Ricci tensor R(mu,nu) not the full curvature Riemann tensor with four indices.

    In 2 or 3 dimensions, Riemann is expressible in terms of Ricci (see section 6.7 in Weinberg's "Gravitation and Cosmology ...."). Thus setting Ricci to zero implies that Riemann = 0 too.

    In 4 dimensions, Riemann is a sum of term proportional to Ricci, another term proportional to Ricci scalar and a part independent of Ricci, called Weyl tensor. Thus, setting Ricci =0 doesn't set the whole Riemann to zero. Examples are Schwarzschild vacuum solution (spacetime around spherical body) or the gravitational wave vacuum solution. They have Ricci = 0 and Ricci scalar = 0 but the Riemann tensor has non-zero components.

    It's weird but Einstein Equations do not fix the full curvature tensor (Riemann) only a contraction of it (Ricci). The part of Riemann that remains free is the Weyl tensor. You have to impose extra conditions to fix that part, and then you get an unique solution. The extra conditions could be requiring a metric that looks in a certain way or has certain symmetries (spherical symmetry in Schwarzschild solution).
    Last edited: Nov 20, 2007
  4. Nov 21, 2007 #3


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    Although Weyl is not determined algebraically from the Einstein field equations, Weyl is not completely free. The Riemann curvature tensor satisfies the Bianchi identity, which relates the derivatives of these curvature tensors.
  5. Nov 21, 2007 #4

    Chris Hillman

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    Ditto robphy.

    Some more detail:

    It helps to know that the Riemann curvature tensor can be decomposed into
    • the scalar part, a four index built from the Ricci scalar,
    • the "semitrace part", a four index tensor built from the traceless Ricci tensor,
    • the completely traceless part, the Weyl tensor, aka conformal curvature tensor.
    (In three dimensions, the completely traceless part vanishes identically. In two dimensions, so does the semitrace part. This phenomenon is purely mathematical and doesn't really affect gtr, which is typically applied to four dimensional Lorentzian manifolds!) The EFE stipulates that the Ricci curvature is proportional to the matter tensor, but this doesn't directly fix the Weyl curvature. However, as robphy said, by applying the once contracted differential Bianchi identity to the EFE, we obtain a differential equation by which, according to gtr, variations in the Ricci tensor (i.e. variations in density and momentum of a configuration of mass-energy) can induce variations in the Weyl tensor. This is why the aspherical variations over time of the distribution of mass and momentum in a configuration of isolated matter give rise to curvature variations which can propagate across a vacuum region in the form of gravitational radiation, and also why concentrating mass-energy in a compact region "curls up" the surrounding vacuum region. There is an excellent discussion of this in the textbook by Carroll, Spacetime and Geometry.
    Last edited: Nov 21, 2007
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