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Geometry of the EFE

  1. Jan 4, 2012 #1
    The stress-energy tensor is associated to a volume density and flux in 4-spacetime and the Einstein tensor seems to represent a three-dimensional curvature (being a one-form with vector values) that acts on and is acted by the stress-energy source.
    If this is correct, I'm not sure what is this 3-curvature associated to in the most accepted cosmological model (L-CDM) that has no spatial curvature.
    Another question is referred to the geometrical meaning of the Einstein tensor, when it is said that it is an average of the Riemann surface curvature (a tensor(1,1)-valued 2-form after raising an index with the inverse metric tensor) over planes, what does it mean exactly?
    I think that you can obtain the Einstein tensor by applying the delta Kronecker to the Riemann 2-form to make a some kind of isotropic average tensor but I'm not sure how to do it.
    Also what is the intuitive geometrical difference between the Ricci and Einstein curvature? I think the Ricci curvature is more related to an average of sectional curvatures (and is also usually defined as the deviation a geodesic ball in curved spacetime has from the euclidean standard ball).
  2. jcsd
  3. Jan 4, 2012 #2
    Maybe the part about the averaging of the Riemann tensor was expressed a little confusingly. What I meant was something like this:
    First we obtain the mixed Riemann tensor form raising an index with the contravariant metric tensor
    and we average it to obtain the Einstein tensor
    [tex]G^a_b=-\frac{1}{4} δ^{ast}_{bij}R^{ij}_{st}=R^{aj}_{bj}-\frac{1}{2} R^{ij}_{ij}δ^a_b[/tex] wich in quadratic form would correspond to the usual LHS of the EFE.
  4. Jan 5, 2012 #3
    So I'm not sure if the lack of response is because what I wrote is maybe wrong, irrelevant or incomprehensible?
    I'm aware it is not the usual treatment found in GR textbooks, but it is discussed in some mathematical texts, I just would like to understand it better.
  5. Jan 6, 2012 #4
    Maybe someone has a different idea of what the Einstein tensor represents geometrically.
  6. Jan 6, 2012 #5


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    Just a quick post to acknowledge that your question is interesting...
    and that I also seek a purely geometrical interpretation of the Einstein tensor
    ...possibly one that is dimensionally- and signature-independent...
    and independent of the equations of motion.

    I suspect that its interpretation would likely involve its differential properties (e.g. Bianchi identity), as well as its algebraic properties. I have also been interested in a related tensor: [itex]L_{ab}=R_{ab}-\frac{1}{6}Rg_{ab}[/itex] (in (3+1)d), which appears in the decomposition of the Riemann [or, similarly, Weyl] tensor (https://www.physicsforums.com/showthread.php?t=153765 ).

    You are probably aware that your expression for the mixed Einstein tensor involves the "double-dual" of Riemann.
  7. Jan 6, 2012 #6
    Thanks robphy, no I was not actually aware of that, I guess you mean:

    That's interesting, cool way to go from a tensor-valued two-form to a vector-valued one-form but it's odd too because without paying it much attention I would have thought it would be another two-form like in the case of the purely antisymmetric differential forms.
    Now that I think of this I'd say some information gets lost in the proccess since we are working in 4 dimensions and therefore the Einstein tensor doesn't determine completely the Riemann curvature, the Weyl tensor completes the picture here.
    The Bianchi identities make sure that by the symmetries of the Riemann tensor and the fact that the covarian derivative of the metric is zero the Eintein tensor must be divergence free.
    As to your comment about dependence on dimensionality, signature and equations of motion: I'm not sure what you intend by independence of dimensions, as you know there is trivially no Einstein tensor in one and two dimensions and it is geometrically different in dimensions 3 and 4, it is hard for me to think geometrically in higher dimensions.
    Similarly I think the Einstein tensor should be signature-dependent for any pseudo-Riemannian manifold (because the Ricci tensor is).
    Maybe it is possible to interpret the Einstein tensor independently of the equations of motion since it seem the Weyl tensor is enough to compute motion in the vacuum solution.
  8. Jan 6, 2012 #7
    Also what is you opinion about my first question in the OP? Can you identify a physical 3-curvature actual analogy of the Einstein tensor? I tend to think about 3-space in the FRW solutions with curved space, but then the favoured solution is the one with flat space.
  9. Jan 7, 2012 #8


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    Look at MisnerThorneWheeler, ch 13.5 (p. 325).
    The Einstein Tensor is the trace of the double-dual of Riemann.
    books.google.com/books?id=w4Gigq3tY1kC&pg=PA326&lpg=PA326&dq=double+dual+riemann+einstein (scroll back to p. 325)

    Concerning dimensionality, I am looking toward larger dimensions for general properties since, as you point out, the lower-dimensional analogues are algebraically trivial and don't suggest generalizations to higher dimensions. It would be interesting, of course, if there was something geometrically special about 4 dimensions, in particular (3+1)d.

    It seems to me that: for a _purely geometrical_ interpretation, one should not use the equations of motion (although it may be that the desired properties of the stress-energy may have identified candidates for what we call the Einstein tensor).

    Sorry, I have no opinion on your first question.
    Last edited: Jan 7, 2012
  10. Jan 7, 2012 #9
    Oh, I see, I misinterpreted what you said in your previous post and though it seemed odd didn't stop to think about it. Thanks for pointing me to that page, it is really interesting.
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