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About scalar curvature

  1. Feb 6, 2010 #1
    Is there any particular reason that scalar curvature is defined [tex] R = g^{ab}R_{ab} [/tex] instead of [tex] R = R^{ab}R_{ab} [/tex] ?

    Do both scalars share the property that they are zero if and only if every component of [itex] R^{a}_{bcd} [/itex] is also zero?
  2. jcsd
  3. Feb 6, 2010 #2
    The Ricci 2-tensor can be zero even though the Riemann 4-tensor isn't. The Ricci scalar gives the expected value of curvature when applied to e.g. a sphere, i.e. 1/r^2. If you are wondering why the Einstein 2-tensor uses that particular combination of the Ricci 2-tensor and Ricci scalar, just look up any derivation of the general theory of relativity.

  4. Feb 6, 2010 #3
    So the scalar [tex] R^{ab}R_{ab} [/tex] can be zero at a point where [tex] R^{a}_{bcd} [/tex] is not, whereas if [tex] g^{ab}R_{ab} [/tex] is zero at a point, then [tex] R^{a}_{bcd} [/tex] is always zero there as well?
  5. Feb 7, 2010 #4
    No, I meant that R_ab can be zero when R_abcd is not. So both your scalars can be zero even when R_abcd is not.

  6. Feb 7, 2010 #5
    Is this true even when all of the components of the metric tensor have the same sign? I ask because I was thinking about this matter in the context of geometric objects embedded in a higher-dimensional Euclidean space - not relativity, where one can have a vector with non-zero components and a zero norm.
  7. Feb 7, 2010 #6
  8. Feb 7, 2010 #7
    The only thing I know is that in einstein's gravity, the ricci 2-tensor can be zero even when the riemann 4-tensor isn't, and I think that the nonzero part of the riemann 4-tensor in that case is then called the weyl 4-tensor or conformal curvature tensor. This happens in empty parts of spacetime in einstein's theory.

    "I just discovered that the other scalar is called "the Kretschmann scalar"."

    This is the "square" of the riemann 4-tensor. But it is not the same as the square of the ricci 2-tensor you had in your original post. In Weinberg's "Gravitation and Cosmology" he states that there are 14 different curvature scalars in 4 spacetime dimensions. Naturally, he doesn't consider e.g. the square of the ricci scalar as a new curvature scalar, otherwise there would be an infinity of them.

  9. Feb 7, 2010 #8
    I assumed that since the Ricci 2-tensor is itself a contraction of the Riemann 4-tensor, contracting it again would produce the same result as contracting the 4-tensor all at once. (But I didn't actually test this.)
  10. Feb 7, 2010 #9
    It's kind of like the difference between a sum of squares and the square of a sum.

  11. Feb 7, 2010 #10
    So - this may be something that I need to find out by experiment, but - outside of relativity, is there any reason that the scalar curvature [tex] g^{ab} R_{ab} [/tex] is better than the Kretschmann scalar [tex] R_{abcd} R ^ {abcd} [/tex]? I ask only because - just looking at them - the Kretschmann one strikes me as a more natural choice. (Again I'm thinking only about situations where the metric is positive definite.)

    Edit: I may have used the term "positive definite" incorrectly. I just meant the kind of metric such that the distance between any two points is always a positive real number.
    Last edited: Feb 7, 2010
  12. Feb 8, 2010 #11
    As a simple test of the vanishing of the Riemann tensor, the Kretschmann would seem to be the best choice. And it is also the simplest one.

    If I understand correctly, the Kretschmann would always be positive in your case (?), so it wouldn't distinguish between positive and negative curvatures.

  13. Feb 8, 2010 #12
    Ah, positive versus negative curvature - I hadn't thought of that. That would be a good thing for a scalar to tell us.
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