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Usefulness of Kretschmann scalar

  1. Nov 20, 2011 #1
    Hello,

    As I'm sure you are aware the Kretschmann scalar (formed by contracting the contravariant and covariant Riemann tensors) has some use in the identification of gravitational singularities. Specifically, because K is essentially the sum of all permutations of R's components, but is itself coordinate invariant, its divergence at a point is sufficient to prove the existence of a true gravitational singularity at that point.

    I am wondering whether this is a necessary condition as well. It seems to me that it is not, since one could imagine a situation in which two terms in the sum diverge in opposite directions. Perhaps I have missed something, though?
     
  2. jcsd
  3. Nov 20, 2011 #2

    Matterwave

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    In fact, singularities are extremely hard to define generally. The divergence of curvature scalars is not always sufficient, nor necessary conditions for us to want to classify a specific space-time as "singular".

    An example of the "insufficient" case, it could be that the divergence of the curvature only blows up "at infinity", so that no observer could ever get to this singularity. On the other hand, there are some spacetimes which have vanishing curvature everywhere but is still singular.

    Whether the curvature becomes unbounded or not is only one of the "indicators" we use to figure out if a singularity exists or not.
     
  4. Nov 21, 2011 #3

    PAllen

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    I've only used it in sense of a rough measure of 'amount of curvature' (no possible scalar is adequate, otherwise we wouldn't need a tensor), that is more useful than R in GR because it is typically non-zero in vacuum regions of GR solutions, while R is identically zero in vacuum regions. I've never run across any theorems using it to draw significant non-obvious conclusions (example of obvious: non-vanishing K implies manifold not flat). (Don't take my not having seen any as a very strong statement).
     
  5. Nov 22, 2011 #4
    Okay, I can see that K diverging at infinity would not be so unexpected. So what you're saying is that if K blows up at some ordinary point in your coordinate system (which I guess might map to infinity in some other coordinate system), that point isn't necessarily a gravitational singularity, but it is a point you might want to investigate in more detail, correct?

    Would the following be sufficient to demonstrate the existence of a singularity:
    1) K diverges at a point p in a coordinate system Q
    2) It is possible to map Q onto some new coordinate system S, with p going to a set of points P.
    3) S is maximally extended.
    4) In S, P is timelike separated from some class of observers.

    PAllen: You mean either Rab (the Ricci tensor) or R (the Ricci scalar), I'm assuming (my fault for not subscripting before). Rabcd (the Riemann tensor) need not be zero in vacuum, I don't think?
     
  6. Nov 22, 2011 #5

    PAllen

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    Since we were discussing scalars, I assumed Ricci scalar was obvious by context.
     
  7. Nov 22, 2011 #6
    Yeah, my bad. Anyways thanks.
     
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