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Singularity of spacetime/singularity of the riemanian tensor

  1. Nov 11, 2015 #1

    I have a question to the singularities of spacetime (where the metric tensor is infinite, but not the coordinate singularities which can be removed be a change of coordinate)
    It's easy to show that a singularity of the riemanian tensor scalar RαβμνRαβμν leads to a singularity of the spacetime. But what about the other way round? Is it possible to proof that each singularity of spacetime is also a singularity of RαβμνRαβμν ?

    Thanks for help.
  2. jcsd
  3. Nov 11, 2015 #2
    I'm not 100% sure, but I think so. A true spacetime singularity (a curvature singularity actually) should pop up on the Kretschmann scalar as that point will not be well-behaved (with the geodesic incompleteness and all).
  4. Nov 11, 2015 #3


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    This is actually not a correct definition of a singularity. A singularity is defined as geodesic incompleteness, meaning that there are geodesics that can't be extended past a finite affine parameter. It is neither necessary nor sufficient for the metric tensor to have a component that blows up, when expressed in some coordinate system, at some set of coordinates.

    This is not quite true. A blowup in a curvature scalar only indicates a curvature singularity if it can be reached along a geodesic in a finite affine parameter.

    No, this is false. A counterexample would be a non-curvature singularity such as a conical singularity. I'm pretty sure it's also possible to have a curvature singularity in which the Kretschmann invariant doesn't blow up, but I don't know of an example.

    We seem to have had multiple discussions and a lot of confusion recently about the definition of singularities in GR. I have a discussion of this in section 6.3.6 of my GR book: http://www.lightandmatter.com/genrel/ .
  5. Nov 11, 2015 #4
    How would this be possible?
  6. Nov 11, 2015 #5


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  7. Nov 11, 2015 #6
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