- #1
- 6,724
- 429
In general, what methods are there for proving that something is a non-curvature singularity?
As a specific example, take the 1+1-dimensional version of the NUT metric, ##ds^2=t^{-1}dt^2-t d\theta^2##, which is discussed in Hawking and Ellis in two places. H&E's treatment is long and meandering, and I have a hard time figuring out what they're getting at.
What I can figure out is that the null geodesics are of the form ##\theta=(\text{const})\pm\ln(\pm t)##, and that ##t## itself qualifies as an affine parameter. This shows that the geodesic specified by this equation doesn't continue to arbitrarily large affine parameters. If the opposite had happened, then clearly we would have had geodesic completeness. But the converse does not seem to be a reliable criterion, because, if I'm understanding H&E correctly, they're able to do a change of coordinates such that a particular null geodesic at ##t>0## is glued end-to-end with another null geodesic at ##t<0##, making it complete. However, they aren't able to find coordinates that accomplish this for all null geodesics at once.
So to prove that a space is geodesically complete, when the singularity is not a curvature singularity, do we have to prove that no coordinate system exists in which all geodesics are complete? That seems like a tall order. How would you prove such a nonexistence result?
As a specific example, take the 1+1-dimensional version of the NUT metric, ##ds^2=t^{-1}dt^2-t d\theta^2##, which is discussed in Hawking and Ellis in two places. H&E's treatment is long and meandering, and I have a hard time figuring out what they're getting at.
What I can figure out is that the null geodesics are of the form ##\theta=(\text{const})\pm\ln(\pm t)##, and that ##t## itself qualifies as an affine parameter. This shows that the geodesic specified by this equation doesn't continue to arbitrarily large affine parameters. If the opposite had happened, then clearly we would have had geodesic completeness. But the converse does not seem to be a reliable criterion, because, if I'm understanding H&E correctly, they're able to do a change of coordinates such that a particular null geodesic at ##t>0## is glued end-to-end with another null geodesic at ##t<0##, making it complete. However, they aren't able to find coordinates that accomplish this for all null geodesics at once.
So to prove that a space is geodesically complete, when the singularity is not a curvature singularity, do we have to prove that no coordinate system exists in which all geodesics are complete? That seems like a tall order. How would you prove such a nonexistence result?