0
$\begingroup$

Types of singularities include curvature singularities and conical singularities. So, for a curvature singularity(black hole) with geodesic incompleteness, is it the same as a physical singularity? If there is something wrong with the following contents, please correct it.

$\endgroup$
  • $\begingroup$ What do you call a physical singularity exactly? $\endgroup$ – Slereah Sep 28 '18 at 9:59
  • $\begingroup$ The classic and very readable paper on this is Geroch, "What is a singularity in general relativity?," Ann Phys 48 (1968) 526. I believe it's possible to find non-paywalled copies online (presumably illegally). $\endgroup$ – user4552 Oct 9 '18 at 20:51
1
$\begingroup$

I think you have the types of singularities slightly mixed up (although possibly I have them mixed up, so watch out!)

All singularities, I think. involve geodesic incompleteness: there's a notion that you can't extend a geodesic to arbitrarily large values of its affine parameter because something goes wrong. So, in particular the singularity theorems say things like 'given some preconditons on causality, energy and gravity, there will be geodesic incompleteness'.

Two subtypes of singularities are curvature singularities, where physical properties blow up as you approach the singularity (and 'become infinite' at the singularity itself) and conical singularities, where things don't blow up as you approach the singularity. (There's another subtype which is the 'I just decided to chop some points out of spacetime' type, but we can ignore that.)

So then I think that it is expected that the singularities at the centre of black holes will be curvature singularities, while cosmic strings, if they exist, would be conical singularities. But the singularity theorems don't actually talk about what sort the singularities they predict are: they just say there will be geodesic incompleteness.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Hi! I am not sure that all the singularities can be revealed through geodesics. Think for instance to a singularity for which only contractions of the riemann tensor with at least a certain number of derivatives diverge. The geodesic equation would not be sensitive to these divergences, I think. Maybe however there is some kind of constraint that can help us out... $\endgroup$ – AoZora Oct 15 '19 at 9:41
  • $\begingroup$ @AoZora If you can extend geodesics through such singularities then. from the GR perspective, they're not really singularities I think, but rather failures of differentiability at some level higher than which the field equations are sensitive to. $\endgroup$ – tfb Oct 15 '19 at 10:15
  • $\begingroup$ well of course these would not be singularities from the geodesic point of view. However we would have some divergent scalar quantity, a curvature singularity.. $\endgroup$ – AoZora Oct 15 '19 at 10:20
  • $\begingroup$ @AoZora: if that scalar is a quantity which GR does not care about then in the sense that it does not alter the solutions to the field equations (and in particular, since it is not finite, make them blow up in some way), then, well, there's no physical significance to such a thing. If it does, then, since the field equations determine the geodesics, the spacetime is geodesically incomplete, as far as I can see. I think you need to be specific about what sort of divergences you mean, and work through the implications. That might be a good topic for a question. $\endgroup$ – tfb Oct 15 '19 at 10:24
  • $\begingroup$ @tbf GR is just an effective espansion. At low energies we can use the Einstein-Hilbert Lagrangian when all higher dimensional operators are small, as they should. If there is a higher dimensional scalar quantity that diverges however, we can build as many divergent higher dimensional operators as we want, making the effective expansion (i.e. GR) meaningless. So the divergence of a scalar quantity, although it may not enter explicitly in the field equations, can make them unusable: you compute a solution which is right up to divergent corrections.. $\endgroup$ – AoZora Oct 15 '19 at 10:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.