# Necessary / sufficient conditions for physical curvature singularity

1. Mar 20, 2015

### binbagsss

For a physical singularity I think it is sufficient that any one scalar quantity blows up,
Why is it not a necessary condition that all blow up?

For a curvature singularity am I correct in thinking that it is a sufficient condition to find a coordinate system in which the metric coefficient no longer blows up at that point?

Is the only necessary condition for a curvature singularity to check that all non scalar quantites are not infinite?

Thanks.

2. Mar 20, 2015

### Matterwave

The necessary and sufficient conditions for a singularity are non-trivial. It is not always necessary or sufficient for any one of the scalar curvatures or tensor curvatures or metric coefficients to "blow up". Indeed, even the definition of a singularity itself is complicated. This is why Hawking and Penrose (and others) worked so hard on their singularity theorems. There are cases where the scalar curvature is 0 everywhere, but the curvature tensor can be singular. Or There can be cases where the scalar curvatures or curvature tensor itself is singular only "at infinity" where no observer can reach.

The best, most general, criterion we have of a singularity, as best as we can figure, is the presence of geodesic incompleteness.

See Wald chapter 9 for details.

3. Apr 20, 2015

### binbagsss

In lecture notes on GR by Sean.M. Carroll he has that a sufficient condition to prove a singularity is that if there exists any (and not all) scalar quantity constructed from the Riemann tensor that goes to infinity at some point, the point is a singularity.

I don't understand how finding a coordinate system in which the metric is no longer singular at some point can not be a sufficient condition to disprove a signularty ? Does anyone have any examples, e.g. a coordinate system in which the singular nature of r=0 dissapears for the Schwarzschild metric,

thanks.

Last edited: Apr 20, 2015
4. Apr 20, 2015

### stevendaryl

Staff Emeritus
Your suggestion is a little complicated to carry out, for the following reason: By definition, any solution of Einstein's Field Equations is a pseudo-Riemannian manifold, which means that for any event $e$ there is a neighborhood that can be described with nonsingular coordinates. So in that sense, there are no singularities. The point (or line, actually $r=0$ is not part of the manifold, strictly speaking. That's why "geodesic incompleteness" is relevant. A timelike geodesic in the neighborhood of $r=0$ will, in a finite amount of proper time, leave the manifold. Or said another way, there is a maximum finite amount of proper time such that the geodesic cannot be extended past that time.

5. Apr 20, 2015

### binbagsss

And so the definition of a curvature singulairty is not a singularity that can be removed by good choice of coordinates?
Instead what's a good definition for curvature and physical singularity?

6. Apr 20, 2015

### Staff: Mentor

Correct. Note that the definitions people have been giving you involve scalars going to infinity. A scalar is an invariant and can't be changed by changing coordinates.