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See the Wikipedia article on Penrose-Hawking singularity theorems.

It says that

A singularity in solutions of the Einstein field equations is one of two things:

1. a situation where matter is forced to be compressed to a point (a space-like singularity)

2. a situation where certain light rays come from a region with infinite curvature (a time-like singularity)

If matter is forced to be compressed to a point, why is this called a space-like singularity?

If certain light rays come from a region with infinite curvature, why is this called a time-like singularity?


marked as duplicate by tparker, John Rennie, Jon Custer, Kyle Kanos, user191954 Sep 6 '18 at 15:30

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  • $\begingroup$ Spacelike and timelike singularities mean that spacelike (resp. Timelike) curves are incomplete and stop at that point. $\endgroup$ – Slereah May 29 '17 at 21:51
  • $\begingroup$ When you say that a curve is geodesically incomplete, do you mean that the curve starts at infinity but ends at a finite point? $\endgroup$ – nightmarish May 29 '17 at 21:55
  • $\begingroup$ It means that, given any point on that curve, if the curve is normalized, it will only have a finite range in either the future or past direction $\endgroup$ – Slereah May 29 '17 at 21:57

The spacelike singularity is a spacelike region where curvature diverges. It is not a point, though it occurs when the radial distance in the metric approaches zero $r~\rightarrow~0$. This is a three dimensional region. The time like singularity is a one dimensional region, think of it a a point evolving in time, where curvature diverges. This has a duality with the spacelike singularity, being one dimensional and orthogonal to the spatial directions of a spacelike singularity.

Think of the Schwarzschild metric with signature $(+,~-,~-,~-)$ $$ ds^2~=~\left(1~-~\frac{2GM}{rc^2}\right)dt^2~-~\left(1~-~\frac{2GM}{rc^2}\right)^{-1}dr^2~-~r^2(d\theta^2~+~sin^2\theta d^2\phi), $$ and see that the metric term $\left(1~-~\frac{2GM}{rc^2}\right)$ flips sign when $r~<~\frac{2GM}{c^2}$. This means the interior of the black hole has signature $(-,~+,~-,~-)$, and in a sense the radial direction serves at a time direction. The $t,~\theta,~\phi$ part of the metric defines the spatial surface.

An infalling observer does not notice anything. It is not as if the radial direction becomes something measured on a clock. However, the observer interior to the black hole will be unable to avoid the singularity. The Penrose conformal diagram illustrates the singularity at the top

enter image description here

You are basically doomed to reach the singularity in a finite time period because it is everywhere in your possible future. In fact if you try to rocket away then since you are not on a geodesic, and geodesics are the maximal path, you in fact seal your fate earlier.

The case for the timelike singularity occurs with charge or angular momentum. I will use the Reissnor-Nordstrom metric since it is simpler to understand than the Kerr metric. In fact it is a modification of the Schwarzschild metric $$ ds^2~=~\left(1~-~\frac{2m}{r}~+~\frac{Q^2}{r^2}\right)dt^2~-~\left(1~-~\frac{2m}{r}~+~\frac{Q^2}{r^2}\right)^{-1}dr^2~-~r^2(d\theta^2~+~sin^2\theta d^2\phi), $$ Since the metric coefficient $\left(1~-~\frac{2m}{r}~+~\frac{Q^2}{r^2}\right)$ is quadratic in $r$ is vanishes at two radii. There is an inner and outer horizon determined by $r_{\pm}~=~m~\pm~\sqrt{m^2~-~Q^2}$. There is then an inner horizon where the metric flips signature again. This means the region is timelike and the region $r~=~0$ is a point in space evolving in time. This is seen in the Penrose diagram

enter image description here

There are subtleties, such as the inner horizon might turn out to be a mass inflation singularity. There is a pile up of null curves there. This appears similar to a Cauchy horizon that could be a sort of singularity. The region with the timelike singularity could be a mathematical fiction.


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