2
$\begingroup$

A curvature invariant is a scalar representation of curvature derived from a curvature tensor. The classic example is the Kretschmann scalar derived from the Riemann curvature, where $K=R_{μνλρ}R^{μνλρ}$.

This is a coordinate independent measure that allows one to discern between coordinate and curvature singularities. For example in the Schwarzschild black hole spacetime the Kretschmann scalar is

$$K=R_{μνλρ}R^{μνλρ}=\dfrac{48M^2}{r^6}$$

where $M$ is the geometrized mass.

Different geometries yield different functions for the curvature invariants and since all classical BH geometries are required to have singularities by the Geodesic Incompleteness theorems of Hawking and Penrose, then the curvature necessarily runs to infinity. For the example above it should be obvious that

$$\lim_{r\to0}K=\lim_{r\to0}\dfrac{48M^2}{r^6}=\infty$$

There are no anti-infinity theorems. If so, can a curvature singularity, as in geodesic incompleteness, actually exist in nature?

$\endgroup$
  • $\begingroup$ What do you mean by "exist"? Normally, "to exist" means "to move in time". However, the BH singularity does not move in the inside time ($r$) and is causally disconnected from the outside time ($t$). Can you clarify what exactly you are asking? $\endgroup$ – safesphere Oct 3 '18 at 5:04
  • $\begingroup$ Can you give a reference for the claim that the curvature necessarily runs to infinity? $\endgroup$ – MBN Oct 3 '18 at 11:12
3
$\begingroup$

Imagine that you have a hand-held detector which has a needle pointer such as an analogue voltmeter: enter image description here

Now imagine a theory that should give you the value to which the needle points. Usually this theory gives you a real value, but you find a point where it gives you an imaginary value as an output of a real physical situation. What do you conclude? That the needle will point to an imaginary value? Surely not. When the theory starts predicting imaginary values, it is simply wrong: it does not predict any actual behavior of the needle.

Now consider the example of the black hole. Quantities such as the curvature scalars are in principle measurable by geodesic deviation. $\infty$ is not a real number. The predictions that are provided at the point of the black hole singularity are thus outside of the real numbers and the theory provides no predictions for the measurements.

Since the theory is simply undefined at the singularity, it does not need to fulfill Einstein equations or any equations whatsoever; its behavior is arbitrary. In other words, the "boundary" of the space-time at $r=0+\varepsilon$ can have absolutely arbitrary boundary conditions. It could start spewing cats and pianos as well as tachyons. On the other hand, if we make a few conservative assumptions such as that momentum-energy is conserved by the action of the singularity, and that exotic matter is not created by it, it actually turns out that the boundary can do whatever it wishes and this will bear no consequence on the space-time outside of the horizon. This is actually the implicit assumption of numerical-relativity evolutions of black holes.

In summary, the $\infty$ is a problem and makes the theory incomplete. A physical quantity with value $\infty$ has no operational definition and thus cannot be physically existent. However, it turns out it is not a practical problem for using the theory for astrophysical purposes as long as we make some basic assumptions about the behavior of the singularity.

$\endgroup$
  • $\begingroup$ You speak of the singularity as if extended in time. For example, "It could start spewing...", etc. However, the time coordinate inside the horizon is $r$ and the length of "existence" of the singularity is zero. Could you please clarify? Also, what exactly do you mean by the boundary conditions? In your example of $r=0+\epsilon$, the spacetime is well defined by geodesics as long as $\epsilon\ne 0$ strictly. $\endgroup$ – safesphere Oct 3 '18 at 16:03
  • $\begingroup$ To a certain degree, this is a figure of speech. The curvature singularity in Schwarzschild space-time is indeed space-like. On the other hand, once the black hole has either charge or any non-zero angular momentum (as described by the Kerr-Newman class), the singularity is time-like and there such wording is completely fine. $\endgroup$ – Void Oct 3 '18 at 18:27
  • $\begingroup$ In the Schwarzschild space-time, your "boundary conditions" are uniquely specified by assuming spherical symmetry and vacuum. However, these are assumptions. The issue of a "boundary condition" in the stricter mathematical sense of the word would in fact be more nuanced because it implies an evolution problem. However, most black holes will have a Cauchy horizon in their interior, well above the singularity, and the boundary condition must be above that in practice. $\endgroup$ – Void Oct 3 '18 at 18:32
  • $\begingroup$ Thanks so much, very helpful. I've upvoted your answer. While in the subject of black holes, may I please solicit your feedback on my answer and how it can be improved? physics.stackexchange.com/questions/426143/… (There is a small inaccuracy on the chart where at the point F the geodesic should turn horizontal.) Thanks again! $\endgroup$ – safesphere Oct 3 '18 at 18:41

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.