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Everyone talks about positive curvature when they talk about the EFE, and for good reason. I'd just like to know what would change in the equation if it accounted for negative curvature of spacetime instead. Would anything change, or is it just a matter of plugging in negative values instead?

I understand it seems strange that anyone would want to know this, but I'm just generally interested. What is different if you consider negative instead of positive curvature? How does one set the metrics to negative (if that makes sense)?

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  • $\begingroup$ This is unclear because “curvature” is not one number. It takes 20 numbers to describe the curvature at a point in 4D spacetime. And you can form a variety of scalars from these Riemann tensor components. $\endgroup$ – G. Smith Jan 28 '20 at 0:19
  • $\begingroup$ Perhaps you are thinking about how Friedmann metrics can have positive, negative, or zero spatial curvature? $\endgroup$ – G. Smith Jan 28 '20 at 0:21
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The negative curvature is the output, not the input. To calculate the curvature invariant (which is called the Kretschmann scalar $K$) you have to compute

$$K=R_{\mu\nu\lambda\rho} \ R^{\mu\nu\lambda\rho}$$

For a rotating black hole (here with $a=M$) you get regions with positive and negative curvature:

Kretschmann scalar for a rotating Kerr black hole in cartesian background coordinates

The bold black lines show the hard crossings between positive and negative curvature, and the colored surfaces in the middle show the inner and outer ergospheres and horizons in cartesian background coordinates.

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  • $\begingroup$ Why is this particular curvature invariant relevant as opposed to the Ricci scalar or one of the other quadratic invariants? $\endgroup$ – G. Smith Jan 28 '20 at 0:15
  • $\begingroup$ The Ricci scalar is 0 everywhere with Schwarzschild and Kerr black holes, although the curvature is not. $\endgroup$ – Gendergaga Jan 28 '20 at 0:16
  • $\begingroup$ My guess is that the OP is thinking about spatial curvature in the Friedmann metric, but the question is unclear. $\endgroup$ – G. Smith Jan 28 '20 at 0:16
  • $\begingroup$ For the curvature invariants of the Friedmann metric see yukterez.net/f/einstein.equations/files/flrw.html $\endgroup$ – Gendergaga Jan 28 '20 at 0:18
  • $\begingroup$ that's a beautiful diagram, thanks for sharing $\endgroup$ – lurscher Jan 28 '20 at 1:35
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The Einstein field equations read,

$$R_{\mu\nu} -\frac12 g_{\mu\nu}R = 8\pi G \, T_{\mu\nu}.$$

There are two main ways we can use them: either to find $g_{\mu\nu}$ given a stress-energy tensor $T_{\mu\nu}$ or computing the stress-energy tensor given a metric.

It's often the case we know the physical scenario at hand, and the stress-energy tensor, so the negative curvature isn't something we put in by hand, it would emerge when solving the equations and finding $g_{\mu\nu}$ if indeed, the spacetime is negatively curved.

There is no issue with starting with a metric with negative curvature. In fact, you'll find quite liberal use of the Einstein field equations in the literature, in the sense of not imposing conditions like continuity, for certain cases of interest.

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