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I found this statement in a discussion about the application of local Lorentz symmetry in spacetime metrics:

Lorentz invariance holds locally in GR, but you're right that it no longer applies globally when gravity gets involved. While in SR, quantities maintain Lorentz (or Poincare) symmetry via Lorentz (or Poincare) transforms, in GR they obey general covariance which is symmetry under arbitrary differentiable and invertible transformations (aka diffeomorphism). If a spacetime was not smooth, and didn't allow local Lorentz symmetry, it would break the principle of equivalence which is the bedrock assumption in GR.

What would it mean that spacetime is not smooth? Are there models or metrics of spacetime where it is not smooth?

vengaq
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2 Answers2

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So generally a spacetime is normally defined as the tuple $(M,g)$ where $M$ is a smooth manifold and $g$ is a smooth, symmetric, non-degenerate, metric on this manifold. By spacetime not being smooth one could in principle mean relaxing one or both of these conditions. This however quickly leads to problems:

Firstly note that in the case that the manifold is not differentiable, defining the tangent bundle on the manifold (and as a result the tensor bundles as well) will be problematic at the points where the manifold is not differentiable. However, these are essential for the physics of our theory in the sense that e.g. the curvature tensors that show up in einsteins equations would not make sense. (As was noted in the linked post in the comments if we are serious about doing this you quickly end up in the realm of distributions and run into a bunch of problems.)

So what if only the metric is not smooth, say only continuous? If we look at Einsteins' equation, these are nonlinear differential equations in the metric components, such that now the same problems of running into (non-linear) use of distributions show up within these equations.

Thomas Tappeiner
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Let first us consider an analogous example model in 2-dimensional space. In this cone-like surface the metric is smooth (i.e. differentiable) almost everywhere.
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Only at one point, the tip of the cone, it is non-differentiable. In other words: the curvature is infinite there. That means, the differential calculus machinery (calculating Christoffel symbols, Riemannian curvature tensor, ...) is not applicable there.

In 4-dimensional spacetime an analogous situation arises when you have an ideal point-like mass. At this point the metric becomes non-differentiable, and the Christoffel symbols and the components of the Riemannian curvature tensor become infinite.

Thomas Fritsch
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  • but would these spacetimes be physically possible? Are there any non-smooth spacetimes that do not directly involve singularities? @ThomasFritsch – vengaq Mar 24 '23 at 17:35
  • @vengaq Yes, such spacetimes are physically possible. The center of a black hole is such a non-smooth singularity. The non-smoothness is always restricted to singular points. Our differential calculus (like Einstein field equations) can only handle spacetimes which are smooth almost everywhere (possibly with the exception of some singular points). – Thomas Fritsch Mar 24 '23 at 22:42
  • but as far as I know, many physicists seriously doubt that these singularities are actually real (e.g. that a black hole is a region with an infinite curvature). So, is there any other scenario of a spacetime with non-smooth structures or regions that could well exist without doubt according to our current knowledge of physics? @ThomasFritsch – vengaq Mar 24 '23 at 23:51
  • @vengaq Well, our current best theory (GR) predicts black holes with a singularity inside. All other theories trying to avoid such singularities currently are just speculations, because they failed to make any experimentally confirmed prediction beyond GR. – Thomas Fritsch Mar 25 '23 at 00:15