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There is an innumerable quantity of questions and answers on this site about spatial, spacetime, and temporal curvature. None of these answer my question.

Unfortunately, many use the terms in different, sometimes confusing ways, that make it unclear what we exactly mean by our universe's spatial curvature, spacetime curvature, and temporal curvature.

Our universe has spacetime curvature, so the spacetime version of the Pythagorean theorem doesn’t hold. Our universe does not have a Minkowski metric. But our universe does not appear to have any measurable spatial curvature, so in only the three spatial dimensions the Pythagorean theorem does hold.

Curved spacetime and geodesics

Very nice explanation.

You need to be cautious about treating a time curvature and spatial curvature separately because this split is not observer-independent. and the answer is that at least two principal curvatures must be non-zero. So you cannot find a geometry/coordinate system where the curvature is only in the time coordinate.

How do spatial curvature and temporal curvature differ?

Now, this is where it gets a little confusing. Our universe has no spatial curvature. But our spacetime does. So the curvature must be in the temporal dimension? But this says we cannot find a coordinate system where the curvature is only in the temporal dimension. So the connection between the statement that our universe and our spacetime is curved, and that there is no spatial curvature, is not trivial.

Our spacetime has intrinsic curvature. But it is not in the spatial dimensions (there is no spatial curvature), so it has to manifest in the temporal dimension?

Just to clarify, one of the answers specifically says our universe does not have spatial curvature (talks about spatial curvature separately), and the other answer talks about possible existing temporal curvature (mentioning that you have to be cautious to treat spatial and temporal curvature separately), but then they both talk about spacetime curvature.

I am looking for a connection between the spatial, temporal, and spacetime curvature.

Question:

  1. What is the connection between spatial, temporal, and spacetime curvature?
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    $\begingroup$ I consider the phrase “temporal curvature” to be nonsense. There is only one time dimension, and one-dimensional (sub)spaces don’t have Riemannian curvature. $\endgroup$ – G. Smith Aug 9 at 16:58
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    $\begingroup$ You should explain if you have had any exposure to Riemannian geometry so that someone knows what kind of answer to write. $\endgroup$ – G. Smith Aug 9 at 17:16
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    $\begingroup$ @G.Smith thank you yes, any answer including Riemannian geometry is fine. $\endgroup$ – Árpád Szendrei Aug 9 at 17:39
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    $\begingroup$ @ACuriousMind " But our universe does not appear to have any measurable spatial curvature", this is what the answer says here: physics.stackexchange.com/questions/527044/… $\endgroup$ – Árpád Szendrei Aug 10 at 15:55
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    $\begingroup$ @ACuriousMind this is why the two answers together are a little bit confusing. One says, and talks about explicitly spatial curvature in our universe (to be zero), and the other one says spatial and temporal shouldn't be treated separately (like you say), but then it talks about temporal curvature. But I will edit to include this. $\endgroup$ – Árpád Szendrei Aug 10 at 16:03
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The spacetime metric of a spatially-flat Friedmann universe — like ours seems to be, on the largest scales — is

$$ds^2=-dt^2+a(t)^2(dx^2+dy^2+dz^2)$$

where the function $a(t)$ is the Friedmann scale factor describing the expansion of space as a function of cosmological time $t$.

You can calculate its 4D Riemann curvature tensor $R_{\mu\nu\lambda\kappa}$ and find that it has various nonzero components involving the first and second time derivatives of $a(t)$. (Even some components where all four indices are spatial are nonzero!) This is an example of spacetime curvature.

Now take a spacelike slice through this spacetime at some constant cosmological time $t_0$.

The metric of this 3D space is

$$ds^2=a(t_0)^2(dx^2+dy^2+dz^2)$$

where the prefactor $a(t_0)^2$ is just some constant that could be absorbed into the coordinates to rescale them.

You can calculate its 3D Riemann curvature tensor and find that every component is zero. (This should be obvious, because it’s just a Euclidean metric.) This is an example of spatial flatness, or zero spatial curvature.

Temporal curvature doesn’t exist because there is only one time dimension and one-dimensional (sub)spaces always have zero Riemannian curvature.

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  • $\begingroup$ Thank you this is a very very interesting answer. I get that the curvature itself is what we call is determined by not one, but many components of the metric, and as I understand more then one component needs to be nonzero to talk about curvature. This is fulfilled in 4D, but not in 3D. But we cannot simply say that the curvature enters because of the temporal component, because there is no such thing as one dimensional subspace curvature. $\endgroup$ – Árpád Szendrei Aug 9 at 21:32
  • $\begingroup$ Good question is what I do not understand how exactly the addition (or the existence) of the 4th dimention (temporal) causes the metric to have more then one nonzero curvature component. I guess somehow the temporal dimension is different from the spatial ones, and the combination of spatial and temporal dimensions causes this curvature. I am not sure if we could say the curvature manifests in the (because of) temporal dimension? $\endgroup$ – Árpád Szendrei Aug 9 at 21:34
  • $\begingroup$ Even if only one component of the Riemann tensor were nonzero (actually, because of symmetries there would have to be at least four) we would say the manifold was curved. $\endgroup$ – G. Smith Aug 9 at 21:35
  • $\begingroup$ The existence of the temporal dimension does not ensure curvature. Minkowski spacetime has zero curvature. $\endgroup$ – G. Smith Aug 9 at 21:36
  • $\begingroup$ To have curvature, you have to have a metric that can’t be made “Pythagorean” (in whatever number of dimensions) by any coordinate transformation. In this case, the function $a(t)$ is the non-Pythagorean thing. $\endgroup$ – G. Smith Aug 9 at 21:40
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I think the essential problem lies in the difference between the mathematical meaning of curvature, and the way in which we actually describe a manifold, or a curved space (or spacetime).

Although we describe the universe as having spacetime curvature (which is mathematically true), curvature refers to the Riemann curvature tensor, which is a rank-4 tensor, meaning that it has $4^4 =256$ components, of which (due to various symmetries) $20$ are independent. This is far too cumbersome for even mathematicians to think about, but what is certainly true is that you cannot separate it nicely into space curvature and time curvature. As @G.Smith says in comments, "temporal curvature" does not make any sense. Time is a single dimension, and a one-dimensional subspace does not have any Riemannian curvature.

In other words, we use the mathematics of curved spacetime, but we don't actually describe anything directly in terms of Riemannian curvature. We do write Einstein's equation for gravity using the Einstein curvature tensor (or Ricci) but since this is zero except in the presence of mass-energy (the source of gravity), it does not directly tell us about the geometry of spacetime; to know that we have to solve Einstein's equation.

When we do solve Einstein's equation, we do not find curvature as such. Instead we find the metric. The metric is much easier to think about than curvature (we can write down a formula from which we could calculate curvature given the metric, but actually we never bother with that horrible calculation).

Rather than think about curvature, we think about scaling distortions in maps. In other words, we choose a coordinate system, and think about how actual or proper quantities appear in those coordinates. Proper quantities are the physical properties which would be measured by an observer moving with the object being measured.

We can compare this to scaling distortions in maps of the surface of the Earth. Any number of different maps are possible. The metric for the map tells us how to compare apparent distances on the map to actual distances as measured by someone on the ground.

So, rather than talk of curvature, talk of scaling distortions in maps. Then your question makes sense. For example, we cannot directly measure scaling distortions in Euclidean geometry in the region of the Earth, because they are too small. But we can, and do, measure scaling distortions in time. Clocks on GPS satellites measure the same unit of time as identical clocks on Earth. They measure exactly one second per second (as required by the general principle of relativity). But they appear on Earth to run at a different rate, because of the scaling distortion in the map used to describe them. Indeed, we can explain Newtonian gravity completely in terms of the scaling distortion of the time component, the scaling distortions of the space components being too small to have any impact.

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  • $\begingroup$ You seem to be saying that by "the universe is spatially flat" we just mean that the spatial components of the curvature are too small to measure. That's not correct. A cosmology can be spatially curved even if all components of the Riemann curvature of spacetime are exactly zero (see the toy examples in my answer), and a critical-density FLRW cosmology without local perturbations is exactly, not just approximately, spatially flat. It comes down to the definition of "spatial curvature" which is not straightforwardly related to the spacetime curvature. $\endgroup$ – benrg Aug 9 at 19:46
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    $\begingroup$ @benrg, No, I am not saying that. I am simply skating over the point because the answer is complicated enough already at the level at which I think it is required. $\endgroup$ – Charles Francis Aug 9 at 19:57
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The notion of "spatial curvature" only makes sense when the spacetime geometry is symmetric enough that there is a natural/preferred foliation of it into spacelike slices. You can then talk about the intrinsic curvature of those slices.

The easiest way to understand why the curvatures can be different is to look at a toy cosmological model, like the "expanding balloon" picture: 3D Euclidean space, with time being distance to the origin. The locus of space"time" points with a given time coordinate in this model is a 2D space of constant positive curvature, but the 3D background space"time" has zero curvature.

A slightly more realistic toy model is the analogous one in 3+1D Minkowski space: the interior of the future light cone of the origin, with time being the (timelike) distance to the origin. The locus of points with a given time coordinate is a 3D space of constant negative curvature. This model is in fact the zero-energy-density or zero-$G$ limit of any expanding FLRW cosmology. As you add energy density, or add gravity, the spacetime becomes positively curved. The spatial slices get an increasing curvature, which reaches zero at the critical density, and is positive at higher densities. The FLRW time coordinate is analogous to the radial coordinate of a polar coordinate system on a curved surface, like the surface of the earth, which is of course where the name "polar" came from. The time coordinate is the latitude, and the position coordinates are the longitude.

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  • $\begingroup$ Thank you so much! $\endgroup$ – Árpád Szendrei Aug 10 at 16:06
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I would add to all of the answers before mine some "food for thought". I would try to show you a visual example of a Riemanian two-manifold (i.e. a regular surface and not exactly a Lorenzian space-time surface), which is a negatively curved surface but has a whole family (in fact two families) that are straight lines.

Take a look at the one-sheeted hyperboloid. It has two families of straight lines on it (the terminology is "it has two transverse foliations of straight lines"). As we know, straight lines are as Euclidean as they come, straight in every sense, be that intrinsic or even extrinsic, as embedded spaces in the hyperboloid as well as in three space. Another term here is "the hyperboloid is a ruled surface". Nevertheless, the hyperboloid, as a two-dimensional manifold is negatively curved. And although on the hyperboloid at every point there are exactly two directions that are straight (flat, Euclidean), the total surface is nevertheless negatively curved!

If you now think of the one-sheeted hyperboloid embedded not in regular Euclidean three space, but in the two plus one Minkowski space, you get a model of one plus one deSitter space which is a type of non-flat spacetime.

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  • $\begingroup$ Thank you do much! $\endgroup$ – Árpád Szendrei Aug 13 at 3:27

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