Perhaps I mean the curvature tensor, not the metric.

I suspect it is not.

  1. I think the metric is a 4-dimensional object whereas the sum of the geodesics is 3-dimensional (probably 3+1 but not 4).

  2. Take Minkowski space. As I understand it, its metric is hyperbolic while its geodesics sum to produce flat, Euclidean, 3-dimensional space.

This leads to a supplementary question. The FLWR formulae define a metric. Cosmology calculators such as iCOSMOS allow you to investigate this. You get an extremely interesting universe if you set all of the densities to zero, when the curvature parameter, K, defaults to 1. This curvature is negative or hyperbolic. A nice picture of a saddle is shown. I assume this curvature refers to the metric or to the 4-dimensional curvature of the universe (are they the same thing?) because I think geodesics in such a universe must have positive curvature (ie the resulting 3-dimensional universe has positive curvature). Is this allowed? The alternative would be to have a curvature-only universe with positive curvature, K=-1, but this doesn't seem to be allowed.

So, following comments and the first answer - is it possible that a FLWR universe with negative curvature can have geodesics with positive curvature?

  • $\begingroup$ This might be better for physics SE? $\endgroup$ Commented Jun 4, 2021 at 16:54
  • $\begingroup$ I'm not sure this question makes sense: in a (pseudo-)Riemannian manifold the metric determines the geodesics, but they are not the same thing and not equivalent. The curvature tensor is a function of the metric but, again, not the same thing. $\endgroup$
    – user38308
    Commented Jun 4, 2021 at 19:33

1 Answer 1


In a Minkowski spacetime, say we use light rays to track the geodesics. They are light-like and reveal the curvature of space only. Now we send light rays in a fixed direction. What will we observe? All rays will be parallel, as you can certainly imagine. We can do this in three perpendicular directions. If they are perpendicular initial they will stay forever (though they can change direction). A huge grid of light rays will emerge. They keep the same distances from each other everywhere.

Now we do the in the curved spacetime we encounter in general relativity. Let's take a positively curved space first. Say a 3d sphere. What if parallel lightrays would speed in this space? In each direction, they seem to converge to one point. This will be seen from each point. The whole of space is made visible in this way. But there will always be three pairs of "converging points" at which the rays intersect. These points are on opposite sides of the space, and the position of these points is dependent on the initial conditions of the rays.

In a negatively curved space all rays will diverge (but stay perpendicular because space is curved). In all directions, the rays will diverge.

Now curvature can be considered globally or locally. The local curvature is determined by the energy-momentum of particles inside the space locally. So the light rays can diverge/converge globally while they can diverge/converge locally. you can indeed say that by looking at the light rays you can say how space is curved. but for the curvature itself (represented by a curvature tensor which gets its form from momentum-energy locally $\Lambda$ globally) you will need to look at the math (if yo want numbers).

  • $\begingroup$ You mean we send light rays in a fixed direction but from different starting points? And are the converging points the infinities in projective geometry? $\endgroup$ Commented Jun 5, 2021 at 0:31
  • $\begingroup$ @C.TowneSpringer Yes. You could choose three perpendicular surfaces (though these surfaces have to be defined first by using other pairs of perpendicular light rays traveling on these surfaces) from where you shoot parallel rays. In each of the directions perpendicular to these surfaces the light rays will converge. Because the surfaces are 2d you can choose a lot of different surfaces. But three suffice. $\endgroup$ Commented Jun 5, 2021 at 9:00

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