How does one imagine the curvature of spacetime in 3D?

Question

I'm learning about general and special relativity, but I can't seem to wrap my head around the concept of spacetime being curved in 3D. The 2D flat drawings are very explanatory, but how does one imagine spacetime and it's curvature in 3D?

By 2D, do you mean the drawings of a sphere lying on a rubber sheet? — Ernie, Apr 02 '17 at 19:21
By 2D I mean these sort of pictures: http://www.esa.int/spaceinimages/Images/2015/09/Spacetime_curvature — Paulo The Cab Driver, Apr 02 '17 at 19:44
@PauloTheCabDriver Those are 2-D sheets stretched into the third dimension, so it's a 3-D diagram. Do you want like a 3-D sphere with 4th dimensional stretching? I don't see how that would make it easier to visualize. — JMac, Apr 03 '17 at 11:48
My question is rather how we can visualize the spacetime in our kosmologic reality. It doesn't need to be easier, but I can imagine that the physical reality differs from that of the analogy, because we live in a 3D world and thus the sheets need to be stacked in a way, am I right? — Paulo The Cab Driver, Apr 03 '17 at 13:14
I have found the same question, with this result: https://www.quora.com/How-do-you-explain-the-space-time-curvature-on-a-3D-plane — Paulo The Cab Driver, Apr 03 '17 at 13:16

score 3 · Accepted Answer · answered Apr 02 '17 at 19:40

The conventional intuitive understanding of curvature is actually not what we usually concern ourselves with in general relativity. You say that,

... The 2D flat drawings are very explanatory...

which suggests to me you imagine curvature as say, being a cylinder since it is formed by a piece of paper wrapped around and hence is curved.

However, this cylinder is embedded in $\mathbb R^3$ and the curvature you see with your own eyes is in fact the extrinsic curvature,

$$K_{ab} = \frac12 \mathcal L_n g_{ab}$$

where $n$ is the normal to the surface, and depends on the embedding. This is not something which is intrinsic to the surface and a cylinder as a manifold in its own right is intrinsically flat.

There are of course many examples of this, but the cylinder seems to be a canonical choice as it will leave you flabbergasted to be told it is really flat.

The notion of intrinsic curvature in general relativity has to do with how data is affected by parallel transportation on a manifold and is computed from a connection which as the name suggests describes how tangent spaces at different points are connected.

score 0 · Answer 2 · answered Apr 02 '17 at 23:32

The heart of curvature${}^{1}$ is the convergence or divergence of "parallel" lines. As JamalS says above, a cylinder will lead you astray, but a sphere is truly curved -- two lines of longitude are parallel at the equator, but intersect at the north pole. In a negatively curved space, "parallel" lines will actually diverge from each other.

You can think of gravity as doing the same thing by playing games with inertia. In flat space, two observers, initially stationary with respect to each other will remain stationary with respect to each other. Put the Earth between them, and their paths will converge at the center of the Earth. Put them in an expanding cosmology, and they'll find themselves farther apart. The initially parallel timelike lines through spacetime will become no longer parallel.

${}^{1}$ Note that curvature can be picked apart a million other ways -- parallel transport around a closed curve, the failure of angles to add up to $180^{\circ}$, etc. Thanks to Euclid's parallel postulate, these all are equivalent ideas. I talk about parallel lines becuause the connection to physics is very obvious.

How does one imagine the curvature of spacetime in 3D?

2 Answers2