I was thinking about how to explain curvature in 3d space, and I decided a good approach was to first describe curvature in 2d space, then generalize a bit. Curvature in 2d is simple, since there is only one non-zero component of the curvature tensor. We can thus think of the curvature of a 2d surface as being a single number.
Let's start with manifolds. Manifolds are a mathematical entity, which we can visualize most easily via an embedding. In particular, we can regard the surface of a 3d object as a 2d manifold. As a mathematical entity, manifolds exist without such an embedding, which is basically a visual aid. But this visual aid makes things easier.
Let's continue with the curvature of a sphere, one of the simplest 2d cases. In GR, when we talk about curvature, we talk about intrinsic curvature. One method of describing curvature is via geodesic deviation. Another method is to consider the angular excess, or angular deficits, of geodesic triangles, but I won't be getting into that here in detail. I will refer the interested reader to the wikipedia on spherical trignometry and "angular excess",
https://en.wikipedia.org/w/index.ph...ry&oldid=1051975245#Area_and_spherical_excess.
I also won't discuss how to draw geodesics in general, I will simply utilize the fact that geodesics on the sphere are great circles. Motivationally, it is sufficient to regard spatial geodesics as curves which are the shortest distance between two nearby points on some manifold. Since we are using embeddings as a visual aid, this is the shortest distance of a curve lying entirely in some surface. This is not the best way, and it actually only works when the points are sufficiently close together. "Antipodal points" are an example of where the idea breaks down when the points are too far apart. There can be occasions where there is more than one straight line between two antipodal points on a surface, and they are not necessarily of the same length. We actually need to add a constraint on the size of the region to make this idea work. And it turns out there are better ways - I will mention that the better way involves a concept called a "connection", but I won't get into more detail than that mention here.
On the sphere, geodesics are simple - they are great circles. If a space is flat, geodesics that start out parallel remain the same distance apart. On the sphere, we can see that space is not flat. If we imagine the set of great circles running through the north and south poles, we can see that these geodesics are all parallel to each other at the equator, as the geodesics go north-south, and they are everywhere orthogonal to the equator, which runs east -west. However, the initially parallel geodesics at the equator do not remain at a constant distance apart, in fact they meet at the north and south poles. This is how we know the surface of a sphere is curved - if it were flat, parallel geodesics would stay at a constant separation.
We want to parameterize the geodesics by some affine parameter. Because we are considering only spatial curvature, we can get away with regarded this affine parameter as time, though it is more rigorous to regard it as an affine parameter. If we regard the affine parameter as time, then the tangent vector at any point p of the geodesic is the velocity u of an object following the purely spatial geodesic. Note we are not considering space-time geodesics, just spatial geodesics, and we are re-purposing time as an affine parameter for ease of exposition. Thus we are not really treating time in the sense that it is used in special relativity, it's simply a parameter that identifies points on our curves. The notation I am using is not really standard, it's been heavily modified with the intent to be easier to grasp. Hopefully this will actually help, rather than cause more confusion, but - time will tell.
With this idea of time as an affine parameter in mind, we can say that the spatial geodesics "accelerate" towards each other. This gives rise to the next section, the geodesic deviation equation. The geodesic equation says that
$$\frac{D^2 x}{dt^2} = -R u^2 x$$
This says that the relative acceleration between objects following these spatial geodesics of the geometry in question (in this case, great circles on the sphere) is proportional to the square of the velocity, u, and the initial separation of the geodesics.
If we had more than two dimensions, we'd need tensor notation and it's associated indices, but with only two dimensions, we can oversimplify things by omitting all the indices, and still have a useful equation.
By dimensional analysis, we can see that R must have units of 1 / distance^2. R is equivalent to the Gaussian curvature, see for instance
https://en.wikipedia.org/w/index.php?title=Curvature&oldid=1047300795#Gaussian_curvature.
We can also use some Newtonian physics to get the same conclusion. The 3-d acceleration vector would be v^2/r, pointed out of the surface, towards the center of the sphere. The component that we are interested in is only the component that lies in the plane of the sphere (or more precisely, it's in the plane tangent to the sphere, the tangent plane). The compnent in the tangent plane is v^2 sin(theta) / r_sphere. We can approximate sin(theta) as (x/r_sphere), so we get that the acceleration of nearby geodesics towards each other is v^2 x / r_sphere^2. We can see this matches the geodesic equation I have previously presented, with R = 1/r_sphere^2.
This procedure illustrates how we can use the geodesic deviation equation to find the intrinsic curvature. We can construct geodesics (great circles) on the sphere, and perform all measurements on the sphere, without considering the geometry in which the sphere is embedded. The sphere has its own intrinsic geometry, the embedding allows us to leverage our knowledge of Euclidean geometry of the globe to understand the non-Euclidean geometry of the surface of the globe, the sphere.
Now that we have describe curvature in 2 dimensions via geodesic deviation, let's talk a little bit about three dimensions. If we have a set of unit length and orthogonal basis vectors at some point p, ##\hat{x}, \hat{y}, \hat{z}##, we can consider the xy plane, the xz plane, and the yz plane as each being a 2 dimensional surface. To be a bit more precise, each vector generates a geodesic, and the set of vectors in the xy plane generate a set of geodesics which lie in some surface. After generating this surface, we can measure it's curvature. We can do the same for the xz and yz surfaces as well.
Each of these generated surfaces can have its own curvature. This gives us three components of curvature for 3d space. This is actually not quite enough in general - but it IS enough if we pick our axes correctly. We can reduce the 6 components in the general case to three, by a proper alignment of our coordinate axes with the underlying geometry, though I won't attempt to prove that here.
The mathematical details underlying this are interesting, but a bit advanced. We take the 4x4x4x4 Riemann tensor, and with a bit of special pleading that works for 3 dimensions, reduce it to a 3x3 symmetric matrix. For instance, we can map R_(12)(12) to R'_33. After this mapping of the Riemann to a 3x3 real matrix, we note that the eigenvectors of any real symmetric matrix are orthogonal. Using these orthogonal eigenvectors, we can diagonalilze the matrix, leaving the eigenvalues as the diagonal elements. In this application , these diagonal eigenvalues are the three curvature parameters that I mentioned, the curavatures of the xy, xy, and yz planes.
