What is really curved, spacetime, or simply the coordinate lines?
Congratulations! You stumbled upon an important question of differential geometry:
How can I know whether the curvature is caused by my choice of coordinates or the space I live in?
As has been mentioned in other answers, the word “curvature” is referred to as either a property of the space, but also a property of the coordinates. Let me call the latter “variation” instead.
To illustrate both cases, imagine:
- Being in “flat” Euclidean space, but using spherical coordinates
- Living on the sphere, using any kind of coordinates.
In the first case, obviously a change to cartesian coordinates eliminates all variation in your coordinates. In the latter, you can choose any representation you want – you will not get variation-free coordinates! For instance, the closer you get to the poles, your coordinates are forced to get “denser”, if they shall stay continuous.
This means it must be caused by the space itself - If coordinates fail to get straight, we say the “Space has curvature”. Curvature is also said to be an “intrinsic property of the space”, meaning exactly that this property does not depend on its representation by coordinates.
To answer your question briefly: No. When saying “spacetime is curved”, we mean “Spacetime has curvature”, and not only “The coordinates vary”.
Some definitions
Note however that the vocabulary is extremely vague. To be more precise, we need to use the mathematical terms: Our “space” or “spacetime” becomes a “Riemannian Manifold”, namely an abstract mathematical set with some nice properties and the ability to measure distances locally. The latter is called the “Metric tensor field”.
“Coordinates” Are actually maps from our Manifold to $\mathbb R^n$, in the case of spacetime $n=4$. Wherever you are, you will find a map giving you a set of real numbers.
Once you introduce a coordinate map, you have a basis for the metric tensor and can represent it by multiple components which are real numbers. That is extremely useful, since we now can easily take derivatives of it (in the directions of our coordinate basis). If these derivatives are zero everywhere, you already know you are in a flat space.
“Curvature” is not so easy to define however. We need to find tools to measure the failure of our coordinate maps to become constant. Luckily, there have been people such as Gauss and Riemann doing the hard work for you.
Gauß' Approach
Gauß' approach is to compare how “circles grow”. If you are on a sphere, the ”perceived radius“ of a circle is slightly larger than the radius corresponding to its circumference / area, so you know you are in a curved space. More precisely, in a Space with positive curvature – the radius can be shorter than expected, as well! Consider a saddle. Since the circle is “stretched”, the circumference and area are larger than expected – this would be an example of negative curvature. A nice mental picture for $n=2$ is that if you tried mounting a sheet of paper, and observe that:
- It rips: Negative curvature
- It fits nicely: Zero curvature
- It squeezes: Positive Curvature
The Problem with Gauß' Approach is although it is intuitive when looking from “outside” at the manifold, determining it from inside the manifold involves taking a limit, and it is not so easy to compute and generalize.
Well, not as easy as the way Riemann did it at least:
Riemann's Approach
Take the sphere: A most famous effect of our world's curvature is the fact you can span a triangle with angles $\frac \pi 2$ only.
Another possibility is parallel transport - if you take a vector and go straight up to the north pole, then straight to your right to the equator, and straight down, your vector shifted by $\frac\pi 2$.
This can be generalized: Take a vector, parallel transport it some distance up, some distance to the right, go back down and back left. In a flat space, the vector wouldn't have changed. In a curved space however, we would observe a shift.
Now note that the notion of “up” and “right” can easily be generalized into the idea of following two coordinate vectors! This is the Idea of the Riemann Tensor: $$R(u,v)w=\nabla_u\nabla_v w - \nabla_v \nabla_u w - \nabla_{[u,v]} w$$ This is essentially implementing the following protocol:
- Take a vector $w$
- Transport it in the direction of the vector (in our case: a coordinate vector) $u$, then $v$
- Transport the same vector in the direction of then $v$, then $u$, plus a correction term that's there for technical reasons
- Observe how the difference in paths made our vector differ.
However, not quite. Since the displacement vector depends on the distance, and we want to define a value of the curvature locally, in this case as a property of the point, shrinking the distance makes the displacement vector goes to zero. So our argument is not quite correct – we are interested in the linear change of said displacement vector when changing the distance.
We can compute the quantity for each pairs of the $n$ coordinates (indices: $\mu, \nu$), and can then observe the $\rho$-component of a unit vector in direction $\sigma$ – let's denote this quantity by $R^\rho{}_{\sigma\mu\nu}$. It has some symmetries, so we actually have $\frac{n^2(n^2-1)}{12}$ independent components (I trust wikipedia on that one). This tensor can be contracted to a smaller one by summing over same $\rho$ and $\mu$, leaving two indices, which can be contraced once again, leaving a scalar $R$, also known as the Ricci Scalar, which is, surprise, in two dimensions twice the gaussian curvature. So Riemannian curvature does seem to capture the right intuition nonetheless!
The equation you saw above can be reduced to first and second partial derivatives of the metric tensor – which is really easy to evaluate (at least if you know the closed form). Remember that the tensor (and obviously derived contractions such as the Ricci scalar) contain a lot of terms; calculating the riemann tensor is a well-beloved exercise for the eager student (or the poor soul willing to pass a class on differential geometry.
Summary
What is meant is the intrinsic curvature of the space, meaning it is independent of the choice of coordinates. There are clever methods of determining whether and to what extend your space deviates from flat euclidean space, namely Gaussian curvature, and, more importantly, the Riemann tensor.
Both, actually. (of course these are completely different, but both are called "curvature")
Coordinates are most definitely curved (that is why they are called curvilinear after all).
But there is a coordinate-independent notion of curvature for the spacetime geometry. This is given by the Riemann curvature tensor.
You probably know that it is equal to zero in flat spacetime. Note how this holds in all coordinate systems - both curvilinear and Gallilean. This is because tensor equations are covariant under coordinate transformations.
Because of this, it is considered a property of spacetime (since it does not depend on coordinates). There is a nice, coordinate-independent way to get a feel of what curvature is: when you parallel-transport a vector along a closed curve, the difference between the original vector and the result of the transformation is nonzero in the presence of curvature.