It is often said that, according to general relativity, spacetime is curved by the presence of matter/energy.

But isn't it simply the coordinate lines of the coordinate system that are curved?


5 Answers 5


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:

  1. Being in “flat” Euclidean space, but using spherical coordinates
  2. 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:

  1. It rips: Negative curvature
  2. It fits nicely: Zero curvature
  3. 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:

  1. Take a vector $w$
  2. Transport it in the direction of the vector (in our case: a coordinate vector) $u$, then $v$
  3. Transport the same vector in the direction of then $v$, then $u$, plus a correction term that's there for technical reasons
  4. 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.


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.

  • $\begingroup$ Great description of how those different Tensors arise, and how you can determine if you have a Gaussian curvature, easily (if you're willing get to do all those calculations) with R. But R is just one scalar number of other possible scalars for the curvature, and R can be zero while you still have curvature. I'm sure you know. Worth mentioning you'd have to calculate a number of scalars for less obvious spaces $\endgroup$
    – Bob Bee
    Nov 5, 2016 at 5:48
  • $\begingroup$ It is a pleasure to read your explanations. To illustrate the triangle with 3 times 90° -if you don't mind- there are sketches like here images.google.de/imgres?imgurl=http://www.solidaritaet.com/… $\endgroup$ Nov 5, 2016 at 6:07
  • 1
    $\begingroup$ This is nice mathematics. But are there scientific evidences that variation-free coordinates cannot be obtained for our universe? $\endgroup$
    – Bob
    Nov 5, 2016 at 22:23
  • 4
    $\begingroup$ "Gauß" in an English language answer seems rather affected. $\endgroup$
    – J...
    Nov 7, 2016 at 14:37
  • 2
    $\begingroup$ @J... You can call it either “German Pride”, or “having a fancy keyboard layout”. ;-) $\endgroup$ Nov 24, 2016 at 23:54

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.

  • $\begingroup$ +1 took me three reads to get your point, shows you how inbuilt assumptions can lead you astray. My earlier comment was way off the mark. Thanks for posting a concise answer. $\endgroup$
    – user108787
    Nov 5, 2016 at 0:06
  • $\begingroup$ +1 for a simple and to the point answer. It occurred to me that I have never thought about what are the quantitative criteria to call a coordinate curved? The Christoffel symbols? $\endgroup$
    – ACat
    Nov 5, 2016 at 10:17
  • $\begingroup$ I am confused because @Moonraker's answer states the opposite of yours. Do you see anything wrong within his or her answer? $\endgroup$
    – Bob
    Nov 5, 2016 at 22:33
  • $\begingroup$ @Bob I do, actually. There is no coordinate transformation to pass from Minkowski metric to the Schwarzschild metric, because these are physically different. Try calculating the Riemann tensor for the Schwarzschild metric and Minkowski metric and you will see that those are different. $\endgroup$ Nov 5, 2016 at 23:53

Regardless of the coordinate system used, matter curves space.
You can choose "Galilean," "Riemmannian," "Einsteinian," etc. coordinate system that you find more useful, but the fact remains that matter curves space.
So, to answer your question, it is space that is curved, not the coordinate lines.


Curved co-ordinates on flat spacetime correspond to accelerating observers, not gravity.

The first physical insight of general relativity is that when you have gravity, you have no globally inertial frames -- contrast this with flat space, where you can always construct a linear co-ordinate system. The second physical insight is that you do have locally inertial frames, specifically the freefalling ones -- this is the "equivalence principle" -- so the manifold you use to model spacetime must necessarily have local flatness. Consequently, (pseudo-)Riemannian manifolds become the right way to model spacetime in general relativity.

This is why Christoffel symbols exist for accelerating observers on flat spacetime too -- they're first-order in the derivatives of the metric, and so can be eliminated by transforming into a flat co-ordinate system where the metric is constant (this is okay because the Christoffel symbols aren't tensors). The Riemann curvature tensor, on the other hand, is second-order in the derivatives of the metric and cannot be eliminated by a co-ordinate transformation.

  • $\begingroup$ Please answer the question, or else it is just a comment. $\endgroup$
    – Bob
    Jan 6, 2019 at 12:30
  • $\begingroup$ @Bob I did answer the question. Curved co-ordinates correspond to acceleration (and are compatible with special relativity), whereas curved spacetime corresponds to actual gravity. That's in the first two lines.The rest of the answer is an explanation for why this makes sense -- did you not understand it? $\endgroup$ Jan 6, 2019 at 12:32
  • $\begingroup$ Curved co-ordinates on flat spacetime correspond to accelerating observers This is false. In general, coordinates don't have anything to do with observers. $\endgroup$
    – user4552
    Jan 6, 2019 at 14:31
  • $\begingroup$ @BenCrowell Ok, then accelerating observers in flat spacetime correspond to curved co-ordinates. But that's not really the point of the answer, or the question. $\endgroup$ Jan 6, 2019 at 20:37

Spacetime curvature is not a physical law, it is simply a very powerful and practical model Einstein introduced for the work with Einstein's field equations.

One main application of curved spacetime is the Schwarzschild metric $$ \mathrm ds^2 = -\left(1 - \frac{2GM}{c^2 r}\right) c^2~\mathrm dt^2 + \frac{1}{1 - \frac{2GM}{c^2 r} }~\mathrm dr^2 + r^2 \left(\mathrm d\Theta^2 + \sin^2 \Theta ~\mathrm d\Phi^2\right)$$ Schwarzschild metric is describing a gravity field which may be represented in the form of curved spacetime.

In contrast, the corresponding Minkowski metric (with flat spacetime) is $$ \mathrm ds^2 = -~ c^2~\mathrm dt^2 + \mathrm dr^2 + r^2 \left(\mathrm d\Theta^2 + \sin^2 \Theta~\mathrm d\Phi^2\right)$$

where $\mathrm dt$ is uncurved time and $\mathrm dr$ is uncurved radial displacement.

Comparing both, you find that in the Schwarzschild metric, time $\mathrm dt$ is multiplied by the constant

$$ \sqrt{1 - \frac{2GM}{c^2 r}}$$ and space $\mathrm dr$ is divided by the same constant. It is exactly this factor which represents the spacetime curvature. The constant is gravitational time dilation. If we set the constant = $C$, we can write Schwarzschild metric shorter in the following way: $$ \mathrm ds^2 = -~c^2 (C~\mathrm dt)^2 + {\left(\frac {\mathrm dr}{C}\right)}^2 + r^2 \left(\mathrm d\Theta^2 + \sin^2 \Theta~\mathrm d\Phi^2\right)$$ Comparing this short form with the equation above of Minkowski metric, the Schwarzschild metric differs from uncurved Minkowski metric only by one coefficient $C$ which is identic with gravitational time dilation. That means that the curved spacetime of Schwarzschild metric may also be described in terms of gravitational time dilation - in absolute, flat space!

So we may describe gravity with flat space coordinates where only gravitational time dilation would act on the flat metric. But as mentioned above, the representation in the form of curved spacetime revealed as far more practical, and it is widely preferred to the description in terms of flat space. But curved spacetime is nothing more than a choice of coordinates.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – rob
    Nov 7, 2016 at 13:12

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