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Curvature of space-time (in General Relativity) is described using the metric tensor. The metric tensor, however, relies on the choice of coordinates, which is totally arbitrary. See for example answers to this question: https://physics.stackexchange.com/a/499297/374314

As the choice of coordinates is arbitrary, can't I just "postulate" to use cartesian coordinates to describe any possible spacetime?

Do I make a mistake by using cartesian coordinates?

Would I diminish the number of possibilities somehow by using Cartesian coordinates?

To the best of my knowledge, there shouldn't be any problem with using (postulating to use) Cartesian coordinates, as the choice of coordinates is totally arbitrary.

Every problem (spacetime) should be possible to describe in Cartesian coordinates.

Right?

If that were right it would be remarkable, because we are calculating so much with different coordinates AND different spacetimes and with only using Cartesian Coordinates we could easier think about the structure of spacetime itself.

Edit: Here, with "using Cartesian coordinates", I mean to globally define a coordinate system and define this to be the most simple one, Cartesian. Coordinates are arbitrary but influence the metric tensor the same as curvature does. I find it very difficult to think about spacetime curvature if I am not allowed to use a stable ground. That's why I want to know whether it's allowed or not to use a global coordinate system. The choice of coordinates and the curvature of spacetime both influence the metric tensor. However, there is only one real spacetime which it describes. I want to chose coordinates which make it possible to only let the curvature define the metric tensor. (To furthermore analyze then, which curvature is possible.)

With using spherical coordinates, for example, in the Schwarzschild solution, we simplify the problem which allows finding the solution easily. However, this comes to the cost of high symmetry of the problem. If I want to understand Einstein's field equations in general, I need (and want) to give up any of those symmetries (and, of course, this comes with the cost of the non-diagonal terms not to vanish)

In this question, Can we just take the underlying set of the spacetime manifold as $\mathbb{R^4}$ for all practical purposes?, which sounds similar to my question - but is similar only in the title,

  1. the topology censorship problem is touched (which is btw not solved yet). This problem asks whether the universal topology can be measured by an observer at all before it collapses. That says nothing about whether I can use Cartesian coordinates or not to describe the curvature. I need not to measure it to simply describe it mathematically.

  2. the OP questions whether manifolds are necessary at all. In contrast, I don't want to question the necessity of manifolds. I only want to use Cartesian coordinates to describe those manifolds to get rid of the double description of curvature.

My question is in contrast, "Can we describe every possible spacetime curvature using a metric tensor and Cartesian coordinates?"

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As the choice of coordinates is arbitrary, can't I just "postulate" to use cartesian coordinates to describe any possible spacetime?

If by cartesian coordinates you mean a set of four coordinates $\in \mathbb{R}^4$ (i.e., no vector space structure), then yes, you can introduce such coordinates in any generic spacetime.

However they might not be very useful as the resulting metric might become complex. This is because the metric will develope non-diagonal terms, which makes calculation much more difficult.

with only using Cartesian Coordinates we could easier think about the structure of spacetime itself.

It is true that certain coordinate systems make calculation easy, but it is not true that these will always be cartesian. Think about problems of spherical symmetry where you intentionally decide to work with spherical coordinates as the latters coordinates are simpler to work with.

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  1. If OP by Cartesian coordinates means a local coordinate system $(x^0,x^1,x^2,x^3)$ [say, in some local open neighborhood $U\subseteq M$ of spacetime] such that the components $g_{\mu\nu}$ of the metric tensor become constant, then that would mean that the Riemann curvature tensor vanishes in $U$. In other words, this ansatz would generically not be able/appropriate to describe curved spacetimes in GR.

  2. Assuming a globally defined coordinate system may unnecessarily open up a whole new can of worms well-known to e.g. cartographers trying to map Earth's surface. Instead differential geometers have long embraced the notion of an atlas of local coordinate systems.

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    $\begingroup$ In the context of GR, I'd assume (pseudo-)Cartesian coordinates not to mean ones where the metric components are constant. Instead I'd think of something akin to the quasi-galilean coordiantes used by Moller (doi.org/10.1016/0003-4916(58)90053-8)? I.e., ones that reduce to the Minkowski metric in Cartesian coordinates $(t,x,y,z)$ in the flat limit. $\endgroup$
    – Eletie
    Sep 24, 2023 at 10:52
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    $\begingroup$ Quasi-Galilean coordinates are indeed more flexible. $\endgroup$
    – Qmechanic
    Sep 24, 2023 at 11:01
  • $\begingroup$ Thank you @Qmechanic for this answer. Indeed I mean your possibility number 2, a globally defined coordinate system: Coordinates are arbitrary but influence the metric tensor. I want to let only the curvature define the metric tensor. I'll write an edit to the original question. $\endgroup$
    – Scibo
    Sep 24, 2023 at 18:27
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Since the question makes no reference to the number of dimensions, you could ask it just as well for a universe that is 2-dimensions of space and 1 of time. If you can't do it even there, then the answer is no, on principle.

So, suppose the universe is actually 2 spatial dimensions and shaped like the surface of the Earth. Now, your exercise reduces to that of making a Cartesian grid to cover the entire planet. Can you visualize what it is that might prevent you from being able to do so? I think it's much easier to see, here, than it is for three spatial dimensions with curvature.

Contrast that to the case where the two dimensional shape is a torus - a donut. There, the answer is yes, you can make a Cartesian grid to cover it.

Toroidal Planet
https://en.wikipedia.org/wiki/Toroidal_planet#See_also

But I don't think you're going to do it with anything that's like a sphere. Balls don't grid.

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    $\begingroup$ Balls grid well, you just need two coordinate patches $\endgroup$
    – Jojo
    Sep 25, 2023 at 5:33
  • $\begingroup$ For the universe being the "shape of the earth", or "shape of a torus" in 3+1 dimensions, you need additional dimensions to describe that, don't you? Or assuming that, at the end of the universe, you spawn at the other side - which is also against all odds, isn't it? $\endgroup$
    – Scibo
    Sep 25, 2023 at 8:00
  • $\begingroup$ Two, but not one. All manifolds grid (by definition), if you're going to allow more than one. So, that's a vacuous condition. It's one we're talking about here. $\endgroup$
    – NinjaDarth
    Sep 25, 2023 at 15:37
  • $\begingroup$ No, you don't need additional dimensions to describe a grid for a torus. You can see it right here. youtube.com/watch?v=okRZNrqPjUk The game-space. That's 2D toroidal, without any reference to any third spatial dimension. By the way - free download of its clone GitHub by my (former) padawan. github.com/LydiaMarieWilliamson/AsteroidCresta $\endgroup$
    – NinjaDarth
    Sep 25, 2023 at 15:40
  • $\begingroup$ More generally: curvature is an intrinsic property of a geometry that makes no reference to any outside dimension (I'll give you an example in a moment). There are "embedding theorems" that lay out which geometries of which intrinsic curvatures embed in higher dimensions, and how many are required. A flat 2D torus requires 4. It embeds in 3D only with curvature. The example: take the 6 distances between New York, Miami, Chicago and Houston. (N-M: 1090, N-C: 740, M-C: 1197, N-H: 1417, M-H: 964, C-H: 925) in miles. Make a 6-rod model to scale and try to connect them up and watch what happens. $\endgroup$
    – NinjaDarth
    Sep 25, 2023 at 17:15
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Why would you want to write it in Cartesian coordinates? Putting aside everything, the statement "... using Cartesian Coordinates we could easier think about the structure of spacetime itself." is completely the opposite of why different coordinates were introduced. Suppose you wanted to understand the causal structure of a black hole in Schwarzschild setting. You could, write it in terms of polar coordinates (not Cartesian), but you will realise that the $r=R_{\text{sch}}$ surface behaves badly, being a coordinate singularity (opposed to the $r=0$ curvature singularity which is not an artefact of bad coordinates). So you switch it with other sorts of coordinates, and the final maximally extended spacetime is in terms of Kruskal coordinates. You can then draw the Penrose diagram for this spacetime to capture all infinities (spacelike $i^{0}$, timelike infinities $i^{\pm }$ and null infinities $\mathcal{I}^{\pm }$). In this conformal setting, you can talk physics very nicely to capture many things (such as the holographic setting of the null infinity, islands in evaporating case, etc.). Now try replicating the same for Cartesian coordinates, and you will see why these coordinates are necessary. Most of the complications of using different metrics for different spacetimes have to do with conformal compactification of the spacetime, but this is a trivial task once you understand why you have to do it. AdS is another geometry where conformal compactification allows physics to be better understood, since you can find a CFT on the conformal boundary dual to the bulk gravitating theory (precisely in the sense of $\mathcal{N}=4$ SYM dual to type IIB string theory AdS$_{5}\times S^{5}$). Try doing these with Cartesian coordinates and you will find the answer.

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I think you have a mixup between global and local properties here. Using differential geometry language, a space-time is a 4-dimensional manifold with a Lorentzian metric.

It is a theorem in differential geometry that given a manifold and a single point on it, you can choose a chart such that at that specific point your chart gives just the standard cartesian coordinates. Away from this point the chart in general has no reason to be cartesian.

Coming from the other direction, if you have a manifold and a chart of it that on some open set is just cartesian coordinates then this implies that on this open set your manifold is isometric to flat space with the standard Lorentz metric.

It can be very useful to consider a chart that is cartesian at a single point but you can't assume that there exists a chart that is cartesian globally.

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  • $\begingroup$ I know that. I only want to use cartesian coordinates. The metric (and therefore, the chart as well) may be whatever it needs to be. I only don't want to change the coordinates. $\endgroup$
    – Scibo
    Sep 25, 2023 at 16:18
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You can choose arbitary coordinates but GR tensor algebra is based on general covariance so that 4D curvature is preserved even if you change your frame of reference. See: https://en.wikipedia.org/wiki/General_covariance.

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