# Curved-spacetime, but why curved coordinates?

I have just started self-studying General Relativity with Landau-Lifshitz' Classical theory of Fields, supplementing it with a bit of extra math here and there ( I have an experimental degree).
He introduces curved spacetime by stating that in a non-inertial reference-frame, the space-time interval ds in curved, since it is described by a general quadratic form:

ds$^{2}$ = g$_{ik}$.dx$^{i}$.dx$^{k}$

That much is obvious to me sinceif one simply takes ds$^{2}$ and graphs it it would sketch a curve. However, it is then stated that the coordinates are curvilinear.

Why is that? Why cant the above equation simply represent a curve, but in orthogonal Cartesian coordinates?

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PeterDonis
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Why cant the above equation simply represent a curve, but in orthogonal Cartesian coordinates?
Because there is no such thing as "orthogonal Cartesian coordinates" if the metric is not the Minkowski metric, i.e., if ##g_{ik} \neq \eta_{ik}##. In other words, the definition of "orthogonal Cartesian coordinates" assumes an inertial reference frame.

• vanhees71
Nugatory
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Why is that? Why cant the above equation simply represent a curve, but in orthogonal Cartesian coordinates?
Consider a two dimensional surface with coordinates ##x^1## and ##x^2##. There is a curve of constant ##x^i## passing through each point. If you want orthogonal you want these two curves to be perpendicular at every point. Latitude and longitude on the surface of the spherical earth are an example, and clearly these can’t be Cartesian.

pervect
Staff Emeritus
I have just started self-studying General Relativity with Landau-Lifshitz' Classical theory of Fields, supplementing it with a bit of extra math here and there ( I have an experimental degree).
He introduces curved spacetime by stating that in a non-inertial reference-frame, the space-time interval ds in curved, since it is described by a general quadratic form:

ds$^{2}$ = g$_{ik}$.dx$^{i}$.dx$^{k}$

That much is obvious to me sinceif one simply takes ds$^{2}$ and graphs it it would sketch a curve. However, it is then stated that the coordinates are curvilinear.

Why is that? Why cant the above equation simply represent a curve, but in orthogonal Cartesian coordinates?
I haven't read the section in question, but my reaction is that the author is likely distinguishing curved space-time (the Riemann tensor being nonzero), from curved coordinates, where the ##g_{ij}## are not
{##g_{00}=-1, g_{11}=g_{22}=g_{33}=1##}, which I will abberviate as ##g_{ij}## = diag(-1,1,1,1).

I'm not sure if you've gotten to the section on the Riemann tensor yet - probably not. But it's possible in to have a perfectly flat space-time with the ##g_{ij}## other than diag(-1,1,1,1). There are many examples - spherical coordinates are one of the most familiar, perhaps. Another example would be the Rindler coordinates of an accelerated frame.

However, it is then stated that the coordinates are curvilinear.

Why is that?
I read sentences below (82.1) in section 82. I interpret as below.

#1
Wiki curniliniar says,
"In geometry, curvilinear coordinates are a coordinate system for Euclidean space in which the coordinate lines may be curved. Commonly used curvilinear coordinate systems include: rectangular, spherical, and cylindrical coordinate systems."
We are familiar with examples of rectangular, spherical and cylindrical coordinates where metric is not constant but function of coordinates, e.g. ##r^2 sin^2\theta##.

#2
The above cases are kind of "disguised" curved space because by simple transformations the geometry reduces to Euclid and physically Galilean in Newtonian mechanics or Minkowsky in SR. But as L-L say in latter part of section 82, GR does not allow such global reduction to Galilean but only infinite small local Galilean space. That means even we apply Cartesian type x,y,z for coordinates, metric g cannot be constant but function of coordinates x,y,z (and t) in general. In this sense GR coordinates resemble curviliniar coordinates in #1. We expect making use of mathematics tools developed in curvilinear coordinates where Euclid space exists for GR where no Galilean space exist.

#3
"Disguised" curved space include rotating system and accelerating system (Rindler). By simple transformations we can go to flat Minkowsky space-time. Curvature R=0 everywhere.
"Real" curved space include gravity of star ( Schwartzshild) where by equivalence principle Minkowsky space-time is realized only locally. R##\neq##0 at least somewhere.

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