# Coordinate systems in General Relativity

In General Relativity spacetime is a four dimensional Lorentzian manifold $M$ with metric tensor $g$. Being a smooth manifold, we can use coordinate systems on spacetime.

And in reality, most traditional GR exposures rely heavily on coordinate systems, so that everything is written in a specific coordinate system $(x^0,x^1,x^2,x^3)$ always. This is so common that one usually referes to the metric tensor just as $g_{\mu\nu}$, that is, by its components in a particular chart.

The trouble is that we don't know what $M$ is! I mean, we don't know at first what manifold spacetime is. This brings a problem: how does one define coordinate systems in the first place if one doesn't know spacetime fully?

A coordinate system is a chart $(U,x)$ on the open set $U\subset M$, being $x : U\to \mathbb{R}^4$ a homeomorphism. How can one build such a coordinate system if one doesn't know what actually is $M$?

I'll give just a simple example. The Schwarzschild metric is usually written as:

$$ds^2=-\left(1-\dfrac{2GM}{r}\right)dt^2+\left(1-\dfrac{2GM}{r}\right)^{-1}dr^2+r^2d\theta^2+r^2\sin^2\theta d\phi^2.$$

Since we don't know what manifold is spacetime, how can we make sense of the coordinate functions $t,r,\theta,\phi : M\to \mathbb{R}$?

This resembles spherical coordinates, but this isn't $\mathbb{R}^3\times \mathbb{R}$ after all!

So on the one hand it seems hard to define coordinate systems because we don't know what manifold is spacetime.

On the other hand, given one observer $\gamma : I\subset \mathbb{R}\to M$, we always have one coordinate system adapted to $\gamma$. In the same way, given a reference frame $e_\mu : M\to TM$ we can also construct a coordinate system adapted to it if I'm not wrong.

So since we don't know what manifold spacetime is, all the relevant coordinate systems for Physics in GR always comes from observers and reference frames?

If not, how to make sense of coordinate systems in GR, considering all these points?

• On the level of smooth manifolds, one can not distinguish, for example, a plane and a wavy surface. What makes them different is the metric, and the metric is the primary object to be determined in GR. The metric is defined locally and can be studied locally. Therefore, one does not care what the chart is, but rather what the metric is in the coordinate system. (in fact in the context of abstract manifold, a single chart does not carry much information, but the transition functions between the charts do. ) If it is still not clear, I can make it a more rigorous answer. – user110373 Mar 24 '17 at 2:49
• user1620696: "The trouble is that we don't know what $M$ is!" -- Perhaps a bit more ... stringently: Considering a set of events ("as such") we don't know its topology (which subsets to call "open", or "neighborhood") to begin with; and considering any non-empty subset $U_a \subseteq M$ together with two (invertible) assignments (of coordinates, without any regard to topology) $$\phi_j : U \rightarrow \mathbb R^n, \qquad \phi_k : U \rightarrow \mathbb R^n,$$ such that $$\phi_j \circ \phi_k^{-1} : \mathbb R^n \rightarrow \mathbb R^n \text{ is NOT a Homeomorphism},$$ [... contd.] – user12262 Mar 24 '17 at 6:23
• ... (i.e. not a homeomorphisms wrt. the "natural topology of $\mathbb R^n$"), then we can't say whether $\phi_j$ is not a homeomorphism, or whether $\phi_k$ is not a homeomorphism, or whether neither is a homeomorphism; much less which assignment of coordinates is a homeom. wrt. $M$. "given a reference frame $$e_{\mu} : M \rightarrow TM$$ we can [...]" -- Being already in trouble since "we don't know what $M$ is!", how could we possibly "know what $TM$ is" ?? (IMHO, that's asking for even more trouble; and a misuse of "reference frame".) – user12262 Mar 24 '17 at 6:24
• Possible duplicate of Global Properties of Spacetime Manifolds – JamalS Mar 24 '17 at 15:17
• @JamalS, although related I don't believe this is a duplicate. I explain the reason: if I understood well, the OP is asking there about topological properties of spacetime and how they relate to the solutions to Einstein's equations. Here the question is another: how does one construct coordinate systems and give meaning to coordinate functions in the context of GR when $M$ isn't known even as a set, let alone as a topological space/manifold? I also am trying to discuss whether or not all GR coordinate systems are construct adapted to some observer or frame of reference. – user1620696 Mar 24 '17 at 15:54

Given a $n+1$ dimensional manifold $M$ one by definition has charts or coordinates that are homeomorphic to $\mathbb{R}^{n+1}$. This is independent of any Lorentzian or Riemaniann metric on $\cal{M}$.

Now if the manifold $M$ admits a Lorentzian metric $g$ then the coordinates use to define the manifold also define the components of the metric.

A common way to define a spacetime even if it is not know the whole manifold $M$ is to work locally. Even if the topology of $M$ is not $\mathbb{R}^{4}$, locally it is. Then using the local (flat) coordinates one define locally a metric and imposing some symmetries and physical conditions one can arrive to relevant metrics such as the Schwarzschild metric.

However, another way to find solutions to Einstein's equations is to see them as an initial value problem. That is given on a hypersurface $\Sigma$ the first and second fundamental form one can determine a Lorentzian manifold $M$ using Einstein's equations.

Regarding your comment of how to make the sense of coordinates. In this case we have chosen an initial n-dimensional $\Sigma$ where we know the coordinates by definition. Then, also one define a lapse function $N$ and shift vector $\beta$ on $\Sigma$. These choices define locally a chart with topology $M={\mathbb{R}} \times {\Sigma}$. The choice of the lapse and shift sometimes are related with physical observers, but others are used for mathematical convenience such as the harmonic slicing. See section 9 This define the $n+1$- dimensional coordinates on $M$. In fact this is the topology of all globally hyperbolic spacetimes.

However, this is only a local characterization. If one wants the total manifold then one is interested in maximal extensions. For example, the chart you used has a well know coordinate singularity at $r=2M$ and therefore one needs to change the chart and coordinates to cover the full spacetime. Notice that the existence of the singularity is responsible for the change in the global topology.

The concept of maximal extension is related to geodesic completeness or the well-posedness of Einstein equations (no curvature singularities). If this two criteria are equivalent is part of the Strong Cosmic Censorship. The global case which correspond to the question of the maximal extension of the space-time as as a well-posed problem of Einstein's equations is an active area of research. The Strong Cosmic Censorship deals with the uniqueness of global solutions.

• yess: "Given a $n+1$ dim. manifold $M$ one by definition ha[s] charts or coordinates that are homeomorphic to $\mathbb R^{n+1}$." -- True. But the OP doesn't necessarily presume manifolds which are given so specificly. What is given instead? (@user1620696 may clarify.) Meanwhile my own take: Given is a set of events together with "physical data" (identities of observers $\gamma$, their coincidence events, their (past) lightcones), and: the assumption or promise that some Lorentzian topology can be determined uniquely for this event set, from this data. Then: By which method, explicitly?? – user12262 Mar 29 '17 at 17:19
• is something like this aip.scitation.org/doi/10.1063/1.523436 what you have in mind? – yess Mar 30 '17 at 3:04
• yess: "is something like this { D.B.Malament, The class of continuous timelike curves determines the topology of spacetime } what you have in mind?" -- Upon taking a closer look at it: yes, indeed, thanks for the suggestion! (From the first sentence: $$\text{Suppose one has two spacetimes } (M,g)\text{ and } (M',g')\text{ together with a bijection ...},$$ as well as from summaries elsewhere, I had been under the wrong impression that Malament's results required detailed knowledge of $g$, and not just of "observational causal relations".) – user12262 Mar 30 '17 at 18:38