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On the initial courses of topology and differential geometry, we learn again and again about charts, and atlas, and transition maps. I feel that transition maps are a very powerful idea, because they claim how and where two neighbouring coordinate patches agree on how to describe geometry.

When working on General Relativity on a spacetime with two or more coordinate patches, I expect some type of "join" conditions on the overlap between the charts, with the pattern:

$$ \text{observable geometric quantities on chart i} = \text{observable geometric quantities on chart j}$$

I wouldn't necessarily expect the metric components to look the same on the transition map, but say, Ricci scalars, or any full contractions of tensors with vectors defined on the transition map region should match on the transition map.

Why would I expect this? we'll we already do such things in the rest of physical systems, when we use boundary conditions to connect different regions with different materials and geometry. The overall allowed solutions should satisfy all boundary conditions. Something akin to this should also be done in General Relativity, but again, I cannot find any examples where something like this is done (at least not explicitly). Yes, there is such thing as boundary conditions on General Relativity, but this is typically limited to asymptotic boundary conditions, and there is no general discussion or even mention whatsoever of boundary conditions between charts.

Am I missing the steps where this is being done? or perhaps I am not looking at any actual problems that require this?

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  • $\begingroup$ I'm not really sure what this question is asking for - yes, if you want to talk about coordinates in the way mathematicians talk rigorously about coordinates, we'd have to talk more about transition maps. But physicists are not mathematicians, and they rarely talk in terms of rigorous differential geometry. Are you looking for physics texts that use the term "transition function" or for a mathematical reinterpretation of what the physicists are doing? (For the latter, see e.g. this answer of mine) $\endgroup$
    – ACuriousMind
    Commented Jul 14 at 10:11
  • $\begingroup$ @ACuriousMind most of my previous question was all over the place and not very focused, I rewrote it almost entirely. $\endgroup$ Commented Jul 20 at 15:09

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There's nothing fancy going on with transition maps. What a mathematician means by a "transition map", in the setting of general relativity, is nothing more or less than a coordinate change map between two overlapping regions of space-time with two distinct coordinate systems.

So if, as you say, you have worked through the coordinate change maps under which the Schwarzschild metric is converted into the Eddington-Finkelstein metric, then you already have a fine understanding of a transition map between different coordinate systems.

To spell it out a bit more, based on this wikipedia article, the Schwarschild metric is defined in a coordinate system with $(t,r,\theta,\phi)$ coordinate. However, the Schwarzschild metric is undefined when $r=0$ and when $r=2GM$. So one may regard this metric as having two disconnected coordinate charts, one with $0<r<2GM$ and the other with $r>2GM$.

On the other hand, the Eddington-Finkelstein metric comes in two versions defined on two separate coordinate systems:

  • The ingoing Eddington-Finkelstein coordinates $(v,r,\theta,\phi)$ have a coordinate change map with the Schwarzschild coordinates of the form $$v = t + r_* = t + r + 2GM \ln|2GM-r| $$ This is simply a transition map between the Schwarzschild coordinates in the region $0<r<2GM$ and the ingoing Eddington-Finkelstein coordinates.
  • The outgoing Eddington-Finkelstein coordinates $(u,r,\theta,\phi)$ have a coordinate change map with the Schwarzschild coordinates of the form $$u = t - r_* = t - r -2GM \ln|2GM-r| $$ This is also simply a transition map between the Schwarzschild coordinates in the region $r>2GM$ and the outgoing Eddington-Finkelstein coordinates.
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  • $\begingroup$ After reading your answer, I pondered why I was still dissatisfied, and it helped me to make a good enough rewrite of the question that is a lot more precise and focused than my earlier attempt. Apologies for being all over the place $\endgroup$ Commented Jul 19 at 20:18

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