My question is regarding Schwarzschild solution. I always heard that coordinate chart is a one to one map from a manifold to real numbers. But when we look inside the black hole using Kruskal coordinates we see that each pair (r,t) in Schwarzschild coordinates appears in 2 distinct places on the diagram. Each point in the interior of black hole with a certain value of (r,t) has its counterpart inside the white hole. How can a ceratin point (r,t,$\phi$,$\theta$ ) specify 2 different places on a manifold? Is this a result of coordinate singularities?
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3$\begingroup$ This is an abuse of notation. They are different charts, defined in different open sets, which one for simplicity decides to use the same notation for. $\endgroup$– GoldJan 23, 2020 at 19:03
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$\begingroup$ There is a thing worse than that. We have an entire line in Kruskal diagram that is t=infinity and r=2GM. So entire line corresponds to a single coordinate. $\endgroup$– Timur9717Jan 23, 2020 at 20:12
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Coordinate charts are always one-to-one from the manifold to some open subset to $\mathbb{R}^n$. This is because we use charts as local homeomorphisms to define what a manifold is (it can be defined in other, more abstract ways, e.g. locally ringed spaces, but let's not get into that).
Let's remark another important point: in general relativity we only know charts, we never ever "touch" the global manifold, and this is because physics knows about local stuff. For example, we don't know about the global topology of the universe, although some people think that it's a Poincarè sphere.
Going back to the question, when we have a change of charts with singularities it can mean one of two things: the manifold has a true (curvature) singularity or the change of charts is extended beyond the range of applicability, i.e. a coordinate singularity. This can easily be seen when we consider the change of charts between cartesian coordinates and polar coordinates: the origin is not really valid point in those coordinates, since one can arbitrarily change the angle(s) but nothing changes.
If this is so, why do we do this? Well because it's useful, and typically it's not a problem, at least when the set of troublesome points is "small enough" (measure zero).
