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The transformation between Schwarzschild coordinates and Kruskal–Szekeres coordinates defined for r > 2GM and < < can be extended, as an analytic function, at least to the first singularity which occurs at =. Thus the above metric is a solution of Einstein's equations throughout this region.
In these coordinate systems, outward (inward) traveling radial light rays (which each follow a null geodesic) define the surfaces of constant "time", while the radial coordinate is the usual area coordinate so that the surfaces of rotation symmetry have an area of 4 π r 2.
An example is the apparent (longitudinal) singularity at the 90 degree latitude in spherical coordinates. An object moving due north (for example, along the line 0 degrees longitude ) on the surface of a sphere will suddenly experience an instantaneous change in longitude at the pole (i.e., jumping from longitude 0 to longitude 180 degrees).
Kruskal–Szekeres coordinates, a chart covering the entire spacetime manifold of the maximally extended Schwarzschild solution and are well-behaved everywhere outside the physical singularity, Eddington–Finkelstein coordinates, an alternative chart for static spherically symmetric spacetimes,
The metric singularity is not a physical one (although there is a real physical singularity at =), as can be shown by using a suitable coordinate transformation (e.g. the Kruskal–Szekeres coordinate system).
A coordinate singularity occurs when an apparent singularity or discontinuity occurs in one coordinate frame, which can be removed by choosing a different frame. An example of this is the apparent singularity at the 90 degree latitude in spherical coordinates. An object moving due north (for example, along the line 0 degrees longitude) on the ...
How can it be seen that the Kruskal-Szekeres metric has an horizon? It could be useful to indicate that. Similarly, in the article on Schwarzschild metric, the horizon seems obvious because of the apparent singularity. But what is the real way to look for an horizon in a metric? (Partial) answer. It depends what kind of horizon you are looking for.
The constant tachyonic geodesic outside is not continued by a constant geodesic inside , but rather continues into a "parallel exterior region" (see Kruskal–Szekeres coordinates). Other tachyonic solutions can enter a black hole and re-exit into the parallel exterior region.