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The Kruskal–Szekeres coordinates also apply to space-time around a spherical object, but in that case do not give a description of space-time inside the radius of the object. Space-time in a region where a star is collapsing into a black hole is approximated by the Kruskal–Szekeres coordinates (or by the Schwarzschild coordinates).
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).
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).
From Wikipedia, the free encyclopedia. Redirect page. Redirect to: Kruskal–Szekeres coordinates
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.
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,
GTG does, however, make different predictions about global solutions. For example, in the study of a point mass, the choice of a "Newtonian gauge" yields a solution similar to the Schwarzschild metric in Gullstrand–Painlevé coordinates. General relativity permits an extension known as the Kruskal–Szekeres coordinates. GTG, on the other ...
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.