Physics Asked on May 8, 2021
Consider then the following reasoning:
Schwarzschild Metric describes the spacetime of a black hole. A gravitational collapse is a mechanism to produce Schwarzschild Black Holes. Conversely, a Schwarzschild metric describes exterior spacetime prior the collapse as well.
Now, I read [1] that this reasoning cannot be applied to Kerr black holes. I mean, Kerr metric does not describe the spacetime of a rotating body, just a rotating black hole. But a star is never a spherical body, and the collapse is never perfectly spherically symmetrical. Therefore the resultant black hole has some rotation, and therefore the spacetime is a Kerr one.
Why cannot we say that the exterior spacetime of a rotating body (not supposed to be a black hole) is described by Kerr solution?
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$[1]$ RAINE.D, THOMAS.E; Black Holes. Imperial College Press. pg 132. 2015.
First of all, the Schwarzschild metric is the most general spherically symmetric vacuum solution to the Einstein field equations. A Schwarzschild black hole is a black hole that, having neither electric charge nor angular momentum, is described by the Schwarzschild metric.
The Kerr metric has a few "problems" and cannot be used to describe realistic stars apart from asymptotically far away. That is because realistic stars have:
A black hole has neither an interior nor it is undergoing gravitational collapse. So the Kerr metric may well describe the spacetime outside a rotating one. Hence the name Kerr black hole.
Answered by SuperCiocia on May 8, 2021
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