The Schwarzschild solution presupposes also a zero cosmological constant.
No, at r=Rs there is no singularity, it can be avoided with a change of coordinates.
According to Birkhoff's theorem, the Schwarzschild metric is the only spherically symmetric solution of the Einstein field equations in vacuum.
As a generalization of Birkhoff's theorem, the Reissner–Nordström metric is the only spherically symmetric solution of the Einstein field equations with an electromagnetic field as source.
The Schwarzschild metric is a particular case (Q→0) of the Reissner–Nordström metric, and then we have a more general solution of the Einstein field equations.
We can't predict the properties of a Schwarzschild black hole, but other people can, see e.g.
Regge T. and Wheeler J. A. Stability of a Schwarzschild singularity Phys. Rev.108,1063,(1957).
Fortunately physicists use both, as Bigun suggested.
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https://en.wikipedia.org/wiki/Quantum_entanglement
Destroy perhaps isn't the right word, but it collapses the superposition.
Destroy perhaps isn't the right word, but it collapses the superposition.
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What's wrong with destroy? I find collapse... way more objectionable.
https://en.wikipedia.org/wiki/Quantum_decoherence
They are just different words for something we really do not understand.
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