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If a single subatomic hypothetical fundamental particle incapable of decay (such as an electron) with a mass exceeding Planck mass possessed angular momentum and it collapsed into a black hole, would it be a Kerr ring black hole?

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marked as duplicate by Hritik Narayan, peterh, honeste_vivere, SRS, Qmechanic Jul 12 '17 at 22:13

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A quick way to think about this is with the definition of the inner and outer horizons of a black hole $$ r_{\pm}~=~m~\pm~\sqrt{m^2~-~a^2}, $$ where we consider the angular momentum parameter at maximum and $a~=~J/Mc$. The mass parameter $m^2~=~GM/c^2$. We now look at this in $n$ units of Planck mass $M_p~=~\hbar c/G$ so that $$ r_{\pm}~=~nM_p~\pm~\sqrt{\frac{n^2G^2M_p^2}{c^4}~-~\frac{J^2}{n^2M_p^2c^2}}. $$ The angular momentum is discrete in integer units $J~=~n'\hbar$ for $n'$ and $n$ independent. The term in the radical must be greater than one so we have the condition that the number of units of angular momentum $n'$ for a black hole with $n$ Planck units of mass is $$ n'~<~\frac{n^2GM_p^2}{\hbar c}~=~n^2. $$ This is the remarkably simple result that $n'~<~n^2$. This is a Regge trajectory result, and it puts an upper bound on the angular momentum of an $n$ Planck mass black hole.

We can modify this for $n'~\rightarrow~n'/2$ to account for fermions. This means a single sub-extremal Planck mass black hole has two possible states with the $m_z~=~\pm 1/2$. The case $n' = 2$ is the extremal case.

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