What happens if $ a^2 > M^2 $ in Kerr metric?

Question

(Boyer-Lindquist coordinates and $ c = G =1 $ taken)

As I know, line element in Kerr metric $ d s^2 = - \left( 1 - \frac{2Mr}{\rho^2} \right) d t^2 - \frac{4 M a r \sin^2 \theta}{\rho^2} d \phi d t + \frac{\rho^2}{\Delta} d r^2 + \rho^2 d \theta^2 + \left( r^2 + a^2 + \frac{2 M r a^2 \sin^2 \theta}{\rho^2} \right) \sin^2 \theta d \phi^2 $ where $ a = \frac{J}{M}, \rho^2 = r^2 + a^2 \cos^2 \theta, \Delta = r^2 - 2 M r + a^2 $ suggests that causality breaks where $ r = M \pm \sqrt{M^2 - a^2} $.

How should I handle when $ a^2 > M^2 $ so $ r $ is complex?

For example, the Sun's mass, $ M_\odot \approx 1.5 \text{km} $ while it's angular momentum, $ J_\odot \approx 4.7 * 10^{19} \text{km}^2 > M_\odot ^2 $.

Will black holes with angular momentum bigger than mass squared not form an event horizon?

Answer 1

If the mass is in the form of a material body like the earth where the spin parameter is $\rm a=J c/(G M)=890 M$, it would have to lose some angular momentum until $\rm a<M$ before it could collapse into a black hole, otherwise it can't since the centrifugal repulsion is larger than the centripetal attraction.

If the mass is in the form of a singularity that would be a naked singularity, and if it is an elementary particles like the electron where $\rm a$ is many orders of magnitudes larger than $\rm M$ and $\rm r=0$ as well the close field should be gravitationally repulsive.

Answer 2

According to Kerr and Penrose, bodies with $ a\ge m $ can not exist in our universe.

They are mathematical objects with naked singularities (ring of diameter $ a $) and are not black holes (no event horizon).

Hoping to have answered your question,

Best regards.