# Would naked singularities have a complex irreducible mass?

The formula for the irreducible mass, also known as the Christodoulou and Ruffini equation, is

$$M_{\rm irr} = \frac{\sqrt{2 M^2-Q^2+2 M \sqrt{M^2-Q^2-a^2}}}{2}$$

where M is the mass equivalent of the total energy, which is composed of the rest mass, the rotational energy and the electromagnetic field energy, with Q for the charge and a for the spin parameter.

When we put in the parameters of a naked singularity, which means

$$a^2+Q^2>M^2$$

we get complex numbers for the irreducible mass (which is the rest mass of the body after all rotation and charge have been extracted via the Penrose process or neutralization).

I've heard many reasons why naked singularities might be forbidden by nature, but the argument with the complex mass never came up.

Would that be an additional argument against naked singularities, or does one have to use another relation when dealing with a naked singularity instead of a black hole? After all, complex rest mass and complex energy are not supposed to exist, which would rule them out in yet another way.

In this reference in footnote 18 it says, quote:

"Since a black ring must be rotating, it is not trivial to divide the mass energy into its irreducible mass and the rotational energy. Here weassume that the same formula for irreducible mass is applicable for a black ring as well."

so if the formula holds, the rest mass of such a ring should be complex as it seems.

• I don't think it's a very compelling argument, since it's purely math based. And formally, they are solution of the equation of motion anyway. – Rexcirus Apr 21 '18 at 9:09
• Why do you think the formula will apply for the case of a naked singularity? And what is the definition of an irreducible mass in that case? The irreducible mass that you mention requires a horizon. – MBN Jun 12 '18 at 10:42