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


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.

  • $\begingroup$ 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. $\endgroup$
    – Rexcirus
    Apr 21, 2018 at 9:09
  • $\begingroup$ 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. $\endgroup$
    – MBN
    Jun 12, 2018 at 10:42
  • $\begingroup$ It's the energy that remains after extracting as much rotational energy as possible via the Penrose process and similar techniques. Since naked singularities would also have ergospheres it would be possible to extract energy, given that they existed at all (nevertheless, we're talking theoretically). The mechanism for extraction stays the same as used in MTW, wherefore you don't need a horizon: i.imgur.com/ijJEcQl.png $\endgroup$
    – Yukterez
    Sep 20, 2019 at 18:04

1 Answer 1


The irreducible mass is defined by the Schwarzschild radius of the external horizon of the black hole: $$\tag{1} r_+ = 2 G M_{\text{irr}} \equiv 2 G M_0. $$ This definition already assumes that there's an horizon.

Super-extremal "black holes" don't have an horizon (they're naked), and they aren't black holes at all. They're just an abomination in spacetime.

  • $\begingroup$ In what coordinates, in the case of an extremal Kerr black hole and in Boyer Lindquist coordinates the irreducible mass is √½ and the outer horizon is at r+=1GM/c²=√(2)GMirr/c² so that's not 2GMirr/c² then. Also, we're talking about theory, not nature and in theory naked singularities are allowed in GR so it must also be possible to extract some energy with the Penrose process (for which you only need an ergosphere, but no horizon), like it is in the case of a normal rotating black hole. So the question still remains, how much energy can one extract, and how much must remain $\endgroup$
    – Yukterez
    Sep 20, 2019 at 15:57
  • $\begingroup$ @Yukterez, good question. I'm unsure of the answer but I believe that the left hand part of (1) could be written as an invariant scalar. $\endgroup$
    – Cham
    Sep 20, 2019 at 20:16

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