Irreducible Mass and Rotational Energy

The irreducible mass packages a black hole’s horizon area into a mass scale that cannot decrease in classical processes satisfying the area theorem. For Kerr–Newman, the remaining mass above this floor is associated with rotation and electromagnetic charge and can, in ideal reversible limits, be extracted. For extremal uncharged Kerr the total rotational reservoir is $1-1/\sqrt2\approx29.3\%$ of the initial mass. This is a sequence-level area bound, distinct from the $20.7\%$ gain bound for one original Penrose decay.

Required background. Spacetime Symmetries and Killing Fields supplies stationary charges. Helpful background. Exact Spacetimes supplies Kerr–Newman horizon data; Causality and Global Structure supplies the event-horizon area theorem; Einstein Equations fixes the classical theory; Black-Hole Geometry and Mechanics supplies first-law and area conventions; Black-Hole Perturbation Theory treats wave extraction rates.

Area defines the irreducible mass

For a four-dimensional black hole, define

$$M_{\mathrm{irr}} =\sqrt{\frac{A_H}{16\pi G^2}}.$$

This definition is chosen so that a Schwarzschild black hole has $M_{\mathrm{irr}}=M$. Under the null energy and global hypotheses of the classical area theorem,

$$\delta A_H\ge0 \quad\Longrightarrow\quad \delta M_{\mathrm{irr}}\ge0.$$

“Irreducible” refers to this classical horizon-area component. It does not mean rest mass of a material object, nor does it survive unchanged when the area theorem’s energy or semiclassical assumptions fail.

For Kerr, with $a=J/M$ and $r_+=GM+\sqrt{G^2M^2-a^2}$,

$$A_H=4\pi(r_+^2+a^2)=8\pi GMr_+.$$

Therefore

$$M_{\mathrm{irr}}^2=\frac{Mr_+}{2G} =\frac12\left(M^2+ \sqrt{M^4-\frac{J^2}{G^2}}\right).$$

Brito, Cardoso, and Pani relate this area variable to Penrose extraction in Brito, Cardoso, and Pani 2020, §4.3.1, pp. 42–44.

Christodoulou’s Kerr mass relation

Solve the preceding equation for $M$. One obtains

$$M^2=M_{\mathrm{irr}}^2 +\frac{J^2}{4G^2M_{\mathrm{irr}}^2}.$$

The first term is the square of the area floor; the second records the rotational contribution to $M^2$. Differentiate:

$$2M\,dM =\left(2M_{\mathrm{irr}} -\frac{J^2}{2G^2M_{\mathrm{irr}}^3}\right)dM_{\mathrm{irr}} +\frac{J}{2G^2M_{\mathrm{irr}}^2}dJ.$$

Using the Kerr horizon identities converts this exactly into

$$dM=\frac{\kappa}{8\pi G}dA_H+\Omega_HdJ.$$

Thus an ideal reversible extraction satisfies $dA_H=0$ and $dM=\Omega_HdJ$. Generic absorption has $dM-\Omega_HdJ\gt0$ and raises $M_{\mathrm{irr}}$, leaving less of the final mass available to future reversible extraction.

The decomposition is nonlinear. It is not correct to call the second term in $M^2$ an energy and subtract it directly from $M$. The extractable mass is the difference between the initial mass and the smallest final mass allowed at fixed nondecreasing $M_{\mathrm{irr}}$.

Maximum rotational reservoir in Kerr

For extremal Kerr, $J=GM^2$ and $r_+=GM$. Hence

$$A_H=8\pi G^2M^2, \qquad M_{\mathrm{irr}}=\frac{M}{\sqrt2}.$$

Imagine an ideal sequence that removes angular momentum reversibly while holding area fixed, ending in Schwarzschild with $J_f=0$. Its final mass is $M_f=M_{\mathrm{irr}}$, so the maximum extractable fraction is

$$\frac{M-M_f}{M} =1-\frac{1}{\sqrt2} \approx0.2929.$$

The $29.3\%$ is a total reservoir bound for an initially extremal Kerr hole. It does not say one particle split reaches it. The original Penrose decay has additional local momentum and escape constraints and reaches only a $20.7\%$ net gain relative to that incident particle in its ideal limit.

For a nonextremal spin, substitute the actual $J/(GM^2)$ into $M_{\mathrm{irr}}$. The reservoir shrinks continuously to zero as $J\to0$.

Charged rotating black holes

With rationalized electric charge, Kerr–Newman obeys

$$M^2= \left(M_{\mathrm{irr}} +\frac{Q^2}{16\pi G M_{\mathrm{irr}}}\right)^2 +\frac{J^2}{4G^2M_{\mathrm{irr}}^2}.$$

This is the Christodoulou–Ruffini relation in the charge convention where the horizon equation contains $GQ^2/(4\pi)$ Christodoulou and Ruffini 1971, pp. 3552–3555. Differentiation gives

$$dM=\frac{\kappa}{8\pi G}dA_H +\Omega_HdJ+\Phi_HdQ.$$

The mass above $M_{\mathrm{irr}}$ now includes coupled rotational and electromagnetic contributions. How much is operationally extractable depends on which charges can be changed and which fields or particles implement the change. Discharging a hole is not the same process as spinning it down, and a fixed-$Q$ bound differs from one permitting $Q$ to leave.

In the extremal Reissner–Nordström case with $J=0$, $M_{\mathrm{irr}}=M/2$ in this convention. Interpreting the remaining half as freely extractable requires a physical discharge channel and controlled backreaction; the algebraic mass relation alone is not an efficiency theorem.

Reversible means an ideal limit

A reversible classical process saturates

$$dM-\Omega_HdJ-\Phi_HdQ=0,$$

so $dA_H=0$ for a nonextremal background. Achieving this requires matter or radiation to cross with vanishing generator energy in a quasistatic limit. Finite fragments, wave packets, self-force, radiation, and capture thresholds usually produce positive generator energy and area growth.

The area theorem provides an upper bound under its hypotheses, not a recipe for reaching it. It also assumes classical general relativity, an appropriate energy condition, and global predictability. Hawking evaporation and generalized entropy belong to semiclassical mechanics and need not preserve the classical area.

Compute a spin reservoir

A Kerr black hole has dimensionless spin $a_*=J/(GM^2)=0.8$. Compute $M_{\mathrm{irr}}/M$ and the ideal maximum fraction removable by a reversible spin-down to Schwarzschild.

Solution

From the Kerr formula,

$$\frac{M_{\mathrm{irr}}^2}{M^2} =\frac12\left(1+\sqrt{1-a_*^2}\right) =\frac12(1+0.6)=0.8.$$

Thus

$$\frac{M_{\mathrm{irr}}}{M}=\sqrt{0.8}\approx0.8944.$$

Holding area fixed while taking $J$ to zero leaves a Schwarzschild mass $M_f=M_{\mathrm{irr}}$, so at most $1-\sqrt{0.8}\approx0.1056$, or $10.56\%$ of the initial mass, is available as rotational energy in the ideal sequence. Any area increase raises the final irreducible mass and reduces the extractable fraction.

Wave Superradiance as an Extraction Channel translates negative horizon generator energy into the frequency condition for bosonic waves.

References

• Brito, Richard, Vitor Cardoso, and Paolo Pani. Superradiance. 2nd ed. Lecture Notes in Physics 971. Cham: Springer, 2020. DOI.
• Christodoulou, Demetrios, and Remo Ruffini. “Reversible Transformations of a Charged Black Hole.” Physical Review D 4 (1971): 3552–3555. DOI.