Overspinning and Overcharging Thought Experiments

An overspinning or overcharging experiment asks whether matter can enter a nearly extremal black hole while changing its conserved charges enough that no member of the stationary black-hole family has the final parameters. The apparent loophole arises when capture and final extremality are tested only to first order. A consistent calculation must use the same perturbation for horizon crossing, radiative and self-force effects, finite size, and the second-order change in mass.

Helpful background. Extremality and horizon generators are developed in Black-Hole Geometry and Mechanics; perturbative fluxes belong to Black-Hole Perturbation Theory; constraint and variational identities use Formulations of General Relativity; global visibility is defined in Causality and Global Structure; and numerical tests require Computational Relativity.

Capture and overextremality inequalities

Use geometrized Einstein–Maxwell units $G=c=4\pi\epsilon_0=1$ through the derivation, so the charge parameter $Q$ has units of length. As a dimensional check, the Kerr extremality bound restored to SI units is $c\lvert J\rvert\le GM^2$, so $cJ/(GM^2)$ is dimensionless. Choose the axial orientation so $J\geq0$. Consider first an uncharged body with energy $E$ and aligned axial angular momentum $L$ incident on a Kerr hole with parameters $(M,J)$. A necessary horizon-crossing condition derived from the future horizon generator is

$$E-\Omega_H L\ge0,$$

for matter satisfying the null energy condition. A naive final-state test substitutes $M_f=M+E$ and $J_f=J+L$ into $J_f\le M_f^2$. Overspinning would require

$$J+L\gt (M+E)^2.$$

For exactly extremal Kerr, $\Omega_H=1/(2M)$ makes the first-order inequalities incompatible, reproducing Wald’s original test-particle result Wald 1974, pp. 548–550, Eqs. (1)–(7). For a nearly extremal hole, truncated expansions can leave a narrow formal window. Charged bodies near Reissner–Nordström give an analogous pair involving $E-\Phi_H q$ and $(M+E)^2\lt (Q+q)^2$.

Why first order is insufficient

The candidate window has the same perturbative size as effects omitted by the test-body model. Conservative self-force shifts the capture threshold; radiation changes $E$ and $L$ before absorption; finite size constrains charge-to-mass and spin; and the body’s own field contributes second-order energy. It is inconsistent to retain a second-order extremality deficit while discarding all second-order backreaction.

The covariant phase-space method packages these effects into variational identities. It assumes a one-parameter Einstein–Maxwell–matter family, a regular initial horizon, matter obeying the null energy condition, suitable asymptotic behavior, and linear stability sufficient for the exterior perturbation to settle to a stationary Kerr–Newman perturbation. This last assumption permits a stationary comparison; it is not a general nonlinear proof that every disturbed black hole settles. For a potentially dangerous, first-order optimal perturbation, the second-order bound is

$$\delta^2M-\Omega_H\delta^2J-\Phi_H\delta^2Q \geq-\frac{\kappa}{8\pi}\delta^2A_{\rm KN}.$$

Here $\delta=d/d\lambda|_{\lambda=0}$ and $\lambda$ labels the physical family. The comparison area $A_{\rm KN}$ is evaluated along the stationary Kerr–Newman family

$$\begin{aligned} M_{\rm KN}(\alpha)&=M+\alpha\delta M,\\ J_{\rm KN}(\alpha)&=J+\alpha\delta J,\\ Q_{\rm KN}(\alpha)&=Q+\alpha\delta Q. \end{aligned}$$

Thus $\delta^2A_{\rm KN}$ means $d^2A_{\rm KN}/d\alpha^2$ at zero: all three comparison charges have zero second derivative. Merely matching their first derivatives would not fix this area variation. The physical family can have nonzero second-order charge changes. The background surface gravity is $\kappa$, and the potential $\Phi_H$ is measured relative to infinity. The comparison area is not an assumed area history of the dynamical hole. Sorce and Wald define this family and derive the bound in Sorce and Wald 2017, §IV, Eqs. (109)–(122), pp. 29–32, PDF. Its use in final-remnant checks appears in Extremality and Final-State Bounds.

A charged example of the missing energy

For a nonrotating Reissner–Nordström background with $0\lt Q\lt M$ (the other sign follows by charge reversal), $r_+=M+\sqrt{M^2-Q^2}$ and $\Phi_H=Q/r_+$. An optimal first-order addition saturates $\delta M=\Phi_H\delta Q$. Choose a nonrotating physical family with $\delta^2Q=0$. Differentiating the comparison identity $M=(r_+^2+Q^2)/(2r_+)$ first gives $\delta r_+=0$. A second derivative, using the affine comparison charges and $A=4\pi r_+^2$, gives $-\kappa\delta^2A_{\rm KN}/(8\pi)=(\delta Q)^2/r_+$. Hence

$$\delta^2M\geq\frac{(\delta Q)^2}{r_+}.$$

The actual Taylor expansion is $M(\lambda)=M+\lambda\delta M+\tfrac12\lambda^2\delta^2M+\cdots$: the factor $1/2$ is essential. This positive quadratic energy is precisely of the order omitted by the naive charge addition. To see the near-extremal competition, write $M^2-Q^2=M^2\epsilon^2$ and count $\lambda$ and $\epsilon$ at the same small order. The bound gives

$$M(\lambda)^2-Q(\lambda)^2 \geq(M\epsilon-\lambda\delta Q)^2+O_3,$$

where $O_3$ denotes terms of total degree at least three in the small parameters. Thus the proposed negative extremality margin becomes a square at the controlled order. If that square is tuned to vanish, second-order analysis alone says nothing about the sign of the omitted higher-order remainder.

Scope and independent checks

This argument does not prove weak cosmic censorship for arbitrary large perturbations. It assumes a regular horizon at the start, controls a perturbative family, and uses a late stationary comparison. It does show why a successful algebraic test-particle window is not a self-consistent horizon-destruction solution.

A numerical experiment should measure fluxes both at the horizon and null infinity, verify constraint convergence, compare the final apparent-horizon area and charges with global balance, and resolve the extremality margin more accurately than all error estimates. A superextremal value computed from coordinate mass or from inconsistent time slices is not evidence of a naked singularity.

Exercise

Take extremal Kerr with $M=1$, $J=1$. Show that the horizon-crossing condition and naive overspinning condition are incompatible at leading order for a small positive $E$.

Solution

Horizon crossing gives $E-\tfrac12L\ge0$, hence $L\le2E$. Overspinning requires $1+L\gt (1+E)^2=1+2E+E^2$, hence $L\gt 2E+E^2$. No $L$ satisfies both. For a nearly extremal hole a truncated window may appear, but then second-order terms are of the same order as the window and cannot be omitted.

References

• Sorce, Jonathan, and Robert M. Wald. “Gedanken Experiments to Destroy a Black Hole II: Kerr–Newman Black Holes Cannot Be Overcharged or Overspun.” Physical Review D 96 (2017): 104014. doi:10.1103/PhysRevD.96.104014.
• Wald, Robert M. “Gedanken Experiments to Destroy a Black Hole.” Annals of Physics 82 (1974): 548–556. doi:10.1016/0003-4916(74)90125-0.