Negative Killing Energy States

A future-directed particle can have negative energy relative to infinity inside an ergoregion because the stationary Killing field there is spacelike. The particle’s locally measured energy remains positive; only the conserved charge $E=-p_a t^a$ changes sign. For matter that crosses a rotating charged horizon, future causality instead constrains the charge associated with the horizon generator: $E-\Omega_H L-q\Phi_H\geq 0$. The distinction is the geometric core of rotational-energy extraction.

Required background. Spacetime Symmetries and Killing Fields supplies conserved charges. Helpful background. Exact Spacetimes supplies Kerr examples; Causality and Global Structure supplies future-directed causal cones; Einstein Equations fixes the background; Black-Hole Geometry and Mechanics supplies the generator and potential; Black-Hole Perturbation Theory develops the wave analogue.

Killing charges and local energy

For a geodesic particle of four-momentum $p^a$, stationarity and axisymmetry give

$$E=-p_a t^a, \qquad L=p_a\phi^a.$$

Both are conserved along geodesic motion because $t^a$ and $\phi^a$ are Killing fields. A local future timelike observer with four-velocity $u^a$ measures

$$E_{\mathrm{local}}=-p_au^a\gt0$$

for every future causal $p^a$. Negative Killing energy therefore is not negative local energy density. It is a negative charge with respect to the asymptotic time-translation symmetry.

Where $t^a$ is future timelike, two future causal vectors have nonpositive inner product, so $-p\cdot t\geq 0$. Negative $E$ is impossible there. At the stationary limit $t^a$ becomes null; inside, it is spacelike and its inner product with future causal $p^a$ can have either sign. This proves that the ergoregion is necessary for negative stationary Killing energy; see Visser 2009, arXiv v3, §§10–11, pp. 29–37.

It is not sufficient to choose an arbitrary negative number. The momentum must still satisfy the mass shell, future direction, and any radial turning- point constraints. Brito, Cardoso, and Pani discuss the ergoregion and the particle Penrose process in Brito, Cardoso, and Pani 2020, Chapter 4, §§4.2–4.3.1, pp. 39–44.

Allowed angular velocities and the energy sign

Consider a circular-direction tangent

$$k^a=t^a+\Omega\phi^a$$

at fixed $r$ and $\theta$. It can be timelike only for $\Omega_-\lt\Omega\lt\Omega_+$, where

$$\Omega_\pm= \frac{-g_{t\phi}\pm \sqrt{g_{t\phi}^2-g_{tt}g_{\phi\phi}}}{g_{\phi\phi}}.$$

Inside a positive-spin Kerr ergoregion, both bounds force co-rotation close enough to the horizon. Yet the canonical angular momentum $L$ can be sufficiently negative that

$$E=-p_t\lt0.$$

Frame dragging of the coordinate angular velocity and negative canonical angular momentum are therefore compatible. Saying “the particle rotates backward” is ambiguous unless one specifies whether it means negative $d\phi/dt$, negative $L$, or rotation relative to a local inertial frame.

A useful local decomposition uses a zero-angular-momentum observer (ZAMO) with

$$n^a=\alpha^{-1}(t^a+\omega\phi^a), \qquad \omega=-\frac{g_{t\phi}}{g_{\phi\phi}}.$$

Here $\alpha\gt0$ is the lapse chosen so that $n^an_a=-1$, and the positive local energy measured by this observer is $E_{\mathrm{ZAMO}}=-p_an^a$. Then

$$E=\alpha E_{\mathrm{ZAMO}}+\omega L.$$

The first term is positive, but the second can make $E$ negative when $L\lt0$ and frame dragging is strong. This formula cleanly separates locally positive energy from the asymptotic symmetry charge.

The horizon-generator inequality

On a stationary rotating horizon,

$$\chi^a=t^a+\Omega_H\phi^a$$

is future null. For an uncharged future causal particle crossing the future horizon,

$$-p_a\chi^a=E-\Omega_HL\ge0.$$

Thus a negative-$E$ particle must carry sufficiently negative $L$. When it is absorbed, the black-hole changes satisfy

$$\delta M=E, \qquad \delta J=L, \qquad \delta M-\Omega_H\delta J\ge0.$$

This is consistent with the first law and area nondecrease. The hole loses mass and angular momentum, but it loses enough angular momentum that the irreducible area need not shrink.

For a particle of electric charge $q$, write the canonical momentum as $p_a=\pi_a+qA_a$, where $\pi^a$ is the future-directed kinetic momentum. With $E=-p_a t^a$, $L=p_a\phi^a$, and the horizon potential $\Phi_H=-\chi^aA_a$, future causality gives the gauge-invariant combination

$$E-\Omega_HL-q\Phi_H\ge0,$$

where $E$ and $\Phi_H$ use the same potential reference at infinity. Indeed, the left side is $-\pi_a\chi^a$, so its sign follows from the two future causal vectors $\pi^a$ and $\chi^a$. A gauge shift changes the canonical energy and potential terms together while leaving the combination invariant. The charged static and neutral rotating wave thresholds are developed separately in Brito, Cardoso, and Pani 2020, Chapter 4, §§4.6–4.7.3, pp. 52–60.

Negative energy does not escape to infinity

At infinity $t^a$ is future timelike, so a future causal free particle has $E\geq 0$. A negative-$E$ geodesic cannot propagate out to that asymptotic region while retaining its conserved charge. In a Penrose split it is the negative-energy fragment that falls through the horizon; the positive-energy fragment can escape with more energy than the incident particle carried.

The statement assumes a stationary background so that $E$ is conserved. During strong backreaction, the exact Killing symmetry and test-particle charge are lost. Global energy balance must then use the appropriate asymptotic charge and radiation flux. The negative-energy construction remains the controlled test limit, not a proof of arbitrary nonlinear efficiency.

An ergoregion without a horizon can also allow negative Killing-energy modes. If reflecting conditions trap them, an instability may result. That is a spectral and boundary-condition question, not a consequence of $g_{tt}\gt0$ alone.

Test the horizon inequality

Take an uncharged Kerr horizon with $\Omega_H\gt0$. A particle crosses it with $E=-1$ in arbitrary energy units. What must be true of its angular momentum? Can a particle with $L=0$ have this negative energy and cross the future horizon?

Solution

Future horizon crossing requires

$$E-\Omega_HL\ge0.$$

With $E=-1$, this gives

$$L\le-\frac{1}{\Omega_H}.$$

The fragment must carry negative angular momentum of sufficient magnitude. For $L=0$, the left side would be $-1$, violating future causality relative to the null generator; such a momentum cannot cross the future horizon. If the allowed negative-$L$ particle is absorbed, both black-hole mass and angular momentum decrease while the inequality keeps the area change nonnegative in the test limit.

The Penrose Particle-Splitting Process uses these charges in a complete local conservation calculation.

References

• Brito, Richard, Vitor Cardoso, and Paolo Pani. Superradiance. 2nd ed. Lecture Notes in Physics 971. Cham: Springer, 2020. DOI.
• Visser, Matt. “The Kerr Spacetime: A Brief Introduction.” In The Kerr Spacetime: Rotating Black Holes in General Relativity. Cambridge: Cambridge University Press, 2009. arXiv.