Can black holes work as windmills?

Introduction

In General Relativity, the Kerr family is a class of exact vacuum solutions to Einstein’s equations, $R_{\mu \nu} = 0$, describing the spacetime geometry around a stationary, uncharged, rotating point-like mass distribution.

The overall structure of a Kerr metric is quite complex, with multiple horizons, extended singularities, closed timelike curves (CTCs), ergosurfaces, etc. These features, resulting from the absence of temporal and spherical symmetry, also provide a more realistic description of astrophysical objects. After all, most things you see in a telescope are spinning. In particular, the Kerr black hole is widely recognized as the last stage in the evolution of sufficiently massive stars.

In 1969, physicist Roger Penrose proposed a theoretical mechanism for extracting energy from a Kerr black hole. This process can only take place in-between the black hole’s event horizon and exterior ergosurface, known as the ergoregion. From the perspective of an asymptotic observer (infinitely far away), the ergoregion forbids static frames of reference due to an extreme Lense-Thirring effect. Those who resist being “dragged” may acquire negative energies. It is the absorption of negative energy particles that enables the Kerr black hole to be spun down, and an effective energy extraction to take place.

The Kerr metric

The Kerr family is a class of axially symmetric stationary exact vacuum solutions to Einstein’s equations. That’s a mouthful which, for asymptotic observers utilizing traditional spherical coordinates $(t,r,\theta,\phi)$, and natural units ($c=G=1$), means

\[g(t,r,\theta,\phi) = g_{ tt } dt^2 + g_{ t \phi } dt d \phi + g_{ \phi t } d \phi dt + g_{ rr } dr^2 + g_{ \theta \theta } d\theta^2 + g_{ \phi \phi } d \phi^2 \;,\]

where

\[g_{tt} = - \left(1 - \frac{ r_{S} r }{ \rho^2 } \right) \;,\] \[g_{t\phi} = g_{\phi t} = \frac{ - r_{S}ra\sin^2\theta }{ \rho^2 }\;,\] \[g_{rr} = \frac{ \rho^2 }{ \Delta }\;,\] \[g_{\theta\theta} = \rho^2\;,\] \[g_{\phi\phi} = \left[(r^2 + a^2)^2 - \Delta a^2 \sin^2\theta\right]\frac{ \sin^2\theta }{ \rho^2 } \;.\]

The notation used is $r_{S} \equiv 2M $, $ \rho^2 \equiv r^2 + a^2\cos^2\theta $, $ \Delta \equiv r^2 - r_S r + {a}^2$, and $ a \equiv J/M $.

The Kerr family is parametrized by 2 quantities, $ M $ and $ J $, usually interpreted as the mass and angular momentum of the central spinning body, respectively. If the Kerr family comprises valid interior solutions, then we really want the bound $ J \leq M $ to be respected. The region $J>M$ of the parameter space is populated not with black holes, but with naked singularities: quite obscene objects which exposes to the world physical pathologies usually hidden inside an event horizon.

We should point out that off-diagonal terms, $g_{t \phi} = g_{\phi t}$, make evident the presence of the Lense-Thirring effect due to the black hole’s rotation in the azimuthal direction, $ \hat{\phi} \equiv \partial_{\phi} / \sqrt{g_{\phi \phi}} $. Additionally, one can visually inspect that all metric components above are explicitly independent of $t$ and $\phi$. This implies that $\partial_{t}$ and $\partial_{\phi}$ are Killing vectors of these solutions — a direct consequence of their stationary and axially symmetric nature.

Last but not least:

• The limit $a \rightarrow 0$, $M$ fixed, results in the Schwarzschild geometry;
• The limit $a$ fixed, $M\rightarrow 0$, results in Minkowski;
• The Kerr family is asymptotically flat, which results in asymptotic observers being inertial;
• Although stationary, the Kerr family is not static, e.g., time reflections flip the rotation’s direction.

Horizons and ergosurfaces

Firstly, notice how the metric component $g_{rr}$ becomes singular at $\Delta=0$. This indicates the presence of horizons at $(t,r_{H}^{\pm},\theta,\phi)$, where

\[r_{H}^{\pm} \equiv \frac{ \left(r_{S} \pm \sqrt{r_{S}^2 - 4a^2}\right) }{ 2 } \;.\]

The outer horizon, $r_{H}^{+}$, is the actual event horizon of the Kerr black hole. The inner one, $r_{H}^{-}$, is a Cauchy horizon: a boundary beyond which the determinism of Einstein’s equations breaks down (see How predictive is General Relativity?).

Secondly, most of the metric components are singular at $\rho^2=0$. This is the ring-shaped singularity at $(t,0,\pi/2,\phi)$. In its vicinity CTCs exist, which leads to violations of causality conditions.

Thirdly, and finally, the metric component $g_{tt}$ vanishes at $\rho^{2}=r_{S}r$. This defines the hypersurfaces $(t,r_{E}^{\pm},\theta,\phi)$, known as static limit and/or infinite redshift surfaces (a.k.a., ergosurfaces), where

\[r_{E}^{\pm} \equiv \frac{ \left(r_{S} \pm \sqrt{r_{S}^2 - 4a^2\cos^2\theta}\right) }{ 2 } \;.\]

Their own definition makes them also a Killing horizon generated by $\partial_{t}$. This is just a fancy way to say that the norm of $\partial_{t}$ vanishes (changes sign) at $r^{\pm}_{E}$. But keep this in mind as Killing horizons explain energies changing signs in the perspective of asymptotic observers!

It should be clear that $r_{E}^{+} \geq r_{H}^{+} > r_{H}^{-} \geq r_{E}^{-}$, with the equalities only holding at the poles $(t,r,0,\phi)$ and $(t,r,\pi,\phi)$, see Figure 1. In particular, the region $r \leq r_{H}^{-}$ is widely contested in the scientific community due to the physical pathologies already mentioned (lack of determinism, causality, etc.). Again, check How predictive is General Relativity? for a more detailed account. Here, we’re going to discuss only the region $r \geq r_{H}^{+}$.

The Lense-Thirring effect

Let’s consider a differentiable curve $\gamma(\tau)$ near a Kerr black hole. The affine parameter $\tau$ represents its proper time, and $\dot{\gamma}(\tau) \equiv d\gamma/d\tau$ its 4-velocity. For our convenience, let’s also assume that, for an asymptotic observer, $\gamma(\tau)$ is static in the radial and zenithal directions, $\dot{\gamma}^{r}=\dot{\gamma}^{\theta}=0$.

The inequality $g\left(\dot{\gamma},\dot{\gamma}\right) < 0$ is sufficient for $\gamma\left(\tau\right)$ to be timelike — this is the kind of curve massive particles follow. But, it also implies the inequalities $\Omega^{-}<\Omega<\Omega^{+}$, where $\Omega \equiv {\dot{\gamma}}^{\phi}/{\dot{\gamma}}^{t}$ is the azimuthal velocity, and

\[\Omega^{\pm} \equiv \omega \pm \sqrt{\omega^2-4 \frac{g_{tt}}{g_{\phi\phi}}} \; ; \; \omega \equiv - \frac{g_{t\phi}}{g_{\phi\phi}} \;.\]

In other words, for an asymptotic observer, the azimuthal velocities of massive particles are always limited to the open interval $(\Omega^{-},\Omega^{+})$. Otherwise, we would have supraluminal speeds, causality violation, etc. Not physically desirable things.

In the exterior region of a Kerr black hole, $r > r_{E}^{+}$, we have $g_{tt} < 0$, $\Omega^{-} < 0$, and $\Omega^{+} > 0$. Massive particles are free to follow $\gamma(\tau)$ at subluminal speeds in the direction of the black hole’s rotation ($\Omega > 0$), against it ($\Omega < 0$), or even remain spatially static ($\Omega = 0$). This is all normal behavior we expect.

Things start to change on the ergosurface though. At $r = r_{E}^{+}$, $g_{tt} = 0$ because we are on the Killing horizon (!) of $\partial_{t}$. As a result, $\Omega^{-} = 0$ and $\Omega^{+} = 2\omega$. Now, every massive particle that follows $\gamma(\tau)$ is forced to do it in the direction of the black hole’s rotation, and spatial static ($\Omega = 0$) is a luxury that only massless particles can enjoy. In other words, light, which follow null trajectories, $g\left(\dot{\gamma},\dot{\gamma}\right) = 0$, can be at a full stop on the ergosurface.

Finally, in the ergoregion $r_{E}^{+} < r < r_{H}^{+}$, we have $g_{tt} > 0$, and $\Omega^{\pm} > 0$. Spatial static is now forbidden to everyone. With no exceptions, every non-spacelike trajectory is forcefully oriented by the black hole’s rotation. Here, asymptotic frames cannot avoid the black hole’s drag just like they cannot avoid the future. This is purely due to the warping of space and time, and represent perhaps the most extreme version of the often tiny Lense-Thirring effect.

Now, we can explicitly calculate $\omega$ for the Kerr family. It is given by

\[\omega = \frac{r_{S}ra}{(r^2 + a^2)^2 - \Delta a^2\sin^2\theta } \;.\]

On the event horizon, $\omega = \omega_{H}$, where $\omega_{H} \equiv a / r_{S}r_{H}^{+}$. Additionally, $\Omega^{\pm}=\omega_{H}$. In other words, everyone is forced to follow the black hole’s rotation at a precisely determined azimuthal speed. Since the properties of particles clearly don’t matter here, $\omega_{H}$ can be seen as intrinsic to the black hole. It is the rotational speed of its event horizon.

Negative energies and momenta

We have just discussed how an extreme Lense-Thirring effect occurs inside the ergosurface of a Kerr black hole. When crossing this spacetime boundary, the azimuthal component of your 4-velocity, $\dot{\gamma}^{\phi}$, is required to change by the ratio $\Omega$, in order to offset the sign change that had just occurred in $\dot{\gamma}^{t}$. The latter is due to the ergosurface coinciding with the Killing horizon (!) of $\partial_{t}$, but the former is to ensure that every observer agrees on the timelike nature of $\gamma$. In other words, that $g(\dot{\gamma},\dot{\gamma}) < 0$ remains a frame-invariant inequality.

A similar analysis can be done for the 4-momentum $p = m \dot{\gamma}$ of a particle with mass $m$ following $\gamma$. Upon entering the ergoregion, the momentum component $p_{t} \equiv g\left(\partial_{t},p\right)$ changes sign, and $p_{\phi}$ must adjust accordingly to maintain the inequality $g \left(p,p\right) < 0$ frame-invariant. The particularly interesting point here is that $p_{t} = - E$ and $p_{\phi} = L$, where $E$ is the energy and $L$ the azimuthal angular momentum of this particle. Thus, asymptotic observers can witness a particle’s energy change sign when it crosses the Kerr ergosurface.

The negative energy condition, $E < 0$, can be expressed as $\Omega < - g_{tt}/g_{t \phi}$. Outside the ergoregion ($r > r_{E}^{+}$), this condition can only be satisfied by spacelike curves, as it implies supraluminal speeds. The same is true at the ergosurface ($r = r_{E}^{+}$), where $g_{tt} = 0$. However, within the ergoregion ($r_{E}^{+} < r < r_{H}^{+}$), we find that $g_{tt} > 0$, and the negative energy condition, clearly intersects with the non-spacelike conditions ($\Omega^{-} \leq \Omega \leq \Omega^{+}$). “Sufficiently slow” particles (${\Omega}^{-} \leq \Omega < - g_{tt}/g_{t \phi}$) — those who resist the Lense-Thirring effect just enough — have negative energies from the perspective of an asymptotic observer. “Sufficiently fast” ones ($- g_{tt}/g_{t \phi} < \Omega \leq \Omega^{+}$) have positive energies, and there are still those in a state of zero energy ($\Omega = - g_{tt} / g_{t \phi}$). Yes, black hole physics can be quite weird!

To conclude this section, consider the Killing vector $k = \partial_{t} + \omega_{H} \partial_{\phi}$. It generates the Killing horizon coinciding with the event horizon of the Kerr black hole. The timelike condition $g \left(k, p\right) < 0$ leads us to the inequality $E > \omega_{H} L$. Thus, negative energy particles also have negative angular momenta — even though their orbital speeds are positive. As we will see next, it is the existence of these particles that enables the energy extraction to take place.

The Penrose process

The Penrose process presents a theoretical mechanism to extract usable energy from a rotating Kerr black hole. It is based on the existence of an ergoregion, which allows for timelike trajectories with negative energies and orbital momenta.

Consider an electron coming from the asymptotic infinity, with initial positive energy $E_{0}$, following the timelike trajectory $\gamma_{1}$, which at some point enters the ergoregion of the Kerr black hole, check Figure 2. Also consider that, due to the gravitational acceleration, a photon will be emitted at some point. We assume that the electron will emit this photon within the ergoregion, and at such a precise angle that this photon will be one of those “slow enough” particles carrying negative energy $\varepsilon < 0$. Let it follow the null trajectory $\gamma_{2}$ that ultimately falls into the event horizon. At the emission event, momentum conservation dictates that the trajectory of the electron needs to change, let it be the timelike $\gamma_{3}$ with positive energy $E_{f}$, that ultimately escapes to the asymptotic infinity.

In this highly theoretical scenario, energy conservation $E_{0} = E_{f} + \varepsilon$ of the electron-photon system dictates $\Delta E = - \varepsilon > 0$, where $\Delta E \equiv E_{f} - E_{0}$. For an asymptotic observer witnessing the whole process, the electron escaped the neighborhood of the Kerr black hole with energy $E_{0} + \Delta E$. Saying it differently, the electron just extracted $\Delta E$ of energy from a Kerr black hole!

In order for the Penrose process to be irreversible, the emitted photon really needs to fall into the event horizon. Additionally, since no lunch is free, this also means that there will be a subtraction $\delta M = \varepsilon$ and $\delta J = L$ (remember, both $\varepsilon$ and $L$ are negative), of the black hole mass and angular momentum in such a way that the inequality $\delta M > \omega_{H} \delta J$ holds.

As a final note, a series of Penrose energy extraction processes can bring the black hole to a thermodynamic equilibrium, defined by $\delta M = \omega_{H} \delta J$. Remarkably, this is recognizable as the 1st Law of Black Hole Thermodynamics, applicable to processes that maintain the area $A$ of the event horizon constant, $\delta A = 0$ (or, equivalently, the Bekenstein-Hawking (BH) entropy constant, $\delta S_{\text{BH}} = 0$).

The thermodynamic equilibrium serves as a maximum limit for Penrose processes: attempting to extract more energy would violate the 2nd Law of Black Hole Thermodynamics, meaning that $A$ cannot shrink, $\delta A \geq 0$ (or, equivalently, $\delta S_{BH} \geq 0$). This limit is reached before $J=0$ is achieved, so Penrose processes cannot fully spin down the Kerr black hole. Putting it differently, Penrose processes do not provide a physical mechanism for the “spacetime decay” $\text{Kerr} \rightarrow \text{Schwarzschild}$.

Conclusions

The Penrose process was the first theoretical example demonstrating that energy extraction from a black hole could be possible. However, as we have just seen, it clearly requires very precise spacetime trajectories. Such a specific set of initial data make Penrose processes unlikely to happen in a natural astrophysical setting.

Less fine-tuned mechanisms have been developed since, among which the Blandford-Znajek process is believed to be the engine powering the brightest objects in the entire Universe: Active Galactic Nuclei (AGNs) inside Quasars.

In the end, rotating black holes appears to function as engines after all, similar to how windmills do.