Reflection, Transmission, and Absorption

A reflected wave can carry more energy than the incident wave without violating conservation: a rotating black hole can absorb negative asymptotic-time energy. This page derives that statement for a minimally coupled massless scalar on a fixed subextremal Kerr exterior. It then distinguishes a dimensionless greybody factor from an absorption cross section, and explains why the low-frequency Schwarzschild cross section approaches the horizon area even though its dominant greybody factor vanishes.

Required background. PDEs and Green Functions for Gravitation supplies conserved Wronskians and one-dimensional scattering.

Helpful background. Black-Hole Geometry and Mechanics supplies horizon generators and Killing charges; Formulations of General Relativity supplies field equations and stress tensors; Events, Intervals, and Minkowski Geometry supplies relativistic energy flux; Black-Hole Perturbation Theory locates the separated master fields.

Scalar normalization and radial current

Let $M$ and $J$ be the physical black-hole mass and angular momentum, with $a=J/M$ and $|a|\lt GM$. Define

$$\begin{gathered} h=r^2+a^2,\qquad \Delta=r^2-2GMr+a^2,\qquad r_+=GM+\sqrt{G^2M^2-a^2},\\ \frac{dx}{dr}=\frac{h}{\Delta},\qquad \Omega_H=\frac{a}{r_+^2+a^2},\qquad k_H=\omega-m\Omega_H. \end{gathered}$$

Fix a real $\omega>0$ and an integer pair $\ell\geq|m|$. Choose a regular scalar spheroidal harmonic with full-sphere normalization

$$Y_{\ell m}(\theta,\phi;a\omega) =\frac{S_{\ell m}(\theta;a\omega)e^{im\phi}}{\sqrt{2\pi}}, \qquad \int |S_{\ell m}|^2\sin\theta\,d\theta=1, \qquad \int |Y_{\ell m}|^2\,d\Omega=1.$$

The separated field is $\Phi=e^{-i\omega t}Y_{\ell m}R(x)/\sqrt h$. For absolute fluxes, take the action and stress tensor to be

$$\begin{aligned} S_{\mathrm m}&=-\frac12\int \sqrt{-g}\, \nabla_\mu\Phi^*\nabla^\mu\Phi\,d^4x,\\ T_{\mu\nu} &=\operatorname{Re}(\nabla_\mu\Phi^*\nabla_\nu\Phi) -\frac12g_{\mu\nu}\nabla_\alpha\Phi^*\nabla^\alpha\Phi. \end{aligned}$$

Thus $\Phi=\phi_1+i\phi_2$ represents two canonically normalized real scalars. A single real field $\operatorname{Re}\Phi$ has half these time-averaged fluxes; all ratios below are unchanged. This choice prevents a hidden factor of two when comparing complex amplitudes with real waves.

Writing $U=R/\sqrt h$, the exact radial equation is

$$\partial_r(\Delta\partial_r U) +\left(\frac{K^2}{\Delta}-\lambda\right)U=0, \qquad K=h\omega-am,\qquad \lambda=A_{0\ell m}+a^2\omega^2-2am\omega.$$

Here $A_{0\ell m}$ is the regular scalar angular eigenvalue. As derived in the scattering-basis construction, the rescaling gives $R''+Q R=0$ with real $Q$, where a prime denotes $d/dx$. Its endpoint limits are $Q\to\omega^2$ at infinity and $Q\to k_H^2$ at the horizon. Subtracting the conjugate radial equation gives

$$j=\frac{R^*R'-RR^{*\prime}}{2i}, \qquad j'=0.$$

For the current $n^\mu=(\Phi^*\nabla^\mu\Phi-\Phi\nabla^\mu\Phi^*)/(2i)$, positive radial flux is outward. Since $\sqrt{-g}=\Sigma\sin\theta$ and $g^{rr}=\Delta/\Sigma$, where $\Sigma=r^2+a^2\cos^2\theta$,

$$\int\sqrt{-g}\,n^r\,d\theta\,d\phi =\frac{\Delta}{2i} \left(U^*\partial_rU-U\partial_rU^*\right)=j.$$

The derivative of the real factor $1/\sqrt h$ cancels in this antisymmetric combination. This is the precise relation between the spacetime current and the radial Wronskian, not a general identification of every master-field current with a probability.

Energy and angular momentum from the stress tensor

Let $\xi=\partial_t$ and $\psi=\partial_\phi$. The conserved energy and axial angular-momentum currents are

$$\mathcal E^\mu=-T^\mu{}_\nu\xi^\nu,\qquad \mathcal L^\mu=T^\mu{}_\nu\psi^\nu.$$

For this mode, $\partial_t\Phi=-i\omega\Phi$ and $\partial_\phi\Phi=im\Phi$. The trace term in $T^\mu{}_\nu$ contributes nothing to its mixed $r,t$ and $r,\phi$ components. Direct substitution therefore gives

$$\begin{gathered} \mathcal E^r=\omega n^r,\qquad \mathcal L^r=m n^r,\\ \int\sqrt{-g}\,\mathcal E^r\,d\theta\,d\phi=\omega j,\qquad \int\sqrt{-g}\,\mathcal L^r\,d\theta\,d\phi=m j. \end{gathered}$$

These are rates per unit stationary time. Energy carries the asymptotic frequency $\omega$ even at the horizon; replacing it by $k_H$ would compute a different charge.

Consider incidence from infinity, with no wave emerging from the past horizon. Assume $k_H\ne0$, nonzero incident amplitude $I$, and finite matching amplitudes:

$$R\sim \begin{cases} T e^{-ik_Hx}, & x\to-\infty,\\ I e^{-i\omega x}+B e^{+i\omega x}, & x\to+\infty. \end{cases}$$

The endpoint diagram and regular ingoing coordinates explain why the first line enters the future horizon for either sign of $k_H$. Evaluating $j$ at the two ends gives

$$\omega(|B|^2-|I|^2)=-k_H|T|^2.$$

Define $F_{\mathrm{inc}}$ and $F_{\mathrm{ref}}$ as positive incident and reflected energy rates. Define horizon rates as positive into the hole, the opposite of the outward radial orientation:

$$\begin{gathered} F_{\mathrm{inc}}=\omega^2|I|^2,\qquad F_{\mathrm{ref}}=\omega^2|B|^2,\\ F_{E,H}=\omega k_H|T|^2,\qquad F_{J,H}=m k_H|T|^2. \end{gathered}$$

Consequently $F_{\mathrm{inc}}=F_{\mathrm{ref}}+F_{E,H}$ and $F_{J,H}=(m/\omega)F_{E,H}$. The physical horizon charge relation is also derived in Teukolsky and Press 1974, pp. 449–450, Eqs. (4.13), (4.16)–(4.26); their metric signature and real-wave averaging differ, so the absolute normalization here follows from the displayed action.

Signed absorption and positive horizon-generator energy

Define the reflection coefficient, greybody factor, and amplification by

$$\mathcal R=\left|\frac BI\right|^2,\qquad \Gamma_{\ell m}=\frac{F_{E,H}}{F_{\mathrm{inc}}} =\frac{k_H}{\omega}\left|\frac TI\right|^2,\qquad Z_{\ell m}=\mathcal R-1=-\Gamma_{\ell m}.$$

The energy balance $\mathcal R+\Gamma_{\ell m}=1$ holds in both frequency bands. For $k_H\gt0$ the greybody factor is nonnegative and at most one. For $k_H\lt0$, a nonzero transmitted field gives $\Gamma_{\ell m}\lt0$ and $\mathcal R\gt1$. The latter is superradiance: for positive frequency it requires $0\lt\omega\lt m\Omega_H$.

There is no negative quantum-event probability here. $\Gamma_{\ell m}$ is a signed energy-flux ratio. An absolute value around $k_H$ would remove the rotational energy extraction from the conservation law.

The horizon generator is $\chi=\xi+\Omega_H\psi$, so its absorbed charge is

$$F_{\chi,H}=F_{E,H}-\Omega_HF_{J,H}=k_H^2|T|^2\geq0.$$

This positivity also follows directly on the future horizon: $\chi^\mu\nabla_\mu\Phi=-ik_H\Phi$ and $\chi^2=0$ imply $T_{\mu\nu}\chi^\mu\chi^\nu=k_H^2|\Phi|^2$. The horizon area measure is $dA_H=(r_+^2+a^2)\sin\theta\,d\theta\,d\phi$. Integration cancels the $1/h$ in $|\Phi|^2$, reproducing $k_H^2|T|^2$ independently of the endpoint Wronskian.

For nonzero transmission, compare the charges rather than the direction of the wave:

Frequency band	$F_{E,H}$	$\Omega_HF_{J,H}$	$F_{\chi,H}$
$k_H\gt0$	Positive	Depends on $m\Omega_H$	Positive
$k_H\lt0$	Negative	Negative	Positive

The last row is compatible with increasing horizon area. If a weak, regular finite wave packet is allowed to backreact and the horizon settles to stationarity, the leading physical-process first law gives

$$\delta M-\Omega_H\delta J=\frac{\kappa}{8\pi G}\delta A_H\geq0,$$

with background surface gravity $\kappa\gt0$. The assumptions and derivation belong to the physical-process first law. A monochromatic calculation on a fixed background supplies flux rates; it does not itself evolve the black hole or prove this relation for a violent finite change.

From greybody factors to a cross section

A greybody factor compares powers in a specified partial wave and has no units. An absorption cross section compares total absorbed power with the energy-flux density of an incident plane wave:

$$\sigma_{\mathrm{abs}} =\frac{\text{absorbed energy per unit time}} {\text{incident energy per unit time per unit area}}.$$

It therefore has units of area. For a Kerr scalar plane wave arriving at angle $\theta_{\mathrm{inc}}$ to the spin axis, the chosen angular normalization gives

$$\sigma_{\mathrm{abs}}(\omega,\theta_{\mathrm{inc}}) =\frac{4\pi^2}{\omega^2} \sum_{\ell=0}^{\infty}\sum_{m=-\ell}^{\ell} |Y_{\ell m}(\theta_{\mathrm{inc}},\phi_{\mathrm{inc}};a\omega)|^2 \Gamma_{\ell m}.$$

The azimuth drops out of the squared harmonic. This formula, including its angular normalization, is given in Macedo et al. 2013, §§ II–III, p. 2, Eqs. (4), (11)–(14), PDF. Kerr needs these directional weights; a spherical degeneracy cannot replace them. A negative individual $\Gamma_{\ell m}$ does not by itself determine the sign of the complete plane-wave sum.

For Schwarzschild, $\Gamma_{\ell m}=\Gamma_\ell$ and the spherical addition theorem gives $\sum_m|Y_{\ell m}|^2=(2\ell+1)/(4\pi)$. Hence

$$\sigma_{\mathrm{abs}}(\omega) =\frac{\pi}{\omega^2}\sum_{\ell=0}^{\infty}(2\ell+1)\Gamma_\ell.$$

One can see where its area factor comes from without memorizing a cross-section formula. In an asymptotic spherical-wave basis, a unit plane wave along the polar axis has the expansion

$$e^{i\omega z} =\sum_{\ell=0}^\infty i^\ell\sqrt{4\pi(2\ell+1)} \,j_\ell(\omega r)Y_{\ell0}(\theta,\phi), \qquad j_\ell(z)\sim\frac{\sin(z-\ell\pi/2)}{z},$$

where $j_\ell$ is a spherical Bessel function. Separating the sine into incoming and outgoing waves shows that a plane-wave amplitude $A_{\mathrm{plane}}$ supplies

$$\left|\frac{I_\ell}{A_{\mathrm{plane}}}\right|^2 =\frac{\pi(2\ell+1)}{\omega^2}.$$

Long-range tortoise-coordinate phases change the asymptotic phase, not this modulus. The incident energy density flux is $\omega^2|A_{\mathrm{plane}}|^2$; the absorbed power in this partial wave is $\Gamma_\ell\omega^2|I_\ell|^2$. Their ratio is exactly the corresponding term in $\sigma_{\mathrm{abs}}$.

Low-frequency scalar absorption by Schwarzschild

Set $r_H=2GM$, the areal horizon radius, and $\epsilon=\omega r_H\ll1$. For the s-wave $\ell=0$, let $f=1-r_H/r$ and $U=R/r$. The exact equation becomes

$$\frac{d}{dr}\left(r^2f\frac{dU}{dr}\right) +\frac{\omega^2r^2}{f}U=0.$$

At zero frequency it integrates to $U=C+D\ln f$. Match this solution to a regular ingoing wave: near the horizon $x=r_H\ln f+O(r_H)$ up to an additive constant. Expanding $R\sim T e^{-i\omega x}$ in the common region $0\lt f\ll1$ and $\epsilon|\ln f|\ll1$ gives, to leading matching order,

$$C\sim\frac{T}{r_H},\qquad D\sim-i\omega T.$$

For example, $f=\epsilon$ satisfies both conditions as $\epsilon\to0$. The expansion is not uniform at $f=0$ itself.

The zero-frequency solution extends to the intermediate region $r_H\ll r\ll\omega^{-1}$. There $\ln f\sim-r_H/r$, so

$$U\sim\frac{T}{r_H}+\frac{i\omega T r_H}{r}.$$

In the same overlap, the leading far-region solution expands as

$$U\sim\frac{I e^{-i\omega r}+B e^{i\omega r}}{r} \sim\frac{I+B}{r}+i\omega(B-I).$$

Matching the constant and inverse-radius terms determines the leading incident amplitude. In particular,

$$\begin{gathered} I\sim\frac{iT}{2\omega r_H},\qquad \Gamma_0=\left|\frac TI\right|^2\sim4(\omega r_H)^2,\\ \sigma_0=\frac{\pi}{\omega^2}\Gamma_0\longrightarrow4\pi r_H^2=A_H. \end{gathered}$$

The individual ratio vanishes, but its plane-wave normalization grows as $\omega^{-2}$, leaving a finite area. Higher partial waves are suppressed in this limit, so the total scalar cross section also tends to $A_H$. This matching and area result are consistent with Das, Gibbons, and Mathur 1997, preprint pp. 1–2, Eqs. (5)–(16), PDF. Their capital $R_H$ is the areal radius used here; their lowercase $r_H$ is an isotropic coordinate radius.

This is a leading low-frequency result for a minimally coupled massless scalar. The matching neglects higher-order terms and long-range phase corrections; it does not determine a resummed finite-frequency transmission formula. Nor is the area limit a field-independent assertion about electromagnetic or gravitational waves.

Thresholds and normalization checks

At $k_H=0$, the horizon flux tends to zero if the matched amplitude has a controlled finite limit. A unit-flux basis is singular there: one must take a limit before dividing by its vanishing normalization. The signed-current reciprocity relations, including their superradiant form, are derived on the scattering-bases page.

For a massive scalar obeying $(\Box_g-\mu^2)\Phi=0$, where $\mu$ is the inverse-length mass parameter, take $\omega\gt\mu$. The open infinity channel has $q_\infty=\sqrt{\omega^2-\mu^2}$. Its incident energy rate is $\omega q_\infty|I|^2$, not $q_\infty^2|I|^2$, and the corresponding signed ratio is $(k_H/q_\infty)|T/I|^2$ in the same radial normalization. Below the propagation threshold there is no incoming plane-wave scattering channel at infinity.

A changed radial field, a spin-weighted Teukolsky variable, or a coupled system requires its actual physical flux form. In particular, the electromagnetic and gravitational flux factors are not obtained by relabeling this scalar $T$. Likewise, classical absorption alone neither selects a quantum state nor determines Hawking occupation numbers.

Exercises

Rotational energy extraction. Let $k_H=-\omega/4$ and $|T/I|^2=1/2$. Find $\Gamma$, $\mathcal R$, $\Omega_HF_{J,H}/F_{\mathrm{inc}}$, and $F_{\chi,H}/F_{\mathrm{inc}}$. Explain how the mass can decrease while the area increases.

Solution

Since $m\Omega_H/\omega=5/4$,

$$\begin{gathered} \Gamma=-\frac18,\qquad \mathcal R=\frac98,\\ \frac{\Omega_HF_{J,H}}{F_{\mathrm{inc}}} =\frac{m\Omega_H}{\omega}\Gamma=-\frac5{32},\qquad \frac{F_{\chi,H}}{F_{\mathrm{inc}}} =\Gamma-\frac54\Gamma=\frac1{32}. \end{gathered}$$

The reflected energy is $12.5\%$ larger than the incident energy. For a weak finite packet satisfying the physical-process assumptions, negative absorbed Killing energy lowers the mass, while the positive combination $\delta M-\Omega_H\delta J$ increases the area. The wave still enters the future horizon.

Changing the radial variable. Suppose $\widehat R=b(r)R$, where $b$ is real, smooth, positive, and approaches constants $b_H$ and $b_\infty$ at the two ends. Determine the conserved current in terms of $\widehat R$ and the absorption formula in terms of its amplitudes. What goes wrong if one uses the uncorrected formula when $b_H=2$ and $b_\infty=3$?

Solution

Antisymmetry cancels the terms proportional to $b'$:

$$\widehat j=\frac{\widehat R^*\widehat R' -\widehat R\widehat R^{*\prime}}{2i}=b^2j, \qquad (b^{-2}\widehat j)'=0.$$

Thus $\widehat j$ itself need not be conserved. Since $\widehat T=b_HT$, $\widehat I=b_\infty I$, and $\widehat B=b_\infty B$,

$$\Gamma=\frac{k_H}{\omega}\frac{b_\infty^2}{b_H^2} \left|\frac{\widehat T}{\widehat I}\right|^2, \qquad \mathcal R=\left|\frac{\widehat B}{\widehat I}\right|^2.$$

The naive absorption formula is $4/9$ of the correct value in the specified example; its missing factor is $9/4$. A common constant rescaling cancels, but unequal endpoint normalizations do not.

References

• Das, S. R., Gibbons, G. W., and Mathur, S. D. (1997). “Universality of Low Energy Absorption Cross Sections for Black Holes.” Physical Review Letters 78, 417–419. DOI. Open PDF.
• Macedo, C. F. B., Leite, L. C. S., Oliveira, E. S., Dolan, S. R., and Crispino, L. C. B. (2013). “Absorption of planar massless scalar waves by Kerr black holes.” Physical Review D 88, 064033. DOI. Open PDF.
• Teukolsky, S. A., and Press, W. H. (1974). “Perturbations of a rotating black hole. III. Interaction of the hole with gravitational and electromagnetic radiation.” Astrophysical Journal 193, 443–461. DOI.