Scalar Fields and Effective Potentials

A scalar wave is the simplest example of black-hole scattering: angular momentum, curvature, and field mass become terms in a one-dimensional effective potential. On Schwarzschild we will derive that potential, reconstruct the energy carried by a real field, and distinguish waves reaching infinity from fields decaying there. On subextremal Kerr the same calculation also separates the angular equation and reveals how rotation can reverse the stationary horizon-energy flux. Throughout, the scalar is neutral, minimally coupled, and classical; the geometry is fixed and no quantum particle interpretation is assumed.

Required background. PDEs and Green Functions for Gravitation supplies hyperbolic equations, characteristics, mode boundary conditions, and conserved Wronskians.

Helpful background. Black-Hole Geometry and Mechanics supplies stationary horizons and Kerr geometry; Formulations of General Relativity supplies continuum field and boundary formulations; Frames, Coframes, and Anholonomy supplies the frame language used later in the chapter.

Klein–Gordon propagation on Schwarzschild

The normalization starts with a real scalar $\Phi$, a nonnegative mass $\mu$, and the action

$$S_\Phi=-\frac12\int \left(\nabla_\alpha\Phi\nabla^\alpha\Phi+\mu^2\Phi^2\right) \sqrt{-g}\,d^4x.$$

Varying $\Phi$ and the inverse metric gives, respectively,

$$\begin{aligned} (\Box-\mu^2)\Phi&=0,\\ T_{\alpha\beta} &=\nabla_\alpha\Phi\nabla_\beta\Phi -\frac12g_{\alpha\beta} \left(\nabla_\gamma\Phi\nabla^\gamma\Phi+\mu^2\Phi^2\right). \end{aligned}$$

In particular, $\nabla^\alpha T_{\alpha\beta}=[(\Box-\mu^2)\Phi]\nabla_\beta\Phi=0$ on solutions. These are the minimally coupled equations, not an arbitrary curvature-coupled scalar model Wald 1984, §4.3, p. 70, Eqs. (4.3.9)–(4.3.10). We calculate this stress tensor but do not evolve its gravitational backreaction.

Let $m=GM\gt 0$ denote the black hole’s mass as a length. In the four-dimensional Schwarzschild exterior,

$$ds^2=-f(r)dt^2+f(r)^{-1}dr^2+r^2d\Omega^2, \qquad f(r)=1-\frac{2m}{r}, \qquad r\gt 2m.$$

The coordinate form of the scalar wave operator is

$$\Box\Phi =-\frac{1}{f}\partial_t^2\Phi +\frac{1}{r^2}\partial_r(r^2f\partial_r\Phi) +\frac{1}{r^2}\nabla_{S^2}^2\Phi.$$

To separate space without making the physical field complex, use real sphere harmonics $Y_{\ell q}$, with $\ell=0,1,\ldots$ and $q=1,\ldots,2\ell+1$. They obey

$$-\nabla_{S^2}^2Y_{\ell q}=L Y_{\ell q}, \qquad L=\ell(\ell+1), \qquad \int_{S^2}Y_{\ell q}Y_{\ell' q'}\,d\Omega =\delta_{\ell\ell'}\delta_{qq'}.$$

The label $q$ here is only a real-basis label, not an azimuthal quantum number. Treat one real mode $\Phi=\Psi(t,x)Y_{\ell q}/r$; arbitrary real fields are sums of these modes. The tortoise coordinate

$$x=r_*=r+2m\ln\left(\frac{r}{2m}-1\right), \qquad \partial_x=f\partial_r$$

runs from $-\infty$ at the horizon to $+\infty$ at spatial infinity. The factor $1/r$ removes the first radial derivative that would otherwise describe spherical spreading. Indeed, its radial contribution satisfies

$$fr\left[\frac{1}{r^2}\partial_r \left(r^2f\partial_r\frac{\Psi}{r}\right)\right] =\partial_x^2\Psi-\frac{ff'}{r}\Psi, \qquad f'=\frac{df}{dr}.$$

Multiplying the separated field equation by $fr$ therefore gives

$$\begin{aligned} \left[-\partial_t^2+\partial_x^2-V_\ell(r)\right]\Psi&=0,\\ V_\ell(r)&=f(r)\left[\frac{L}{r^2}+\frac{2m}{r^3}+\mu^2\right]. \end{aligned}$$

Each potential term has a different origin. The first is the angular-gradient energy; the second comes from differentiating the spherical rescaling in a curved metric; the last is the redshifted rest-mass term. All have dimensions of inverse length squared. Setting $m=0$ recovers the flat radial Klein–Gordon equation. Setting $\ell=\mu=0$ does not remove Schwarzschild scattering: $V_0=2mf/r^3$ remains. Unlike an electromagnetic charge monopole, a scalar monopole can propagate.

Physical energy and radial rescaling

An effective potential describes the rescaled field, not a new matter stress tensor. To connect it to measurable flux, let $\xi=\partial_t$ and define the outward Killing-energy current $J^\alpha=-T^\alpha{}_{\beta}\xi^\beta$. On a Schwarzschild $t$ slice, the future unit normal is $n^\alpha=f^{-1/2}\xi^\alpha$ and the volume measure is $r^2f^{-1/2}dr\,d\Omega$. Hence, on a finite radial interval,

$$\begin{aligned} E_{\rm phys} &=\int T_{\alpha\beta}n^\alpha\xi^\beta\,d\Sigma =\int_{x_1}^{x_2}\mathcal E_{\rm phys}\,dx,\\ \mathcal E_{\rm phys} &=\frac12\left[ \Psi_t^2+\left(\Psi_x-\frac{f}{r}\Psi\right)^2 +f\left(\frac{L}{r^2}+\mu^2\right)\Psi^2\right],\\ P_{\rm phys} &=-r^2\int_{S^2}T^r{}_t\,d\Omega =-\Psi_t\left(\Psi_x-\frac{f}{r}\Psi\right). \end{aligned}$$

Subscripts $t$ and $x$ denote partial derivatives. The angular-gradient integral is $\int|\nabla_{S^2}Y_{\ell q}|^2d\Omega=L$. Positive $P_{\rm phys}$ means energy moves toward larger $r$, including when it is evaluated near the inner endpoint; absorbed horizon energy has the opposite sign. The conservation law gives $dE_{\rm phys}/dt=P_{\rm phys}(x_1)-P_{\rm phys}(x_2)$.

The wave equation has its own convenient energy density and power,

$$\mathcal E_{\rm master} =\frac12(\Psi_t^2+\Psi_x^2+V_\ell\Psi^2), \qquad P_{\rm master}=-\Psi_t\Psi_x.$$

Multiplication by $\Psi_t$ proves $\partial_t\mathcal E_{\rm master}+\partial_xP_{\rm master}=0$. But the two densities are not equal pointwise. Writing $b=f/r$, direct expansion gives

$$\begin{aligned} \mathcal E_{\rm phys} &=\mathcal E_{\rm master}-\frac12\partial_x(b\Psi^2),\\ P_{\rm phys}&=P_{\rm master}+\frac12\partial_t(b\Psi^2),\\ E_{\rm phys}&=E_{\rm master} -\left[\frac{f\Psi^2}{2r}\right]_{x_1}^{x_2}. \end{aligned}$$

The endpoint term vanishes for smooth compactly supported data on the tortoise line, or under explicitly imposed falloffs that make it vanish. Those are sufficient conditions for the integrated energies to agree. They do not justify dropping the term on an arbitrary finite interval. Orthogonality makes integrated energies additive over the real harmonics; it does not remove pointwise cross terms in the original field.

For $\mu^2\geq0$ the displayed Schwarzschild physical energy is nonnegative. It nevertheless degenerates as the stationary Killing field becomes null at the horizon. Conservation alone proves neither nondegenerate derivative control nor decay; those require the additional arguments in Energy Estimates and Test-Field Stability.

Monochromatic waves and conserved flux

Now let $\Psi(t,x)=\operatorname{Re}[\psi(x)e^{-i\omega t}]$ with real $\omega\gt 0$. The complex function $\psi$ is an amplitude used to represent one real wave. Its equation is

$$\psi''+(\omega^2-V_\ell)\psi=0, \qquad {}'=\frac{d}{dx}.$$

Subtract its conjugate equation to obtain

$$W=\psi^*\psi'-\psi(\psi')^*, \qquad W'=0, \qquad j=\frac{W}{2i}=\operatorname{Im}(\psi^*\psi').$$

The physical normalization follows from $\langle\operatorname{Re}(Ae^{-i\omega t})\operatorname{Re}(Be^{-i\omega t})\rangle=\tfrac12\operatorname{Re}(AB^*)$, where brackets mean a period average. In particular, $\langle\Psi\Psi_t\rangle=0$, so the extra rescaling term in the power averages to zero:

$$\langle P_{\rm phys}\rangle =\frac{\omega}{2}\operatorname{Im}(\psi^*\psi') =\frac{\omega}{4i}W.$$

For $\psi=Ae^{\pm ikx}$ with $k\gt 0$, the auxiliary current is $j=\pm k|A|^2$ and the averaged energy power is $\pm\omega k|A|^2/2$. Thus $j$ alone is not the physical energy power. A real scalar has no nontrivial continuous phase symmetry and no corresponding classical U(1) particle-number current. Its complex-mode Wronskian remains a useful calculation tool without assigning particles to the wave.

There is also a precise limit to the conjugate-Wronskian argument. For a complex $\omega$ in the Schwarzschild equation with the same real, frequency-independent potential,

$$W'=-2i\operatorname{Im}(\omega^2)|\psi|^2.$$

It is $\omega^2$ being real that conserves this Wronskian, so purely imaginary frequencies are an exception to the slogan “complex frequency destroys it.” They do not supply a stationary oscillation with the period average used above. Generic quasinormal frequencies require an analytically continued boundary problem, not a positive conserved real-wave flux.

Horizon and massive infinity

At the future horizon $V_\ell\to0$. With $v=t+x$, the ingoing radial factor $\psi\sim T_H e^{-i\omega x}$ makes the full mode proportional to $e^{-i\omega v}Y_{\ell q}/r$. Smoothness is imposed in these horizon-penetrating coordinates, not in a static orthonormal frame. For a regular monochromatic ingoing branch its limiting outward power is $-\omega^2|T_H|^2/2$. The mass term is redshifted away in this radial limit, not removed from the full field equation.

At infinity a massive field is qualitatively different from a massless one:

$$V_\ell(r)=\mu^2-\frac{2m\mu^2}{r}+\frac{L}{r^2}+O(r^{-3}).$$

For real $\omega\gt \mu$, define $k_\infty=\sqrt{\omega^2-\mu^2}\gt 0$. The $1/r$ term is long-range: plane waves with only the phase $k_\infty x$ omit a phase that grows logarithmically. To leading asymptotic order use

$$\begin{aligned} \Theta(r)&=k_\infty x+\frac{m\mu^2}{k_\infty}\ln\left(\frac{r}{r_0}\right),\\ \psi_{\rm out/in}&\sim e^{\pm i\Theta(r)}, \qquad r_0\gt 0. \end{aligned}$$

Indeed, $\partial_x\Theta=k_\infty+m\mu^2/(k_\infty r)+O(r^{-2})$; its square cancels the $2m\mu^2/r$ term in $\omega^2-V_\ell$. The remaining amplitude and phase corrections are subleading at fixed nonzero $k_\infty$. The arbitrary reference length $r_0$ changes a constant phase, not a flux. Since $x=r+2m\ln(r/(2m))+O(r^{-1})$, the logarithmic coefficient written with the leading phase $k_\infty r$ instead is $m(2\omega^2-\mu^2)/k_\infty$.

The table summarizes the asymptotic factors for unit amplitudes. Fluxes refer to the single real harmonic above.

Boundary and regime	Radial factor	Averaged outward power
Future horizon, real $\omega\gt 0$	$e^{-i\omega x}$	$-\omega^2/2$
Infinity, $\omega\gt \mu$, incoming	$e^{-i\Theta}$	$-\omega k_\infty/2$
Infinity, $\omega\gt \mu$, outgoing	$e^{+i\Theta}$	$+\omega k_\infty/2$
Infinity, $0\lt \omega\lt \mu$, decaying	$e^{-\kappa_\infty x}(r/r_0)^{m\mu^2/\kappa_\infty}$	$0$

In the last row $\kappa_\infty=\sqrt{\mu^2-\omega^2}\gt 0$. For a complex quasibound frequency, the locally decaying branch is chosen by $\operatorname{Re}\kappa_\infty\gt 0$; there is then no stationary period-averaged flux formula. The power multiplying the exponential is important in both cases. None of these infinity expansions is uniform at $\omega^2=\mu^2$, and setting $k_\infty=0$ in the traveling-wave formulas does not solve the threshold problem. The long-range powers agree with the Schwarzschild limit of Dolan 2007, §II.C, p. 084001-4, Eq. (30), PDF, after restoring $GM$ and changing from his exponential in $r$ to ours in $x$.

For unit incidence from infinity and no outgoing horizon branch, write

$$\psi\sim \begin{cases} T_H e^{-i\omega x},&x\to-\infty,\\ e^{-i\Theta}+R_\infty e^{+i\Theta},&x\to+\infty. \end{cases}$$

Constancy of $j$ gives the energy balance

$$k_\infty(1-|R_\infty|^2)=\omega|T_H|^2, \qquad \mathscr R=|R_\infty|^2, \qquad \mathscr A=\frac{\omega}{k_\infty}|T_H|^2=1-\mathscr R.$$

Here $\mathscr R$ and $\mathscr A$ are reflected and absorbed energy fractions. For $\mu=0$, $k_\infty=\omega$ and the familiar unweighted amplitude identity follows. For $\mu\gt 0$ it generally does not. The limiting group speed at infinity is $k_\infty/\omega$, another way to see why equal amplitudes need not carry equal power.

A scattering problem fixes the incident wave. A quasinormal problem removes that incident wave and continues the outgoing condition to a specified complex-frequency sheet. A quasibound problem instead selects decay at infinity and ingoing behavior at the horizon. These are different boundary problems, not three names for one spectrum. In particular, the decaying row of the table is a local infinity condition, not evidence that a global real-frequency bound mode exists. See Quasinormal Boundary Conditions and Resonances for continuation and Massive Boson Quasi-Bound States for the confined problem.

The potential gives intuition but not scattering coefficients by inspection. Where $\omega^2-V_\ell$ is positive, a slowly varying approximation gives an oscillatory local wave; where it is negative, it gives exponential behavior. At a turning point that approximation fails. Tunneling, reflection, and any resonance still depend on the full equation and both boundaries.

Massive scalar separation on Kerr

Let $a=J_{\rm BH}/M$, $|a|\lt m=GM$, and use Boyer–Lindquist coordinates on $r\gt r_+$. Define

$$\begin{aligned} \Delta&=r^2-2mr+a^2, &\Sigma&=r^2+a^2\cos^2\theta,\\ r_\pm&=m\pm\sqrt{m^2-a^2}, &\Omega_H&=\frac{a}{r_+^2+a^2}. \end{aligned}$$

The separation is visible in the coordinate operator itself. Multiplying the Klein–Gordon equation by $\Sigma$ gives

$$\begin{aligned} 0={}&\Bigg\{ \partial_r(\Delta\partial_r) +\frac{1}{\sin\theta}\partial_\theta(\sin\theta\partial_\theta) -\mu^2(r^2+a^2\cos^2\theta)\\ &-\left[\frac{(r^2+a^2)^2}{\Delta}-a^2\sin^2\theta\right]\partial_t^2 -\frac{4mar}{\Delta}\partial_t\partial_\phi\\ &+\left(\frac{1}{\sin^2\theta}-\frac{a^2}{\Delta}\right)\partial_\phi^2 \Bigg\}\Phi. \end{aligned}$$

Take the real part of $R(r)S(\theta)e^{-i\omega t+im_z\phi}$, where $m_z\in\mathbb Z$ is now the azimuthal number. Stationarity and axial symmetry replace $\partial_t$ and $\partial_\phi$ by $-i\omega$ and $im_z$. The remaining terms separate because $\Sigma\mu^2=\mu^2r^2+\mu^2a^2\cos^2\theta$. With a separation constant $A_{\ell m_z}$, the angular equation is

$$\frac{1}{\sin\theta}\frac{d}{d\theta} \left(\sin\theta\frac{dS}{d\theta}\right) +\left[a^2(\omega^2-\mu^2)\cos^2\theta -\frac{m_z^2}{\sin^2\theta}+A_{\ell m_z}\right]S=0.$$

Require $S e^{im_z\phi}$ smooth at both poles. For real $\omega$, choose $2\pi\int_0^\pi|S|^2\sin\theta\,d\theta=1$; the angular eigenvalue is real and $\ell\geq|m_z|$ labels the branch tending to $\ell(\ell+1)$ as $a\to0$. The radial equation is

$$\begin{aligned} \frac{d}{dr}\left(\Delta\frac{dR}{dr}\right) +\left(\frac{K^2}{\Delta}-\mu^2r^2-\lambda\right)R&=0,\\ K&=(r^2+a^2)\omega-am_z,\\ \lambda&=A_{\ell m_z}+a^2\omega^2-2am_z\omega. \end{aligned}$$

The angular parameter is $a^2(\omega^2-\mu^2)$, but the radial shift contains $a^2\omega^2$, not that angular parameter. In the Schwarzschild limit, set $a=0$, $A_{\ell m_z}=L$, and $R=\psi/r$ to recover the potential derived above.

These massive equations follow directly from Dolan 2007, §II.C, p. 084001-4, Eqs. (25)–(28), PDF. His angular $\Lambda_{\ell m}$ is our $A_{\ell m_z}$, not our $\lambda$. He uses the opposite metric signature, so $\Box_{\rm here}=-\Box_{\rm Dolan}$ and his $(\Box_{\rm Dolan}+\mu^2)\Phi=0$ becomes our equation by multiplication by $-1$; his geometrized $M$ is our $GM$. The spin-zero Teukolsky equation is the massless specialization, not the source of the additional massive terms.

Kerr horizon energy

For the following stationary flux calculation keep $\omega\gt 0$ real. The Kerr tortoise coordinate satisfies $dx/dr=(r^2+a^2)/\Delta$, and the horizon generator is $\chi=\partial_t+\Omega_H\partial_\phi$. A mode has frequency

$$k_H=\omega-m_z\Omega_H$$

along that generator. This is not the locally measured frequency of a static observer, since no such timelike observer can remain at the horizon.

Use regular coordinates $v=t+x$ and $\widetilde\phi=\phi+\int a\,dr/\Delta$. The transformed radial factor is $R_{\rm reg}=\exp(i\omega x-im_z\int a\,dr/\Delta)R$. Future-horizon smoothness selects the regular branch of $R_{\rm reg}$ and, for nonzero $k_H$, implies

$$R\sim T_H^{(R)}e^{-ik_Hx} \ \propto\ \left(\frac{r-r_+}{r_0}\right)^{-i\sigma}, \qquad \sigma=\frac{(r_+^2+a^2)k_H}{r_+-r_-}.$$

Only the leading power is displayed; full regularity concerns the reconstructed field in $(v,r,\theta,\widetilde\phi)$. The formula assumes a nonextremal horizon. At $k_H=0$ the two traveling-wave powers coalesce and the regular solution must be distinguished from the logarithmic second branch separately.

For real frequency and the regular angular eigenfunction, the radial equation conserves

$$\begin{aligned} W_\Delta&=\Delta\left(R^*\frac{dR}{dr}-R\frac{dR^*}{dr}\right), &\frac{dW_\Delta}{dr}&=0,\\ j_\Delta&=\frac{W_\Delta}{2i}, &\langle P_E\rangle&=-\int\Sigma\sin\theta\, \langle T^r{}_t\rangle\,d\theta\,d\phi =\frac{\omega}{2}j_\Delta. \end{aligned}$$

The factor $\Sigma\sin\theta=\sqrt{-g}$ matters. For example, $T^r{}_t=(\Delta/\Sigma)\partial_r\Phi\partial_t\Phi$; performing the real-wave time average and the normalized angular integral yields the last equality. The angular-momentum power defined with $+T^r{}_{\phi}$ is similarly $\langle P_J\rangle=m_z j_\Delta/2$.

Let $\mathcal C_H=r_+^2+a^2$. With the raw radial amplitude $T_H^{(R)}$ defined above, the absorbed rates per unit advanced time $v$, positive into the hole, are

$$\begin{aligned} \langle\dot E_H\rangle &=\frac12\omega k_H\mathcal C_H|T_H^{(R)}|^2,\\ \langle\dot J_H\rangle &=\frac12m_z k_H\mathcal C_H|T_H^{(R)}|^2,\\ \langle\dot E_H-\Omega_H\dot J_H\rangle &=\frac12k_H^2\mathcal C_H|T_H^{(R)}|^2\geq0. \end{aligned}$$

The last line also follows directly from $T_{\alpha\beta}\chi^\alpha\chi^\beta=(\chi^\alpha\nabla_\alpha\Phi)^2\geq0$ on the horizon. Generator energy is nonnegative, while stationary Killing energy is negative in the superradiant range $0\lt \omega\lt m_z\Omega_H$. There is no contradiction: these are different contractions of the same stress tensor.

The normalization agrees with Dolan 2007, §II.B, p. 084001-3, Eqs. (14)–(20), PDF after translating the mixed stress sign and taking one real field. His half-normalized complex field represents two canonical real scalars; the period-averaged stress of its real part at real frequency is half its stationary complex-field stress. The later Reflection, Transmission, and Absorption calculation also uses that complex-field normalization, so its absolute fluxes are twice ours for the same complex representative; flux ratios agree.

There is a second useful translation. Later scattering calculations use the rescaled radial field $u=\sqrt{r^2+a^2}\,R$. Its horizon amplitude is $T_H^{(u)}=\sqrt{\mathcal C_H}\,T_H^{(R)}$, and therefore absorbs the explicit $\mathcal C_H$ in the rates above. For $\omega\gt\mu$, at infinity $R\sim r^{-1}(A_{\rm in}e^{-i\Theta}+A_{\rm out}e^{i\Theta})$, with the same leading massive Coulomb phase and $j_\Delta\to k_\infty(|A_{\rm out}|^2-|A_{\rm in}|^2)$. A raw amplitude square is not a flux until these radial, angular, and real-versus-complex normalizations are fixed.

Kerr’s coefficients depend on $\omega$, including linearly through $am_z\omega$, and its angular eigenvalue is generally complex at complex frequency. The Schwarzschild imaginary-frequency Wronskian exception therefore cannot be transferred automatically. Separation is also not a proof of a self-adjoint resonance problem or a global stability theorem. Rotational amplification and confinement are developed in Superradiant Frequency Condition; the present calculation establishes the signs and normalizations they must use.

Common pitfalls

A potential is not an invariant landscape. Its form depends on the radial coordinate and field rescaling. Compare reconstructed fields and boundary-normalized flux ratios, not just the heights of differently defined potentials.

Conserved does not mean positive. A Wronskian is a signed radial bilinear. The Schwarzschild physical energy has a positivity property under the assumptions above, while Kerr stationary energy need not; neither conservation statement alone supplies decay.

A mode is not a finite packet. Monochromatic scattering solutions are useful generalized solutions, usually with infinite total energy on an entire stationary slice. Their averaged powers describe flux per unit time. Finite-energy evolution requires suitable initial data or wave packets, and quantum particles require additional state-dependent structure.

Exercises

A barrier without angular momentum. Recover the curvature term in $V_\ell$ from the radial rescaling. Then locate the maximum of the massless $\ell=0$ potential. Diagnose formulas that omit $2m/r^3$ or place $\mu^2$ outside $f$.

Solution

The radial differentiation displayed above leaves $ff'/r=2mf/r^3$. For $\ell=\mu=0$,

$$V_0=\frac{2m}{r^3}-\frac{4m^2}{r^4}, \qquad \frac{dV_0}{dr}=\frac{2m(8m-3r)}{r^5}.$$

The derivative changes from positive to negative at

$$r_{\rm peak}=\frac{8m}{3}, \qquad V_0(r_{\rm peak})=\frac{27}{1024m^2}.$$

Thus omitting curvature wrongly deletes the entire massless monopole barrier. Moving $\mu^2$ outside $f$ gets the infinity threshold right but the horizon limit wrong: it leaves a nonzero mass potential there. Both errors fail direct substitution. The finite-multipole peak at $8m/3$ is also not the photon orbit at $3m$; the latter controls the leading large-angular-momentum potential, not this monopole.

Amplitude versus absorbed energy. A massive Schwarzschild wave has $\omega=5\mu/4$ with $\mu\gt 0$ and reflected fraction $|R_\infty|^2=2/5$. Find $k_\infty$, $|T_H|^2$, and the absorbed energy fraction. Does the Coulomb phase change the limiting flux weights?

Solution

The dispersion relation and flux balance give

$$k_\infty=\frac{3\mu}{4}, \qquad |T_H|^2=\frac{k_\infty}{\omega}\left(1-\frac25\right) =\frac9{25}, \qquad \mathscr A=\frac{\omega}{k_\infty}|T_H|^2=\frac35.$$

The absorbed fraction is not the horizon amplitude squared. The Coulomb phase has $\partial_x\Theta\to k_\infty$, so it changes the asymptotic phase matching but not these limiting weights. The inference uses measured reflection as input; it does not calculate a reflection coefficient from the potential.

A constant field and an endpoint term. Let $\mu=\ell=0$ and take $\Phi=C Y_{0,1}$. Here $C$ is constant and the normalized harmonic is $Y_{0,1}=1/\sqrt{4\pi}$. Compute the physical and master energies between $r=3m$ and $r=6m$. Explain why they need not agree.

Solution

All derivatives of $\Phi$ vanish, so its stress tensor and physical energy vanish. But its master field is $\Psi=Cr$, giving $\Psi_x=Cf$ and

$$\mathcal E_{\rm master}=\frac12C^2f, \qquad E_{\rm master}=\frac12C^2\int_{3m}^{6m}dr =\frac32mC^2.$$

Precisely the same amount occurs in the endpoint term:

$$\left[\frac{f\Psi^2}{2r}\right]_{3m}^{6m} =\left[\frac{C^2(r-2m)}2\right]_{3m}^{6m} =\frac32mC^2, \qquad E_{\rm phys}=0.$$

This exact finite-annulus example tests the rescaling identity. It is not compactly supported data and does not have finite master energy on the whole tortoise line, so it does not contradict equality under the falloff assumptions stated earlier.

References

• Dolan, Sam R. “Instability of the Massive Klein–Gordon Field on the Kerr Spacetime.” Physical Review D 76 (2007): 084001. DOI. Open PDF.
• Wald, Robert M. General Relativity. Chicago: University of Chicago Press, 1984. Publisher.