Chandrasekhar Transformations and Isospectrality

Why can two unequal potentials produce the same ringing frequencies? For vacuum Schwarzschild gravitational perturbations, a first-order Darboux transformation, also called a Chandrasekhar transformation, maps axial solutions to polar solutions. Away from its exceptional frequencies, it preserves the quasinormal boundary problem and gives equal real-frequency reflection and transmission probabilities. The algebraically special frequency requires a separate analysis: the differential map loses its inverse, and a continued boundary solution can differ from the wave suggested by its leading exponential. We derive the ordinary equivalence first, then explain this exception and what changes at a reflecting boundary.

Required background. PDEs and Green Functions for Gravitation supplies operator domains, Wronskians, and boundary-value reasoning.

Helpful background. Black-Hole Geometry and Mechanics fixes the Schwarzschild endpoints; Formulations of General Relativity supplies the reduced field equations; Presymplectic Potentials and Currents prepares flux comparisons; Black-Hole Perturbation Theory supplies the master-field conventions.

Partner potentials on the Schwarzschild exterior

Work on the vacuum exterior $r\gt2m$, with $m=GM\gt0$, $x=r_*$, $f=1-2m/r$ and $D=d/dx=f\,d/dr$. For a gravitational multipole $\ell\geq2$, set

$$\lambda=\frac{(\ell-1)(\ell+2)}2, \qquad \alpha=\frac{\lambda(\lambda+1)}{3m}.$$

The two radial equations have the form

$$H\psi=\omega^2\psi,\qquad H=-D^2+V.$$

Their potentials are

$$V_{\mathrm{RW}} =f\left[\frac{\ell(\ell+1)}{r^2}-\frac{6m}{r^3}\right],$$ $$V_{\mathrm Z} =\frac{2f\left[\lambda^2(\lambda+1)r^3 +3\lambda^2mr^2+9\lambda m^2r+9m^3\right]} {r^3(\lambda r+3m)^2}.$$

These are the potentials of the Regge–Wheeler and Zerilli master equations. Multiplying a homogeneous master field by a nonzero constant does not change its potential; identifying the field with a metric perturbation or a physical energy flux still requires the normalization given on those pages.

Define the real auxiliary function, conventionally called a superpotential,

$$W(r)=\alpha+\frac{3m f}{r(\lambda r+3m)},$$

which has dimensions of inverse length. Direct differentiation gives

$$V_{\mathrm{RW}}=W^2-DW-\alpha^2, \qquad V_{\mathrm Z}=W^2+DW-\alpha^2.$$

For example, substitute $W=\alpha+q$ with $q=3mf/[r(\lambda r+3m)]$: the two expressions become $2\alpha q+q^2\mp f\,dq/dr$. Collecting their numerators and using $\ell(\ell+1)=2(\lambda+1)$ reproduces the explicit potentials above. Every term has dimensions of inverse length squared. This is the Schwarzschild Darboux pair in Glampedakis, Johnson and Kennefick 2017, §§ III.A–B, pp. 3–4, eqs. (15)–(29), PDF.

Define

$$A=D+W,\qquad A^\dagger=-D+W.$$

The dagger is the formal adjoint with respect to $dx$; it becomes an operator adjoint only after a domain and boundary form have been specified. The two Schrödinger operators factor as

$$H_{\mathrm{RW}}=-D^2+V_{\mathrm{RW}}=A^\dagger A-\alpha^2,$$ $$H_{\mathrm Z}=-D^2+V_{\mathrm Z}=AA^\dagger-\alpha^2.$$

Associativity now gives the differential identities

$$AH_{\mathrm{RW}}=H_{\mathrm Z}A,\qquad A^\dagger H_{\mathrm Z}=H_{\mathrm{RW}}A^\dagger.$$

This is the mechanism: applying $A$ before or after the RW operator gives the same result, provided the operator on the other side is the Zerilli operator. It is an identity of differential expressions, before imposing endpoint conditions.

Differential map and its inverse

The intertwining identity therefore maps

$$\psi_{\mathrm Z}=A\psi_{\mathrm{RW}}$$

to a Zerilli solution at the same $\omega$. Applying $A^\dagger$ gives

$$A^\dagger\psi_{\mathrm Z} =(H_{\mathrm{RW}}+\alpha^2)\psi_{\mathrm{RW}} =(\omega^2+\alpha^2)\psi_{\mathrm{RW}}.$$

Thus, when $\omega^2+\alpha^2\ne0$,

$$\psi_{\mathrm{RW}} =\frac{A^\dagger\psi_{\mathrm Z}}{\omega^2+\alpha^2}.$$

Thus $A$ is a bijection between the two local, two-dimensional solution spaces at a nonexceptional frequency. This does not yet say that it maps the particular solutions selected by a boundary problem.

The raw field $A\psi_{\mathrm{RW}}$ has one extra inverse-length dimension. A frequency-dependent normalization, specified below, restores the original dimension. Nor does the transformation identify physical sources: if $(H_{\mathrm{RW}}-\omega^2)\psi_{\mathrm{RW}}=S_{\mathrm{RW}}$, its image has source $AS_{\mathrm{RW}}$, generally not the polar source produced by the same stress tensor. A prescribed source changes excitation amplitudes and residues; it does not change the homogeneous resonance frequencies of an unchanged operator with unchanged boundary conditions.

Boundary amplitudes and Wronskians

Both potentials decay exponentially as $x\to-\infty$ and as $O(x^{-2})$ as $x\to+\infty$. They are short-range for the present scattering problem, and $W\to\alpha$ at both ends. For nonzero frequency, the endpoint multipliers are

$$A e^{+i\omega x}\sim(\alpha+i\omega)e^{+i\omega x}, \qquad A e^{-i\omega x}\sim(\alpha-i\omega)e^{-i\omega x}.$$

The retarded endpoint solutions, often called Jost solutions, are selected as the uniquely normalized decaying solutions for $\operatorname{Im}\omega\gt0$, then analytically continued in frequency:

Solution	Endpoint	Unit asymptotic form in the upper half-plane
$F$	Horizon, $x\to-\infty$	$e^{-i\omega x}$
$G$	Infinity, $x\to+\infty$	$e^{+i\omega x}$

The first is ingoing at the future horizon; the second is outgoing to future null infinity. In the lower half-plane, adding a subdominant exponential need not change the leading asymptotic form. Consequently the continued solution, not just that leading form, defines the quasinormal boundary condition. The infinity solution has a branch point at zero frequency; fix a continuation sheet, and specify a lateral value when approaching its negative-imaginary-axis cut. This prescription and its anomalous normalizations are explained in Maassen van den Brink 2000, § II.A, pp. 3–4, eqs. (2.1)–(2.3), PDF.

Where the endpoint solutions are regular and $\alpha\pm i\omega\ne0$, their unit-normalized transforms are

$$F_{\mathrm Z}=\frac{AF_{\mathrm{RW}}}{\alpha-i\omega}, \qquad G_{\mathrm Z}=\frac{AG_{\mathrm{RW}}}{\alpha+i\omega}.$$

First establish these relations in the upper half-plane, where the decaying endpoint solution is unique, and then continue the identities on the chosen sheet. For any two RW solutions at the same frequency, their equation gives

$$\mathcal W[Au,Av] =(\omega^2+\alpha^2)\mathcal W[u,v], \qquad \mathcal W[u,v]=uDv-vDu.$$

The exercise derives this determinant identity without assuming constant $W$. Dividing by the two endpoint multipliers yields the stronger normalized result

$$\mathcal W[F_{\mathrm Z},G_{\mathrm Z}] =\mathcal W[F_{\mathrm{RW}},G_{\mathrm{RW}}].$$

A zero of this Wronskian means that two nonzero endpoint solutions are proportional: one solution obeys both boundary conditions. Thus the ordinary quasinormal frequencies, including zero multiplicities in regular local bases, agree wherever the map is invertible. At singular normalizations, inspect the complete Green function. For the convention $(D^2+\omega^2-V)\mathcal G(x,y;\omega)=\delta(x-y)$, it is

$$\mathcal G(x,y;\omega) =\frac{F(x_<,\omega)G(x_>,\omega)}{\mathcal W[F,G]}, \qquad x_<=\min(x,y),\quad x_>=\max(x,y).$$

Its derivative has jump $+1$ at $x=y$. Multiplying either endpoint solution by a frequency-dependent factor multiplies numerator and denominator together; a pole of a chosen basis alone is not sufficient evidence of a resonance. See Mode Green Functions and Wronskians for the full construction.

Reflection and transmission at real frequency

Take $\omega\gt0$ real and incidence from infinity. The unit-horizon solution has the expansion

$$F_{\mathrm{RW}}\sim A_{\mathrm{in}}e^{-i\omega x} +A_{\mathrm{out}}e^{+i\omega x} \qquad (x\to+\infty).$$

Dividing by $A_{\mathrm{in}}$ produces a unit incident amplitude. Thus $R_{\mathrm{RW}}=A_{\mathrm{out}}/A_{\mathrm{in}}$ and $T_{\mathrm{RW}}=1/A_{\mathrm{in}}$. The normalized Darboux map leaves $A_{\mathrm{in}}$ unchanged but multiplies $A_{\mathrm{out}}$ by the ratio of outgoing and ingoing endpoint factors:

$$R_{\mathrm Z} =\frac{\alpha+i\omega}{\alpha-i\omega}R_{\mathrm{RW}}, \qquad T_{\mathrm Z}=T_{\mathrm{RW}}.$$

Because $\alpha$ is real, that ratio is a phase. The reflection amplitudes need not be equal, but their absolute squares are; the transmission amplitudes agree in this unit-endpoint convention. This is the scattering relation of Glampedakis, Johnson and Kennefick 2017, Appendix A, p. 10, eqs. (A4)–(A11), PDF.

One can check the probabilities without a metric reconstruction. The real potential conserves the radial current

$$j=\frac{\overline\psi D\psi-\psi D\overline\psi}{2i}, \qquad Dj=0.$$

For a unit incident wave it is $\omega(|R|^2-1)$ at infinity and $-\omega|T|^2$ at the horizon, hence $|R|^2+|T|^2=1$. These are master-equation flux ratios. Absolute gravitational powers add the master-field and tensor-harmonic normalizations described in Metric Reconstruction and Radiative Observables.

Isospectrality and the algebraically special exception

The inverse fails at both $\omega=\pm i\alpha$. The damped algebraically special choice is

$$\omega_*^2=-\alpha^2, \qquad \omega_*=-i\alpha$$

For $\ell=2$, $\lambda=2$ and $\omega_*=-2i/m$. To see the failure explicitly, integrate $D\log\chi_{\mathrm{RW}}=-W$:

$$\chi_{\mathrm{RW}} =\left(\lambda+\frac{3m}{r}\right)e^{-\alpha x}, \qquad \chi_{\mathrm Z}=\chi_{\mathrm{RW}}^{-1}.$$

Then

$$A\chi_{\mathrm{RW}}=0,\qquad A^\dagger\chi_{\mathrm Z}=0,$$

and factorization gives $H_{\mathrm{RW}}\chi_{\mathrm{RW}}=-\alpha^2 \chi_{\mathrm{RW}}$ and $H_{\mathrm Z}\chi_{\mathrm Z}=-\alpha^2\chi_{\mathrm Z}$. The other RW solution can be constructed without dividing by $\omega^2+\alpha^2$:

$$\eta_{\mathrm{RW}}(x) =\chi_{\mathrm{RW}}(x)\int_{x_0}^{x} \frac{dy}{\chi_{\mathrm{RW}}(y)^2}, \qquad A\eta_{\mathrm{RW}}=\chi_{\mathrm Z}.$$

Indeed $\mathcal W[\chi_{\mathrm{RW}},\eta_{\mathrm{RW}}]=1$. The map kills one independent RW solution and maps the other to the polar kernel: its rank is now one. This is the reduction-of-order structure of Maassen van den Brink 2000, § II.B, p. 4, eqs. (2.5)–(2.6), PDF.

Why the leading exponential is insufficient

The prefactor in $\chi_{\mathrm{RW}}$ tends to a finite nonzero constant at each end, so it has an $e^{-\alpha x}$ leading factor at both ends. Calling it “ingoing at both ends” from this observation would skip the frequency continuation. At $\omega_*$, the two horizon Frobenius exponents differ by the integer

$$N=4m\alpha=\frac{4\lambda(\lambda+1)}3.$$

A logarithmic term or its cancellation can then change which linear combination is obtained by continuing the ingoing solution. In the RW equation the logarithmic obstruction cancels, but the continued horizon solution is not $\chi_{\mathrm{RW}}$: a subleading contribution survives. For the Zerilli equation the unit-normalized horizon solution has a simple pole in frequency. Its finite rescaled limit is proportional to $\chi_{\mathrm Z}$. The horizon analysis establishing these facts is given in Maassen van den Brink 2000, § III, pp. 5–6, PDF.

For vacuum Schwarzschild and each $\ell\geq2$, with the continuation just specified, the detailed result at $\omega_*$ is:

• The RW equation has no quasinormal mode and no total-transmission mode.
• The Zerilli solution $\chi_{\mathrm Z}$ is a simple quasinormal mode and also a total-transmission mode from the horizon toward infinity. Here total transmission means membership in the continued horizon-outgoing (from the hole) and infinity-outgoing solution lines.

The coincidence of these two polar classifications is possible because the finite rescaled ingoing-horizon line coincides with the opposite horizon line at this anomalous frequency. It is not a statement about simultaneous oppositely directed real-frequency fluxes. At this frequency the left and right continuations of the infinity solution agree. The proof distinguishes the exact RW connection coefficients and then transforms the resulting boundary lines; see Maassen van den Brink 2000, §§ V.B–C, p. 7, PDF.

The Green function makes the consequence especially clear. $F_{\mathrm Z}=AF_{\mathrm{RW}}/(\alpha-i\omega)$ has a simple pole at $\omega_*$, while $G_{\mathrm Z}$ is finite and nonzero there. The unit-normalized Wronskian equals the RW one, which is finite and nonzero at that frequency. Thus this pole remains in the polar Green kernel; it does not cancel as a harmless basis normalization. Equivalently, after rescaling $F_{\mathrm Z}$ to a finite solution, the corresponding Wronskian vanishes. This is the retarded Green-function pole established in the same source, not a conclusion obtainable from the kernel’s leading exponential alone.

Neither kernel is square-integrable on the full tortoise line. Nor does the algebraic degeneracy at $+i\alpha$ imply an instability: an upper-half-plane retarded mode would decay at both ends, whereas the positive RW and Zerilli potentials forbid a negative eigenvalue $\omega^2=-\alpha^2$ for such a square-integrable solution.

Finite boundary conditions

An identity of differential expressions is not an equality of closed operators with specified domains. In the asymptotically flat problem, the continued endpoint argument above establishes the nonexceptional boundary correspondence. A finite limit of $W$ alone would not do so.

For a concrete test, impose a Robin condition at a finite $x_b$: $D\psi_{\mathrm{RW}}=h\psi_{\mathrm{RW}}$, where $h$ has dimensions of inverse length. Write $\phi=A\psi_{\mathrm{RW}}$ and $W_b=W(x_b)$. The radial equation and factorization give, at the boundary,

$$\phi=(h+W_b)\psi_{\mathrm{RW}},\qquad D\phi=\left[W_b(h+W_b)-(\omega^2+\alpha^2)\right]\psi_{\mathrm{RW}}.$$

Away from the exceptional frequencies, eliminating the RW value without division gives the equivalent transformed boundary condition

$$(h+W_b)D\phi =\left[W_b(h+W_b)-(\omega^2+\alpha^2)\right]\phi.$$

If $h+W_b\ne0$, this is a Robin condition with parameter

$$h_{\mathrm Z}=W_b-\frac{\omega^2+\alpha^2}{h+W_b}.$$

It is generally frequency-dependent and different from $h$. If $h+W_b=0$ at a nonexceptional frequency, the transformed condition is Dirichlet, $\phi=0$. Conversely, an original Dirichlet condition $\psi_{\mathrm{RW}}=0$ gives $D\phi=W_b\phi$, not $\phi=0$. The map can therefore relate differently specified boundary problems without proving equality for the same reflecting condition in both sectors.

Changing the asymptotic geometry, coupling matter dynamically, or modifying the field equations likewise requires a new operator and boundary analysis. The nonexceptional Schwarzschild result is powerful precisely because all three ingredients—differential equation, normalization and endpoint condition—can be checked explicitly.

Exercises

1. Verify the Wronskian multiplier for two solutions of the Regge–Wheeler radial equation at the same frequency.
2. At a finite Schwarzschild boundary $r_b=3m$, take $\ell=2$, $\omega=1/m$, and the RW Neumann condition $D\psi_{\mathrm{RW}}=0$. Does $A$ produce a polar Neumann solution? Determine the actual polar Robin parameter.

Solution

Write $Au=u'+Wu$ and $Av=v'+Wv$, where primes denote $D$. Using $u''=(V_{\mathrm{RW}}-\omega^2)u$ gives the map on the solution and its derivative:

$$\begin{pmatrix}Au\\(Au)'\end{pmatrix} = \begin{pmatrix} W&1\\ V_{\mathrm{RW}}-\omega^2+W'&W \end{pmatrix} \begin{pmatrix}u\\u'\end{pmatrix}.$$

The same matrix acts on $(v,v')$. Since the Wronskian is the determinant of the two solution columns, its multiplier is the determinant of this matrix. Thus $\mathcal W[Au,Av]$ equals

$$\left(W^2-W'-V_{\mathrm{RW}}+\omega^2\right)(uv'-u'v).$$

The factorization identity $V_{\mathrm{RW}}=W^2-W'-\alpha^2$ reduces the parenthesis to $\omega^2+\alpha^2$. Therefore $\mathcal W[Au,Av]=(\omega^2+\alpha^2)\mathcal W[u,v]$. Its vanishing at $\omega_*^2=-\alpha^2$ is precisely why the exceptional mode needs separate analysis.

For the boundary question, $\lambda=2$, $\alpha=2/m$ and $W_b=55/(27m)$. Hence the transformed parameter is

$$h_{\mathrm Z} =\frac{55}{27m}-\frac{5/m^2}{55/(27m)} =-\frac{124}{297m}.$$

This is not zero, so the image of a nontrivial RW Neumann solution is not a polar Neumann solution. A shared local frequency and an invertible differential map do not ensure the same boundary condition.

References

• Glampedakis, K., Johnson, A. D., and Kennefick, D. (2017). “The Darboux transformation in black hole perturbation theory.” Physical Review D 96, 024036. DOI. Open PDF.
• Maassen van den Brink, A. (2000). “Analytic treatment of black-hole gravitational waves at the algebraically special frequency.” Physical Review D 62, 064009. DOI. Open PDF.