Maxwell Fields and Gauge-Invariant Radiation

The electromagnetic potential is redundant; the electric and magnetic fields are not. On a fixed Schwarzschild background, two gauge-invariant master fields describe the electric- and magnetic-parity sectors, and both obey the same wave equation. This page derives their normalization, reconstructs the field strength, and computes the energy they carry. On Kerr, useful variables are tetrad components of the field strength, whose radiative interpretation requires a specified frame. These are classical source-free test fields: the metric does not respond to their stress–energy here.

Required background. PDEs and Green Functions for Gravitation supplies hyperbolic equations, boundary data, mode reduction, and flux identities.

Helpful background. Black-Hole Geometry and Mechanics supplies horizon generators and stationary energy; Formulations of General Relativity supplies continuum gauge and boundary formulations; Frames, Coframes, and Anholonomy supplies the tetrad language used for radiative scalars.

Electromagnetic gauge invariance

Use the canonical real Maxwell action. Its normalization fixes every energy factor below:

$$S_{\mathrm{EM}}=-\frac14\int F_{\mu\nu}F^{\mu\nu}\sqrt{-g}\,d^4x, \qquad F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu.$$

The potential exists locally; a possible magnetic-charge sector needs more than one potential patch. Variation and the identity $d^2A=0$ give

$$\begin{gathered} \nabla_\mu F^{\mu\nu}=0,\qquad \nabla_{[\rho}F_{\mu\nu]}=0,\\ T_{\mu\nu}=F_{\mu\alpha}F_\nu{}^\alpha -\frac14g_{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}. \end{gathered}$$

These curved-spacetime equations are given in Wald 1984, §4.3, p. 70, Eqs. (4.3.12)–(4.3.14). Wald uses Gaussian electromagnetic units: for the same $F$, our stress tensor is $4\pi$ times his. Only the source-free equations are used here.

The transformation $A_\mu\mapsto A_\mu+\partial_\mu\Lambda$ leaves $F$ unchanged because mixed derivatives of a scalar commute. This is electromagnetic gauge freedom, distinct from coordinate gauge freedom. In perturbation theory about a background with $F^{(0)}=0$, the first-order field strength is also coordinate-gauge invariant: its change is $\mathcal L_\xi F^{(0)}=0$. Its energy is quadratic in that first-order field. Neglecting this quadratic stress as a source for the metric is the test-field approximation.

Maxwell’s equations and antisymmetry give $\nabla_\mu T^\mu{}_\nu=0$. For a Killing field $\xi$, symmetry of $T$ then gives the conserved current

$$\mathcal J^\mu_{(\xi)}=-T^\mu{}_\nu\xi^\nu,\qquad \nabla_\mu\mathcal J^\mu_{(\xi)} =-T^{\mu\nu}\nabla_{(\mu}\xi_{\nu)}=0.$$

On a spacelike slice with future unit normal $N$, its energy density is $-N_\mu\mathcal J^\mu_{(\xi)}=T_{\mu\nu}N^\mu\xi^\nu$. It is nonnegative when $\xi$ is future timelike. The minus sign in the raised current is important: positive outward stationary energy is $-T^r{}_t$.

Schwarzschild parity fields

Let $m=GM\gt0$ and work in the exterior $r\gt2m$:

$$ds^2=-f\,dt^2+f^{-1}dr^2+r^2\gamma_{AB}dx^A dx^B, \qquad f=1-\frac{2m}{r},\qquad \frac{dr_*}{dr}=f^{-1}.$$

Here $\gamma_{AB}$ is the unit-sphere metric, $D_A$ its derivative, and $a,b\in\{t,r\}$ are orbit-space indices. For the real-field energy calculation choose a real orthonormal basis $Y_{\ell q}$, $q=1,\ldots,2\ell+1$, of scalar spherical harmonics. It consists of real linear combinations of the usual complex $Y_{\ell m_z}$; $q$ is not an azimuthal quantum number. Suppress the mode label temporarily and set

$$D^A D_A Y=-LY,\qquad L=\ell(\ell+1),\qquad \int_{S^2}Y^2\,d\Omega=1,\qquad \ell\ge1.$$

With $\epsilon_{\theta\phi}=\sin\theta$ on the unit sphere, the odd vector harmonic is $X_A=\epsilon_A{}^B D_BY$. The useful identities are

$$\begin{gathered} D^A X_A=0,\qquad D_AX_B-D_BX_A=L\epsilon_{AB}Y,\\ \int_{S^2}D_AYD^AY\,d\Omega =\int_{S^2}X_AX^A\,d\Omega=L. \end{gathered}$$

The last identity follows by integration by parts on the closed sphere. Under the antipodal parity map, $D_AY$ has parity $(-1)^\ell$, whereas $X_A$ has parity $(-1)^{\ell+1}$. These are the electric/even and magnetic/odd sectors, not two different species of photon.

Odd sector

Write $A_a=0$, $A_A=\Psi^{(-)}(t,r)X_A$. A scalar gauge parameter supplies an angular gradient but no axial harmonic, so $\Psi^{(-)}$ is gauge invariant. Taking its curl gives $F_{tA}=\partial_t\Psi^{(-)}X_A$, $F_{rA}=\partial_r\Psi^{(-)}X_A$, and $F_{AB}=L\Psi^{(-)}\epsilon_{AB}Y$.

Now use the density form of Maxwell’s equation, $\partial_\mu(\sqrt{-g}F^{\mu A})=0$, with $\sqrt{-g}=r^2\sin\theta$. The angular identity above leaves

$$-\frac1f\partial_t^2\Psi^{(-)} +\partial_r\!\left(f\partial_r\Psi^{(-)}\right) -\frac{L}{r^2}\Psi^{(-)}=0.$$

The other divergence components vanish because $D^AX_A=0$; the Faraday identity already holds because $F=dA$.

Even sector

Write $A_a=a_a(t,r)Y$ and $A_A=v(t,r)D_AY$. A gauge parameter $\Lambda=\lambda(t,r)Y$ changes $a_a$ by $\partial_a\lambda$ and $v$ by $\lambda$, leaving $p_a=a_a-\partial_av$ invariant. Define the coefficient $e(t,r)$, not the full angular field, by

$$F_{tr}=eY,\qquad e=\partial_t p_r-\partial_r p_t, \qquad F_{aA}=-p_aD_AY,\qquad F_{AB}=0.$$

The $t$ and $r$ components of the Maxwell divergence relate the angular fields to this radial electric coefficient:

$$\partial_r(r^2e)+\frac{L}{f}p_t=0,\qquad -\partial_t(r^2e)-Lf p_r=0.$$

The first is the Gauss constraint; the second supplies the electric field’s time evolution. Together they reconstruct $p_a$. Since $L\ne0$, choose the exact normalization

$$\Psi^{(+)}=\frac{r^2e}{L},\qquad p_t=-f\partial_r\Psi^{(+)},\qquad p_r=-f^{-1}\partial_t\Psi^{(+)}.$$

Substitution into $e=\partial_t p_r-\partial_r p_t$ gives precisely the odd equation with $\Psi^{(-)}$ replaced by $\Psi^{(+)}$. The angular Maxwell equation is also satisfied: $\partial_t(p_t/f)-\partial_r(fp_r)=0$ becomes equality of mixed derivatives. Thus the reduction preserves the constraints; it is not merely an equation for one selected component.

Reconstruction and the common potential

The table collects everything needed to recover the measurable field from either master variable. The omitted components follow by antisymmetry.

Component	Odd sector	Even sector
$F_{tr}$	$0$	$L\Psi^{(+)}Y/r^2$
$F_{tA}$	$\partial_t\Psi^{(-)}X_A$	$f\partial_r\Psi^{(+)}D_AY$
$F_{rA}$	$\partial_r\Psi^{(-)}X_A$	$f^{-1}\partial_t\Psi^{(+)}D_AY$
$F_{AB}$	$L\Psi^{(-)}\epsilon_{AB}Y$	$0$

Multiplying the derived equation by $f$ and using $\partial_{r_*}=f\partial_r$ gives

$$\left[-\partial_t^2+\partial_{r_*}^2-V_\ell^{\mathrm{EM}}\right] \Psi^{(\pm)}=0,\qquad V_\ell^{\mathrm{EM}}=f\frac{L}{r^2}.$$

The potential has dimension length$^{-2}$, reduces to the centrifugal barrier in flat spacetime, and vanishes at the horizon. Unlike the minimally coupled scalar potential, it has no $2mf/r^3$ term; compare Scalar Fields and Effective Potentials. The equality of the physical massless parity equations is also obtained from the Proca system in Rosa and Dolan 2012, §II.C, Eqs. (19)–(26), preprint pp. 4–5, PDF. Their potential coefficients use a different harmonic normalization, so their amplitudes must not be inserted directly into the flux formula below.

There is a geometric reason for the equality. For two-forms in four dimensions, define

$$(*F)_{\mu\nu}=\frac12\epsilon_{\mu\nu\alpha\beta}F^{\alpha\beta}, \qquad \epsilon_{tr\theta\phi}=r^2\sin\theta,\qquad **F=-F.$$

With this Lorentzian Hodge convention, the table gives $F^{(+)}[\Psi]=*F^{(-)}[\Psi]$ on solutions. Vacuum Maxwell equations are invariant under this electric–magnetic duality, and the stress tensor is unchanged. One degree of freedom per parity accounts for the two transverse polarizations. A complete real field is the sum over both parities and all real harmonics.

Energy carried by a master field

Keep one normalized real harmonic and one parity for the moment. On a Schwarzschild $t$ slice, the Killing field is $\xi=\partial_t$, the future normal is $N=f^{-1/2}\partial_t$, and the volume element is $r^2f^{-1/2}dr\,d\Omega$. Direct substitution of the reconstruction table into $T_{\mu\nu}$ gives

$$\begin{aligned} E&=\int \frac{r^2}{f}T_{tt}\,dr\,d\Omega =\int\mathcal E\,dr_*,\\ \mathcal E&=\frac{L}{2}\left[ (\partial_t\Psi)^2+(\partial_{r_*}\Psi)^2 +V_\ell^{\mathrm{EM}}\Psi^2\right],\\ P&=-r^2\int_{S^2}T^r{}_t\,d\Omega =-L\,\partial_t\Psi\,\partial_{r_*}\Psi. \end{aligned}$$

For example, the odd sector has $T^r{}_t=f\partial_r\Psi\partial_t\Psi\,X_AX^A/r^2$; the even sector replaces $X_AX^A$ by $D_AYD^AY$. Their sphere integrals are both $L$. The radial electric or magnetic component supplies the potential energy in $\mathcal E$. This derives the normalization from physical fields, rather than assigning a convenient energy to an abstract wave equation.

Multiplication of that equation by $L\partial_t\Psi$ verifies $\partial_t\mathcal E+\partial_{r_*}P=0$. Between two fixed radii,

$$\frac{d}{dt}\int_{x_1}^{x_2}\mathcal E\,dx =P(t,x_1)-P(t,x_2),\qquad x=r_*.$$

For smooth finite-energy data this energy is nonnegative outside the horizon; orthogonality makes the integrated energy and power additive over real modes and parities. This identity alone is not a decay theorem or a horizon-nondegenerate bound; those distinctions belong to Energy Estimates and Test-Field Stability.

At large radius an outgoing packet has leading behavior $\Psi=H(u)$, $u=t-r_*$, giving $P=L[H'(u)]^2\ge0$. At the future horizon, an ingoing packet has leading behavior $\Psi=G(v)$, $v=t+r_*$, giving $P=-L[G'(v)]^2$. Thus the positive absorbed power is $-P$ at the inner boundary. Regularity is tested in the advanced coordinate $v$, not by requiring each Schwarzschild component in the table to stay finite. In particular the even field satisfies

$$F^{(v,r)}_{rA} =F^{(t,r)}_{rA}-f^{-1}F_{tA} =-(\partial_r\Psi)_vD_AY.$$

The apparent $f^{-1}$ singularities cancel for smooth $\Psi(v,r)$.

For steady scattering take a real frequency $\omega\gt0$ and the real field $\Psi=\operatorname{Re}[\psi(r_*)e^{-i\omega t}]$. With a prime denoting $d/dr_*$, define

$$\begin{gathered} \psi''+(\omega^2-V_\ell^{\mathrm{EM}})\psi=0,\qquad W=\psi^*\psi'-\psi(\psi')^*,\qquad W'=0,\\ \langle P\rangle =\frac{L\omega}{2}\operatorname{Im}(\psi^*\psi') =\frac{L\omega}{4i}W. \end{gathered}$$

The factor one half is a real-wave time average. If $\psi\sim A_{\mathrm{out}}e^{i\omega r_*}+A_{\mathrm{in}}e^{-i\omega r_*}$ at infinity, then

$$\langle P\rangle_\infty =\frac{L\omega^2}{2} \left(|A_{\mathrm{out}}|^2-|A_{\mathrm{in}}|^2\right).$$

The conjugate Wronskian is conserved when $\omega^2$ is real. The stationary period average above requires real $\omega\gt0$ and cannot be used for a complex-frequency quasinormal mode. In the stated action normalization $\Psi$ is dimensionless, so $E$ has dimension length$^{-1}$ and $P$ length$^{-2}$.

The nonradiative monopoles

For $\ell=0$, both $D_AY$ and $X_A$ vanish: division by $L$ is unavailable. The remaining spherically symmetric source-free field is

$$F=\frac{C_E}{r^2}\,dt\wedge dr +C_B\sin\theta\,d\theta\wedge d\phi,\qquad \partial_t C_E=\partial_r C_E =\partial_t C_B=\partial_r C_B=0.$$

The constants are electric and magnetic Coulomb coefficients, proportional to the corresponding conserved fluxes. The divergence equation fixes the electric coefficient and the Faraday identity fixes the magnetic one. Neither describes an outgoing photon, and $T^r{}_t=0$. They are nonzero, gauge-invariant fields without radiative power.

In particular $\int_{S^2}F=4\pi C_B$; a nonzero value is incompatible with a globally smooth potential on the sphere, since the integral of an exact two-form there is zero. Patchwise potentials accommodate this magnetic sector even though there is no magnetic source in the exterior itself. If one assumes a single global potential, one has already restricted $C_B=0$; that assumption does not exclude the electric Coulomb field.

Energy at a rotating horizon

On subextremal Kerr let $a=J_{\mathrm{BH}}/M$, $|a|\lt GM$, and $r_+=GM+\sqrt{(GM)^2-a^2}$. The future horizon generator normalized by stationary time at infinity is

$$\chi=\partial_t+\Omega_H\partial_\phi,\qquad \Omega_H=\frac{a}{r_+^2+a^2}.$$

For a real field regular at that horizon, set $w_\alpha=\chi^\mu F_{\mu\alpha}$. Antisymmetry gives $w_\alpha\chi^\alpha=0$, so $w$ has nonnegative squared norm: the space orthogonal to a null vector contains transverse spacelike directions and that null direction, but no timelike direction. Hence

$$T_{\mu\nu}\chi^\mu\chi^\nu=w_\alpha w^\alpha\ge0.$$

To identify which energy this measures, use a horizon parameter $v_H$ with $\chi=\partial_{v_H}$ at fixed co-rotating angle and area element $dA_H$. For a finite horizon segment define absorbed stationary energy and angular momentum by

$$\begin{aligned} E_H&=\int_H T(\partial_t,\chi)\,dv_H\,dA_H,\\ J_H&=-\int_H T(\partial_\phi,\chi)\,dv_H\,dA_H,\\ E_H-\Omega_H J_H &=\int_H T(\chi,\chi)\,dv_H\,dA_H\ge0. \end{aligned}$$

The last line is nonnegative horizon-generator energy, not a claim that $E_H$ must be nonnegative. The stationary Killing field need not be timelike in the ergoregion. Its absorbed energy can be negative during rotational superradiance, consistently with the last inequality. The electromagnetic flux conversion is developed in Brito, Cardoso, and Pani 2020 edition, §4.7.2, pp. 58–60, PDF; the Superradiant Frequency Condition explains the role of $\omega-m_z\Omega_H$. No change of the background black-hole parameters is evolved in this fixed-metric calculation.

Newman–Penrose radiation and its normalization

Order a null tetrad as $(\ell,n,m,\bar m)$ with $\ell\!\cdot n=-1$, $m\!\cdot\bar m=1$, and other products zero. Define

$$\begin{aligned} \phi_0&=F_{\mu\nu}\ell^\mu m^\nu,\\ \phi_1&=\frac12F_{\mu\nu}(\ell^\mu n^\nu+\bar m^\mu m^\nu),\\ \phi_2&=F_{\mu\nu}\bar m^\mu n^\nu. \end{aligned}$$

These quantities are electromagnetic-gauge invariant, but depend on the chosen tetrad. Under $\ell\mapsto b\ell$, $n\mapsto b^{-1}n$, $m\mapsto e^{i\alpha}m$, with $b\gt0$,

$$\phi_0\mapsto b e^{i\alpha}\phi_0,\qquad \phi_1\mapsto\phi_1,\qquad \phi_2\mapsto b^{-1}e^{-i\alpha}\phi_2.$$

For the specific Kinnersley tetrad, the separated spin-$+1$ and spin-$-1$ variables are, respectively,

$$\psi_{+1}=\phi_0,\qquad \psi_{-1}=\rho_{NP}^{-2}\phi_2 =(r-ia\cos\theta)^2\phi_2,\qquad \rho_{NP}=-\frac1{r-ia\cos\theta}.$$

See Teukolsky 1973, Table 1, p. 641, PDF and the Derivation of the Teukolsky Equation for the operator and separation. The rescaling belongs to this fixed principal tetrad, not to an arbitrary spin-boosted one. Ingoing and outgoing interpretations refer to appropriate asymptotic limits, not to globally one-way propagation. The middle component includes the Coulomb sector but is not exclusively a stationary-charge variable. Raw Kinnersley components are also not horizon-regular components; use the frame conversion on the linked tetrad page before diagnosing a divergence.

To fix the luminosity factor, choose an asymptotic orthonormal frame $(U,e_{\hat r},e_{\hat\theta},e_{\hat\phi})$, with $U$ future timelike, $e_{\hat r}$ outward, and the spatial orientation $(\hat r,\hat\theta,\hat\phi)$. The Kinnersley normalization tends to

$$\ell=U+e_{\hat r},\qquad n=\frac12(U-e_{\hat r}),\qquad m=\frac1{\sqrt2}(e_{\hat\theta}+i e_{\hat\phi}).$$

For the real field define $E_{\hat i}=F_{\hat i\hat0}$ and $B_{\hat i}=\epsilon_{\hat i\hat j\hat k}F_{\hat j\hat k}/2$. At leading outgoing radiative order, $B_{\hat\phi}=E_{\hat\theta}$ and $B_{\hat\theta}=-E_{\hat\phi}$. Direct contraction gives

$$\phi_2=\frac{E_{\hat\theta}-iE_{\hat\phi}}{\sqrt2},\qquad \phi_0=0,\qquad S_{\hat r}=(\mathbf E\times\mathbf B)_{\hat r}=2|\phi_2|^2.$$

Consequently, for a smooth outgoing radiative asymptotic expansion,

$$\frac{dE}{du\,d\Omega} =\lim_{r\to\infty}[-r^2T^r{}_t] =\lim_{r\to\infty}2r^2|\phi_2|^2,$$

where the limit is taken at fixed retarded time $u$. The radiative transverse orthonormal fields are $O(r^{-1})$; coordinate angular components such as $F_{t\theta}$ are instead $O(1)$. A stationary radial Coulomb field is $O(r^{-2})$ in the orthonormal frame and contributes no such luminosity.

This factor agrees with Teukolsky 1973, §V, Eq. (5.13), p. 644, PDF after translating both signature and units. Keeping covariant $F$ and the contravariant tetrad fixed while replacing his metric by its negative leaves these Maxwell scalars unchanged. Our covariant stress is $4\pi$ times his Gaussian-unit stress, whereas the raised radial index reverses sign: $-T^r{}_{t,\mathrm{here}}=4\pi T^r{}_{t,\mathrm{Teukolsky}}$. Thus his $r^2|\phi_2|^2/(2\pi)$ becomes the displayed $2r^2|\phi_2|^2$.

These are instantaneous relations for real $F$, even though $\phi_2$ is complex. For periodic radiation, average $2|\phi_2(t)|^2$ itself. Do not take the real part of $\phi_2$ and then apply a universal extra one half: its real and imaginary parts already encode two real polarizations. A finite tetrad boost changes the coefficient relating $|\phi_2|^2$ to luminosity; electromagnetic gauge invariance does not remove this frame dependence.

With sources, energy exchange with charged matter must be included. Perturbing an already charged gravitating background also couples Maxwell and metric perturbations. Neither problem is solved by the source-free test-field potential above. The classical flux calculation likewise does not specify quantum photon states or particle creation.

Exercises

Gauge oscillations and charge. Take $a_a=\partial_a\lambda$ and $v=\lambda$ for an arbitrarily oscillating $\lambda(t,r)$. Is this an electromagnetic wave? Contrast it with the monopole having $C_E\ne0$, $C_B=0$.

Solution

The gauge-invariant combination is $p_a=\partial_a\lambda-\partial_a\lambda=0$. Hence $e=0$ and all components of $F$ vanish: the oscillating potential is pure gauge, with zero stress and zero flux. The electric monopole instead has $F_{tr}=C_E/r^2\ne0$ and nonzero energy density but $T^r{}_t=0$. Therefore gauge invariance is necessary for an electromagnetic observable, not sufficient for radiation. The monopole is not represented by setting $L=0$ in the definition of $\Psi^{(+)}$; that definition divided by $L$.

The energy of a dipole pulse. Far from the hole, take one odd real $\ell=1$ mode with $Y=\sqrt{3/(4\pi)}\cos\theta$ and outgoing profile $H(u)=A\exp[-u^2/(2\tau^2)]$, where $A$ is dimensionless and $\tau\gt0$. Find $X_\phi$, verify the sign of $F_{\theta\phi}$, and integrate the leading radiated energy over all retarded times.

Solution

Here $X_\theta=0$ and $X_\phi=-\sin\theta\,\partial_\theta Y=\sqrt{3/(4\pi)}\sin^2\theta$. Therefore $F_{\theta\phi}=\partial_\theta(HX_\phi)=2H\sin\theta\,Y$, agreeing with $L=2$ in the table. The pulse power is $P=2[H'(u)]^2$, so

$$E_{\mathrm{rad}}=\frac{2A^2}{\tau^4} \int_{-\infty}^{\infty}u^2e^{-u^2/\tau^2}\,du =\frac{\sqrt\pi A^2}{\tau}.$$

The dimensions are correct, and shortening a fixed-amplitude pulse raises its energy. This is an asymptotic outgoing waveform, not an exact Gaussian solution at finite radius through the Schwarzschild potential. For an incident pulse, reflection and absorption determine which outgoing profile actually emerges.

A boost and circular polarization. At leading outgoing order, let $E_{\hat\theta}=C\cos(\omega u)/r$ and $E_{\hat\phi}=C\sin(\omega u)/r$, with real $C$. Compute the luminosity per solid angle in the Kinnersley normalization. Then change to the symmetric null legs $\ell'=(U+e_{\hat r})/\sqrt2$, $n'=(U-e_{\hat r})/\sqrt2$. Does the physical answer change?

Solution

In the first frame,

$$\phi_2=\frac{C}{\sqrt2r}e^{-i\omega u},\qquad \frac{dE}{du\,d\Omega}=2r^2|\phi_2|^2=C^2.$$

The flux is constant: the electric vector rotates while its magnitude stays fixed. Taking only the real part of $\phi_2$ would discard one polarization. The new frame is the boost $b=1/\sqrt2$, so $\phi_2'=\sqrt2\phi_2$. Its flux formula is $r^2|\phi_2'|^2=C^2$, with no leading factor two. The observable is unchanged; the numerical value of a tetrad component and the conversion factor change together.

References

• Brito, Richard, Vitor Cardoso, and Paolo Pani. Superradiance: New Frontiers in Black Hole Physics. Lecture Notes in Physics 971. Cham: Springer, 2020. DOI. Open PDF, arXiv:1501.06570v8, updated 2021.
• Rosa, João G., and Sam R. Dolan. “Massive Vector Fields on the Schwarzschild Spacetime: Quasinormal Modes and Bound States.” Physical Review D 85 (2012): 044043. DOI. Open PDF, arXiv:1110.4494v2.
• Teukolsky, Saul A. “Perturbations of a Rotating Black Hole. I. Fundamental Equations for Gravitational, Electromagnetic, and Neutrino-Field Perturbations.” Astrophysical Journal 185 (1973): 635–647. DOI. Open PDF.
• Wald, Robert M. General Relativity. Chicago: University of Chicago Press, 1984. Publisher.