# Dynamical Implications of the Kerr Multipole Moments for Spinning Black
Holes

Trevor Scheopner  Mani L. Bhaumik Institute for Theoretical Physics,
University of California at Los Angeles, Los Angeles, CA 90095, USA  Justin
Vines  Mani L. Bhaumik Institute for Theoretical Physics, University of
California at Los Angeles, Los Angeles, CA 90095, USA

###### Abstract

Previously the linearized stress tensor of a stationary Kerr black hole has
been used to determine some of the values of gravitational couplings for a
spinning black hole to linear order in the Riemann tensor in the action
(worldline or quantum field theory). In particular, the couplings on operators
containing derivative structures of the form $(S\cdot\nabla)^{n}$ acting on
the Riemann tensor were fixed, with $S^{\mu}$ the spin vector of the black
hole. In this paper we find that the Kerr solution determines all of the
multipole moments in the sense of Dixon of a stationary spinning black hole
and that these multipole moments determine all linear in $R$ couplings. For
example, additional couplings beyond the previously mentioned are fixed on
operators containing derivative structures of the form
$S^{2n}(p\cdot\nabla)^{2n}$ acting on the Riemann tensor with $p^{\mu}$ the
momentum vector of the black hole. These additional operators do not
contribute to the three-point amplitude, and so do not contribute to the
linearized stress tensor for a stationary black hole. However, we find that
they do contribute to the Compton amplitude. Additionally, we derive formal
expressions for the electromagnetic and gravitational Compton amplitudes of
generic spinning bodies to all orders in spin in the worldline formalism and
evaluated expressions for these amplitudes to $\mathcal{O}(S^{3})$ in
electromagnetism and $\mathcal{O}(S^{5})$ in gravity.

###### Contents

  1. I Introduction
    1. I.1 General Overview
    2. I.2 Summary of Method and Results
    3. I.3 Notation
  2. II Electromagnetic MPD Equations
  3. III Dixon’s Multipole Moments
    1. III.1 Moments of a scalar field
    2. III.2 Moments of a general tensor field
    3. III.3 Moments of a conserved vector field
    4. III.4 Moments of the current density in Minkowski space
  4. IV Root-Kerr Multipole Moments
    1. IV.1 From Kerr-Newman to Root-Kerr
    2. IV.2 Mechanical Properties of the Stationary Root-Kerr Solution
    3. IV.3 Charge and Current density
    4. IV.4 Stationary Multipole Moments
    5. IV.5 Dynamical Multipole Moments
  5. V Electromagnetic Compton Amplitude
    1. V.1 Formal Classical Compton
    2. V.2 Compton Amplitude through Spin Cubed
  6. VI Gravitational MPD Equations
  7. VII Kerr Multipole Moments
    1. VII.1 Moments of the Energy Momentum Tensor
    2. VII.2 Source of the Kerr Metric
    3. VII.3 Stationary Multipole Moments of Kerr
    4. VII.4 Dynamical Multipole Moments of Kerr
  8. VIII Gravitational Compton Amplitude
    1. VIII.1 Formal Classical Compton
    2. VIII.2 Compton Amplitude through Spin to the Fifth
  9. IX Conclusion

##  I Introduction

###  I.1 General Overview

The observation of gravitational waves by the LIGO/Virgo collaboration [1, 2]
began a new era of gravitational physics, with implications for astronomy,
cosmology, and possibly particle physics. Physical black holes and neutron
stars generically carry significant spin angular momentum which affects their
dynamics during mergers in binary systems and the gravitational wave signals
they emit. These spin effects will play an increasingly important role in
signal analysis as gravitational wave detectors become more sensitive [3, 4,
5] and also lead to rich theoretical structure for generic bodies, but
especially so for black holes.

The study of the dynamics of generic spinning bodies in general relativity has
a long history [6, 7, 8, 9, 10, 11, 12]. Multiple successful field theoretic
and worldline based approaches exist for the study of spinning bodies in both
the post-Newtonian (PN) approximation [13, 14, 15, 16, 17, 18, 19, 20, 21, 22,
23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41,
42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60,
61, 62, 63, 64, 65, 66] and the post-Minkowskian (PM) approximation [67, 68,
69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87,
88, 89, 90, 91, 92, 93, 94, 95, 87, 96, 97, 98, 99, 100, 101, 102, 103, 104,
105, 106, 107, 108]. The electromagnetic [109, 110, 111, 112, 113, 114, 115]
and non-abelian gauge theory [116, 117] cases are very similar in structure to
gravity and can be used to develop helpful insights for the harder
gravitational problem.

In both the field theoretic and worldline approaches when considering only the
minimal Poincaré degrees of freedom, the interaction of the body with gravity
is characterized by a tower of effective field theory operators in the action,
each carrying a Wilson coefficient, some number of powers of the spin of the
body, and some number of powers of the Riemann tensor and its derivatives. For
generic bodies these Wilson coefficients take arbitrary values. We will
specialize our interest in this paper exclusive to spinning black holes. This
restriction in principle determines the values of all such Wilson
coefficients. However, presently the values for these Wilson coefficients on
linear and quadratic in Riemann tensor are only partly known.

The coefficients for operators of the form
$(S\cdot\nabla)^{n}R_{\cdot\cdot\cdot\cdot}$ were fixed in Ref. [48]. Such
operators are precisely those which contribute to the three point amplitude.
There are possible operators which are linear in the Riemann tensor but not of
this form, such as those of the form
$S^{2n}(p\cdot\nabla)^{2n}R_{\cdot\cdot\cdot\cdot}$, whose Wilson coefficients
cannot be determined from the three point amplitude. Using the equations of
motion in the action, one can see that such operators contribute at order
$R^{2}$ for scattering processes. We find that for a black hole the
coefficients for all operators which are linear in the Riemann curvature can
be fixed by matching against the multipole moments of the Kerr solution.
Several proposals [93, 95, 100, 101] have appeared in the literature to fix
the coefficients on quadratic in Riemann operators based on a shift-symmetry
principle which is already true of the linear in Riemann results. Refs. [118,
119] find that the Compton amplitude derived by solving the Teukolsky equation
agrees with the shift-symmetry principle through $\mathcal{O}(S^{4})$ but that
tension with the shift-symmetry begins at $\mathcal{O}(S^{5})$ (though the
results from the Teukolsky equation involve a subtle analytic continuation
between the black-hole and naked-singularity regimes). The couplings we find
based on consideration of multipole moments can be made consistent with spin-
exponentiation, shift-symmetry, or the Teukolsky results through
$\mathcal{O}(S^{5})$. As well, they can be made simultaneously consistent with
spin-exponentiation and the Teukolsky results through $\mathcal{O}(S^{5})$.
(Beginning at $\mathcal{O}(S^{5})$ one helicity combination develops a
spurious pole in the the spin-exponentiated amplitude; when we say that we can
match spin-exponentiation at $\mathcal{O}(S^{5})$ or beyond we only mean that
we match to the helicity combination without a spurious pole after
$\mathcal{O}(S^{4})$.)

###  I.2 Summary of Method and Results

In Dixon’s landmark papers Ref. [11, 12] on the worldline formalism, among
other results, he proves that there is a unique way to define the multipole
moments of the current density or stress tensor for an extended body in
general relativity so that those multipole moments can be made into a
generating function for the current density/stress tensor in the usual way and
so that those multipole moments are fully reduced (i.e. contain no
interdependencies between moments of different orders). The definitions of
these moments are highly nontrivial and only coincide with the “naive” moments
(from integrating powers of displacement against the current density/stress
tensor over a spatial slice) for a body in uniform motion in flat spacetime.
From the Kerr solution, following Israel’s analysis [120], we compute the
stress tensor which acts as its source (in the maximally causally extended
spacetime) and from that source we compute the multipole moments of a spinning
black hole using Dixon’s definitions of the multipole moments. Those multipole
moments can then be used to determine a stress tensor, which in turn can be
used to determine an action for the spinning black hole, up to couplings to
operators which are quadratic in the Riemann tensor. The action we find is put
in dynamical mass function form in (216) and the specific dynamical mass
function we find for spinning black holes is given in (295).

The dynamical mass function we find contains all of the equivalent black hole
couplings identified in Ref. [48], which can be found by comparison to the
three-point amplitude (the stationary stress tensor), as well as many new
terms. Dixon’s formalism specifies the unique way to lift those naive moments
to proper multipole moments, and that lifting fixes the additional couplings
in (295) relative to Ref. [48]. This lifting is only able to fix linear in
Riemann couplings in the action because no information about higher order in
Riemann operators is contained in the stationary moments. We find that the
dynamical mass function in (295) can be made consistent with the spin-
exponentiation proposed by Ref. [77] and shift-symmetry proposed by Refs. [93,
95, 100, 101] through $\mathcal{O}(S^{4})$ for the appropriate choice of
quadratic in Riemann couplings. As well, it can be made consistent with any
one of the three principles of spin-exponentiation, shift-symmetry, or the
Teukolsky equation results found in Ref [118] through $\mathcal{O}(S^{5})$ and
made consistent with any pair of them except the combination of shift-symmetry
and the Teukolsky equation, which are incompatible. To facilitate this
analysis, we find a formal expression for the gravitational Compton amplitude
for a generic spinning body to all orders in spin and explicitly compute that
amplitude in terms of all possible Wilson coefficients in the action through
$\mathcal{O}(S^{5})$. We find that there is one linearly independent structure
in the amplitude at $\mathcal{O}(S^{2})$, with one more appearing at
$\mathcal{O}(S^{3})$, seven more at $\mathcal{O}(S^{4})$, and eleven more at
$\mathcal{O}(S^{5})$.

In order to understand these gravitational results, it is instructive to first
follow all of the same steps of analysis for a $\sqrt{\text{Kerr}}$ particle
in electromagnetism. In section II we review the worldline formalism with a
dynamical mass function for electromagnetism. In section III we review the
basics of Dixon’s theory of multipole moments and specialize his results to
the current-density in flat spacetime. In section IV we use Dixon’s formalism
to compute the multipole moments of a $\sqrt{\text{Kerr}}$ particle and from
those moments we compute the necessary dynamical mass function for such a
particle, up to corrections which are quadratic in the field strength. Our
electromagnetic analysis culminates in section V in which we compute the
electromagnetic Compton amplitude for a generic spinning body to all orders in
spin in terms of its dynamical mass function. We then specialize that all
orders result to cubic order in spin by enumerating all possible operators in
the action and study the spin-exponentiation and shift-symmetry properties of
the resultant amplitude. We find that for electromagnetism, it is possible to
simultaneously demand spin-exponentiation, shift-symmetry, and consistency
with the $\sqrt{\text{Kerr}}$ multipole moment based dynamical mass function
through $\mathcal{O}(S^{3})$.

In the second half of the paper, we perform the same analysis for gravity. We
begin in VI by reviewing the worldline formalism with a dynamical mass
function in general relativity. In section VII we use Dixon’s formalism to
compute the multipole moments of a Kerr particle and from those moments we
compute the necessary dynamical mass function for such a particle, up to
corrections which are quadratic in the Riemann tensor. Then in VIII we derive
a formal expression for the gravitational Compton amplitude for a generic
spinning body to all orders in spin, which is unfortunately much more complex
than the corresponding electromagnetic formula. We then specialize that all
orders result to quintic order in spin by enumerating all possible operators
in the action and study the requirements imposed on that amplitude by spin-
exponentiation, shift-symmetry, and matches to the Teukolsky equation. We find
that for gravity, the multipole moment based dynamical mass function is
consistent with the combination of spin-exponentiation and the Teukolsky
equation at $\mathcal{O}(S^{5})$ and that requiring these fixes all available
coefficients in the dynamical mass function at this order.

###  I.3 Notation

We call the spacetime manifold $\mathbb{T}$. Beginning alphabet Greek letter
indices $\alpha,\beta,\gamma,\delta,...$ are used for spacetime indices at a
generic event $X\in\mathbb{T}$. The tangent space to $\mathbb{T}$ at $X$ is
written $T_{X}(\mathbb{T})$. Late alphabet Greek letter indices
$\mu,\nu,\rho,\kappa,...$ are used for spacetime indices at a particular event
of interest $Z\in\mathbb{T}$. Indices are symmetrized using parentheses and
antisymmetrized with brackets, both with the typical symmetry factors:

$M^{(\alpha\beta)}=\frac{M^{\alpha\beta}+M^{\beta\alpha}}{2},\hskip
21.52771ptM^{[\alpha\beta]}=\frac{M^{\alpha\beta}-M^{\beta\alpha}}{2}.$ (1)

For a generic vector $v^{\alpha}$, we define:

$|v|=\sqrt{|g_{\alpha\beta}v^{\alpha}v^{\beta}|},\hskip
21.52771pt\hat{v}^{\alpha}=\frac{v^{\alpha}}{|v|}.$ (2)

Let $\zeta(s,Z,v)$ be a geodesic with affine parameter $s$ so that
$\zeta^{\mu}(0,Z,v)=z^{\mu}$ and $\frac{d\zeta^{\mu}}{ds}(0,Z,v)=v^{\mu}$.
Then, the exponential map is defined by the event:

$\exp_{Z}(v)=\zeta(1,Z,v).$ (3)

Consider the geodesic $\zeta(s)$ with affine parameter $s$ so that
$\zeta^{\mu}(0)=z^{\mu}$ and $\zeta^{\alpha}(1)=x^{\alpha}$ (for $Z$ and $X$
sufficiently close for one only such geodesic to exist). Then, Synge’s
worldfunction $\sigma(Z,X)$ is defined by:

$\sigma(Z,X)=\frac{1}{2}\int_{0}^{1}g_{\alpha\beta}(\zeta)\frac{d\zeta^{\alpha}}{ds}\frac{d\zeta^{\beta}}{ds}ds.$
(4)

Instead viewing $\sigma$ as a functional or the path, under variation with
respect to the path $\zeta$, we find:

$\delta\sigma=\frac{d\zeta_{\alpha}}{ds}(1)\delta
x^{\alpha}-\frac{d\zeta_{\mu}}{ds}(0)\delta z^{\mu}.$ (5)

We place indices on $\sigma$ to indicate covariant derivatives with
$\alpha,\beta,...$ indices for $x$ and $\mu,\nu,...$ indices for $z$:

$\sigma_{\alpha}=\nabla_{\alpha}\sigma=\frac{\partial\sigma}{\partial
x^{\alpha}}=\frac{d\zeta_{\alpha}}{ds}(1),\hskip
21.52771pt\sigma_{\mu}=\nabla_{\mu}\sigma=\frac{\partial\sigma}{\partial
z^{\mu}}=-\frac{d\zeta_{\mu}}{ds}(0).$ (6)

If more that two indices of the same type were to be placed on $\sigma$ the
order would be important due to the noncommutativity of covariant derivatives,
however we will have no need to do this. As well, we introduce the inverse
matrix $\sigma^{-1}$:

$\sigma^{-\alpha}{}_{\mu}\sigma^{\mu}{}_{\beta}=\delta^{\alpha}_{\beta}.$ (7)

We will describe the motion of the spinning body so that the worldline
$z^{\mu}(\lambda)$ tracks the center of momentum of the body with $\lambda$
the worldline time parameter. $u^{\mu}(\lambda)$ will be a smooth one
parameter family of future oriented timelike unit vectors which later will be
specialized to be $\hat{p}^{\mu}(\lambda)$, the unit vector in the direction
of the linear momentum $p^{\mu}(\lambda)$ of the body. We let
$\Sigma(\lambda)$ be the Cauchy slice formed by shooting out geodesics based
at $z^{\mu}(\lambda)$ which are orthogonal to $u^{\mu}(\lambda)$. Explicitly:

$\Sigma(\lambda)=\\{X\in\mathbb{T}:u^{\mu}(\lambda)\sigma_{\mu}(Z(\lambda),X)=0\\}.$
(8)

Let $\tau(X)$ be the value of $\lambda$ so that $X\in\Sigma(\lambda)$. Let
$w_{1}^{\alpha}(X)$ be any vector field satisfying:

$\lambda=\tau(X)\implies\lambda+\delta\lambda=\tau(\exp_{X}(w_{1}\delta\lambda))+\mathcal{O}(\delta\lambda^{2}).$
(9)

That is, if each point of $\Sigma(\lambda)$ is displaced by
$w_{1}^{\alpha}\delta\lambda$ then it produces a point in
$\Sigma(\lambda+\delta\lambda)$ for sufficiently small $\delta\lambda$.
Automatically:

$w_{1}^{\alpha}\nabla_{\alpha}\tau=1.$ (10)

Let $D^{4}x$ be the invariant spacetime volume measure:

$D^{4}x=\sqrt{-\det g}d^{4}x.$ (11)

Let $d\Sigma_{\alpha}$ be the future oriented invariant volume measure on
$\Sigma$. For any scalar function $f(X)$ then:

$\int_{\mathbb{T}}f(X)D^{4}x=\int_{-\infty}^{\infty}\int_{\Sigma(\lambda)}f(X)w_{1}^{\alpha}d\Sigma_{\alpha}d\lambda.$
(12)

Because $\lambda$ is an arbitrary parameter for the worldline, it is useful to
introduce the einbein $\mathtt{e}(\lambda)$ to help manage reparameterization
invariance manifestly. The einbein is an arbitrary function defined so that
under a smooth monotone increasing reparameterization
$\lambda^{\prime}=\lambda^{\prime}(\lambda)$:

$\mathtt{e}^{\prime}(\lambda^{\prime})=\frac{d\lambda}{d\lambda^{\prime}}\mathtt{e}(\lambda).$
(13)

The reparameterization invariant worldline measure $D\lambda$ is then defined
by:

$D\lambda=\mathtt{e}(\lambda)d\lambda.$ (14)

We also define the reparameterization invariant version of the vector field
$w_{1}^{\alpha}(X)$:

$w^{\alpha}(X)=\frac{w_{1}^{\alpha}(X)}{\mathtt{e}(\tau(X))}.$ (15)

Then, we have:

$\int_{\mathbb{T}}f(X)d\mathbb{T}=\int_{-\infty}^{\infty}\int_{\Sigma(\lambda)}f(X)w^{\alpha}d\Sigma_{\alpha}D\lambda.$
(16)

Selection of a worldline parameter then amounts to choosing
$\mathtt{e}(\lambda)$ as an arbitrary function.

##  II Electromagnetic MPD Equations

We begin with an analysis of the $\sqrt{\text{Kerr}}$ electromagnetic particle
in flat spacetime to develop a road-map for the gravitational analysis which
follows. The motion of a generic spinning body under the influence of
electromagnetism in Minkowski space is described by the electromagnetic flat
space Mathisson Papapetrou Dixon (MPD) equations [12]. It is well established
[12, 121, 122, 76, 123, 107] that the electromagnetic MPD equations can be
derived from a variational principle through an action $\mathcal{S}$ of the
form:

$\mathcal{S}[z,p,\Lambda,S,\alpha,\beta]=\int_{-\infty}^{\infty}\left(p_{\mu}\dot{z}^{\mu}+qA_{\mu}\dot{z}^{\mu}+\frac{1}{2}\epsilon_{\mu\nu\rho\sigma}u^{\mu}S^{\nu}\Omega^{\rho\sigma}-\frac{\alpha}{2}(p^{2}+\mathcal{M}^{2})+\beta
p\cdot S\right)d\lambda.$ (17)

In this action, $z^{\mu}(\lambda)$ is the center of momentum worldline of the
body, its conjugate is the linear momentum carried by the body
$p_{\mu}(\lambda)$, $q$ is the charge of the body, $A_{\mu}$ is the vector
potential, $u^{\mu}=\hat{p}^{\mu}$, $\alpha=\frac{\mathtt{e}}{\mathcal{M}}$ is
a Lagrange multiplier which reinforces reparameterization invariance, and
$S^{\mu}$ is the spin vector of the body. The spin vector is defined so that
it becomes the angular momentum vector of the body in its center of momentum
frame. Automatically, then $S\cdot p=0$ and so $\beta$ is a Lagrange
multiplier included to enforce this constraint. Also appearing is
${\Lambda^{\mu}}_{A}(\lambda)$, a tetrad tracking the orientation of the body,
which satisfies:

$\eta^{\mu\nu}=\Lambda^{\mu}{}_{A}\Lambda^{\nu}{}_{B}\eta^{AB},\hskip
21.52771pt\eta_{AB}=\eta_{\mu\nu}\Lambda^{\mu}{}_{A}\Lambda^{\nu}{}_{B}.$ (18)

Without loss of generality, we take:

${\Lambda^{\mu}}_{0}(\lambda)=u^{\mu}(\lambda),\hskip
21.52771pt{\Lambda^{\mu}}_{3}(\lambda)=\hat{S}^{\mu}(\lambda).$ (19)

(If these are set as initial conditions, they are maintained dynamically
automatically.) Capital Latin indices $A,B,C,D,...$ are always used for
Lorentz indices in the default frame of the body so that ${\Lambda^{\mu}}_{A}$
represents the Lorentz transformation the body has undergone in its motion
relative to an arbitrary default orientation. The angular velocity tensor of
the body is defined by:

$\Omega^{\mu\nu}=\eta^{AB}{\Lambda^{\mu}}_{A}\frac{d{\Lambda^{\nu}}_{B}}{d\lambda}.$
(20)

Finally, $\mathcal{M}(z,u,S)$, called the dynamical mass function of the body,
encodes the free mass of the body and all of its nonminimal couplings to
electromagnetism. In particular, it takes the form:

$\mathcal{M}^{2}(z,u,S)=m^{2}+\mathcal{O}(q\mathcal{F})$ (21)

where $m$ is the mass of the body in vacuum.

For variations of the action it is useful to define the antisymmetric tensor:

$\delta\theta^{\mu\nu}=\eta^{AB}{\Lambda^{\mu}}_{A}\delta\Lambda^{\nu}{}_{B}.$
(22)

Then, the variation of the above action gives:

$\displaystyle\delta\mathcal{S}=$
$\displaystyle\int_{-\infty}^{\infty}\left(\delta
z^{\mu}\left(-\dot{p}_{\mu}+q\mathcal{F}_{\mu\nu}\dot{z}^{\nu}-\mathtt{e}\frac{\partial\mathcal{M}}{\partial
z^{\mu}}\right)\right.$ $\displaystyle\left.+\delta
p_{\mu}\left(\dot{z}^{\mu}-\mathtt{e}u^{\mu}-\frac{\mathtt{e}}{|p|}\frac{\partial\mathcal{M}}{\partial
u^{\nu}}(\eta^{\mu\nu}+u^{\mu}u^{\nu})+\beta
S^{\mu}+\frac{1}{2|p|}(\delta^{\mu}_{\alpha}+u^{\mu}u_{\alpha})\epsilon^{\alpha\nu\rho\sigma}S_{\nu}\Omega_{\rho\sigma}\right)\right.$
$\displaystyle\left.+\frac{1}{2}\delta\theta^{\rho\sigma}\left(-\frac{d}{d\lambda}\left(\epsilon_{\mu\nu\rho\sigma}u^{\mu}S^{\nu}\right)+\epsilon_{\mu\nu\rho\alpha}u^{\mu}S^{\nu}\Omega^{\alpha}{}_{\sigma}-\epsilon_{\mu\nu\sigma\alpha}u^{\mu}S^{\nu}\Omega^{\alpha}{}_{\rho}\right)\right.$
$\displaystyle\left.+\delta
S^{\mu}\left(-\frac{1}{2}\epsilon_{\mu\nu\rho\sigma}u^{\nu}\Omega^{\rho\sigma}-\mathtt{e}\frac{\partial\mathcal{M}}{\partial
S^{\mu}}+\beta
p_{\mu}\right)-\frac{\delta\alpha}{2}\left(p^{2}+\mathcal{M}^{2}\right)+\delta\beta
p\cdot S\right)d\lambda.$ (23)

Using the $\delta S^{\mu}$ variation to solve for the angular velocity tensor,
one can then determine the value of $\beta$. That value of $\beta$ can then be
used to simplify the spin and trajectory equations of motion. Explicitly,
these give:

$\displaystyle\Omega^{\mu\nu}$
$\displaystyle=\dot{u}^{\mu}u^{\nu}-u^{\mu}\dot{u}^{\nu}+\mathtt{e}\epsilon^{\mu\nu\rho\sigma}u_{\rho}\frac{\partial\mathcal{M}}{\partial
S^{\sigma}}$ (24) $\displaystyle\beta$
$\displaystyle=-\frac{\mathtt{e}}{\mathcal{M}}u^{\mu}\frac{\partial\mathcal{M}}{\partial
S^{\mu}}$ (25) $\displaystyle\dot{S}^{\mu}$ $\displaystyle=u^{\mu}\dot{u}\cdot
S+\mathtt{e}\epsilon^{\mu\nu\rho\sigma}u_{\nu}S_{\rho}\frac{\partial\mathcal{M}}{\partial
S^{\sigma}}$ (26) $\displaystyle\dot{z}^{\mu}$
$\displaystyle=\mathtt{e}u^{\mu}+\frac{\mathtt{e}}{\mathcal{M}}\frac{\partial\mathcal{M}}{\partial
u_{\mu}}+\frac{\mathtt{e}}{\mathcal{M}}u^{\mu}u^{\nu}\frac{\partial\mathcal{M}}{\partial
u^{\nu}}+\frac{\mathtt{e}}{\mathcal{M}}S^{\mu}u^{\nu}\frac{\partial\mathcal{M}}{\partial
S^{\nu}}+\frac{1}{\mathcal{M}^{2}}\epsilon^{\mu\nu\rho\sigma}S_{\nu}u_{\rho}\dot{p}_{\sigma}.$
(27)

In order to determine the trajectory evolution explicitly we must insert the
momentum equation of motion into (27). To simplify, it is useful to introduce
the dual field strength:

${}^{\star}\mathcal{F}_{\mu\nu}=\frac{1}{2}\epsilon^{\mu\nu\rho\sigma}\mathcal{F}_{\rho\sigma}\implies\mathcal{F}_{\mu\nu}=-\frac{1}{2}\epsilon_{\mu\nu\rho\sigma}{}^{\star}\mathcal{F}^{\rho\sigma}.$
(28)

Simplifying finally gives the electromagnetic MPD equations of motion for the
spinning body:

$\displaystyle\left(1-\frac{q}{\mathcal{M}^{2}}{}^{\star}{\mathcal{F}}^{\alpha\beta}u_{\alpha}S_{\beta}\right)\frac{\dot{z}^{\mu}}{\mathtt{e}}=$
$\displaystyle\
u^{\mu}+\frac{1}{\mathcal{M}}\frac{\partial\mathcal{M}}{\partial
u_{\mu}}+u^{\mu}\frac{u^{\nu}}{\mathcal{M}}\frac{\partial\mathcal{M}}{\partial
u^{\nu}}+S^{\mu}\frac{u^{\nu}}{\mathcal{M}}\frac{\partial\mathcal{M}}{\partial
S^{\nu}}+\frac{1}{\mathcal{M}^{2}}\epsilon^{\mu\nu\rho\sigma}u_{\nu}S_{\rho}\frac{\partial\mathcal{M}}{\partial
z^{\sigma}}$ $\displaystyle\
\phantom{u^{\mu}}+\frac{q}{\mathcal{M}^{2}}{}^{\star}{\mathcal{F}}^{\mu\nu}S_{\nu}+\frac{q}{\mathcal{M}^{3}}\left(S^{\rho}\frac{\partial\mathcal{M}}{\partial
u^{\rho}}+S^{2}u^{\rho}\frac{\partial\mathcal{M}}{\partial
S^{\rho}}\right){}^{\star}{\mathcal{F}}^{\mu\nu}u_{\nu}$ (29)
$\displaystyle\dot{p}_{\mu}=$ $\displaystyle\
q\mathcal{F}_{\mu\nu}\dot{z}^{\nu}-\mathtt{e}\frac{\partial\mathcal{M}}{\partial
z^{\mu}}$ (30) $\displaystyle\dot{S}^{\mu}=$ $\displaystyle\
u^{\mu}\dot{u}\cdot
S+\mathtt{e}\epsilon^{\mu\nu\rho\sigma}u_{\nu}S_{\rho}\frac{\partial\mathcal{M}}{\partial
S^{\sigma}}.$ (31)

For solving these equations of motion we will always choose $\lambda$ so that
$\mathtt{e}=1$.

To understand how the dynamical mass function relates to the multipole moments
of the body, we will need to study the current produced by our action. Define
the $\mathcal{Q}_{n}$ moments:

$\mathcal{Q}_{n}^{\rho_{1}...\rho_{n}\mu\nu}=\mathcal{Q}_{n}^{(\rho_{1}...\rho_{n})[\mu\nu]}=\frac{\partial\mathcal{M}}{\partial\partial^{n}_{\rho_{1}...\rho_{n}}\mathcal{F}_{\mu\nu}}$
(32)

Then, our action produces a formal distributional expression for $J^{\mu}$:

$J^{\mu}(X)=\frac{\delta\mathcal{S}}{\partial
A_{\mu}}=\int_{-\infty}^{\infty}\left(q\dot{z}^{\mu}\delta(X-Z)-2\mathtt{e}\sum_{n=0}^{\infty}(-1)^{n}\mathcal{Q}_{n}^{\rho_{1}...\rho_{n}\mu\nu}\partial^{n+1}_{\rho_{1}...\rho_{n}\nu}\delta(X-Z)\right)d\lambda$
(33)

##  III Dixon’s Multipole Moments

In this section we summarize some of the ingredients and results of Dixon’s
definition of multipole moments [11]. The multipole moments of the current
density (in the case of electromagnetism) and of the energy-momentum tensor
(in the case of gravity) directly enter the equations of motion of the body
and can be computed from the stationary fields produced by the body when
isolated. We find that these multipole moments determine all linear in
$\mathcal{F}$ (for electromagnetism) or linear in $R$ (for gravity) operators
in the dynamical mass function. The necessary matching is similar to the
three-point amplitude matching performed in Ref. [76] in the worldline or in
Ref. [80] in the field theory. However, using Dixon’s multipole construction
we are able to extract more physical information from the stationary
$\sqrt{\text{Kerr}}$ or Kerr solutions than is contained in the three-point
amplitude, which allows the determination of an increased number of Wilson
coefficients.

Following Dixon’s discussion, we first consider how multipole moments are
defined for a generic scalar field, then a generic tensor field. Then, we see
how the general multipole moments of a vector field are constrained in
complicated ways if that vector field satisfies the continuity equation. This
leads to Dixon’s definition of the reduced multipole moments of a conserved
vector field which we apply to the current density. In this section, in all
but the final subsection, we keep the spacetime generic in anticipation of
applying our analysis to gravity. In the final subsection, we specialize to
Minkowski space in preparation for electromagnetic calculations.

###  III.1 Moments of a scalar field

Using the exponential map we can take functions on $\mathbb{T}$ and turn them
into functions on $T_{z}(\mathbb{T})$. For a scalar field $\phi(X)$
($X\in\mathbb{T}$) we may define the function $\phi^{\prime}(Z,v)$ ($v\in
T_{Z}(\mathbb{T})$) by:

$\phi^{\prime}(Z,v)=\phi(\exp_{Z}(v)).$ (34)

Because $\phi^{\prime}(Z,v)$ is a function on the flat tangent space
$T_{Z}(\mathbb{T})$, it is simple to define the Fourier transform
$\widetilde{\phi}(Z,k)$ and inverse Fourier transform:

$\widetilde{\phi}(Z,k)=\int e^{-ik\cdot
v}\phi^{\prime}(Z,v)\frac{D^{4}v}{(2\pi)^{2}},\hskip
21.52771pt\phi^{\prime}(Z,v)=\int e^{ik\cdot
v}\widetilde{\phi}(Z,k)\frac{D^{4}k}{(2\pi)^{2}}$ (35)

where the invariant tangent space measure and Fourier space measure are
defined as:

$D^{4}v=\sqrt{-\det g(Z)}d^{4}v,\hskip
21.52771ptD^{4}k=\frac{d^{4}k}{\sqrt{-\det g(Z)}}.$ (36)

Using that $v^{\mu}=-\sigma^{\mu}(Z,X)$ we can express the inverse Fourier
transform for our original scalar function as:

$\phi(X)=\int
e^{-ik_{\mu}\sigma^{\mu}(Z,X)}\widetilde{\phi}(Z,k)\frac{D^{4}k}{(2\pi)^{2}}$
(37)

It immediately follows that for scalar functions $\phi(X)$ and $\psi(X)$ and
their associated $\phi^{\prime}(Z,v),\psi^{\prime}(Z,v)$:

$\int_{\Sigma(\lambda)}\psi^{*}(X)\phi(X)w^{\alpha}d\Sigma_{\alpha}=\int\sum_{n=0}^{\infty}\frac{i^{n}}{n!}k_{\mu_{1}}...k_{\mu_{n}}\widetilde{\psi}^{*}(Z,k)\int_{\Sigma}\sigma^{\mu_{1}}...\sigma^{\mu_{n}}\phi(X)w^{\alpha}d\Sigma_{\alpha}\frac{D^{4}k}{(2\pi)^{2}}.$
(38)

Define then the moments $F^{\mu_{1}...\mu_{n}}_{n}$ and the moment generating
function $F$ associated to the scalar function $\phi$:

$\displaystyle
F_{n}^{\mu_{1}...\mu_{n}}[\phi](\lambda)=\int_{\Sigma}(-\sigma^{\mu_{1}})...(-\sigma^{\mu_{n}})\phi(X)w^{\alpha}d\Sigma_{\alpha}$
(39) $\displaystyle
F[\phi](\lambda,k)=\sum_{n=0}^{\infty}\frac{(-i)^{n}}{n!}F_{n}^{\mu_{1}...\mu_{n}}k_{\mu_{1}}...k_{\mu_{n}}.$
(40)

Automatically the moments of $\phi$ satisfy the conditions:

$F^{\mu_{1}...\mu_{n}}_{n}=F^{(\mu_{1}...\mu_{n})}_{n},\hskip
21.52771ptu_{\mu_{1}}F_{n}^{\mu_{1}...\mu_{n}}=0$ (41)

where for this last condition it is useful to remember equation (8). The
moment generating function determines how $\phi(X)$ behaves against test
functions and thus determines $\phi(X)$ completely according to:

$\int_{\Sigma(\lambda)}\psi^{*}(X)\phi(X)w^{\alpha}d\Sigma_{\alpha}=\int\widetilde{\psi}^{*}(Z,k)F(\lambda,k)\frac{D^{4}k}{(2\pi)^{2}}.$
(42)

###  III.2 Moments of a general tensor field

Suppose we now consider a tensor field
$\phi^{\alpha_{1}...\alpha_{m}}{}_{\beta_{1}...\beta_{n}}(X)$. Using Synge’s
world function and the exponential map we can translate this to a tensor
function on the tangent space at $z$ by:

$\phi^{\prime\mu_{1}...\mu_{m}}{}_{\nu_{1}...\nu_{n}}(Z,v)=(-\sigma^{\mu_{1}}{}_{\alpha_{1}})...(-\sigma^{\mu_{m}}{}_{\alpha_{m}})(-\sigma^{-\beta_{1}}{}_{\nu_{1}})...(-\sigma^{-\beta_{n}}{}_{\nu_{n}})\phi^{\alpha_{1}...\alpha_{m}}{}_{\beta_{1}...\beta_{n}}(\exp_{Z}(v)).$
(43)

The Fourier transform and its inverse may now be defined as:

$\displaystyle\widetilde{\phi}^{\mu_{1}...\mu_{m}}{}_{\nu_{1}...\nu_{n}}(Z,k)$
$\displaystyle=\int\frac{e^{-ik\cdot
v}}{(2\pi)^{2}}\phi^{\prime\mu_{1}...\mu_{m}}{}_{\nu_{1}...\nu_{n}}(Z,v)D^{4}v$
(44) $\displaystyle\phi^{\prime\mu_{1}...\mu_{m}}{}_{\nu_{1}...\nu_{n}}(Z,v)$
$\displaystyle=\int\frac{e^{ik\cdot
v}}{(2\pi)^{2}}\widetilde{\phi}^{\mu_{1}...\mu_{m}}{}_{\nu_{1}...\nu_{n}}(Z,k)D^{4}k$
(45)

For a given $\phi$, suppose we define the lower rank tensor $\varphi$ by
contracting two indices:

$\varphi^{\alpha_{1}...\alpha_{m-1}}{}_{\beta_{1}...\beta_{n-1}}=\phi^{\alpha_{1}...\alpha_{m-1}\alpha_{m}}{}_{\beta_{1}...\beta_{n-1}\alpha_{m}}\implies\varphi^{\prime\mu_{1}...\mu_{m-1}}{}_{\nu_{1}...\nu_{n-1}}=\phi^{\prime\mu_{1}...\mu_{m-1}\mu_{m}}{}_{\nu_{1}...\nu_{n-1}\mu_{m}}.$
(46)

This property only holds because the upper index bitensor propagator in (43)
$(-\sigma^{\mu}{}_{\alpha})$ is the matrix inverse of the lower index bitensor
propagator $(-\sigma^{-\beta}{}_{\nu})$. Alternatively, suppose we consider
the moments of a vector field $\phi_{\beta}$ which is itself the gradient of a
scalar field $\varphi$:

$\phi_{\beta}=\nabla_{\beta}\varphi\implies\widetilde{\phi}_{\nu}=ik_{\nu}\widetilde{\varphi}.$
(47)

This property only holds if the lower index bitensor propagator is
$-\sigma^{-\beta}{}_{\nu}$. Therefore while one could have imagined other
bitensor propagators to use for transporting the components of $\phi$, such as
parallel transport, the choice in (43) is unique in satisfying both (46) and
(47). Unfortunately, the expression given in (43) has the property that
translation to a tensor function on tangent space does not commute with
raising/lowering indices of the original tensor field. This causes no real
inconvenience for our calculations but one should be aware that
$\phi^{\prime\mu}\neq g^{\mu\nu}(Z)\phi^{\prime}_{\nu}$ if $\phi^{\prime\mu}$
is defined as above from $\phi^{\mu}$ and $\phi^{\prime}_{\mu}$ is defined as
above from $\phi_{\mu}$.

For a generic $\phi$, the moments and moment generating function of $\phi$ are
defined by:

$\displaystyle
F_{N}^{\mu_{1}...\mu_{N}\rho_{1}...\rho_{m}}{}_{\nu_{1}...\nu_{n}}[\phi](\lambda)$
$\displaystyle=\int_{\Sigma}(-\sigma^{\mu_{1}})....(-\sigma^{\mu_{N}})\phi^{\prime\rho_{1}...\rho_{m}}{}_{\nu_{1}...\nu_{n}}w^{\alpha}d\Sigma_{\alpha}$
(48) $\displaystyle
F^{\rho_{1}...\rho_{m}}{}_{\nu_{1}...\nu_{n}}[\phi](\lambda,k)$
$\displaystyle=\sum_{N=0}^{\infty}\frac{(-i)^{N}}{N!}F_{N}^{\mu_{1}...\mu_{N}\rho_{1}...\rho_{m}}{}_{\nu_{1}...\nu_{n}}k_{\mu_{1}}...k_{\mu_{N}}.$
(49)

Automatically the moments of $\phi$ satisfy the conditions:

$F_{N}^{\mu_{1}...\mu_{N}\rho_{1}...\rho_{m}}{}_{\nu_{1}...\nu_{n}}=F_{N}^{(\mu_{1}...\mu_{N})\rho_{1}...\rho_{m}}{}_{\nu_{1}...\nu_{n}},\hskip
21.52771ptu_{\mu_{1}}F_{N}^{\mu_{1}...\mu_{N}\rho_{1}...\rho_{m}}{}_{\nu_{1}...\nu_{n}}=0.$
(50)

As well, $\phi$’s behavior against test functions is determined by:

$\int_{\Sigma}\psi^{*\beta_{1}...\beta_{n}}{}_{\alpha_{1}...\alpha_{m}}\phi^{\alpha_{1}...\alpha_{m}}{}_{\beta_{1}...\beta_{n}}w^{\gamma}d\Sigma_{\gamma}=\int\widetilde{\psi}^{*\nu_{1}...\nu_{n}}{}_{\mu_{1}...\mu_{m}}F^{\mu_{1}...\mu_{m}}{}_{\nu_{1}...\nu_{n}}(\lambda,k)\frac{D^{4}k}{(2\pi)^{2}}.$
(51)

Thus, just as in the scalar case the moment generating function determines the
original tensor field.

###  III.3 Moments of a conserved vector field

If we consider a vector field $\phi^{\alpha}(X)$ with moments as defined above
then a brief calculation reveals:

$\int_{\mathbb{T}}\psi^{*}\nabla_{\alpha}\phi^{\alpha}d\mathbb{T}=-\int_{-\infty}^{\infty}\int\widetilde{\psi}^{*}(z,k)\sum_{n=1}^{\infty}\frac{(-i)^{n}}{(n-1)!}k_{\mu_{1}}...k_{\mu_{n}}F^{(\mu_{1}...\mu_{n})}_{n-1}\frac{D^{4}k}{(2\pi)^{2}}D\lambda$
(52)

If $\phi^{\alpha}$ is a conserved vector field so that
$\nabla_{\alpha}\phi^{\alpha}=0$, then the left hand side is 0 for all $\psi$.
One may then wish based on this to conclude that
$F^{(\mu_{1}...\mu_{n})}_{n-1}$ is 0 for each $n\geq 1$. However, this is not
a valid deduction. $\widetilde{\psi}^{*}(Z,k)$ at each fixed $\lambda$
determines the function $\psi(X)$ through the inverse Fourier transform and so
$\widetilde{\psi}^{*}(Z(\lambda_{1}),k)$ and
$\widetilde{\psi}^{*}(Z(\lambda_{2}),k)$ for $\lambda_{1}\neq\lambda_{2}$ are
not independent of each other. Because of this, one cannot conclude that the
integrand above at each $\lambda$ must be individually 0 as they may conspire
to cancel at different $\lambda$ for arbitrary $\psi^{*}(X)$. Consequently, it
is difficult to conclude anything explicit about the moments of
$\phi^{\alpha}$ from the condition $\nabla_{\alpha}\phi^{\alpha}=0$.

Because of the mentioned difficulty, Dixon helpfully introduced an alternate
set of reduced moments for a conserved vector field. For us the vector field
of interest will always be $J^{\alpha}$. The goal of this reduced set of
moments is precisely to produce a set which does not have the entanglements of
the naive moments for a conserved vector field. Due to the absence of
entanglements between the moments, when using Dixon’s moments the it is valid
to equate integrands moment-by-moment. To arrive at Dixon’s reduced multipole
moments, which for a conserved vector field are written
$m^{\lambda_{1}...\lambda_{n}\mu}_{n}$, we first need to introduce a few
building blocks. Define:

$\displaystyle\Theta^{\kappa\lambda}_{n}(Z,X)$
$\displaystyle=(n+1)\int_{0}^{1}{\sigma^{\kappa}}_{\alpha}(Z,\zeta(t))\sigma^{\alpha\lambda}(Z,\zeta(t))t^{n}dt$
(53) $\displaystyle\mathfrak{q}_{n}^{\lambda_{1}...\lambda_{n}\mu\nu}$
$\displaystyle=(-1)^{n}\int_{\Sigma}\sigma^{\lambda_{1}}...\sigma^{\lambda_{n}}\Theta^{\mu\nu}_{n-1}J^{\alpha}d\Sigma_{\alpha}\hskip
21.52771pt(n\geq 1)$ (54)
$\displaystyle\mathfrak{j}_{n}^{\lambda_{1}...\lambda_{n}\mu}$
$\displaystyle=(-1)^{n}\int_{\Sigma}\sigma^{\lambda_{1}}...\sigma^{\lambda_{n}}{\sigma^{\mu}}_{\alpha}J^{\alpha}w^{\beta}d\Sigma_{\beta}$
(55) $\displaystyle Q_{n}^{\lambda_{1}...\lambda_{n}\mu\nu}$
$\displaystyle=\mathfrak{j}_{n+1}^{\lambda_{1}...\lambda_{n}[\mu\nu]}+\frac{1}{n+1}\mathfrak{q}_{n+1}^{\lambda_{1}...\lambda_{n}[\mu\nu]\kappa}\frac{\dot{z}^{\kappa}}{\mathtt{e}}$
(56) $\displaystyle m_{n}^{\lambda_{1}...\lambda_{n}\mu}$
$\displaystyle=\frac{2n}{n+1}Q^{(\lambda_{1}...\lambda_{n})\mu}_{n-1}\hskip
21.52771pt(n\geq 1).$ (57)

The $m^{\lambda_{1}...\lambda_{n}\mu}_{n}$ moments (called the reduced
moments) will be the actual moments of interest. The other quantities defined
are useful intermediate pieces for calculation. The reduced moments
automatically satisfy:

$\displaystyle
m_{n}^{\lambda_{1}...\lambda_{n}\mu}=m_{n}^{(\lambda_{1}...\lambda_{n})\mu},\hskip
21.52771ptm^{(\lambda_{1}...\lambda_{n}\mu)}_{n}=0,$ (58) $\displaystyle
u_{\lambda_{1}}m^{\lambda_{1}...\lambda_{n-1}[\lambda_{n}\mu]}_{n}=0\hskip
21.52771pt(n\geq 2).$ (59)

Dixon finds that beyond these conditions, the reduced moments are not
restricted by the conservation of $J^{\alpha}$ and that they are independent
of each other for different values of $n$. It is useful to define the
$0^{\text{th}}$ moment:

$m^{\mu}_{0}=q\frac{\dot{z}^{\mu}}{\mathtt{e}}$ (60)

where $q$ is the total charge of the body defined by:

$q=\int_{\Sigma}J^{\alpha}d\Sigma_{\alpha}.$ (61)

Due to the conservation of $J^{\alpha}$, $q$ is independent of $\lambda$.
Dixon finds also that the reduced moments are independent of this
$0^{\text{th}}$ moment. Then, define the reduced moment generating function:

$M^{\mu}(\lambda,k)=\sum_{n=0}^{\infty}\frac{(-i)^{n}}{n!}m^{\lambda_{1}...\lambda_{n}\mu}_{n}k_{\lambda_{1}}...k_{\lambda_{n}}.$
(62)

Like the naive moments, the reduced moment generating functions determine the
behavior of $J^{\alpha}$ against test functions. In particular, for an
arbitrary vector field $A_{\alpha}(X)$:

$\int_{\Sigma}A^{*}_{\alpha}(X)J^{\alpha}(X)w^{\beta}d\Sigma_{\beta}=\int\widetilde{A}^{*}_{\mu}(Z,k)M^{\mu}(\lambda,k)\frac{D^{4}k}{(2\pi)^{2}}.$
(63)

The moment generating function automatically satisfies:

$M^{\mu}k_{\mu}=\frac{q}{\mathtt{e}}\dot{z}\cdot k.$ (64)

This implies that for the gradient of a scalar function, using the definition
of the reduced moment generating function:

$\int_{\Sigma}J^{\alpha}\nabla_{\alpha}fw^{\beta}d\Sigma_{\beta}=\frac{q}{\mathtt{e}}\frac{d}{d\lambda}f(Z).$
(65)

For $f$ which decay sufficiently quickly for no boundary term to be necessary
under integration by parts we immediately have:

$0=\int_{\mathbb{T}}f\nabla_{\alpha}J^{\alpha}d\mathbb{T}=-q\left.f(Z)\right|^{\lambda\to\infty}_{\lambda\to-\infty}$
(66)

which is true without placing any restrictions on the reduced moments of
$J^{\alpha}$ beyond $m_{0}^{\mu}$. Dixon proved [11] that these reduced
moments are the unique set of moments which are independent of each other for
different $n$, have only $m_{0}^{\mu}$ restricted by the conservation law, and
satisfy the index symmetry conditions in equations (58) and (59).

Through (63), the current density is determined in terms of the reduced
multipole moments. Explicitly comparing that behavior against test functions
to (33) and using crucially that the reduced multipole moments are unique and
contain no interdependencies, we can identify:

$m_{n}^{\rho_{1}...\rho_{n}\mu}=-2\,n!\mathcal{Q}_{n-1}^{(\rho_{1}...\rho_{n})\mu}$
(67)

which gives the reduced multipole moments from the couplings in the action.
Alternatively, this can be nicely inverted using the index symmetry conditions
of both quantities to find:

$\mathcal{Q}_{n}^{\rho_{1}...\rho_{n}\mu\nu}=-\frac{1}{n!}\frac{1}{n+2}m_{n+1}^{\rho_{1}...\rho_{n}[\mu\nu]}.$
(68)

This allows the direct determination of the coupling of the body to the field
strength in the action from its exact reduced multipole moments.

###  III.4 Moments of the current density in Minkowski space

We now specialize our calculations to Minkowski space. We can represent an
arbitrary element $v^{\mu}$ of $T_{z}(\mathbb{T})$ by a vector
$Y=y^{A}\vec{\iota}_{A}=y^{A}\Lambda^{\mu}{}_{A}\vec{e}_{\mu}$ where
$\vec{\iota}_{A}$ are local Minkowskian basis vectors
($\vec{\iota}_{A}\cdot\vec{\iota}_{B}=\eta_{AB}$) and $\vec{e}_{\mu}$ are
coordinate basis vectors. Thus:

$v^{\mu}={\Lambda^{\mu}}_{A}y^{A}.$ (69)

Just as in section II we continue to always choose the tetrad so that
$\Lambda^{\mu}{}_{0}=u^{\mu}$. In flat space:

$\sigma(Z,X)=\frac{1}{2}(x-z)^{2},\hskip
21.52771pt\sigma^{\mu}=-(x^{\mu}-z^{\mu}),\hskip
21.52771pt{\sigma^{\mu}}_{\alpha}=-\delta^{\mu}_{\alpha}\implies\Theta^{\mu\nu}_{n}=\eta^{\mu\nu}$
(70)

We always use lowercase beginning Latin alphabet indices $a,b,c,...$ for
values $1,2,3$ on the tangent space. Then, we have:

$X\in\Sigma(\lambda)\implies\exists\ y^{a}\in\mathbb{R}^{3}\ :\
x^{\mu}=z^{\mu}+{\Lambda^{\mu}}_{a}y^{a}.$ (71)

Using the $y^{a}$ coordinates and the definition of $\tau(X)$ we may identify
explicit flat space expressions for $w^{\alpha}$ and $d\Sigma_{\alpha}$:

$\displaystyle\lambda=\tau(z+\lambda u)\implies
1=(\dot{z}^{\alpha}+{\dot{\Lambda}^{\alpha}}{}_{a}y^{a})\nabla_{\alpha}\tau$
(72) $\displaystyle
w^{\alpha}=\frac{\dot{z}^{\alpha}+{\dot{\Lambda}^{\alpha}}{}_{a}y^{a}}{\mathtt{e}},\hskip
21.52771pt\hskip 21.52771ptd\Sigma_{\alpha}=-u_{\alpha}d^{3}y.$ (73)

With these the $\text{q}_{n}$ and $\text{j}_{n}$ moments become:

$\displaystyle\mathfrak{q}^{\lambda_{1}...\lambda_{n}\mu\nu}_{n}$
$\displaystyle=\eta^{\mu\nu}{\Lambda^{\lambda_{1}}}_{A_{1}}...{\Lambda^{\lambda_{n}}}_{A_{n}}\int_{\Sigma}y^{A_{1}}...y^{A_{n}}(-J\cdot
u)d^{3}y$ (74) $\displaystyle\mathfrak{j}^{\lambda_{1}...\lambda_{n}\mu}_{n}$
$\displaystyle={\Lambda^{\lambda_{1}}}_{A_{1}}...{\Lambda^{\lambda_{n}}}_{A_{n}}{\Lambda^{\mu}}_{B}\int_{\Sigma}y^{A_{1}}...y^{A_{n}}{\Lambda_{\nu}}^{B}J^{\nu}\frac{-u\cdot\dot{z}-u\cdot\dot{\Lambda}\cdot
y}{\mathtt{e}}d^{3}y.$ (75)

Using the explicit formula for the worldline velocity $\dot{z}^{\mu}$ in (29),
$u\cdot\dot{z}=-\mathtt{e}$. As well, due to $\eqref{eq:angvelsol}$,
$\dot{\Lambda}=\mathcal{O}(\mathcal{F})$. Therefore:

$\mathfrak{j}^{\lambda_{1}...\lambda_{n}\mu}_{n}={\Lambda^{\lambda_{1}}}_{A_{1}}...{\Lambda^{\lambda_{n}}}_{A_{n}}{\Lambda^{\mu}}_{B}\int_{\Sigma}y^{A_{1}}...y^{A_{n}}{\Lambda_{\nu}}^{B}J^{\nu}d^{3}y+\mathcal{O}(\mathcal{F}).$
(76)

It is useful to define one last class of moments:

$\displaystyle K_{n}^{a_{1}...a_{n}B}$
$\displaystyle=\int_{\Sigma}y^{a_{1}}...y^{a_{n}}{\Lambda_{\nu}}^{B}J^{\nu}d^{3}y$
(77) $\displaystyle\implies\mathfrak{q}^{\lambda_{1}...\lambda_{n}\mu\nu}_{n}$
$\displaystyle=\eta^{\mu\nu}{\Lambda^{\lambda_{1}}}_{a_{1}}...{\Lambda^{\lambda_{n}}}_{a_{n}}K^{a_{1}...a_{n}0}_{n}$
(78) $\displaystyle\implies\mathfrak{j}^{\lambda_{1}...\lambda_{n}\mu}_{n}$
$\displaystyle={\Lambda^{\lambda_{1}}}_{a_{1}}...{\Lambda^{\lambda_{n}}}_{a_{n}}{\Lambda^{\mu}}_{B}K^{a_{1}...a_{n}B}_{n}+\mathcal{O}(\mathcal{F}).$
(79)

For a stationary body in its center of momentum frame, the $K_{n}$ moments
coincide with the naive moments of the stationary current density. By
following the chain of definitions from (53) to (57) these $K_{n}$ moments
determine the full reduced moments $m_{n}$. Therefore, computing the naive
moments of a stationary body allows the determination of the (coupling
independent part of the) full set of reduced moments.

##  IV Root-Kerr Multipole Moments

In this section we determine the linear in $\mathcal{F}$ couplings in the
dynamical mass function to all orders in spin for a $\sqrt{\text{Kerr}}$
particle. We begin by reviewing how the vector potential which defines the
$\sqrt{\text{Kerr}}$ particle arises from the Kerr-Newman solution and
computing some basic mechanical properties of the $\sqrt{\text{Kerr}}$ fields.
Then, we use the $\sqrt{\text{Kerr}}$ fields to determine the charge and
current densities which are the source of the $\sqrt{\text{Kerr}}$ solution
using analysis which parallels the calculation of the source of the Kerr
metric performed in Ref. [120]. From these charge and current densities we are
then able to determine the $K_{n}$ moments for $\sqrt{\text{Kerr}}$ and
consequently determine the $m_{n}$ reduced multipole moments up to corrections
of $\mathcal{O}(\mathcal{F})$. Comparing these reduced moments to those
produced by the current density produced by (33) fully determines all linear
in $\mathcal{F}$ terms in the dynamical mass function for a
$\sqrt{\text{Kerr}}$ particle.

###  IV.1 From Kerr-Newman to Root-Kerr

The Kerr-Newman solution for a stationary charged spinning black hole is
defined by the line element and vector potential:

$\displaystyle-d\tau^{2}=$ $\displaystyle\
-\frac{r^{2}-2Gmr+a^{2}+\frac{q^{2}G}{4\pi}}{r^{2}+a^{2}\cos^{2}\theta}(dt-a\sin^{2}\theta
d\varphi)^{2}+\frac{\sin^{2}\theta}{r^{2}+a^{2}\cos^{2}\theta}((r^{2}+a^{2})d\varphi-
adt)^{2}$ $\displaystyle\
+(r^{2}+a^{2}\cos^{2}\theta)\left(\frac{dr^{2}}{r^{2}-2Gmr+a^{2}+\frac{q^{2}G}{4\pi}}+d\theta^{2}\right)$
(80) $\displaystyle A_{\mu}dx^{\mu}=$ $\displaystyle\ \
\frac{q}{4\pi}\frac{-rdt+ar\sin^{2}\theta
d\varphi}{r^{2}+a^{2}\cos^{2}\theta}.$ (81)

The $G\to 0$ limit of the Kerr-Newman metric produces the line element:

$-d\tau^{2}=-dt^{2}+\frac{r^{2}+a^{2}\cos^{2}\theta}{r^{2}+a^{2}}dr^{2}+(r^{2}+a^{2}\cos^{2}\theta)d\theta^{2}+(r^{2}+a^{2})\sin^{2}\theta
d\varphi^{2}.$ (82)

This is just the Minkowski line element in a particular choice of coordinates.
The associated spatial metric $g^{(3)}_{ab}$ implied by this line element is
the natural metric of oblate-spheroidal coordinates, related to Cartesian
coordinates by:

$\displaystyle x=\sqrt{r^{2}+a^{2}}\sin\theta\cos\varphi,\hskip
21.52771pty=\sqrt{r^{2}+a^{2}}\sin\theta\sin\varphi,\hskip
21.52771ptz=r\cos\theta$ (83)

and with corresponding coordinate basis vectors:

$\displaystyle\vec{e}_{r}=\frac{\partial\vec{x}}{\partial r},\hskip
21.52771pt\vec{e}_{\theta}=\frac{\partial\vec{x}}{\partial\theta},\hskip
21.52771pt\vec{e}_{\varphi}=\frac{\partial\vec{x}}{\partial\varphi},\hskip
21.52771ptg^{(3)}_{ab}=\vec{e}_{a}\cdot\vec{e}_{b}.$ (84)

The vector potential of this system defines the vector potential of a
stationary $\sqrt{\text{Kerr}}$ particle:

$\displaystyle\phi$
$\displaystyle=-A_{0}=\frac{q}{4\pi}\frac{r}{r^{2}+a^{2}\cos^{2}\theta}$ (85)
$\displaystyle\vec{A}$
$\displaystyle=g^{(3)ab}A_{a}\vec{e}_{b}=\frac{q}{4\pi}\frac{ar}{(r^{2}+a^{2})(r^{2}+a^{2}\cos^{2}\theta)}\vec{e}_{\varphi}.$
(86)

These potentials the produce the electric and magnetic fields:

$\displaystyle\vec{E}$
$\displaystyle=\frac{q}{4\pi}\frac{(r^{2}+a^{2})(r^{2}-a^{2}\cos^{2}\theta)}{(r^{2}+a^{2}\cos^{2}\theta)^{3}}\vec{e}_{r}-\frac{q}{4\pi}\frac{2ra^{2}\sin\theta\cos\theta}{(r^{2}+a^{2}\cos^{2}\theta)^{3}}\vec{e}_{\theta}$
(87) $\displaystyle\vec{B}$
$\displaystyle=\frac{q}{4\pi}\frac{2ar(r^{2}+a^{2})\cos\theta}{(r^{2}+a^{2}\cos^{2}\theta)^{3}}\vec{e}_{r}+\frac{q}{4\pi}\frac{a(r^{2}-a^{2}\cos^{2}\theta)\sin\theta}{(r^{2}+a^{2}\cos^{2}\theta)^{3}}\vec{e}_{\theta}.$
(88)

Away from any singularities, this magnetic field satisfies
$\nabla\times\vec{B}=0$ and so can be expressed in terms of a magnetic scalar
potential $\vec{B}=-\nabla\psi$. In general coordinates in Minkowski space,
the condition that the magnetic field (away from any current sources) be given
by both a magnetic scalar potential and vector potential is:

$-\sqrt{\det g^{(3)}}g^{(3)ab}\frac{\partial\psi}{\partial
x^{b}}=\epsilon^{abc}\frac{\partial A_{c}}{\partial x^{b}}.$ (89)

Using the stationary $\sqrt{\text{Kerr}}$ vector potential, one may use this
to explicitly compute the associated magnetic scalar potential. Doing so
gives:

$\psi=\frac{q}{4\pi}\frac{a\cos\theta}{r^{2}+a^{2}\cos^{2}\theta}.$ (90)

Defining $\vec{a}=a\vec{e}_{z}$, in oblate-spheroidal coordinates we have the
identity:

$|\vec{x}-i\vec{a}|=r-ia\cos\theta.$ (91)

Consequently, the stationary $\sqrt{\text{Kerr}}$ particle is equivalently
defined [124] by producing the electric and magnetic scalar potentials or
electric and magnetic fields given by:

$\phi+i\psi=\frac{q}{4\pi|\vec{x}-i\vec{a}|}\implies\vec{E}+i\vec{B}=-\nabla\left(\frac{q}{4\pi|\vec{x}-i\vec{a}|}\right).$
(92)

###  IV.2 Mechanical Properties of the Stationary Root-Kerr Solution

Consider a surface of constant $r=R$ for integration. If we take $R\to\infty$
this becomes the spatial boundary. The surface area element is:

$d\vec{\mathbb{A}}=(r^{2}+a^{2})\sin\theta\vec{e}_{r}d\theta d\varphi.$ (93)

Then, by Gauss’ law the total charge of the $\sqrt{\text{Kerr}}$ particle is
given by:

$\int_{\mathbb{S}}\rho d^{3}\vec{x}=\int_{\partial\mathbb{S}}\vec{E}\cdot
d\vec{\mathbb{A}}=\lim_{R\to\infty}\int_{0}^{\pi}\int_{0}^{2\pi}\frac{q}{4\pi}\frac{R^{2}-a^{2}\cos^{2}\theta}{(R^{2}+a^{2}\cos^{2}\theta)^{2}}(R^{2}+a^{2})\sin\theta
d\varphi d\theta=q$ (94)

confirming our consistent use of the symbol $q$ for the total charge of the
distribution.

The magnetic dipole moment of a charge distribution is by definition:

$\vec{\mu}=\frac{1}{2}\int_{\mathbb{S}}\vec{x}\times\vec{J}d^{3}\vec{x}.$ (95)

Because the distribution is stationary, rewriting
$\vec{J}=\nabla\times\vec{B}$ and integrating by parts sufficiently many times
this is equivalently:

$\vec{\mu}=-\frac{3}{2}\int_{\partial\mathbb{S}}\vec{A}\times
d\vec{\mathbb{A}}.$ (96)

With the same surface of constant $r=R$:

$\vec{\mu}=-\frac{3}{2}\lim_{R\to\infty}\int_{0}^{\pi}\int_{0}^{2\pi}\frac{qa}{4\pi}\frac{R}{R^{2}+a^{2}\cos^{2}\theta^{2}}\vec{e}_{\varphi}\times\vec{e}_{r}\sin\theta
d\varphi d\theta=q\vec{a}.$ (97)

So, $\vec{a}$ is the magnetic dipole moment per unit charge.

Like for a point charge, the total energy in the electromagnetic field for
this system is divergent. To regulate this instead of integrating all the way
down to $r=0$ for the total energy, we integrate down to a cutoff
$r=\varepsilon$. Then, the total energy is:

E
$\displaystyle=\int_{\mathbb{S}}\frac{|\vec{E}|^{2}+|\vec{B}|^{2}}{2}d^{3}\vec{x}=\frac{q^{2}}{16\pi\varepsilon}\left(1+\frac{a}{\varepsilon}\arctan\left(\frac{a}{\varepsilon}\right)+\frac{\varepsilon}{a}\arctan\left(\frac{a}{\varepsilon}\right)\right).$
(98)

The Poynting vector for the particle is:

$\vec{\wp}=\vec{E}\times\vec{B}=\frac{q^{2}}{16\pi^{2}}\frac{a\vec{e}_{\varphi}}{(r^{2}+a^{2}\cos^{2}\theta)^{3}}.$
(99)

This leads to the total linear momentum:

$\vec{p}=\int_{\mathbb{S}}\vec{\wp}d^{3}\vec{x}=0.$ (100)

So, the stationary $\sqrt{\text{Kerr}}$ fields given are in fact in the center
of momentum frame. Consequently, the energy found before is the invariant mass
of the distribution:

$m=\frac{q^{2}}{16\pi\varepsilon}\left(1+\frac{a}{\varepsilon}\arctan\left(\frac{a}{\varepsilon}\right)+\frac{\varepsilon}{a}\arctan\left(\frac{a}{\varepsilon}\right)\right).$
(101)

The Poynting vector we found leads to the total angular momentum (which
because we are in the center of momentum frame is the spin angular momentum):

$\displaystyle\vec{S}$
$\displaystyle=\int_{\mathbb{S}}\vec{x}\times\vec{\wp}d^{3}\vec{x}=\frac{q^{2}}{16\pi\varepsilon}\left(1+\frac{a}{\varepsilon}\arctan\left(\frac{a}{\varepsilon}\right)+\frac{2\varepsilon}{a}\arctan\left(\frac{a}{\varepsilon}\right)-\frac{\varepsilon^{2}}{a^{2}}+\frac{\varepsilon^{3}}{a^{3}}\arctan\left(\frac{a}{\varepsilon}\right)\right)\vec{a}.$
(102)

While both the spin and the mass of the solution are divergent in
$\varepsilon$, their ratio has a finite $\varepsilon$ to 0 limit. Therefore,
the physical value of the ratio of the spin to the mass is:

$\frac{\vec{S}}{m}=\vec{a}.$ (103)

Immediately it follows that for this distribution:

$\vec{\mu}=\frac{q}{m}\vec{S}.$ (104)

Thus, the $\sqrt{\text{Kerr}}$ particle has a gyromagnetic ratio of exactly
$2$.

###  IV.3 Charge and Current density

The electric and magnetic fields as well as scalar and vector potential given
diverge whenever $r^{2}+a^{2}\cos^{2}\theta=0$ and nowhere else. For $r>0$ the
potentials and their derivatives are continuous and so there are no surface
charges or currents for $r>0$. Because the divergence of the right hand side
in (87) is 0 away from any poles, there is no charge density for $r>0$.
Because the curl of the right hand side in (88) is 0 away from any poles,
there is no current density for $r>0$. Consequently, the source for the
$\sqrt{\text{Kerr}}$ particle must only have support for $r=0$. For $r=0$ the
oblate spheroidal coordinates reduce to the disk in the $xy$ plane with center
at the origin and radius $a$. In order to approach this disk from above we
must take $r\to 0$ with $\theta<\frac{\pi}{2}$ and in order to approach from
below we must take $r\to 0$ with $\theta>\frac{\pi}{2}$. It is useful to
define the coordinate $\chi=\theta$ for $z>0$ and $\chi=\pi-\theta$ for $z<0$.
This way, two points approaching the disk, one from above and one from below,
will limit to the same point in space when their limiting values of $\chi$ and
$\varphi$ coincide. The disk is then parameterized by the coordinates $\chi$
and $\varphi$ by:

$\displaystyle x=a\sin\chi\cos\varphi,\hskip
21.52771pty=a\sin\chi\sin\varphi,\hskip 21.52771ptz=0.$ (105)

We can also use the cylindrical radial coordinate
$\text{r}=\sqrt{x^{2}+y^{2}}=a\sin\chi$.

The surface charge density inside the disk is given by:

$\sigma_{\text{disk}}=\vec{e}_{z}\cdot(\vec{E}(0,\chi,\varphi)-\vec{E}(0,\pi-\chi,\varphi))=-\frac{qa}{2\pi(a^{2}-\text{r}^{2})^{\frac{3}{2}}}.$
(106)

We can integrate this from $\chi=0$ to $\chi=\frac{\pi}{2}-\varepsilon$ to get
the total charge inside the disk:

$Q_{\text{disk}}=\int_{\text{disk}}\sigma_{\text{disk}}d\mathbb{A}=q\left(1-\frac{1}{\sin\varepsilon}\right).$
(107)

The surface charge density cannot be extended all the way to the ring at
$\chi=\frac{\pi}{2}$ as it has a nonintegrable divergence. We know from before
that the total charge of the distribution is $q$ and so there must be a
charged ring at $\chi=\frac{\pi}{2}$ so that the total charge between this
ring and the disk is $q$. This ring is precisely the ring on which
$r^{2}+a^{2}\cos^{2}\theta=0$ and so where the potentials and fields are
divergent. The charge density on the ring must be azimuthally symmetric
because the fields are. So:

$q=Q_{\text{disk}}+Q_{\text{ring}}=q-\frac{q}{\sin\varepsilon}+\int_{0}^{2\pi}\lambda_{\text{ring}}ad\varphi\implies\lambda_{\text{ring}}=\frac{q}{2\pi
a\sin\varepsilon}.$ (108)

Thus the total charge density of the stationary $\sqrt{\text{Kerr}}$ particle
is:

$\rho=-\frac{qa}{2\pi(a^{2}-\text{r}^{2})^{\frac{3}{2}}}\delta(z)\vartheta(a\cos\varepsilon-\text{r})+\frac{q}{2\pi
a\sin\varepsilon}\delta(z)\delta(\text{r}-a).$ (109)

The surface current density inside the disk is:

$\displaystyle\vec{K}_{\text{disk}}$
$\displaystyle=\vec{e}_{z}\times(\vec{B}(0,\chi,\varphi)-\vec{B}(0,\pi-\chi,\varphi))=\sigma_{\text{disk}}\frac{\text{r}}{a}\vec{\iota}_{\varphi}$
(110)

where $\vec{\iota}_{\varphi}$ is the unit vector in the direction of
$\vec{e}_{\varphi}$. This is precisely the surface current density produced by
rigidly rotating the surface charge density of the disk with angular velocity
$\vec{\omega}$ given by:

$\vec{\omega}=\frac{1}{a}\vec{e}_{z}.$ (111)

Rotating the ring with this same angular velocity produces a current in the
ring of:

$\vec{I}_{\text{ring}}=\lambda_{\text{ring}}\vec{\iota}_{\varphi}=\frac{q}{2\pi
a\sin\varepsilon}\vec{\iota}_{\varphi}.$ (112)

These currents produce a magnetic dipole moment:

$\displaystyle\vec{\mu}$
$\displaystyle=\frac{1}{2}\int_{\text{disk}}\vec{x}\times\vec{K}_{\text{disk}}d\mathbb{A}+\frac{1}{2}\int_{\text{ring}}\vec{x}\times\vec{I}_{\text{ring}}ds=q\vec{a}\left(1-\frac{1}{2}\sin\varepsilon\right).$
(113)

In the $\varepsilon\to 0$ limit this produces exactly the magnetic dipole
moment we found before. Therefore, the current density for the stationary
$\sqrt{\text{Kerr}}$ distribution is:

$\vec{J}=\rho\frac{\vec{a}\times\vec{x}}{a^{2}}$ (114)

and the stationary $\sqrt{\text{Kerr}}$ particle is exactly a charged disk and
ring with the charge distribution given by equation (109) and rotating with
the angular velocity $\frac{1}{a}$ about the central axis of the disk. These
charge and current densities are precisely analogous to the mass density and
energy-momentum tensor found in Ref. [120] for a stationary Kerr black hole.

###  IV.4 Stationary Multipole Moments

We can now compute the $K_{n}$ moments explicitly for a stationary
$\sqrt{\text{Kerr}}$ particle:

$K^{a_{1}...a_{n}0}_{n}=\int_{\mathbb{S}}x^{a_{1}}...x^{a_{n}}\rho(\vec{x})d^{3}\vec{x},\hskip
21.52771ptK^{a_{1}...a_{n}b}_{n}=\int_{\mathbb{S}}x^{a_{1}}...x^{a_{n}}J^{b}(\vec{x})d^{3}\vec{x}.$
(115)

By using (114), we find immediately:

$K^{a_{1}...a_{n}b}_{n}=\frac{1}{a^{2}}{\epsilon^{b}}_{cd}a^{c}K^{da_{1}...a_{n}0}_{n+1}$
(116)

and so only the multipole moments of $\rho$ need to be computed in order to
determine the moments of the full distribution. Further, due to the full
symmetrization of $K_{n}^{a_{1}...a_{n}0}$, for an arbitrary 3-vector $k_{a}$:

$K^{a_{1}...a_{n}0}_{n}=\frac{1}{n!}\frac{\partial^{n}}{\partial
k_{a_{1}}...\partial
k_{a_{n}}}\left(K_{n}^{b_{1}...b_{n}0}k_{b_{1}}...k_{b_{n}}\right).$ (117)

The charge density of the $\sqrt{\text{Kerr}}$ particle is localized to the
$xy$ plane and is rotationally symmetric about the $z$ axis. So, it can be
written as:

$\rho(\vec{x})=\sigma(\text{r})\delta(z)$ (118)

where in particular for $\sqrt{\text{Kerr}}$ :

$\sigma(\text{r})=-\frac{qa}{2\pi(a^{2}-\text{r}^{2})^{\frac{3}{2}}}\vartheta(a\cos\varepsilon-\text{r})+\frac{q}{2\pi
a\sin\varepsilon}\delta(\text{r}-a).$ (119)

Then:

$K_{n}^{a_{1}...a_{n}0}k_{a_{1}}...k_{a_{n}}=(k_{x}^{2}+k_{y}^{2})^{\frac{n}{2}}\int_{0}^{2\pi}\cos^{n}\varphi
d\varphi\int_{0}^{\infty}\sigma(\text{r})\text{r}^{n+1}d\text{r}.$ (120)

If $n$ is odd the azimuthal integral gives 0, so we only need to consider even
$n$. For even $n$ the azimuthal integral is:

$\int_{0}^{2\pi}\cos^{2n}\varphi d\varphi=2\pi\frac{(2n)!}{4^{n}n!^{2}}.$
(121)

With the integration variable $x=\frac{1}{a}\sqrt{a^{2}-\text{r}^{2}}$ the
radial integral becomes:

$\int_{0}^{\infty}\sigma(\text{r})\text{r}^{2n+1}d\text{r}=\frac{qa^{2n}}{2\pi}\left(1+\int_{\sin\varepsilon}^{1}\frac{1-(1-x^{2})^{n}}{x^{2}}dx\right).$
(122)

Here it is safe to take the $\varepsilon\to 0$ limit to give:

$\int_{0}^{\infty}\sigma(\text{r})\text{r}^{2n+1}d\text{r}=\frac{qa^{2n}}{2\pi}\frac{4^{n}n!^{2}}{(2n)!}.$
(123)

Thus the nonzero $K_{n}$ moments become:

$\displaystyle K_{2n}^{a_{1}...a_{2n}0}v_{a_{1}}...v_{a_{2n}}$
$\displaystyle=q|\vec{a}\times\vec{v}|^{2n}$ (124) $\displaystyle
K_{2n+1}^{a_{1}...a_{2n+1}b}v_{a_{1}}...v_{a_{2n+1}}$
$\displaystyle=q|\vec{a}\times\vec{v}|^{2n}(\vec{a}\times\vec{v})^{b}.$ (125)

###  IV.5 Dynamical Multipole Moments

Now we return the $K_{n}$ moments to the definition of the reduced multipole
moments. For a generic body allowed to respond to external fields, the
multipole moments will in general depend on those external fields and so the
general moments in external fields cannot be fully determined from the
stationary moments. It is unclear at this time what is the appropriate
response for a $\sqrt{\text{Kerr}}$ body to external fields. So, for an exact
dynamical $\sqrt{\text{Kerr}}$ body:

$K_{n}^{a_{1}...a_{n}b}=K_{n\text{
stat}}^{a_{1}...a_{n}b}+\mathcal{O}(\mathcal{F})$ (126)

where $K_{n\text{ stat}}^{a_{1}...a_{n}b}$ are the $K_{n}$ moments we found
for a stationary $\sqrt{\text{Kerr}}$ particle. To express the reduced moments
it is useful to define the projector $\perp^{\mu}{}_{\nu}$ which projects
4-vectors into the plane of the disk:

$\perp^{\mu}{}_{\nu}=\delta^{\mu}_{\nu}+u^{\mu}u_{\nu}-\hat{a}^{\mu}\hat{a}_{\nu}.$
(127)

We also write $k_{\perp}^{\mu}$ as a shorthand for
$k_{\perp}^{\mu}=\perp^{\mu}{}_{\nu}k^{\nu}$ and $k_{\perp}$ without an index
as a shorthand for $\sqrt{|k_{\mu}\perp^{\mu}{}_{\nu}k^{\nu}|}$. Using the
definition of the reduced multipole moments for arbitrary 4-vectors $k_{\mu}$
and $v_{\mu}$ we find:

$\displaystyle
k_{\lambda_{1}}...k_{\lambda_{2n}}v_{\mu}m^{\lambda_{1}...\lambda_{2n}\mu}_{2n}$
$\displaystyle=qa^{2n}k_{\perp}^{2n}u\cdot v-qa^{2n}k_{\perp}^{2n-2}(u\cdot
k)(k_{\perp}\cdot v)+\mathcal{O}(\mathcal{F})$ (128) $\displaystyle
k_{\lambda_{1}}...k_{\lambda_{2n+1}}v_{\mu}m^{\lambda_{1}...\lambda_{2n+1}\mu}_{2n+1}$
$\displaystyle=qa^{2n}k^{2n}_{\perp}\epsilon_{\mu\nu\rho\sigma}u^{\mu}a^{\nu}k^{\rho}v^{\sigma}+\mathcal{O}(\mathcal{F}).$
(129)

Returning these to (68), we find:

$\displaystyle\mathcal{Q}_{2n}^{\rho_{1}...\rho_{2n}\mu\nu}\partial^{2n}_{\rho_{1}...\rho_{2n}}\mathcal{F}_{\mu\nu}$
$\displaystyle=\frac{q}{(2n+1)!}a^{2n}\partial_{\perp}^{2n}{}^{\star}\mathcal{F}_{\mu\nu}a^{\mu}u^{\nu}+\mathcal{O}(\mathcal{F}^{2})$
(130)
$\displaystyle\mathcal{Q}_{2n+1}^{\rho_{1}...\rho_{2n+1}\mu\nu}\partial^{2n+1}_{\rho_{1}...\rho_{2n+1}}\mathcal{F}_{\mu\nu}$
$\displaystyle=-\frac{q}{(2n+2)!}a^{2n+2}\partial_{\perp}^{2n}\partial_{\perp}^{\mu}\mathcal{F}_{\mu\nu}u^{\nu}+\mathcal{O}(\mathcal{F}^{2}).$
(131)

With the definition of $\mathcal{Q}_{n}$, these produce the dynamical mass
function:

$\mathcal{M}=m+q\frac{\sinh(a\partial_{\perp})}{a\partial_{\perp}}{}^{\star}\mathcal{F}_{\mu\nu}a^{\mu}u^{\nu}+q\frac{1-\cosh(a\partial_{\perp})}{\partial_{\perp}^{2}}\partial_{\perp}^{\mu}\mathcal{F}_{\mu\nu}u^{\nu}+\mathcal{O}(\mathcal{F}^{2}).$
(132)

There is no subtlety in defining the square root or inverse of the
differential operator here because once the trigonometric functions are series
expanded only positive even powers of $\partial_{\perp}$ survive. We will see
later that for $\sqrt{\text{Kerr}}$ particles, at least at low orders in spin,
the squared dynamical mass function $\mathcal{M}^{2}$ may be a simpler object
when $\mathcal{O}(\mathcal{F})^{2}$ operators are considered. So, going
forward we will express the dynamical mass function as:

$\mathcal{M}^{2}=m^{2}+2mq\frac{\sinh(a\partial_{\perp})}{a\partial_{\perp}}{}^{\star}\mathcal{F}_{\mu\nu}a^{\mu}u^{\nu}+2mq\frac{1-\cosh(a\partial_{\perp})}{\partial_{\perp}^{2}}\partial_{\perp}^{\mu}\mathcal{F}_{\mu\nu}u^{\nu}+\mathcal{O}(\mathcal{F}^{2}).$
(133)

This $\sqrt{\text{Kerr}}$ dynamical mass function is our primary result for
electromagnetism. When acting on a field strength which in the neighborhood of
the body is a vacuum solution ($\partial_{\nu}\mathcal{F}^{\mu\nu}=0$) of
Maxwell’s equations, this reduces to:

$\displaystyle\mathcal{M}^{2}=m^{2}+2mq\frac{\sin(a\Delta)}{a\Delta}{}^{\star}{\mathcal{F}}_{\mu\nu}a^{\mu}u^{\nu}+2mq\frac{1-\cos(a\Delta)}{(a\Delta)^{2}}(a\cdot\partial)\mathcal{F}_{\mu\nu}a^{\mu}u^{\nu}+\mathcal{O}(\mathcal{F}^{2})$
(134)

where the $\Delta$ differential operator is defined by:

$a\Delta=\sqrt{(a\cdot\partial)^{2}-a^{2}(u\cdot\partial)^{2}}.$ (135)

Again there is no subtlety in defining the square root or inverse of the
differential operator. In general (134) and (133) are not equivalent, however
we show that for computing Compton amplitudes they are interchangeable.

The dynamical mass function in (134) very nearly matches the couplings in Ref.
[48] (if the analysis done there for gravity is done for electromagnetism). In
particular, phrased as a dynamical mass function, the couplings in Ref. [48]
determine:

$\mathcal{M}^{2}=m^{2}+2mq\frac{\sin(a\cdot\partial)}{a\cdot\partial}{}^{\star}{\mathcal{F}}_{\mu\nu}a^{\mu}u^{\nu}+2mq\frac{1-\cos(a\cdot\partial)}{a\cdot\partial}\mathcal{F}_{\mu\nu}a^{\mu}u^{\nu}+\mathcal{O}(\mathcal{F}^{2})$
(136)

Equation (134) becomes this result with the replacement:

$a\Delta\rightarrow a\cdot\partial.$ (137)

The $(u\cdot\partial)$ terms present in $a\Delta$ do not contribute to the
three point amplitude and so $a\Delta$ and $a\cdot\partial$ are
indistinguishable in a three point matching calculation. For this reason, the
analyses of Refs. [76, 80] were insensitive to the couplings on terms of the
form $S^{2n}(u\cdot\partial)^{2n}\mathcal{F}_{\cdot\cdot}$. The first of such
terms is present in Ref. [125] with an undetermined Wilson coefficient.
Similarly, in Ref. [48] $(u\cdot\partial)$ are not considered because if one
uses the equations of motion in the action, they can be shuffled into order
$\mathcal{F}^{2}$ terms. One can see this quickly on the lowest order such
term, for example:

$\displaystyle\mathtt{e}a^{2}(u\cdot\partial)^{2}{}^{\star}\mathcal{F}_{\mu\nu}a^{\mu}u^{\nu}$
$\displaystyle=a^{2}a^{\mu}u^{\nu}u^{\rho}\dot{z}^{\sigma}\partial_{\sigma}\partial_{\rho}{}^{\star}\mathcal{F}_{\mu\nu}+\mathcal{O}(\mathcal{F}^{2})$
(138)
$\displaystyle=\frac{d}{d\lambda}\left(a^{2}a^{\mu}u^{\nu}u^{\rho}\partial_{\rho}{}^{\star}\mathcal{F}_{\mu\nu}\right)-\frac{d}{d\lambda}(a^{2}a^{\mu}u^{\nu}u^{\rho})\partial_{\rho}{}^{\star}\mathcal{F}_{\mu\nu}+\mathcal{O}(\mathcal{F}^{2})$
(139)
$\displaystyle=\frac{d}{d\lambda}\left(a^{2}a^{\mu}u^{\nu}u^{\rho}\partial_{\rho}{}^{\star}\mathcal{F}_{\mu\nu}\right)+\mathcal{O}(\mathcal{F}^{2}).$
(140)

For three-point amplitudes, (133), (134), and (136) are interchangeable. For
Compton amplitudes, (133) and (134) are interchangeable but distinct from
(136). The worldline evolution of all three are distinct. Importantly, by the
definition of Dixon’s moments (63) holds for any test function vector field
$A_{\alpha}(X)$, not only the physically relevant vector potential and that it
holds without using the solution to the equations of motion. We have shown
that for a $\sqrt{\text{Kerr}}$ particle the only dynamical mass function for
which (63) holds without using the electromagnetic MPD equations is (133). If
one shuffles away $(u\cdot\partial)$ operators in favor of $\mathcal{F}^{2}$
so that the linear in $\mathcal{F}$ dynamical mass function is given by (136),
then (63) will not be true identically. In this sense, the advantage of (133)
is that it uniquely provides at each $\lambda$ the physically correct
multipole moments for $\sqrt{\text{Kerr}}$ .

##  V Electromagnetic Compton Amplitude

In this section we formally compute the $\sqrt{\text{Kerr}}$ Compton amplitude
to all orders in spin up to contact terms. Those contact terms are determined
by $\mathcal{F}^{2}$ operators in the dynamical mass function which the
multipole analysis is insensitive to. We begin by computing a formal
expression for the all orders in spin Compton amplitude for a generic charged
spinning body in terms of the dynamical mass function. Next we consider that
generic Compton amplitude explicitly to order $S^{3}$. At $\mathcal{O}(S^{1})$
there is a single Wilson coefficient in the action and it can be determined by
matching the $\mathcal{O}(S^{1})$ three-point amplitude or Compton amplitude.
At $\mathcal{O}(S^{2})$ there are 5 new Wilson coefficients in the action. One
of them can be determined by matching the $\mathcal{O}(S^{2})$ three-point
amplitude while the other 4 can be determined uniquely by matching the
$\mathcal{O}(S^{2})$ Compton amplitude. At $\mathcal{O}(S^{3})$ there are 8
new Wilson coefficients in the action. One of them can be determined by
matching the $\mathcal{O}(S^{3})$ three-point amplitude while the other 7
appear only as contact terms in the Compton amplitude. Among those 7, there
are only 6 linearly independent structures appearing in the
$\mathcal{O}(S^{3})$ Compton amplitude, and so there is one linear combination
of Wilson coefficients that the Compton amplitude is independent of at this
order in spin.

Once we have the results for a generic body through $\mathcal{O}(S^{3})$, we
specialize our interest to $\sqrt{\text{Kerr}}$ . Requiring the exponentiation
of spin structure as found in Ref. [77] through $\mathcal{O}(S^{2})$ for the
helicity-preserving amplitude (which develops a spurious pole starting at
$\mathcal{O}(S^{3})$) and through $\mathcal{O}(S^{3})$ for the helicity-
reversing amplitude (which has no such spurious pole) fixes all Wilson
coefficients through $\mathcal{O}(S^{2})$ and 4 of the 8 new operators at
$\mathcal{O}(S^{3})$. Alternatively, requiring the helicity-preserving
amplitude to have the shift-symmetry described in Refs. [93, 95, 100, 101]
through $\mathcal{O}(S^{3})$ fixes 3 of the 8 new operators at
$\mathcal{O}(S^{3})$. The spin-exponentiation and shift-symmetry are
consistent with each other, and together fix 6 of the 8 new operators at
$\mathcal{O}(S^{3})$ (as they share one redundant condition). The dynamical
mass function (134) determines 2 Wilson coefficients at $\mathcal{O}(S^{3})$,
one of which is fixed by the three-point amplitude and the other of which is
independent of and consistent with both spin-exponentiation and shift-
symmetry. Together then Dixon’s multipole moments, spin-exponentiation, and
shift-symmetry fix 7 of the 8 Wilson coefficients at $\mathcal{O}(S^{3})$,
which is the maximum amount possible by using the Compton amplitude (due to
the presence of a linearly independent combination of Wilson coefficients that
the amplitude is independent of).

###  V.1 Formal Classical Compton

We consider Compton scattering of an incoming photon with polarization
$\mathcal{E}_{1}$ and momentum $k_{1}$ off of a massive spinning charged body
with initial momentum $mv$ ($v\cdot v=-1$) and initial spin $s$ to an outgoing
photon with polarization $\mathcal{E}_{2}$ and momentum $k_{2}$ and perturbed
massive body. For a plane wave vector potential with strength $\epsilon$:

$A_{\mu}(X)=\epsilon\mathcal{E}_{\mu}e^{ik\cdot x},\hskip
21.52771pt\mathcal{F}_{\mu\nu}(X)=\epsilon f_{\mu\nu}e^{ik\cdot x},\hskip
21.52771ptf_{\mu\nu}=ik_{\mu}\mathcal{E}_{\nu}-ik_{\nu}\mathcal{E}_{\mu}.$
(141)

Because the tree level Compton amplitude is $\mathcal{O}(q^{2})$, it depends
only on the $\mathcal{O}(q)$ and $\mathcal{O}(q^{2})$ pieces of the dynamical
mass function. Consequently, we will only be concerned with operators in
$\mathcal{M}$ which are linear or quadratic in $\mathcal{F}$ and so we
consider an $\mathcal{M}$ of the form:

$\mathcal{M}^{2}(z,u,S)=m^{2}+q\delta\mathcal{M}_{1}^{2}(z,u,S)+q^{2}\delta\mathcal{M}_{2}^{2}(z,u,S)+\mathcal{O}(q^{3}\mathcal{F}^{3})$
(142)

where $\delta\mathcal{M}_{1}^{2}$ is of the form:

$\delta\mathcal{M}_{1}^{2}(z,u,S)=\sum_{n=0}^{\infty}T_{n}^{\rho_{1}...\rho_{n}\mu\nu}(u,S)\partial^{n}_{\rho_{1}...\rho_{n}}\mathcal{F}_{\mu\nu}(z)$
(143)

for some functions $T_{n}^{\rho_{1}...\rho_{n}\mu\nu}(u,S)$ satisfying:

$T_{n}^{\rho_{1}...\rho_{n}\mu\nu}=T_{n}^{(\rho_{1}...\rho_{n})[\mu\nu]}$
(144)

and $\delta\mathcal{M}_{2}^{2}$ is of the form:

$\delta\mathcal{M}_{2}^{2}(z,u,S)=\sum_{n=0}^{\infty}\sum_{l=0}^{\infty}V_{nl}^{\rho_{1}...\rho_{n}\mu\nu|\kappa_{1}...\kappa_{l}\tau\omega}(u,S)\partial^{n}_{\rho_{1}...\rho_{n}}\mathcal{F}_{\mu\nu}(z)\partial^{l}_{\kappa_{1}...\kappa_{l}}\mathcal{F}_{\tau\omega}(z)$
(145)

for some functions $V_{nl}^{...}(u,S)$ satisfying:

$\displaystyle
V_{nl}^{\rho_{1}...\rho_{n}\mu\nu|\kappa_{1}...\kappa_{l}\tau\omega}$
$\displaystyle=V_{nl}^{(\rho_{1}...\rho_{n})[\mu\nu]|(\kappa_{1}...\kappa_{l})[\tau\omega]}$
(146) $\displaystyle
V_{nl}^{\rho_{1}...\rho_{n}\mu\nu|\kappa_{1}...\kappa_{l}\tau\omega}$
$\displaystyle=V_{ln}^{\kappa_{1}...\kappa_{l}\tau\omega|\rho_{1}...\rho_{n}\mu\nu}$
(147)

Maxwell’s equations together with the flat space electromagnetic MPD equations
with the described initial conditions will produce solutions of the form:

$\displaystyle z^{\mu}(\lambda)$ $\displaystyle=v^{\mu}\lambda+q\epsilon\delta
z^{\mu}(\lambda)+\mathcal{O}(q^{2})$ (148) $\displaystyle p_{\mu}(\lambda)$
$\displaystyle=mv_{\mu}+q\epsilon\delta p_{\mu}(\lambda)+\mathcal{O}(q^{2})$
(149) $\displaystyle S^{\mu}(\lambda)$ $\displaystyle=s^{\mu}+q\epsilon\delta
S^{\mu}(\lambda)+\mathcal{O}(q^{2})$ (150) $\displaystyle J^{\mu}(X)$
$\displaystyle=qJ^{\mu}_{\text{stat}}(X)+q^{2}\epsilon\delta
J(X)+\mathcal{O}(q^{3})$ (151) $\displaystyle A_{\mu}(X)$
$\displaystyle=\epsilon\mathcal{E}_{1\mu}e^{ik\cdot
x}+qA^{\text{stat}}_{\mu}(X)+q^{2}\epsilon\delta
A_{\mu}(X)+\mathcal{O}(q^{3})$ (152)

where $qJ^{\mu}_{\text{stat}}$ is the current density produced by the
stationary spinning body in the absence of the incoming photon $(\epsilon\to
0)$ and $qA^{\text{stat}}_{\mu}$ is the (Lorenz gauge) vector potential
produced by that current density. The equation of motion perturbations will be
oscillatory from solving the electromagnetic MPD equations. In particular,
their solutions take the form:

$\delta z^{\mu}=\delta\widetilde{z}^{\mu}e^{ik_{1}\cdot v\lambda},\hskip
21.52771pt\delta u^{\mu}=\delta\widetilde{u}^{\mu}e^{ik_{1}\cdot
v\lambda},\hskip 21.52771pt\delta
S^{\mu}=\delta\widetilde{S}^{\mu}e^{ik_{1}\cdot v\lambda}$ (153)

for constant vectors
$\delta\widetilde{z},\delta\widetilde{u},\delta\widetilde{S}$. ($\delta p$ is
determined from $\delta u$ and $\delta\mathcal{M}_{1}^{2}$ because
$p^{\mu}=\mathcal{M}u^{\mu}$.)

The $\mathcal{O}(q^{2}\epsilon)$ piece of the vector potential, $\delta
A_{\mu}$, determines the linear in $\epsilon$ outgoing electromagnetic field
and thus determines the tree level Compton amplitude. From the Lorenz-gauge
Maxwell equation, $\delta A_{\mu}$ satisfies:

$-\partial^{2}\delta A^{\mu}=\delta J^{\mu}$ (154)

and limits to 0 in the asymptotic past (when the appropriate $i\varepsilon$ is
used for the oscillatory solutions). Therefore, it is determined by the
delayed Green’s function solution to the wave equation and given by:

$\delta A^{\mu}(X)=\int_{\mathbb{S}}\frac{\delta
J^{\mu}(t-|\vec{x}-\vec{x}\mkern 2.0mu\vphantom{x}^{\prime}|,\vec{x}\mkern
2.0mu\vphantom{x}^{\prime})}{4\pi|\vec{x}-\vec{x}\mkern
2.0mu\vphantom{x}^{\prime}|}d^{3}\vec{x}\mkern 2.0mu\vphantom{x}^{\prime}$
(155)

where:

$t=-v\cdot x,\hskip 21.52771pt\vec{x}^{\mu}=x^{\mu}+v^{\mu}v\cdot x.$ (156)

Thus, $\delta A$ is determined by $\delta J$. The current perturbation is
determined by the equation of motion perturbations $\delta z,\delta p,\delta
S$. From (32) we may identify:

$\mathcal{Q}_{n}^{\rho_{1}...\rho_{n}\mu\nu}=\frac{q}{2m}\left(1-\frac{q\delta\mathcal{M}_{1}^{2}}{2m^{2}}\right)T_{n}^{\rho_{1}...\rho_{n}\mu\nu}+\frac{q^{2}}{m}\sum_{l=0}^{\infty}V^{\rho_{1}...\rho_{n}\mu\nu|\kappa_{1}...\kappa_{l}\tau\omega}_{nl}\partial^{l}_{\kappa_{1}...\kappa_{l}}\mathcal{F}_{\tau\omega}+\mathcal{O}(\mathcal{F}^{2}).$
(157)

Define the Fourier transform of the current:

$\widetilde{J}^{\mu}(k)=\int\frac{e^{-ik\cdot
x}}{(2\pi)^{2}}J^{\mu}(X)d^{4}X.$ (158)

In terms of the $\mathcal{Q}_{n}$ the exact Fourier transform of $J^{\mu}(X)$
is:

$\widetilde{J}^{\mu}(k_{2})=\int_{-\infty}^{\infty}\left(q\dot{z}^{\mu}+2\sum_{n=0}^{\infty}(-i)^{n+1}\mathcal{Q}_{n}^{\rho_{1}...\rho_{n}\mu\nu}k_{2\rho_{1}}...k_{2\rho_{n}}k_{2\nu}\right)\frac{e^{-ik_{2}\cdot
z}}{(2\pi)^{2}}d\lambda$ (159)

For a plane wave vector potential it is useful to introduce the function:

$\mathcal{N}(u,S,k,\mathcal{E})=\sum_{n=0}^{\infty}\frac{i^{n}}{2m}T_{n}^{\rho_{1}...\rho_{n}\mu\nu}(u,S)k_{\rho_{1}}...k_{\rho_{n}}f_{\mu\nu}$
(160)

so that if we evaluate $\delta\mathcal{M}_{1}^{2}$ on a plane wave vector
potential:

$\left.\delta\mathcal{M}_{1}^{2}\right|_{\text{plane
wave}}=2m\epsilon\mathcal{N}(u,S,k,\mathcal{E})e^{ik\cdot z}.$ (161)

Similarly, for a pair of plane waves it is useful to introduce the function:

$\mathcal{P}(u,S,k,\mathcal{E},k^{\prime},\mathcal{E}^{\prime})=\sum_{n=0}^{\infty}\sum_{l=0}^{\infty}\frac{i^{l-n}}{m}V_{nl}^{\rho_{1}...\rho_{n}\mu\nu|\kappa_{1}...\kappa_{l}\tau\omega}(u,S)k^{\prime}_{\rho_{1}}...k^{\prime}_{\rho_{n}}f^{\prime*}_{\mu\nu}k_{\kappa_{1}}...k_{\kappa_{l}}f_{\tau\omega}.$
(162)

As a shorthand, we write:

$\mathcal{N}_{1}=\mathcal{N}(v,s,k_{1},\mathcal{E}_{1}),\hskip
21.52771pt\mathcal{N}_{2}=\mathcal{N}(v,s,k_{2},\mathcal{E}_{2}),\hskip
21.52771pt\mathcal{P}_{12}=\mathcal{P}(v,s,k_{1},\mathcal{E}_{1},k_{2},\mathcal{E}_{2}).$
(163)

Expanding the definition of the $\mathcal{Q}_{n}$ moments and returning the
result to the Fourier transform of the current produces:

$\delta\widetilde{J}^{\mu}(k_{2})=H^{\mu}(k_{2})\frac{\delta(k_{1}\cdot
v-k_{2}\cdot v)}{2\pi}$ (164)

where:

$\displaystyle H^{\mu}(k_{2})=$ $\displaystyle-
ik_{2}\cdot\delta\widetilde{z}v^{\mu}+ik_{2}\cdot
v\delta\widetilde{z}^{\mu}+2\sum_{n=0}^{\infty}(-i)^{n+1}\delta\widetilde{\mathcal{Q}}_{n}^{\rho_{1}...\rho_{n}\mu\nu}k_{2\rho_{1}}...k_{2\rho_{n}}k_{2\nu}$
(165)
$\displaystyle\delta\widetilde{\mathcal{Q}}^{\rho_{1}...\rho_{n}\mu\nu}_{n}=$
$\displaystyle\
\frac{1}{2m}\left(-\frac{\mathcal{N}_{1}}{m}+\delta\widetilde{u}^{\sigma}\frac{\partial}{\partial
v^{\sigma}}+\delta\widetilde{S}^{\sigma}\frac{\partial}{\partial
s^{\sigma}}\right)T_{n}^{\rho_{1}...\rho_{n}\mu\nu}(v,s)$ $\displaystyle\hskip
21.52771pt\hskip
21.52771pt+\sum_{l=0}^{\infty}\frac{i^{l}}{m}V_{nl}^{\rho_{1}...\rho_{n}\mu\nu|\kappa_{1}...\kappa_{l}\tau\omega}k_{1\kappa_{1}}...k_{1\kappa_{l}}f_{1\tau\omega}.$
(166)

Returning our current perturbation to $\delta A$ produces:

$\delta A^{\mu}=\frac{e^{i\omega(r-t)}}{4\pi
r}H^{\mu}(k_{2})+\mathcal{O}\left(\frac{1}{r^{2}}\right)$ (167)

where

$r=\sqrt{x^{2}+(x\cdot v)^{2}},\hskip
21.52771ptn^{\mu}=v^{\mu}+\frac{x^{\mu}+(x\cdot v)v^{\mu}}{r},\hskip
21.52771pt\omega=-k_{1}\cdot v,\hskip 21.52771ptk_{2}^{\mu}=\omega n^{\nu}.$
(168)

Therefore, the canonically normalized Compton amplitude is:

$\mathcal{A}_{\text{canonical}}=\mathcal{E}^{*}_{2\mu}H^{\mu}(k_{2}).$ (169)

The covariant (Feynman) normalized Compton amplitude can be obtained by
multiplying by the usual factor of $\sqrt{(2\text{E}_{1})(2\text{E}_{2})}$.
Because the calculation is performed in the classical limit and in a frame in
which the initial body is at rest, this normalization factor simply becomes
$2m$. Thus:

$\mathcal{A}=2m\mathcal{E}^{*}_{2\mu}H^{\mu}(k_{2}).$ (170)

Expanding the electromagnetic MPD equations to linear order in $\epsilon$
produces the solutions:

$\displaystyle\delta\widetilde{u}^{\mu}=$
$\displaystyle\frac{-if^{\mu\nu}_{1}v_{\nu}}{mk_{1}\cdot
v}-\frac{\mathcal{N}_{1}}{m}\left(v^{\mu}+\frac{k_{1}^{\mu}}{k_{1}\cdot
v}\right)$ (171) $\displaystyle\delta\widetilde{S}^{\mu}=$ $\displaystyle
v^{\mu}\delta\widetilde{u}\cdot s-\frac{i}{k_{1}\cdot
v}\epsilon^{\mu\nu\rho\sigma}v_{\nu}s_{\rho}\frac{\partial\mathcal{N}_{1}}{\partial
s^{\sigma}}$ (172) $\displaystyle\delta\widetilde{z}^{\mu}=$
$\displaystyle\frac{-i}{k_{1}\cdot
v}\left(\delta\widetilde{u}^{\mu}+\frac{\delta^{\mu}_{\alpha}+v^{\mu}v_{\alpha}}{m^{2}}{}^{\star}{f}_{1}^{\alpha\beta}s_{\beta}+\frac{\eta^{\mu\nu}+v^{\mu}v^{\nu}}{m}\frac{\partial\mathcal{N}_{1}}{\partial
v^{\nu}}\right.$ $\displaystyle\hskip 21.52771pt\hskip 21.52771pt\hskip
21.52771pt\hskip
21.52771pt\left.+\frac{s^{\mu}}{m}v^{\nu}\frac{\partial\mathcal{N}_{1}}{\partial
s^{\nu}}+\frac{i}{m^{2}}\mathcal{N}_{1}\epsilon^{\mu\nu\rho\sigma}v_{\nu}s_{\rho}k_{1\sigma}\right).$
(173)

Simplifying the Compton amplitude allows it to be expressed in terms of these
solutions as:

$\mathcal{A}=2m\left(f^{*}_{2\mu\nu}\delta\widetilde{z}^{\mu}v^{\nu}+ik_{2}\cdot\delta\widetilde{z}\mathcal{N}_{2}^{*}-\delta\widetilde{u}^{\sigma}\frac{\partial\mathcal{N}_{2}^{*}}{\partial
v^{\sigma}}-\delta\widetilde{S}^{\sigma}\frac{\partial\mathcal{N}_{2}^{*}}{\partial
s^{\sigma}}+\frac{\mathcal{N}_{1}\mathcal{N}_{2}^{*}}{m}-\mathcal{P}_{12}\right).$
(174)

This gives the formal tree-level electromagnetic Compton amplitude for an
arbitrary dynamical mass function.

Using either the dynamical mass function in (133) or in (134) determines the
$\mathcal{N}$ function to be:

$\displaystyle\mathcal{N}(u,S,k,\mathcal{E})=$ $\displaystyle\
\frac{\sin\left(\frac{1}{m}\sqrt{S^{2}(u\cdot k)^{2}-(S\cdot
k)^{2}}\right)}{\sqrt{S^{2}(u\cdot k)^{2}-(S\cdot
k)^{2}}}{}^{\star}{f}_{\mu\nu}S^{\mu}u^{\nu}$ $\displaystyle\ \ \ +ik\cdot
S\frac{1-\cos\left(\frac{1}{m}\sqrt{S^{2}(u\cdot k)^{2}-(S\cdot
k)^{2}}\right)}{S^{2}(u\cdot k)^{2}-(S\cdot k)^{2}}f_{\mu\nu}S^{\mu}u^{\nu}$
(175)

however the multipole analysis is unable to determine the $\mathcal{P}$
function. Thus, this determines the Compton amplitude through equations
(171)-(174) up to contact terms. Because the electromagnetic Compton amplitude
only depends on the linear in $\mathcal{F}$ dynamical mass function through
the $\mathcal{N}$ function and (133) and (134) determine the same
$\mathcal{N}$ function, they are interchangeable for Compton amplitudes.

###  V.2 Compton Amplitude through Spin Cubed

In this subsection we explicitly compute the Compton amplitude for a generic
dynamical mass function through order $\mathcal{O}(S^{3})$. Requiring a match
to the spin-exponentiated result for a $\sqrt{\text{Kerr}}$ particle fixes all
Wilson coefficients on $\mathcal{F}^{1}$ and $\mathcal{F}^{2}$ operators
through $\mathcal{O}(S^{2})$. The implied values of the Wilson coefficients on
$\mathcal{F}^{1}$ operators match those determined by (134). For the helicity-
conserving Compton amplitude, the spin-exponentiation cannot be continued past
$\mathcal{O}(S^{2})$ due to spurious poles. However, the helicity-reversing
Compton amplitude has a perfectly healthy exponentiation at
$\mathcal{O}(S^{3})$. The helicity-conserving amplitude can instead be
required to satisfy the shift symmetry at $\mathcal{O}(S^{3})$ (which is
automatic for lower orders which satisfy spin-exponentiation). We find that at
$\mathcal{O}(S^{3})$ requiring spin-exponentiation for the helicity-reversing
Compton and shift symmetry for the helicity-conserving Compton are consistent
with each other and consistent with (134). However, we find that these three
requirements still leave one remaining free parameter in the
$\mathcal{O}(\mathcal{F}^{2}S^{3})$ piece of the dynamical mass function.

We consider only effects in the dynamical mass function which introduce no
additional length scales beyond $\frac{S}{m}$. Consequently, we only consider
terms for which the number of powers of spin equals the number of derivatives
on the vector potential(s). As well, we only consider terms which are parity
symmetric and not proportional to the field equations (no
$\partial_{\nu}\mathcal{F}^{\mu\nu}$ or $\partial^{2}\mathcal{F}^{\mu\nu}$
terms). (Terms proportional to $\partial_{\nu}\mathcal{F}^{\mu\nu}$ or
$\partial^{2}\mathcal{F}_{\mu\nu}$ do not contribute to the electromagnetic
Compton amplitude because they evaluate to 0 in the $\mathcal{N}$ function.)
The most general such $\delta\mathcal{M}_{1}^{2}$ to order $S^{3}$ is:

$\displaystyle\delta\mathcal{M}_{1}^{2}=$ $\displaystyle\
2C_{1}{}^{\star}\mathcal{F}_{\mu\nu}S^{\mu}u^{\nu}+\frac{C_{2}}{m}(S\cdot\partial)\mathcal{F}_{\mu\nu}S^{\mu}u^{\nu}$
$\displaystyle\ \
-\frac{C_{3}}{3m^{2}}(S\cdot\partial)^{2}{}^{\star}\mathcal{F}_{\mu\nu}S^{\mu}u^{\nu}+\frac{E_{3}}{3m^{2}}S^{2}(u\cdot\partial)^{2}{}^{\star}\mathcal{F}_{\mu\nu}S^{\mu}u^{\nu}+\mathcal{O}(S^{4})$
(176)

for some Wilson coefficients $C_{1}$, $C_{2}$, $C_{3}$, $E_{3}$. The most
general such $\delta\mathcal{M}_{2}^{2}$ to order $S^{3}$ is:

$\displaystyle\delta\mathcal{M}_{2}^{2}=$ $\displaystyle\
\frac{D_{2a}}{m^{2}}(\mathcal{F}_{\mu\nu}S^{\mu}u^{\nu})^{2}+\frac{D_{2b}}{m^{2}}({}^{\star}\mathcal{F}_{\mu\nu}S^{\mu}u^{\nu})^{2}+\frac{D_{2c}}{m^{2}}S^{2}\mathcal{F}^{\mu\nu}\mathcal{F}_{\mu\nu}+\frac{D_{2d}}{m^{2}}S^{2}u^{\mu}\mathcal{F}_{\mu\nu}\mathcal{F}^{\nu\rho}u_{\rho}$
$\displaystyle\ \
+\frac{D_{3a}}{m^{3}}{}^{\star}\mathcal{F}_{\mu\nu}S^{\mu}u^{\nu}(S\cdot\partial)\mathcal{F}_{\rho\sigma}S^{\rho}u^{\sigma}+\frac{D_{3b}}{m^{3}}S^{2}{}^{\star}\mathcal{F}^{\mu\nu}u_{\mu}\partial_{\nu}\mathcal{F}_{\rho\sigma}S^{\rho}u^{\sigma}$
$\displaystyle\ \
+\frac{D_{3c}}{m^{3}}S^{2}{}^{\star}\mathcal{F}^{\mu\nu}u_{\mu}(S\cdot\partial)\mathcal{F}_{\nu\rho}u^{\rho}+\frac{D_{3d}}{m^{3}}S^{2}{}^{\star}\mathcal{F}^{\mu\nu}S_{\mu}(u\cdot\partial)\mathcal{F}_{\nu\rho}u^{\rho}$
$\displaystyle\ \
+\frac{D_{3e}}{m^{3}}{}^{\star}\mathcal{F}^{\mu\nu}S_{\mu}(S\cdot\partial)\mathcal{F}_{\nu\rho}S^{\rho}+\frac{D_{3f}}{m^{3}}S^{2}{}^{\star}\mathcal{F}^{\mu\nu}(S\cdot\partial)\mathcal{F}_{\mu\nu}+\mathcal{O}(S^{4})$
(177)

for some Wilson coefficients $D_{2a}$, $D_{2b}$, $D_{2c}$, $D_{2d}$ for
quadratic-in-spin terms and $D_{3a}$, $D_{3b}$, $D_{3c}$, $D_{3d}$, $D_{3e}$,
$D_{3f}$ for cubic in spin. These lead to the $\mathcal{N}$ and $\mathcal{P}$
functions:

$\displaystyle\mathcal{N}=$ $\displaystyle\
\frac{C_{1}}{m}{}^{\star}f_{\mu\nu}S^{\mu}u^{\nu}+\frac{iC_{2}}{2m^{2}}k\cdot
Sf_{\mu\nu}S^{\mu}u^{\nu}+\frac{C_{3}}{6m^{3}}(k\cdot
S)^{2}{}^{\star}f_{\mu\nu}S^{\mu}u^{\nu}-\frac{E_{3}}{6m^{3}}S^{2}(k\cdot
u)^{2}{}^{\star}f_{\mu\nu}S^{\mu}u^{\nu},$ (178) $\displaystyle\mathcal{P}=$
$\displaystyle\
\frac{D_{2a}}{m^{3}}f^{\prime*}_{\mu\nu}S^{\mu}u^{\nu}f_{\rho\sigma}S^{\rho}u^{\sigma}+\frac{D_{2b}}{m^{3}}{}^{\star}f^{\prime*}_{\mu\nu}S^{\mu}u^{\nu}{}^{\star}f_{\rho\sigma}S^{\rho}u^{\sigma}+\frac{D_{2c}}{m^{3}}S^{2}f^{\prime*}_{\mu\nu}f^{\mu\nu}+\frac{D_{2d}}{m^{3}}S^{2}u^{\mu}f^{\prime*}_{\mu\nu}f^{\nu\rho}u_{\rho}$
$\displaystyle\ \
+\frac{iD_{3a}}{2m^{4}}{}^{\star}f^{\prime*}_{\mu\nu}S^{\mu}u^{\nu}(S\cdot
k)f_{\rho\sigma}S^{\rho}u^{\sigma}-\frac{iD_{3a}}{2m^{4}}{}^{\star}f_{\mu\nu}S^{\mu}u^{\nu}(S\cdot
k^{\prime})f^{\prime*}_{\rho\sigma}S^{\rho}u^{\sigma}$ $\displaystyle\ \
+\frac{iD_{3b}}{2m^{4}}S^{2}{}^{\star}f^{\prime*}_{\mu\nu}u^{\mu}k^{\nu}f_{\rho\sigma}S^{\rho}u^{\sigma}-\frac{iD_{3b}}{2m^{4}}S^{2}{}^{\star}f_{\mu\nu}u^{\mu}k^{\prime\nu}f^{\prime*}_{\rho\sigma}S^{\rho}u^{\sigma}$
$\displaystyle\ \
+\frac{iD_{3c}}{2m^{4}}S^{2}{}^{\star}f^{\prime*\mu\nu}u_{\mu}(S\cdot
k)f_{\nu\rho}u^{\rho}-\frac{iD_{3c}}{2m^{4}}S^{2}{}^{\star}f^{\mu\nu}u_{\mu}(S\cdot
k^{\prime})f^{\prime*}_{\nu\rho}u^{\rho}$ $\displaystyle\ \
+\frac{iD_{3d}}{2m^{4}}S^{2}{}^{\star}f^{\prime*\mu\nu}S_{\mu}(u\cdot
k)f_{\nu\rho}u^{\rho}-\frac{iD_{3d}}{2m^{4}}S^{2}{}^{\star}f^{\mu\nu}S_{\mu}(u\cdot
k^{\prime})f^{\prime*}_{\nu\rho}u^{\rho}$ $\displaystyle\ \
+\frac{iD_{3e}}{2m^{4}}{}^{\star}f^{\prime*\mu\nu}S_{\mu}(S\cdot
k)f_{\nu\rho}S^{\rho}-\frac{iD_{3e}}{2m^{4}}{}^{\star}f^{\mu\nu}S_{\mu}(S\cdot
k^{\prime})f^{\prime*}_{\nu\rho}S^{\rho}$ $\displaystyle\ \
+\frac{iD_{3f}}{2m^{4}}S^{2}{}^{\star}f^{\prime*\mu\nu}(S\cdot
k)f_{\mu\nu}-\frac{iD_{3f}}{2m^{4}}S^{2}{}^{\star}f^{\mu\nu}(S\cdot
k^{\prime})f^{\prime*}_{\mu\nu}.$ (179)

We find it advantageous to express our results for the Compton amplitude in a
basis of definite-helicity/circularly polarized/(anti-)self-dual states for
the incoming and outgoing electromagnetic waves, or “photons,” while
manifesting special covariance and gauge invariance (ultimately). For
concreteness (initially), we can work in a particular Lorentz frame,
associated to inertial Cartesian coordinates $x^{\mu}=(t,x,y,z)$, such that
the charged massive spinning particle’s initial velocity $v$ and the two
photon wavevectors, $k_{1}$ incoming and $k_{2}$ outgoing (both future-
pointing), are given by

$v^{\mu}=(1,0,0,0),\qquad k_{1}^{\mu}=\omega(1,0,0,1),\qquad
k_{2}^{\mu}=\omega(1,\sin\theta,0,\cos\theta).$ (180)

Then $\theta$ is the photon scattering angle in the $z$-$x$-plane, and
$\omega=-v\cdot k_{1}=-v\cdot k_{2}$ is the waves’ angular frequency. The
“momentum transfer” (per $\hbar$) $q=k_{2}-k_{1}$ squares to

$q^{2}=(k_{2}-k_{1})^{2}=-2k_{1}\cdot
k_{2}=2\omega^{2}(1-\cos\theta)=4\omega^{2}\sin^{2}\frac{\theta}{2},$ (181)

vanishing at forward scattering, $\theta=0$. In choosing a particular basis of
definite-helicity (complex null) polarization vectors, $\mathcal{E}_{1\pm}$
incoming, $\mathcal{E}_{2\pm}$ outgoing, with
$k_{n}\cdot\mathcal{E}_{n\pm}=0=\mathcal{E}_{n\pm}^{2}$, it is natural to fix
the gauge freedom $\mathcal{E}_{n}\to\mathcal{E}_{n}+\alpha k_{n}$ by imposing
$v\cdot\mathcal{E}=0$. Up to little group transformations ($\mathcal{E}\to
e^{2i\varphi}\mathcal{E}$), this determines

$\mathcal{E}_{1\sigma_{1}}^{\mu}=\frac{1}{\sqrt{2}}\left(0,1,i\sigma_{1},0\right),\qquad\mathcal{E}_{2\sigma_{2}}^{\mu}=\frac{1}{\sqrt{2}}\left(0,\cos\theta,i\sigma_{2},-\sin\theta\right),$
(182)

for helicities $\sigma_{1}=\pm 1$ and $\sigma_{2}=\pm 1$, with complex
conjugates $\mathcal{E}_{n\pm}^{*\mu}=\mathcal{E}_{n\mp}^{\mu}$, normalized as
$\mathcal{E}_{n\sigma_{n}}\cdot\mathcal{E}_{n\sigma_{n}}^{*}=1$. This frame
(180) and polarization basis (or gauge) (182) are just as in [126] and in
[118]. The spinless Compton amplitude is given simply by the contraction of
the ingoing and conjugate-outgoing polarization vectors (only) in this
$v\cdot\mathcal{E}=0$ gauge:

$\mathcal{A}^{(0)}_{\sigma_{1}\sigma_{2}}=-2\mathcal{E}_{1\sigma_{1}}\cdot\mathcal{E}_{2\sigma_{2}}^{*}=-\sigma_{1}\sigma_{2}-\cos\theta.$
(183)

In analyzing the helicity-preserving amplitudes
$\mathcal{A}_{\pm\pm}\propto\mathcal{E}_{1\pm}^{\mu}\mathcal{E}_{2\pm}^{*\nu}$,
it is useful to define as in [75]111In [75] as in most Amplitudes literature,
our _helicity-preserving_ amplitudes
$\mathcal{A}_{\pm\pm}\propto\mathcal{E}_{1+}^{\mu}\mathcal{E}_{2+}^{\pm\nu}$
are called “opposite-helicity amplitudes $\mathsf{A}_{\pm\mp}$”, while our
_helicity-reversing_ amplitudes
$\mathcal{A}_{\pm\mp}\propto\mathcal{E}_{1\pm}^{\mu}\mathcal{E}_{2\mp}^{*\nu}$
are called “same-helicity amplitudes $\mathsf{A}_{\pm\pm}$”, due to differing
conventions (essentially $k_{n}\leftrightarrow-k_{n}$ entailing
$\sigma_{n}\leftrightarrow-\sigma_{n}$).  a complex null vector $w$ orthogonal
to both $k_{1}$ and $k_{2}$. The conditions $w^{2}=k_{1}\cdot w=k_{2}\cdot
w=0$ determine $w$ up to an overall normalization, which we fix by setting
$v\cdot w=-\omega$, and up to a binary choice of branch
($w\propto|1\rangle[2|$ or $w\propto|2\rangle[1|$) to be correlated with the
photons’ helicities. For our $++$ (or $--$) case, the appropriate $w$ is given
by

$\displaystyle w^{\mu}$
$\displaystyle=\frac{\omega}{4\omega^{2}-q^{2}}\Big{(}2\omega(k_{1}^{\mu}+k_{2}^{\mu})-q^{2}v^{\mu}-2i\epsilon^{\mu}{}_{\nu\rho\sigma}v^{\nu}k_{1}^{\rho}k_{2}^{\sigma}\Big{)}$
(184)

(or the complex conjugate $w^{*\mu}$). In the frame of (180),

$w^{\mu}=\omega\Big{(}1,\tan\frac{\theta}{2},i\tan\frac{\theta}{2},1\Big{)},$
(185)

noting $4\omega^{2}-q^{2}=4\omega^{2}\cos^{2}\frac{\theta}{2}$ along with
(181). The complex null direction $\propto w$ provides an alternative
$(+)$-helicity polarization direction for the incoming $k_{1}$ photon, as well
as an alternative conjugate $(+)$-helicity polarization direction for the
outgoing $k_{2}$ photon. We see that we can recover the normalized orthogonal-
to-$v$ polarization vectors $\mathcal{E}_{1+}$ and $\mathcal{E}_{2+}^{*}$ of
(182) from $w$ via $\mathcal{E}_{n}\to\mathcal{E}_{n}+\alpha k_{n}$ shifts and
rescalings:

$\mathcal{E}_{1+}^{\mu}=\frac{w^{\mu}-k_{1}^{\mu}}{\sqrt{2}\omega\tan\frac{\theta}{2}},\qquad\mathcal{E}_{2+}^{*\mu}=\frac{k_{2}^{\mu}-w^{\mu}}{\sqrt{2}\omega\tan\frac{\theta}{2}},$
(186)

As in [100] (modulo conventions), let us define the vectors

$\check{k}_{1}^{\mu}=k_{1}^{\mu}-w^{\mu},\qquad\check{k}_{2}^{\mu}=k_{2}^{\mu}-w^{\mu},$
(187)

proportional to those in (186), which, along with $w$ as in (184), provide a
relatively compact way to express the $++$ Compton amplitude.

The complex field-strength amplitudes
$f^{\pm}_{\mu\nu}=2ik^{\phantom{\pm}}_{[\mu}\mathcal{E}^{\pm}_{\nu]}$ are
invariant under $\mathcal{E}\to\mathcal{E}+\alpha k$, they transform under the
little groups like the $\mathcal{E}$s, and they are self-dual (or anti-self-
dual),

${}^{\star}f^{\pm}_{\mu\nu}=\frac{1}{2}\epsilon_{\mu\nu}{}^{\kappa\lambda}f^{\pm}_{\kappa\lambda}=\pm
if^{\pm}_{\mu\nu},$ (188)

for states of helicity $+1$ (or $-1$), while the complex conjugates are
reversed: ${}^{\star}f_{\mu\nu}^{{\pm}{*}}=\mp if_{\mu\nu}^{{\pm}{*}}$. For
one way to see this, we can construct $f^{+}_{1\mu\nu}$ from the
$\mathcal{E}^{+}_{1\mu}$ of (182), using (186) with (184),

$\displaystyle f^{+}_{1\mu\nu}$
$\displaystyle=2ik^{\phantom{\pm}}_{1[\mu}\mathcal{E}_{1\nu]}^{+}=\frac{\sqrt{2}i}{\omega\tan\frac{\theta}{2}}k_{1[\mu}w_{\nu]}$
(189)
$\displaystyle=\frac{\sqrt{2}i}{\omega^{2}\sin\theta}k_{1[\mu}\Big{(}v_{\nu]}(k_{1}\cdot
k_{2})+\omega
k_{2\nu]}-i\epsilon_{\nu]\rho\alpha\beta}v^{\rho}k_{1}^{\alpha}k_{2}^{\beta}\Big{)}$
$\displaystyle=\frac{\sqrt{2}i}{\omega^{2}\sin\theta}\Big{(}\delta_{[\mu}{}^{\alpha}\delta_{\nu]}{}^{\beta}-\frac{i}{2}\epsilon_{\mu\nu}{}^{\alpha\beta}\Big{)}k_{1\alpha}\big{(}v_{\beta}(k_{1}\cdot
k_{2})+\omega k_{2\beta}\big{)}$
$\displaystyle=\frac{\sqrt{2}i}{\omega^{2}\sin\theta}2{}^{\,+}\mathcal{G}_{\mu\nu}{}^{\alpha\beta}k_{1\alpha}2v_{[\beta}k_{2\gamma]}k_{1}^{\gamma},$

where we have used
$0=5k_{1[\mu}\epsilon_{\nu\rho\alpha\beta]}v^{\rho}k_{1}^{\alpha}k_{2}^{\beta}=2k_{1[\mu}\epsilon_{\nu]\rho\alpha\beta}v^{\rho}k_{1}^{\alpha}k_{2}^{\beta}-\epsilon_{\mu\nu\alpha\beta}k_{1}^{\alpha}(v^{\beta}(k_{1}\cdot
k_{2})+\omega
k_{2}^{\beta})=2k_{1[\mu}\epsilon_{\nu]vk_{1}k_{2}}-\epsilon_{\mu\nu\alpha\beta}k_{1}^{\alpha}\,2v^{[\beta}k_{2}^{\gamma]}k_{1\gamma}$
and $2\tan\frac{\theta}{2}\cos^{2}\frac{\theta}{2}=\sin\theta$, and where we
recognize the (anti-)self-dual ((A)SD) projector,

${}^{\pm}\mathcal{G}_{\mu\nu}{}^{\kappa\lambda}=\frac{1}{2}\delta_{[\mu}{}^{\kappa}\delta_{\nu]}{}^{\lambda}\mp\frac{i}{4}\epsilon_{\mu\nu}{}^{\kappa\lambda}={}^{\pm}\mathcal{G}_{\mu\nu}{}^{\alpha\beta}{}^{\pm}\mathcal{G}_{\alpha\beta}{}^{\kappa\lambda}=\mp
i\,{}^{\star}{}^{\pm}\mathcal{G}_{\mu\nu}{}^{\kappa\lambda},$ (190)

which maps a 2-form (or any tensor) $A_{\mu\nu}$ onto its (A)SD part
${}^{\pm}\\!A_{\mu\nu}$:

${}^{\pm}\\!A_{\mu\nu}={}^{\pm}\mathcal{G}_{\mu\nu}{}^{\kappa\lambda}A_{\kappa\lambda}=\frac{1}{2}\big{(}A_{[\mu\nu]}\mp
i\,{}^{\star}\\!A_{[\mu\nu]}\big{)},\qquad{}^{\star}{}^{\pm}\\!A_{\mu\nu}=\pm
i{}^{\pm}\\!A_{\mu\nu}.$ (191)

Note the useful identity

${}^{\pm}\mathcal{G}_{\mu\nu}{}^{(\alpha}{}_{\rho}{}^{\pm}\mathcal{G}_{\kappa\lambda}{}^{\beta)\rho}=\frac{1}{4}{}^{\pm}\mathcal{G}_{\mu\nu\kappa\lambda}g^{\alpha\beta},$
(192)

or
${}^{\pm}\mathcal{G}_{\mu\nu(\alpha}{}^{\gamma}{}^{\,\pm}\\!A_{\beta)\gamma}=\frac{1}{4}{}^{\pm}\\!A_{\mu\nu}g_{\alpha\beta}$
and thus
${}^{\pm}\mathcal{G}_{\mu\nu}{}^{\rho\sigma}p_{\sigma}{}^{\pm}\\!A_{\rho\tau}p^{\tau}=\frac{1}{4}p^{2}{}^{\,\pm}\\!A_{\mu\nu}$,
following from $\epsilon_{\alpha\beta\gamma\delta}\epsilon^{\mu\nu\rho\sigma}$
$=$
$-24\delta_{\alpha}{}^{[\mu}\delta_{\beta}{}^{\nu}\delta_{\gamma}{}^{\rho}\delta_{\delta}{}^{\sigma]}$
and
$\delta_{\omega}{}^{[\lambda}\delta_{\alpha}{}^{\mu}\delta_{\beta}{}^{\nu}\delta_{\gamma}{}^{\rho}\delta_{\delta}{}^{\sigma]}=0$.
Collecting (189) along with its conjugate outgoing versions, also following
from (184) and (186), we have

$\displaystyle f^{+}_{1\mu\nu}$
$\displaystyle=+2ik^{\phantom{+}}_{1[\mu}\mathcal{E}_{1\nu]}^{+\phantom{*}}=\frac{4\sqrt{2}i}{\omega^{2}\sin\theta}{}^{\,+}\mathcal{G}_{\mu\nu}{}^{\alpha\beta}k_{1\alpha}v_{[\beta}k_{2\gamma]}k_{1}^{\gamma}=\frac{\sqrt{2}i}{\omega\tan\frac{\theta}{2}}k_{1[\mu}w_{\nu]\,},$
$\displaystyle f^{+*}_{2\mu\nu}$
$\displaystyle=-2ik^{\phantom{+}}_{2[\mu}\mathcal{E}_{2\nu]}^{+*}=\frac{4\sqrt{2}i}{\omega^{2}\sin\theta}{}^{\,-}\mathcal{G}_{\mu\nu}{}^{\alpha\beta}k_{2\alpha}v_{[\beta}k_{1\gamma]}k_{2}^{\gamma}=\frac{\sqrt{2}i}{\omega\tan\frac{\theta}{2}}k_{2[\mu}w_{\nu]\,},$
$\displaystyle f^{-*}_{2\mu\nu}$
$\displaystyle=-2ik^{\phantom{+}}_{2[\mu}\mathcal{E}_{2\nu]}^{-*}=\frac{4\sqrt{2}i}{\omega^{2}\sin\theta}{}^{\,+}\mathcal{G}_{\mu\nu}{}^{\alpha\beta}k_{2\alpha}v_{[\beta}k_{1\gamma]}k_{2}^{\gamma}=\frac{\sqrt{2}i}{\omega\tan\frac{\theta}{2}}k^{\phantom{*}}_{2[\mu}w^{*}_{\nu]\,},$
(193)

noting
$w^{*\mu}=-w^{\mu}+\omega\dfrac{2\omega(k_{1}+k_{2})^{\mu}-q^{2}v^{\mu}}{4\omega^{2}-q^{2}}$
and

$\omega^{2}\sin\theta=\sqrt{4\omega^{4}\sin^{2}\frac{\theta}{2}\cos^{2}\frac{\theta}{2}=\frac{q^{2}}{4}(4\omega^{2}-q^{2})=-\epsilon^{\mu}{}_{vk_{1}k_{2}}\epsilon_{\mu}{}^{vk_{1}k_{2}}\propto[12]\langle
12\rangle[2|v|1\rangle[1|v|2\rangle}.$ (194)

In simplifying the helicity-basis Compton amplitudes
$\mathcal{A}_{+\pm}\propto f_{1\mu\nu}^{+}f_{2\alpha\beta}^{\pm*}$, the (A)SD
properties ${}^{\star}f_{1\mu\nu}^{+}=+if_{1\mu\nu}$ and
${}^{\star}f_{2\mu\nu}^{\pm*}=\mp if_{2\mu\nu}^{\pm*}$ can be used immediately
within (before differentiation of) the $\mathcal{N}(f)$ and
$\mathcal{P}(f,f^{\prime})$ functions in (178). These functions completely
determine the amplitudes via (171)–(174). They can finally be evaluated
directly in terms of the complex null $w^{\mu}(k_{1},k_{2},v)$ by using the
extreme equalities of (193), noting e.g.
$\epsilon_{vak_{1}w}=i\omega\check{k}_{1}\cdot a$,
$\epsilon_{vak_{2}w}=-i\omega\check{k}_{2}\cdot a$,
$\epsilon_{vk_{1}k_{2}w}=i\omega q^{2}/2$ following from (184), recalling
$\check{k}_{1}=k_{1}-w$ and $\check{k}_{2}=k_{2}-w$; they otherwise depend
only on $k_{1}^{\mu}$, $k_{2}^{\mu}$, $v^{\mu}$ and the initial spin
$s^{\mu}=ma^{\mu}$.

For the ++ amplitudes $\mathcal{A}^{(n)}_{++}\propto f_{1}^{+}f_{2}^{+*}a^{n}$
at $n$th order in spin, we find

$\displaystyle\mathcal{A}_{++}^{(0)}$
$\displaystyle=-\frac{4\omega^{2}-q^{2}}{2\omega^{2}},$ (195a)
$\displaystyle\mathcal{A}_{++}^{(1)}$
$\displaystyle=-\frac{4\omega^{2}-q^{2}}{2\omega^{2}}\bigg{\\{}C_{1}(\check{k}_{1}+\check{k}_{2})\cdot
a-(C_{1}-1)^{2}w\cdot a\bigg{\\}},$ (195b)
$\displaystyle\mathcal{A}_{++}^{(2)}$
$\displaystyle=-\frac{4\omega^{2}-q^{2}}{2\omega^{2}}\Bigg{\\{}\frac{C_{2}}{2}[(\check{k}_{1}+\check{k}_{2})\cdot
a]^{2}-\Big{(}(C_{1}-1)C_{2}+D_{2d}\Big{)}(w\cdot a)^{2}$ (195c)
$\displaystyle\qquad\qquad\qquad\quad+(C_{1}-1)(C_{1}-C_{2})\,w\cdot
a\,(\check{k}_{1}+\check{k}_{2})\cdot a$
$\displaystyle\quad+\bigg{[}\Big{(}(C_{1}-1)(2C_{1}-C_{2})+2D_{2d}-D_{2a}-D_{2b}\Big{)}\frac{2\omega^{2}}{q^{2}}+C_{1}^{2}-C_{2}\bigg{]}\check{k}_{1}\cdot
a\,\check{k}_{2}\cdot a\Bigg{\\}},$ $\displaystyle\mathcal{A}_{++}^{(3)}$
$\displaystyle=-\frac{4\omega^{2}-q^{2}}{2\omega^{2}}\Bigg{\\{}\frac{C_{3}}{6}[(\check{k}_{1}+\check{k}_{2})\cdot
a]^{3}+\hat{\mathcal{A}}_{++}^{(3)(C_{2}-1,C_{1}-1)}$
$\displaystyle\qquad+\bigg{(}\,^{\\!}\frac{4}{3}+2F_{\text{i}}-\frac{F_{\text{ii}}}{4}+\frac{F_{\text{iii}}}{2}\bigg{)}(w\cdot
a)^{3}+\bigg{(}F_{\text{i}}-\frac{C_{3}-1}{6}\bigg{)}(w\cdot
a)^{2}(\check{k}_{1}+\check{k}_{2})\cdot a$
$\displaystyle\qquad+\bigg{[}F_{\text{ii}}\frac{\omega^{2}}{q^{2}}w\cdot
a+\bigg{(}F_{\text{iii}}\frac{\omega^{2}}{q^{2}}-\frac{C_{3}-1}{2}\bigg{)}(\check{k}_{1}+\check{k}_{2})\cdot
a\bigg{]}\check{k}_{1}\cdot a\,\check{k}_{2}\cdot a\Bigg{\\}},$ (195d) where
$\displaystyle F_{\text{i}}$
$\displaystyle=\frac{E_{3}-1}{6}-\frac{D_{3b}+D_{3c}+D_{3e}}{2},\qquad\quad$
$\displaystyle F_{\text{ii}}$ $\displaystyle=3-2D_{3a}+4D_{3c}+4D_{3d},$
$\displaystyle F_{\text{iii}}$
$\displaystyle=-1-\frac{2}{3}E_{3}-D_{3a}+2D_{3b}+2D_{3c},$ (195e)

and $\hat{\mathcal{A}}_{++}^{(3)(C_{2}-1,C_{1}-1)}$ vanishes when
$C_{2}=C_{1}=1$ (which for $\sqrt{\text{Kerr}}$ are fixed by lower-order-in-
spin pieces of the Compton amplitude, or by the three-point amplitude). Here
we have used

$\omega^{2}a^{2}=(w\cdot a)^{2}-\frac{4\omega^{2}}{q^{2}}\check{k}_{1}\cdot
a\,\check{k}_{2}\cdot a,$ (196)

resulting from
$0=(u^{[\lambda}k_{1}^{\mu}k_{2}^{\nu}w^{\rho}a^{\sigma]})^{2}$, to eliminate
$a^{2}$ in favor of $(w\cdot a)^{2}$. This leads to the relatively compact
expressions (195) —– paralleling the parametrization (385) of the black
hole–graviton $++$ Compton amplitude at fifth order in spin as formulated in
[118] —– but it makes the amplitude appear to have (in addition to the
physical poles at $v\cdot k_{1}=v\cdot k_{2}$ $=$ $-\omega=0$) unphysical
poles: firstly, explicitly, at $q^{2}=4\omega^{2}\sin^{2}\frac{\theta}{2}=0$ —
at forward scattering, $\theta=0$ — the would-be physical pole corresponding
to an internal photon of momentum $q=k_{2}-k_{1}$ going on shell, but on which
the residue must be zero because the three-photon amplitude vanishes; and
secondly, hidden inside $w$ (and $\check{k}_{1}$ and $\check{k}_{2}$) in
(184), at $4\omega^{2}-q^{2}=4\omega^{2}\cos^{2}\frac{\theta}{2}=0$ — at back-
scattering, $\theta=\pi$ — “the spurious pole.” However, no unphysical poles
are actually present, for arbitrary values of the Wilson coefficients, as can
be made manifest by using (196) to eliminate factors of $(w\cdot a)^{2}$ in
favor of $a^{2}$.

The helicity-preserving amplitude $\mathcal{A}_{++}$ is well expressed in
terms of the spin component $w\cdot a$ along the complex null
$w^{\mu}(k_{1},k_{2},u)$ of (184) because of its symmetry $w\leftrightarrow
w^{*}$ under $k_{1}\leftrightarrow k_{2}$ [with $-2\omega=(k_{1}+k_{2})\cdot
v$].

Turning to the helicity-reversing amplitude $\mathcal{A}_{+-}$, the
appropriate symmetry is reflected by a vector $x^{\mu}(k_{1},k_{2},v)$ with
$x\leftrightarrow-x$ under $k_{1}\leftrightarrow k_{2}$ (modulo any component
along $v$). An apt choice is $x(k_{1},k_{2},v)\cdot a\propto
w(k_{2},-k_{1},v)\cdot a$:

$\displaystyle w\cdot a\,$ $\displaystyle=\Big{(}\omega(k_{1}+k_{2})\cdot
a+i\epsilon_{\mu\nu\rho\sigma}v^{\mu}k_{1}^{\nu}k_{2}^{\rho}a^{\sigma}\Big{)}\frac{2\omega}{4\omega^{2}-q^{2}}\,,$
$\displaystyle x\cdot a\,$ $\displaystyle=\Big{(}\omega(k_{2}-k_{1})\cdot
a+i\epsilon_{\mu\nu\rho\sigma}v^{\mu}k_{1}^{\nu}k_{2}^{\rho}a^{\sigma}\Big{)}\,\frac{2\omega}{q^{2}}=w\cdot
a+\frac{4\omega^{2}}{q^{2}}(k_{1}-w)\cdot a$
$\displaystyle=\frac{2\omega}{q^{2}}\Big{(}\omega\,q\cdot a+ik_{1}\times
k_{2}\cdot a\Big{)}\,,$ (197)

coinciding with “$w_{O}\cdot a$” from [126] or $\propto$ “$w_{++}\cdot a$”
from [100], with $x\cdot a\propto\langle 1|av|1\rangle$, in contrast to
$w\cdot a\propto[2|a|1\rangle$. The identity (196) for $(w\cdot a)^{2}$
translates into

$\displaystyle\frac{q^{2}}{4\omega^{2}}\Big{(}(x\cdot
a)^{2}-\omega^{2}a^{2}\Big{)}$
$\displaystyle=\dfrac{-q^{2}}{4\omega^{2}-q^{2}}(k_{1}\cdot a-x\cdot
a)(k_{2}\cdot a+x\cdot a)$ (198) $\displaystyle=(k_{1}\cdot a)(k_{2}\cdot
a)-(x\cdot a)(q\cdot a)-\omega^{2}a^{2}=(aya)$

for $(x\cdot a)^{2}$ and defines a convenient quadratic $(aya)$ in the spin.
With this, we find

$\displaystyle\mathcal{A}_{+-}$
$\displaystyle=\frac{q^{2}}{2\omega^{2}}\Bigg{\\{}1-C_{1}\,q\cdot
a+(C_{1}^{2}-1)x\cdot a$ (199)
$\displaystyle\qquad\;\;\,+\frac{C_{2}}{2}(q\cdot
a)^{2}-(C_{1}-1)(C_{1}+C_{2})k_{1}\cdot a\,k_{2}\cdot a$
$\displaystyle\qquad\qquad+\bigg{[}\Big{(}D_{2b}-D_{2a}-(C_{1}-1)(2C_{1}+C_{2})\Big{)}\frac{2\omega^{2}}{q^{2}}+(C_{2}-1)C_{1}\bigg{]}(aya)$
$\displaystyle\qquad\qquad-\Big{(}4D_{2c}+D_{2d}+(C_{1}-1)C_{1}\Big{)}\omega^{2}a^{2}$
$\displaystyle\qquad\;\;\,-\frac{C_{3}}{6}(q\cdot a)^{3}+(C_{3}-1)x\cdot
a\,k_{1}\cdot a\,k_{2}\cdot
a+\hat{\mathcal{A}}_{+-}^{(3)(C_{2}-1,C_{1}-1)}\phantom{\bigg{|}}$
$\displaystyle\qquad\qquad-\bigg{(}(1+D_{3a})\frac{\omega^{2}}{q^{2}}+\frac{C_{3}-1}{2}\bigg{)}(aya)q\cdot
a-\bigg{(}\frac{3}{4}+E_{3}+D_{3b}+D_{3d}\bigg{)}\omega^{2}a^{2}x\cdot a$
$\displaystyle\qquad\qquad-\bigg{(}\frac{1+E_{3}}{6}+\frac{D_{3b}+D_{3c}-D_{3e}}{2}+2D_{3f}+\frac{C_{3}-1}{6}\bigg{)}\omega^{2}a^{2}q\cdot
a+\mathcal{O}(a^{4})\Bigg{\\}},$

where again $\hat{\mathcal{A}}_{+-}^{(3)(C_{2}-1,C_{1}-1)}$ vanishes when
$C_{2}=C_{1}=1$.

At linear order in spin, $C_{1}$ is determined by the $\mathcal{O}(S^{1})$
amplitude. At $\mathcal{O}(S^{2})$, all 5 new operators, $C_{2},D_{2a,b,c,d}$
contribute linearly independent structures to the amplitude. At
$\mathcal{O}(S^{3})$ there are 8 new operators $C_{3},D_{3a,b,c,d,e,f},E_{3}$
but only 7 linearly independent structures in the Compton. In particular, the
Compton amplitude is independent of the value of the linear combination:

$Z=-6E_{3}+D_{3f}+2C_{1}D_{3b}-2(1+C_{1})D_{3c}+4D_{3d}$ (200)

These amplitudes produce the spin-exponentiation of Ref. [77] through order
$\mathcal{O}(S^{2})$ if and only if:

$C_{1}=C_{2}=1,\hskip 21.52771ptD_{2a}=D_{2b}=D_{2c}=D_{2d}=0.$ (201)

For these values of the Wilson coefficients, we recover the spin-exponentiated
amplitudes:

$\displaystyle\mathcal{A}_{++}$
$\displaystyle=\mathcal{A}^{(0)}_{++}\exp\left(a\cdot(\check{k}_{1}+\check{k}_{2})\right)+\mathcal{O}(S^{3})$
(202) $\displaystyle\mathcal{A}_{+-}$
$\displaystyle=\mathcal{A}^{(0)}_{+-}\exp\left(a\cdot(k_{1}-k_{2})\right)+\mathcal{O}(S^{3})$
(203)

For the equal helicity amplitude the spin-exponentiation cannot continue past
$\mathcal{O}(S^{2})$ due to the spurious pole in $\cos(\theta/2)$. There is no
such trouble for the opposite helicity amplitude. If one demands the
continuance of the spin-exponentiation for the opposite helicity amplitude
through $\mathcal{O}(S^{3})$, it fixes $C_{3}$ and 3 of the $D_{3a,b,c,d,e,f}$
coefficients. In particular it determines:

$\displaystyle C_{3}=1,\hskip 21.52771ptD_{3a}=-1,\hskip
21.52771ptD_{3d}=-D_{3b}-E_{3}-\frac{3}{4},$ $\displaystyle
D_{3f}=-\frac{1}{12}-\frac{E_{3}}{12}-\frac{D_{3b}}{4}-\frac{D_{3c}}{4}+\frac{D_{3e}}{4}.$
(204)

The value $C_{3}=1$ is consistent with (134). Equation (134) fixes the value
of $E_{3}=1$ which is possible while continuing the spin-exponentiation but
not demanded by it.

It is also interesting to study the shift symmetry condition identified in
Refs. [93, 95, 100, 101]. In order for the same helicity Compton amplitude to
maintain shift symmetry through $\mathcal{O}(S^{3})$ according to the criteria
of Ref. [93], some of the $C_{3}$, $E_{3}$, $D_{3a,b,c,d,e,f}$ are fixed. In
particular:

$\displaystyle C_{3}=1,\hskip
21.52771ptD_{3c}=\frac{E_{3}}{3}+\frac{1+D_{3a}}{2}-D_{3b},\hskip
21.52771ptD_{3d}=D_{3b}-\frac{E_{3}}{3}-\frac{5}{4}.$ (205)

Thus, at this order the shift symmetry is also consistent with (134).

The conditions necessary to maintain opposite helicity spin-exponentiation,
shift symmetry, and match (134) are consistent with each other at
$\mathcal{O}(S^{3})$ and lead to the combined set of conditions:

$\displaystyle C_{3}=1,\hskip 21.52771ptE_{3}=1,\hskip
21.52771ptD_{3a}=-1,\hskip 21.52771ptD_{3b}=-\frac{1}{12}$ $\displaystyle
D_{3c}=\frac{5}{12},\hskip 21.52771ptD_{3d}=-\frac{5}{3},\hskip
21.52771ptD_{3f}=\frac{D_{3e}-1}{4}.$ (206)

Thus, at $\mathcal{O}(S^{3})$ there is a one parameter family of dynamical
mass functions (as $D_{3e}$ is fully undetermined) satisfying all of these
constraints.

Following the decomposition of Ref. [93], the same helicity amplitude in terms
of $D_{3e}$ through $\mathcal{O}(S^{3})$ may be written as:

$\mathcal{A}_{++}=e^{a\cdot(k_{1}+k_{2})}\sum_{n=0}^{3}\frac{\bar{I}_{n}}{n!}+\mathcal{O}(S^{4})$
(207)

with:

$\displaystyle\bar{I}_{0}=-2\cos^{2}\frac{\theta}{2},$
$\displaystyle\bar{I}_{1}=-2a\cdot w\bar{I}_{0},$
$\displaystyle\bar{I}_{2}=\left(2a\cdot w\right)^{2}\bar{I}_{0},$
$\displaystyle\bar{I}_{3}=-(3D_{3e}+1)(a\cdot
w)^{2}a\cdot(k_{1}+k_{2})\bar{I}_{0}.$ (208)

Thus we can identify $c_{0}^{(3)}$ in equation (3.9b) of Ref. [93] with
$-2-6D_{3e}$.

We can also compare to the recent work of [127] on $\sqrt{\text{Kerr}}$
amplitudes from higher-spin gauge interactions. To that end, following the
lead of [127], instead of using (196) to eliminate $a^{2}\omega^{2}$ leaving
$(w\cdot a)^{2}$ as we did in (195), we can express the ++ amplitude in terms
of both $(w\cdot a)^{2}$ and $\omega^{2}a^{2}$ while eliminating
$q^{2}/\omega^{2}$ using (196). Defining $k_{\pm}=k_{2}\pm k_{1}$,

$\displaystyle k_{+}$
$\displaystyle\,=\,k_{1}+k_{2\,},\qquad\;\,\check{k}_{1}+\check{k}_{2}\,=\,k_{+\\!}-2w_{\,},$
(209) $\displaystyle q\,=\,k_{-}$
$\displaystyle\,=\,k_{2}-k_{1}\;=\;\check{k}_{2}-\check{k}_{1\,},\quad\;\;\;\;\;\;[(k_{+\\!}-2w)\cdot
a]^{2}-(q\cdot a)^{2}=4_{\,}\check{k}_{1}\cdot a\,\check{k}_{2}\cdot a_{\,},$

and replacing $\omega^{2}/q^{2}$ with $[(w\cdot
a)^{2}-\omega^{2}a^{2}]/(4_{\,}\check{k}_{1}\cdot a\,\check{k}_{2}\cdot a)$,
our amplitude (195) becomes

$\displaystyle\frac{\mathcal{A}_{++}}{\mathcal{A}^{(0)}_{++}}$
$\displaystyle=1+C_{1}(k_{+\\!}-2w)\cdot a-(C_{1}-1)^{2}w\cdot a$
$\displaystyle\qquad+\frac{1}{2}[(k_{+\\!}-2w)\cdot
a]^{2}\frac{C_{1}^{2}+C_{2}}{2}+\frac{C_{2}-C_{1}^{2}}{4}(q\cdot a)^{2}$
$\displaystyle\qquad\qquad\qquad\qquad\;\;\,+(C_{1}-1)\bigg{[}(C_{1}-C_{2})k_{+\\!}\cdot
a\,w\cdot a+\frac{C_{2}-2C_{1}}{2}\Big{(}\omega^{2}a^{2}+(w\cdot
a)^{2}\Big{)}\bigg{]}$
$\displaystyle\qquad\qquad-\frac{D_{2a}+D_{2b}}{2}(w\cdot
a)^{2}+\frac{D_{2a}+D_{2b}-2D_{2d}}{2}\omega^{2}a^{2}\phantom{\bigg{|}}$
$\displaystyle\qquad+\frac{1}{6}(k_{+\\!}-2w)\cdot a\Big{[}(k_{+\\!}-4w)\cdot
a\,k_{+\\!}\cdot a\frac{3+C_{3}}{4}+3\frac{C_{3}-1}{4}(q\cdot
a)^{2}\Big{]}+\hat{\mathcal{A}}_{++}^{(3)(C_{2}-1,C_{1}-1)}$
$\displaystyle\qquad\qquad+\frac{1-D_{3a}-2D_{3e}}{4}(w\cdot
a)^{2}k_{+\\!}\cdot
a+\frac{-15+12(D_{3b}-D_{3d})-4E_{3}}{12}\omega^{2}a^{2}w\cdot
a\phantom{\Bigg{|}}$
$\displaystyle\qquad\qquad+\frac{3+3D_{3a}-6(D_{3b}+D_{3c})+2E_{3}}{12}\omega^{2}a^{2}k_{+\\!}\cdot
a+\mathcal{O}(a^{4}).$ (210)

Similarly, using (V.2),

$\displaystyle\frac{\mathcal{A}_{+-}}{\mathcal{A}_{+-}^{(0)}}$
$\displaystyle=1-C_{1\,}q\cdot a+(C_{1}^{2}-1)x\cdot a$
$\displaystyle\qquad+\frac{1}{2}(q\cdot
a)^{2}\frac{C_{1}^{2}+C_{2}}{2}+\frac{C_{2}-C_{1}^{2}}{4}(k_{+\\!}\cdot
a)^{2}-(C_{2}-1)C_{1\,}q\cdot a\,x\cdot a$
$\displaystyle\qquad\qquad+\frac{1}{2}\Big{(}(1-C_{1})(C_{2}+2C_{1})-D_{2a}+D_{2b}\Big{)}(x\cdot
a)^{2}$
$\displaystyle\qquad\qquad+\frac{1}{2}\Big{(}2C_{1}-C_{2}-C_{1}C_{2}+D_{2a}-D_{2b}-8D_{2c}-2D_{2d}\Big{)}\omega^{2}a^{2}$
$\displaystyle\qquad-\frac{1}{6}(q\cdot
a)^{3}\frac{3+C_{3}}{4}+\frac{C_{3}-1}{4}\Big{[}(q\cdot a)^{2}x\cdot
a-\frac{1}{2}(k_{+\\!}\cdot a)^{2}(q-2x)\cdot
a+\frac{4}{3}\omega^{2}a^{2}q\cdot a\Big{]}\phantom{\bigg{|}}$
$\displaystyle\qquad\qquad-\frac{1+D_{3a}}{4}(x\cdot a)^{2}q\cdot
a-\frac{3+4(D_{3b}+D_{3d}+E_{3})}{4}\omega^{2}a^{2}x\cdot
a+\hat{\mathcal{A}}_{+-}^{(3)(C_{2}-1,C_{1}-1)}$
$\displaystyle\qquad\qquad+\frac{1+3D_{3a}-6(D_{3b}+D_{3c}-D_{3e})-24D_{3f}-2E_{3}}{12}\omega^{2}a^{2}q\cdot
a+\mathcal{O}(a^{4}).\phantom{\bigg{|}}$ (211)

With $C_{1}=C_{2}=C_{3}=1$,

$\displaystyle\frac{\mathcal{A}_{++}}{\mathcal{A}^{(0)}_{++}}=1+(k_{+\\!}-2w)\cdot
a+\frac{1}{2}[(k_{+\\!}-2w)\cdot
a]^{2}+\frac{D_{2a}+D_{2b}}{2}\big{[}\omega^{2}a^{2}-(w\cdot
a)^{2}\big{]}-D_{2d\,}\omega^{2}a^{2}$
$\displaystyle\qquad\qquad\;\,+\frac{1}{6}(k_{+\\!}-4w)\cdot
a\,(k_{+\\!}-2w)\cdot a\,k_{+\\!}\cdot a+\frac{1-D_{3a}-2D_{3e}}{4}(w\cdot
a)^{2}k_{+\\!}\cdot a\phantom{\bigg{|}}$
$\displaystyle\quad+\frac{-15+12(D_{3b}-D_{3d})-4E_{3}}{12}\omega^{2}a^{2}w\cdot
a+\frac{3+3D_{3a}-6(D_{3b}+D_{3c})+2E_{3}}{12}\omega^{2}a^{2}k_{+\\!}\cdot a$
$\displaystyle+\mathcal{O}(a^{4}),$ (212)

and

$\displaystyle\frac{\mathcal{A}_{+-}}{\mathcal{A}_{+-}^{(0)}}$
$\displaystyle=1-q\cdot a+\frac{1}{2}(q\cdot
a)^{2}+\frac{D_{2a}-D_{2b}}{2}\big{[}\omega^{2}a^{2}-(x\cdot
a)^{2}\big{]}+(4D_{2c}+D_{2d})\omega^{2}a^{2}$
$\displaystyle\quad-\frac{1}{6}(q\cdot a)^{3}-\frac{1+D_{3a}}{4}(x\cdot
a)^{2}q\cdot a-\frac{3+4(D_{3b}+D_{3d}+E_{3})}{4}\omega^{2}a^{2}x\cdot a$
$\displaystyle\qquad\quad+\frac{1+3D_{3a}-6(D_{3b}+D_{3c}-D_{3e})-24D_{3f}-2E_{3}}{12}\omega^{2}a^{2}q\cdot
a+\mathcal{O}(a^{4}).\phantom{\Bigg{|}}$ (213)

This matches (6.76) of [127],

$\displaystyle\frac{\mathcal{A}_{++}}{\mathcal{A}^{(0)}_{++}}$
$\displaystyle=1+(k_{+\\!}-2w)\cdot a+\frac{1}{2}[(k_{+\\!}-2w)\cdot
a]^{2}+2\delta\big{[}\omega^{2}a^{2}-(w\cdot a)^{2}\big{]}$ (214)
$\displaystyle\quad+\frac{1}{6}(k_{+\\!}-2w)\cdot a\Big{[}(k_{+\\!}-4w)\cdot
a\,k_{+\\!}\cdot
a+4\omega^{2}a^{2}\Big{]}+\frac{4}{3}\delta\big{[}\omega^{2}a^{2}-(w\cdot
a)^{2}\big{]}k_{+\\!}\cdot a+\mathcal{O}(a^{4}),$

and $\mathcal{A}_{+-}/\mathcal{A}_{+-}^{(0)}=e^{-q\cdot a}+\mathcal{O}(a^{4})$
if

$\displaystyle D_{2a}$ $\displaystyle=D_{2b}=2\delta,\qquad$ $\displaystyle
D_{2c}$ $\displaystyle=D_{2d}=0,$ (215) $\displaystyle D_{3a}$
$\displaystyle=-1,\qquad$ $\displaystyle D_{3b}$
$\displaystyle=-\frac{5+4E_{3}}{12},\qquad$ $\displaystyle D_{3c}$
$\displaystyle=\frac{-11+8E_{3}-32\delta}{12},$ $\displaystyle D_{3d}$
$\displaystyle=-\frac{1+2E_{3}}{3},\qquad$ $\displaystyle D_{3e}$
$\displaystyle=1+\frac{8}{3}\delta,\qquad$ $\displaystyle D_{3f}$
$\displaystyle=\frac{3-E_{3}+8\delta}{6}.$

##  VI Gravitational MPD Equations

We now turn our attention to gravity. We find that the line of analysis is
directly analogous to that of electromagnetism and the resulting dynamical
mass function is very similar. The motion of a generic spinning body in
general relativity is described by the MPD equations [6, 7, 9, 10, 11, 12]. It
is well established [12, 121, 122, 76, 128, 123, 129, 48] that the MPD
equations can be derived from a variational principle through an action
$\mathcal{S}$ of the form:

$\mathcal{S}[z,p,\Lambda,S,\alpha,\beta]=\int_{-\infty}^{\infty}\left(p_{\mu}\dot{z}^{\mu}+\frac{1}{2}\widetilde{\epsilon}_{\mu\nu\rho\sigma}u^{\mu}S^{\nu}\Omega^{\rho\sigma}-\frac{\alpha}{2}(p^{2}+\mathcal{M}^{2})+\beta
p\cdot S\right)d\lambda$ (216)

where the pseudotensor Levi-Civita symbol is defined by:

$\widetilde{\epsilon}_{\mu\nu\rho\sigma}=\epsilon_{\mu\nu\rho\sigma}\sqrt{-\det
g}$ (217)

in terms of the purely numerical antisymmetric Levi-Civita symbol
$\epsilon_{\mu\nu\rho\sigma}$ which has $\epsilon_{0123}=1$. In curved
spacetime, the ${\Lambda^{\mu}}_{A}(\lambda)$ tetrad satisfies:

$g^{\mu\nu}(Z)=\Lambda^{\mu}{}_{A}\Lambda^{\nu}{}_{B}\eta^{AB},\hskip
21.52771pt\eta_{AB}=g_{\mu\nu}(Z)\Lambda^{\mu}{}_{A}\Lambda^{\nu}{}_{B}.$
(218)

Just as in the case of electromagnetism, we take:

${\Lambda^{\mu}}_{0}(\lambda)=u^{\mu}(\lambda),\hskip
21.52771pt{\Lambda^{\mu}}_{3}(\lambda)=\hat{S}^{\mu}(\lambda).$ (219)

The angular velocity tensor of the body is defined by:

$\Omega^{\mu\nu}=\eta^{AB}{\Lambda^{\mu}}_{A}\frac{D{\Lambda^{\nu}}_{B}}{D\lambda}$
(220)

where $\frac{D}{D\lambda}$ indicates covariant $\lambda$ differentiation:

$\frac{D{\Lambda^{\mu}}_{A}}{D\lambda}=\frac{d{\Lambda^{\mu}}_{A}}{d\lambda}+\Gamma^{\mu}{}_{\rho\sigma}\dot{z}^{\rho}\Lambda^{\sigma}{}_{A}.$
(221)

The dynamical mass function $\mathcal{M}(z,u,S)$ now encodes the mass of the
body and all of its nonminimal couplings to gravity and in particular takes
the form:

$\mathcal{M}^{2}(z,u,S)=m^{2}+\mathcal{O}(R)$ (222)

where $m$ is the mass of the body in vacuum and $R_{\mu\nu\rho\sigma}$ is the
Riemann tensor.

For variations of the action, it is useful to define the covariant variations:

$\displaystyle\Delta p_{\mu}$ $\displaystyle=\delta
p_{\mu}-{\Gamma^{\rho}}_{\sigma\mu}p_{\rho}\delta z^{\sigma}$ (223)
$\displaystyle\Delta S^{\mu}$ $\displaystyle=\delta
S^{\mu}+{\Gamma^{\mu}}_{\rho\sigma}S^{\rho}\delta z^{\sigma}$ (224)
$\displaystyle\Delta{\Lambda^{\mu}}_{A}$
$\displaystyle=\delta{\Lambda^{\mu}}_{A}+\Gamma^{\mu}{}_{\rho\sigma}\delta
z^{\rho}\Lambda^{\sigma}{}_{A}$ (225)

and the antisymmetric tensor:

$\Delta\theta^{\mu\nu}=\eta^{AB}{\Lambda^{[\mu}}_{A}\Delta\Lambda^{\nu]}{}_{B}.$
(226)

This definition leads to the identity:

$\Delta\Omega^{\mu\nu}=\frac{D}{D\lambda}\Delta\theta^{\mu\nu}+{\Omega^{\mu}}_{\rho}\Delta\theta^{\rho\nu}-\Omega^{\nu}{}_{\rho}\Delta\theta^{\rho\mu}+R^{\mu\nu}{}_{\rho\sigma}\dot{z}^{\rho}\delta
z^{\sigma}.$ (227)

Then, the variation of the above action gives:

$\displaystyle\delta\mathcal{S}=\int_{-\infty}^{\infty}$
$\displaystyle\left(\delta
z^{\mu}\left(-\frac{Dp_{\mu}}{D\lambda}-R^{\star}_{\mu\nu\rho\sigma}\dot{z}^{\nu}u^{\rho}S^{\sigma}-\mathtt{e}\nabla_{\mu}\mathcal{M}\right)\right.$
$\displaystyle+\Delta
p_{\mu}\left(\dot{z}^{\mu}-\mathtt{e}u^{\mu}-\frac{\mathtt{e}}{|p|}\frac{\partial\mathcal{M}}{\partial
u^{\nu}}(g^{\mu\nu}+u^{\mu}u^{\nu})+\beta S^{\mu}\right.$ $\displaystyle\hskip
21.52771pt\hskip 21.52771pt\hskip
21.52771pt\left.+\frac{1}{2|p|}\widetilde{\epsilon}^{\mu\nu\rho\sigma}S_{\nu}\Omega_{\rho\sigma}+\frac{u^{\mu}}{2|p|}\widetilde{\epsilon}_{\alpha\beta\rho\sigma}u^{\alpha}S^{\beta}\Omega^{\rho\sigma}\right)$
$\displaystyle+\frac{1}{2}\Delta\theta^{\rho\sigma}\left(-\frac{D}{D\lambda}\left(\widetilde{\epsilon}_{\mu\nu\rho\sigma}u^{\mu}S^{\nu}\right)+\widetilde{\epsilon}_{\mu\nu\rho\alpha}u^{\mu}S^{\nu}\Omega^{\alpha}{}_{\sigma}-\widetilde{\epsilon}_{\mu\nu\sigma\alpha}u^{\mu}S^{\nu}\Omega^{\alpha}{}_{\rho}\right)$
$\displaystyle\left.+\Delta
S^{\mu}\left(-\frac{1}{2}\widetilde{\epsilon}_{\mu\nu\rho\sigma}u^{\nu}\Omega^{\rho\sigma}-\mathtt{e}\frac{\partial\mathcal{M}}{\partial
S^{\mu}}+\beta
p_{\mu}\right)-\frac{\delta\alpha}{2}\left(p^{2}+\mathcal{M}^{2}\right)+\delta\beta
p\cdot S\right)d\lambda$ (228)

where the right-dual of the Riemann tensor is defined by:

$R^{\star}_{\mu\nu\rho\sigma}=\frac{1}{2}\widetilde{\epsilon}_{\rho\sigma}{}^{\alpha\beta}R_{\mu\nu\alpha\beta}.$
(229)

Using the $\delta S^{\mu}$ variation to solve for the angular velocity tensor,
one can then determine the value of $\beta$. That value of $\beta$ can then be
used to simplify the spin and trajectory equations of motion. Explicitly,
these give:

$\displaystyle\Omega^{\mu\nu}$
$\displaystyle=\frac{Du^{\mu}}{D\lambda}u^{\nu}-u^{\mu}\frac{Du^{\nu}}{D\lambda}+\mathtt{e}\widetilde{\epsilon}^{\mu\nu\rho\sigma}u_{\rho}\frac{\partial\mathcal{M}}{\partial
S^{\sigma}}$ (230) $\displaystyle\beta$
$\displaystyle=-\frac{\mathtt{e}}{\mathcal{M}}u^{\mu}\frac{\partial\mathcal{M}}{\partial
S^{\mu}}$ (231) $\displaystyle\frac{DS^{\mu}}{D\lambda}$
$\displaystyle=u^{\mu}\frac{Du^{\nu}}{D\lambda}S_{\nu}+\mathtt{e}\epsilon^{\mu\nu\rho\sigma}u_{\nu}S_{\rho}\frac{\partial\mathcal{M}}{\partial
S^{\sigma}}$ (232) $\displaystyle\dot{z}^{\mu}$
$\displaystyle=\mathtt{e}u^{\mu}+\frac{\mathtt{e}}{\mathcal{M}}(g^{\mu\nu}+u^{\mu}u^{\nu})\frac{\partial\mathcal{M}}{\partial
u^{\nu}}+\frac{\mathtt{e}}{\mathcal{M}}S^{\mu}u^{\nu}\frac{\partial\mathcal{M}}{\partial
S^{\nu}}+\frac{1}{\mathcal{M}^{2}}\widetilde{\epsilon}^{\mu\nu\rho\sigma}S_{\nu}u_{\rho}\frac{Dp_{\sigma}}{D\lambda}.$
(233)

In order to determine the trajectory evolution explicitly we must insert the
momentum equation of motion into (233). To simplify, it is useful to introduce
the two sided dual Riemann tensor:

${}^{\star}R^{\star}_{\mu\nu\rho\sigma}=\frac{1}{2}\widetilde{\epsilon}_{\mu\nu\alpha\beta}R^{\star\alpha\beta}{}_{\rho\sigma}$
(234)

Simplifying finally gives the gravitational MPD equations of motion for the
spinning body:

$\displaystyle\left(1+\frac{{}^{\star}R^{\star}_{uSuS}}{\mathcal{M}^{2}}\right)\frac{\dot{z}^{\mu}}{\mathtt{e}}=$
$\displaystyle\
u^{\mu}+\frac{g^{\mu\nu}+u^{\mu}u^{\nu}}{\mathcal{M}}\frac{\partial\mathcal{M}}{\partial
u^{\nu}}+S^{\mu}\frac{u^{\nu}}{\mathcal{M}}\frac{\partial\mathcal{M}}{\partial
S^{\nu}}+\frac{1}{\mathcal{M}^{2}}\widetilde{\epsilon}^{\mu\nu\rho\sigma}u_{\nu}S_{\rho}\nabla_{\sigma}\mathcal{M}$
$\displaystyle-\frac{{}^{\star}R^{\star\mu
SuS}}{\mathcal{M}^{2}}-\frac{1}{\mathcal{M}^{3}}\left(S^{\nu}\frac{\partial\mathcal{M}}{\partial
u^{\nu}}+S^{2}u^{\nu}\frac{\partial\mathcal{M}}{\partial
S^{\nu}}\right){}^{\star}R^{\star\mu uuS}$ (235)
$\displaystyle\frac{Dp_{\mu}}{D\lambda}=$ $\displaystyle\
-R^{\star}_{\mu\nu\rho\sigma}\dot{z}^{\nu}u^{\rho}S^{\sigma}-\mathtt{e}\nabla_{\mu}\mathcal{M}$
(236) $\displaystyle\frac{DS^{\mu}}{D\lambda}=$ $\displaystyle\
u^{\mu}S_{\nu}\frac{Du^{\nu}}{D\lambda}+\mathtt{e}\widetilde{\epsilon}^{\mu\nu\rho\sigma}u_{\nu}S_{\rho}\frac{\partial\mathcal{M}}{\partial
S^{\sigma}}.$ (237)

Vectors such as $u$ and $S$ are used as indices to indicate contractions with
them (${}^{\star}R^{\star\mu
uuS}={}^{\star}R^{\star\mu\nu\rho\sigma}u_{\nu}u_{\rho}S_{\sigma}$). For
solving these equations of motion we will always choose $\lambda$ so that
$\mathtt{e}=1$.

To understand how the dynamical mass function relates to the multipole moments
of the body, we will need to study the energy-momentum tensor produced by our
action. The energy-momentum tensor will be given by:

$T_{\mu\nu}=-\frac{2}{\sqrt{-\det g}}\frac{\delta\mathcal{S}}{\delta
g^{\mu\nu}}.$ (238)

In varying the metric, we have $\delta z^{\mu}=0$, $\delta p_{\mu}=0$, $\delta
S^{\mu}=0$, and $\Delta\theta^{\mu\nu}=0$. ${\delta\Lambda}^{\mu}{}_{A}$
cannot be made 0 as its variation is related to the metric variation on the
worldline. It is useful to introduce the DeWitt index shuffling operator [123]
${\hat{G}}^{\alpha}{}_{\beta}$, which acts on tensors according to:

$\displaystyle\hat{G}^{\alpha}{}_{\beta}F^{\mu_{1}...\mu_{m}}{}_{\nu_{1}...\nu_{n}}=$
$\displaystyle\delta^{\mu_{1}}_{\beta}F^{\alpha\mu_{2}...\mu_{m}}{}_{\nu_{1}...\nu_{n}}+...+\delta^{\mu_{m}}_{\beta}F^{\mu_{1}...\mu_{m-1}\alpha}{}_{\nu_{1}...\nu_{n}}$
$\displaystyle\ \
-\delta^{\alpha}_{\nu_{1}}F^{\mu_{1}...\mu_{m}}{}_{\beta\nu_{2}...\nu_{n}}-...-\delta^{\alpha}_{\nu_{n}}F^{\mu_{1}...\mu_{m}}{}_{\nu_{1}...\nu_{n-1}\beta}$
(239)

so that:

$\nabla_{\rho}F^{\mu_{1}...\mu_{m}}{}_{\nu_{1}...\nu_{n}}=\partial_{\rho}F^{\mu_{1}...\mu_{m}}{}_{\nu_{1}...\nu_{n}}+\Gamma^{\beta}{}_{\alpha\rho}\hat{G}^{\alpha}{}_{\beta}F^{\mu_{1}...\mu_{m}}{}_{\nu_{1}...\nu_{n}}.$
(240)

For expressing the energy-momentum tensor simply it is important to use the
scalarity of the dynamical mass function. In particular, requiring it to be a
scalar function of $u_{\mu}$, $S^{\mu}$, $g_{\mu\nu}$, and symmetric covariant
derivatives of $R_{\mu\nu\rho\sigma}$ implies it is invariant under a small
diffeomorphism. This is only true if:

$\frac{\partial\mathcal{M}}{\partial
g^{\alpha\beta}}=\frac{1}{2}\frac{\partial\mathcal{M}}{\partial
u^{\alpha}}u_{\beta}-\frac{1}{2}S_{\alpha}\frac{\partial\mathcal{M}}{\partial
S^{\beta}}-\frac{1}{2}\sum_{n=0}^{\infty}\frac{\partial\mathcal{M}}{\partial\nabla^{n}_{(\lambda_{1}...\lambda_{n})}R_{\mu\nu\rho\sigma}}\hat{G}_{\alpha\beta}\nabla^{n}_{(\lambda_{1}...\lambda_{n})}R_{\mu\nu\rho\sigma}.$
(241)

With these ingredients, the variation of the action with respect to the metric
becomes:

$\displaystyle\delta\mathcal{S}$
$\displaystyle=\int_{-\infty}^{\infty}\left(-\frac{1}{2}p_{\mu}\dot{z}_{\nu}\delta
g^{\mu\nu}+\frac{1}{2}\widetilde{\epsilon}_{\mu\nu\rho\sigma}u^{\mu}S^{\nu}\dot{z}_{\alpha}\nabla^{\sigma}\delta
g^{\rho\alpha}\right.$ $\displaystyle\hskip
21.52771pt\left.+\mathtt{e}\sum_{n=0}^{\infty}\frac{\partial\mathcal{M}}{\partial\nabla^{n}_{(\lambda_{1}...\lambda_{n})}R_{\mu\nu\rho\sigma}}\left(\frac{1}{2}\delta
g^{\alpha\beta}\hat{G}_{\alpha\beta}\nabla^{n}_{(\lambda_{1}...\lambda_{n})}R_{\mu\nu\rho\sigma}-\delta(\nabla^{n}_{(\lambda_{1}...\lambda_{n})}R_{\mu\nu\rho\sigma})\right)\right)d\lambda$
(242)

For computing variations of derivatives of the Riemann curvature, a strategy
from Ref. [130] is helpful. Consider a tensor field
$F^{\lambda_{1}...\lambda_{n}\mu\nu\rho\sigma}$ with the same index symmetries
as $\nabla^{n}_{(\lambda_{1}...\lambda_{n})}R_{\mu\nu\rho\sigma}$ which decays
to 0 sufficiently quickly at infinity for no surface terms to be necessary
upon the relevant integrations by parts we will perform. Then, a short
calculation gives:

$\displaystyle\int_{\mathbb{T}}F^{\lambda_{1}...\lambda_{n}\mu\nu\rho\sigma}\delta(\nabla^{n}_{\lambda_{1}...\lambda_{n}}R_{\mu\nu\rho\sigma})D^{4}x$
$\displaystyle\hskip
21.52771pt=\int_{\mathbb{T}}\left(F^{\lambda_{1}...\lambda_{n}\mu\nu\rho\sigma}\delta\Gamma^{\beta}{}_{\alpha\lambda_{n}}\hat{G}^{\alpha}{}_{\beta}\nabla^{n-1}_{\lambda_{1}...\lambda_{n-1}}R_{\mu\nu\rho\sigma}-\nabla_{\lambda_{n}}F^{\lambda_{1}...\lambda_{n}\mu\nu\rho\sigma}\delta(\nabla^{n-1}_{\lambda_{1}...\lambda_{n-1}}R_{\mu\nu\rho\sigma})\right)D^{4}x.$
(243)

The final term is now in the same form as the initial variational problem, but
of a lower rank. Applying this formula iteratively allows all derivatives on
the Riemann tensor to eventually be pushed past the variation, resulting in:

$\displaystyle\int_{\mathbb{T}}F^{\lambda_{1}...\lambda_{n}\mu\nu\rho\sigma}\delta(\nabla^{n}_{\lambda_{1}...\lambda_{n}}R_{\mu\nu\rho\sigma})D^{4}x$
$\displaystyle\hskip
21.52771pt=\int_{\mathbb{T}}\left(\sum_{k=0}^{n-1}(-1)^{k}\nabla^{k}_{\lambda_{n}...\lambda_{n-k+1}}F^{\lambda_{1}...\lambda_{n}\mu\nu\rho\sigma}\delta\Gamma^{\beta}{}_{\alpha\lambda_{n-k}}\hat{G}^{\alpha}{}_{\beta}\nabla^{n-k-1}_{\lambda_{1}...\lambda_{n-k-1}}R_{\mu\nu\rho\sigma}\right.$
$\displaystyle\hskip 21.52771pt\hskip 21.52771pt\hskip
21.52771pt\left.+(-1)^{n}\nabla^{n}_{\lambda_{1}...\lambda_{n}}F^{\lambda_{1}...\lambda_{n}\mu\nu\rho\sigma}\delta
R_{\mu\nu\rho\sigma}\right)D^{4}x.$ (244)

Now using:

$\displaystyle\delta\Gamma^{\rho}{}_{\mu\nu}$
$\displaystyle=\frac{1}{2}g_{\mu\alpha}g_{\nu\beta}\nabla^{\rho}\delta
g^{\alpha\beta}-\frac{1}{2}g_{\mu\sigma}\nabla_{\nu}\delta
g^{\rho\sigma}-\frac{1}{2}g_{\nu\sigma}\nabla_{\mu}\delta g^{\rho\sigma}$
(245) $\displaystyle\delta R^{\rho}{}_{\mu\sigma\nu}$
$\displaystyle=\nabla_{\sigma}\delta\Gamma^{\rho}{}_{\nu\mu}-\nabla_{\nu}\delta\Gamma^{\rho}{}_{\sigma\mu}$
(246)

we are able to arrive at:

$\displaystyle\int
F^{\lambda_{1}...\lambda_{n}\mu\nu\rho\sigma}\delta(\nabla^{n}_{(\lambda_{1}...\lambda_{n})}R_{\mu\nu\rho\sigma})D^{4}x$
$\displaystyle\hskip 8.61108pt=\int\left(\sum_{k=0}^{n-1}(-1)^{k}\delta
g^{\alpha\beta}g_{\alpha\lambda_{n-k}}\nabla^{\gamma}\nabla^{k}_{(\lambda_{n-k+1}...\lambda_{n})}F^{\lambda_{1}...\lambda_{n}\mu\nu\rho\sigma}\hat{G}_{[\gamma\beta]}(\nabla^{n-k-1}_{(\lambda_{1}...\lambda_{n-k-1})}R_{\mu\nu\rho\sigma})\right.$
$\displaystyle\hskip 8.61108pt\hskip 21.52771pt+\sum_{k=0}^{n-1}(-1)^{k}\delta
g^{\alpha\beta}g_{\alpha\lambda_{n-k}}\nabla^{k}_{(\lambda_{n-k+1}...\lambda_{n})}F^{\lambda_{1}...\lambda_{n}\mu\nu\rho\sigma}g^{\gamma\tau}\hat{G}_{[\tau\beta]}(\nabla_{\gamma}\nabla^{n-k-1}_{(\lambda_{1}...\lambda_{n-k-1})}R_{\mu\nu\rho\sigma})$
$\displaystyle\hskip 8.61108pt\hskip
21.52771pt+\sum_{k=0}^{n-1}2(-1)^{k}\delta
g^{\alpha\beta}g_{\alpha\lambda_{n-k}}\nabla^{k}_{(\lambda_{n-k+1}...\lambda_{n})}F^{\lambda_{1}...\lambda_{n}\mu\nu\rho\sigma}\nabla_{\beta}\nabla^{n-k-1}_{(\lambda_{1}...\lambda_{n-k-1})}R_{\mu\nu\rho\sigma}$
$\displaystyle\hskip 8.61108pt\hskip 21.52771pt+(-1)^{n}\delta
g^{\alpha\beta}\nabla^{n}_{(\lambda_{1}...\lambda_{n})}F^{\lambda_{1}...\lambda_{n}\mu\nu\rho\sigma}g_{\alpha\mu}R_{\beta\nu\rho\sigma}+2(-1)^{n}\nabla^{2}_{(\mu\nu)}\nabla^{n}_{(\lambda_{1}...\lambda_{n})}F^{\lambda_{1}...\lambda_{n}\mu}{}_{\alpha}{}^{\nu}{}_{\beta}\delta
g^{\alpha\beta}$ $\displaystyle\hskip 8.61108pt\hskip
21.52771pt\left.+\frac{1}{2}\delta
g^{\alpha\beta}F^{\lambda_{1}...\lambda_{n}\mu\nu\rho\sigma}\hat{G}_{\alpha\beta}(\nabla^{n}_{\lambda_{1}...\lambda_{n}}R_{\mu\nu\rho\sigma})\right)D^{4}x.$
(247)

Define the gravitational $\mathcal{Q}_{n}$ moments:

$\mathcal{Q}_{n}^{\lambda_{1}...\lambda_{n}\mu\nu\rho\sigma}=\frac{\partial\mathcal{M}}{\partial\nabla^{n}_{(\lambda_{1}...\lambda_{n})}R_{\mu\nu\rho\sigma}}.$
(248)

As well, define the scalar Dirac delta distribution:

$\delta(X,Z)=\frac{\delta(x-z)}{\sqrt{-\det g}}$ (249)

and the $\Phi_{n}$ fields:

$\Phi_{n}^{\lambda_{1}...\lambda_{n}\mu\nu\rho\sigma}(X)=\int_{-\infty}^{\infty}\mathcal{Q}_{n}^{\lambda_{1}...\lambda_{n}\mu\nu\rho\sigma}\delta(X,Z)D\lambda.$
(250)

Then, formally as a distributional expression the energy-momentum tensor
becomes:

$T_{\alpha\beta}=\int_{-\infty}^{\infty}\left(\dot{z}_{(\alpha}p_{\beta)}\delta(X,Z)+\nabla^{\gamma}(\dot{z}_{(\alpha}\widetilde{\epsilon}_{\beta)\gamma\rho\sigma}u^{\rho}S^{\sigma}\delta(X,Z))\right)d\lambda+\sum_{n=0}^{\infty}\Psi^{(n)}_{\alpha\beta}$
(251)

with:

$\displaystyle\Psi^{(n)}_{\alpha\beta}=$ $\displaystyle\
\sum_{k=0}^{n-1}(-1)^{k}\nabla^{k}_{\lambda_{n-k+1}...\lambda_{n}}\Phi_{n}^{\lambda_{1}...\lambda_{n}\mu\nu\rho\sigma}g_{\lambda_{n-k}(\alpha}\nabla_{\beta)}\nabla^{n-k-1}_{\lambda_{1}...\lambda_{n-k-1}}R_{\mu\nu\rho\sigma}$
$\displaystyle+\sum_{k=0}^{n-1}2(-1)^{k}\nabla_{\gamma}\left(\nabla^{k}_{\lambda_{n-k+1}...\lambda_{n}}\Phi_{n}^{\lambda_{1}...\lambda_{n}\mu\nu\rho\sigma}g_{\lambda_{n-k}(\alpha}g_{\beta)\delta}\hat{G}^{[\gamma\delta]}(\nabla^{n-k-1}_{\lambda_{1}...\lambda_{n-k-1}}R_{\mu\nu\rho\sigma})\right)$
$\displaystyle+2(-1)^{n}\nabla^{n}_{\lambda_{1}...\lambda_{n}}\Phi_{n}^{\lambda_{1}...\lambda_{n}\mu\nu\rho\sigma}g_{\mu(\alpha}R_{\beta)\nu\rho\sigma}+4(-1)^{n}\nabla^{2}_{\mu\nu}\nabla^{n}_{\lambda_{1}...\lambda_{n}}\Phi^{\lambda_{1}...\lambda_{n}}_{n}{}^{\mu}{}_{(\alpha}{}^{\nu}{}_{\beta)}.$
(252)

##  VII Kerr Multipole Moments

In this section we perform the analysis of sections III and IV but for the