# The tensorial representation of the distributional stress–energy quadrupole
and its dynamics

Jonathan Gratus1,2,3,∗, Spyridon Talaganis1,4

###### Abstract

We investigate stress-energy tensors constructed from the covariant
derivatives of delta functions on a worldline. Since covariant derivatives are
used all the components transform as tensors. We derive the dynamical
equations for the components, up to quadrupole order. The components do,
however, depend in a non-tensorial way, on a choice of a vector along the
worldline. We also derive a number of important results about general
multipoles, including that their components are unique, and all multipoles can
be written using covariant derivatives. We show how the components of a
multipole are related to standard moments of a tensor field, by parallelly
transporting that tensor field.

1 Physics department, Lancaster University, Lancaster LA1 4YB, UK,
2 The Cockcroft Institute Daresbury Laboratory, Daresbury, Warrington WA4 4AD,
UK.
3<EMAIL_ADDRESS>orcid:0000-0003-1597-6084
4<EMAIL_ADDRESS>orcid:0000-0003-0113-7546
∗ Corresponding author.

##  1 Introduction

Finite size objects can be approximated using moments. This makes sense when
the object is viewed from a distance. The usual definition of moments involves
integrating over a spatial hypersurface of an integrand which involves
coordinates. For example one may define the quadrupole of a rank (2,0) tensor
$S^{\mu\nu}$ as $\int_{\text{space}}z^{a}z^{b}S^{\mu\nu}d^{3}z$, where
$a,b=1,2,3$ and $\mu,\nu=0,\ldots,3$. Such an object will not, in general, be
tensorial. Instead it will be highly dependent on the choice of coordinates,
and there will be no simple way of transforming the expression from one
coordinate system to another. In addition it is necessary to choose a
foliation of spacelike hypersurfaces over which one can perform the
integration. By contrast one can represent an extended object by a
distribution (in the Schwartz sense) over a worldline, which may represent the
“centre” of the object. This distribution is tensorial in that it acts on
appropriate test tensors to give a number. Using this action one can find the
transformation rules for the moments. In this article such distributions will
be called multipoles. The components of the multipoles are intimately related
to the standard moments, as we show below.

There are many extended objects one may wish to approximate. For example a
scalar field concentrated at one point in space can be represented by a scalar
multipole, while the current of an extended charge may be represented by a
vector valued multipole. In this article we are primary interested in
modelling, as a multipole, the stress-energy tensor of an extended massive
object. Unlike the current which can be used directly as a source for
electromagnetism, the distributional stress-energy tensor cannot be used
directly in Einstein’s equations. This is because, unlike Maxwell’s equations,
Einstein’s equations are not linear. Instead one can use them as a source for
linearised gravity.

Even if one is not concerned with the effect on gravity, considering the
stress-energy multipole is useful. This is because the constraint that it must
be divergenceless tells us information about the dynamics of the moments. For
the monopole, there is just a single component, which is constant and
corresponds to the total mass. For the dipole there are 10 components whose
dynamics are completely determined by the Mathisson-Papapetrou-Tulczyjew-Dixon
ODEs [1, 2]. By contrast for the quadrupole, as well as the 40 components
which have ODEs, there are 20 free components [3]. One may consider the
dynamics of these free components to be constitutive relations. They could
either be posited as part of the model or derived by considering the
underlying matter which makes up the extended body. For example, we would
expect different results for dust than for a neutron star.

Multipole distributions can be represented in a number of ways. In [3] we
discuss the advantages and disadvantages of using two different expressions to
represent a multipole over a worldline. These are labelled the Ellis [4] and
Dixon [5, 6, 7, 8, 9] representations. The Ellis representation uses partial
derivatives and can be applied to multipoles over any line on any manifold.
However, it requires complicated coordinate transformation rules for the
components, involving derivatives of the Jacobi matrix and integration over
the world line [3, 10]. The Dixon representation uses covariant derivatives
and has the significant advantage that the components are tensors. Another
advantage is that the multipole naturally splits into different orders of
“poles”. That is, it is a sum of a monopole, a dipole, a quadrupole and so on.
We refer to this in this as the Dixon split.

The price one pays for this is that the manifold needs to have a connection,
and one must choose a vector along the worldline, which we call the Dixon
vector. In general changing the Dixon vector will result in a complicated
transformation, which not only mixes orders but involves higher derivatives of
the components.

With regards to multipoles over timelike worldlines in general relativity,
these constraints are not a problem. Spacetime is endowed with a connection,
usually the Levi-Civita connection, and there is a preferred vector over the
worldline given by it’s tangent. Other possible choices of the Dixon vector
are discussed in the conclusion.

The key result of this article, given in section 4, is the derivation of the
dynamics of the components of the Dixon stress-energy quadrupole. This is the
generalisation of the Mathisson-Papapetrou-Tulczyjew-Dixon equations. These
are derived from the divergenceless of the stress-energy tensors and shows how
the quadrupole couples to the curvature and derivatives of the curvature.
Similar equations have been derived by Steinhoff and Puetzfeld [11]. However
their method leads to an implicit equation for the dynamics. By contrast our
equations are clearer with the time derivatives of the relevant components
given explicitly. The method, which involves commuting covariant derivatives
can be extended to arbitrary order multipole. However, as noted in [3], the
equations do not completely determine the dynamics of the quadrupole and must
be augmented with 20 constitutive relations.

In his work, Dixon makes two conjectures for the dynamics of the components of
a quadrupole, which we summarise in section 5. We can compare these to our
dynamical equations and see that neither of them couple to the curvature. Thus
they do not correspond to the divergenceless condition and are not the
generalisation of the Mathisson-Papapetrou–Tulczyjew–Dixon equations for the
quadrupole.

$\Sigma(\sigma)$$C(\sigma)$$\,\,N_{{{\mu}}}$$H_{\sigma,V}(s)$$V^{\mu}$ Figure
1: The worldline $C$ (black), with the Dixon geodesic hypersurface
$\Sigma(\sigma)$ (grey) which intersects the worldline at $C(\sigma)$.
Emanating from $C(\sigma)$ is the geodesic $H_{\sigma,V}(s)$ (red) which lies
inside $\Sigma(\sigma)$. The tangent to $H_{\sigma,V}$ at $C(\sigma)$ is the
vector $V^{\mu}$ (blue), which is orthogonal to the Dixon vector $N_{{{\mu}}}$
(green). All points in the Dixon geodesic hypersurface $\Sigma(\sigma)$ can be
reached by a geodesic like $H_{\sigma,V}(s)$.

In addition to deriving the dynamical equations for stress-energy quadrupole,
in this article we also establish important results about the Dixon
representation of multipoles in general. In particular we show that all
multipoles can be represented as Dixon multipoles and that both the Dixon
split and the components are unique. We show how the components can be
extracted from a multipole by letting it act on appropriate test tensors. All
of these results are valid for an arbitrary tensor multipole of arbitrary
order. Thus as well as the stress-energy multipole, they can also be applied
to electromagnetic current multipoles. The uniqueness of the splitting and the
components is an essential step in the derivation of the dynamical equations,
as it is applied to the divergence of the stress-energy tensor.

This article is arranged as follows. In section 2 we give a summary of the
stress-energy tensor. In section 3 we give the properties of arbitrary
multipoles as described above, namely: the Dixon split (section 3.4), the
formula for the components (section 3.5) and the demonstration that all
multipoles can be written as Dixon multipoles (section 3.7). For this we
define a natural coordinate system, which is adapted to the worldline and the
Dixon vector (section 3.2). We also define the Dixon geodesic hypersurface
which is the hypersurface generated from all geodesics which emanate from the
worldline in a direction orthogonal to the Dixon vector (figure 1). We make
explicit the two point tensor associated with parallel transport (section 3.3)
and give the Taylor expansion of a test tensor which respects parallel
transport and the Dixon vector (section 3.6).

As stated the definition of the spatial moments of a extended object is in
terms of the integral over space with respect to some coordinate system.
Fortunately, once we have chosen a Dixon vector, one can use Dixon geodesic
hypersurfaces, and the adapted coordinate system. Alternatively one can
parallel transport the tensor to the worldline. Thus the moments defined this
way are tensorial objects with tensorial transformation properties. Although
they are still non-tensorially dependent on the choice of the Dixon vector.
There is a natural way of linking these moments with the components of the
multipole distribution. This is done by “squeezing” a regular tensor. This is
the process of reducing a tensor’s spatial extent while keeping the quantity
of matter constant. This is demonstrated in section 3.8.

In section 4 we derive the dynamical equations for the components of the
stress-energy quadrupole. In section 5 we compare these the equations proposed
by Dixon. Finally in section 6 we conclude.

##  2 The stress-energy distribution

We shall use the notation as defined in [3]. Let ${\cal M}$ be a spacetime
with metric $g_{{{\mu}}{{\nu}}}$, signature $(-,+,+,+)$, and the Levi-Civita
connection $\nabla_{{{\mu}}}$ with Christoffel symbol
$\Gamma^{{{\mu}}}_{{{\nu}}{{\rho}}}$. Here Greek indices
${{\mu}},{{\nu}}=0,1,2,3$ and Latin indices $a,b=1,2,3$. Let $C:{\cal
I}\to{\cal M}$ where ${\cal I}\subset{\mathbb{R}}$ is the worldline of the
source111Even using proper time in Minkowski space, one cannot assume that
${\cal I}={\mathbb{R}}$ since it is possible to accelerate to lightlike
infinity in finite proper time. with components $C^{{{\mu}}}(\sigma)$. At this
point we do not assume that $\sigma$ is proper time. Here we consider stress-
energy tensors $T^{{{\mu}}{{\nu}}}$ which are non-zero only on the worldline
$C^{{{\mu}}}(\sigma)$, where it has Dirac–$\delta$ like properties. Such
stress-energy tensors are called distributional.

Since we are dealing with distributions it is most convenient to consider
$T^{{{\mu}}{{\nu}}}$ as a tensor density222An integral over ${\cal M}$ must
contain the measure ${\omega}$. There is therefore the following choice: one
can choose $T^{{{\mu}}{{\nu}}}$ or ${{\cal\phi}}_{{{\mu}}{{\nu}}}$ to be a
density of weight 1, or put ${\omega}$ explicitly in the integrand. Here we
have chosen to make $T^{{{\mu}}{{\nu}}}$ a density. of weight 1. Thus
${\omega}^{-1}T^{{{\mu}}{{\nu}}}$ is a tensor, where

$\displaystyle{\omega}=\sqrt{-\det(g_{{{\mu}}{{\nu}}})}\,.$ (1)

The definition of the covariant derivative of a tensor
${Y}^{{{\mu}}{{\nu}}\cdots}$ density of weight 1 is given by

$\displaystyle\nabla_{{\mu}}{Y}^{{{\nu}}{{\rho}}\cdots}$
$\displaystyle={\omega}\nabla_{{\mu}}({\omega}^{-1}{Y}^{{{\nu}}{{\rho}}\cdots})=-\Gamma^{{\kappa}}_{{{\mu}}{{\kappa}}}\,{Y}^{{{\nu}}{{\rho}}\cdots}+\partial_{{\mu}}{Y}^{{{\nu}}{{\rho}}\cdots}+\Gamma^{{\nu}}_{{{\mu}}{{\kappa}}}{Y}^{{{\kappa}}{{\rho}}\cdots}+\Gamma^{{\rho}}_{{{\mu}}{{\kappa}}}{Y}^{{{\nu}}{{\kappa}}\cdots}+\cdots\,,$
(2)

where $\Gamma^{{{\nu}}}_{{{\mu}}{{\rho}}}$ are the Christoffel symbols. In
this article the term stress-energy tensor always refers to a stress-energy
tensor density of weight 1, even if not explicitly stated. In addition the
symbol $T^{{{\mu}}{{\nu}}}$ always refers to a distributional stress-energy
tensor density of weight 1 over the worldline $C$.

As already stated the Dixon representation depends crucially on a choice of a
vector field $N_{{{\mu}}}(\sigma)$ along the worldline $C$, called the Dixon
vector. The only constraint on the choice of $N_{{{\mu}}}(\sigma)$ is that it
is not orthogonal to the worldline $C$, $N_{{{\mu}}}\,{\dot{C}}^{{{\mu}}}\neq
0$. In section 4, we need to project out the spatial components. Thus we scale
$N_{{{\mu}}}$ so that

$\displaystyle N_{{{\mu}}}\,{\dot{C}}^{{{\mu}}}=1\,.$ (3)

As long as the worldline $C$ is timelike, a natural choice of the Dixon vector
is ${\dot{C}}^{{{\mu}}}$, i.e.
$N_{{{\mu}}}=-g_{{{\mu}}{{\nu}}}\,{\dot{C}}^{{{\nu}}}$ but this is not the
only choice. Having chosen $N_{{{\mu}}}$, the Dixon representation of a
multipole is given [6, Equation (1.9)][8, Equation (4.18), (7.4), (7.5)] by

$\displaystyle
T^{{{\mu}}{{\nu}}}=\sum_{r=0}^{k}{\frac{1}{r!}}\nabla_{{{\rho}}_{1}}\cdots\nabla_{{{\rho}}_{r}}\int_{\cal
I}\xi^{{{\mu}}{{\nu}}{{\rho}}_{1}\ldots{{\rho}}_{r}}(\sigma)\,\delta^{(4)}\big{(}x-C(\sigma)\big{)}\,d\sigma\,.$
(4)

Tulczyjew [2] calls this the canonical form, in the case when
$N_{{{\mu}}}=-{\dot{C}}_{{{\mu}}}$.

Since $T^{{{\mu}}{{\nu}}}$ is a stress-energy tensor, we have the symmetry of
the indices

$\displaystyle T^{{{\mu}}{{\nu}}}=T^{{{\nu}}{{\mu}}}\,,$ (5)

which leads to

$\displaystyle\xi^{{{\mu}}{{\nu}}{{\rho}}_{1}\ldots{{\rho}}_{r}}=\xi^{{{\nu}}{{\mu}}{{\rho}}_{1}\ldots{{\rho}}_{r}}\,.$
(6)

We demand that the components
$\xi^{{{\mu}}{{\nu}}{{\rho}}_{1}\ldots{{\rho}}_{k}}$ are orthogonal to the
vector $N_{{{\mu}}}$,

$\displaystyle N_{{{\rho}}_{j}}\
\xi^{{{\mu}}{{\nu}}{{\rho}}_{1}\ldots{{\rho}}_{k}}=0$ (7)

for $j=1,\ldots,k$. The covariant derivatives do not commute. Instead they
give rise to curvature terms and lower the number of derivatives. We therefore
make the minimal choice and impose
$\xi^{{{\mu}}{{\nu}}{{\rho}}_{1}\ldots{{\rho}}_{k}}$ are symmetric in the
relevant indices.

$\displaystyle\xi^{{{\mu}}{{\nu}}{{\rho}}_{1}\ldots{{\rho}}_{k}}=\xi^{{{\mu}}{{\nu}}({{\rho}}_{1}\ldots{{\rho}}_{k})}\,.$
(8)

Since $T^{{{\mu}}{{\nu}}}$ is a tensor density this enables us to throw the
covariant derivative over onto the test tensor
${{\cal\phi}}_{{{\mu}}{{\nu}}}$, giving

$\displaystyle\int_{\cal M}T^{{{\mu}}{{\nu}}}\ {{\cal\phi}}_{{{\mu}}{{\nu}}}\
d^{4}x$ $\displaystyle=\sum_{r=0}^{k}(-1)^{r}{\frac{1}{r!}}\int_{\cal
I}\xi^{{{\mu}}{{\nu}}{{\rho}}_{1}\ldots{{\rho}}_{r}}(\sigma)\,\big{(}\nabla_{{{\rho}}_{1}}\cdots\nabla_{{{\rho}}_{r}}{{\cal\phi}}_{{{\mu}}{{\nu}}}\big{)}\big{|}_{C(\sigma)}\
d\sigma\,.$ (9)

This follow since if $v^{{\mu}}$ is a vector density of weight 1 then from (2)
$\nabla_{{\mu}}\,v^{{\mu}}=\partial_{{\mu}}\,v^{{\mu}}$.

At the quadrupole level the stress-energy distribution becomes

$\displaystyle T^{\mu\nu}$ $\displaystyle=\int_{\cal
M}\xi^{\mu\nu}\,\delta^{(4)}(z-C)\,d\sigma+\nabla_{\rho}\,\int_{\cal
M}\xi^{\mu\nu\rho}\,\delta^{(4)}(z-C)\,d\sigma+\tfrac{1}{2}\nabla_{\rho}\nabla_{\sigma}\,\int_{\cal
M}\xi^{\mu\nu\rho\sigma}\,\delta^{(4)}(z-C)\,d\sigma$ (10)

where from (7)

$\displaystyle N_{\rho}\xi^{\mu\nu\rho}=0\qquad\textup{and}\qquad
N_{\rho}\xi^{\mu\nu\rho\sigma}=0\,,$ (11)

and from (6) and (8)

$\displaystyle\xi^{\mu\nu}=\xi^{\nu\mu},\quad\xi^{\mu\nu\rho}=\xi^{\nu\mu\rho},\quad\xi^{\mu\nu\rho\sigma}=\xi^{\nu\mu\rho\sigma}\quad\textup{and}\quad\xi^{\mu\nu\rho\sigma}=\xi^{\mu\nu\sigma\rho}\,.$
(12)

In this article we assume that $T^{\mu\nu}$ is divergenceless, i.e.

$\displaystyle\nabla_{\mu}\,T^{\mu\nu}=0\,.$ (13)

This gives rise to dynamical equations for the components $\xi^{\nu\mu}$,
$\xi^{\nu\mu\rho}$ and $\xi^{\mu\nu\sigma\rho}$, which we give below in
theorem 21.

##  3 Properties of the Dixon representation of arbitrary distributions over
worldlines

In this section, general details of multipoles are presented, which are needed
to analyse the stress-energy distribution. The most important result we will
use, is that the Dixon components are unique (section 3.5). This is needed so
that when we take the divergence of the stress-energy tensor and write that as
a Dixon distribution, we know that all the terms must vanish.

Since the results are true for all tensor distributions, not simply the
stress-energy tensor, we have chosen to derive the results for an arbitrary
tensor of rank $(m,0)$ and order $n$.

There are a number of concepts we need to define in order to show the
uniqueness of the components. First we need to establish the Dixon geodesic
hypersurfaces, the adapted coordinate system and the radial vector (section
3.2), the notation for parallel transport (section 3.3) and the Dixon split
(section 3.4).

The next step is to show that all multipoles are Dixon multipoles (section
3.7). For this we need to be able to take a Taylor expansion of the test
tensor (section 3.6). Since we are dealing with tensors it is necessary to
transport the tensors around. There is no unique way of transporting tensors
and different choices will lead to different Taylor expansions. The natural
choice in this case is to use parallel transport along the geodesics emanating
from the worldline.

The final subsection of this section relates the moments of a regular
distribution with the components of a multipole. This is achieved by squeezing
the distribution.

An arbitrary tensor density distribution of rank $(m,0)$, weight 1, and order
$n$ with support on $C$ is given by

$\displaystyle
J^{\mu_{1}\cdots\mu_{m}}=\sum_{k=0}^{N}\frac{1}{k!}\,\nabla_{\rho_{1}}\ldots\nabla_{\rho_{k}}\int_{\cal
I}\zeta^{\mu_{1}\cdots\mu_{m}\rho_{1}\cdots\rho_{k}}\delta^{(4)}\big{(}x-C(\sigma)\big{)}\,d\sigma\,,$
(14)

where

$\displaystyle
N_{\rho_{j}}\zeta^{\mu_{1}\cdots\mu_{m}\rho_{1}\cdots\rho_{k}}=0$ (15)

for $j=1,\dots,k$ and

$\displaystyle\zeta^{\mu_{1}\cdots\mu_{m}\rho_{1}\cdots\rho_{k}}=\zeta^{\mu_{1}\cdots\mu_{m}(\rho_{1}\cdots\rho_{k})}\,.$
(16)

Unlike $\xi^{\mu\nu\rho_{1}\cdots\rho_{k}}$ we do not assume any symmetry on
the indices $\mu_{1},\ldots,\mu_{m}$. For convenience when dealing with
arbitrary tensors we replace the indices $\mu_{1}\cdots\mu_{m}$ with the
symbol ${\underline{\mu}}$ so that
$J^{\mu_{1}\cdots\mu_{m}}=J^{{\underline{\mu}}}$ and
$\zeta^{\mu_{1}\cdots\mu_{m}\rho_{1}\cdots\rho_{k}}=\zeta^{{\underline{\mu}}\rho_{1}\cdots\rho_{k}}$.

Let us introduce the notation for the symmetric sum of multiple covariant
derivatives

$\displaystyle\nabla^{k}_{\rho_{1}\cdots\rho_{k}}=\nabla_{(\rho_{1}}\ldots\nabla_{\rho_{k})}\,.$
(17)

The result of applying a test tensor $\phi_{{\underline{\mu}}}$ is given by

$\displaystyle\int_{\cal
M}J^{\underline{\mu}}\,\phi_{{\underline{\mu}}}\,d^{4}x=\sum_{k=0}^{n}\frac{1}{k!}\,(-1)^{k}\int_{\cal
I}\zeta^{{\underline{\mu}}\rho_{1}\cdots\rho_{k}}\,(\nabla^{k}_{\rho_{1}\cdots\rho_{k}}\phi_{{\underline{\mu}}})|_{C(\sigma)}\,d\sigma\,.$
(18)

Although the commutator of two covariant derivatives gives rise to curvature
terms, these are of a lower order so we can always write the distribution
using the symmetric sum of indices.

The key advantage of imposing the constraints (15), (16) is that they give
rise to unique components $\zeta^{{\underline{\mu}}\rho_{1}\cdots\rho_{k}}$.
We will see this below is section 3.5.

###  3.1 Notation and results for covariant derivatives

Given a vector $V$ and a tensor $S^{{\underline{\nu}}}_{{\underline{\mu}}}$
introduce the notation333We write vectors in the usual index notation as
$V^{{\mu}}$. However when a vector is an argument to a function or is a
subscript we will drop the index and just write $V$.

$\displaystyle\nabla_{V}S^{{\underline{\nu}}}_{{\underline{\mu}}}=V^{\rho}\nabla_{\rho}S^{{\underline{\nu}}}_{{\underline{\mu}}}$
(19)

and the notation

$\displaystyle\nabla^{r}_{V}S^{{\underline{\nu}}}_{{\underline{\mu}}}=V^{\rho_{1}}\cdots
V^{\rho_{k}}\nabla^{r}_{\rho_{1}\cdots\rho_{k}}S^{{\underline{\nu}}}_{{\underline{\mu}}}\,.$
(20)

Observe that in general
$\nabla^{r}_{V}S^{{\underline{\nu}}}_{{\underline{\mu}}}\neq(\nabla_{V})^{r}S^{{\underline{\nu}}}_{{\underline{\mu}}}$.
However they do coincide when we have a geodesic.

######  Lemma 1.

If $H(s)$ is a geodesic and $\phi_{{\underline{\mu}}}$ is a tensor then

$\displaystyle\nabla^{r}_{\dot{H}}\phi_{{\underline{\mu}}}=(\nabla_{\dot{H}})^{r}\phi_{{\underline{\mu}}}\,.$
(21)

###### Proof.

$\displaystyle\nabla^{r}_{\dot{H}}\phi_{{\underline{\mu}}}$
$\displaystyle={\dot{H}}^{\rho_{1}}\cdots{\dot{H}}^{\rho_{r}}\nabla^{r}_{\rho_{1}\cdots\rho_{r}}\phi_{{\underline{\mu}}}={\dot{H}}^{\rho_{1}}\cdots{\dot{H}}^{\rho_{r}}\nabla_{\rho_{r}}\cdots\nabla_{\rho_{1}}\phi_{{\underline{\mu}}}={\dot{H}}^{\rho_{1}}\cdots{\dot{H}}^{\rho_{r-1}}\nabla_{\dot{H}}\big{(}\nabla_{\rho_{r-1}}\cdots\nabla_{\rho_{1}}\phi_{{\underline{\mu}}}\big{)}$
$\displaystyle=\nabla_{\dot{H}}\big{(}{\dot{H}}^{\rho_{1}}\cdots{\dot{H}}^{\rho_{r-1}}\nabla_{\rho_{r-1}}\cdots\nabla_{\rho_{1}}\phi_{{\underline{\mu}}}\big{)}$
$\displaystyle\qquad-\Big{(}(\nabla_{\dot{H}}{\dot{H}}^{\rho_{1}}){\dot{H}}^{\rho_{2}}\cdots{\dot{H}}^{\rho_{r-1}}+\cdots+{\dot{H}}^{\rho_{1}}\cdots{\dot{H}}^{\rho_{r-2}}(\nabla_{\dot{H}}{\dot{H}}^{\rho_{r-1}})\Big{)}\nabla_{\rho_{r-1}}\cdots\nabla_{\rho_{1}}\phi_{{\underline{\mu}}}$
$\displaystyle=\nabla_{\dot{H}}\big{(}{\dot{H}}^{\rho_{1}}\cdots{\dot{H}}^{\rho_{r-1}}\nabla_{\rho_{r-1}}\cdots\nabla_{\rho_{1}}\phi_{{\underline{\mu}}}\big{)}$
$\displaystyle=\cdots=(\nabla_{\dot{H}})^{r}\phi_{{\underline{\mu}}}\,.$

∎

###  3.2 The Dixon geodesic hypersurface, Dixon adapted coordinate system and
the radial vector

Given $\sigma\in{\cal I}$, the set of vectors which are perpendicular to
$N_{\mu}$ are denoted

$\displaystyle{N^{\perp}}(\sigma)={\left\\{\textup{vectors $V^{\mu}$ at the
point $C(\sigma)$ }\big{|}\,N_{\mu}\,V^{\mu}=0\right\\}}\,.$ (22)

Given $V^{\mu}\in{N^{\perp}}(\sigma)$ let $H_{\sigma,V}(s)$ be the geodesic
satisfying

$\displaystyle
H_{\sigma,V}(0)=C(\sigma)\qquad\textup{and}\qquad{\dot{H}}^{\mu}_{\sigma,V}(0)=V^{\mu}\,,$
(23)

see figure 1. Note that the parameter $s$ is not normalised, so there is no
constraint on the value of
$g_{\mu\nu}{\dot{H}}_{\sigma,V}^{\mu}(s){\dot{H}}_{\sigma,V}^{\nu}(s)=g_{\mu\nu}V^{\mu}V^{\nu}$.
The domain of $H_{\sigma,V}(s)$ is distinct from the domain ${\cal I}$ of
$C(\sigma)$. It will always contain the initial value $0$. Although it may not
go all the way to $\pm\infty$ it will go to the edge of the Dixon tube which
is defined below.

It is useful to label the point $P(\sigma,V)\in{\cal M}$ reached from
$C(\sigma)$ travelling along the geodesic $H_{\sigma,V}(s)$ a parameter
distance $1$, that is

$\displaystyle P(\sigma,V)=H_{\sigma,V}(1)\,.$ (24)

This gives the points for the geodesic $H_{\sigma,V}$ as

$\displaystyle H_{\sigma,V}(s)=H_{\sigma,sV}(1)=P(\sigma,s\,V)\,.$ (25)

All the points $P(\sigma,V)$, which are uniquely defined by $(\sigma,V)$ form
a neighbourhood of $C$. We call this the Dixon tube, ${\cal
U}_{\textup{DT}}\subset{\cal M}$. Clearly $C\in{\cal U}_{\textup{DT}}$. There
may be points where two different geodesics $H_{\sigma,V}$ and
$H_{\sigma^{\prime},V^{\prime}}$, with $\sigma\neq\sigma^{\prime}$ intersect.
However these will be at some distance from $C$ and outside of ${\cal
U}_{\textup{DT}}$. In addition there may be points which are unreachable from
$C$. The set ${\left\\{(\sigma,V^{\mu})\,\big{|}\,\sigma\in{\cal I},\
V^{\mu}\in{N^{\perp}}(\sigma),\ P(\sigma,V)\in{\cal
U}_{\textup{DT}}\right\\}}$, is diffeomorphic to ${\cal U}_{\textup{DT}}$.

Since we are dealing with Schwartz distributions we demand that all test forms
have compact support which lie in ${\cal U}_{\textup{DT}}$. This is not a
significant restriction as any other test function can be written, using
partitions of unity, as the sum of two test tensors, one with support inside
${\cal U}_{\textup{DT}}$ and another test tensor with support away from $C$.
This second test tensor, when acted upon by distributions on $C$ will always
give zero.

To define the Dixon adapted coordinate system we require a frame along $C$,
${\left\\{e_{0},e_{1},e_{2},e_{3}\right\\}}$ where $e^{\mu}(\sigma)$ is a
vector at the point $C(\sigma)$. We set $(e_{0})^{\mu}={\dot{C}}^{\mu}$ and
require $(e_{1})^{\mu},(e_{2})^{\mu},(e_{1})^{\mu}\in{N^{\perp}}(\sigma)$.
Thus for any $\hat{U}^{\mu}\in{N^{\perp}}(\sigma)$ we can decompose it in
terms of this basis, giving
$\hat{U}^{\mu}={U}^{1}(e_{1})^{\mu}+{U}^{2}(e_{2})^{\mu}+{U}^{3}(e_{3})^{\mu}$.

The Dixon adapted coordinate system $(\sigma,z^{1},z^{2},z^{3})$, is given on
the Dixon tube such that

$\displaystyle\sigma|_{P(\sigma^{\prime},V)}=P^{0}(\sigma^{\prime},V)=\sigma^{\prime}\qquad\textup{and}\qquad
z^{a}|_{P(\sigma^{\prime},V)}=P^{a}(\sigma^{\prime},V)=V^{a}\,.$ (26)

We set $z^{0}=\sigma$ so that we can label the coordinates of a point $p$ by
$(p^{0},p^{1},p^{2},p^{3})$. We use Latin indices $a,b,\ldots=1,2,3$ and use
the summation convention over Latin indices to sum from 1 to 3. Any reference
in the article to adapted coordinates, or whenever Latin indices are used we
always mean the Dixon adapted coordinate system.

######  Lemma 2.

In the Dixon adapted coordinate system

$\displaystyle H_{\sigma,V}^{0}(s)=\sigma,\quad
H_{\sigma,V}^{a}(s)=s\,V^{a},\quad{\dot{H}}_{\sigma,V}^{0}=0\quad\textup{and}\quad{\dot{H}}_{\sigma,V}^{a}(s)=V^{a}\,.$
(27)

###### Proof.

Equation (27.1) follows444(27.1) refers to the first equation in (27). from
$H_{\sigma^{\prime},V}^{0}(s)=H_{\sigma^{\prime},sV}^{0}(1)=P(\sigma^{\prime},sV)^{0}=\sigma^{\prime}$.
(27.2) follows from
$H_{\sigma,V}^{a}(s)=H_{\sigma,sV}^{a}(1)=P(\sigma,sV)^{a}=sV^{a}$. For (27.3)
and (27.4)

$\displaystyle{\dot{H}}_{\sigma,V}^{0}=\frac{d}{ds}(\sigma)=0\qquad\textup{and}\qquad{\dot{H}}_{\sigma,V}^{a}=\frac{d}{ds}({H_{\sigma,sV}^{a}(s)})=\frac{d}{ds}(s\,V^{a})=V^{a}\,.$

∎

The radial vector field, $R^{\mu}$, is a vector field on ${\cal M}$ given by

$\displaystyle R^{\mu}|_{P(\sigma,V)}={\dot{H}}^{\mu}_{\sigma,V}(1)\,.$ (28)

It is a key ingredient for the Dixon split. In [3], the radial vector was not
defined completely. Instead, only some of the necessary properties of the
radial vector field were given, in order to give the Dixon split up to
quadrupole order. Here the specific radial vector field is defined in order to
give the Dixon split to arbitrary order.

######  Lemma 3.

The radial vector has the properties that

$\displaystyle
R^{\mu}|_{C}=0,\qquad\nabla_{U_{1}}R^{\mu}=U_{1},\qquad{U_{1}^{\rho_{1}}U_{2}^{\rho_{2}}\cdots
U_{r}^{\rho_{r}}}\nabla^{r}_{\rho_{1}\cdots\rho_{r}}R^{\mu}=0\,,$ (29)

for $r\geq 2$, $U^{\rho_{i}}_{i}\in{N^{\perp}}(\sigma)$.

In the Dixon adapted coordinate system

$\displaystyle R^{0}=0\qquad\textup{and}\qquad R^{a}=z^{a}\,.$ (30)

###### Proof.

Clearly $R^{\mu}|_{C}=0$.

Fix $\sigma$ and $V$ and let $H=H_{\sigma,V}$ so that $H(s)=P(\sigma,sV)$ and
$s{\dot{H}}^{\mu}(s)=R^{\mu}|_{H(s)}$. Then

$\displaystyle\nabla_{{\dot{H}}(s)}R^{\mu}=\nabla_{{\dot{H}}(s)}\big{(}s{\dot{H}}^{\mu}(s)\big{)}={\dot{H}}^{\mu}(s)\,.$

Hence setting $V=U_{1}$ and $s=0$ we get (29.2).

For $r\geq 2$ we have from (21)

$\displaystyle\nabla^{r}_{{\dot{H}}}R^{\mu}$
$\displaystyle=(\nabla_{{\dot{H}}})^{r}R^{\mu}=(\nabla_{{\dot{H}}})^{r-1}(\nabla_{{\dot{H}}}R^{\mu})=(\nabla_{{\dot{H}}})^{r-1}{\dot{H}}^{\mu}=0\,.$

Hence setting $s=0$ we have $\nabla^{r}_{V}R^{\mu}=0$. Now setting $V$ equal
to combinations of sums $V=U_{1}\pm U_{2}\pm\cdots\pm U_{r}$, and then taking
the sums, we are left with (29.3).

From (27.3) we have (30.1). From (27.4) and (26.2)
$R^{a}|_{P(\sigma,V)}={\dot{H}}_{\sigma,V}^{a}=V^{a}=z^{a}|_{P(\sigma,V)}$
giving (30.2) ∎

We use the Dixon vector to define the Dixon geodesic hypersurface
$\Sigma(\sigma)\subset{\cal U}_{\textup{DT}}$ as the set

$\displaystyle\Sigma(\sigma)={\left\\{P(\sigma,V)\in{\cal
U}_{\textup{DT}}\,\big{|}\,V^{\mu}\in{N^{\perp}}(\sigma)\right\\}}\,.$ (31)

This is given in figure 1.

###  3.3 Parallel transport

The Dixon geodesic hypersurfaces (31) are constructed from geodesics emanating
from the worldline, with initial tangent orthogonal to the Dixon vector. It is
necessary to parallel transport tensors along these geodesics. We therefore
define a two point tensor $\Pi^{\mu}_{\nu}|^{p}_{q}$ where $p$ and $q$ lie
along the same geodesic. This tensor can be used to parallel transport tensors
along the geodesics and hence defined on the Dixon geodesics hypersurfaces.
This is seen in lemmas 5, 6 and 7 below.

Given $\sigma\in{\cal I}$ and $V^{\mu}\in{N^{\perp}}(\sigma)$ and given
$s_{0},s_{1}\in{\mathbb{R}}$ such that $P(\sigma,s_{0}V)\in{\cal
U}_{\textup{DT}}$ and $P(\sigma,s_{1}V)\in{\cal U}_{\textup{DT}}$, let

$\displaystyle\Pi^{\mu}_{\nu}\big{|}^{P(\sigma,s_{0}V)}_{P(\sigma,s_{1}V)}:{\left\\{\textup{vectors
at $P(\sigma,s_{0}V)$}\right\\}}\longrightarrow{\left\\{\textup{vectors at
$P(\sigma,s_{1}V)$}\right\\}}$ (32)

be the two point tensor satisfying the differential equation

$\displaystyle\frac{d}{ds}\Pi^{\mu}_{\nu}\big{|}^{P(\sigma,s_{0}V)}_{P(\sigma,sV)}=-\Gamma^{\mu}_{{\lambda}\rho}|_{P(\sigma,sV)}\,{\dot{H}}_{\sigma,V}^{\rho}(s)\
\Pi^{{\lambda}}_{\nu}\big{|}^{P(\sigma,s_{0}V)}_{P(\sigma,sV)}\qquad\textup{and}\qquad\Pi^{\mu}_{\nu}\big{|}^{P(\sigma,s_{0}V)}_{P(\sigma,s_{0}V)}=\delta^{\mu}_{\nu}\,.$
(33)

Observe that

$\displaystyle\Pi^{\mu}_{\nu}\big{|}^{P(\sigma,s_{0}V)}_{P(\sigma,s_{1}V)}\,\Pi^{\nu}_{\rho}|^{P(\sigma,s_{1}V)}_{P(\sigma,s_{2}V)}$
$\displaystyle=\Pi^{\mu}_{\rho}|^{P(\sigma,s_{0}V)}_{P(\sigma,s_{2}V)}\,\qquad\textup{and}\qquad\Pi^{\mu}_{\nu}\big{|}^{P(\sigma,s_{0}V)}_{P(\sigma,s_{1}V)}\,\Pi^{\nu}_{\rho}|^{P(\sigma,s_{1}V)}_{P(\sigma,s_{0}V)}=\delta^{\mu}_{\rho}$
(34)

so that from (34.2)

$\displaystyle\Pi^{\mu}_{\nu}\big{|}^{P(\sigma,s_{1}V)}_{P(\sigma,s_{1}V)}=\delta^{\mu}_{\nu}\,.$
(35)

From the definition (33) it may look like the parallel transport is dependent
on the choice of parameterisation of the geodesic $H_{\sigma,V}$. We see here
that this is not the case.

######  Lemma 4.

If $s_{0}V^{\mu}=\hat{s}_{0}\hat{V}^{\mu}$ and
$s_{1}V^{\mu}=\hat{s}_{1}\hat{V}^{\mu}$ then

$\displaystyle\Pi^{\mu}_{\nu}\big{|}^{P(\sigma,s_{0}V)}_{P(\sigma,s_{1}V)}=\Pi^{\mu}_{\nu}\big{|}^{P(\sigma,\hat{s}_{0}\hat{V})}_{P(\sigma,\hat{s}_{1}\hat{V})}\,.$
(36)

I.e. given $p,q\in{\cal U}_{\textup{DT}}$ along the same $H_{\sigma,V}$, then
$\Pi^{\mu}_{\nu}\big{|}^{p}_{q}$ is independent of the choice of
parameterisation of $H_{\sigma,V}$.

###### Proof.

Let $V^{\mu}=\kappa\hat{V}^{\mu}$ so that $s_{0}=\hat{s}_{0}/\kappa$ and
$s_{1}=\hat{s}_{1}/\kappa$. Setting $\frac{d}{ds}=\kappa\,\frac{d}{d\hat{s}}$
then (33.1) becomes

$\displaystyle\kappa\frac{d}{d\hat{s}}\Pi^{\mu}_{\nu}\big{|}^{P(\sigma,s_{0}V)}_{P(\sigma,\hat{s}V)}=-\Gamma^{\mu}_{{\lambda}\rho}|_{P(\sigma,\hat{s}V)}\,\kappa\frac{d}{d\hat{s}}H_{\sigma,V}^{\rho}(\hat{s})\
\Pi^{{\lambda}}_{\nu}\big{|}^{P(\sigma,s_{0}V)}_{P(\sigma,\hat{s}V)}\,,$

and the $\kappa$’s cancel. Setting $V^{\mu}\to\kappa\hat{V}^{\mu}$ and
likewise for $s_{0}$ and $s_{1}$ we see the ODE (33.1) is independent of
scaling of $s,V,s_{0},s_{1}$. The initial condition follows from (35). ∎

######  Lemma 5.

Given $(\sigma,V)$, a vector $\hat{U}^{\mu}$ the point $P(\sigma,s_{0}V)$, and
vector field $U^{\mu}(s)$ along $H_{\sigma,V}$ then

$\displaystyle\nabla_{{\dot{H}}_{\sigma,V}}U^{\mu}=0\qquad\textup{and}\qquad
U^{\mu}|_{P(\sigma,s_{0}V)}=\hat{U}^{\mu}\,,$ (37)

if and only if

$\displaystyle
U^{\mu}|_{P(\sigma,sV)}=\Pi^{\mu}_{\nu}\big{|}^{P(\sigma,s_{0}V)}_{P(\sigma,sV)}\
\hat{U}^{\nu}\,.$ (38)

###### Proof.

Let $H(s)=H_{\sigma,V}(s)$ then

$\displaystyle\nabla_{{\dot{H}}(s_{1})}\big{(}\Pi^{\mu}_{\nu}\big{|}^{P(\sigma,s_{0}V)}_{P(\sigma,sV)}\
\hat{U}^{\nu}\big{)}$
$\displaystyle=\frac{d}{ds}\big{(}\Pi^{\mu}_{\nu}\big{|}^{P(\sigma,s_{0}V)}_{P(\sigma,sV)}\
\hat{U}^{\nu}\big{)}\Big{|}_{s=s_{1}}+\Gamma^{\mu}_{\nu\rho}|_{H(s_{1})}\
{\dot{H}}^{\rho}(s_{1})\ U^{\nu}(s_{1})$
$\displaystyle=\frac{d}{ds}\big{(}\Pi^{\mu}_{\nu}\big{|}^{P(\sigma,s_{0}V)}_{P(\sigma,sV)}\big{)}\Big{|}_{s=s_{1}}\
\hat{U}^{\nu}+\Gamma^{\mu}_{\nu\rho}|_{H(s_{1})}\ {\dot{H}}^{\rho}(s_{1})\
U^{\nu}(s_{1})$
$\displaystyle=-\Gamma^{\mu}_{\sigma\rho}|_{H(s)}\,{\dot{H}}^{\rho}(s)\
\Pi^{\sigma}_{\nu}\big{|}^{P(\sigma,s_{0}V)}_{P(\sigma,sV)}\
\Big{|}_{s=s_{1}}\hat{U}^{\nu}+\Gamma^{\mu}_{\nu\rho}|_{H(s_{1})}\
{\dot{H}}^{\rho}(s_{1})\ U^{\nu}(s_{1})$
$\displaystyle=-\Gamma^{\mu}_{\sigma\rho}|_{H(s_{1})}\,{\dot{H}}^{\rho}(s_{1})\
\Pi^{\sigma}_{\nu}\big{|}^{P(\sigma,s_{0}V)}_{P(\sigma,s_{1}V)}\
\hat{U}^{\nu}+\Gamma^{\mu}_{\nu\rho}|_{H(s_{1})}\ {\dot{H}}^{\rho}(s_{1})\
U^{\nu}(s_{1})$ $\displaystyle=0\,.$

Hence result. ∎

######  Lemma 6.

Given a 1–form $\hat{\phi}_{\mu}$ at the point $P(\sigma,s_{0}V)$, then

$\displaystyle\nabla_{{\dot{H}}_{\sigma,V}}\phi_{\mu}=0\quad\textup{and}\quad\phi_{\mu}|_{P(\sigma,s_{0}V)}=\hat{\phi}_{\mu}\,,$
(39)

if and only if

$\displaystyle\phi_{\mu}|_{P(\sigma,sV)}=\Pi^{\nu}_{\mu}\big{|}^{P(\sigma,sV)}_{P(\sigma,s_{0}V)}\
\hat{\phi}_{\nu}\,.$ (40)

###### Proof.

Given a vector $\hat{U}\in T_{H(s_{0})}{\cal M}$ and a vector field $U(s)$
along $H$ then such that (38) then

$\displaystyle\nabla_{{\dot{H}}(s_{1})}\big{(}\Pi^{\nu}_{\mu}\big{|}^{P(\sigma,sV)}_{P(\sigma,s_{0}V)}\
\hat{\phi}_{\nu}\big{)}\big{|}_{s_{1}}U^{\mu}(s_{1})$
$\displaystyle=\nabla_{{\dot{H}}(s_{1})}\big{(}\Pi^{\nu}_{\mu}\big{|}^{P(\sigma,sV)}_{P(\sigma,s_{0}V)}\
\hat{\phi}_{\nu}\big{)}\big{|}_{s_{1}}U^{\mu}(s_{1})+\Pi^{\nu}_{\mu}(s_{1},s_{0})\
\hat{\phi}_{\nu}\nabla_{{\dot{H}}(s_{1})}U^{\mu}$
$\displaystyle=\nabla_{{\dot{H}}(s_{1})}\big{(}\Pi^{\nu}_{\mu}\big{|}^{P(\sigma,sV)}_{P(\sigma,s_{0}V)}\
\hat{\phi}_{\nu}U^{\mu}(s)\big{)}\big{|}_{s_{1}}$
$\displaystyle=\frac{d}{ds}\big{(}\Pi^{\nu}_{\mu}\big{|}^{P(\sigma,sV)}_{P(\sigma,s_{0}V)}\
\hat{\phi}_{\nu}\ \Pi^{\mu}_{\rho}\big{|}^{P(\sigma,s_{0}V)}_{P(\sigma,sV)}\
\hat{U}^{\rho}\big{)}\big{|}_{s_{1}}=\frac{d}{ds}\big{(}\hat{\phi}_{\nu}\hat{U}^{\nu}\big{)}\big{|}_{s_{1}}=0\,.$

Hence (40) if and only if (39). ∎

We need to extend the results of lemma 6 for tensors with arbitrary number of
indices, i.e. $\phi_{\underline{\mu}}$. Let

$\displaystyle\Pi^{\underline{\mu}}_{\underline{\nu}}\big{|}^{P(\sigma,s_{0}V)}_{P(\sigma,sV)}=\Pi^{\mu_{1}}_{\nu_{1}}\big{|}^{P(\sigma,s_{0}V)}_{P(\sigma,sV)}\cdots\Pi^{\mu_{m}}_{\nu_{m}}\big{|}^{P(\sigma,s_{0}V)}_{P(\sigma,sV)}\,.$
(41)

######  Lemma 7.

Given a tensor $\hat{\phi}_{\underline{\mu}}$ at the point at the point
$P(\sigma,s_{0}V)$, then

$\displaystyle\nabla_{{\dot{H}}_{\sigma,V}}\phi_{\underline{\mu}}=0\quad\textup{and}\quad\phi_{\underline{\mu}}|_{P(\sigma,s_{0}V)}=\hat{\phi}_{\underline{\mu}}\,,$
(42)

if and only if

$\displaystyle\phi_{\underline{\mu}}|_{P(\sigma,sV)}=\Pi^{\underline{\nu}}_{\underline{\mu}}\big{|}^{P(\sigma,sV)}_{P(\sigma,s_{0}V)}\
\hat{\phi}_{\underline{\nu}}\,.$ (43)

###### Proof.

Set $\phi_{\underline{\mu}}$ to be the outer product of 1–forms
$\phi_{\underline{\mu}}=\phi^{1}_{\mu_{1}}\cdots\phi^{m}_{\mu_{m}}$. Then
apply lemma 6. ∎

######  Lemma 8.

Let $\phi_{\underline{\mu}}$ be a 1–form field over $H_{\sigma,V}$. Then

$\displaystyle\frac{d^{k}}{ds^{k}}\Big{(}\Pi^{\underline{\mu}}_{\nu}\big{|}^{P(\sigma,sV)}_{P(\sigma,s_{0}V)}\phi_{\underline{\mu}}|_{P(\sigma,sV)}\Big{)}=\nabla_{{\dot{H}}_{\sigma,V}}^{k}\phi_{\nu}|_{P(\sigma,s_{0}V)}\,.$
(44)

###### Proof.

Proof by induction on $r$. Clearly true when $r=0$ and $r=1$, Let
$H(s)=H_{\sigma,V}(s)$. Assume true for $r$,

$\displaystyle\frac{d^{r+1}}{ds^{r+1}}\Big{(}\Pi^{\underline{\mu}}_{\underline{\nu}}\big{|}^{P(\sigma,s_{0}V)}_{P(\sigma,sV)}\phi_{\underline{\mu}}|_{P(\sigma,sV)}\Big{)}$
$\displaystyle=\frac{d}{ds_{2}}\bigg{(}\frac{d^{r}}{ds_{1}^{r}}\Big{(}\Pi^{\underline{\mu}}_{\underline{\nu}}\big{|}^{P(\sigma,s_{0}V)}_{P(\sigma,s_{1}V)}\phi_{\underline{\mu}}|_{P(\sigma,s_{1}V)}\Big{)}\Big{|}_{s_{1}=s_{2}}\bigg{)}\bigg{|}_{s_{2}=s_{0}}$
$\displaystyle=\frac{d}{ds_{2}}\bigg{(}\frac{d^{r}}{ds_{1}^{r}}\Big{(}\Pi^{\underline{\mu}}_{\underline{\rho}}\big{|}^{P(\sigma,s_{0}V)}_{P(\sigma,s_{2}V)}\
\Pi^{\underline{\rho}}_{\underline{\nu}}\big{|}^{P(\sigma,s_{2}V)}_{P(\sigma,s_{1}V)}\
\phi_{\underline{\mu}}|_{P(\sigma,s_{1}V)}\Big{)}\Big{|}_{s_{1}=s_{2}}\bigg{)}\bigg{|}_{s_{2}=s_{0}}$
$\displaystyle=\frac{d}{ds_{2}}\bigg{(}\Pi^{\underline{\mu}}_{\underline{\rho}}\big{|}^{P(\sigma,s_{0}V)}_{P(\sigma,s_{2}V)}\
\frac{d^{r}}{ds_{1}^{r}}\Big{(}\Pi^{\underline{\rho}}_{\underline{\nu}}\big{|}^{P(\sigma,s_{2}V)}_{P(\sigma,s_{1}V)}\
\phi_{\underline{\mu}}|_{P(\sigma,s_{1}V)}\Big{)}\Big{|}_{s_{1}=s_{2}}\bigg{)}\bigg{|}_{s_{2}=s_{0}}$
$\displaystyle=\frac{d}{ds_{2}}\bigg{(}\Pi^{\underline{\mu}}_{\underline{\rho}}\big{|}^{P(\sigma,s_{0}V)}_{P(\sigma,s_{2}V)}\
\nabla_{{\dot{H}}}^{r}\phi_{\underline{\mu}}|_{P(\sigma,s_{2}V)}\bigg{)}\bigg{|}_{s_{2}=s_{0}}$
$\displaystyle=\nabla_{\dot{H}}\nabla^{r}_{{\dot{H}}}\phi_{\underline{\nu}}|_{P(\sigma,s_{0}V)}=\nabla_{\dot{H}}(\nabla_{{\dot{H}}})^{r}\phi_{\underline{\nu}}|_{P(\sigma,s_{0}V)}=\nabla^{r+1}_{\dot{H}}\phi_{\underline{\nu}}|_{P(\sigma,s_{0}V)}\,.$

∎

It is necessary to define a collection of tensor fields
$\overline{\boldsymbol{\Pi}}^{{\\{{\underline{\nu}}\\}}}_{\underline{\mu}}$,
which are used to define the test tensors needed to extract the components
$\zeta^{{\underline{\mu}}\rho_{1}\cdots\rho_{k}}$ in section 3.5. These
tensors are parallel and form a basis for tensors. These will be necessary to
define a Taylor expansion of the test tensors and for extracting the
components of a multipole.

$\displaystyle\overline{\boldsymbol{\Pi}}^{{\\{{\underline{\nu}}\\}}}_{\underline{\mu}}|_{P(\sigma,V)}=\Pi^{\underline{\nu}}_{\underline{\mu}}\big{|}^{P(\sigma,V)}_{C(\sigma)}\,.$
(45)

The reason for placing the ${\underline{\nu}}$ in curly brackets is because
these are tensor indices referring to the point $C(\sigma)$. Therefore the
covariant derivative $\nabla_{\rho}$ does not produce the Christoffel symbols
$\Gamma^{\nu_{i}}_{\rho\sigma}$. I.e.

$\displaystyle\nabla_{\rho}\overline{\boldsymbol{\Pi}}^{{\\{{\underline{\nu}}\\}}}_{\underline{\mu}}$
$\displaystyle=\partial_{\rho}\overline{\boldsymbol{\Pi}}^{{\\{{\underline{\nu}}\\}}}_{\underline{\mu}}-\Gamma^{\sigma}_{\rho\mu_{1}}\,\overline{\boldsymbol{\Pi}}^{{\\{{\underline{\nu}}\\}}}_{\sigma\mu_{2}\cdots\mu_{m}}-\cdots-\Gamma^{\sigma}_{\rho\mu_{m}}\,\overline{\boldsymbol{\Pi}}^{{\\{{\underline{\nu}}\\}}}_{\mu_{1}\cdots\mu_{m-1}\sigma}\,.$
(46)

Hence from lemma 7,

$\displaystyle\nabla_{{\dot{H}}_{\sigma,V}}\overline{\boldsymbol{\Pi}}^{{\\{{\underline{\nu}}\\}}}_{\underline{\mu}}=0\,.$
(47)

######  Lemma 9.

In adapted coordinates

$\displaystyle\nabla^{k}_{a_{1}\cdots
a_{k}}\overline{\boldsymbol{\Pi}}^{{\\{{\underline{\nu}}\\}}}_{\underline{\mu}}=0\,.$
(48)

###### Proof.

Fix $\sigma$ and $V^{\mu}\in{N^{\perp}}$ and let $H=H_{\sigma,V}$ so that
$H(s)=P(\sigma,sV)$. Since $H$ is a geodesic

$\displaystyle V^{a_{1}}\cdots V^{a_{k}}\nabla^{k}_{a_{1}\cdots
a_{k}}\overline{\boldsymbol{\Pi}}^{{\\{{\underline{\nu}}\\}}}_{\underline{\mu}}\big{|}_{C(\sigma)}=\nabla_{V}^{k}\overline{\boldsymbol{\Pi}}^{{\\{{\underline{\nu}}\\}}}_{\underline{\mu}}\big{|}_{C(\sigma)}=\nabla_{{\dot{H}}}^{k}\overline{\boldsymbol{\Pi}}^{{\\{{\underline{\nu}}\\}}}_{\underline{\mu}}\big{|}_{C(\sigma)}=0\,.$

Now setting $V^{\mu}$ equal to combinations of sums $V=U^{\mu}_{1}\pm
U^{\mu}_{2}\pm\cdots\pm U^{\mu}_{r}$, and then taking the sums, we are left
with

$\displaystyle U_{1}^{a_{1}}\cdots U_{k}^{a_{k}}\nabla^{k}_{a_{1}\cdots
a_{k}}\overline{\boldsymbol{\Pi}}^{{\\{{\underline{\nu}}\\}}}_{\underline{\mu}}\big{|}_{C(\sigma)}=0\,,$

since $\nabla^{k}_{a_{1}\cdots a_{k}}$ is symmetric in $a_{1},\ldots,a_{k}$.
Since this is true for all vectors $V^{\mu}=U^{\mu}_{1},\dots,U^{\mu}_{r}$ at
$C(\sigma)$ we have (48)

∎

###  3.4 The Dixon split

Given an arbitrary Dixon multipole $J^{\underline{\mu}}$, or order $n$, let

$\displaystyle
J^{{\underline{\mu}}}=\sum_{k=0}^{n}J^{{\underline{\mu}}}_{(k)}\,,$ (49)

where

$\displaystyle
J^{{\underline{\mu}}}_{(k)}=\frac{1}{k!}\,\nabla_{\rho_{1}\ldots\rho_{k}}\int\zeta^{{\underline{\mu}},\rho_{1}\cdots\rho_{k}}\delta^{(4)}(z-C)\,d\sigma\,.$
(50)

I.e.

$\displaystyle
J^{{\underline{\mu}}}_{(k)}[\phi_{{\underline{\mu}}}]=(-1)^{k}\frac{1}{k!}\,\int\zeta^{{\underline{\mu}},\rho_{1}\cdots\rho_{k}}\nabla^{k}_{\rho_{1}\ldots\rho_{k}}\phi_{{\underline{\mu}}}\,d\sigma\,.$
(51)

We call the set ${\left\\{J^{{\underline{\mu}}}_{(k)}\right\\}}$ the Dixon
split of $J^{{\underline{\mu}}}$. It is the first step towards the extraction
of the components.

######  Lemma 10.

$\displaystyle
J^{{\underline{\mu}}}_{(k)}[\nabla^{r}_{R}\,\phi_{{\underline{\mu}}}]=\begin{cases}\displaystyle\frac{k!}{(k-r)!}\
J^{{\underline{\mu}}}_{(k)}[\phi_{{\underline{\mu}}}]&\quad\textup{for}\quad
r\leq k\\\ 0&\quad\textup{for}\quad r>k\,.\end{cases}$ (52)

###### Proof.

This follows from two observations

$\displaystyle\nabla^{r+1}_{R}\phi_{\underline{\mu}}=-r\nabla^{r}_{R}\phi_{\underline{\mu}}+\nabla^{r}_{R}(\nabla_{R}\phi_{\underline{\mu}})\,,$
(53)

and

$\displaystyle
J^{{\underline{\mu}}}_{(k)}[\nabla_{R}\phi]=k\,J^{{\underline{\mu}}}_{(k)}[\phi]\,.$
(54)

Equation (53) follows since in the Dixon adapted coordinates.

$\displaystyle\nabla^{r}_{R}(\nabla_{R}\phi_{\underline{\mu}})$
$\displaystyle=\nabla^{r}_{R}(z^{b}\nabla_{b}\phi_{\underline{\mu}})$
$\displaystyle=z^{a_{1}}\dots
z^{a_{r}}\nabla_{a_{1}}\dots\nabla_{a_{r}}(z^{b}\nabla_{b}\phi_{\underline{\mu}})$
$\displaystyle=z^{a_{1}}\dots
z^{a_{r}}\nabla_{a_{1}}\dots\nabla_{a_{r-1}}(\delta^{b}_{a_{r}}\nabla_{b}\phi_{\underline{\mu}}+z^{b}\nabla_{a_{r}}\nabla_{b}\phi_{\underline{\mu}})$
$\displaystyle=z^{a_{1}}\dots
z^{a_{r}}\nabla_{a_{1}}\dots\nabla_{a_{r-1}}(\nabla_{a_{r}}\phi_{\underline{\mu}}+z^{b}\nabla_{a_{r}}\nabla_{b}\phi_{\underline{\mu}})$
$\displaystyle=z^{a_{1}}\dots
z^{a_{r}}\nabla_{a_{1}}\dots\nabla_{a_{r}}\phi_{\underline{\mu}}+z^{a_{1}}\dots
z^{a_{r}}\nabla_{a_{1}}\dots\nabla_{a_{r-1}}(z^{b}\nabla_{a_{r}}\nabla_{b}\phi_{\underline{\mu}})$
$\displaystyle=\nabla^{r}_{R}\phi_{\underline{\mu}}+z^{a_{1}}\dots
z^{a_{r}}\nabla_{a_{1}}\dots\nabla_{a_{r-1}}(z^{b}\nabla_{a_{r}}\nabla_{b}\phi_{\underline{\mu}})$
$\displaystyle=r\nabla^{r}_{R}\phi_{\underline{\mu}}+z^{a_{1}}\dots
z^{a_{r}}z^{b}\nabla_{a_{1}}\dots\nabla_{a_{r}}\nabla_{b}\phi_{\underline{\mu}}$
$\displaystyle=r\nabla^{r}_{R}\phi_{\underline{\mu}}+\nabla^{r+1}_{R}\phi_{\underline{\mu}}\,.$

Equation (54) follows from

$\displaystyle
J_{(k)}^{{\underline{\mu}}}[\nabla_{R}\,\phi_{\underline{\mu}}]$
$\displaystyle=J_{(k)}^{{\underline{\mu}}}[z^{b}\nabla_{b}\phi_{\underline{\mu}}]=(-1)^{k}\frac{1}{k!}\int\zeta^{{\underline{\mu}}a_{1}\dots
a_{k}}\nabla^{k}_{a_{1}\cdots
a_{k}}\Big{(}z^{b}\nabla_{b}\phi_{\underline{\mu}}\Big{)}\,d\sigma$
$\displaystyle=(-1)^{k}\frac{1}{k!}\int\zeta^{{\underline{\mu}}a_{1}\dots
a_{k}}\nabla_{a_{1}}\dots\nabla_{a_{k}}\Big{(}z^{b}\nabla_{b}\phi_{\underline{\mu}}\Big{)}\,d\sigma$
$\displaystyle=(-1)^{k}\frac{1}{k!}\int\zeta^{{\underline{\mu}}a_{1}\dots
a_{k}}\nabla_{a_{1}}\dots\nabla_{a_{k-1}}\Big{(}\delta_{a_{k}}^{b}\nabla_{b}\phi_{\underline{\mu}}+z^{b}\nabla_{a_{k}}\nabla_{b}\phi_{\underline{\mu}}\Big{)}\,d\sigma$
$\displaystyle=(-1)^{k}\frac{1}{k!}\int\zeta^{{\underline{\mu}}a_{1}\dots
a_{k}}\nabla_{a_{1}}\dots\nabla_{a_{k-1}}\Big{(}\nabla_{a_{k}}\phi_{\underline{\mu}}+z^{b}\nabla_{a_{k}}\nabla_{b}\phi_{\underline{\mu}}\Big{)}$
$\displaystyle=(-1)^{k}\frac{1}{k!}\int\zeta^{{\underline{\mu}}a_{1}\dots
a_{k}}\Big{(}\nabla_{a_{1}}\dots\nabla_{a_{k}}\phi_{\underline{\mu}}+\nabla_{a_{1}}\dots\nabla_{a_{k-1}}(z^{b}\nabla_{a_{k}}\nabla_{b}\phi_{\underline{\mu}})\Big{)}\,d\sigma$
$\displaystyle=(-1)^{k}\frac{1}{k!}\int\zeta^{{\underline{\mu}}a_{1}\dots
a_{k}}\Big{(}k\nabla_{a_{1}}\dots\nabla_{a_{k}}\phi_{\underline{\mu}}+z^{b}\nabla_{a_{1}}\dots\nabla_{a_{k}}\nabla_{b}\phi_{\underline{\mu}}\Big{)}\,d\sigma$
$\displaystyle=(-1)^{k}\frac{1}{(k-1)!}\int\zeta^{{\underline{\mu}}a_{1}\dots
a_{k}}\nabla_{a_{1}}\dots\nabla_{a_{k}}\phi_{\underline{\mu}}\,d\sigma$
$\displaystyle=(-1)^{k}\frac{1}{(k-1)!}\int\zeta^{{\underline{\mu}}a_{1}\dots
a_{k}}\nabla_{a_{1}\cdots a_{k}}\phi_{\underline{\mu}}\,d\sigma$
$\displaystyle=kJ_{(k)}^{{\underline{\mu}}}[\phi_{\underline{\mu}}]\,.$

We now show (52) by induction. Trivial for $r=0$. Assume true for $r$, from
(53) and (54)

$\displaystyle
J^{{\underline{\mu}}}_{(k)}[\nabla^{r+1}_{R}\,\phi_{{\underline{\mu}}}]$
$\displaystyle=-rJ^{{\underline{\mu}}}_{(k)}[\nabla^{r}_{R}\phi_{{\underline{\mu}}}]+J^{{\underline{\mu}}}_{(k)}[\nabla^{r}_{R}(\nabla_{R}\phi_{{\underline{\mu}}})]=\frac{k!}{(k-r)!}\big{(}-rJ^{{\underline{\mu}}}_{(k)}[\phi_{{\underline{\mu}}}]+J^{{\underline{\mu}}}_{(k)}[\nabla_{R}\phi_{{\underline{\mu}}}]\big{)}$
$\displaystyle=\frac{k!}{(k-r)!}\big{(}-rJ^{{\underline{\mu}}}_{(k)}[\phi_{{\underline{\mu}}}]+kJ^{{\underline{\mu}}}_{(k)}[\phi_{{\underline{\mu}}}]\big{)}=\frac{k!(k-r)}{(k-r)!}J^{{\underline{\mu}}}_{(k)}[\phi_{{\underline{\mu}}}]=\frac{k!}{(k-(r+1))!}J^{{\underline{\mu}}}_{(k)}[\phi_{{\underline{\mu}}}]\,.$

In the case when $r=k+1$ then

$\displaystyle
J^{{\underline{\mu}}}_{(k)}[\nabla^{k+1}_{R}\,\phi_{{\underline{\mu}}}]$
$\displaystyle=\frac{k!(k-k)}{(k-r)!}J^{{\underline{\mu}}}_{(k)}[\phi_{{\underline{\mu}}}]=0\,.$

Hence (52) holds for all $r$. ∎

######  Theorem 11.

$\displaystyle
J^{{\underline{\mu}}}_{(r)}[\phi_{{\underline{\mu}}}]={\sum_{k=r}^{n}}\frac{(-1)^{k-r}}{(k-r)!\
{r!}}J^{{\underline{\mu}}}[\nabla_{R}^{k}\phi_{{\underline{\mu}}}]\,.$ (55)

###### Proof.

$\displaystyle{\sum_{k=r}^{n}}\frac{(-1)^{k-r}}{(k-r)!\
{r!}}J^{{\underline{\mu}}}[\nabla_{R}^{k}\phi_{{\underline{\mu}}}]$
$\displaystyle={\sum_{k=r}^{n}}\frac{(-1)^{k-r}}{(k-r)!\
{r!}}\sum_{\ell=0}^{n}J^{{\underline{\mu}}}_{(\ell)}[\nabla_{R}^{k}\phi_{{\underline{\mu}}}]={\sum_{k=r}^{n}}\sum_{\ell=0}^{n}\frac{(-1)^{k-r}}{(k-r)!\
{r!}}\frac{\ell!}{(\ell-k)!}J^{{\underline{\mu}}}_{(\ell)}[\phi_{{\underline{\mu}}}]$
$\displaystyle=\sum_{\ell=0}^{n}{\sum_{k=r}^{n}}(-1)^{k-r}\binom{k}{r}\binom{\ell}{k}J^{{\underline{\mu}}}_{(\ell)}[\phi_{{\underline{\mu}}}]=\sum_{\ell=0}^{n}\Bigg{(}{\sum_{k=0}^{n}}(-1)^{k-r}\binom{k}{r}\binom{\ell}{k}\Bigg{)}J^{{\underline{\mu}}}_{(\ell)}[\phi_{{\underline{\mu}}}]$
$\displaystyle=\sum_{\ell=0}^{n}\delta_{r\ell}J^{{\underline{\mu}}}_{(\ell)}[\phi_{{\underline{\mu}}}]=J^{{\underline{\mu}}}_{(r)}[\phi_{{\underline{\mu}}}]\,,$

where we have used the standard result for the sums of products of binomial
coefficients. ∎

Although the Dixon split, given by (49), is defined via the coordinate system,
theorem 11 shows that the Dixon split is actually independent of the
coordinate system, once the Dixon vector is chosen.

######  Corollary 12.

If $J^{{\underline{\mu}}}=0$ then
$J^{{\underline{\mu}}}[\nabla^{r}_{R}\phi_{{\underline{\mu}}}]=0$ for all $r$.

###### Proof.

Follows trivially since if $J^{{\underline{\mu}}}=0$, then
$J^{{\underline{\mu}}}_{(k)}=0$ for all $0\leq k\leq n$. ∎

###  3.5 Extraction of the Dixon components

In this section we show how we can extract the Dixon components
$\zeta^{{\underline{\mu}}a_{1}\dots a_{k}}$, in an adapted coordinate system,
by applying the distribution to particular test tensors.

Let $\psi_{0}:{\mathbb{R}}\to{\mathbb{R}}$ be any test function such that

$\displaystyle\int_{\mathbb{R}}\psi_{0}(z)\,dz=1,\qquad\psi_{0}(0)=1\,,$

and $\psi_{0}(z)$ is flat in an interval about $0$.

In Dixon adapted coordinates, given $\sigma_{0}\in{\cal I}$ and $\epsilon>0$,
choose a set of indices ${\underline{\nu}}$. Let the tensor
$(\phi_{\sigma_{0},\epsilon}^{{\underline{\nu}},a_{1}\cdots
a_{r}})_{\underline{\mu}}$ be given by

$\displaystyle(\phi_{\sigma_{0},\epsilon}^{{\underline{\nu}},a_{1}\cdots
a_{r}})_{\underline{\mu}}|_{(\sigma,z^{1},\ldots,z^{3})}=\frac{(-1)^{k}}{\epsilon}z^{a_{1}}\cdots
z^{a_{r}}\psi_{0}\Big{(}\frac{\sigma-\sigma_{0}}{\epsilon}\Big{)}\psi_{0}(z^{1})\psi_{0}(z^{2})\psi_{0}(z^{3})\overline{\boldsymbol{\Pi}}^{{\\{{\underline{\nu}}\\}}}_{{\underline{\mu}}}\,.$
(56)

######  Theorem 13.

$\displaystyle\zeta^{{\underline{\nu}}a_{1}\cdots
a_{k}}(\sigma_{0})=\lim_{\epsilon\to
0}J^{{\underline{\mu}}}_{(k)}[(\phi_{\sigma_{0},\epsilon}^{{\underline{\nu}},a_{1}\cdots
a_{r}})_{\underline{\mu}}]\,.$ (57)

###### Proof.

Since $\psi_{0}(z^{a})$ is flat about $z^{a}=0$, then
$\nabla_{b}\psi_{0}(z^{a})=0$. Since $\zeta^{{\underline{\nu}}a_{1}\cdots
a_{k}}$ is totally symmetric in $a_{1}\ldots a_{k}$ we have from (48)
$\nabla^{k}_{a_{1}\cdots
a_{k}}\overline{\boldsymbol{\Pi}}^{{\\{{\underline{\nu}}\\}}}_{\underline{\mu}}=0$.
Also $\nabla_{b}z^{a}=\partial_{b}z^{a}=\delta^{a}_{b}$.

$\displaystyle\lim_{\epsilon\to 0}J^{{\underline{\mu}}}_{(k)}$
$\displaystyle[(\phi_{\sigma_{0},\epsilon}^{{\underline{\nu}},a_{1}\cdots
a_{r}})_{\underline{\mu}}]=\lim_{\epsilon\to
0}\frac{1}{k!\,\epsilon}\int_{\cal I}\zeta^{{\underline{\mu}}b_{1}\cdots
b_{k}}\nabla^{k}_{b_{1}\cdots b_{k}}\bigg{(}z^{a_{1}}\cdots
z^{a_{k}}\psi_{0}\Big{(}\frac{\sigma-\sigma_{0}}{\epsilon}\Big{)}\psi_{0}(z^{1})\psi_{0}(z^{2})\psi_{0}(z^{3})\overline{\boldsymbol{\Pi}}^{{\\{{\underline{\nu}}\\}}}_{{\underline{\mu}}}\,d\sigma\bigg{)}$
$\displaystyle=\lim_{\epsilon\to 0}\frac{1}{k!\,\epsilon}\int_{\cal
I}\zeta^{{\underline{\mu}}b_{1}\cdots b_{k}}\nabla^{k}_{b_{1}\cdots
b_{k}}\big{(}z^{a_{1}}\cdots
z^{a_{k}}\big{)}\psi_{0}\Big{(}\frac{\sigma-\sigma_{0}}{\epsilon}\Big{)}\psi_{0}(z^{1})\psi_{0}(z^{2})\psi_{0}(z^{3})\overline{\boldsymbol{\Pi}}^{{\\{{\underline{\nu}}\\}}}_{{\underline{\mu}}}\big{|}_{z=0}\,d\sigma$
$\displaystyle=\lim_{\epsilon\to 0}\frac{1}{k!\,\epsilon}\int_{\cal
I}\zeta^{{\underline{\mu}}b_{1}\cdots
b_{k}}\partial_{b_{1}}\ldots\partial_{b_{k}}\big{(}z^{a_{1}}\cdots
z^{a_{k}}\big{)}\psi_{0}\Big{(}\frac{\sigma-\sigma_{0}}{\epsilon}\Big{)}\psi_{0}(0)^{3}\delta^{{\underline{\nu}}}_{{\underline{\mu}}}\big{|}_{z=0}\,d\sigma$
$\displaystyle=\lim_{\epsilon\to 0}\frac{1}{\epsilon}\int_{\cal
I}\zeta^{{\underline{\nu}}a_{1}\cdots
a_{k}}\psi_{0}\Big{(}\frac{\sigma-\sigma_{0}}{\epsilon}\Big{)}\,d\sigma=\zeta^{{\underline{\nu}}a_{1}\cdots
a_{k}}(\sigma_{0})\,,$

as
$\epsilon^{-1}\psi_{0}(\epsilon^{-1}(\sigma-\sigma_{0}))\to\delta(\sigma-\sigma_{0})$
as $\epsilon\to 0$. ∎

######  Corollary 14.

$\displaystyle J^{{\underline{\mu}}}=0\qquad\textup{if and only
if}\qquad\zeta^{{\underline{\mu}}a_{1}\cdots a_{k}}=0\,,$ (58)

and hence the Dixon components are unique.

###### Proof.

Assuming $J^{{\underline{\mu}}}=0$ then from corollary 12 we have
$J^{{\underline{\mu}}}_{(k)}=0$, hence from (57) all the components are zero.
∎

###  3.6 The Taylor expansion with respect to Dixon geodesic hypersurfaces

Given a test tensor $\phi_{{\underline{\mu}}}$, we need to take its Taylor
expansion, in a way that respects the Dixon geodesic hypersurfaces. We define
the tensor field

$\displaystyle\phi^{(k)}_{\underline{\mu}}|_{P(\sigma,V)}$
$\displaystyle=\Pi^{\underline{\nu}}_{\underline{\mu}}\big{|}_{C(\sigma)}^{P(\sigma,V)}(\nabla_{V}^{k}\phi_{\underline{\nu}}|_{C(\sigma)})\,,$
(59)

by replacing $V$ with $sV$ this becomes

$\displaystyle\phi^{(k)}_{\underline{\mu}}|_{P(\sigma,sV)}$
$\displaystyle=\Pi^{\underline{\nu}}_{\underline{\mu}}\big{|}_{C(\sigma)}^{P(\sigma,sV)}(\nabla_{sV}^{k}\phi_{\underline{\nu}}|_{C(\sigma)})=s^{k}\,\Pi^{\underline{\nu}}_{\underline{\mu}}\big{|}_{C(\sigma)}^{P(\sigma,sV)}(\nabla_{V}^{k}\phi_{\underline{\nu}}|_{C(\sigma)})\,.$
(60)

In adapted coordinates
$H_{\sigma,V}(1)=P(\sigma,V)=(\sigma,z^{1},\ldots,z^{3})$ we have

$\displaystyle\phi^{(k)}_{\underline{\mu}}|_{(\sigma,z^{1},\ldots,z^{3})}=z^{a_{1}}\ldots
z^{a_{k}}\overline{\boldsymbol{\Pi}}^{{\\{{\underline{\nu}}\\}}}_{\underline{\mu}}\big{(}\nabla_{a_{1}\cdots
a_{k}}^{k}\phi_{\underline{\nu}}|_{C(\sigma)}\big{)}\,,$ (61)

since

$\displaystyle\phi^{(k)}_{\underline{\mu}}|_{(\sigma,z^{1},\ldots,z^{3})}$
$\displaystyle=\phi^{(k)}_{\underline{\mu}}|_{P(\sigma,V)}=\Pi^{\underline{\nu}}_{\underline{\mu}}\big{|}_{C(\sigma)}^{P(\sigma,V)}\nabla_{{V}}^{k}\phi_{\underline{\nu}}|_{C(\sigma)}=\overline{\boldsymbol{\Pi}}^{{\\{{\underline{\nu}}\\}}}_{\underline{\mu}}\big{(}\nabla_{{V}}^{k}\phi_{\underline{\nu}}|_{C(\sigma)}\big{)}=V^{a_{1}}\ldots
V^{a_{k}}\overline{\boldsymbol{\Pi}}^{{\\{{\underline{\nu}}\\}}}_{\underline{\mu}}\big{(}\nabla_{a_{1}\cdots
a_{k}}^{k}\phi_{\underline{\nu}}|_{C(\sigma)}\big{)}$
$\displaystyle=z^{a_{1}}\ldots
z^{a_{k}}\overline{\boldsymbol{\Pi}}^{{\\{{\underline{\nu}}\\}}}_{\underline{\mu}}\big{(}\nabla_{a_{1}\cdots
a_{k}}^{k}\phi_{\underline{\nu}}|_{C(\sigma)}\big{)}\,.$

The Taylor expansion, with respect to the Dixon geodesic hypersurfaces, of a
test tensor $\phi_{\underline{\mu}}$ is given by

$\displaystyle\phi_{\underline{\mu}}|_{P(\sigma,sV)}$
$\displaystyle=\sum_{k=0}^{N}\frac{1}{k!}\,\phi^{(k)}_{\underline{\mu}}|_{P(\sigma,sV)}+O(s^{N+1})\,.$
(62)

######  Lemma 15.

The order $O(s^{N+1})$, as $s\to 0$, in (62) is correct.

###### Proof.

Taking the Taylor expansion about $C(\sigma)$ along each geodesic
$H_{\sigma,V}(s)$ and using (44) we have

$\displaystyle\Pi^{\underline{\mu}}_{\underline{\nu}}\big{|}^{C(\sigma)}_{P(\sigma,sV)}\phi_{\underline{\mu}}|_{P(\sigma,sV)}$
$\displaystyle=\sum_{k=0}^{N}\frac{s^{k}}{k!}\,\frac{d^{k}}{ds^{k}}\Big{|}_{s=0}\Big{(}\Pi^{\underline{\mu}}_{\underline{\nu}}\big{|}^{C(\sigma)}_{P(\sigma,sV)}\phi_{\underline{\mu}}|_{P(\sigma,sV)}\Big{)}+O(s^{N+1})$
$\displaystyle=\sum_{k=0}^{N}\frac{s^{k}}{k!}\,\nabla_{{V}}^{k}\phi_{\underline{\nu}}|_{C(\sigma)}+O(s^{N+1})\,.$

Hence

$\displaystyle\phi_{\underline{\mu}}|_{P(\sigma,sV)}$
$\displaystyle=\sum_{k=0}^{N}\frac{s^{k}}{k!}\,\Pi^{\underline{\nu}}_{\underline{\mu}}\big{|}_{C(\sigma)}^{P(\sigma,sV)}\nabla_{{V}}^{k}\phi_{\underline{\nu}}|_{C(\sigma)}+O(s^{N+1})=\sum_{k=0}^{N}\frac{1}{k!}\,\phi^{(k)}_{\underline{\mu}}|_{P(\sigma,sV)}+O(s^{N+1})\,.$

∎

######  Lemma 16.

$\displaystyle\nabla^{r}_{R}\phi^{(k)}_{\underline{\mu}}=\frac{k!}{(k-r)!}\,\phi^{(k)}_{\underline{\mu}}\,.$
(63)

###### Proof.

From (48) and (61)

$\displaystyle\nabla^{r}_{R}\phi^{(k)}_{\underline{\mu}}$
$\displaystyle=z^{b_{1}}\ldots z^{b_{r}}\nabla^{r}_{b_{1}\cdots
b_{r}}\phi^{(k)}_{\underline{\mu}}$ $\displaystyle=z^{b_{1}}\ldots
z^{b_{r}}\nabla^{r}_{b_{1}\cdots b_{r}}\Big{(}z^{a_{1}}\ldots
z^{a_{k}}\overline{\boldsymbol{\Pi}}^{{\\{{\underline{\nu}}\\}}}_{\underline{\mu}}\big{(}\nabla_{a_{1}\cdots
a_{k}}^{k}\phi_{\underline{\nu}}|_{C(\sigma)}\big{)}\Big{)}$
$\displaystyle=z^{b_{1}}\cdots
z^{b_{r}}\partial_{b_{1}}\ldots\partial_{b_{r}}\big{(}z^{a_{1}}\ldots
z^{a_{k}}\big{)}\overline{\boldsymbol{\Pi}}^{{\\{{\underline{\nu}}\\}}}_{\underline{\mu}}\big{(}\nabla_{a_{1}\cdots
a_{k}}^{k}\phi_{\underline{\nu}}|_{C(\sigma)}\big{)}$
$\displaystyle=\frac{k!}{(k-r)!}\,\big{(}z^{a_{1}}\ldots
z^{a_{k}}\big{)}\overline{\boldsymbol{\Pi}}^{{\\{{\underline{\nu}}\\}}}_{\underline{\mu}}\big{(}\nabla_{a_{1}\cdots
a_{k}}^{k}\phi_{\underline{\nu}}|_{C(\sigma)}\big{)}=\frac{k!}{(k-r)!}\,\phi^{(k)}_{\underline{\mu}}\,.$

∎

###  3.7 All multipoles are Dixon multipoles

In the previous subsection we showed that the components of a Dixon multipole
are unique and can be extracted using particular text functions given by (55).
In this section we show that the all multipoles can be written as a Dixon
multipole. Thus we see that the Dixon multipole is merely a representation of
a multipole. Thus if a multipole is defined via (14) with respect to one Dixon
vector, $N_{\mu}$, then it is guaranteed that it can be written with respect
to another Dixon vector $\hat{N}_{\mu}$. A similar result is available in [12]
but here an explicit formula is given, which in addition respects the
constraint (3).

We assume we are given an arbitrary multipole $Y^{{\underline{\mu}}}$, i.e. an
operator which takes test tensors $\phi_{{\underline{\mu}}}$ to give a number
$Y^{{\underline{\mu}}}[\phi_{{\underline{\mu}}}]$. In [3] we show how to
construct an arbitrary distribution. This is by taking a collection of
monopoles, and allowing finite sums, derivatives, contractions and products
with scalar fields. In [3] we showed that all multipoles could be expressed as
Ellis multipoles. Thus another interpretation is that $Y^{{\underline{\mu}}}$
is given in the Ellis representation. An alternative is that
$Y^{{\underline{\mu}}}$ is given to us in the Dixon representation, but with a
different Dixon vector $N_{\mu}$. Even in this case it is not obvious that the
multipole can be represented as a Dixon multipole with the new Dixon vector.

Our goal is to establish $Y^{{\underline{\mu}}}$ can be written as a Dixon
multipole, but the only information we can use are the values
$Y^{{\underline{\mu}}}[\phi_{{\underline{\mu}}}]$ for particular
$\phi_{{\underline{\mu}}}$’s.

The result that all multipoles can be written as Dixon multipoles, should not
be surprising as if one were to count the number of components in the Ellis
representation, it is the same as the Dixon representation. However one should
be careful, as the components in the Ellis representation, also have the same
number of components, but they are not unique. So simply counting number of
components is not sufficient.

We can deduce the order of $Y^{{\underline{\mu}}}$ by the statement [3, eqn.
(115)]: The order of $Y^{{\underline{\mu}}}$ is the smallest $n$ such that

$\displaystyle
Y^{{\underline{\mu}}}[\lambda^{n+1}\phi_{{\underline{\mu}}}]=0\quad\text{for
all tensors $\phi_{{\underline{\mu}}}$ and all scalar fields $\lambda$ such
that $\lambda|_{C(\sigma)}=0$.}$ (64)

Having established the order of $Y^{{\underline{\mu}}}$ we can now perform the
Dixon split. I.e. using (55) we can create the multipoles
$Y^{{\underline{\mu}}}_{(k)}$

$\displaystyle
Y^{{\underline{\mu}}}_{(r)}[\phi_{{\underline{\mu}}}]={\sum_{k=r}^{n}}\frac{(-1)^{k-r}}{(k-r)!\
{r!}}Y^{{\underline{\mu}}}[\nabla_{R}^{k}\phi_{{\underline{\mu}}}]\,.$ (65)

We can now use (57) to calculate the components of $Y^{{\underline{\mu}}}$,

$\displaystyle\theta^{{\underline{\nu}}a_{1}\cdots
a_{k}}(\sigma_{0})=\lim_{\epsilon\to
0}Y^{{\underline{\mu}}}_{(k)}[(\phi_{\sigma_{0},\epsilon}^{{\underline{\nu}},a_{1}\cdots
a_{r}})_{\underline{\mu}}]\,.$ (66)

We emphasise here that $Y^{{\underline{\mu}}}$ was not defined using
$\theta^{{\underline{\nu}}a_{1}\cdots a_{k}}$, via (14), instead
$Y^{{\underline{\mu}}}$ was given to us and we have used various test
functions to calculate the components.

It is now necessary to establish that (14) gives us the correct distribution.
To do this we define a new distribution as

$\displaystyle
J^{{\underline{\mu}}}=\sum_{k=0}^{N}\frac{1}{k!}\,\nabla^{k}_{\rho_{1}\ldots\rho_{k}}\int_{\cal
I}\theta^{{\underline{\mu}}\rho_{1}\cdots\rho_{k}}\delta^{(4)}\big{(}x-C(\sigma)\big{)}\,d\sigma\,.$
(67)

It is then necessary to show $J^{{\underline{\mu}}}=Y^{{\underline{\mu}}}$.

######  Theorem 17.

$\displaystyle J^{{\underline{\mu}}}=Y^{{\underline{\mu}}}\,.$ (68)

Hence all all multipoles over $C$ are Dixon multipoles over $C$.

###### Proof.

$\displaystyle J^{{\underline{\mu}}}_{(k)}[\phi_{\underline{\mu}}]$
$\displaystyle=\frac{(-1)^{k}}{k!}\int_{\cal
I}\theta^{{\underline{\mu}}a_{1}\cdots a_{k}}(\sigma_{0})(\nabla_{a_{1}\cdots
a_{k}}^{k}\phi_{\underline{\nu}})|_{C(\sigma_{0})}\ d\sigma_{0}$
$\displaystyle=\frac{(-1)^{k}}{k!}\int_{\cal I}\lim_{\epsilon\to
0}Y^{{\underline{\mu}}}_{(k)}[(\phi_{\sigma_{0},\epsilon}^{{\underline{\nu}},a_{1}\cdots
a_{r}})_{\underline{\mu}}](\nabla_{a_{1}\cdots
a_{k}}^{k}\phi_{\underline{\nu}})|_{C(\sigma_{0})}\ d\sigma_{0}$
$\displaystyle=\frac{(-1)^{k}}{k!}\lim_{\epsilon\to
0}Y^{{\underline{\mu}}}_{(k)}\Big{[}\int_{\cal
I}(\phi_{\sigma_{0},\epsilon}^{{\underline{\nu}},a_{1}\cdots
a_{r}})_{\underline{\mu}}(\nabla_{a_{1}\cdots
a_{k}}^{k}\phi_{\underline{\nu}})|_{C(\sigma_{0})}\ d\sigma_{0}\Big{]}$
$\displaystyle=\frac{1}{k!}\lim_{\epsilon\to
0}Y^{{\underline{\mu}}}_{(k)}\Big{[}\int_{\cal
I}\frac{1}{\epsilon}z^{a_{1}}\cdots
z^{a_{r}}\Big{(}\psi_{0}\Big{(}\frac{\sigma-\sigma_{0}}{\epsilon}\Big{)}\psi_{0}(z^{1})\psi_{0}(z^{2})\psi_{0}(z^{3})\overline{\boldsymbol{\Pi}}^{{\\{{\underline{\nu}}\\}}}_{{\underline{\mu}}}\Big{)}(\nabla_{a_{1}\cdots
a_{k}}^{k}\phi_{\underline{\nu}})|_{C(\sigma_{0})}\ d\sigma_{0}\Big{]}$
$\displaystyle=\lim_{\epsilon\to
0}\frac{1}{k!\,\epsilon}Y^{{\underline{\mu}}}_{(k)}\Big{[}\int_{\cal
I}\psi_{0}\Big{(}\frac{\sigma-\sigma_{0}}{\epsilon}\Big{)}\psi_{0}(z^{1})\psi_{0}(z^{2})\psi_{0}(z^{3})z^{a_{1}}\cdots
z^{a_{r}}\overline{\boldsymbol{\Pi}}^{{\\{{\underline{\nu}}\\}}}_{{\underline{\mu}}}(\nabla_{a_{1}\cdots
a_{k}}^{k}\phi_{\underline{\nu}})|_{C(\sigma_{0})}\ d\sigma_{0}\Big{]}$
$\displaystyle=\lim_{\epsilon\to
0}\frac{1}{k!\,\epsilon}Y^{{\underline{\mu}}}_{(k)}\Big{[}\int_{\cal
I}\psi_{0}\Big{(}\frac{\sigma-\sigma_{0}}{\epsilon}\Big{)}\psi_{0}(z^{1})\psi_{0}(z^{2})\psi_{0}(z^{3})\phi^{(k)}_{\underline{\mu}}|_{C(\sigma_{0})}d\sigma_{0}\Big{]}$
$\displaystyle=\lim_{\epsilon\to
0}\frac{1}{k!\,\epsilon}Y^{{\underline{\mu}}}_{(k)}\Big{[}\int_{\cal
I}\psi_{0}\Big{(}\frac{\sigma-\sigma_{0}}{\epsilon}\Big{)}\phi^{(k)}_{\underline{\mu}}|_{C(\sigma_{0})}d\sigma_{0}\Big{]}$
$\displaystyle=\frac{1}{k!}Y^{{\underline{\mu}}}_{(k)}\Big{[}\lim_{\epsilon\to
0}\epsilon^{-1}\int_{\cal
I}\psi_{0}\Big{(}\frac{\sigma-\sigma_{0}}{\epsilon}\Big{)}\phi^{(k)}_{\underline{\mu}}|_{C(\sigma_{0})}d\sigma_{0}\Big{]}$
$\displaystyle=\frac{1}{k!}Y^{{\underline{\mu}}}_{(k)}[\phi^{(k)}_{\underline{\mu}}]=\frac{1}{k!}{\sum_{\ell=k}^{n}}\frac{(-1)^{\ell-k}}{(\ell-k)!\
{k!}}Y^{{\underline{\mu}}}[\nabla_{R}^{\ell}\phi^{(k)}_{{\underline{\mu}}}]$
$\displaystyle=\frac{1}{k!}{\sum_{\ell=k}^{k}}\frac{(-1)^{\ell-k}}{(\ell-k)!\
{k!}}\,\frac{k!}{(k-\ell)!}\,Y^{{\underline{\mu}}}[\phi^{(k)}_{{\underline{\mu}}}]=\frac{1}{k!}Y^{{\underline{\mu}}}[\phi^{(k)}_{{\underline{\mu}}}]\,.$

Hence

$\displaystyle
J^{{\underline{\mu}}}[\phi_{\underline{\mu}}]=\sum_{k=0}^{N}J^{{\underline{\mu}}}_{(k)}[\phi_{\underline{\mu}}]=\sum_{k=0}^{N}\frac{1}{k!}Y^{{\underline{\mu}}}\Big{[}\phi^{(k)}_{\underline{\mu}}\Big{]}=Y^{{\underline{\mu}}}\Big{[}\sum_{k=0}^{N}\frac{1}{k!}\phi^{(k)}_{\underline{\mu}}\Big{]}=Y^{{\underline{\mu}}}[\phi_{\underline{\mu}}]\,.$

Since for any arbitrary distribution $Y^{{\underline{\mu}}}$ we can construct
using (65), (66) and (67) $J^{{\underline{\mu}}}$, then (68) implies all
multipoles over $C$ are Dixon multipoles over $C$.

Observe that between lines 7 and 8 of this derivation we swapped
$\lim_{\epsilon\to 0}$ and $Y^{{\underline{\mu}}}_{(k)}$. This is permitted
since $Y^{{\underline{\mu}}}_{(k)}$ consists of only a finite number of
derivatives and is hence continuous and

$\displaystyle\lim_{\epsilon\to 0}\epsilon^{-1}\int_{\cal
I}\psi_{0}\Big{(}\frac{\sigma-\sigma_{0}}{\epsilon}\Big{)}\phi^{(k)}_{\underline{\mu}}|_{C(\sigma_{0})}d\sigma_{0}\to\phi^{(k)}_{\underline{\mu}}|_{C(\sigma)}\,,$

in the LF-topology. ∎

######  Corollary 18.

Given a multipole $J^{{\underline{\mu}}}$ defined by (14) with respect to one
Dixon vector $N_{\mu}$, and given a different Dixon vector $\hat{N}_{\mu}$,
then there exist unique components $\hat{\zeta}^{{\underline{\mu}}a_{1}\cdots
a_{k}}$ so that we can write $J^{{\underline{\mu}}}$ with respect to
$\hat{N}_{\mu}$.

###### Proof.

Using theorem 17 and corollary 14. ∎

###  3.8 Parallel squeezed tensors

So far, moments have been discussed as components of the distribution. There
is also the intuitive notion of moments in terms of integrals over space,
multiplied by one or more coordinates.

In general relativity such objects are not covariant as they depend on both
the choice of the spatial hypersurface and the choice of the coordinates on
that hypersurface. The other issue is that the tensor has to be integrated,
and this requires transporting the tensors to the same point.

One choice, given in [3] is to use an adapted coordinate system. This is
coordinate dependent. However by squeezing the tensor field, one can produce a
well defined distribution. This corresponds to the Ellis representation of the
distribution.

Dixon [8] gives a natural choice of moments. This uses the Dixon geodesic
hypersurfaces, the geodesic coordinates and parallel transport. Thus all the
necessary structure is given to us uniquely, once we have chosen the Dixon
vector.

In this section we give a method of squeezing a tensor field, such that the
coefficients of the expansion are the Dixon components of the distribution.

Let $U^{\underline{\mu}}$ be a tensor density (of weight 1) on ${\cal M}$.
From this we can construct the squeezed tensor density
$U_{\epsilon}^{\underline{\mu}}\in\Gamma TM$ given by

$\displaystyle
U_{\epsilon}^{\underline{\mu}}|_{P(\sigma,V)}=\epsilon^{-3}\,\Pi^{\underline{\mu}}_{\underline{\nu}}\big{|}^{P(\sigma,\epsilon^{-1}V}_{P(\sigma,V)}\,U^{\underline{\nu}}|_{P(\sigma,\epsilon^{-1}V)}\,.$
(69)

Let

$\displaystyle\xi^{{\underline{\mu}}\rho_{1}\cdots\rho_{k}}(\sigma)=(-1)^{k}\int_{{N^{\perp}}(\sigma)}\,V^{\rho_{1}}\cdots
V^{\rho_{k}}\,\Pi^{\underline{\mu}}_{\underline{\nu}}\big{|}^{P(\sigma,V)}_{C(\sigma)}\
U^{\underline{\nu}}|_{P(\sigma,V)}\,d^{3}V\,.$ (70)

These components clearly satisfy (14) and (15). In Dixon geodesics coordinates
this becomes

$\displaystyle\xi^{{\underline{\mu}}\rho_{1}\cdots\rho_{k}}(\sigma)=(-1)^{k}\int_{\Sigma(\sigma)}\,z^{\rho_{1}}\cdots
z^{\rho_{k}}\,U^{\underline{\mu}}|_{(\sigma,z^{1},z^{2},z^{3})}\,d^{3}z\,.$
(71)

######  Lemma 19.

$\displaystyle\int_{\cal
M}U_{\epsilon}^{\underline{\mu}}\,\phi_{\underline{\mu}}\,d^{4}x$
$\displaystyle=\sum_{k=0}^{N}(-1)^{k}\frac{\epsilon^{k}}{k!}\int_{\cal
I}\,\xi^{{\underline{\mu}}\rho_{1}\cdots\rho_{k}}\,\nabla^{k}_{\rho_{1}\cdots\rho_{k}}\phi_{\underline{\mu}}\,d\sigma+O(\epsilon^{N+1})\,.$
(72)

Thus

$\displaystyle
U^{\underline{\mu}}_{\epsilon}=\sum_{k=0}^{N}\frac{\epsilon^{k}}{k!}\nabla^{k}_{\rho_{1}\cdots\rho_{k}}\int_{\cal
I}d\sigma\,\xi^{{\underline{\mu}}\rho_{1}\cdots\rho_{k}}\,\delta^{(4)}(x-C)+O(\epsilon^{N+1})\,.$
(73)

###### Proof.

From (60) and (62) we have

$\displaystyle\Pi^{\underline{\mu}}_{\underline{\nu}}\big{|}^{C(\sigma)}_{P(\sigma,sV)}\phi_{\underline{\mu}}|_{P(\sigma,sV)}$
$\displaystyle=\sum_{k=0}^{N}\frac{s^{k}}{k!}\,\nabla_{V}^{k}\phi_{\underline{\nu}}|_{C(\sigma)}+O(s^{N+1})\,.$

In Dixon geodesic coordinates $(\sigma,{z^{1}},z^{2},z^{3})$. We set the
vector field $V^{\mu}=\epsilon\,W^{\mu}$ and do a Taylor expansion in
$\epsilon$.

$\displaystyle\int_{\cal M}$ $\displaystyle
U_{\epsilon}^{\underline{\mu}}\,\phi_{\underline{\mu}}\,d\sigma\,d^{3}z$
$\displaystyle=\int_{\cal
I}d\sigma\int_{\Sigma(\sigma)}U_{\epsilon}^{\underline{\mu}}\,\phi_{\underline{\mu}}\,d^{3}z$
$\displaystyle=\int_{\cal
I}d\sigma\int_{{N^{\perp}}(\sigma)}U_{\epsilon}^{\underline{\mu}}|_{P(\sigma,V)}\,\phi_{\underline{\mu}}|_{P(\sigma,V)}\,d^{3}V$
$\displaystyle=\epsilon^{-3}\int_{\cal
I}d\sigma\int_{{N^{\perp}}(\sigma)}\Pi^{\underline{\mu}}_{\underline{\nu}}\big{|}^{P(\sigma,\epsilon^{-1}V)}_{P(\sigma,V)}\,U^{\underline{\nu}}|_{P(\sigma,\epsilon^{-1}V)}\
\phi_{\underline{\mu}}|_{P(\sigma,V)}\,d^{3}V$
$\displaystyle=\epsilon^{-3}\int_{\cal
I}d\sigma\int_{{N^{\perp}}(\sigma)}\Pi^{\underline{\mu}}_{\underline{\lambda}}\big{|}^{C(\sigma)}_{P(\sigma,V)}\
\Pi^{\underline{\lambda}}_{\underline{\nu}}\big{|}^{P(\sigma,\epsilon^{-1}V)}_{C(\sigma)}\
\,U^{\underline{\nu}}|_{P(\sigma,\epsilon^{-1}V)}\
\phi_{\underline{\mu}}|_{P(\sigma,V)}\,d^{3}V$ $\displaystyle=\int_{\cal
I}d\sigma\int_{{N^{\perp}}(\sigma)}\Pi^{\underline{\lambda}}_{\underline{\nu}}\big{|}^{P(\sigma,W)}_{C(\sigma)}\,U^{\underline{\nu}}|_{P(\sigma,W)}\,\Pi^{\underline{\mu}}_{\underline{\lambda}}\big{|}^{C(\sigma)}_{P(\sigma,\epsilon
W)}\ \phi_{\underline{\mu}}|_{P(\sigma,\epsilon W)}\,d^{3}W$
$\displaystyle=\int_{\cal
I}d\sigma\int_{{N^{\perp}}(\sigma)}\Pi^{\underline{\lambda}}_{\underline{\nu}}\big{|}^{P(\sigma,W)}_{C(\sigma)}\,U^{\underline{\nu}}|_{P(\sigma,W)}\,d^{3}W\bigg{(}\sum_{k=0}^{N}\frac{\epsilon^{k}}{k!}\nabla_{W}^{k}\phi_{\underline{\lambda}}|_{C(\sigma)}\bigg{)}+O(\epsilon^{N+1})$
$\displaystyle=\sum_{k=0}^{N}\frac{\epsilon^{k}}{k!}\int_{\cal
I}d\sigma\int_{{N^{\perp}}(\sigma)}\Pi^{\underline{\lambda}}_{\underline{\nu}}\big{|}^{P(\sigma,W)}_{C(\sigma)}\,U^{\underline{\nu}}|_{P(\sigma,W)}\,W^{\rho_{1}}\cdots
W^{\rho_{k}}\,d^{3}W\big{(}\nabla^{k}_{\rho_{1}\cdots\rho_{k}}\phi_{\underline{\lambda}}|_{C(\sigma)}\big{)}+O(\epsilon^{N+1})$
$\displaystyle=\sum_{k=0}^{N}(-1)^{k}\frac{\epsilon^{k}}{k!}\int_{\cal
I}d\sigma\,\xi^{{\underline{\lambda}}\rho_{1}\cdots\rho_{k}}(\sigma)\big{(}\nabla^{k}_{\rho_{1}\cdots\rho_{k}}\phi_{\underline{\lambda}}|_{C(\sigma)}\big{)}+O(\epsilon^{N+1})\,.$

∎

Thus we have shown the link between the components of the multipoles and the
moments of a regular distribution.

##  4 Dynamical equations in Dixon language

In this section we derive the key result of this article, namely the dynamical
equations of the stress-energy quadrupole, in the Dixon representation. These
follow from the divergenceless condition (13). As stated this has the
advantage that they are tensorial. The tensorial expression is given below in
theorem 21. We first derive the equation in Dixon adapted coordinates.

######  Theorem 20.

In Dixon adapted coordinates the divergenceless condition (13) corresponds to
the following dynamical equations for the components

$\displaystyle\nabla_{0}\xi^{\mu 0bc}$ $\displaystyle=-2\xi^{\mu(bc)}\,,$ (74)
$\displaystyle\nabla_{0}\xi^{\mu 0a}$ $\displaystyle=-\xi^{\mu
a}\,+\tfrac{1}{2}\xi^{\mu 0bc}\,R^{a}{}_{b0c}+\xi^{\rho 0ac}\,R^{\mu}{}_{\rho
0c}+\tfrac{1}{2}\xi^{\rho cba}\,R^{\mu}{}_{\rho cb}+\tfrac{1}{6}\xi^{\mu
dbc}\,R^{a}{}_{bdc}\,,$ (75) $\displaystyle\nabla_{0}\xi^{\mu 0}$
$\displaystyle=\xi^{\rho 0b}\,R^{\mu}{}_{\rho 0b}\,+\tfrac{1}{2}\xi^{\rho
ab}\,R^{\mu}{}_{\rho ab}\,+\tfrac{1}{2}\nabla_{0}\big{(}\xi^{\mu
0bc}\,R^{0}{}_{b0c}\big{)}\,+\tfrac{1}{2}\xi^{\rho
0bc}\,(\nabla_{c}R^{\mu}{}_{\rho 0b})+\tfrac{1}{6}\nabla_{0}\big{(}\xi^{\mu
abc}\,R^{0}{}_{bac}\big{)}\,$ $\displaystyle\qquad-\tfrac{1}{3}\xi^{\rho
abc}\,(\nabla_{c}R^{\mu}{}_{\rho ab})\,,$ (76)

together with the constraint

$\displaystyle\xi^{\mu(abc)}$ $\displaystyle=0\,.$ (77)

###### Proof.

For this proof, all integrals are implicitly over ${\cal I}$. Let

$\displaystyle
F^{\mu}=\nabla_{\sigma}T^{\sigma\mu}=F_{(1)}^{\mu}+F_{(2)}^{\mu}+F_{(3)}^{\mu}\,,$

where

$\displaystyle
F_{(1)}^{\mu}=\nabla_{\nu}\int\xi^{\mu\nu}\delta(x-C)d\sigma,\quad
F_{(2)}^{\mu}=\nabla_{\nu}\nabla_{\rho}\int\xi^{\mu\nu\rho}\delta(x-C)d\sigma,\quad
F_{(3)}^{\mu}=\tfrac{1}{2}\nabla_{\nu}\nabla^{2}_{\rho\sigma}\int\xi^{\mu\nu\rho\sigma}\delta(x-C)d\sigma\,.$
(78)

Manipulating $F^{\mu}_{(1)}$, $F^{\mu}_{(2)}$ and $F^{\mu}_{(3)}$ in turn.

$\displaystyle F^{\mu}_{(1)}[\phi_{\mu}]$
$\displaystyle=-\int\xi^{\mu\nu}\,\nabla_{\nu}\,\phi_{\mu}\,d\sigma=-\int\xi^{\mu
a}\,\nabla_{a}\,\phi_{\mu}\,d\sigma-\int\xi^{\mu
0}\,\nabla_{0}\phi_{\mu}\,d\sigma$ $\displaystyle=-\int\xi^{\mu
a}\,\nabla_{a}\phi_{\mu}\,d\sigma+\int(\nabla_{0}\xi^{\mu
0})\,\phi_{\mu}\,d\sigma\,,$

which using (52) in lemma 10 gives

$\displaystyle F^{\mu}_{(1)}[\nabla^{3}_{R}\phi_{\mu}]$
$\displaystyle=F^{\mu}_{(1)}[\nabla^{2}_{R}\phi_{\mu}]=0,$ (79) $\displaystyle
F^{\mu}_{(1)}[\nabla_{R}\phi_{\mu}]$ $\displaystyle=-\int\xi^{\mu
a}\,\nabla_{a}\phi_{\mu}\,d\sigma,$ (80)
$\displaystyle\qquad\textup{and}\qquad F^{\mu}_{(1)}[\phi_{\mu}]$
$\displaystyle=F^{\mu}_{(1)}[\nabla_{R}\phi_{\mu}]+\int(\nabla_{0}\xi^{\mu
0})\,\phi_{\mu}\,d\sigma\,.$ (81)

For $F^{\mu}_{(2)}[\phi_{\mu}]$ we have

$\displaystyle F_{(2)}^{\mu}\big{[}\phi_{\mu}\big{]}$
$\displaystyle=\int\xi^{\mu\nu\rho}\,\nabla_{\rho}\nabla_{\nu}\,\phi_{\mu}\,d\sigma=\int\xi^{\mu
0b}\,\nabla_{b}\nabla_{0}\,\phi_{\mu}\,d\sigma+\int\xi^{\mu
ab}\,\nabla_{b}\nabla_{a}\,\phi_{\mu}\,d\sigma$ $\displaystyle=\int\xi^{\mu
0b}\,\nabla_{0}\nabla_{b}\,\phi_{\mu}\,d\sigma-\int\xi^{\mu
0b}\,R^{\rho}{}_{\mu
0b}\,\phi_{\rho}\,d\sigma+\int\xi^{\mu(ab)}\,\nabla_{a}\nabla_{b}\,\phi_{\mu}\,d\sigma-\tfrac{1}{2}\int\xi^{\mu
ab}\,R^{\rho}{}_{\mu ab}\,\phi_{\rho}\,d\sigma$
$\displaystyle=-\int(\nabla_{0}\xi^{\mu
0b})\nabla_{b}\,\phi_{\mu}\,d\sigma-\int\xi^{\mu 0b}\,R^{\rho}{}_{\mu
0b}\,\phi_{\rho}\,d\sigma+\int\xi^{\mu(ab)}\,\nabla_{ab}\phi_{\mu}\,d\sigma-\tfrac{1}{2}\int\xi^{\mu
ab}\,R^{\rho}{}_{\mu ab}\,\phi_{\rho}\,d\sigma\,.$

Using equation (52) we can split $F^{\mu}_{(2)}[\phi_{\mu}]$ to give

$\displaystyle F_{(2)}^{\mu}\big{[}\nabla^{3}_{R}\phi_{\mu}\big{]}$
$\displaystyle=0$ (82) $\displaystyle
F_{(2)}^{\mu}\big{[}\nabla^{2}_{R}\phi_{\mu}\big{]}$
$\displaystyle=2\int\xi^{\mu(ab)}\,\nabla_{ab}\phi_{\mu}\,d\sigma$ (83)
$\displaystyle F_{(2)}^{\mu}\big{[}\nabla_{R}\phi_{\mu}\big{]}$
$\displaystyle=\tfrac{1}{2}F_{(2)}^{\mu}\big{[}\nabla^{2}_{R}\phi_{\mu}\big{]}-\int(\nabla_{0}\xi^{\mu
0a})\nabla_{a}\,\phi_{\mu}\,d\sigma$ (84) $\displaystyle
F_{(2)}^{\mu}\big{[}\phi_{\mu}\big{]}$
$\displaystyle=F_{(2)}^{\mu}\big{[}\nabla_{R}\phi_{\mu}\big{]}-\int\xi^{\rho
0b}\,R^{\mu}{}_{\rho 0b}\,\phi_{\mu}\,d\sigma-\tfrac{1}{2}\int\xi^{\rho
ab}\,R^{\mu}{}_{\rho ab}\,\phi_{\mu}\,d\sigma\,.$ (85)

Now consider $F^{\mu}_{(3)}[\phi_{\mu}]$.

$\displaystyle F_{(3)}^{\mu}\big{[}\phi_{\mu}\big{]}$
$\displaystyle=-\frac{1}{2}\int\xi^{\mu\nu\rho\sigma}\,\nabla_{\sigma}\nabla_{\rho}\nabla_{\nu}\,\phi_{\mu}\,d\sigma=-\frac{1}{2}\int\xi^{\mu
abc}\,\nabla_{c}\nabla_{b}\nabla_{a}\,\phi_{\mu}\,d\sigma-\frac{1}{2}\int\xi^{\mu
0bc}\,\nabla_{c}\nabla_{b}\nabla_{0}\,\phi_{\mu}\,d\sigma\,.$ (86)

The second term on the right hand side of (86) gives

$\displaystyle\int\xi^{\mu
0bc}\,\nabla_{c}\nabla_{b}\nabla_{0}\,\phi_{\mu}\,d\sigma$
$\displaystyle=\int\xi^{\mu
0bc}\,\nabla_{c}\nabla_{0}\nabla_{b}\,\phi_{\mu}\,d\sigma-\int\xi^{\mu
0bc}\,\nabla_{c}(R^{\rho}{}_{\mu 0b}\,\phi_{\rho})\,d\sigma$
$\displaystyle=\int\xi^{\mu
0bc}\,\nabla_{0}\nabla_{c}\nabla_{b}\,\phi_{\mu}\,d\sigma-\int\xi^{\mu
0bc}\,\Big{(}R^{\rho}{}_{b0c}\nabla_{\rho}\,\phi_{\mu}+R^{\rho}{}_{\mu
0c}\nabla_{b}\,\phi_{\rho}+\nabla_{c}R^{\rho}{}_{\mu
0b}\,\phi_{\rho}+R^{\rho}{}_{\mu 0b}\,\nabla_{c}\phi_{\rho}\Big{)}\,d\sigma$
$\displaystyle=\int\xi^{\mu
0bc}\,\nabla_{0}\nabla_{c}\nabla_{b}\,\phi_{\mu}\,d\sigma-\int\xi^{\mu
0bc}\,\Big{(}R^{0}{}_{b0c}\nabla_{0}\,\phi_{\mu}+R^{d}{}_{b0c}\nabla_{d}\,\phi_{\mu}+R^{\rho}{}_{\mu
0c}\nabla_{b}\,\phi_{\rho}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad+\nabla_{c}R^{\rho}{}_{\mu
0b}\,\phi_{\rho}+R^{\rho}{}_{\mu 0b}\,\nabla_{c}\phi_{\rho}\Big{)}\,d\sigma$
$\displaystyle=-\int\nabla_{0}(\xi^{\mu
0bc})\,\nabla^{2}_{cb}\,\phi_{\mu}\,d\sigma+\int\nabla_{0}\Big{(}\xi^{\mu
0bc}\,R^{0}{}_{b0c}\Big{)}\,\phi_{\mu}\,d\sigma$
$\displaystyle\qquad-\int\xi^{\mu
0bc}\,\Big{(}R^{d}{}_{b0c}\nabla_{d}\,\phi_{\mu}+R^{\rho}{}_{\mu
0c}\nabla_{b}\,\phi_{\rho}+\nabla_{c}R^{\rho}{}_{\mu
0b}\,\phi_{\rho}+R^{\rho}{}_{\mu 0b}\,\nabla_{c}\phi_{\rho}\Big{)}\,d\sigma$
$\displaystyle=-\int\nabla_{0}(\xi^{\mu
0bc})\,\nabla^{2}_{cb}\,\phi_{\mu}\,d\sigma+\int\Big{(}\nabla_{0}\big{(}\xi^{\mu
0bc}\,R^{0}{}_{b0c}\big{)}\,+\xi^{\rho 0bc}\,(\nabla_{c}R^{\mu}{}_{\rho
0b})\Big{)}\phi_{\mu}\,d\sigma$ $\displaystyle\qquad-\int\Big{(}\xi^{\mu
0bc}\,R^{a}{}_{b0c}+2\xi^{\rho 0ac}\,R^{\mu}{}_{\rho
0c}\Big{)}\nabla_{a}\phi_{\mu}\,d\sigma\,.$

For the first term on the right hand side of (86), it is necessary to
symmetrise the $abc$ of $\xi^{abc}$.

$\displaystyle\int\xi^{\mu
abc}\,\nabla_{c}\nabla_{b}\nabla_{a}\,\phi_{\mu}\,d\sigma$
$\displaystyle=\int\xi^{\mu
abc}\,\nabla_{c}\nabla_{a}\nabla_{b}\,\phi_{\mu}\,d\sigma-\int\xi^{\mu
abc}\,\nabla_{c}(R^{\rho}{}_{\mu ab}\,\phi_{\rho})\,d\sigma$
$\displaystyle=\int\xi^{\mu
abc}\,\nabla_{c}\nabla_{a}\nabla_{b}\,\phi_{\mu}\,d\sigma-\int\xi^{\mu
abc}\,\Big{(}(\nabla_{c}R^{\rho}{}_{\mu ab})\phi_{\rho}+R^{\rho}{}_{\mu
ab}\,\nabla_{c}\phi_{\rho}\Big{)}\,d\sigma$ $\displaystyle=\int\xi^{\mu
abc}\,\nabla_{c}\nabla_{a}\nabla_{b}\,\phi_{\mu}\,d\sigma-\int\xi^{\rho
cba}\,\Big{(}(\nabla_{a}R^{\mu}{}_{\rho cb})\phi_{\mu}+R^{\mu}{}_{\rho
cb}\,\nabla_{a}\phi_{\mu}\Big{)}\,d\sigma\,.$

Continuing

$\displaystyle\int\xi^{\mu
abc}\,\nabla_{c}\nabla_{b}\nabla_{a}\,\phi_{\mu}\,d\sigma$
$\displaystyle=\int\xi^{\mu
abc}\,\nabla_{a}\nabla_{c}\nabla_{b}\,\phi_{\mu}\,d\sigma-\int\xi^{\mu
abc}\,\Big{(}R^{\rho}{}_{bac}\nabla_{\rho}\,\phi_{\mu}+R^{\rho}{}_{\mu
ac}\nabla_{b}\,\phi_{\rho}+(\nabla_{c}R^{\rho}{}_{\mu
ab})\phi_{\rho}+R^{\rho}{}_{\mu ab}\,\nabla_{c}\phi_{\rho}\Big{)}\,d\sigma$
$\displaystyle=\int\xi^{\mu
abc}\,\nabla_{a}\nabla_{c}\nabla_{b}\,\phi_{\mu}\,d\sigma-\int\xi^{\mu
abc}\,\Big{(}R^{\rho}{}_{bac}\nabla_{\rho}\,\phi_{\mu}+2R^{\rho}{}_{\mu
ac}\nabla_{b}\,\phi_{\rho}+(\nabla_{c}R^{\rho}{}_{\mu
ab})\phi_{\rho}\Big{)}\,d\sigma$ $\displaystyle=\int\xi^{\mu
abc}\,\nabla_{a}\nabla_{c}\nabla_{b}\,\phi_{\mu}\,d\sigma-\int\xi^{\mu
abc}\,\Big{(}R^{0}{}_{bac}\nabla_{0}\,\phi_{\mu}+R^{d}{}_{bac}\nabla_{d}\,\phi_{\mu}+2R^{\rho}{}_{\mu
ac}\nabla_{b}\,\phi_{\rho}+(\nabla_{c}R^{\rho}{}_{\mu
ab})\phi_{\rho}\Big{)}\,d\sigma$ $\displaystyle=\int\xi^{\mu
abc}\,\nabla_{a}\nabla_{c}\nabla_{b}\,\phi_{\mu}\,d\sigma+\int\nabla_{0}\Big{(}\xi^{\mu
abc}\,R^{0}{}_{bac}\Big{)}\,\phi_{\mu}\,d\sigma$
$\displaystyle\quad-\int\xi^{\mu
abc}\,\Big{(}R^{d}{}_{bac}\nabla_{d}\,\phi_{\mu}+2R^{\rho}{}_{\mu
ac}\nabla_{b}\,\phi_{\rho}+(\nabla_{c}R^{\rho}{}_{\mu
ab})\phi_{\rho}\Big{)}\,d\sigma$ $\displaystyle=\int\xi^{\mu
abc}\,\nabla_{a}\nabla_{c}\nabla_{b}\,\phi_{\mu}\,d\sigma+\int\Big{(}-\xi^{\mu
dbc}\,R^{a}{}_{bdc}-2\xi^{\rho bac}\,R^{\mu}{}_{\rho
bc}\Big{)}\nabla_{a}\,\phi_{\mu}\,d\sigma$
$\displaystyle\qquad+\int\Big{(}\nabla_{0}\big{(}\xi^{\mu
abc}\,R^{0}{}_{bac}\big{)}\,-\xi^{\rho abc}\,(\nabla_{c}R^{\mu}{}_{\rho
ab})\Big{)}\phi_{\mu}\,d\sigma\,.$

Hence

$\displaystyle 3\int\xi^{\mu abc}\,$
$\displaystyle\nabla_{c}\nabla_{b}\nabla_{a}\,\phi_{\mu}\,d\sigma$
$\displaystyle=\int\xi^{\mu
abc}\,\nabla_{c}\nabla_{b}\nabla_{a}\,\phi_{\mu}\,d\sigma$
$\displaystyle\qquad+\int\xi^{\mu
abc}\,\nabla_{c}\nabla_{a}\nabla_{b}\,\phi_{\mu}\,d\sigma-\int\xi^{\rho
cba}\,\Big{(}(\nabla_{a}R^{\mu}{}_{\rho cb})\phi_{\mu}+R^{\mu}{}_{\rho
cb}\,\nabla_{a}\phi_{\mu}\Big{)}\,d\sigma$ $\displaystyle\qquad+\int\xi^{\mu
abc}\,\nabla_{a}\nabla_{c}\nabla_{b}\,\phi_{\mu}\,d\sigma+\int\Big{(}-\xi^{\mu
dbc}\,R^{a}{}_{bdc}-2\xi^{\rho bac}\,R^{\mu}{}_{\rho
bc}\Big{)}\nabla_{a}\,\phi_{\mu}\,d\sigma$
$\displaystyle\qquad+\int\Big{(}\nabla_{0}\big{(}\xi^{\mu
abc}\,R^{0}{}_{bac}\big{)}\,-\xi^{\rho abc}\,(\nabla_{c}R^{\mu}{}_{\rho
ab})\Big{)}\phi_{\mu}\,d\sigma$
$\displaystyle=3\int\xi^{\mu(abc)}\,\nabla_{(abc)}\phi_{\mu}\,d\sigma+\int\Big{(}-\xi^{\rho
cba}\,R^{\mu}{}_{\rho cb}-\xi^{\mu dbc}\,R^{a}{}_{bdc}-2\xi^{\rho
bac}\,R^{\mu}{}_{\rho bc}\Big{)}\nabla_{a}\,\phi_{\mu}\,d\sigma$
$\displaystyle\qquad+\int\Big{(}-\xi^{\rho cba}\,(\nabla_{a}R^{\mu}{}_{\rho
cb})+\nabla_{0}\big{(}\xi^{\mu abc}\,R^{0}{}_{bac}\big{)}\,-\xi^{\rho
abc}\,(\nabla_{c}R^{\mu}{}_{\rho ab})\Big{)}\phi_{\mu}\,d\sigma$
$\displaystyle=3\int\xi^{\mu(abc)}\,\nabla_{(abc)}\phi_{\mu}\,d\sigma+\int\Big{(}-\xi^{\mu
dbc}\,R^{a}{}_{bdc}-3\xi^{\rho bac}\,R^{\mu}{}_{\rho
bc}\Big{)}\nabla_{a}\,\phi_{\mu}\,d\sigma$
$\displaystyle\qquad+\int\Big{(}\nabla_{0}\big{(}\xi^{\mu
abc}\,R^{0}{}_{bac}\big{)}\,-2\xi^{\rho abc}\,(\nabla_{c}R^{\mu}{}_{\rho
ab})\Big{)}\phi_{\mu}\,d\sigma\,.$

Hence (86) becomes

$\displaystyle F_{(3)}^{\mu}\big{[}\phi_{\mu}\big{]}$
$\displaystyle=\tfrac{1}{2}\int\nabla_{0}(\xi^{\mu
0bc})\,\nabla^{2}_{cb}\,\phi_{\mu}\,d\sigma-\tfrac{1}{2}\int\Big{(}\nabla_{0}\big{(}\xi^{\mu
0bc}\,R^{0}{}_{b0c}\big{)}\,+\xi^{\rho 0bc}\,(\nabla_{c}R^{\mu}{}_{\rho
0b})\Big{)}\phi_{\mu}\,d\sigma$
$\displaystyle\qquad+\tfrac{1}{2}\int\Big{(}\xi^{\mu
0bc}\,R^{a}{}_{b0c}+2\xi^{\rho 0ac}\,R^{\mu}{}_{\rho
0c}\Big{)}\nabla_{a}\phi_{\mu}\,d\sigma$
$\displaystyle\qquad-\tfrac{1}{2}\int\xi^{\mu(abc)}\,\nabla^{3}_{abc}\phi_{\mu}\,d\sigma+\tfrac{1}{6}\int\Big{(}\xi^{\mu
dbc}\,R^{a}{}_{bdc}+3\xi^{\rho bac}\,R^{\mu}{}_{\rho
bc}\Big{)}\nabla_{a}\,\phi_{\mu}\,d\sigma$
$\displaystyle\qquad+\tfrac{1}{6}\int\Big{(}-\nabla_{0}\big{(}\xi^{\mu
abc}\,R^{0}{}_{bac}\big{)}\,+2\xi^{\rho abc}\,(\nabla_{c}R^{\mu}{}_{\rho
ab})\Big{)}\phi_{\mu}\,d\sigma$
$\displaystyle=-\tfrac{1}{2}\int\xi^{\mu(abc)}\,\nabla^{3}_{abc}\phi_{\mu}\,d\sigma+\tfrac{1}{2}\int\nabla_{0}(\xi^{\mu
0bc})\,\nabla^{2}_{cb}\,\phi_{\mu}\,d\sigma$
$\displaystyle\qquad+\int\Big{(}\tfrac{1}{2}\xi^{\mu
0bc}\,R^{a}{}_{b0c}+\xi^{\rho 0ac}\,R^{\mu}{}_{\rho 0c}R^{\mu}{}_{\rho
cb}+\tfrac{1}{6}\xi^{\mu dbc}\,R^{a}{}_{bdc}+\tfrac{1}{2}\xi^{\rho
bac}\,R^{\mu}{}_{\rho bc}\Big{)}\nabla_{a}\,\phi_{\mu}\,d\sigma$
$\displaystyle\qquad+\int\Big{(}-\tfrac{1}{2}\nabla_{0}\big{(}\xi^{\mu
0bc}\,R^{0}{}_{b0c}\big{)}\,-\tfrac{1}{2}\xi^{\rho
0bc}\,(\nabla_{c}R^{\mu}{}_{\rho 0b})-\tfrac{1}{6}\nabla_{0}\big{(}\xi^{\mu
abc}\,R^{0}{}_{bac}\big{)}\,+\tfrac{1}{3}\xi^{\rho
abc}\,(\nabla_{c}R^{\mu}{}_{\rho ab})\Big{)}\phi_{\mu}\,d\sigma\,.$

Splitting $F^{\mu}_{(3)}[\phi_{\mu}]$ gives

$\displaystyle F^{\mu}_{(3)}[\nabla^{3}_{R}\phi_{\mu}]$
$\displaystyle=-3\int\xi^{\mu(abc)}\,\nabla^{3}_{abc}\phi_{\mu}\,d\sigma\,,$
(87) $\displaystyle F^{\mu}_{(3)}[\nabla^{2}_{R}\phi_{\mu}]$
$\displaystyle=\tfrac{1}{3}F^{\mu}_{(3)}[\nabla^{3}_{R}\phi_{\mu}]+\int\nabla_{0}(\xi^{\mu
0bc})\,\nabla^{2}_{cb}\,\phi_{\mu}\,d\sigma\,,$ (88) $\displaystyle
F^{\mu}_{(3)}[\nabla_{R}\phi_{\mu}]$
$\displaystyle=\tfrac{1}{2}F^{\mu}_{(3)}[\nabla^{2}_{R}\phi_{\mu}]+\int\Big{(}\tfrac{1}{2}\xi^{\mu
0bc}\,R^{a}{}_{b0c}+\xi^{\rho 0ac}\,R^{\mu}{}_{\rho 0c}+\tfrac{1}{6}\xi^{\mu
dbc}\,R^{a}{}_{bdc}+\tfrac{1}{2}\xi^{\rho bac}\,R^{\mu}{}_{\rho
bc}\Big{)}\nabla_{a}\,\phi_{\mu}\,d\sigma\,,$ (89) $\displaystyle
F^{\mu}_{(3)}[\phi_{\mu}]$
$\displaystyle=F^{\mu}_{(3)}[\nabla_{R}\phi_{\mu}]+\int\Big{(}-\tfrac{1}{2}\nabla_{0}\big{(}\xi^{\mu
0bc}\,R^{0}{}_{b0c}\big{)}\,-\tfrac{1}{2}\xi^{\rho
0bc}\,(\nabla_{c}R^{\mu}{}_{\rho 0b})-\tfrac{1}{6}\nabla_{0}\big{(}\xi^{\mu
abc}\,R^{0}{}_{bac}\big{)}\,$
$\displaystyle\qquad\qquad\qquad\qquad+\tfrac{1}{3}\xi^{\rho
abc}\,(\nabla_{c}R^{\mu}{}_{\rho ab})\Big{)}\phi_{\mu}\,d\sigma\,.$ (90)

Since $F^{\mu}=0$, then from corollary 12,

$\displaystyle 0$
$\displaystyle=F^{\mu}[\nabla^{3}_{R}\phi_{\mu}]=F^{\mu}_{(3)}[\nabla^{3}_{R}\phi_{\mu}]\,,$
(91) $\displaystyle 0$
$\displaystyle=F^{\mu}[\nabla^{2}_{R}\phi_{\mu}]=F^{\mu}_{(3)}[\nabla^{2}_{R}\phi_{\mu}]+F^{\mu}_{(2)}[\nabla^{2}_{R}\phi_{\mu}]\,,$
(92) $\displaystyle 0$
$\displaystyle=F^{\mu}[\nabla_{R}\phi_{\mu}]=F^{\mu}_{(3)}[\nabla_{R}\phi_{\mu}]+F^{\mu}_{(2)}[\nabla_{R}\phi_{\mu}]+F^{\mu}_{(1)}[\nabla_{R}\phi_{\mu}]\,,$
(93) $\displaystyle 0$
$\displaystyle=F^{\mu}[\phi_{\mu}]=F^{\mu}_{(3)}[\phi_{\mu}]+F^{\mu}_{(2)}[\phi_{\mu}]+F^{\mu}_{(1)}[\phi_{\mu}]\,.$
(94)

From (91) and (87) we get (77).

From (92), (83) and (88) we can calculate $\nabla_{0}\xi^{\mu 0ab}$

$\displaystyle
0=F^{\mu}_{(3)}[\nabla^{2}_{R}\phi_{\mu}]+F^{\mu}_{(2)}[\nabla^{2}_{R}\phi_{\mu}]$
$\displaystyle=2\int\xi^{\mu(ab)}\,\nabla_{ab}\phi_{\mu}\,d\sigma+\int\nabla_{0}(\xi^{\mu
0ab})\,\nabla^{2}_{ab}\,\phi_{\mu}\,d\sigma\,,$

which gives (74).

From (93), (80), (84) and (89) we can calculate $\nabla_{0}\xi^{\mu 0a}$

$\displaystyle 0$
$\displaystyle=F^{\mu}[\nabla_{R}\phi_{\mu}]=F^{\mu}_{(3)}[\nabla_{R}\phi_{\mu}]+F^{\mu}_{(2)}[\nabla_{R}\phi_{\mu}]+F^{\mu}_{(1)}[\nabla_{R}\phi_{\mu}]$
$\displaystyle=-\int\xi^{\mu
a}\,\nabla_{a}\phi_{\mu}\,d\sigma+\tfrac{1}{2}F_{(2)}^{\mu}\big{[}\nabla^{2}_{R}\phi_{\mu}\big{]}-\int(\nabla_{0}\xi^{\mu
0a})\nabla_{a}\,\phi_{\mu}\,d\sigma$
$\displaystyle\quad+\tfrac{1}{2}F^{\mu}_{(3)}[\nabla^{2}_{R}\phi_{\mu}]+\int\Big{(}\tfrac{1}{2}\xi^{\mu
0bc}\,R^{a}{}_{b0c}+\xi^{\rho 0ac}\,R^{\mu}{}_{\rho 0c}++\tfrac{1}{6}\xi^{\mu
dbc}\,R^{a}{}_{bdc}+\tfrac{1}{2}\xi^{\rho bac}\,R^{\mu}{}_{\rho
bc}\Big{)}\nabla_{a}\,\phi_{\mu}\,d\sigma$ $\displaystyle=\int\Big{(}-\xi^{\mu
a}\,-(\nabla_{0}\xi^{\mu 0a})+\tfrac{1}{2}\xi^{\mu
0bc}\,R^{a}{}_{b0c}+\xi^{\rho 0ac}\,R^{\mu}{}_{\rho 0c}+\tfrac{1}{6}\xi^{\mu
dbc}\,R^{a}{}_{bdc}+\tfrac{1}{2}\xi^{\rho bac}\,R^{\mu}{}_{\rho
bc}\Big{)}\nabla_{a}\,\phi_{\mu}\,d\sigma\,,$

which gives (75).

From (94), (81), (85) and (90) we have

$\displaystyle 0$
$\displaystyle=F^{\mu}[\phi_{\mu}]=F^{\mu}_{(3)}[\phi_{\mu}]+F^{\mu}_{(2)}[\phi_{\mu}]+F^{\mu}_{(1)}[\phi_{\mu}]$
$\displaystyle=F^{\mu}_{(1)}[\nabla_{R}\phi_{\mu}]+\int(\nabla_{0}\xi^{\mu
0})\,\phi_{\mu}\,d\sigma$
$\displaystyle\quad+F_{(2)}^{\mu}\big{[}\nabla_{R}\phi_{\mu}\big{]}-\int\xi^{\rho
0b}\,R^{\mu}{}_{\rho 0b}\,\phi_{\mu}\,d\sigma-\tfrac{1}{2}\int\xi^{\rho
ab}\,R^{\mu}{}_{\rho ab}\,\phi_{\mu}\,d\sigma$
$\displaystyle\quad+F^{\mu}_{(3)}[\nabla_{R}\phi_{\mu}]+\int\Big{(}-\tfrac{1}{2}\nabla_{0}\big{(}\xi^{\mu
0bc}\,R^{0}{}_{b0c}\big{)}\,-\tfrac{1}{2}\xi^{\rho
0bc}\,(\nabla_{c}R^{\mu}{}_{\rho 0b})-\tfrac{1}{6}\nabla_{0}\big{(}\xi^{\mu
abc}\,R^{0}{}_{bac}\big{)}\,$
$\displaystyle\qquad\qquad\qquad\qquad+\tfrac{1}{3}\xi^{\rho
abc}\,(\nabla_{c}R^{\mu}{}_{\rho ab})\Big{)}\phi_{\mu}\,d\sigma$
$\displaystyle=\int\Big{(}(\nabla_{0}\xi^{\mu 0})\,-\xi^{\rho
0b}\,R^{\mu}{}_{\rho 0b}\,-\tfrac{1}{2}\xi^{\rho ab}\,R^{\mu}{}_{\rho
ab}\,-\tfrac{1}{2}\nabla_{0}\big{(}\xi^{\mu
0bc}\,R^{0}{}_{b0c}\big{)}\,-\tfrac{1}{2}\xi^{\rho
0bc}\,(\nabla_{c}R^{\mu}{}_{\rho 0b})-\tfrac{1}{6}\nabla_{0}\big{(}\xi^{\mu
abc}\,R^{0}{}_{bac}\big{)}\,$
$\displaystyle\qquad\qquad\qquad\qquad+\tfrac{1}{3}\xi^{\rho
abc}\,(\nabla_{c}R^{\mu}{}_{\rho ab})\Big{)}\phi_{\mu}\,d\sigma\,,$

which gives (76)

∎

As stated the key advantage with using the Dixon representation is that the
components are tensors. Therefore writing (74)-(77) using spacetime indices,
results in tensor equations which are valid for all coordinate systems. We
introduce the spatial projection tensor

$\displaystyle{\pi}^{\rho}_{\alpha}=\delta^{\rho}_{\alpha}-{\dot{C}}^{\rho}\,N_{\alpha}\,.$
(95)

######  Theorem 21.

The divergenceless condition (13) corresponds to the following tensor
equations for the components

$\displaystyle\nabla_{\dot{C}}(N_{\nu}\,\xi^{\mu\nu\rho\sigma})$
$\displaystyle=-2\,{\pi}^{\rho}_{\beta}\,{\pi}^{\sigma}_{\alpha}\,\xi^{\mu(\beta\alpha)}\,,$
(96) $\displaystyle\nabla_{\dot{C}}(N_{\nu}\,\xi^{\mu\nu\rho})$
$\displaystyle={\pi}^{\rho}_{\alpha}\big{(}-\xi^{\mu\alpha}+\tfrac{1}{2}N_{\nu}{\dot{C}}^{\beta}\,\xi^{\mu\nu\lambda\sigma}R^{\alpha}{}_{\lambda\beta\sigma}+(N_{\nu}{\dot{C}}^{\beta}+\tfrac{1}{2}{\pi}^{\beta}_{\nu})\,\xi^{\lambda\nu\sigma\alpha}R^{\mu}{}_{\lambda\beta\sigma}+\tfrac{1}{6}{\pi}^{\beta}_{\nu}\,\xi^{\alpha\nu\lambda\sigma}R^{\mu}{}_{\lambda\beta\sigma}\big{)}\,,$
(97) $\displaystyle\nabla_{\dot{C}}(N_{\nu}\,\xi^{\mu\nu})$
$\displaystyle=(N_{\nu}{\dot{C}}^{\beta}+\tfrac{1}{2}{\pi}^{\beta}_{\nu})\xi^{\rho\nu\lambda}R^{\mu}{}_{\rho\beta\lambda}+\nabla_{\dot{C}}\big{(}(\tfrac{1}{2}N_{\nu}{\dot{C}}^{\beta}+\tfrac{1}{6}{\pi}^{\beta}_{\nu})N_{\zeta}\,\xi^{\mu\nu\lambda\sigma}\,R^{\zeta}{}_{\lambda\beta\sigma}\big{)}$
$\displaystyle\qquad\qquad+(\tfrac{1}{2}N_{\nu}{\dot{C}}^{\beta}-\tfrac{1}{3}{\pi}^{\beta}_{\nu})\,\xi^{\rho\nu\lambda\sigma}\,(\nabla_{\lambda}R^{\mu}{}_{\rho\beta\sigma})\,,$
(98)

together with the constraint

$\displaystyle{\pi}^{\beta}_{\nu}\,{\pi}^{\alpha}_{\rho}\,{\pi}^{\lambda}_{\sigma}\,\xi^{\mu(\nu\rho\sigma)}=0\,.$
(99)

###### Proof.

The equations (74)-(77) are replaced by (96)-(99) as follows. $\nabla_{0}$ is
replaced by $\nabla_{\dot{C}}$. Each lower index $0$ it is necessary to
contract with ${\dot{C}}^{\beta}$ and each upper index $0$ it is necessary to
contract with $N_{\nu}$. Each spatial index $a,b,\ldots$ it is necessary to
project out using ${\pi}^{\beta}_{\nu}$. However if the spatial index is one
the third or fourth index of $\xi^{\rho\nu a}$ or $\xi^{\rho\nu ab}$ the
projection is not necessary.

As a result, in the adapted coordinate system where
${\dot{C}}^{\beta}=\delta^{\beta}_{0}$ and $N_{\nu}=\delta^{0}_{\nu}$ then
(96)-(99) become (74)-(77). However, (96)-(99) are clearly tensorial equations
and therefore true for all coordinate systems. ∎

As was observed in [3], the equations for the components arising from the
divergenceless condition of the stress-energy tensor are insufficient to
completely determine the dynamics of the components. To see this, from (12) we
have (10+40+100=150) components. From (11) this reduces to (150-10-40=100)
components. Using either (77) or (99) gives us 40 constraints, hence there are
60 components. Equations (74)-(76) or (96)-(98) give 40 ODEs. Hence there are
20 free components. These still have to be determined via constitutive
relations from a model of the underlying material.

These are similar to the equations have been found by Steinhoff and Puetzfeld
[11]. However their equations are implicit. On the right (98) we see that
there is a covariant derivative of $\xi^{\rho\nu\lambda\sigma}$. However this
term can be expanded out to identified
$\nabla_{\dot{C}}\xi^{\rho\nu\lambda\sigma}$. One can then substitute in (96)
and the appropriate constitutive relations.

######  Corollary 22.

As a simple consistency check we see if we reduce to a dipole. Set

$\displaystyle\xi^{\rho\nu\lambda\sigma}=0,\quad\xi^{\rho\nu\lambda}=X^{\lambda}\dot{C}^{\rho}\dot{C}^{\nu}+S^{\lambda(\rho}\dot{C}^{\nu)}\quad\textup{and}\quad\xi^{\rho\nu}=2P^{(\rho}\dot{C}^{\nu)}-2m\dot{C}^{\rho}\dot{C}^{\nu}\,,$
(100)

where $N_{\nu}=-{\dot{C}}_{\nu}$, $m$ is the rest mass, $X^{{{\mu}}}$ is the
displacement vector, $P^{{{\mu}}}$ is the rate of change of the displacement
vector and $S^{{{\mu}}{{\nu}}}$ is the spin tensor satisfy

$\displaystyle X_{{{\mu}}}\,{\dot{C}}^{{{\mu}}}=0,\quad
P_{{{\mu}}}\,{\dot{C}}^{{{\mu}}}=0,\quad{\dot{C}}_{{{\mu}}}\,S^{{{\mu}}{{\nu}}}=0\quad\textup{and}\quad
S^{{{\mu}}{{\nu}}}+S^{{{\nu}}{{\mu}}}=0\,.$ (101)

Then we get the Mathisson-Papapetrou-Tulczyjew-Dixon for a dipole along a
geodesics.

$\displaystyle\dot{m}=0\,,\quad\nabla_{\dot{C}}{X^{{{\mu}}}}=-P^{{{\mu}}}\,,\quad\nabla_{\dot{C}}{P^{{{\mu}}}}=\tfrac{1}{2}R^{{{\mu}}}{}_{{{\nu}}{{\rho}}{{\kappa}}}\,{\dot{C}}^{{\nu}}\,S^{{{\kappa}}{{\rho}}}+R^{{{\mu}}}{}_{{{\nu}}{{\rho}}{{\kappa}}}\,{\dot{C}}^{{\nu}}\,{\dot{C}}^{{\rho}}\,X^{{\kappa}}\quad\textup{and}\quad\nabla_{\dot{C}}{S^{{{\mu}}{{\nu}}}}$
$\displaystyle=0\,,$ (102)

###### Proof.

Since $C$ is a geodesic and $N_{\nu}=-{\dot{C}}_{\nu}$ then
$\nabla_{\dot{C}}N_{\nu}=0$. From (100) we have

$\displaystyle\nabla_{\dot{C}}(N_{\nu}\,\xi^{\mu\nu\rho})=\nabla_{\dot{C}}\Big{(}N_{\nu}\big{(}X^{\rho}\dot{C}^{\mu}\dot{C}^{\nu}+S^{\rho(\mu}\dot{C}^{\nu)}\big{)}\Big{)}=\nabla_{\dot{C}}\big{(}X^{\rho}\dot{C}^{\mu}+\tfrac{1}{2}S^{\rho\mu}\big{)}=\dot{C}^{\mu}\nabla_{\dot{C}}X^{\rho}+\tfrac{1}{2}\nabla_{\dot{C}}S^{\rho\mu}$

While from (97) we have

$\displaystyle\nabla_{\dot{C}}(N_{\nu}\,\xi^{\mu\nu\rho})$
$\displaystyle=-{\pi}^{\rho}_{\alpha}\xi^{\mu\alpha}=2{\pi}^{\rho}_{\alpha}\big{(}m\dot{C}^{\mu}\dot{C}^{\alpha}-P^{(\mu}\dot{C}^{\alpha)}\big{)}=-P^{\rho}\dot{C}^{\mu}$

giving

$\displaystyle\dot{C}^{\mu}\nabla_{\dot{C}}X^{\rho}+\tfrac{1}{2}\nabla_{\dot{C}}S^{\rho\mu}=-P^{\rho}\dot{C}^{\mu}$

Projecting this out with respect to ${\dot{C}}_{\mu}$ and $\pi_{\mu}^{\alpha}$
gives (102.2) and (102.4).

From (100.3) we have

$\displaystyle{{\nabla}}_{\dot{C}}(N_{\nu}\xi^{\mu\nu})={{\nabla}}_{\dot{C}}\Big{(}N_{\nu}\big{(}P^{(\mu}\dot{C}^{\nu)}-2m\dot{C}^{\mu}\dot{C}^{\nu}\big{)}\Big{)}={{\nabla}}_{\dot{C}}P^{\mu}-2\dot{C}^{\mu}{{\nabla}}_{\dot{C}}m={{\nabla}}_{\dot{C}}P^{\mu}-2\dot{C}^{\mu}\dot{m}{{\nabla}}_{\dot{C}}P^{\mu}$

Substituting (100.1) into (98) gives

$\displaystyle{{\nabla}}_{\dot{C}}(N_{\nu}\xi^{\mu\nu})$
$\displaystyle=(N_{\nu}{\dot{C}}^{\beta}+\tfrac{1}{2}{\pi}^{\beta}_{\nu})\xi^{\rho\nu\lambda}R^{\mu}{}_{\rho\beta\lambda}=(N_{\nu}{\dot{C}}^{\beta}+\tfrac{1}{2}{\pi}^{\beta}_{\nu})\left(X^{\lambda}\dot{C}^{\rho}\dot{C}^{\nu}+S^{\lambda(\rho}\dot{C}^{\nu)}\right)R^{\mu}{}_{\rho\beta\lambda}$
$\displaystyle=\dot{C}^{\beta}X^{\lambda}\dot{C}^{\rho}R^{\mu}{}_{\rho\beta\lambda}+\frac{1}{2}N_{\nu}\dot{C}^{\beta}\dot{C}^{\nu}S^{\lambda\rho}R^{\mu}{}_{\rho\beta\lambda}+\frac{1}{2}N_{\nu}\dot{C}^{\beta}\dot{C}^{\rho}S^{\lambda\nu}R^{\mu}{}_{\rho\beta\lambda}$
$\displaystyle\qquad\qquad+\tfrac{1}{2}{\pi}^{\beta}_{\nu}\left(X^{\lambda}\dot{C}^{\rho}\dot{C}^{\nu}+S^{\lambda(\rho}\dot{C}^{\nu)}\right)R^{\mu}{}_{\rho\beta\lambda}$
$\displaystyle=X^{\lambda}\dot{C}^{\beta}\dot{C}^{\rho}R^{\mu}{}_{\rho\beta\lambda}+\frac{1}{2}\dot{C}^{\beta}S^{\lambda\rho}R^{\mu}{}_{\rho\beta\lambda}+\tfrac{1}{2}{\pi}^{\beta}_{\nu}\left(X^{\lambda}\dot{C}^{\rho}\dot{C}^{\nu}+S^{\lambda(\rho}\dot{C}^{\nu)}\right)R^{\mu}{}_{\rho\beta\lambda}$
$\displaystyle=X^{\lambda}\dot{C}^{\beta}\dot{C}^{\rho}R^{\mu}{}_{\rho\beta\lambda}+\frac{1}{2}\dot{C}^{\beta}S^{\lambda\rho}R^{\mu}{}_{\rho\beta\lambda}+\frac{1}{4}\dot{C}^{\rho}S^{\lambda\beta}R^{\mu}{}_{\rho\beta\lambda}$
$\displaystyle=X^{\lambda}\dot{C}^{\beta}\dot{C}^{\rho}R^{\mu}{}_{\rho\beta\lambda}+\frac{1}{4}\dot{C}^{\rho}S^{\lambda\beta}R^{\mu}{}_{\rho\beta\lambda}+\frac{1}{4}\dot{C}^{\rho}S^{\lambda\beta}R^{\mu}{}_{\rho\beta\lambda}$
$\displaystyle=X^{\lambda}\dot{C}^{\beta}\dot{C}^{\rho}R^{\mu}{}_{\rho\beta\lambda}+\frac{1}{2}\dot{C}^{\rho}S^{\lambda\beta}R^{\mu}{}_{\rho\beta\lambda}\,.$

hence

$\displaystyle{{\nabla}}_{\dot{C}}P^{\mu}-2\dot{C}^{\mu}\dot{m}=X^{\lambda}\dot{C}^{\beta}\dot{C}^{\rho}R^{\mu}{}_{\rho\beta\lambda}+\frac{1}{2}\dot{C}^{\rho}S^{\lambda\beta}R^{\mu}{}_{\rho\beta\lambda}$

Projecting this out with respect to ${\dot{C}}_{\mu}$ and $\pi_{\mu}^{\alpha}$
gives (102.1) and (102.3). ∎

##  5 Comparison with Dixon’s results

In his work, Dixon makes two conjectures for the dynamics of the components of
a quadrupole. Neither of these equations correspond to the divergenceless
condition. Thus they are not the generalisation of the Mathisson-
Papapetrou–Tulczyjew–Dixon equations for the quadrupole.

Recall from [3], that we can identify $\xi^{\mu\nu}=I^{\mu\nu}$,
$\xi^{\mu\nu\kappa}=-I^{\kappa\mu\nu}$ and
$\xi^{\mu\nu\kappa\lambda}=I^{\kappa\lambda\mu\nu}$. Using [9, eqn. (1.37)]
and [6, eqn. (2.4)] we can555In [3], we identified $J^{\kappa\lambda\mu\nu}$
incorrectly. write
$J^{\kappa\lambda\mu\nu}=I^{[\kappa[\lambda\mu]\nu]}=\tfrac{1}{4}(I^{\kappa\lambda\mu\nu}-I^{\mu\lambda\kappa\nu}-I^{\kappa\nu\mu\lambda}+I^{\mu\nu\kappa\lambda})$.

In [7, eqns. (7.34)-(7.37)] Dixon proposes a simple rotational dynamics. Here
he introduces a connection (7.18), ${\stackrel{{\scriptstyle
M}}{{\nabla}}_{{\dot{C}}}}$ which he writes as
$\stackrel{{\scriptstyle(m)}}{{\frac{\delta}{ds}}}$

$\displaystyle{\stackrel{{\scriptstyle
M}}{{\nabla}}_{{\dot{C}}}}B^{\kappa}=\nabla_{\dot{C}}B^{\kappa}-(\dot{u}^{\kappa}\,u^{\lambda}-\dot{u}^{\lambda}\,u^{\kappa})B^{\lambda}\,,$
(103)

where $u^{\kappa}$ is described as the body’s dynamical velocity. Using this
connection we can define a rotation tensor $\chi\Omega_{\lambda\kappa}$ where
(7.34)

$\displaystyle{\stackrel{{\scriptstyle
M}}{{\nabla}}_{{\dot{C}}}}B^{\kappa}=\chi\Omega^{\kappa}{}_{\lambda}\,B^{\lambda},\quad\Omega_{(\kappa\lambda)}=0\quad\textup{and}\quad\Omega^{\kappa}{}_{\lambda}\,u^{\lambda}=0\,.$
(104)

From this the dynamical equation for a non-rotating quadrupole is given by
(7.36)

$\displaystyle{\stackrel{{\scriptstyle
M}}{{\nabla}}_{{\dot{C}}}}J^{\kappa\lambda\mu\nu}=-\chi\Omega^{\kappa}{}_{\rho}J^{\lambda\rho\mu\nu}+\chi\Omega^{\lambda}{}_{\rho}J^{\kappa\rho\mu\nu}-\chi\Omega^{\nu}{}_{\rho}J^{\kappa\lambda\rho\mu}+\chi\Omega^{\mu}{}_{\rho}J^{\kappa\lambda\rho\nu}\,.$
(105)

It is clear that such a non-rotating quadrupole would not satisfy (96)–(99)
and therefore not correspond to a divergenceless stress-energy tensor.

By contrast in [8, eqn. (4.11)] Dixon proposes a non-dynamical equation based
on symmetry. In our language this becomes

$\displaystyle\xi^{\mu(\nu\rho\sigma)}=0\,.$ (106)

This is very similar to the constraint (99) but without the projections.
Interestingly such a constraint gives the same number of free components as
the dynamical equations arising from the divergenceless condition. Again such
conditions on $J^{\mu\nu\rho\sigma}$ do not correspond to the divergenceless
condition.

##  6 Discussion and Conclusion

The key result of this article is derivation of the dynamics of the Dixon
quadrupole, as given in section 4. It is clear looking at the method used to
derive the equations, that this could be extended to arbitrary order
multipoles, giving rise to higher covariant derivatives of the curvature. In
section 5 we compared these equations with two which had be conjectured by
Dixon. We observe that since Dixon’s equations do not couple to curvature as
they were not compatible with the divergenceless property of the stress-energy
tensor.

In order to get to derive the dynamical equations it was necessary to
establish many properties of general Dixon multipoles, presented in section 3.
These include the fact that all multipoles can be represented as Dixon
multipoles, that the components were unique and that they can be extracted
using particular test tensors. We concluded this section showing the link
between the moments of regular tensors and multipoles.

$\,\,N_{{{\mu}}}$ (a) The Dixon vector given by the tangent to the worldline.

$q$$\,\,N_{{{\mu}}}$ (b) The Dixon vector given by the backward lightcone of a
distant observer event $q$.

Figure 2: The Dixon vector $N_{{{\mu}}}$ and the corresponding Dixon geodesic
hypersurfaces, for two different scenarios.

As noted throughout this article, all the results depend on the choice of the
Dixon vector. There are multiple choices of such Dixon vectors, and there may
not be a natural one. In figure 2 we see two such choice. In figure 2(a) we
show the Dixon vector given by the tangent of the worldline. This is a natural
choice if one is interested in the multipole dynamics as observed by the
particle itself.

Alternatively, one may wish to model the dynamics of the multipole as we, as
distant observers, see it. As seen in figure 2(b), this involves constructing
the backward lightcone from each event $q$ in our worldline. This give rise to
a lightlike Dixon vector which points from the multipole worldline, $C$ to
$q$. The corresponding Dixon geodesic hypersurface is then tangent to backward
lightcone. It does not, however, coincide with the backward lightcone. For a
very distant object the two would be very close and this may be a sufficiently
good approximation. However if the object and the observer were closer this
discrepancy may become important. It would then be necessary to use the Ellis
representation of multipoles and a coordinate system adapted to backward
lightcones. As was observed, the Dixon vector $N_{{{\mu}}}$ in figure 2(b) is
lightlike. This does not affect any of the calculation as the only constraint
on the Dixon vector is (3).

As noted in section 4, there are 20 free components of quadrupole. The
equations for these components, also known as constitutive relations, will
need to use additional information. This could be knowledge of the underlying
matter which makes up the extended object, or they could arise from
observation. Now that the tensorial expression of the dynamical equations for
the moments are known, it will help in establishing the constitutive relations
for different objects.

Once the additional equations have been chosen, one could ask how several
quadrupoles gravitationally interact. As stated in [3] this interaction can
only be perturbative, where the quadrupoles interact via gravitational waves.
There is still the issue of gravitational radiation reaction. However, this
can be avoided if we only consider the action of one quadrupoles by the fields
generated by all the other quadrupoles. This is similar to the electromagnetic
case considered in [13].

## Acknowledgements

JG is grateful for the support provided by STFC (the Cockcroft Institute
ST/P002056/1 and ST/V001612/1). ST would like to thank the Faculty of Science
and Technology, Lancaster University for their support.

We would like to thank Alex Warwick, Lancaster University, for reading the
manuscript and making suggestions.

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