Conserved charges of the Kerr black hole revisited

Okan Günel 111email<EMAIL_ADDRESS>and Özgür Sarıoğlu 222email:
<EMAIL_ADDRESS>
Department of Physics, Faculty of Arts and Sciences,

Middle East Technical University, 06800, Ankara, Turkey

Abstract

We revisit the Kerr black hole as cast in the Boyer-Lindquist, Kerr-Schild and
Weyl canonical coordinates, and calculate its total mass/energy and total
angular momentum by using linearized gravity along with its background Killing
isometries. We argue that the integration of the relevant gravitational flux
does not depend on the geometry of the closed and simply connected spatial
boundary provided it is also piecewise smooth.

Dedicated to the memory of Stanley Deser (March 19, 1931 – April 21, 2023).

###### Contents

  1. 1 Introduction
  2. 2 Conserved charges in linearized gravity
  3. 3 Kerr black hole in Boyer-Lindquist coordinates
  4. 4 Kerr black hole in Kerr-Schild coordinates
  5. 5 Kerr black hole in Weyl canonical coordinates
  6. 6 Conclusions
  7. A The 2-form potential $\ell$ in Weyl coordinates

##  1 Introduction

We are obviously living in a new era of gravitational physics ever since the
first successful experimental detection of gravitational waves from the
collision of two black holes [1] and the first image of a black hole
distinguishing its horizon structure [2] in the way predicted by General
Relativity. The gravitational wave observations so far are thought to consist
of the products of collisions of compact astrophysical rotating objects. Since
collisions of stationary black holes seem to form a rather large class in this
catalogue, it is only natural that we heavily rely on the celebrated Kerr
solution [3] in the analysis and understanding of the basic physical
properties of these collision events.

The Kerr solution has, of course, been thoroughly studied over the years and
its basic physical and geometrical properties are well understood (See e.g.
[4] and [5] for a concise introduction and starting point.). Specifically the
determination of its conserved gravitational charges, in particular its total
mass/energy and total angular momentum, is standard textbook material [6].
Here we want to _revisit_ the particular method based on linearized gravity
and the use of background Killing isometries [7], that is in fact closely
related to the celebrated Arnowitt-Deser-Misner (ADM) formulation [8] of
General Relativity. We do this in three separate coordinate systems: the
ubiquitous Boyer-Lindquist coordinates [9], the Kerr-Schild form [10] and the
lesser-known Weyl canonical coordinates [5, 11]. Our motivation in doing so is
foremost to examine how the integration of the gravitational flux through “the
boundary at infinity” depends on the geometry of the latter. For the three
poses of the Kerr solution studied, “the cube at infinity” for the Kerr-Schild
form and “the cylinder at infinity” for the Weyl canonical coordinates are
certainly not smooth geometries as for “the sphere at infinity” for the Boyer-
Lindquist case. As a by-product, we also explicitly verify that the whole
procedure is indeed background gauge invariant.

The organization of the paper is as follows. In section 2, we start by giving
a concise, self-contained description of how gravitational charges are defined
using background Killing isometries. We explicitly calculate the total
mass/energy and the total angular momentum of the Kerr black hole cast in the
Boyer-Lindquist, Kerr-Schild and Weyl canonical coordinates in sections 3,
4, 5, respectively. To our knowledge, the discussion presented in sections
4 and 5 is novel. We finally conclude with a brief summary and some remarks
in section 6. Finally, we explicitly relegate some unwieldy formulas needed
for the discussion in section 5 to the appendix A.

##  2 Conserved charges in linearized gravity

Let us briefly recapitulate how gravitational charges that use background
Killing isometries are defined. (Please refer to [7] for details.) Let
$h_{ab}$ denote the “deviation” of an asymptotically flat metric from its flat
“background” $\bar{g}_{ab}$111Note that this discussion in fact holds for
asymptotically maximally symmetric spacetimes but we simply work with a
vanishing cosmological constant in what follows., i.e. $h_{ab}\equiv
g_{ab}-\bar{g}_{ab}$, where $\bar{g}_{ab}$ is the Ricci flat background with
curvature tensors $\bar{R}_{ab}=0=\bar{R}$. It is also assumed that the
deviation $h_{ab}$ goes to zero “sufficiently fast” as one approaches the
background $\bar{g}_{ab}$, which is typically located at “the boundary at
infinity”. It can be shown [7] that one can construct a conserved vector
current $J^{a}:=G_{L}^{ab}\bar{\xi}_{b}$, with $\bar{\nabla}_{a}J^{a}=0$, out
of the linearized Einstein tensor $G_{L}^{ab}$ and the background Killing
vector $\bar{\xi}^{a}$, that are well-defined and smooth in the geometry
described by $\bar{g}_{ab}$. Then the following gives a conserved and
background gauge invariant gravitational charge

$Q(\bar{\xi})=\frac{1}{8\pi}\int_{\Sigma}d^{3}x\,\sqrt{|\gamma|}\,n_{a}\,J^{a}=\frac{1}{8\pi}\oint_{\partial\Sigma}d^{2}x\,\sqrt{|q|}\,{n}_{[a}\,r_{b]}\,\ell^{ab}\,,$
(2.1)

where $\ell_{ab}$ is the potential 2-form of the conserved vector current
$J^{a}:=\bar{\nabla}_{b}\ell^{ab}$.

Here we assume that the background admits a foliation

$\bar{g}_{ab}:=-n_{a}n_{b}+\gamma_{ab}=-n_{a}n_{b}+r_{a}r_{b}+q_{ab}\,,$ (2.2)

where the induced non-degenerate metric $\gamma_{ab}$ on the $3$-dimensional
spacelike hypersurface $\Sigma$ has an everywhere well-defined timelike normal
$n^{a}$ and $\gamma:=\det{\gamma_{ab}}$. We also assume that the hypersurface
$\Sigma$ has a boundary $\partial\Sigma$ with a metric $q_{ab}$, with
$q:=\det{q_{ab}}$; $n^{a}$ and the spacelike vector $r^{a}$ are mutually
orthogonal unit vectors to $\partial\Sigma$, with $n^{a}n_{a}=-1$ and
$r^{a}r_{a}=1$.

For the conventions adopted in this work, one has [7]

$\ell^{ab}(\bar{\xi}):=\bar{\xi}_{c}\bar{\nabla}^{[a}h^{b]c}+\bar{\xi}^{[b}\bar{\nabla}_{c}h^{a]c}+h^{c[b}\bar{\nabla}_{c}\bar{\xi}^{a]}+\bar{\xi}^{[a}\bar{\nabla}^{b]}h+\frac{1}{2}h\bar{\nabla}^{[a}\bar{\xi}^{b]}\,,$
(2.3)

where all raising and lowering of indices are done with respect to
$\bar{g}_{ab}$, $\bar{\nabla}$ indicates the covariant derivative with respect
to the background metric and the antisymmetrization of indices is done with
‘weight 1’ i.e. $A_{[ab]}\equiv\tfrac{1}{2}(A_{ab}-A_{ba})$.

All of these are of course well-known. Our aim here is to carefully apply
these to the celebrated Kerr metric written in three separate coordinates, and
show how “delicate” the integration of the properly weighted 2-form potential
$\ell$ on the two-dimensional boundary $\partial\Sigma$ (2.1) can be. For
warming up, we start our discussion by first considering the Kerr black hole
in the ubiquitous Boyer-Lindquist coordinates.

##  3 Kerr black hole in Boyer-Lindquist coordinates

The Kerr black hole [3] in Boyer-Lindquist coordinates [9, 4] reads

$\displaystyle ds^{2}$ $\displaystyle=$
$\displaystyle-\left(1-\frac{2Mr}{\Sigma}\right)dt^{2}-\frac{4aMr\sin^{2}{\theta}}{\Sigma}\,dt\,d\phi+\frac{\Sigma}{\Delta}\,dr^{2}+\Sigma\,d\theta^{2}$
(3.1)
$\displaystyle+\left(r^{2}+a^{2}+\frac{2a^{2}Mr\sin^{2}{\theta}}{\Sigma}\right)\sin^{2}{\theta}\,d\phi^{2}\,,$

where the metric functions are given by

$\Sigma(r,\theta):=r^{2}+a^{2}\cos^{2}{\theta}\qquad\mbox{and}\qquad\Delta(r):=r^{2}+a^{2}-2Mr\,.$
(3.2)

The relevant flat background is found by setting $M=0,a=0$ in (3.1):

$\bar{g}_{ab}={\rm diag}(-1,1,r^{2},r^{2}\sin^{2}\theta)\,,$ (3.3)

which admits the foliation (2.2) with

$\displaystyle\gamma_{ab}={\rm
diag}(0,1,r^{2},r^{2}\sin^{2}\theta)\quad\mbox{and}\quad q_{ab}={\rm
diag}(0,0,r^{2},r^{2}\sin^{2}\theta)\,,$ (3.4) $\displaystyle
n^{a}=(-\partial_{t})^{a}\quad\mbox{and}\quad r^{a}=(\partial_{r})^{a}\,.$

Clearly $\gamma_{ab}$ is the metric on the 3-dimensional Euclidean space
$\Sigma$ in spherical polar coordinates and $q_{ab}$ is the metric on a
2-sphere $S_{2}$ of radius $r$. The 2-dimensional boundary $\partial\Sigma$ is
found when one takes $r\to\infty$. The background clearly has two globally
defined, timelike $\bar{\xi}^{a}=\left(-\partial_{t}\right)^{a}$ and spacelike
$\bar{\zeta}^{a}=\left(\partial_{\phi}\right)^{a}$, Killing vectors.

The relevant components of the 2-form potential (2.3) for each Killing vector
are

$\displaystyle\ell^{tr}(\bar{\xi})=\frac{1}{2r^{3}}\left(\frac{2a^{2}Mr^{2}\sin^{2}{\theta}\Sigma^{\prime}}{\Sigma^{2}}+\left(\frac{2r^{2}}{\Delta}+1\right)\Sigma-r\Sigma^{\prime}+a^{2}-r^{2}\right)\,,$
(3.5)
$\displaystyle\ell^{tr}(\bar{\zeta})=\frac{aM\sin^{2}{\theta}\left(r\Sigma^{\prime}+\Sigma\right)}{\Sigma^{2}}\,,$
(3.6)

in terms of the metric functions (3.2) where a prime indicates derivative with
respect to the $r$ coordinate. A careful integration gives

$\oint_{S_{2}}r^{2}d\Omega\,{n}_{[a}\,r_{b]}\,\ell^{ab}(\bar{\xi})=\frac{4\pi}{3}\left(\frac{r\left(a^{2}+3r^{2}\right)}{\Delta}+\frac{2a^{3}+3M\left(a^{2}-r^{2}\right)\arctan{(a/r)}+3ar(M-r)}{ar}\right),$
(3.7)

where we have used $d\Omega:=\sin{\theta}\,d\theta\,d\phi$, that leads to

$\displaystyle
Q(\bar{\xi})=\frac{1}{8\pi}\lim_{r\to\infty}\oint_{S_{2}}r^{2}d\Omega\,{n}_{[a}\,r_{b]}\,\ell^{ab}(\bar{\xi})=M\,,$
(3.8)

the total mass of the Kerr black hole. Similarly, one finds

$\oint_{S_{2}}r^{2}d\Omega\,{n}_{[a}\,r_{b]}\,\ell^{ab}(\bar{\zeta})=8\pi
Mr\arctan{(a/r)},$ (3.9)

which yields

$\displaystyle
Q(\bar{\zeta})=\frac{1}{8\pi}\lim_{r\to\infty}\oint_{S_{2}}r^{2}d\Omega\,{n}_{[a}\,r_{b]}\,\ell^{ab}(\bar{\zeta})=aM\,,$
(3.10)

the total angular momentum of the Kerr black hole.

At hindsight, what we did was to first integrate the properly furbished 2-form
potential $\ell$ on a constant radius 2-sphere $S_{2}$ and later take the
radius $r\to\infty$ limit. It is perhaps worthy of considering the “short-cut”
of taking the $r\to\infty$ limit of the properly furbished 2-form potential
$\ell$ first and then integrate the result on “the sphere at infinity” later.
This interchange is in general clearly not up to par with mathematical rigor,
but it should be possible for a “smooth” boundary $\partial\Sigma$ and a
“smooth” potential $\ell$. Thus, e.g.

$\displaystyle\frac{1}{8\pi}\oint
d\Omega\left(\lim_{r\to\infty}r^{2}\,{n}_{[a}\,r_{b]}\,\ell^{ab}(\bar{\xi})\right)=\frac{1}{8\pi}\oint
d\Omega\,(2M)=M\,,$ $\displaystyle\frac{1}{8\pi}\oint
d\Omega\left(\lim_{r\to\infty}r^{2}\,{n}_{[a}\,r_{b]}\,\ell^{ab}(\bar{\zeta})\right)=\frac{1}{8\pi}\oint
d\Omega\,(3aM\sin^{2}{\theta})=aM\,,$ (3.11)

which “validates” our hunch for the shortcut.

##  4 Kerr black hole in Kerr-Schild coordinates

Let us consider the Kerr black hole in its Kerr-Schild form [10, 12, 13]:

$\displaystyle
ds^{2}=g_{ab}dx^{a}dx^{b}=\left(\eta_{ab}+\frac{2M}{U}\lambda_{a}\lambda_{b}\right)dx^{a}dx^{b}\,,$
(4.1)

where $\lambda^{a}$ is null and geodesic with respect to both the full metric
$g_{ab}$ and the flat Minkowski metric $\eta_{ab}$. Explicitly, in flat
coordinates $(t,x,y,z)$, one has

$\displaystyle\lambda=\lambda_{a}dx^{a}=dt+\frac{r(xdx+ydy)+a(ydx-
xdy)}{r^{2}+a^{2}}+\frac{zdz}{r}$ (4.2)

and

$\displaystyle U=r+\frac{a^{2}z^{2}}{r^{3}}\,,$ (4.3)

where $r$ is not a coordinate but a function of $(x,y,z)$ and defined by

$\displaystyle\frac{x^{2}+y^{2}}{r^{2}+a^{2}}+\frac{z^{2}}{r^{2}}=1\,.$ (4.4)

Note that the Eddington-like coordinates [9] $(\bar{t},r,\theta,\bar{\phi})$
can be obtained from (4.1) by the coordinate transformation

$\displaystyle t$ $\displaystyle=$ $\displaystyle\bar{t}\,.$ $\displaystyle x$
$\displaystyle=$
$\displaystyle\sqrt{r^{2}+a^{2}}\,\sin{\theta}\,\cos{\left(\bar{\phi}+\arctan{(a/r)}\right)}\,,$
$\displaystyle y$ $\displaystyle=$
$\displaystyle\sqrt{r^{2}+a^{2}}\,\sin{\theta}\,\sin{\left(\bar{\phi}+\arctan{(a/r)}\right)}\,,$
(4.5) $\displaystyle z$ $\displaystyle=$ $\displaystyle r\cos{\theta}\,,$

A final time-dependent coordinate transformation is required to arrive at the
Kerr metric in Boyer-Lindquist coordinates $(t,r,\theta,\phi)$ as in (3.1)

$\displaystyle d\bar{t}$ $\displaystyle=$ $\displaystyle
dt+\frac{2Mr}{\Delta}\,dr\,,$ $\displaystyle d\bar{\phi}$ $\displaystyle=$
$\displaystyle d\phi+\frac{a}{\Delta}\,dr\,.$

One advantage of the Kerr-Schild form (4.1) is that the background metric is
clearly the flat Minkowski metric $\eta_{ab}$ in the usual Cartesian
coordinates. This background admits the foliation (2.2) with

$\displaystyle\gamma_{ab}={\rm diag}(0,1,1,1)\,,\quad
n^{a}=(-\partial_{t})^{a}\,,$ (4.6)

where $\gamma_{ab}$ is the metric on the 3-dimensional flat Euclidean space in
the Cartesian coordinates. Now arises the question about the integration on
the boundary of the space. The boundary relevant for the charge definition
(2.1) can be thought of as “the cube at infinity” in Cartesian coordinates.
This can be described best by first considering a cube centered at the origin
and of side length $2L$, and then taking the $L\to\infty$ limit.

Any such cube of finite side length has six faces and the integration on each
face must be carried out with the following normal vectors and the
corresponding two-dimensional metrics in mind:

$\displaystyle
x_{\pm}^{a}=\big{(}\pm\partial_{x}\big{)}^{a}\quad\mbox{and}\quad\big{(}q_{x}\big{)}_{ab}={\rm
diag}(0,0,1,1)\,,$ $\displaystyle
y_{\pm}^{a}=\big{(}\pm\partial_{y}\big{)}^{a}\quad\mbox{and}\quad\big{(}q_{y}\big{)}_{ab}={\rm
diag}(0,1,0,1)\,,$ (4.7) $\displaystyle
z_{\pm}^{a}=\big{(}\pm\partial_{z}\big{)}^{a}\quad\mbox{and}\quad\big{(}q_{z}\big{)}_{ab}={\rm
diag}(0,1,1,0)\,.$

“The cube at infinity” presumably has its faces placed at the $L\to\infty$
limits of the coordinates along the direction of the normal vectors. Thus the
integration on the boundary can be thought of as the sum of six separate
integrals on each face of “the cube at infinity”

$\displaystyle Q(\bar{\xi})=$
$\displaystyle\frac{1}{8\pi}\lim_{L\to\infty}\left(\int_{-L}^{L}\int_{-L}^{L}dy\,dz\,{n}_{[a}\,x_{b]}\,\ell^{ab}(\bar{\xi})\rvert_{x=L}-\int_{-L}^{L}\int_{-L}^{L}dy\,dz\,{n}_{[a}\,x_{b]}\,\ell^{ab}(\bar{\xi})\rvert_{x=-L}\right.$
$\displaystyle\qquad\qquad+\int_{-L}^{L}\int_{-L}^{L}dx\,dz\,{n}_{[a}\,y_{b]}\,\ell^{ab}(\bar{\xi})\rvert_{y=L}-\int_{-L}^{L}\int_{-L}^{L}dx\,dz\,{n}_{[a}\,y_{b]}\,\ell^{ab}(\bar{\xi})\rvert_{y=-L}$
$\displaystyle\qquad\qquad\left.+\int_{-L}^{L}\int_{-L}^{L}dx\,dy\,{n}_{[a}\,z_{b]}\,\ell^{ab}(\bar{\xi})\rvert_{z=L}-\int_{-L}^{L}\int_{-L}^{L}dx\,dy\,{n}_{[a}\,z_{b]}\,\ell^{ab}(\bar{\xi})\rvert_{z=-L}\right)\,.$
(4.8)

Using (4.6) and (4.7), this explicitly becomes

$\displaystyle Q(\bar{\xi})=$
$\displaystyle\frac{1}{8\pi}\lim_{L\to\infty}\left(\int_{-L}^{L}\int_{-L}^{L}dy\,dz\,\ell^{tx}(\bar{\xi})\rvert_{x=L}-\int_{-L}^{L}\int_{-L}^{L}dy\,dz\,\ell^{tx}(\bar{\xi})\rvert_{x=-L}\right.$
$\displaystyle\qquad\qquad+\int_{-L}^{L}\int_{-L}^{L}dx\,dz\,\ell^{ty}(\bar{\xi})\rvert_{y=L}-\int_{-L}^{L}\int_{-L}^{L}dx\,dz\,\ell^{ty}(\bar{\xi})\rvert_{y=-L}$
$\displaystyle\qquad\qquad\left.+\int_{-L}^{L}\int_{-L}^{L}dx\,dy\,\ell^{tz}(\bar{\xi})\rvert_{z=L}-\int_{-L}^{L}\int_{-L}^{L}dx\,dy\,\ell^{tz}(\bar{\xi})\rvert_{z=-L}\right)\,,$
(4.9)

keeping $a$ and $M$ fixed, and the definition of $r$ (4.4) in mind. The
relevant timelike Killing vector of the background is still
$\bar{\xi}^{a}=\big{(}-\partial_{t}\big{)}^{a}$, which generates three
nontrivial components for the 2-form potential $\ell$ (2.3)

$\displaystyle\ell^{tx}(\bar{\xi})=$
$\displaystyle\frac{Mr^{2}}{(a^{2}+r^{2})^{3}(a^{2}z^{2}+r^{4})^{3}}\Big{(}2r^{13}x+a^{9}yz^{4}-2a^{8}rxz^{4}-2a^{7}r^{2}yz^{2}(r^{2}-3z^{2})\Big{.}$
$\displaystyle\quad-a^{6}r^{3}xz^{2}\big{(}4r^{2}+3(x^{2}+y^{2})\big{)}+a^{5}r^{4}y\big{(}r^{4}+z^{2}(4x^{2}+4y^{2}+5z^{2})\big{)}$
$\displaystyle\quad-a^{4}r^{5}x(r^{2}+z^{2})(2r^{2}-x^{2}-y^{2}-2z^{2})+2a^{3}r^{8}y(3r^{2}-z^{2})$
(4.10a)
$\displaystyle\quad\Big{.}+a^{2}r^{9}x(5x^{2}+5y^{2}+4z^{2})+ar^{10}y\big{(}5r^{2}-4(x^{2}+y^{2}+z^{2})\big{)}\Big{)}\,,$
$\displaystyle\ell^{ty}(\bar{\xi})=$
$\displaystyle\frac{Mr^{2}}{(a^{2}+r^{2})^{3}(a^{2}z^{2}+r^{4})^{3}}\Big{(}2r^{13}y-a^{9}xz^{4}-2a^{8}ryz^{4}+2a^{7}r^{2}xz^{2}(r^{2}-3z^{2})\Big{.}$
$\displaystyle\quad-a^{6}r^{3}yz^{2}\big{(}4r^{2}+3(x^{2}+y^{2})\big{)}-a^{5}r^{4}x\big{(}r^{4}+z^{2}(4x^{2}+4y^{2}+5z^{2})\big{)}$
$\displaystyle\quad-a^{4}r^{5}y(r^{2}+z^{2})(2r^{2}-x^{2}-y^{2}-2z^{2})+2a^{3}r^{8}x(z^{2}-3r^{2})$
(4.10b)
$\displaystyle\quad\Big{.}+a^{2}r^{9}y(5x^{2}+5y^{2}+4z^{2})+ar^{10}x\big{(}4(x^{2}+y^{2}+z^{2})-5r^{2}\big{)}\Big{)}\,,$
$\displaystyle\ell^{tz}(\bar{\xi})=$
$\displaystyle\frac{Mr^{3}z}{(a^{2}+r^{2})^{2}(a^{2}z^{2}+r^{4})^{3}}\Big{(}2r^{10}-a^{6}z^{2}(x^{2}+y^{2}-2z^{2})\Big{.}$
(4.10c)
$\displaystyle\quad\Big{.}+a^{4}\big{(}r^{4}(3x^{2}+3y^{2}+4z^{2})+r^{2}z^{2}(x^{2}+y^{2}+2z^{2})\big{)}+a^{2}r^{6}(2r^{2}+5x^{2}+5y^{2}+4z^{2})\Big{)}\,.$

To arrive at the expressions (4.10), we have utilized

$\displaystyle\frac{\partial r}{\partial x}=\frac{x}{U}\,,\qquad\frac{\partial
r}{\partial y}=\frac{y}{U}\,,\qquad\frac{\partial r}{\partial
z}=\frac{z}{U}\Big{(}1+\frac{a^{2}}{r^{2}}\Big{)}\,,$ (4.11)
$\displaystyle\frac{\partial U}{\partial
x}=\frac{x}{U}\Big{(}1-\frac{3a^{2}z^{2}}{r^{4}}\Big{)}\,,\quad\frac{\partial
U}{\partial
y}=\frac{y}{U}\Big{(}1-\frac{3a^{2}z^{2}}{r^{4}}\Big{)}\,,\quad\frac{\partial
U}{\partial
z}=\frac{z}{U}\Big{(}1-\frac{3a^{2}z^{2}}{r^{4}}\Big{)}\Big{(}1+\frac{a^{2}}{r^{2}}\Big{)}+\frac{2a^{2}z}{r^{3}}\,,$
(4.12)

that follow from the definitions of $r$ (4.4) and $U$ (4.3), respectively.

Next we substitute (4.10) into (4). Notice that the variable $r$ (4.4) is
symmetric in $x$ and $y$ by definition. Thus, we can interchange the latter in
the integration for the terms that include, e.g., $\ell^{ty}$. A careful
consideration along the $yz$\- and $xz$-surfaces leads to

$\displaystyle Q(\bar{\xi})=$
$\displaystyle\frac{M}{4\pi}\lim_{L\to\infty}\bigg{(}\int_{-L}^{L}\int_{-L}^{L}dx\,dz\,\frac{2\,r(x,L,z)^{3}L}{(r^{2}+a^{2})^{3}(r^{4}+a^{2}z^{2})^{3}}\left(2r^{12}-2a^{8}z^{4}-a^{6}r^{2}z^{2}\big{(}4r^{2}+3(x^{2}+L^{2})\big{)}\right.$
$\displaystyle\qquad\qquad\qquad\left.-a^{4}r^{4}(r^{2}+z^{2})(2r^{2}-x^{2}-L^{2}-2z^{2})+a^{2}r^{8}(5x^{2}+5L^{2}+4z^{2})\right)$
$\displaystyle+\int_{-L}^{L}\int_{-L}^{L}dx\,dy\,\frac{r(x,y,L)^{3}L}{(r^{2}+a^{2})^{2}(r^{4}+a^{2}L^{2})^{3}}\left(2r^{10}-a^{6}L^{2}(x^{2}+y^{2}-2L^{2})\right.$
(4.13)
$\displaystyle\qquad\left.+a^{4}r^{2}\big{(}r^{2}(3x^{2}+3y^{2}+4L^{2})+L^{2}(x^{2}+y^{2}+2L^{2})\big{)}+a^{2}r^{6}(2r^{2}+5x^{2}+5y^{2}+4L^{2})\right)\bigg{)}\,,$

where we have implicitly indicated the relevant $y=\pm L$ and $z=\pm L$
substitutions in $r$ (4.4) by writing it with its pertinent arguments. Using
(4.4) to eliminate $x^{2}$ and $x^{2}+y^{2}$ combinations in the two
integrands of (4.13), respectively, we arrive at

$\displaystyle Q(\bar{\xi})=$
$\displaystyle\frac{M}{4\pi}\lim_{L\to\infty}\bigg{(}\int_{-L}^{L}\int_{-L}^{L}dx\,dz\,\frac{2\,r(x,L,z)^{3}L}{(r^{2}+a^{2})(r^{4}+a^{2}z^{2})^{3}}\left(2r^{8}+a^{4}z^{2}(z^{2}-3r^{2})+a^{2}r^{4}(r^{2}-z^{2})\right)$
$\displaystyle\qquad\qquad+\int_{-L}^{L}\int_{-L}^{L}dx\,dy\,\frac{r(x,y,L)L}{\left(r^{4}+a^{2}L^{2}\right)^{3}}\left(2r^{8}-a^{4}L^{2}(r^{2}-L^{2})-a^{2}r^{4}(L^{2}-3r^{2})\right)\bigg{)}\,$
(4.14)

where the first integrand can be factorized to write

$\displaystyle Q(\bar{\xi})$ $\displaystyle=I_{xy}+I_{xz}\,,$ (4.15)
$\displaystyle I_{xy}$
$\displaystyle:=\frac{M}{4\pi}\lim_{L\to\infty}\bigg{(}\int_{-L}^{L}\int_{-L}^{L}dx\,dy\,\frac{r(x,y,L)L}{\left(r^{4}+a^{2}L^{2}\right)^{3}}\left(2r^{8}-a^{4}L^{2}(r^{2}-L^{2})-a^{2}r^{4}(L^{2}-3r^{2})\right)\bigg{)}\,,$
$\displaystyle I_{xz}$
$\displaystyle:=\frac{M}{4\pi}\lim_{L\to\infty}\bigg{(}\int_{-L}^{L}\int_{-L}^{L}dx\,dz\,\frac{2\,r(x,L,z)^{3}L}{(r^{4}+a^{2}z^{2})}\left(\frac{r^{2}(r^{4}-3a^{2}z^{2})}{(r^{4}+a^{2}z^{2})^{2}}+\frac{1}{r^{2}+a^{2}}\right)\,.$

Since the solution of $r$ (4.4) is not as simple as in the usual spherical
coordinates, it is painful to take these integrals in the Euclidean
coordinates.

For dealing with $I_{xy}$, we first note that the relevant root of
$r=r(x,y,L)$ (4.4) is

$\displaystyle
r=\frac{\sqrt{\sqrt{\left(-a^{2}+L^{2}+x^{2}+y^{2}\right)^{2}+4a^{2}L^{2}}-a^{2}+L^{2}+x^{2}+y^{2}}}{\sqrt{2}}\,.$
(4.16)

This is to be substituted into $I_{xy}$ in (4.15). Note, however, that the
resultant expression is an analytic function of the rotation parameter $a$ and
that we are to first do the integration on the $x$ and $z$ variables, then set
$x=\pm L$ and $z=\pm L$ and next take the $L\to\infty$ limit, keeping the
parameter $a$ finite and fixed the whole time. Thus the whole integral
$I_{xy}$ must be an analytic function of the parameter $a$, or better yet,
$u:=a/L$ about $u=0$ for fixed $a$. Specifically, we find

$\frac{2L}{\left(L^{2}+x^{2}+y^{2}\right)^{3/2}}+\frac{2u^{2}L^{3}\left(3\left(x^{2}+y^{2}\right)-2L^{2}\right)}{\left(L^{2}+x^{2}+y^{2}\right)^{7/2}}+\frac{3u^{4}L^{5}\left(8L^{4}-40L^{2}\left(x^{2}+y^{2}\right)+15\left(x^{2}+y^{2}\right)^{2}\right)}{4\left(L^{2}+x^{2}+y^{2}\right)^{11/2}}$

for the integrand of $I_{xy}$ to quintic order about $u=0$. We next integrate
this $u$-series term by term in both $x$ and $y$ over the square $x\in(-L,L)$
and $y\in(-L,L)$ to arrive at

$\displaystyle I_{xy}=\frac{M}{108\pi}\,\lim_{u\to
0}\,\left(36\pi+\sqrt{3}\left(-32+u^{2}\right)u^{2}\right)=\frac{M}{3}\,.$
(4.17)

As for $I_{xz}$, we similarly have

$\displaystyle
r=\frac{\sqrt{\sqrt{\left(-a^{2}+L^{2}+x^{2}+z^{2}\right)^{2}+4a^{2}z^{2}}-a^{2}+L^{2}+x^{2}+z^{2}}}{\sqrt{2}}$
(4.18)

for the root of $r=r(x,L,z)$ (4.4), and the corresponding series expansion in
$u$ is

$\frac{4L}{\left(L^{2}+x^{2}+z^{2}\right)^{3/2}}+\frac{4u^{2}L^{3}\left(x^{2}+L^{2}-4z^{2}\right)}{\left(L^{2}+x^{2}+z^{2}\right)^{7/2}}+\frac{9u^{4}L^{5}\left(8z^{4}-12z^{2}\left(x^{2}+L^{2}\right)+\left(x^{2}+L^{2}\right)^{2}\right)}{2\left(L^{2}+x^{2}+z^{2}\right)^{11/2}}$

for the integrand of $I_{xz}$ to quintic order about $u=0$. We integrate term
by term in both $x$ and $z$ over the square $x\in(-L,L)$ and $z\in(-L,L)$, and
get

$\displaystyle I_{xz}=\frac{M}{108\pi}\,\lim_{u\to
0}\,\left(72\pi+\sqrt{3}\left(32-u^{2}\right)u^{2}\right)=\frac{2M}{3}\,.$
(4.19)

Hence the total mass $Q(\bar{\xi})$ is

$\displaystyle Q(\bar{\xi})=M\,,$ (4.20)

in accordance with (3.8).

The azimuthal (rotational) Killing vector reads

$\displaystyle\bar{\zeta}^{a}=\left(x\,\partial{{}_{y}}-y\,\partial_{x}\right)^{a}\,.$
(4.21)

in the Kerr-Schild coordinates. Using the same foliation (4.6) and the six
boundary faces (4.7), we end up with $Q(\bar{\zeta})$ similar to (4). This
time the three nontrivial components of the 2-form potential $\ell$ (2.3) are

$\displaystyle\ell^{tx}(\bar{\zeta})=$
$\displaystyle\frac{Mr^{2}}{(a^{2}+r^{2})^{3}(a^{2}z^{2}+r^{4})^{3}}\Big{(}a^{8}yz^{4}-2a^{6}r^{2}yz^{2}\left(r^{2}-2z^{2}\right)-3a^{5}r^{3}xz^{2}\left(x^{2}+y^{2}\right)\Big{.}$
$\displaystyle\quad+a^{4}r^{4}y\left(r^{4}-6r^{2}z^{2}+z^{2}\left(4x^{2}+4y^{2}+5z^{2}\right)\right)+a^{3}r^{5}x\left(r^{2}-z^{2}\right)\left(x^{2}+y^{2}\right)\Big{.}$
$\displaystyle\quad\Big{.}+2a^{2}r^{6}y\left(r^{4}-3r^{2}z^{2}+z^{2}\left(x^{2}+y^{2}+z^{2}\right)\right)+3ar^{9}x\left(x^{2}+y^{2}\right)\Big{.}$
(4.22a)
$\displaystyle\quad+r^{10}y\left(r^{2}-2\left(x^{2}+y^{2}+z^{2}\right)\right)\Big{)}\,,$
$\displaystyle\ell^{ty}(\bar{\zeta})=$
$\displaystyle-\frac{Mr^{2}}{(a^{2}+r^{2})^{3}(a^{2}z^{2}+r^{4})^{3}}\Big{(}a^{8}xz^{4}-2a^{6}r^{2}xz^{2}\left(r^{2}-2z^{2}\right)+3a^{5}r^{3}yz^{2}\left(x^{2}+y^{2}\right)\Big{.}$
$\displaystyle\quad+a^{4}r^{4}x\left(r^{4}-6r^{2}z^{2}+z^{2}\left(4x^{2}+4y^{2}+5z^{2}\right)\right)-a^{3}r^{5}y\left(r^{2}-z^{2}\right)\left(x^{2}+y^{2}\right)\Big{.}$
$\displaystyle\quad\Big{.}+2a^{2}r^{6}x\left(r^{4}-3r^{2}z^{2}+z^{2}\left(x^{2}+y^{2}+z^{2}\right)\right)-3ar^{9}y\left(x^{2}+y^{2}\right)\Big{.}$
(4.22b)
$\displaystyle\quad+r^{10}x\left(r^{2}-2\left(x^{2}+y^{2}+z^{2}\right)\right)\Big{)}\,,$
$\displaystyle\ell^{tz}(\bar{\zeta})=$
$\displaystyle-\frac{aMr^{3}(x^{2}+y^{2})z(-3r^{4}+a^{2}z^{2})}{(r^{4}+a^{2}z^{2})^{3}}\,,$
(4.22c)

where we have again used (4.11) and (4.12). We next substitute (4.22) into
$Q(\bar{\zeta})$. Once again using the symmetry $r(x,y)=r(y,x)$ (4.4), we can
add the integrals on the $xz$\- and $yz$-surfaces (in complete analogy to how
(4.15) was found) to finally arrive at

$\displaystyle Q(\bar{\zeta})$
$\displaystyle=\frac{aM}{4\pi}\left(J_{xy}+J_{xz}\right)\,,$ (4.23)
$\displaystyle J_{xy}$
$\displaystyle:=\lim_{L\to\infty}\bigg{(}\int_{-L}^{L}\int_{-L}^{L}dx\,dy\,\frac{r(x,y,L)L\left(r^{2}+a^{2}\right)\left(r^{2}-L^{2}\right)\left(3r^{4}-a^{2}L^{2}\right)}{\left(r^{4}+a^{2}L^{2}\right)^{3}}\bigg{)}\,,$
$\displaystyle J_{xz}$
$\displaystyle:=\lim_{L\to\infty}\bigg{(}\int_{-L}^{L}\int_{-L}^{L}dx\,dz\,\frac{2\,r(x,L,z)^{3}L\left(r^{2}-z^{2}\right)\left(3r^{6}+a^{2}r^{4}-3a^{4}z^{2}-a^{2}r^{2}z^{2}\right)}{\left(r^{2}+a^{2}\right)\left(r^{4}+a^{2}z^{2}\right)^{3}}\bigg{)}\,.$

We calculate $J_{xy}$ similar to how we have calculated $I_{xy}$. After the
substitution of (4.16), the series expansion of the integrand of $J_{xy}$
about $u=0$ is

$\displaystyle\frac{3L\left(x^{2}+y^{2}\right)}{\left(L^{2}+x^{2}+y^{2}\right)^{5/2}}-\frac{5\left(L^{3}\left(x^{2}+y^{2}\right)\left(4L^{2}-3\left(x^{2}+y^{2}\right)\right)\right)}{2\left(L^{2}+x^{2}+y^{2}\right)^{9/2}}u^{2}$
$\displaystyle+\frac{21L^{5}\left(x^{2}+y^{2}\right)\left(8L^{4}-20L^{2}\left(x^{2}+y^{2}\right)+5\left(x^{2}+y^{2}\right)^{2}\right)}{8\left(L^{2}+x^{2}+y^{2}\right)^{13/2}}u^{4}+O(u^{5})\,.$

Similarly, the substitution of (4.18) gives the series expansion of the
integrand of $J_{xz}$ as

$\displaystyle\frac{6L\left(L^{2}+x^{2}\right)}{\left(L^{2}+x^{2}+z^{2}\right)^{5/2}}+\frac{5L^{3}\left(L^{2}+x^{2}\right)\left(L^{2}+x^{2}-6z^{2}\right)}{\left(L^{2}+x^{2}+z^{2}\right)^{9/2}}u^{2}$
$\displaystyle+\frac{21L^{5}\left(L^{2}+x^{2}\right)\left(-16z^{2}\left(L^{2}+x^{2}\right)+\left(L^{2}+x^{2}\right)^{2}+16z^{4}\right)}{4\left(L^{2}+x^{2}+z^{2}\right)^{13/2}}u^{4}+O(u^{5})\,.$

The integration of these on the relevant squares then leads to

$\displaystyle J_{xy}=\frac{aM}{2430\pi}\,\lim_{u\to
0}\,\left(810(\pi-\sqrt{3})-495\sqrt{3}u^{2}+82\sqrt{3}u^{4}\right)=\frac{aM}{3}-\frac{aM\sqrt{3}}{3\pi}\,,$
(4.24) $\displaystyle J_{xz}=\frac{aM}{2430\pi}\,\lim_{u\to
0}\,\left(810(2\pi+\sqrt{3})+495\sqrt{3}u^{2}-82\sqrt{3}u^{4}\right)=\frac{2aM}{3}+\frac{aM\sqrt{3}}{3\pi}\,.$
(4.25)

giving the total angular momentum

$\displaystyle Q(\bar{\zeta})=aM\,,$ (4.26)

in accordance with (3.10).

##  5 Kerr black hole in Weyl canonical coordinates

The Kerr black hole can also be written in the Weyl canonical coordinates [5,
11]

$\displaystyle
ds^{2}=-e^{2U}(dt+A\,d\phi)^{2}+e^{-2U}\left(e^{2k}\left(d\rho^{2}+dz^{2}\right)+\rho^{2}d\phi^{2}\right)\,,$
(5.1)

where

$\displaystyle
e^{2U}=\frac{\left(R_{+}+R_{-}\right)^{2}-4M^{2}+\frac{a^{2}}{M^{2}-a^{2}}\left(R_{+}-R_{-}\right)^{2}}{\left(R_{+}+R_{-}+2M\right)^{2}+\frac{a^{2}}{M^{2}-a^{2}}\left(R_{+}-R_{-}\right)^{2}}\,,$
(5.2) $\displaystyle
e^{2k}=\frac{\left(R_{+}+R_{-}\right)^{2}-4M^{2}+\frac{a^{2}}{M^{2}-a^{2}}\left(R_{+}-R_{-}\right)^{2}}{4R_{+}R_{-}}\,,$
(5.3) $\displaystyle
A=\frac{2aM\left(M+\frac{R_{+}+R_{-}}{2}\right)\left(1-\frac{\left(R_{+}-R_{-}\right)^{2}}{4\left(M^{2}-a^{2}\right)}\right)}{\frac{1}{4}\left(R_{+}+R_{-}\right)^{2}-M^{2}+a^{2}\frac{\left(R_{+}-R_{-}\right)^{2}}{4\left(M^{2}-a^{2}\right)}}\,,$
(5.4) $\displaystyle
R_{\pm}=\sqrt{\rho^{2}+\left(z\pm\sqrt{M^{2}-a^{2}}\right)^{2}}=r-M\pm\sqrt{M^{2}-a^{2}}\,\cos{\theta}\,.$
(5.5)

The transformation back to Boyer-Lindquist coordinates $(t,r,\theta,\phi)$
(3.1) is given by

$\displaystyle\rho=\sqrt{r^{2}-2Mr+a^{2}}\,\sin{\theta}\,,\qquad
z=(r-M)\cos{\theta}\,.$ (5.6)

The flat background can be obtained at the limit of $M=0,\,a=0$ in (5.1):

$\displaystyle\bar{g}_{ab}={\rm diag}(-1,1,\rho^{2},1)\,,$ (5.7)

and it admits the foliation (2.2) with

$\gamma_{ab}={\rm diag}(0,1,\rho^{2},1)\,,\quad
n^{a}=\left(-\partial_{t}\right)^{a}\,.$ (5.8)

$\gamma_{ab}$ is the metric on the 3-dimensional Euclidean space in
cylindrical polar coordinates. For this particular case, let us first consider
a right-angled circular cylinder which has its axis of symmetry on the
$z$-axis, of radius $R$ and of height $2L$. There are three faces in total,
the two top and bottom caps/disks and the side surface, with the following
normal vectors and the corresponding two-dimensional metrics, respectively:

$\displaystyle
z_{\pm}^{a}=\left(\pm\partial_{z}\right)^{a}\quad\mbox{and}\quad\big{(}q_{z}\big{)}_{ab}={\rm
diag}\big{(}0,1,\rho^{2},0\big{)}\,,$ (5.9)
$\displaystyle\rho^{a}=\left(\partial_{\rho}\right)^{a}\quad\mbox{and}\quad\big{(}q_{\rho}\big{)}_{ab}={\rm
diag}\big{(}0,0,\rho^{2},1\big{)}\,.$ (5.10)

The boundary is then “the cylinder at infinity” found by taking the
$R\to\infty$ and $L\to\infty$ limits simultaneously along the directions of
the normal vectors. Finally, the background again has two globally defined,
timelike $\bar{\xi}^{a}=\left(-\partial_{t}\right)^{a}$ and spacelike
$\bar{\zeta}^{a}=\left(\partial_{\phi}\right)^{a}$, Killing vectors, and the
integration on the boundary amounts to three separate integrals on the three
faces of “the cylinder at infinity”

$\displaystyle Q(\bar{\xi})$
$\displaystyle=\frac{1}{8\pi}\lim_{L\to\infty}\bigg{(}\int_{0}^{2\pi}\int_{0}^{R}d\rho\,d\phi\,\rho\,{n}_{[a}\,z_{+\,b]}\ell^{ab}(\bar{\xi})\rvert_{z=L}+\int_{0}^{2\pi}\int_{0}^{R}d\rho\,d\phi\,\rho\,{n}_{[a}\,z_{-\,b]}\ell^{ab}(\bar{\xi})\rvert_{z=-L}\bigg{)}$
$\displaystyle\qquad\quad+\frac{1}{8\pi}\lim_{R\to\infty}\int_{0}^{2\pi}\int_{-L}^{L}dz\,d\phi\,\rho\,{n}_{[a}\,\rho_{b]}\ell^{ab}(\bar{\xi})\rvert_{\rho=R}\,,$
$\displaystyle=\frac{1}{4}\lim_{\begin{subarray}{c}L\to\infty\\\
R\to\infty\end{subarray}}\int_{0}^{R}d\rho\,\rho\Big{(}\ell^{tz}(\bar{\xi})\rvert_{z=L}-\ell^{tz}(\bar{\xi})\rvert_{z=-L}\Big{)}+\frac{1}{4}\lim_{\begin{subarray}{c}R\to\infty\\\
L\to\infty\end{subarray}}\int_{-L}^{L}dz\,R\,\ell^{t\rho}(\bar{\xi})\rvert_{\rho=R}\,,$
(5.11)

where $a$ and $M$ is to be kept fixed, and we used the fact that all relevant
parts are independent of $\phi$. We relegate the explicit and unsavory forms
of the two nontrivial components of the 2-form potential $\ell$ (2.3) for the
two Killing vectors $\bar{\xi}$ and $\bar{\zeta}$ to appendix A. We calculate
the integrals separately as we did in section 4 by defining

$\displaystyle Q(\bar{\xi})$ $\displaystyle=I_{\rho}+I_{z}\,,$ (5.12)
$\displaystyle I_{\rho}$
$\displaystyle:=\frac{1}{4}\lim_{\begin{subarray}{c}L\to\infty\\\
R\to\infty\end{subarray}}\int_{0}^{R}d\rho\,\rho\Big{(}\ell^{tz}(\bar{\xi})\rvert_{z=L}-\ell^{tz}(\bar{\xi})\rvert_{z=-L}\Big{)}\,,$
$\displaystyle I_{z}$
$\displaystyle:=\frac{1}{4}\lim_{\begin{subarray}{c}R\to\infty\\\
L\to\infty\end{subarray}}\int_{-L}^{L}dz\,R\,\ell^{t\rho}(\bar{\xi})\rvert_{\rho=R}\,.$

In Kerr-Schild coordinates, the integrands of $I_{xy}$, $I_{xz}$ (4.15) and
$J_{xy}$, $J_{xz}$ (4.23) depended only on the rotation parameter $a$, and the
mass parameter $M$ came out as an overall factor. However, as can be seen from
the relevant components of the 2-form potential $\ell$ in appendix A, the
integrands of $I_{\rho}$ and $I_{z}$ (5.12) explicitly depend on both $a$ and
$M$. Moreover, there is also the complication that arises from taking the two
separate limits $R\to\infty$ and $L\to\infty$. In order to have better control
on the geometry and these limits, we set $L=kR$, where $k>0$ is a _finite
chubbiness parameter_ between the length $L$ and the radius $R$ of “the
cylinder at infinity”, just before the limiting step and take $R\to\infty$
later. In analogy to the discussion in section 4, we define

$\displaystyle u_{L}:=\frac{a}{L}\;,v_{L}:=\frac{M}{L}\,,\qquad
u_{R}:=\frac{a}{R}\;,v_{R}:=\frac{M}{R}$ (5.13)

as new (small) parameters which will allow us to expand the integrands of
$I_{\rho}$ and $I_{z}$ (5.12) in a (double) Taylor series expansion222For
convenience, recall that a function $f(u,v)$ which is analytic about the point
$(u,v)=(0,0)$ can be expanded in a double Taylor series as
$f(u,v)=\sum_{m\geq 0}\sum_{n\geq
0}\frac{1}{m!}\frac{1}{n!}\left(\frac{\partial^{m+n}f}{\partial
u^{m}\,\partial v^{n}}\right)_{(0,0)}\,u^{m}\,v^{n}\,.$ to evaluate $I_{\rho}$
and $I_{z}$ (5.12).

Thus, the expansion of the integrand of $I_{\rho}$ is

$\frac{4L^{2}\rho}{\left(L^{2}+\rho^{2}\right)^{3/2}}v_{L}+\frac{4L^{3}\rho}{\left(L^{2}+\rho^{2}\right)^{2}}\Big{(}1+\frac{L^{2}}{L^{2}+\rho^{2}}\Big{)}v_{L}^{2}$

to cubic order terms. Integrating this expression term by term on
$\rho\in[0,R)$ and back-substituting $v_{L}$ later, we obtain

$\displaystyle I_{\rho}=\frac{1}{4}\lim_{\begin{subarray}{c}L\to\infty\\\
R\to\infty\end{subarray}}M\left(4-\frac{4L}{\sqrt{L^{2}+R^{2}}}+\frac{3MR^{2}}{L\left(L^{2}+R^{2}\right)}+\frac{MR^{2}L}{\left(L^{2}+R^{2}\right)^{2}}\right)\,.$
(5.14)

Finally, letting $L=kR$, we can see the effect of the chubbiness parameter $k$
on the result

$\displaystyle
I_{\rho}=\frac{1}{4}\lim_{R\to\infty}M\left(4-\frac{4k}{\sqrt{k^{2}+1}}+\frac{3M}{R\,k\left(k^{2}+1\right)}+\frac{Mk}{R\left(k^{2}+1\right)^{2}}\right)=M-M\frac{k}{\sqrt{k^{2}+1}}\,.$
(5.15)

Similarly, the expansion of the integrand of $I_{z}$ is

$\frac{2R^{3}}{\left(R^{2}+z^{2}\right)^{3/2}}v_{R}+\frac{R^{4}}{\left(R^{2}+z^{2}\right)^{2}}\Big{(}\frac{5}{2}+\frac{2z^{2}}{R^{2}+z^{2}}\Big{)}v_{R}^{2}$

to cubic order. Term by term integration of this expression on $z\in(-L,L)$
and back-substitution of $v_{R}$ gives

$\displaystyle I_{z}=\frac{1}{4}\lim_{\begin{subarray}{c}R\to\infty\\\
L\to\infty\end{subarray}}M\left(\frac{4L}{\sqrt{L^{2}+R^{2}}}+\frac{ML^{3}}{\left(L^{2}+R^{2}\right)^{2}}+\frac{2ML}{L^{2}+R^{2}}+\frac{3M}{R}\arctan{(L/R)}\right)\,.$
(5.16)

Finally, with $L=kR$, we have

$\displaystyle
I_{z}=\frac{1}{4}\lim_{R\to\infty}M\left(\frac{4k}{\sqrt{k^{2}+1}}+\frac{Mk^{3}}{R\left(k^{2}+1\right)^{2}}+\frac{2Mk}{R\left(k^{2}+1\right)}+\frac{3M}{R}\arctan{k}\right)=M\frac{k}{\sqrt{k^{2}+1}}\,.$
(5.17)

Substituting (5.15) and (5.17) in (5.12), the total mass $Q(\bar{\xi})$ is

$\displaystyle Q(\bar{\xi})=M\,,$ (5.18)

as expected.

We proceed similarly for the Killing vector $\bar{\zeta}$ to calculate the
angular momentum. We first write

$\displaystyle Q(\bar{\zeta})$ $\displaystyle=J_{\rho}+J_{z}\,,$ (5.19)
$\displaystyle J_{\rho}$
$\displaystyle:=\frac{1}{4}\lim_{\begin{subarray}{c}L\to\infty\\\
R\to\infty\end{subarray}}\int_{0}^{R}d\rho\,\rho\,\Big{(}\ell^{tz}(\bar{\zeta})\rvert_{z=L}-\ell^{tz}(\bar{\zeta})\rvert_{z=-L}\Big{)}\,,$
$\displaystyle J_{z}$
$\displaystyle:=\frac{1}{4}\lim_{\begin{subarray}{c}R\to\infty\\\
L\to\infty\end{subarray}}\int_{-L}^{L}dz\,R\,\ell^{t\rho}(\bar{\zeta})\rvert_{\rho=R}\,.$

Analogously, a series expansion of the integrand of $J_{\rho}$ gives

$\frac{6L^{3}\rho^{3}}{\left(L^{2}+\rho^{2}\right)^{5/2}}u_{L}v_{L}-\frac{8L^{4}\rho^{3}}{\left(L^{2}+\rho^{2}\right)^{3}}u_{L}v_{L}^{2}$

whereas that of $J_{z}$ is

$\frac{3R^{6}}{\left(R^{2}+z^{2}\right)^{5/2}}u_{R}v_{R}-\frac{4R^{7}}{\left(R^{2}+z^{2}\right)^{3}}u_{R}v_{R}^{2}$

both to quartic order. Integrating on $\rho\in[0,R)$ and $z\in(-L,L)$,
respectively, and back-substituting, we get

$\displaystyle J_{\rho}=\frac{1}{4}\lim_{\begin{subarray}{c}L\to\infty\\\
R\to\infty\end{subarray}}2aM\left(2-\frac{2L}{\sqrt{L^{2}+R^{2}}}-\frac{LR^{2}}{\left(L^{2}+R^{2}\right)^{3/2}}-\frac{MR^{4}}{L\left(L^{2}+R^{2}\right)^{2}}\right)\,,$
$\displaystyle J_{z}=\frac{1}{4}\lim_{\begin{subarray}{c}R\to\infty\\\
L\to\infty\end{subarray}}aM\left(\frac{4L}{\sqrt{L^{2}+R^{2}}}+\frac{2LR^{2}}{\left(L^{2}+R^{2}\right)^{3/2}}-\frac{2LMR^{2}}{\left(L^{2}+R^{2}\right)^{2}}-\frac{3LM}{L^{2}+R^{2}}-\frac{3M}{R}\arctan{(L/R)}\right)\,.$

Finally, letting $L=kR$, we have

$\displaystyle J_{\rho}$
$\displaystyle=\frac{1}{4}\lim_{R\to\infty}2aM\left(2-2\frac{k}{\sqrt{k^{2}+1}}-\frac{k}{\left(k^{2}+1\right)^{3/2}}-\frac{M}{R\,k\left(k^{2}+1\right)^{2}}\right)$
(5.20)
$\displaystyle=aM\left(1-\frac{k}{\sqrt{k^{2}+1}}-\frac{1}{2}\frac{k}{\left(k^{2}+1\right)^{3/2}}\right)\,,$
$\displaystyle J_{z}$
$\displaystyle=\frac{1}{4}\lim_{R\to\infty}aM\left(4\frac{k}{\sqrt{k^{2}+1}}+\frac{2k}{\left(k^{2}+1\right)^{3/2}}-\frac{2Mk}{R\left(k^{2}+1\right)^{2}}-\frac{3Mk}{R\left(k^{2}+1\right)}-\frac{3M}{R}\arctan{k}\right)$
$\displaystyle=aM\left(\frac{k}{\sqrt{k^{2}+1}}+\frac{1}{2}\frac{k}{\left(k^{2}+1\right)^{3/2}}\right)\,.$

Substituting (5.20) in (5.19), the total angular momentum $Q(\bar{\zeta})$ is

$\displaystyle Q(\bar{\zeta})=aM\,,$ (5.21)

as expected.

##  6 Conclusions

In this work, we have concisely reviewed the construction of asymptotic
conserved gravitational charges using background Killing vectors. One of the
advantages of this description is that the spacetime whose charges are to be
calculated does not need to have a Killing vector, rather it is enough if the
spacetime it asymptotes to admits one itself. Using this method, we have
calculated the total mass/energy and the total angular momentum of the Kerr
black hole with its timelike and rotational Killing vectors in the Boyer-
Lindquist, Kerr-Schild and Weyl canonical coordinates in sections 3, 4, 5,
respectively, and shown that they all agree, explicitly verifying the
background gauge independence of this prescription. Even though the
calculations are harder for the Kerr-Schild and Weyl canonical coordinates, we
have overcome the difficulties involved by introducing appropriate series
expansions in suitably chosen “small parameters”. We would like to emphasize
that, to our knowledge, the presentations given in sections 4 and 5 are
original.

One of our main motivations was to examine how the gravitational flux
integration through “the asymptotic boundary at infinity” depended on the
geometry of the integration surface. Obviously, “the cube at infinity” for the
Kerr-Schild form and “the cylinder at infinity” for the Weyl canonical
coordinates are not smooth geometries as for “the sphere at infinity” for the
Boyer-Lindquist coordinates. Hence, we were unable to determine the charges
for the former via the shortcut of evaluating the limit of the relevant
integrand at the spatial boundary and integrating the result on the boundary
surface later, unlike the case for the Boyer-Lindquist coordinates. Moreover,
for the calculation of the total angular momentum in the case of the Kerr-
Schild form with its boundary as a “cube at infinity”, note that the
contribution of $J_{xy}$ (4.24) is less than _half_ of that of $J_{xz}$
(4.25), which lets us deduce that the relevant gravitational flux is more
abundant along the rotation axis compared to the other axes. In Weyl canonical
coordinates, we introduced a chubbiness parameter $k$ while describing the
spatial boundary. Although $k$ has an effect on the contribution of the
relevant gravitational fluxes along different directions (5.20), it is
remarkable that the total charges remain independent of $k$.

For the case of the Kerr-Schild form, if we were to choose a spatial boundary
as a “generic rectangular prism at infinity” with three unequal side lengths,
say $(L_{x},L_{y},L_{z})$, instead of one (equal) side length $L$ “cube at
infinity”, we would expect to recover the correct calculations again by using
three series expansions along the respective pairwise side surfaces in the
three “small parameters” $(a/L_{x},a/L_{y},a/L_{z})$. In retrospect, we
conjecture, at least heuristically, that provided the closed and simply
connected spatial boundary is also piecewise smooth, i.e. a finite union of
smooth surfaces, the integration of the relevant gravitational flux, apart
from obvious complications resulting in the pertinent calculations, does not
depend on the geometry of the boundary.

Finally, as advocates of the use of background Killing isometries for the
determination of gravitational charges, the details of our calculations should
have at least some pedagogical value for those who want to appreciate the
delicate technicalities involved.

Acknowledgments
It is a pleasure to thank Ulf Lindström for a careful reading of this
manuscript.

##  Appendix A The 2-form potential $\ell$ in Weyl coordinates

For completeness sake, here we list the relevant non-vanishing components of
$\ell(\bar{\xi})$ and $\ell(\bar{\zeta})$ that are needed in determining the
gravitational charges in the Weyl canonical coordinates of section 5.

$\displaystyle\ell^{t\rho}(\bar{\xi})$
$\displaystyle=\frac{1}{8\rho^{3}}\bigg{(}\tfrac{\rho^{4}\left(\frac{a^{2}(R_{-}-R_{+})^{2}}{M^{2}-a^{2}}+(2M+R_{-}+R_{+})^{2}\right)}{R_{-}^{3}R_{+}}+\tfrac{\rho^{4}\left(\frac{a^{2}(R_{-}-R_{+})^{2}}{M^{2}-a^{2}}+(2M+R_{-}+R_{+})^{2}\right)}{R_{-}R_{+}^{3}}$
$\displaystyle+\rho^{2}\left(\tfrac{\frac{a^{2}(R_{-}-R_{+})^{2}}{M^{2}-a^{2}}+(2M+R_{-}+R_{+})^{2}}{R_{-}R_{+}}-4\right)-\tfrac{2\rho^{4}\left(\left(a^{2}-M^{2}\right)(R_{-}+R_{+})(2M+R_{-}+R_{+})+a^{2}(R_{-}-R_{+})^{2}\right)}{R_{-}^{2}R_{+}^{2}\left(a^{2}-M^{2}\right)}$
$\displaystyle+\tfrac{8\left(\rho^{2}\left(a^{2}(R_{-}-R_{+})^{2}-\left(a^{2}-M^{2}\right)(2M+R_{-}+R_{+})^{2}\right)^{2}-a^{2}M^{2}(2M+R_{-}+R_{+})^{2}\left(4a^{2}-4M^{2}+(R_{-}-R_{+})^{2}\right)^{2}\right)}{\left(a^{2}(R_{-}-R_{+})^{2}-\left(a^{2}-M^{2}\right)(2M+R_{-}+R_{+})^{2}\right)\left(4a^{2}\left(M^{2}-R_{-}R_{+}\right)-4M^{4}+M^{2}(R_{-}+R_{+})^{2}\right)}$
$\displaystyle+\tfrac{8a^{2}M^{2}\rho^{2}(2M+R_{-}+R_{+})^{2}\left(4a^{2}-4M^{2}+(R_{-}-R_{+})^{2}\right)^{2}\left(2a^{2}\left(R_{-}^{2}+R_{+}^{2}\right)-M^{2}(R_{-}+R_{+})^{2}\right)}{R_{-}R_{+}\left(M^{2}(2M+R_{-}+R_{+})^{2}-4a^{2}(M+R_{-})(M+R_{+})\right)\left(4a^{2}\left(M^{2}-R_{-}R_{+}\right)-4M^{4}+M^{2}(R_{-}+R_{+})^{2}\right)^{2}}$
$\displaystyle-\tfrac{8\rho^{4}\left(M^{2}(2M+R_{-}+R_{+})^{2}-4a^{2}(M+R_{-})(M+R_{+})\right)\left(2a^{2}\left(R_{-}^{2}+R_{+}^{2}\right)-M^{2}(R_{-}+R_{+})^{2}\right)}{R_{-}R_{+}\left(4a^{2}\left(M^{2}-R_{-}R_{+}\right)-4M^{4}+M^{2}(R_{-}+R_{+})^{2}\right)^{2}}$
$\displaystyle+\tfrac{8a^{2}M^{2}\rho^{2}(2M+R_{-}+R_{+})^{2}\left(4a^{2}-4M^{2}+(R_{-}-R_{+})^{2}\right)^{2}\left(2a^{2}\left(M(R_{-}+R_{+})+R_{-}^{2}+R_{+}^{2}\right)-M^{2}(R_{-}+R_{+})(2M+R_{-}+R_{+})\right)}{R_{-}R_{+}\left(a^{2}(R_{-}-R_{+})^{2}-\left(a^{2}-M^{2}\right)(2M+R_{-}+R_{+})^{2}\right)^{2}\left(4a^{2}\left(M^{2}-R_{-}R_{+}\right)-4M^{4}+M^{2}(R_{-}+R_{+})^{2}\right)}$
$\displaystyle-\tfrac{8\rho^{4}\left(M^{2}(2M+R_{-}+R_{+})^{2}-4a^{2}(M+R_{-})(M+R_{+})\right)^{2}\left(2a^{2}\left(M(R_{-}+R_{+})+R_{-}^{2}+R_{+}^{2}\right)-M^{2}(R_{-}+R_{+})(2M+R_{-}+R_{+})\right)}{R_{-}R_{+}\left(a^{2}(R_{-}-R_{+})^{2}-\left(a^{2}-M^{2}\right)(2M+R_{-}+R_{+})^{2}\right)^{2}\left(4a^{2}\left(M^{2}-R_{-}R_{+}\right)-4M^{4}+M^{2}(R_{-}+R_{+})^{2}\right)}$
$\displaystyle+\tfrac{8a^{2}M^{2}\rho^{2}(2M+R_{-}+R_{+})\left(4a^{2}-4M^{2}+(R_{-}-R_{+})^{2}\right)\left(4a^{2}(R_{-}+R_{+})-4M^{2}(R_{-}+R_{+})-4M(R_{-}-R_{+})^{2}-(R_{-}-R_{+})^{2}(R_{-}+R_{+})\right)}{R_{-}R_{+}\left(M^{2}(2M+R_{-}+R_{+})^{2}-4a^{2}(M+R_{-})(M+R_{+})\right)\left(4a^{2}\left(M^{2}-R_{-}R_{+}\right)-4M^{4}+M^{2}(R_{-}+R_{+})^{2}\right)}$
$\displaystyle-\tfrac{8\rho^{2}\left(4a^{2}\left(-\left(\rho^{2}\left(M(R_{-}+R_{+})+R_{-}^{2}+R_{+}^{2}\right)\right)-R_{-}R_{+}(M+R_{-})(M+R_{+})\right)+M^{2}(2M+R_{-}+R_{+})\left(R_{-}R_{+}(2M+R_{-}+R_{+})+2\rho^{2}(R_{-}+R_{+})\right)\right)}{R_{-}R_{+}\left(4a^{2}\left(M^{2}-R_{-}R_{+}\right)-4M^{4}+M^{2}(R_{-}+R_{+})^{2}\right)}$
$\displaystyle+4\left(\rho^{2}-\tfrac{\rho^{2}\left(a^{2}(R_{-}-R_{+})^{2}-\left(a^{2}-M^{2}\right)(2M+R_{-}+R_{+})^{2}\right)^{2}-a^{2}M^{2}(2M+R_{-}+R_{+})^{2}\left(4a^{2}-4M^{2}+(R_{-}-R_{+})^{2}\right)^{2}}{\left(a^{2}(R_{-}-R_{+})^{2}-\left(a^{2}-M^{2}\right)(2M+R_{-}+R_{+})^{2}\right)\left(4a^{2}\left(M^{2}-R_{-}R_{+}\right)-4M^{4}+M^{2}(R_{-}+R_{+})^{2}\right)}\right)\bigg{)}\,,$
$\displaystyle\vspace{1cm}\ell^{tz}(\bar{\xi})$
$\displaystyle=\frac{1}{8}\bigg{(}\tfrac{\left(z-\sqrt{M^{2}-a^{2}}\right)\left(\frac{a^{2}(R_{-}-R_{+})^{2}}{M^{2}-a^{2}}+(2M+R_{-}+R_{+})^{2}\right)}{R_{-}^{3}R_{+}}+\tfrac{\left(\sqrt{M^{2}-a^{2}}+z\right)\left(\frac{a^{2}(R_{-}-R_{+})^{2}}{M^{2}-a^{2}}+(2M+R_{-}+R_{+})^{2}\right)}{R_{-}R_{+}^{3}}$
$\displaystyle-\tfrac{\frac{2a^{2}(R_{-}-R_{+})\left(\frac{z-\sqrt{M^{2}-a^{2}}}{R_{-}}-\frac{\sqrt{M^{2}-a^{2}}+z}{R_{+}}\right)}{M^{2}-a^{2}}+2(2M+R_{-}+R_{+})\left(\frac{z-\sqrt{M^{2}-a^{2}}}{R_{-}}+\frac{\sqrt{M^{2}-a^{2}}+z}{R_{+}}\right)}{R_{-}R_{+}}$
$\displaystyle+\tfrac{8\left(\frac{R_{+}\left(2a^{2}-M^{2}\right)\left(\sqrt{M^{2}-a^{2}}-z\right)}{R_{-}}-\frac{R_{-}\left(2a^{2}-M^{2}\right)\left(\sqrt{M^{2}-a^{2}}+z\right)}{R_{+}}+2M^{2}z\right)}{\rho^{2}\left(a^{2}(R_{-}-R_{+})^{2}-\left(a^{2}-M^{2}\right)(2M+R_{-}+R_{+})^{2}\right)\left(4a^{2}\left(M^{2}-R_{-}R_{+}\right)-4M^{4}+M^{2}(R_{-}+R_{+})^{2}\right)^{2}}\times$
$\displaystyle-\tfrac{8a^{2}(R_{-}-R_{+})\left(R_{-}\left(\sqrt{M^{2}-a^{2}}+z\right)+R_{+}\left(\sqrt{M^{2}-a^{2}}-z\right)\right)+\left(a^{2}-M^{2}\right)(2M+R_{-}+R_{+})\left(R_{-}\left(\sqrt{M^{2}-a^{2}}+z\right)+R_{+}\left(z-\sqrt{M^{2}-a^{2}}\right)\right)}{\rho^{2}R_{-}R_{+}\left(a^{2}(R_{-}-R_{+})^{2}-\left(a^{2}-M^{2}\right)(2M+R_{-}+R_{+})^{2}\right)^{2}\left(4a^{2}\left(M^{2}-R_{-}R_{+}\right)-4M^{4}+M^{2}(R_{-}+R_{+})^{2}\right)}\times$
$\displaystyle+\tfrac{16}{\rho^{2}R_{-}R_{+}\left(a^{2}(R_{-}-R_{+})^{2}-\left(a^{2}-M^{2}\right)(2M+R_{-}+R_{+})^{2}\right)\left(4a^{2}\left(M^{2}-R_{-}R_{+}\right)-4M^{4}+M^{2}(R_{-}+R_{+})^{2}\right)}\times$
$\displaystyle\scalebox{0.7}{$\Big{.}\left(a^{2}(R_{-}-R_{+})\left(R_{-}\left(\sqrt{M^{2}-a^{2}}+z\right)+R_{+}\left(\sqrt{M^{2}-a^{2}}-z\right)\right)+\left(a^{2}-M^{2}\right)(2M+R_{-}+R_{+})\left(R_{-}\left(\sqrt{M^{2}-a^{2}}+z\right)+R_{+}\left(z-\sqrt{M^{2}-a^{2}}\right)\right)\right)\Big{)}$}\bigg{)}\,,$

$\displaystyle\ell^{t\rho}(\bar{\zeta})$ $\displaystyle=\tfrac{aM}{2\rho
R_{-}R_{+}\left(M^{2}(2M+R_{-}+R_{+})^{2}-4a^{2}(M+R_{-})(M+R_{+})\right)^{2}}\times$
$\displaystyle\bigg{(}\scalebox{0.8}{$-2R_{-}R_{+}(2M+R_{-}+R_{+})\left(4a^{2}-4M^{2}+(R_{-}-R_{+})^{2}\right)\left(M^{2}(2M+R_{-}+R_{+})^{2}-4a^{2}(M+R_{-})(M+R_{+})\right)$}$
$\displaystyle\scalebox{0.8}{$+2\rho^{2}(2M+R_{-}+R_{+})\left(4a^{2}-4M^{2}+(R_{-}-R_{+})^{2}\right)\left(2a^{2}\left(M(R_{-}+R_{+})+R_{-}^{2}+R_{+}^{2}\right)-M^{2}(R_{-}+R_{+})(2M+R_{-}+R_{+})\right)$}\bigg{)}\,,$
$\displaystyle\ell^{tz}(\bar{\zeta})$
$\displaystyle=\tfrac{aM}{2R_{-}R_{+}\left(M^{2}(2M+R_{-}+R_{+})^{2}-4a^{2}(M+R_{-})(M+R_{+})\right)^{2}}\times$
$\displaystyle\bigg{(}\scalebox{0.8}{$-2(R_{-}-R_{+})(2M+R_{-}+R_{+})\left(M^{2}(2M+R_{-}+R_{+})^{2}-4a^{2}(M+R_{-})(M+R_{+})\right)\left(R_{-}\left(\sqrt{M^{2}-a^{2}}+z\right)+R_{+}\left(\sqrt{M^{2}-a^{2}}-z\right)\right)$}$
$\displaystyle\scalebox{0.7}{$\left(R_{+}\left(\sqrt{M^{2}-a^{2}}-z\right)\left(2a^{2}(M+R_{+})-M^{2}(2M+R_{-}+R_{+})\right)-R_{-}\left(\sqrt{M^{2}-a^{2}}+z\right)\left(2a^{2}(M+R_{-})-M^{2}(2M+R_{-}+R_{+})\right)\right)$}\bigg{)}\,.$

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