# Circular geodesic orbits in the equatorial plane of a charged rotating disc
of dust

David Rumler<EMAIL_ADDRESS>Theoretisch-Physikalisches Institut,
Friedrich-Schiller-Universität Jena, Max-Wien-Platz 1, D-07743 Jena, Germany

###### Abstract

Equatorial circular geodesic orbits of neutral test particles in the exterior
spacetime of the charged rotating disc of dust are analyzed in dependence of
its specific charge and a relativity parameter. The charged rotating disc of
dust is an axisymmetric, stationary solution of the Einstein-Maxwell equations
in terms of a post-Newtonian expansion. In particular, photon, marginally
bound and marginally stable orbits are discussed. It turns out that general
formulae in closed form for angular velocity, specific energy and specific
angular momentum of the test particles can be derived, which hold for any
(exterior) asymptotically flat, axisymmetric, stationary and reflection
symmetric (electro-)vacuum spacetime. Furthermore, circular motion in the
exterior spacetime of the charged rotating disc of dust is qualitatively very
similar to that around a Kerr-Newman black hole for sufficiently large radii,
but differs strongly in the respective closer vicinity.

##  I Introduction

Investigating the motion of test particles is a very direct and instructive
method for analyzing the geometric structure of a given spacetime within the
framework of general relativity. It is well known that freely falling neutral
test particles travel along geodesics.

The motion of test particles also plays a role in astrophysics, for instance
in the accretion of matter by a central object and in the (indirect) formation
of black hole shadows by circular photon orbits.

Various aspects of geodesic motion around Kerr and Kerr-Newman black holes
have been studied, see, e.g., the publications by Bardeen et al. [1] and
Dadhich and Kale [2].111Also (electrically) charged test particle orbits in
the Kerr-Newman spacetime have been explored (see, e.g., [3, 4]). However,
those orbits are no longer geodesic. Charged test particles generally move
under the influence of the gravitational and the electromagnetic field (if
present).

Meinel and Kleinwächter [5] (see also [6]) and Ansorg [7] analyzed (timelike)
geodesic motion in the gravitational field of the uncharged rotating disc of
dust [8, 9, 6].

In the present paper circular orbits of neutral test particles in the
equatorial plane of an isolated, axisymmetric, stationary, reflection
symmetric, charged, rotating body are investigated first. Remarkably, even
without knowing the explicit solution of the rotating body, energy, angular
momentum and angular velocity of the test particles can be specified in closed
form solely as functions of the metric and their radial derivatives.

The general examination of equatorial circular orbits of neutral test
particles is followed by a specialization on the spacetime of the charged
rotating disc of dust. This disc is rigidly rotating, has constant specific
charge and is an axisymmetric, stationary solution of the Einstein-Maxwell
equations expressed in terms of a post-Newtonian expansion up to tenth order
[10, 11]. A detailed discussion of equatorial circular orbits in the spacetime
of the charged rotating disc of dust is provided.

##  II Charged rotating disc of dust

Within general relativity an infinitesimally thin equilibrium disc
configuration is considered. The disc is made of dust (i.e. a perfect fluid
with vanishing pressure), it carries a constant specific (electric) charge
$\epsilon\in\left[-1,1\right]$ (electric charge density over baryonic mass
density) and it is rotating rigidly around the axis of symmetry with a
constant angular velocity $\Omega$. It is, furthermore, axisymmetric,
stationary and it obeys reflection symmetry [10, 11, 12].

Using axisymmetry and stationarity the coupled Einstein-Maxwell equations in
electro-vacuum can be reduced to the Ernst equations [13]. Together with
reflection symmetry, a well-defined boundary value problem to the Ernst
equations for the charged rotating disc of dust can be formulated (see [10,
14]). It was solved by means of a post-Newtonian expansion up to eighth order
by Palenta and Meinel [10] and up to tenth by Breithaupt et al. [11].

The corresponding line element of the disc is expressed globally in terms of
Weyl-Lewis-Papapetrou coordinates222Units with $c=G=4\pi\epsilon_{0}=1$ are
used.,

$\mathrm{d}s^{2}=f^{-1}\left[h\left(\mathrm{d}\rho^{2}+\mathrm{d}\zeta^{2}\right)+\rho^{2}\mathrm{d}\varphi^{2}\right]-f\left(\mathrm{d}t+a\,\mathrm{d}\varphi\right)^{2}\,,$
(1)

and the electromagnetic four-potential is given by

$A_{a}=(0,0,A_{\varphi},A_{t})\,.$ (2)

Introducing a relativity parameter $g\coloneqq\sqrt{\gamma}$, where

$\gamma\coloneqq 1-\sqrt{f_{c}}\,,\quad\text{with}\quad f_{c}\coloneqq
f\left(\rho=0,\zeta=0\right)\,,$ (3)

was originally used by Bardeen and Wagoner [15], the disc solution can be
specified by:

$\displaystyle f=1+\sum_{k=1}^{10}f_{2k}g^{2k}\,,\quad
h=1+\sum_{k=2}^{10}h_{2k}g^{2k}\,,$ (4) $\displaystyle
a^{*}=\sum_{k=1}^{10}a^{*}_{2k+1}g^{2k+1}\,,$ (5) $\displaystyle
A_{\varphi}^{*}=\sum_{k=1}^{10}A_{\varphi\,2k+1}^{*}g^{2k+1}\,,\quad
A_{t}=\sum_{k=1}^{10}A_{t\,2k}g^{2k}\,.$ (6)

The coefficient functions $f_{2k}$, $h_{2k}$, $a^{*}_{2k+1}$, … only depend on
the elliptic coordinates $\eta\in[-1,1]$ and $\nu\in[0,\infty]$, which are
defined via

$\rho=\rho_{0}\sqrt{\left(1-\eta^{2}\right)(1+\nu^{2})}\,,\quad\zeta=\rho_{0}\eta\nu\,.$
(7)

Here, * denotes the normalization by suitable powers of the disc’s coordinate
radius $\rho_{0}$ in order to obtain dimensionless quantities, such as
$a^{*}=\frac{a}{\rho_{0}}$ and $A_{\varphi}^{*}=\frac{A_{\varphi}}{\rho_{0}}$.

Note that the relativity parameter can equivalently be expressed in in terms
of the redshift $z_{c}$ of a photon traveling from the center of the disc to
spatial infinity:

$g^{2}=\frac{z_{c}}{1+z_{c}}\,.$ (8)

Therefore, the relativity parameter connects the regime of Newtonian physics,
$g\ll 1$, with the ultra-relativistic regime, $g\to 1$, (where black hole
formation is assumed) and the parameter space of the disc solution is spanned
by $g\in[0,1]$, $\epsilon\in[0,1]$ and $\rho_{0}$. Without loss of generality
only positive charges are considered and the coordinate radius $\rho_{0}$
serves as a scaling parameter. All physical quantities studied in this paper
are functions of $g$, $\epsilon$ and $\rho_{0}$ only.

The specific charge $\epsilon$, furthermore, directly influences the disc’s
rotation speed. As is evident from the Newtonian limit (where each dust
particle within the disc is in an equilibrium of gravitational, electric and
centrifugal force), disc configurations with $\epsilon=0$ have a maximum
angular velocity and those with $\epsilon=1$ are static.

##  III Circular geodesic orbits in a general spacetime

We first consider the exterior spacetime of an isolated, axisymmetric,
stationary, reflection symmetric, charged, rotating body. These
qualifications, besides a non-zero charge, are expected to be fulfilled for
most astrophysical bodies. (Note that the special cases of zero charge and
vanishing rotation are included.)

A freely falling uncharged test particle moves through spacetime along a
(timelike) curve determined by the geodesic equation,

$\frac{\mathrm{d}^{2}x^{i}}{\mathrm{d}\tau^{2}}+\Gamma^{i}_{mn}\frac{\mathrm{d}x^{m}}{\mathrm{d}\tau}\frac{\mathrm{d}x^{n}}{\mathrm{d}\tau}=0\,,$
(9)

where $\tau$ is the proper time of the test particle and $\Gamma^{i}_{mn}$ are
the Christoffel symbols.

The body is arranged so that the axis of rotation coincides with the
$\zeta$-axis and its equatorial plane is defined by $\zeta=0$. We restrict the
motion of the test particles to this plane, i.e. $\zeta=0$,
$\frac{\mathrm{d}\zeta}{\mathrm{d}\tau}=0$ and
$\frac{\mathrm{d}^{2}\zeta}{\mathrm{d}\tau^{2}}=0$. (In order to keep the
notation simple, the dependency on the coordinates is suppressed and, e.g.,
$g_{ab}$ is written instead of $g_{ab}(\rho,\zeta=0)$ in the following.) Let
the geodesics, furthermore, be circular orbits.

Due to stationarity and axisymmetry the metric $g_{ab}$ of the exterior
spacetime of the body can be expressed in terms of Weyl-Lewis-Papapetrou
coordinates, see eq. 1.

The circular motion of a test particle with constant angular velocity
$\Omega_{\text{tp}}=\frac{\mathrm{d}\varphi}{\mathrm{d}t}$ (as seen from
infinity) is also stationary and the four-velocity is therefore, given by

$\displaystyle\left(u_{\text{tp}}\right)^{i}$
$\displaystyle=\frac{\mathrm{d}x^{i}}{\mathrm{d}\tau}=\left(0,0,\Omega_{\text{tp}}e^{-U^{\prime}},e^{-U^{\prime}}\right)\,,$
(10) $\displaystyle\left(u^{\prime}_{\text{tp}}\right)^{i}$
$\displaystyle=\left(0,0,0,e^{-U^{\prime}}\right)\,,$ (11)

where $f=e^{2U}$ and the primes denote the frame of reference co-rotating to
the test particle, with $\varphi^{\prime}=\varphi-\Omega_{\text{tp}}t$.333Note
that in the literature listed on the uncharged and charged rotating disc of
dust, the primes refer to the co-rotating frame of the rigidly rotating disc.
Notation-wise this, however, does not cause a problem, as no reference to the
co-rotating frame of the disc is made here.

Using eq. 11, the only non-vanishing component of the geodesic equation (9) is
obtained for $i=\rho$:

$g^{\prime}_{tt,\rho}=0\,.$ (12)

The $\zeta$-component vanishes due to the restriction of the motion to the
equatorial plane of the body and reflection symmetry of the metric. In other
words, setting $\zeta=0$, $\frac{\mathrm{d}\zeta}{\mathrm{d}\tau}=0$ and
$\frac{\mathrm{d}^{2}\zeta}{\mathrm{d}\tau^{2}}=0$ is actually solving the
$\zeta$-component of the geodesic equation.

Note that the transformation law for a transition to the co-rotating frame of
the test particle reads in case of $g_{tt}$:

$g^{\prime}_{tt}=g_{tt}+2\Omega_{\text{tp}}g_{\varphi
t}+\Omega_{\text{tp}}^{2}g_{\varphi\varphi}\,.$ (13)

Eq. (12) is therefore equivalent to

$g_{\varphi\varphi,\rho}\Omega_{\text{tp}}^{2}+2g_{\varphi
t,\rho}\Omega_{\text{tp}}+g_{tt,\rho}=0\,.$ (14)

This immediately gives the solution for the angular velocity of the test
particle:

$\Omega_{1/2}=\frac{-g_{\varphi t,\rho}\pm\sqrt{g_{\varphi
t,\rho}^{2}-g_{\varphi\varphi,\rho}g_{tt,\rho}}}{g_{\varphi\varphi,\rho}}\,,$
(15)

where (in general) one of the solutions corresponds to a prograde and the
other one to a retrograde orbit.

The two symmetries, axisymmetry and stationarity, give rise to two conserved
quantities associated with the motion of the test particle. Axisymmetry
represented by the spacelike Killing vector
$\boldsymbol{\eta}=\frac{\partial}{\partial\varphi}$ implies a conserved
specific angular momentum $L=\eta_{i}\left(u_{\text{tp}}\right)^{i}$ and
stationarity with a timelike Killing vector
$\boldsymbol{\xi}=\frac{\partial}{\partial t}$ ensures the conservation of the
specific energy $E=-\xi_{i}\left(u_{\text{tp}}\right)^{i}$. With the four-
velocity from eq. 10 one obtains the specific angular momentum

$L_{1/2}=\frac{g_{\varphi\varphi}\Omega_{1/2}+g_{\varphi
t}}{\sqrt{-g_{\varphi\varphi}\Omega_{1/2}^{2}-2g_{\varphi
t}\Omega_{1/2}-g_{tt}}}$ (16)

and the specific energy

$E_{1/2}=-\frac{g_{\varphi
t}\Omega_{1/2}+g_{tt}}{\sqrt{-g_{\varphi\varphi}\Omega_{1/2}^{2}-2g_{\varphi
t}\Omega_{1/2}-g_{tt}}}$ (17)

of the test particle. Prograde orbits are defined by the positive specific
angular momentum solution and retrograde orbits by the negative one.

It should be stressed that eqs. 15, 16 and 17 hold for any (exterior)
asymptotically flat, axisymmetric, stationary and reflection symmetric
(electro-)vacuum spacetime and the solutions could be derived in a closed form
without knowledge of the concrete spacetime metric.

Circular orbits only exist for

$g_{\varphi\varphi}\Omega_{1/2}^{2}+2g_{\varphi t}\Omega_{1/2}+g_{tt}\leq
0\,,$ (18)

according to eqs. 16 and 17. In the limiting case of equality, specific
angular momentum and specific energy diverge. This characterizes a photon
orbit for prograde and retrograde motion respectively. Prograde or retrograde
circular motion of uncharged test particles is therefore only possible for
radii larger than that of the associated photon orbit.444For charged test
particles this innermost boundary of circular orbits generally no longer
coincides with the photon orbit (as a result of the electromagnetic
interaction between the rotating body and the test particle), see e.g. [4].

There is an equivalent way of deriving eqs. 15, 16 and 17 via the Lagrange
formalism. Generally, the Lagrangian describing the (timelike) motion of a
test particle in a spacetime furnished with the metric $g_{ab}$ is

$\mathcal{L}=\frac{1}{2}g_{ab}\dot{x}^{a}\dot{x}^{b}\,,$ (19)

where $\dot{x}^{a}=\frac{\mathrm{d}x^{a}}{\mathrm{d}\tau}$. The resulting
Euler-Lagrange equation is equivalent to the geodesic equation, i.e. the
extremal path between two points is also straightest possible path.

As the Lagrangian corresponding to the above defined rotating body does not
depend on $\varphi$ and $t$ ($g_{ab}$ is axisymmetric and stationary), there
are two conserved quantities:

$\displaystyle L$
$\displaystyle\coloneqq\frac{\partial\mathcal{L}}{\partial\dot{\varphi}}=g_{\varphi\varphi}\dot{\varphi}+g_{\varphi
t}\dot{t}\,,$ (20) $\displaystyle E$
$\displaystyle\coloneqq-\frac{\partial\mathcal{L}}{\partial\dot{t}}=-g_{\varphi
t}\dot{\varphi}-g_{tt}\dot{t}\,.$ (21)

$L$ and $E$ coincide with the specific angular momentum and the specific
energy, defined previously via the Killing vectors.

From eqs. 20 and 21, the angular velocity of the test particle
$\Omega_{\text{tp}}=\frac{\mathrm{d}\varphi}{\mathrm{d}t}$ (as seen from
infinity) can be deduced:

$\Omega_{\text{tp}}=-\frac{g_{tt}L+g_{\varphi t}E}{g_{\varphi
t}L+g_{\varphi\varphi}E}\,.$ (22)

As above, we restrict the motion of the test particle to the equatorial plane
of the body. Inserting $\dot{\varphi}$ and $\dot{t}$, using eqs. 20 and 21,
into the normalization condition of the four-velocity for timelike curves,
$g_{ab}\dot{x}^{a}\dot{x}^{b}=-1$, leads to

$\frac{1}{2}g_{\rho\rho}\left(\frac{\mathrm{d}\rho}{\mathrm{d}\tau}\right)^{\\!2}+\mathcal{U}=0\,,$
(23)

where

$\mathcal{U}\coloneqq-\frac{1}{2\rho^{2}}\left(g_{tt}L^{2}+2g_{\varphi
t}LE+g_{\varphi\varphi}E^{2}\right)+\frac{1}{2}$ (24)

is an effective potential. The motion of the test particle in the equatorial
plane can therefore also be described by a motion in an effective potential.

A useful relation that was applied to $\mathcal{U}$ and that is also employed
in subsequent calculations is

$g_{\varphi t}^{2}-g_{\varphi\varphi}g_{tt}=\rho^{2}\,.$ (25)

Differentiation of eq. 23 with respect to the particles proper time $\tau$
leads to the equation of motion:

$g_{\rho\rho}\frac{\mathrm{d}^{2}\rho}{\mathrm{d}\tau^{2}}+\frac{1}{2}g_{\rho\rho,\rho}\left(\frac{\mathrm{d}\rho}{\mathrm{d}\tau}\right)^{\\!2}+\mathcal{U}_{,\rho}=0\,.$
(26)

One can verify that eq. 26 corresponds to the $\rho$-component of the geodesic
equation.555For not necessarily circular motion, the $\rho$-component of the
geodesic equation is given by:
$g_{\rho\rho}\frac{\mathrm{d}^{2}\rho}{\mathrm{d}\tau^{2}}+\frac{1}{2}g_{\rho\rho,\rho}\left(\frac{\mathrm{d}\rho}{\mathrm{d}\tau}\right)^{\\!2}-\frac{1}{2}\left(g_{\varphi\varphi,\rho}\Omega_{\text{tp}}^{2}+2g_{\varphi
t,\rho}\Omega_{\text{tp}}+g_{tt,\rho}\right)\left(\frac{\mathrm{d}t}{\mathrm{d}\tau}\right)^{\\!2}\\\
=0$.

Circular orbits must fulfill the requirements
$\frac{\mathrm{d}\rho}{\mathrm{d}\tau}=0$ and
$\frac{\mathrm{d}^{2}\rho}{\mathrm{d}\tau^{2}}=0$. According to eqs. 23 and
26, the conditions for circular orbits (in the equatorial plane) are therefore
$\mathcal{U}=0$ and $\mathcal{U_{,\rho}}=0$. These conditions imply:

$\displaystyle g_{tt}L^{2}+2g_{\varphi
t}LE+g_{\varphi\varphi}E^{2}-\rho^{2}=0\,,$ (27) $\displaystyle
g_{tt,\rho}L^{2}+2g_{\varphi
t,\rho}LE+g_{\varphi\varphi,\rho}E^{2}-2\rho=0\,.$ (28)

Notice that the above equations agree with those derived in [7].

Solving eqs. 27 and 28 reveals the specific angular momentum and the specific
energy of the test particle:

$\displaystyle\left|L_{1/2}\right|$
$\displaystyle=\left|l_{1/2}\right|\frac{\rho}{\sqrt{g_{tt}l_{1/2}^{2}+2g_{\varphi
t}l_{1/2}+g_{\varphi\varphi}}}\,,$ (29) $\displaystyle\left|E_{1/2}\right|$
$\displaystyle=\frac{\rho}{\sqrt{g_{tt}l_{1/2}^{2}+2g_{\varphi
t}l_{1/2}+g_{\varphi\varphi}}}\,,$ (30)

where

$l_{1/2}\coloneqq\frac{L_{1/2}}{E_{1/2}}=-\frac{2g_{\varphi t}-\rho g_{\varphi
t,\rho}\pm\rho\sqrt{g_{\varphi
t,\rho}^{2}-g_{\varphi\varphi,\rho}g_{tt,\rho}}}{2g_{tt}-\rho g_{tt,\rho}}\,.$
(31)

The sign of $E_{1/2}$ follows from eq. 21 (where $\dot{t}>0$) and the sign of
$L_{1/2}$ is then automatically given by $L_{1/2}=l_{1/2}E_{1/2}$. Clearly,
the angular velocity (see eq. 22) can also be expressed in terms of $l_{1/2}$:

$\Omega_{1/2}=-\frac{g_{tt}l_{1/2}+g_{\varphi t}}{g_{\varphi
t}l_{1/2}+g_{\varphi\varphi}}\,.$ (32)

It can be shown explicitly that the two sets of equations for specific angular
momentum, specific energy and angular velocity, i.e. eqs. 15, 16 and 17 and
eqs. 32, 29, 30 and 31, are indeed equivalent. In particular, the following
relation holds:

$\displaystyle g_{tt}l_{1/2}^{2}+2g_{\varphi t}l_{1/2}+g_{\varphi\varphi}$
$\displaystyle=\frac{-\rho^{2}}{\left(-g_{\varphi
t}\Omega_{1/2}-g_{tt}\right)^{2}}\left(g_{\varphi\varphi}\Omega_{1/2}^{2}+2g_{\varphi
t}\Omega_{1/2}+g_{tt}\right)\,.$ (33)

Eq. (32) can furthermore be easily converted to

$l_{1/2}=-\frac{g_{\varphi\varphi}\Omega_{1/2}+g_{\varphi t}}{g_{\varphi
t}\Omega_{1/2}+g_{tt}}\,.$ (34)

The advantage of the second approach (using the Lagrange formalism in
conjunction with the effective potential) is that the stability of the
circular orbits can be analyzed directly on the basis of the effective
potential. According to eq. 23, non-circular motion is only possible for
$\mathcal{U}<0$. For given $L$ and $E$, circular orbits are therefore stable
(with respect to radial and angular perturbations) if and only if the extremum
of $\mathcal{U}$ is a minimum, i.e. $\mathcal{U}_{,\rho\rho}>0$. Using
$\mathcal{U}=0$ and $\mathcal{U}_{,\rho}=0$, $\mathcal{U}_{,\rho\rho}>0$ is
equivalent to

$g_{tt,\rho\rho}L^{2}+2g_{\varphi
t,\rho\rho}LE+g_{\varphi\varphi,\rho\rho}E^{2}-2<0\,.$ (35)

For $\mathcal{U}_{,\rho\rho}<0$ circular orbits are thus unstable.666A
discussion of the stability of circular orbits in stationary, axisymmetric
spacetimes can also be found in [16] and [17]. Moreover, if $E^{2}<1$ circular
orbits are bound, i.e. the test particle cannot reach infinity.777For
$E^{2}>1$ there can also be additional bound orbits, see, e.g., [7].

Inserting the metric functions of the uncharged rotating disc of dust at the
rim into the above formulae (of the first or second approach) reveals the same
expressions for $\Omega_{1/2}$, $L_{1/2}$ and $E_{1/2}$ (with $1/2=+/-$) as in
[5]. Also the corresponding results for equatorial circular geodesic orbits in
the (exterior) Kerr [1] and Kerr-Newman spacetime [2] can be obtained from the
above formulae (using either approach).888Note that $\rho\geq 0$.

Figure 1: Normalized angular velocities $R_{0}\Omega_{+}$ (top) and
$R_{0}\Omega_{-}$ (bottom) of test particles moving along
(prograde/retrograde) circular geodesic orbits, shown for $g=0.7$ and
$\epsilon\in\\{0,1/3,2/3,1\\}$.

Figure 2: Normalized specific angular momenta $\frac{L_{+}}{R_{0}}$ (top) and
$\frac{L_{-}}{R_{0}}$ (bottom) of test particles moving along
(prograde/retrograde) circular geodesic orbits, shown for $g=0.7$ and
$\epsilon\in\\{0,1/3,2/3,1\\}$.

Figure 3: Specific energies $E_{+}$ (top) and $E_{-}$ (bottom) of test
particles moving along (prograde/retrograde) circular geodesic orbits, shown
for $g=0.7$ and $\epsilon\in\\{0,1/3,2/3,1\\}$.

Figure 4: Normalized angular velocities $R_{0}\Omega_{+}$ (top) and
$R_{0}\Omega_{-}$ (bottom) of test particles moving along
(prograde/retrograde) circular geodesic orbits, shown for $\epsilon=0.7$ and
$g\in\\{0.4,0.5,0.6,0.7\\}$.

Figure 5: Normalized specific angular momenta $\frac{L_{+}}{R_{0}}$ (top) and
$\frac{L_{-}}{R_{0}}$ (bottom) of test particles moving along
(prograde/retrograde) circular geodesic orbits, shown for $\epsilon=0.7$ and
$g\in\\{0.4,0.5,0.6,0.7\\}$.

Figure 6: Specific energies $E_{+}$ (top) and $E_{-}$ (bottom) of test
particles moving along (prograde/retrograde) circular geodesic orbits, shown
for $\epsilon=0.7$ and $g\in\\{0.4,0.5,0.6,0.7\\}$.

##  IV Circular geodesic orbits in the spacetime of the charged rotating disc
of dust

Now circular geodesic orbits in the equatorial plane ($\eta=0$) of the charged
rotating disc of dust are to be discussed. By evaluating eqs. 15, 16 and 17 or
eqs. 32, 29, 30 and 31 with the metric of the disc, using eqs. 4 and 5, the
angular velocity $\Omega_{1/2}=\Omega_{+/-}$, the specific angular momentum
$L_{1/2}=L_{+/-}$ and the specific energy $E_{1/2}=E_{+/-}$ can be expressed
in terms of a post-Newtonian expansion, see appendix A. Here “$+$” denotes
direct and “$-$” retrograde orbits. The angular velocity of the disc,
$\Omega$, is non-negative.

It should be noted that the angular velocity $\Omega_{+/-}$ and the specific
angular momentum $L_{+/-}$ are odd functions and the specific energy $E_{+/-}$
is an even function in $g$. This ensures that $\Omega_{+/-}$ and $L_{+/-}$
change their sign and $E_{+/-}$ remains unchanged under a change of sense of
rotation of the disc (for $g\to-g$).999For further (general) details, see [10]
and [14].

In the Newtonian limit (i.e. up to and including
$\mathcal{O}\left(g^{2}\right)$) prograde and retrograde orbits coincide, as
the angular momentum of disc, $J$, does not contribute to the Newtonian
gravitational field.101010Indeed, the angular momentum of the disc starts at
order $g^{3}$, see [18]. In particular, it holds $\Omega_{+}=-\Omega_{-}$,
$L_{+}=-L_{-}$ and $E_{+}=E_{-}$.

Since the angular momentum of the disc vanishes in the limiting case of
$\epsilon=1$ (see [18]), prograde and retrograde orbits also coincide for
$\epsilon=1$.

It was further confirmed that the simultaneous limit $\epsilon\to 0$,
$\rho\to\rho_{0}$ leads to the same results for $\Omega_{\pm}$ and $L_{\pm}$,
$E_{\pm}$, in terms of a series expansion in $g$, as in [5]. Especially
noteworthy is the fact that $\Omega_{+}=\Omega$ in this limiting case. The
reason is that the $\rho$-component of the geodesic equation of a test
particle, eq. 12, also represents a boundary condition of the uncharged
rotating disc of dust (see [6]).

Figs. 1 to 6 show, for a fixed proper disc radius
$R_{0}\coloneqq\int_{0}^{\rho_{0}}\\!\sqrt{g_{\rho\rho}}\,\mathrm{d}\rho$ (see
also [14]), the radial dependence of the angular velocity $\Omega_{\pm}$, the
specific angular momentum $L_{\pm}$ and the specific energy $E_{\pm}$ for
different values of the specific charge $\epsilon$ and the relativity
parameter $g$. Apart from the region directly adjacent to the rim of the disc,
the qualitative behavior of the curves is very similar to the ones
corresponding to the (exterior) spacetime of a Kerr-Newman black hole.

As can be seen in figs. 1, 2 and 3, reducing the specific charge $\epsilon$
from $1$ to $0$ results (for fixed $R_{0}$) in a decrease in the absolute
values of $L_{+}$ and $E_{+}$ and an increase in the absolute values of
$\Omega_{-}$, $L_{-}$ and $E_{-}$. $\Omega_{+}$ shows a mixed behavior (for
sufficiently large radii, however, it increases). Furthermore, raising the
angular momentum of a Kerr-Newman black hole leads to the same qualitative
change of the absolute values as for the charged rotating disc of dust by
decreasing $\epsilon$ (with the exception of $\Omega_{+}$). This is plausible,
since the angular momentum of the disc comes with the global prefactor
$\sqrt{1-\epsilon^{2}}$, see [18].111111Increasing the charge parameter of the
Kerr-Newman black hole qualitatively coincides with an increase of the
specific charge of the disc in case of $\Omega_{-}$, $L_{-}$ and $E_{-}$ (as
well as $\Omega_{+}$ for sufficiently large radii).

On the other hand, the absolute values of $R_{0}\Omega_{\pm}$ and
$\frac{L_{\pm}}{R_{0}}$ increase and $E_{+}$ decreases with the relativity
parameter $g$, see figs. 4, 5 and 6. $E_{-}$ increases close to the rim and
decreases for sufficiently large radii.

Figure 7: Radii of photon, marginally bound and marginally stable orbits,
$\frac{\rho_{-}^{\text{ph}}}{\rho_{0}}$,
$\frac{\rho_{\pm}^{\text{mb}}}{\rho_{0}}$ and
$\frac{\rho_{\pm}^{\text{ms}}}{\rho_{0}}$, as functions of $\epsilon$ (top),
for $g=0.6$, and $g$ (bottom), for $\epsilon=0.8$.

Interesting in the context of (equatorial) circular motion of neutral test
particles are also the radii of photon orbits ($g^{\prime}_{tt}=0$),
marginally bound orbits ($E^{2}=1$) and marginally stable orbits
($\mathcal{U}_{,\rho\rho}=0$). They are shown as functions of $\epsilon$ and
$g$ in fig. 7.121212Note that circular orbits with radii smaller than that of
the photon orbit / marginally bound orbit / marginally stable orbit are
forbidden / unbound / unstable and those with larger radii are permitted /
bound / stable. The corresponding radii of the retrograde orbits decrease and
those of the prograde orbits increase with $\epsilon\in\left[0,1\right]$. In
the case of $\epsilon=1$ they coincide. For growing $g$ both the radii of
retrograde and prograde orbits increase and they merge in the Newtonian limit
($g\ll 1$). For sufficiently large values of $\epsilon$ and $g$ there are both
retrograde and prograde photon, marginally bound and marginally stable orbits.
However, in contrast to the (exterior) spacetime of a Kerr-Newman black hole,
for sufficiently small $g$ there are no prograde photon orbits (as shown in
fig. 7) and for even smaller $g$ there are also no retrograde photon orbits
(each depending on $\epsilon$).131313In the Newtonian theory of gravity there
are in general no photon orbits. In other words, there are disc configurations
for which both prograde and retrograde circular motion is possible arbitrarily
close to the rim of the disc, i.e. no orbits are forbidden. For $g\to 0$ all
orbits are bound. Moreover, in comparison to the uncharged rotating disc of
dust, for $\epsilon>0$ not all prograde orbits are bound (and for sufficiently
large $\epsilon$ and $g$ there are prograde photon orbits). Note that for
$g=0.7$ there is no ergosphere yet.

Figure 8: Convergence estimate of the functions $R_{0}\Omega_{+}$,
$\frac{L_{+}}{R_{0}}$, $E_{+}$ (top) and $R_{0}\Omega_{-}$,
$\frac{L_{-}}{R_{0}}$, $E_{-}$ (bottom), displayed for $g=0.7$, $\epsilon=0$
and $\frac{\rho}{\rho_{0}}=2$.

Finally, the convergence behavior of the normalized angular velocity
$R_{0}\Omega_{\pm}$, the normalized specific angular momentum
$\frac{L_{\pm}}{R_{0}}$ and the specific energy $E_{\pm}$ is illustrated in
fig. 8. As an example, to estimate the convergence of the specific energy, the
resulting change by adding the $\mathcal{K}$th order to the post-Newtonian
expansion at order $\mathcal{K}-1$ relative to the expansion solution at
$\mathcal{K}$th order is calculated:

$\displaystyle\delta{E_{\pm}}_{\,|\,\mathcal{K}}\coloneqq\frac{\left|{E_{\pm}}_{\,|\,\mathcal{K}}-{E_{\pm}}_{\,|\,\mathcal{K}-1}\right|}{\left|{E_{\pm}}_{\,|\,\mathcal{K}}\right|}\,,$
(36)
$\displaystyle\text{where}\quad{E_{\pm}}_{\,|\,\mathcal{K}}\coloneqq\sum_{k=1}^{\mathcal{K}}E_{\pm
2k}g^{2k}\,.$ (37)

The convergence behavior of $R_{0}\Omega_{\pm}$ and $\frac{L_{\pm}}{R_{0}}$ is
computed analogously. Generally, the quantities corresponding to prograde
circular motion exhibit a significantly better convergence behavior than those
of retrograde circular motion. It should be noted that $\Omega_{\pm}$ and
$L_{\pm}$ are only exact up to ninth order.

##  V Conclusions and outlook

Within general relativity various aspects of circular motion of uncharged test
particles in the equatorial plane of the charged rotating disc of dust were
analyzed. Angular velocity, specific angular momentum and specific energy of
the test particles were discussed in dependence of the specific charge and the
relativity parameter. Also the radii of photon, marginally bound and
marginally stable orbits were determined. In addition, formulae for angular
velocity, specific angular momentum and specific energy of the test particles
were derived that can be applied to the (exterior) spacetime of any isolated,
axisymmetric, stationary, reflection symmetric, charged, rotating body.

It can be concluded that for sufficiently large radii (at least
$\rho>\rho_{+}^{\text{ms}}$), the qualitative behavior of equatorial circular
motion around the disc is very similar to that in the exterior spacetime of a
Kerr-Newman black hole. In the close vicinity of the disc, however, circular
motion shows a very distinct behavior from the one near the event horizon of a
black hole.141414This is consistent with the conclusion drawn in [7] about
equatorial circular motion around the uncharged rotating disc of dust.

Both methods presented can be generalized to charged test particles. General
formulae to be obtained for charged test particles can then also be applied to
the charged rotating disc of dust. Similar to the transition from neutral to
charged test particles in the case of the Kerr-Newman spacetime, this should
reveal all kinds of new facets.

Furthermore, both methods can also be adapted to null geodesics. In this way
also circular photon orbits in the equatorial plane of the disc can be studied
in detail. Another possibility is to investigate circular motion (of either
neutral/charged test particles or photons) in the interior of the disc.
Without further ado, general formulae for the exterior can also be used for
the interior spacetime of the disc, since the metric of the disc is globally
written in terms of Weyl-Lewis-Papapetrou coordinates. Clearly, circular
motion in the interior of the disc is quite different from that in the
exterior.

##  Appendix A Post-Newtonian expansion of the normalized angular velocity,
the normalized specific angular momentum and the specific energy of neutral
test particles

The post-Newtonian expansion of the normalized angular velocity
$R_{0}\Omega_{\pm}$ (up to first order), the normalized specific angular
momentum $\frac{L_{\pm}}{R_{0}}$ (up to first order) and the specific energy
$E_{\pm}$ (up to second order) of uncharged test particles moving along
equatorial circular geodesics is as follows:

$\displaystyle R_{0}\Omega_{+}$
$\displaystyle=\frac{\sqrt{\left(\pi-2\arctan\\!\left(\nu\right)\right)\left(1+\nu^{2}\right)-2\nu}\,g}{\sqrt{\pi}\,\sqrt{1+\nu^{2}}}$
$\displaystyle\quad+\frac{1}{\pi^{2}\sqrt{\left(\pi-2\arctan\\!\left(\nu\right)\right)\left(1+\nu^{2}\right)-2\nu}\,\sqrt{1+\nu^{2}}}$
$\displaystyle\quad\cdot\left[\left(\frac{13}{4}\left(1+\nu^{2}\right)\left(\left(\epsilon^{2}-\frac{36}{13}\right)\nu^{2}+\frac{23\epsilon^{2}}{39}+\frac{2}{13}\right)\arctan\\!\left(\nu\right)-9\nu^{3}-\frac{11\nu}{2}\right.\right.$
$\displaystyle\quad\left.\left.+\,\frac{13\epsilon^{2}\nu^{3}}{4}+\frac{15\nu\,\epsilon^{2}}{4}\right)\pi^{\frac{3}{2}}-\frac{15}{8}\left(1+\nu^{2}\right)\left(\left(\epsilon^{2}-\frac{8}{5}\right)\nu^{2}+\frac{17\epsilon^{2}}{45}-\frac{2}{15}\right)\pi^{\frac{5}{2}}\right.$
$\displaystyle\quad\left.+\,\left(\arctan\\!\left(\nu\right)\left(1+\nu^{2}\right)+\nu\right)\sqrt{\pi}\,\left(\left(\nu^{2}\left(\epsilon^{2}+6\right)-\epsilon^{2}-2\right)\arctan\\!\left(\nu\right)+\left(\epsilon^{2}+6\right)\nu\right)\vphantom{\frac{2}{15}}\right.$
$\displaystyle\quad\left.-\,3\sqrt{1-\epsilon^{2}}\,\sqrt{\left(\pi-2\arctan\\!\left(\nu\right)\right)\left(1+\nu^{2}\right)-2\nu}\,\sqrt{1+\nu^{2}}\left(\vphantom{\frac{2}{15}}\pi\,\nu^{2}-2\arctan\\!\left(\nu\right)\nu^{2}\right.\right.$
$\displaystyle\quad\left.\left.+\,\frac{\pi}{3}-2\nu-\frac{2\arctan\left(\nu\right)}{3}\right)\pi+\frac{5\pi^{3/2}}{6}\left(\left(\pi-2\arctan\\!\left(\nu\right)\right)\left(1+\nu^{2}\right)-2\nu\right)\right]g^{3}+\mathcal{O}\left(g^{5}\right)\,,$
$\displaystyle R_{0}\Omega_{-}$
$\displaystyle=-\frac{\sqrt{\left(\pi-2\arctan\\!\left(\nu\right)\right)\left(1+\nu^{2}\right)-2\nu}\,g}{\sqrt{\pi}\,\sqrt{1+\nu^{2}}}$
$\displaystyle\quad+\frac{1}{\pi^{2}\sqrt{\left(\pi-2\arctan\\!\left(\nu\right)\right)\left(1+\nu^{2}\right)-2\nu}\,\sqrt{1+\nu^{2}}}$
$\displaystyle\quad\cdot\left[-\left(\left(\frac{13}{4}\left(1+\nu^{2}\right)\left(\left(\epsilon^{2}-\frac{36}{13}\right)\nu^{2}+\frac{23\epsilon^{2}}{39}+\frac{2}{13}\right)\arctan\\!\left(\nu\right)-9\nu^{3}-\frac{11\nu}{2}\right.\right.\right.$
$\displaystyle\quad\left.\left.\left.+\,\frac{13\epsilon^{2}\nu^{3}}{4}+\frac{15\nu\,\epsilon^{2}}{4}\right)\pi^{\frac{3}{2}}-\frac{15}{8}\left(1+\nu^{2}\right)\left(\left(\epsilon^{2}-\frac{8}{5}\right)\nu^{2}+\frac{17\epsilon^{2}}{45}-\frac{2}{15}\right)\pi^{\frac{5}{2}}\right.\right.$
$\displaystyle\quad\left.\left.+\left(\arctan\\!\left(\nu\right)\left(1+\nu^{2}\right)+\nu\right)\sqrt{\pi}\,\left(\left(\nu^{2}\left(\epsilon^{2}+6\right)-\epsilon^{2}-2\right)\arctan\\!\left(\nu\right)+\left(\epsilon^{2}+6\right)\nu\right)\vphantom{\frac{2}{15}}\right)\right.$
$\displaystyle\quad\left.-\,3\sqrt{1-\epsilon^{2}}\,\sqrt{\left(\pi-2\arctan\\!\left(\nu\right)\right)\left(1+\nu^{2}\right)-2\nu}\,\sqrt{1+\nu^{2}}\left(\vphantom{\frac{2}{15}}\pi\,\nu^{2}-\,2\arctan\\!\left(\nu\right)\nu^{2}\right.\right.$
$\displaystyle\quad\left.\left.+\frac{\pi}{3}-2\nu-\frac{2\arctan\left(\nu\right)}{3}\right)\pi-\frac{5\pi^{3/2}}{6}\left(\left(\pi-2\arctan\\!\left(\nu\right)\right)\left(1+\nu^{2}\right)-2\nu\right)\right]g^{3}+\mathcal{O}\left(g^{5}\right)\,,$
$\displaystyle\frac{L_{+}}{R_{0}}$
$\displaystyle=\frac{\sqrt{\left(\pi-2\arctan\\!\left(\nu\right)\right)\left(1+\nu^{2}\right)-2\nu}\,\sqrt{1+\nu^{2}}\,g}{\sqrt{\pi}}$
$\displaystyle\quad-\frac{3}{2\pi^{\frac{27}{2}}\left(\left(\pi-2\arctan\\!\left(\nu\right)\right)\left(1+\nu^{2}\right)-2\nu\right)}$
$\displaystyle\quad\cdot\left[\frac{5\pi^{12}}{4}\left(-\frac{8}{15}\left(\nu^{2}\left(\epsilon^{2}+2\right)-\epsilon^{2}+6\right)\left(1+\nu^{2}\right)\arctan^{2}\\!\left(\nu\right)+\left(\pi\left(-\frac{26\epsilon^{2}}{15}+\frac{8}{3}\right)\nu^{4}\right.\right.\right.$
$\displaystyle\quad\left.\left.\left.+\,\left(-\frac{32}{15}-\frac{16\epsilon^{2}}{15}\right)\nu^{3}+\pi\left(\frac{20}{3}-\frac{124\epsilon^{2}}{45}\right)\nu^{2}-\frac{64\nu}{15}+\pi\left(4-\frac{46\epsilon^{2}}{45}\right)\right)\arctan\left(\nu\right)\right.\right.$
$\displaystyle\quad\left.\left.+\,\pi^{2}\left(\epsilon^{2}-\frac{16}{15}\right)\nu^{4}+\pi\left(-\frac{26\epsilon^{2}}{15}+\frac{8}{3}\right)\nu^{3}+\left(\left(-\frac{34}{15}+\frac{62\epsilon^{2}}{45}\right)\pi^{2}-\frac{16}{15}-\frac{8\epsilon^{2}}{15}\right)\nu^{2}\right.\right.$
$\displaystyle\quad\left.\left.+\,\left(-2\epsilon^{2}+4\right)\pi\nu+\pi^{2}\left(\frac{17\epsilon^{2}}{45}-\frac{6}{5}\right)\right)\sqrt{1+\nu^{2}}\,\sqrt{\left(\pi-2\arctan\\!\left(\nu\right)\right)\left(1+\nu^{2}\right)-2\nu}\right.$
$\displaystyle\quad\left.+\,\left(4\left(\left(1+\nu^{2}\right)^{2}\arctan\\!\left(\nu\right)+\nu^{3}+\frac{5\nu}{3}\right)\left(\left(1+\nu^{2}\right)\arctan\\!\left(\nu\right)+\nu\right)\pi^{\frac{25}{2}}\right.\right.$
$\displaystyle\quad\left.\left.+\,\left(-4\left(1+\nu^{2}\right)^{2}\arctan\\!\left(\nu\right)+\pi\,\nu^{4}+2\pi\,\nu^{2}-4\nu^{3}+\pi-\frac{16\nu}{3}\right)\pi^{\frac{27}{2}}\left(1+\nu^{2}\right)\right)\sqrt{1-\epsilon^{2}}\right.$
$\displaystyle\quad\left.+\,\frac{5\pi^{13}}{9}\left(\left(\pi-2\arctan\\!\left(\nu\right)\right)\left(1+\nu^{2}\right)-2\nu\right)^{3/2}\sqrt{1+\nu^{2}}\right]g^{3}+\mathcal{O}\left(g^{5}\right)\,,$
$\displaystyle\frac{L_{-}}{R_{0}}$
$\displaystyle=-\frac{\sqrt{\left(\pi-2\arctan\\!\left(\nu\right)\right)\left(1+\nu^{2}\right)-2\nu}\,\sqrt{1+\nu^{2}}\,g}{\sqrt{\pi}}$
$\displaystyle\quad-\frac{3}{2\pi^{\frac{27}{2}}\left(\left(\pi-2\arctan\\!\left(\nu\right)\right)\left(1+\nu^{2}\right)-2\nu\right)}$
$\displaystyle\quad\cdot\left[-\frac{5\pi^{12}}{4}\left(-\frac{8}{15}\left(\nu^{2}\left(\epsilon^{2}+2\right)-\epsilon^{2}+6\right)\left(1+\nu^{2}\right)\arctan^{2}\\!\left(\nu\right)+\left(\pi\left(-\frac{26\epsilon^{2}}{15}+\frac{8}{3}\right)\nu^{4}\right.\right.\right.$
$\displaystyle\quad\left.\left.\left.+\,\left(-\frac{32}{15}-\frac{16\epsilon^{2}}{15}\right)\nu^{3}+\pi\left(\frac{20}{3}-\frac{124\epsilon^{2}}{45}\right)\nu^{2}-\frac{64\nu}{15}+\pi\left(4-\frac{46\epsilon^{2}}{45}\right)\right)\arctan\left(\nu\right)\right.\right.$
$\displaystyle\quad\left.\left.+\,\pi^{2}\left(\epsilon^{2}-\frac{16}{15}\right)\nu^{4}+\pi\left(-\frac{26\epsilon^{2}}{15}+\frac{8}{3}\right)\nu^{3}+\left(\left(-\frac{34}{15}+\frac{62\epsilon^{2}}{45}\right)\pi^{2}-\frac{16}{15}-\frac{8\epsilon^{2}}{15}\right)\nu^{2}\right.\right.$
$\displaystyle\quad\left.\left.+\,\left(-2\epsilon^{2}+4\right)\pi\nu+\pi^{2}\left(\frac{17\epsilon^{2}}{45}-\frac{6}{5}\right)\right)\sqrt{1+\nu^{2}}\,\sqrt{\left(\pi-2\arctan\\!\left(\nu\right)\right)\left(1+\nu^{2}\right)-2\nu}\right.$
$\displaystyle\quad\left.+\,\left(4\left(\left(1+\nu^{2}\right)^{2}\arctan\\!\left(\nu\right)+\nu^{3}+\frac{5\nu}{3}\right)\left(\left(1+\nu^{2}\right)\arctan\\!\left(\nu\right)+\nu\right)\pi^{\frac{25}{2}}\right.\right.$
$\displaystyle\quad\left.\left.+\,\left(-4\left(1+\nu^{2}\right)^{2}\arctan\\!\left(\nu\right)+\pi\,\nu^{4}+2\pi\,\nu^{2}-4\nu^{3}+\pi-\frac{16\nu}{3}\right)\pi^{\frac{27}{2}}\left(1+\nu^{2}\right)\right)\sqrt{1-\epsilon^{2}}\right.$
$\displaystyle\quad\left.-\,\frac{5\pi^{13}}{9}\left(\left(\pi-2\arctan\\!\left(\nu\right)\right)\left(1+\nu^{2}\right)-2\nu\right)^{3/2}\sqrt{1+\nu^{2}}\right]g^{3}+\mathcal{O}\left(g^{5}\right)\,,$
$\displaystyle E_{+}$
$\displaystyle=1+\frac{\nu\left(\pi\nu-2\arctan\\!\left(\nu\right)\nu-2\right)g^{2}}{\pi}$
$\displaystyle\quad-\frac{3}{\sqrt{1+\nu^{2}}\,\pi^{\frac{11}{2}}}\left[\frac{\pi^{4}}{3}\sqrt{\left(\pi-2\arctan\\!\left(\nu\right)\right)\left(1+\nu^{2}\right)-2\nu}\,\left(1+\nu^{2}\right)\right.$
$\displaystyle\quad\left.\cdot\,\left(\left(-6\nu^{2}-2\right)\arctan\\!\left(\nu\right)+3\pi\,\nu^{2}+\pi-6\nu\right)\sqrt{1-\epsilon^{2}}+\frac{15}{16}\left(\left(\left(-\frac{16}{9}-\frac{8\epsilon^{2}}{15}\right)\nu^{4}\right.\right.\right.$
$\displaystyle\quad\left.\left.\left.+\,\left(-\frac{32}{15}+\frac{16\epsilon^{2}}{45}\right)\nu^{2}-\frac{16}{15}+\frac{8\epsilon^{2}}{45}\right)\arctan^{2}\\!\left(\nu\right)+\left(\left(\left(\frac{152}{45}-\frac{26\epsilon^{2}}{15}\right)\nu^{4}\right.\right.\right.\right.$
$\displaystyle\quad\left.\left.\left.\left.+\,\left(\frac{64}{15}-\frac{68\epsilon^{2}}{27}\right)\nu^{2}-\frac{58\epsilon^{2}}{135}+\frac{56}{45}\right)\pi-\frac{16\nu}{15}\left(\left(\frac{10}{3}+\epsilon^{2}\right)\nu^{2}-\frac{\epsilon^{2}}{3}+2\right)\right)\arctan\\!\left(\nu\right)\right.\right.$
$\displaystyle\quad\left.\left.+\,\left(1+\nu^{2}\right)\left(\left(-\frac{56}{45}+\epsilon^{2}\right)\nu^{2}+\frac{23\epsilon^{2}}{135}-\frac{16}{45}\right)\pi^{2}+\left(-\frac{26}{15}\epsilon^{2}\nu^{3}-\frac{214}{135}\nu\,\epsilon^{2}+\frac{152}{45}\nu^{3}+\frac{8}{3}\nu\right)\pi\right.\right.$
$\displaystyle\quad\left.\left.-\,\frac{8}{15}\left(\frac{10}{3}+\epsilon^{2}\right)\nu^{2}\right)\pi^{\frac{7}{2}}\sqrt{1+\nu^{2}}\right]g^{4}+\mathcal{O}\left(g^{6}\right)\,,$
$\displaystyle E_{-}$
$\displaystyle=1+\frac{\nu\left(\pi\nu-2\arctan\\!\left(\nu\right)\nu-2\right)g^{2}}{\pi}$
$\displaystyle\quad-\frac{45}{16\sqrt{1+\nu^{2}}\,\pi^{\frac{11}{2}}}\left[-\frac{16}{15}\sqrt{1-\epsilon^{2}}\,\left(1+\nu^{2}\right)\left(\pi\,\nu^{2}-2\arctan\\!\left(\nu\right)\nu^{2}+\frac{\pi}{3}-2\nu\right.\right.$
$\displaystyle\quad\left.\left.-\,\frac{2\arctan\\!\left(\nu\right)}{3}\right)\pi^{4}\sqrt{\left(\pi-2\arctan\\!\left(\nu\right)\right)\left(1+\nu^{2}\right)-2\nu}\right.$
$\displaystyle\quad\left.+\,\left(\left(\left(-\frac{16}{9}-\frac{8\epsilon^{2}}{15}\right)\nu^{4}+\left(-\frac{32}{15}+\frac{16\epsilon^{2}}{45}\right)\nu^{2}-\frac{16}{15}+\frac{8\epsilon^{2}}{45}\right)\arctan^{2}\\!\left(\nu\right)\right.\right.$
$\displaystyle\quad\left.\left.+\,\left(\left(\left(\frac{152}{45}-\frac{26\epsilon^{2}}{15}\right)\nu^{4}+\left(\frac{64}{15}-\frac{68\epsilon^{2}}{27}\right)\nu^{2}-\frac{58\epsilon^{2}}{135}+\frac{56}{45}\right)\pi\right.\right.\right.$
$\displaystyle\quad\left.\left.\left.-\,\frac{16\nu}{15}\left(\left(\frac{10}{3}+\epsilon^{2}\right)\nu^{2}-\frac{\epsilon^{2}}{3}+2\right)\right)\arctan\\!\left(\nu\right)+\left(1+\nu^{2}\right)\left(\left(-\frac{56}{45}+\epsilon^{2}\right)\nu^{2}+\frac{23\epsilon^{2}}{135}-\frac{16}{45}\right)\pi^{2}\right.\right.$
$\displaystyle\quad\left.\left.+\,\left(-\frac{26}{15}\epsilon^{2}\nu^{3}-\frac{214}{135}\nu\,\epsilon^{2}+\frac{152}{45}\nu^{3}+\frac{8}{3}\nu\right)\pi-\frac{8}{15}\left(\frac{10}{3}+\epsilon^{2}\right)\nu^{2}\right)\pi^{\frac{7}{2}}\sqrt{1+\nu^{2}}\right]g^{4}+\mathcal{O}\left(g^{6}\right)\,.$

###### Acknowledgements.

This work has been funded by the Deutsche Forschungsgemeinschaft (DFG) under
Grant No. 406116891 within the Research Training Group RTG 2522/1. The author
would like to thank Reinhard Meinel and Lars Maiwald for valuable discussions
and Martin Breithaupt for the provided disc solution in terms of the post-
Newtonian expansion up to tenth order.

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