# Triangular solution to the planar elliptic three-body problem in the
parametrized post-Newtonian formalism

Yuya Nakamura<EMAIL_ADDRESS>Hideki Asada
<EMAIL_ADDRESS>Graduate School of Science and Technology, Hirosaki
University, Hirosaki 036-8561, Japan

###### Abstract

A triangular solution [Phys. Rev. D 107, 044005 (2023)] has recently been
found to the planar circular three-body problem in the parametrized post-
Newtonian (PPN) formalism, for which they focus on a class of fully
conservative theories characterized by the Eddington-Robertson parameters
$\beta$ and $\gamma$. The present paper extends the PPN triangular solution to
quasi-elliptic motion, for which the shape of the triangular configuration
changes with time at the PPN order. The periastron shift due to the PPN
effects is also obtained.

###### pacs:

04.25.Nx, 45.50.Pk, 95.10.Ce, 95.30.Sf

##  I Introduction

The three-body problem is among the classical ones in physics, which led to a
study of the chaos Goldstein . Particular solutions, notably Euler’s collinear
solution and Lagrange’s equilateral one Danby ; Marchal  represent regular
orbits, which have attracted a lot of interest e.g. Asada ; Torigoe ; Seto ;
Schnittman ; Connors .

Nordtvedt Nordtvedt  pointed out that the position of the triangular points is
very sensitive to the ratio between the gravitational mass and the inertial
one in gravitational experimental tests, though the post-Newtonian (PN) terms
are partly considered.

For the restricted three-body problem in the PN approximation, Krefetz Krefetz
and Maindl Maindl  found the PN triangular configuration for a general mass
ratio between two masses. These studies were extended to the PN three-body
problem for general masses Yamada2010 ; Yamada2011 ; Ichita2011 ; Yamada2012 ;
Yamada2015 ; Yamada2016 , where the PN counterparts for Euler’s collinear
Yamada2010 ; Yamada2011  and Lagrange’s equilateral solutions Ichita2011 ;
Yamada2012  were found. It should be noted that the PN triangular solutions
are not necessarily equilateral for general mass ratios and they are
equilateral only for either the equal mass case or two test masses. The
stability of the PN solution and the radiation reaction at 2.5PN order were
also studied Yamada2015 ; Yamada2016 .

In a scalar-tensor theory of gravity, a collinear configuration for three-body
problem was discussed Zhou . In addition to such fully classical treatments, a
possible quantum gravity correction to the Lagrange points was argued
Battista2015a ; Battista2015b .

Moreover, the recent discovery of a relativistic hierarchical triple system
including a neutron star Ransom  has sparked renewed interest in the
relativistic three-body problem and the related gravitational experiments
Archibald ; Will2018 ; Voisin .

In the PPN formalism, collinear and triangular solutions to the planar
circular three-body problem have recently been found Nakamura2023 , where they
focus on a class of fully conservative theories characterized by the
Eddington-Robertson parameters $\beta$ and $\gamma$, because the two
parameters are the most important ones; $\beta$ measures how much nonlinearity
there is in the superposition law for gravity and $\gamma$ measures how much
space curvature is produced by unit rest mass Will ; Poisson  . See e.g.
Klioner  for the celestial mechanics in this class of PPN theories.

In the Newtonian gravity, triangular solutions are not only to the circular
three-body problem but also to the elliptic one Danby ; Roy . Can a
(quasi-)elliptic orbit of triangular solutions be found in PPN case? A point
is that the PPN force seems to be too complicated to admit elliptic orbits for
a triple system. The main purpose of the present paper is to find it in the
class of fully conservative theories.

This paper is organized as follows. In Section II, basic methods and equations
are presented. Section III discusses the PPN triangular solution to the planar
elliptic three-body problem. Section V summarizes this paper. Throughout this
paper, $G=c=1$. $A,B$ and $C\in\\{1,2,3\\}$ label three masses.

##  II Basic methods and equations

###  II.1 Newtonian planar elliptic triangular solution

Let us begin by briefly summarizing the triangular solution to the Newtonian
planar elliptic three-body problem Danby ; Roy . A homothetic solution is
possible and it represents the Lagrange equilateral solution in elliptic
motion. See e.g. Section 5 of Reference Danby  for more detail.

The equation of motion (EOM) for three masses ($M_{A}$ at the position
$\bm{R}_{A}$) reads

$\displaystyle
M_{A}\bm{a}_{A}=\sum_{B=1}^{N}\frac{M_{A}M_{B}}{(R_{AB})^{2}}\bm{n}_{AB},$ (1)

where $\bm{a}_{A}$ denotes the acceleration of the $A$-th mass,
$\bm{R}_{AB}\equiv\bm{R}_{A}-\bm{R}_{B}$, $R_{AB}\equiv|\bm{R}_{AB}|$, and
$\bm{n}_{AB}\equiv\bm{R}_{AB}/R_{AB}$.

By taking the cross product of $\bm{R}_{1}$ and Eq. (1) for $A=1$, we obtain

$\displaystyle\bm{R}_{1}\times\bm{R}_{2}\left(\frac{1}{(R_{12})^{3}}-\frac{1}{(R_{31})^{3}}\right)=0,$
(2)

where the coordinate center is chosen as the center of mass (COM) of
$\sum_{A}M_{A}\bm{R}_{A}=0$. For a triangular configuration,
$\bm{R}_{1}\nparallel\bm{R}_{2}$. From Eq. (2), we thus obtain
$R_{12}=R_{23}$. By cyclic arguments, we obtain an equilateral solution Danby
; Roy .

In elliptic motion, the arm length $R_{A}$ becomes
$R_{1}=af_{N}\sqrt{\nu_{2}^{2}+\nu_{2}\nu_{3}+\nu_{3}^{2}}$,
$R_{2}=af_{N}\sqrt{\nu_{3}^{2}+\nu_{3}\nu_{1}+\nu_{1}^{2}}$, and
$R_{3}=af_{N}\sqrt{\nu_{1}^{2}+\nu_{1}\nu_{2}+\nu_{2}^{2}}$, where the total
mass is $M\equiv\sum_{A}M_{A}$, the mass ratio is defined as $\nu_{A}\equiv
M_{A}/M$, $a$ is some constant, and $f_{N}$ denotes the dilation factor Danby
; Roy ; Ichita2011 ; Yamada2012 . In circular motion, $f_{N}=1$, while $f_{N}$
is a function of time in elliptic motion Danby ; Roy .

From the total energy and angular momentum, an elliptic orbit is obtained as
Danby ; Roy

$\displaystyle
f_{\rm{N}}=\frac{\mathcal{A}_{\rm{N}}(1-e_{\rm{N}}^{2})}{1+e_{\rm{N}}\cos\theta},$
(3)

where $\theta$ denotes the true anomaly, $e_{\rm{N}}$ is the eccentricity of
the elliptic orbit as
$e_{\rm{N}}=\sqrt{1+2L_{\rm{N}}^{2}\mathcal{E}_{\rm{N}}M^{-2}\mu^{-3}}$ for
the total energy $\mathcal{E}_{\rm{N}}$, the total angular momentum
$L_{\rm{N}}$ and $\mu\equiv M(\nu_{1}\nu_{2}+\nu_{2}\nu_{3}+\nu_{3}\nu_{1})$,
and $\mathcal{A}_{\rm{N}}\equiv-\mu M/(2a\mathcal{E}_{\rm{N}})$. Here,
$\theta=0$ is chosen as the periastron.

For the simplicity, we refer to $A\equiv a\mathcal{A}_{\rm{N}}$ as the semi-
major axis and $P\equiv a\mathcal{A}_{\rm{N}}(1-e_{\rm{N}}^{2})$ as the semi-
latus rectum. For instance, the semi-major axis for the elliptic orbit of
$M_{1}$ is
$a\mathcal{A}_{\rm{N}}\sqrt{\nu_{2}^{2}+\nu_{2}\nu_{3}+\nu_{3}^{2}}$.

From the total angular momentum, the angular velocity $\omega_{\rm{N}}$ of the
triangular configuration is obtained as

$\displaystyle\omega_{\rm{N}}=$
$\displaystyle(1+e_{\rm{N}}\cos\theta)^{2}\sqrt{\frac{M}{P^{3}}}.$ (4)

###  II.2 EOM in the PPN formalism

In a class of fully conservative theories including only the Eddington-
Robertson parameters $\beta$ and $\gamma$, the PPN EOM becomes Will ; Poisson

$\displaystyle\bm{a}_{A}=$ $\displaystyle-\sum_{B\neq
A}\frac{M_{B}}{R_{AB}^{2}}\bm{n}_{AB}$ $\displaystyle-\sum_{B\neq
A}\frac{M_{B}}{R_{AB}^{2}}\bigg{\\{}\gamma
v_{A}^{2}-2(\gamma+1)(\bm{v}_{A}\cdot\bm{v}_{B})$
$\displaystyle~{}~{}~{}~{}~{}+(\gamma+1)v_{B}^{2}-\frac{3}{2}(\bm{n}_{AB}\cdot\bm{v}_{B})^{2}-\bigg{(}2\gamma+2\beta+1\bigg{)}\frac{M_{A}}{R_{AB}}$
$\displaystyle~{}~{}~{}~{}~{}-2(\gamma+\beta)\frac{M_{B}}{R_{AB}}\bigg{\\}}\bm{n}_{AB}$
$\displaystyle+\sum_{B\neq
A}\frac{M_{B}}{R_{AB}^{2}}\bigg{\\{}\bm{n}_{AB}\cdot[2(\gamma+1)\bm{v}_{A}-(2\gamma+1)\bm{v}_{B}]\bigg{\\}}(\bm{v}_{A}-\bm{v}_{B})$
$\displaystyle+\sum_{B\neq A}\sum_{C\neq
A,B}\frac{M_{B}M_{C}}{R_{AB}^{2}}\bigg{[}\frac{2(\gamma+\beta)}{R_{AC}}+\frac{2\beta-1}{R_{BC}}$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}-\frac{1}{2}\frac{R_{AB}}{R_{BC}^{2}}(\bm{n}_{AB}\cdot\bm{n}_{BC})\bigg{]}\bm{n}_{AB}$
$\displaystyle-\frac{1}{2}(4\gamma+3)\sum_{B\neq A}\sum_{C\neq
A,B}\frac{M_{B}M_{C}}{R_{AB}R_{BC}^{2}}\bm{n}_{BC}+O(c^{-4}),$ (5)

where $\bm{v}_{A}$ denotes the velocity of the $A$-th mass.

Figure 1:  Schematic figure for the PPN triangular configuration of three
masses. The inequilateral triangle is characterized by $\varepsilon_{AB}$. In
the Newtonian limit, $\varepsilon_{AB}$ vanishes and $R_{AB}$ becomes
$af_{\rm{N}}$.

##  III PPN planar elliptic triangular solution

###  III.1 PPN planar elliptic orbit

In order to obtain a PPN solution as a perturbation around the Newtonian
equilateral elliptic solution, we assume a quasi-common dilation as
$R_{AB}=af(1+\varepsilon_{AB})$ for three masses, where $\varepsilon_{AB}$
denotes a PPN distortion. The perfectly common dilation occurs at the Newton
order, whereas the dilation is not common by $\varepsilon_{AB}$. See also
Figure 1.

In the same way as deriving Eq. (2), we take the cross product of $\bm{R}_{1}$
and Eq. (5) for $M_{1}$ to obtain

$\displaystyle\ell_{1}^{2}\frac{d}{dt}\bigg{(}f^{2}\omega\bigg{)}(\bm{\lambda}\times\bm{\rho})$
$\displaystyle=$
$\displaystyle(\bm{\lambda}\times\bm{\rho})\left\\{-\frac{\sqrt{3}}{2}\frac{M}{f_{\rm{N}}a}\nu_{2}\nu_{3}\right.$
$\displaystyle\left.\times\bigg{[}3(\varepsilon_{12}-\varepsilon_{31})+\frac{M}{2a}(\nu_{3}-\nu_{2})\bigg{(}\frac{1}{f_{\rm{N}}}-\frac{1}{\mathcal{A}_{\rm{N}}}\bigg{)}\right.$
$\displaystyle~{}~{}~{}~{}\left.+\frac{3}{8}a^{2}\\{\dot{f}_{\rm{N}}(1+3\nu_{1})+\sqrt{3}f_{\rm{N}}\omega_{\rm{N}}(1-\nu_{1}-2\nu_{2})\\}\right.$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\left.\times\\{\dot{f}_{\rm{N}}(1-\nu_{1}-2\nu_{2})+\sqrt{3}f_{\rm{N}}\omega_{\rm{N}}(1-\nu_{1})\\}\right.$
$\displaystyle~{}~{}~{}~{}~{}\left.-\frac{M}{4f_{\rm{N}}a}(\nu_{2}-\nu_{3})(8\beta-3)\bigg{]}\right.$
$\displaystyle-\frac{\sqrt{3}}{4}Ma\nu_{2}\bigg{(}\nu_{3}\frac{\dot{f}_{\rm{N}}}{f_{\rm{N}}}+\frac{\omega_{\rm{N}}}{\sqrt{3}}(\nu_{1}-\nu_{2}-1)\bigg{)}$
$\displaystyle~{}~{}~{}~{}~{}\left.\times\bigg{(}(4\gamma+3+\nu_{2}-\nu_{1})f_{\rm{N}}-\sqrt{3}\nu_{3}f_{\rm{N}}\omega_{\rm{N}}\bigg{)}\right.$
$\displaystyle\left.+\frac{\sqrt{3}}{4}Ma\nu_{3}\bigg{(}\nu_{2}\frac{\dot{f}_{\rm{N}}}{f_{\rm{N}}}-\frac{\omega_{\rm{N}}}{\sqrt{3}}(\nu_{1}-\nu_{3}-1)\bigg{)}\right.$
$\displaystyle~{}~{}~{}~{}~{}\left.\times\bigg{(}(4\gamma+3+\nu_{3}-\nu_{1})\dot{f}_{\rm{N}}+\sqrt{3}\nu_{2}f_{\rm{N}}\omega_{\rm{N}}\bigg{)}\right\\}$
$\displaystyle+O(c^{-4}),$ (6)

where $\ell_{1}\equiv a\sqrt{\nu_{2}^{2}+\nu_{2}\nu_{3}+\nu_{3}^{2}}$ Danby ;
Roy ; Yamada2012 ; Nakamura2023 , and we introduce an orthonormal basis
$\bm{\lambda}$ and $\bm{\rho}$. Here, $\bm{\lambda}\equiv\bm{R}_{1}/R_{1}$,
and $\bm{\rho}$ is the 90 degree rotation of $\bm{\lambda}$. It is more
convenient to use the orthonormal basis than $\bm{R}_{1}$ and $\bm{R}_{2}$,
because the right-hand side of Eq. (5) relies upon not only the positions but
also the velocities. In elliptic motion, the velocity is not always orthogonal
to the position vector, though it is in circular motion.

From the PPN total angular momentum, we find

$\displaystyle\frac{d}{dt}\bigg{(}f^{2}\omega\bigg{)}$ $\displaystyle=$
$\displaystyle-\frac{M\dot{f}_{\rm{N}}\omega}{4a}\frac{13\nu_{1}\nu_{2}\nu_{3}-8\\{(\gamma+1)\eta-\zeta\\}}{\eta}+O(c^{-4}),$
(7)

where dot denotes the time derivative and we denote
$\eta\equiv\nu_{1}\nu_{2}+\nu_{2}\nu_{3}+\nu_{3}\nu_{1}$ and
$\zeta\equiv\nu_{1}^{2}\nu_{2}^{2}+\nu_{2}^{2}\nu_{3}^{2}+\nu_{3}^{2}\nu_{1}^{2}$.
Eq. (7) is reduced to $d(f_{N}\omega_{\rm{N}}^{2})/dt=0$ in the Newtonian
limit, which recovers the Newtonian case of the planar elliptic triangular
solution. It follows that Eq. (7) can be derived also from the sum of Eq. (6)
for $A=1,2,3$.

By substituting Eq. (7) into the left-hand side of Eq. (6), we obtain

$\displaystyle\varepsilon_{12}-\varepsilon_{31}$ $\displaystyle=$
$\displaystyle\frac{M}{8A}(\nu_{3}-\nu_{2})(3\nu_{1}+1)$
$\displaystyle-\frac{M}{12P}(\nu_{3}-\nu_{2})(9\nu_{1}+8\beta-2)(1+e_{\rm{N}}\cos\theta)$
$\displaystyle+\frac{M}{4P}(\nu_{3}-\nu_{2})(3\nu_{1}-1)(1+e_{\rm{N}}\cos\theta)^{2}$
$\displaystyle-\frac{\sqrt{3}e_{\rm{N}}M}{72\nu_{2}\nu_{3}P}\sin\theta(1+e_{\rm{N}}\cos\theta)$
$\displaystyle~{}~{}~{}\times\bigg{[}34\nu_{1}\nu_{2}\nu_{3}+16\nu_{1}(\nu_{2}^{2}+\nu_{3}^{2})$
$\displaystyle~{}~{}~{}~{}~{}+9\nu_{2}\nu_{3}\\{1-3\nu_{1}^{2}+(\nu_{2}-\nu_{3})^{2}\\}$
$\displaystyle~{}~{}~{}~{}~{}-\frac{4(\nu_{2}^{2}+\nu_{2}\nu_{3}+\nu_{3}^{2})(13\nu_{1}\nu_{2}\nu_{3}+8\zeta)}{\eta}\bigg{]}+O(c^{-4}),$
(8)

where Eq. (3) is used for $f_{N}$. By cyclic arguments,
$\varepsilon_{23}-\varepsilon_{12}$ and $\varepsilon_{31}-\varepsilon_{23}$
are obtained.

Following Reference Yamada2012 , the gauge fixing is chosen as
$\varepsilon_{12}+\varepsilon_{23}+\varepsilon_{31}=0$, which keeps the
triangular area. From this gauge fixing and Eq. (8), we obtain

$\displaystyle\varepsilon_{12}$ $\displaystyle=$
$\displaystyle\frac{M}{24A}[3\\{\nu_{1}(\nu_{3}-2\nu_{2})+\nu_{3}(1+\nu_{2})\\}-1]$
$\displaystyle-\frac{M}{36P}[2(4\beta-1)(3\nu_{3}-1)$
$\displaystyle~{}~{}~{}~{}+9\\{\nu_{1}(\nu_{3}-\nu_{2})+\nu_{2}(\nu_{3}-\nu_{1})\\}](1+e_{\rm{N}}\cos\theta)$
$\displaystyle+\frac{M}{12P}[1-3(\nu_{3}^{2}+2\nu_{1}\nu_{2})](1+e_{\rm{N}}\cos\theta)^{2}$
$\displaystyle-\frac{e_{\rm{N}}M\sqrt{3}}{108P}(\nu_{1}-\nu_{2})\sin\theta(1+e_{\rm{N}}\cos\theta)$
$\displaystyle~{}~{}~{}\times\bigg{[}\frac{8\nu_{3}(1-\nu_{3})}{\nu_{1}\nu_{2}}+27\nu_{3}-1$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}+\frac{2(\nu_{1}\nu_{2}-\nu_{3}^{2})(13\nu_{1}\nu_{2}\nu_{3}+8\zeta)}{\nu_{1}\nu_{2}\nu_{3}\eta}\bigg{]}+O(c^{-4}).$
(9)

It is worthwhile to mention that $\gamma$ makes no contribution to Eq. (9),
while $\beta$ is included in it. This is because the space curvature by unit
rest mass has no relevance with asymmetry characterized by $\varepsilon_{12}$.

We can obtain also $\varepsilon_{23}$ and $\varepsilon_{31}$ cyclically. We
thus obtain the PPN triangular quasi-elliptic solution. Note that this
solution does not follow a perfectly elliptic motion, though the Newtonian
orbit as the zeroth-order solution is elliptic. The PPN solution does not
stand in a long time scale owing to the periastron shift by PPN effects as
shown below. Namely, the above solution is obtained in the sense of an
osculating orbit Danby ; Roy .

Figure 2:  $\varepsilon_{12},\varepsilon_{23},\varepsilon_{31}$ for
$\nu_{1}=1/2$, $\nu_{2}=1/3$, $\nu_{3}=1/6$ and $M/a=0.01$ in elliptic motion
with $e_{\rm N}=0.5$. The horizontal axis denotes $\theta$ from a periastron
($\theta=0$) to the next periastron ($\theta=2\pi$).

###  III.2 Periastron shift

After direct calculations, the PPN expression of the total energy Baker1978 ;
Baker1979  for the PPN planar quasi-elliptic triangular solution can be
rewritten as

$\displaystyle\bigg{(}\frac{du}{d\theta}\bigg{)}^{2}+G(u)\frac{du}{d\theta}=F(u),$
(10)

where $u\equiv 1/f$. $F(u)$ and $G(u)$ are functions of $u$, which are too
long to write down in this paper.

The periastron shift is

$\displaystyle\theta_{\rm{PPN}}=\int_{u_{\rm{min}}}^{u_{\rm{max}}}du\frac{1}{\left(\cfrac{du}{d\theta}\right)}-\pi,$
(11)

where $u_{\rm{max}}$ and $u_{\rm{min}}$ correspond to the periastron and
periapsis, respectively.

In the same way as the post-Newtonian calculations of the periastron shift
Will ; Poisson , by using Eq(10) for Eq. (11), we obtain the periastron shift
at the PPN order as

$\displaystyle\theta_{\rm{PPN}}$ $\displaystyle=$ $\displaystyle\frac{\pi
M}{36P\eta}\bigg{[}18\nu_{1}\nu_{2}\nu_{3}(9-2\beta)+\eta(65-44\beta+72\gamma)+36\zeta\bigg{]}$
$\displaystyle+O(c^{-4}).$ (12)

The periastron shift per orbital period is $2\theta_{\rm{PPN}}$. In GR
($\beta=\gamma=1$), Eq. (12) becomes

$\displaystyle\theta_{\rm{PPN}}=$ $\displaystyle\frac{\pi
M}{36P\eta}(126\nu_{1}\nu_{2}\nu_{3}+93\eta+36\zeta)+O(c^{-4}).$ (13)

In the test particle limit of a third mass ($\nu_{3}\to 0$), Eq. (12)
disagrees with that of a binary system, because the restricted three-body
dynamics does not equal to the binary dynamics Danby ; Roy ; Yamada2012 . See
e.g. Eq. (66) in Reference Will  and Eq. (13.51) in Poisson  for the PPN
periastron shift formula of a binary case.

##  IV Conclusion

We found a PPN triangular solution to the planar elliptic three-body problem
in a class of fully conservative theories. The distortion function
$\varepsilon_{AB}$ of a triangular solution depends on $\beta$ but not on
$\gamma$. It follows that, in the circular limit, the present solution
recovers the PPN triangular circular solution in Reference Nakamura2023 . In
the limit of $e_{\rm{N}}\to 0$, Eq. (9) agrees with Eq. (41) in Nakamura2023 .

The periastron shift of the PPN triangular solution was also obtained. Because
of the three-body interactions, the periastron shift in the PPN triangular
solution is different from that of a binary system.

It is left for future to study the stability of the present solution.

##  V Acknowledgments

We thank Yuuiti Sendouda and Marcus Werner for encouraging comments. This work
was supported in part by Japan Science and Technology Agency (JST) SPRING,
Grant Number, JPMJSP2152 (Y.N.), and in part by Japan Society for the
Promotion of Science (JSPS) Grant-in-Aid for Scientific Research, No. 20K03963
(H.A.).

## References

  * (1) H. Goldstein, Classical Mechanics (Addison-Wesley, MA, 1980).
  * (2) J. M. A. Danby, Fundamentals of Celestial Mechanics (William-Bell, VA, 1988).
  * (3) A. E. Roy, Orbital Motion (Adam Hilger, Bristol, 1982).
  * (4) C. Marchal, The Three-Body Problem (Elsevier, Amsterdam, 1990).
  * (5) H. Asada, Phys. Rev. D 80 064021 (2009).
  * (6) Y. Torigoe, K. Hattori and H. Asada, Phys. Rev. Lett. 102, 251101 (2009).
  * (7) N. Seto, T. Muto, Phys. Rev. D 81 103004 (2010).
  * (8) J. D. Schnittman, Astrophys. J. 724 39 (2010).
  * (9) M. Connors, P. Wiegert and C. Veillet, Nature 475, 481 (2011).
  * (10) K. Nordtvedt, Phys. Rev. 169 1014 (1968).
  * (11) E. Krefetz, Astron. J 72, 471 (1967).
  * (12) T. I. Maindl, Completing the Inventory of the Solar System, Astronomical Society of the Pacific Conference Proceedings, edited by T.W. Rettig and J.M. Hahn, (Astronomical Society of the Pacific, San Francisco, 1996), 107, 147.
  * (13) K. Yamada, H. Asada, Phys. Rev. D 82, 104019 (2010).
  * (14) K. Yamada, H. Asada, Phys. Rev. D 83, 024040 (2011).
  * (15) T. Ichita, K. Yamada, H. Asada, Phys. Rev. D 83, 084026 (2011).
  * (16) K. Yamada and H. Asada, Phys. Rev. D 86, 124029 (2012).
  * (17) K. Yamada, T. Tsuchiya and H. Asada Phys. Rev. D 91, 124016 (2015).
  * (18) K. Yamada and H. Asada, Phys. Rev. D 93, 084027 (2016).
  * (19) T. Y. Zhou, W. G. Cao, and Y. Xie, Phys. Rev. D 93, 064065 (2016).
  * (20) E. Battista, S. Dell’Agnello, G. Esposito, and J. Simo, Phys. Rev. D 91, 084041 (2015); Erratum, Phys. Rev. D 93, 049902(E) (2016).
  * (21) E. Battista, S. Dell’Agnello, G. Esposito, L. Di Fiore, J. Simo, and A. Grado, Phys. Rev. D 92, 064045 (2015); Erratum, Phys. Rev. D 93, 109904(E) (2016).
  * (22) S. M. Ransom, I. H. Stairs, A. M. Archibald, J. W. T. Hessels, D. L. Kaplan and et al. Nature, 505, 520 (2014).
  * (23) Anne M. Archibald, Nina V. Gusinskaia, Jason W. T. Hessels, Adam T. Deller, David L. Kaplan and et al. Nature, 559, 73 (2018).
  * (24) C. M.Will, Nature, 559, 40 (2018).
  * (25) G. Voisin, I. Cognard, P. C. C. Freire, et al., Astron. Astrophys. 638, A24 (2020).
  * (26) C. M. Will, Living Rev. Relativity, 17, 4 (2014).
  * (27) E. Poisson, and C. M. Will, Gravity, (Cambridge Univ. Press, UK. 2014).
  * (28) S. A. Klioner and M. H. Soffel, Phys. Rev. D 62, 024019 (2000).
  * (29) B. M. Barker and R. F. O’Connell, Phys. Lett. A 68, 289 (1978).
  * (30) B. M. Barker and R. F. O’Connell, J. Math. Phys. 20, 1427 (1979).
  * (31) C. W. Misner, K. S. Thorne, and J. A. Wheeler, Gravitation (Freeman, New York, 1973).
  * (32) L. D. Landau and E. M. Lifshitz, The Classical Theory of Fields (Pergamon, New York, 1962).
  * (33) Yuya Nakamura and Hideki Asada, Phys. Rev. D 107, 044005 (2023).