Collinear and triangular solutions to the coplanar and circular three-body problem in the parametrized post-Newtonian formalism

Let us begin with briefly mentioning the Euler’s collinear solution for the circular three-body problem in Newton gravity Danby ; Marchal , for which each mass $M_{A}$ ($A=1,2,3$) at $\bm{x}_{A}$ is orbiting around the common center of mass (COM) at $\bm{x}_{G}$, and the orbital velocity and acceleration are denoted as $\bm{v}_{A}$ and $\bm{a}_{A}$, respectively. In this section, we suppose that three masses are always aligned, for which it is convenient to use the corotating frame with a constant angular velocity $\omega$ on the orbital plane chosen as the $x-y$ plane.

Without loss of generality, we assume $x_{1}>x_{2}>x_{3}$ for $\bm{x}_{A}\equiv(x_{A},0)$. Let $R_{A}$ denote the relative position of each mass $M_{A}$ from the COM at $\bm{x}_{G}\equiv(x_{G},0)$. Namely, $R_{A}=x_{A}-x_{G}$. Note that $|R_{A}|\neq|\bm{x}_{A}|$ unless $x_{G}=0$ is chosen. We define the relative vector between masses as $\bm{R}_{AB}\equiv\bm{x}_{A}-\bm{x}_{B}$, for which the relative length is $R_{AB}=|\bm{R}_{AB}|$. See Figure 1 for a configuration of the Euler’s collinear solution.

The coordinate origin $x=0$ is chosen between $M_{1}$ and $M_{3}$, such that $R_{1}>R_{2}>R_{3}$, $R_{1}>0$ and $R_{3}<0$. By taking account of this sign convention, the equation of motion becomes

We define the distance ratio as $z\equiv R_{23}/R_{12}$, which plays a key role in the following calculations. Note that $z>0$ by definition. We subtract Eq. (2) from Eq. (1) and Eq. (3) from Eq. (2). By combining the results including the same angular velocity $\omega$, we obtain a fifth-order equation for $z$ as

for which there exists the only positive root Danby ; Marchal . In order to obtain Eq. (4), we do not have to specify the coordinate origin e.g. $x_{G}=0$. This is because Eq. (4) does not refer to any coordinate system. Once Eq. (4) is solved for $z$, we can obtain $\omega$ by substituting $z$ into any of Eqs. (1)-(3).

In a class of fully conservative theories including only the Eddington-Robertson parameters $\beta$ and $\gamma$, the equation of motion is Will ; Poisson

where

For three aligned masses, Eq. (5) becomes the force-balance equation as

where we define $\ell\equiv R_{31}$, the mass ratio $\nu_{A}\equiv M_{A}/M$ for $M\equiv\sum_{A}M_{A}$, and

and

By rearranging Eq. (5) for the collinear configuration by the same way as in subsection II.A, we find a seventh-order equation for $z$ as

where the coefficients are

It follows that Eq. (11) recovers the PN collinear configuration by Eq. (13) of Reference Yamada2011 if and only if $\beta=\gamma=1$. The uniqueness is because the number of the parameters $\beta$, $\gamma$ is two for eight coefficients $A_{0},\cdots,A_{7}$.

From Eq. (7) for $z$ obtained above, the angular velocity $\omega_{PPN}$ of the PPN collinear configuration is obtained as

where $\omega_{N}=(F_{N}/\ell)^{1/2}$ is the Newtonian angular velocity. The subscript $N$ denotes the Newtonian case.

In this subsection, we suppose that the three masses are in coplanar and circular motion with keeping the same separation between the masses, namely $R_{AB}=a$ for a constant $a$.

It is convenient to choose the coordinate origin as the COM,

for which the equation of motion for each mass in the equilateral triangle configuration takes a compact form as Danby

See e.g. Eq. (8.6.5) in Reference Danby for the derivation of Eq. (22). A triangular configuration is a solution, if the Newtonian angular velocity $\omega_{N}$ satisfies

The orbital radius $\ell_{A}$ of each mass around the COM is Danby

We suppose again that three masses in circular motion are in a triangular configuration with a constant angular velocity $\omega$. By noting that a vector in the orbital plane can be expressed as a linear combination of $\bm{x}_{1}$ and $\bm{v}_{1}$, Eq.(5) becomes

where Eq. (23) is used and

By a cyclic permutation, we obtain the similar equations for $M_{2}$ and $M_{3}$.

The second and third terms in the right-had side of Eq. (27) are the PPN forces. The second term is parallel to $\bm{x}_{1}$, whereas the third term is parallel to $\bm{v}_{1}$ Note that $\bm{v}_{1}$ is not parallel to $\bm{x}_{1}$ in circular motion.

The location of the COM in the fully conservative theories of PPN Baker1978 ; Baker1979 remains the same as that in the PN approximation of general relativity MTW ; LL

where $E$ is defined as

This coincidence allows us to obtain the PPN orbital radius $\ell^{PPN}_{A}$ around the COM by straightforward calculations. The orbital radius of $M_{1}$ is formally obtained as

and the similar expressions of $\ell^{PPN}_{2}$ and $\ell^{PPN}_{3}$ for the orbital radius of $M_{2}$ and $M_{3}$ are obtained.

Unless the second term of the right-hand side in Eq. (31) vanishes, the difference between $\ell^{PPN}_{1}$ and $\ell_{1}$ would make our computations rather complicated. However, it vanishes because $\omega_{N}$ satisfies Eq. (23). As a result, the PPN orbital radius remains the same as the Newtonian one. Namely, $\ell^{PPN}_{A}=\ell_{A}$.

First, we discuss a condition for an equilateral configuration.

For Eq. (27) to hold, the coefficient of the velocity vector $\bm{v_{1}}$ must vanish, because there are no other terms including $\bm{v_{1}}$. The coefficient is proportional to $\nu_{2}\nu_{3}(\nu_{2}-\nu_{3})$. The same thing is true also of $M_{2}$ and $M_{3}$. For any value of $\beta$, therefore, the equilateral configuration in the PPN gravity can be present if and only if three finite masses are equal or two test masses orbit around one finite mass.

Note that one can find a very particular value of $\beta$ satisfying

which leads to the vanishing coefficient of the velocity vector $\bm{v_{1}}$. However, this choice is very unlikely, because the particular value of $\beta$ is dependent on the mass ratio $\nu_{1}$ and it is not universal. Hence, this case will be ignored.

Next, let us consider a PPN triangle configuration for general masses. For this purpose, we introduce a nondimensional parameter $\varepsilon_{AB}$ at the PPN order, such that each side length of the PPN triangle can be expressed as

The equilateral case is achieved by assuming $\varepsilon_{AB}=0$ for every masses. See Figure 2 for the PPN triangular configuration.

In order to fix the degree of freedom corresponding to a scale transformation, we follow Reference Yamada2012 to suppose that the arithmetic mean of the three side lengths is unchanged as

The left-hand side of Eq. (34) is $a$ in the Newtonian case, which leads to

This is a gauge fixing in $\varepsilon_{AB}$.

In terms of $\varepsilon_{AB}$, Eq. (27) is rearranged as

where

By a cyclic permutation, the equations for $M_{2}$ and $M_{3}$ can be obtained.

A triangular equilibrium configuration can exist if and only if the two conditions (A) and (B) are simultaneously satisfied; (A) Each mass satisfies Eq. (36), and (B) the configuration is unchanged in time.

Eq. (36) is the equation of motion for $M_{1}$. To be more accurate, therefore, $\omega$ in Eq. (36) should be denoted as $\omega_{1}$. Similarly, we introduce $\omega_{2}$ and $\omega_{3}$ in the equations of motion for $M_{2}$ and $M_{3}$, respectively. Then, Condition (B) means $\omega_{1}=\omega_{2}=\omega_{3}$.

Condition (A) is equivalent to Condition (A2); The coefficient of $\bm{v}_{A}$ in the equation of motion vanishes as

From Eqs. (38)-(40) and the gauge fixing as $\varepsilon_{12}+\varepsilon_{23}+\varepsilon_{31}=0$, we obtain

and

Therefore, the PPN triangle is inequilateral depending on $\beta$ via $\varepsilon_{AB}$ but not on $\gamma$. This suggests that also the PPN Lagrange points corresponding to $L_{4}$ and $L_{5}$ are sensitive to $\beta$ but are free from $\gamma$, as shown in Section IV.

It follows that Eqs. (41)-(43) recover the PN counterpart of Eq. (26)-(28) of Reference Yamada2012 if and only if $\beta=1$. The uniqueness is because the PPN parameter is only $\beta$ for three equations as Eqs. (41)-(43).

Condition (B) is satisfied, if $\omega_{1}=\omega_{2}=\omega_{3}\equiv\omega_{PPN}$, where $\omega_{PPN}$ means the angular velocity of the PPN configuration. By substituting Eqs. (41) and (43) into Eq. (36), $\omega_{PPN}$ is obtained as

where, by using Eq. (28), the PPN correction $\delta_{\omega}$ is

There is a symmetry among $M_{1},M_{2},M_{3}$ in the second line of Eq. (45), which means that $\delta_{\omega}$ is the same for all bodies. Condition (B) is thus satisfied.

In this section, we discuss PPN modifications of the Lagrange points that are originally defined in the restricted three-body problem in Newton gravity. We choose $\nu_{A}=1-\nu$, $\nu_{B}=\nu$ and $\nu_{C}=0$, where $\nu$ is the the mass ratio of the secondary object (a planet).

First, we seek PPN corrections to $L_{1},L_{2}$ and $L_{3}$. There are three choices of how to correspond $M_{1},M_{2}$ and $M_{3}$ to the Sun, a planet and a test mass in the collinear configuration. Indeed the three choices lead to the Lagrange points $L_{1}$, $L_{2}$ and $L_{3}$.

We consider the collinear solution by Eq. (11). We denote the physical root for Eq. (11) as $z=z_{N}(1+\varepsilon)$ for the Newtonian root $z_{N}$ with using a small parameter $\varepsilon$ ($|\varepsilon|\ll 1$) at the PPN order. We substitute $z$ into Eq. (11) and rearrange it to obtain $\varepsilon$ as

where $O(\varepsilon^{2})$ is discarded because of being at the 2PN order, and $A^{N}_{k}$ and $A^{PPN}_{k}$ denote the Newtonian and PPN parts of $A_{k}$, respectively, as $A_{k}=A^{N}_{k}+\varepsilon A^{PPN}_{k}$ ($A^{N}_{0}=0$ and $A^{N}_{7}=0$ because there are no counterparts in the Newtonian case).

Eq. (46) is used for calculating the PPN corrections to $L_{1},L_{2}$ and $L_{3}$. The PPN displacement from the Newtonian Lagrange point $L_{1}$ is thus obtained as

where $M_{1}$, $M_{2}$ and $M_{3}$ are chosen as a planet, a test mass and the Sun, respectively.

Similarly, the PPN displacement from the Newtonian Lagrange point $L_{2}$ becomes

where $M_{1}$, $M_{2}$ and $M_{3}$ are chosen as the Sun, a planet and a test mass, respectively. The PPN displacement from the Newtonian Lagrange point $L_{3}$ is

where $M_{1}$, $M_{2}$ and $M_{3}$ are chosen as a planet, the Sun and a test mass, respectively. Here, a value of $z_{N}$ depends on $L_{1}$, $L_{2}$ or $L_{3}$, which is given by Eq. (4).

Next, we discuss PPN corrections to the Lagrange points $L_{4}$ and $L_{5}$, for which we consider the PPN triangular solution. Let $a$ denote the orbital separation between the primary object and the secondary one, which equals to $R_{12}=\ell(1+\varepsilon_{12})$. Therefore, $\ell=a(1-\varepsilon_{12})+O(a\varepsilon^{2})$, where $\varepsilon^{2}$ denotes the second order in $\varepsilon_{AB}$. By using this for $R_{23}$ and $R_{31}$, we obtain $R_{23}=a(1+\varepsilon_{23}-\varepsilon_{12})+O(a\varepsilon^{2})$, and $R_{31}=a(1+\varepsilon_{31}-\varepsilon_{12})+O(a\varepsilon^{2})$.

The PPN displacement from the Newtonian Lagrange point $L_{4}$ (and $L_{5}$) with respect to the Sun is obtained as