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UMPA, Ecole Normale Supérieure de Lyon
Email: [email protected]


Velocity Syzygies and Bounding Syzygy Moments in the Planar Three-Body Problem

Alexei Tsygvintsev
Abstract.

We consider the Newtonian planar three-body problem, defining a syzygy (velocity syzygy) as a configuration where the positions (velocities) of the three bodies become collinear. We demonstrate that if the total energy is negative, every collision-free solution has an infinite number of velocity syzygies. Specifically, the velocities of the three bodies become parallel within every interval of time containing three consecutive syzygies. Using comparison theory for matrix Riccati equations, we derive new upper and lower bounds on the moments when syzygies occur.

Key words and phrases:
dynamical systems, celestial mechanics, three-body problem, syzygies

1. Introduction and Preliminary Results

The Newtonian planar three-body problem has been a subject of extensive study in celestial mechanics. It involves predicting the motion of three bodies moving under their mutual gravitational attraction. Understanding syzygies, where the positions or velocities of the three bodies become collinear, provides significant insight into the dynamics of the system.

Let P1P_{1}, P2P_{2}, and P3P_{3} be three points in the plane with strictly positive masses m1,m2,m3m_{1},m_{2},m_{3} and Euclidean coordinates (xi,yi)2(x_{i},y_{i})\in\mathbb{R}^{2}, i=1,2,3i=1,2,3. The Newtonian three-body problem [2] can be formulated as follows:

z¨1=m2z21|z21|3m3z13|z13|3,z¨2=m3z32|z32|3m1z21|z21|3,z¨3=m1z13|z13|3m2z32|z32|3,\ddot{z}_{1}=m_{2}\frac{z_{21}}{|z_{21}|^{3}}-m_{3}\frac{z_{13}}{|z_{13}|^{3}},\quad\ddot{z}_{2}=m_{3}\frac{z_{32}}{|z_{32}|^{3}}-m_{1}\frac{z_{21}}{|z_{21}|^{3}},\quad\ddot{z}_{3}=m_{1}\frac{z_{13}}{|z_{13}|^{3}}-m_{2}\frac{z_{32}}{|z_{32}|^{3}}\,, (1.1)

where zk=xk+iykz_{k}=x_{k}+iy_{k}\in\mathbb{C}, k=1,2,3k=1,2,3 and zkl=zkzlz_{kl}=z_{k}-z_{l}.

Given that the total linear momentum (which is conserved) is zero, one can always set

kmkz˙k=0,kmkzk=0,\sum\limits_{k}m_{k}\dot{z}_{k}=0,\quad\sum\limits_{k}m_{k}z_{k}=0\,, (1.2)

by placing the center of mass of the bodies at the origin.

Let tzi(t)t\mapsto z_{i}(t), i=1,2,3i=1,2,3 be any collision-free solution of equations (1.1) defined for tI=[0,a)t\in I=[0,a), a>0a>0, and determined by a set of initial conditions (zi(0),z˙i(0))(z_{i}(0),\dot{z}_{i}(0)), i=1,2,3i=1,2,3.

Definition 1.1.

We say that the three bodies P1,P2,P3P_{1},P_{2},P_{3} form a syzygy (velocity syzygy) at the moment t0It_{0}\in I if the complex triplet (z1,z2,z3)(t0)(z_{1},z_{2},z_{3})(t_{0}) ((z˙1,z˙2,z˙3)(t0)(\dot{z}_{1},\dot{z}_{2},\dot{z}_{3})(t_{0})) is collinear.

Our first result concerns the existence of velocity syzygies.

Theorem 1.1.

Let t(z1(t),z2(t),z3(t))t\mapsto(z_{1}(t),z_{2}(t),z_{3}(t)), t[0,+)t\in[0,+\infty), be a zero angular momentum, collision-free solution to the three-body problem (1.1) with negative energy. Then, it has an infinite number of velocity syzygies.

Proof.

After introducing the new variables wi=miziw_{i}=m_{i}z_{i}, i=1,2,3i=1,2,3, the relations (1.2) yield

iwi=iw˙i=0.\sum\limits_{i}w_{i}=\sum\limits_{i}\dot{w}_{i}=0. (1.3)

Writing

wk=Xk+iYk,Xk=mkxk,Yk=mkyk,k=1,2,3,w_{k}=X_{k}+iY_{k},\quad X_{k}=m_{k}x_{k},\quad Y_{k}=m_{k}y_{k},\quad k=1,2,3\,, (1.4)

and using (1.3), one derives from the equations (1.1) the following 2×22\times 2 matrix equation:

X¨=AX,X=(X1Y1X2Y2),A=(m2ρ3m13ρ2m1ρ32m2ρ31m1ρ3m32ρ1),\ddot{X}=AX,\quad X=\begin{pmatrix}X_{1}&Y_{1}\\ X_{2}&Y_{2}\end{pmatrix},\quad A=\begin{pmatrix}-m_{2}\rho_{3}-m_{13}\rho_{2}&m_{1}\rho_{32}\\ m_{2}\rho_{31}&-m_{1}\rho_{3}-m_{32}\rho_{1}\end{pmatrix}\,, (1.5)

where

ρ1=1|z32|3,ρ2=1|z13|3,ρ3=1|z21|3,mij=mi+mj,ρij=ρiρj.\rho_{1}=\frac{1}{|z_{32}|^{3}},\quad\rho_{2}=\frac{1}{|z_{13}|^{3}},\quad\rho_{3}=\frac{1}{|z_{21}|^{3}},\quad m_{ij}=m_{i}+m_{j},\quad\rho_{ij}=\rho_{i}-\rho_{j}\,. (1.6)

We define the determinants

Δ1(t)=|X1Y1X2Y2|(t),Δ2(t)=|X˙1Y˙1X˙2Y˙2|(t),t[0,+).\Delta_{1}(t)=\left|\begin{array}[]{ll}X_{1}&Y_{1}\\ X_{2}&Y_{2}\end{array}\right|(t),\quad\Delta_{2}(t)=\left|\begin{array}[]{ll}\dot{X}_{1}&\dot{Y}_{1}\\ \dot{X}_{2}&\dot{Y}_{2}\end{array}\right|(t),\quad t\in[0,+\infty)\,. (1.7)

It is sufficient to show that Δ2\Delta_{2} has an infinite number of zeros for t0t\geq 0. As demonstrated by Montgomery [1], and later independently by the author in [4, 5], the solution t(z1(t),z2(t),z3(t))t\mapsto(z_{1}(t),z_{2}(t),z_{3}(t)), for t[0,+)t\in[0,+\infty), has an infinite number of consecutive syzygies. That is, there exist ti0t_{i}\geq 0, i=1,2,i=1,2,\dots, with ti+1>tit_{i+1}>t_{i} such that Δ1(ti)=0\Delta_{1}(t_{i})=0. As shown in [5, p. 6833], in the zero angular momentum case:

Δ˙1(t)24Δ1(t)Δ2(t)0,t0,\dot{\Delta}_{1}(t)^{2}-4\Delta_{1}(t)\Delta_{2}(t)\geq 0,\quad\forall t\geq 0\,, (1.8)

where equality occurs only if the positions zi(t)z_{i}(t) and velocities z˙i(t)\dot{z}_{i}(t) for i=1,2,3i=1,2,3 are parallel.

Let ti<ti+1<ti+2t_{i}<t_{i+1}<t_{i+2} be three consecutive syzygy moments. By Rolle’s theorem, there exist τ(ti,ti+1)\tau\in(t_{i},t_{i+1}) and η(ti+1,ti+2)\eta\in(t_{i+1},t_{i+2}) such that τ<η\tau<\eta and Δ˙1(τ)=Δ˙1(η)=0\dot{\Delta}_{1}(\tau)=\dot{\Delta}_{1}(\eta)=0. Without loss of generality, we can assume that Δ1(τ)>0\Delta_{1}(\tau)>0 and Δ1(η)<0\Delta_{1}(\eta)<0. Indeed, if for some i0i\geq 0, Δ1(ti)=Δ˙1(ti)=0\Delta_{1}(t_{i})=\dot{\Delta}_{1}(t_{i})=0, then according to (1.8), the corresponding solution is a straight-line one (with all positions and velocities lying on the same line) and will result in a triple collision since the energy is negative.

Thus, by the Intermediate Value Theorem and (1.8), Δ2\Delta_{2} has a zero in the interval [τ,η][\tau,\eta]. The proof is complete. ∎

2. Bounds on Syzygy Moments in the Zero Angular Momentum Case

It is well-established that in the case of negative energy and zero angular momentum, every collision-free solution in the interval [0,+)[0,+\infty) has an infinite number of syzygies [1, 4, 5]. In the author’s previous works [4, 5], an upper bound was determined for the moment when the very first syzygy occurs, expressed as a function of the energy value and the masses alone. In this section, we refine our results by providing both upper and lower bounds that depend on the initial positions and velocities of the bodies, resulting in significantly more precise estimates.

First, we present a preliminary result from the comparison theory for matrix Riccati equations, developed by Eschenburg and Heintze in 1990 [6], which will be utilised later in the proof of our main result.

Let EE be a finite-dimensional real vector space equipped with an inner product (,)(\cdot,\cdot), and let S(E)S(E) denote the space of self-adjoint linear endomorphisms of EE. Consider the Riccati differential equation with a smooth coefficient curve R:(0,t0)S(E)R:(0,t_{0})\to S(E):

B˙+B2+R=0,\dot{B}+B^{2}+R=0, (2.1)

where B:(0,t0)S(E)B:(0,t_{0})\to S(E) is the solution.

Given two smooth coefficient curves R1,R2:(0,t0)S(E)R_{1},R_{2}:(0,t_{0})\to S(E) with R1R2R_{1}\geq R_{2} (i.e., R1R2R_{1}-R_{2} is positive semidefinite), we can compare the solutions B1B_{1} and B2B_{2} of the Riccati equations:

B˙1+B12+R1=0,\dot{B}_{1}+B_{1}^{2}+R_{1}=0, (2.2)

and

B˙2+B22+R2=0,\dot{B}_{2}+B_{2}^{2}+R_{2}=0, (2.3)

subject to appropriate initial conditions.

Theorem 2.1 ([6]).

Let R1,R2:(0,t0)S(E)R_{1},R_{2}:(0,t_{0})\to S(E) be smooth coefficient curves with R1R2R_{1}\geq R_{2}. For j=1,2j=1,2, let Bj:(0,tj)S(E)B_{j}:(0,t_{j})\to S(E) be the solution to the Riccati equation corresponding to RjR_{j}, with maximal tj(0,+]t_{j}\in(0,+\infty]. If the difference U:=B2B1U:=B_{2}-B_{1} has a continuous extension to 0 with U(0)0U(0)\geq 0, then t1t2t_{1}\leq t_{2} and B1B2B_{1}\leq B_{2} on (0,t1)(0,t_{1}).

We now assume that at the instant t=0t=0, the configuration of the three-body problem is different from a syzygy, i.e., the positions of the three bodies are not parallel. In this case, the matrix C0=X˙0X01C_{0}=\dot{X}_{0}X_{0}^{-1}, where X0=X(0)X_{0}=X(0) and X˙0=X˙(0)\dot{X}_{0}=\dot{X}(0), is well-defined since det(X0)0\det(X_{0})\neq 0. In the zero angular momentum case, as shown in [5], both eigenvalues of C0C_{0} are real. We denote the spectrum of C0C_{0} as:

Spec(C0)={λ1,λ2},\text{Spec}(C_{0})=\{\lambda_{1},\lambda_{2}\}\subset\mathbb{R}\,, (2.4)

and define

πs=min{λ1,λ2}.\pi_{s}=\min\{\lambda_{1},\lambda_{2}\}\,. (2.5)

Finally, we suppose that during the motion, all mutual distances of the bodies are bounded above and below by the constants α,β>0\alpha,\beta>0, with α<β\alpha<\beta, i.e.,

α|zij(t)|β,i<j,\alpha\leq|z_{ij}(t)|\leq\beta,\quad\forall\,i<j\,, (2.6)

and we define

θα=Mα3/2,θβ=Mβ3/2,M=m1+m2+m3.\theta_{\alpha}=\frac{\sqrt{M}}{\alpha^{3/2}},\quad\theta_{\beta}=\frac{\sqrt{M}}{\beta^{3/2}},\quad M=m_{1}+m_{2}+m_{3}\,. (2.7)
Theorem 2.2.

Let Ts>0T_{s}>0 be the moment in time when the first syzygy occurs. Then the following bounds hold:

1θαarccot(πsθα)Ts1θβarccot(πsθβ).\frac{1}{\theta_{\alpha}}\mathrm{arccot}\left(-\frac{\pi_{s}}{\theta_{\alpha}}\right)\leq T_{s}\leq\frac{1}{\theta_{\beta}}\mathrm{arccot}\left(-\frac{\pi_{s}}{\theta_{\beta}}\right)\,. (2.8)
Proof.

Let tϕi(t)t\mapsto\phi_{i}(t), i=1,2,3i=1,2,3, be arbitrary positive smooth functions on [0,+)[0,+\infty). We consider a linear system of second order

Z¨=𝒜Z,Z=(Z11Z12Z21Z22),𝒜(ϕ1,ϕ2,ϕ3)=(m2ϕ3m13ϕ2m1ϕ32m2ϕ31m1ϕ3m32ϕ1),\ddot{Z}=\mathcal{A}Z,\quad Z=\begin{pmatrix}Z_{11}&Z_{12}\\ Z_{21}&Z_{22}\end{pmatrix},\quad\mathcal{A}(\phi_{1},\phi_{2},\phi_{3})=\begin{pmatrix}-m_{2}\phi_{3}-m_{13}\phi_{2}&m_{1}\phi_{32}\\ m_{2}\phi_{31}&-m_{1}\phi_{3}-m_{32}\phi_{1}\end{pmatrix}\,, (2.9)

where ϕij=ϕiϕj\phi_{ij}=\phi_{i}-\phi_{j}, mij=mi+mjm_{ij}=m_{i}+m_{j}.

If (Z11,Z12)=(X1,Y1)(Z_{11},Z_{12})=(X_{1},Y_{1}), (Z21,Z22)=(X2,Y2)(Z_{21},Z_{22})=(X_{2},Y_{2}), and ϕi=ρi\phi_{i}=\rho_{i} for i=1,2,3i=1,2,3, then the equations (2.9) and (1.5) coincide.

It is a straightforward computation to verify that for any solution ZZ of the system (2.9), the following identity holds:

1m1|Z11Z12Z˙11Z˙12|+1m2|Z21Z22Z˙21Z˙22|+1m3|Z11+Z21Z12+Z22Z˙11+Z˙21Z˙12+Z˙22|=k,k,\frac{1}{m_{1}}\begin{vmatrix}Z_{11}&Z_{12}\\ \dot{Z}_{11}&\dot{Z}_{12}\end{vmatrix}+\frac{1}{m_{2}}\begin{vmatrix}Z_{21}&Z_{22}\\ \dot{Z}_{21}&\dot{Z}_{22}\end{vmatrix}+\frac{1}{m_{3}}\begin{vmatrix}Z_{11}+Z_{21}&Z_{12}+Z_{22}\\ \dot{Z}_{11}+\dot{Z}_{21}&\dot{Z}_{12}+\dot{Z}_{22}\end{vmatrix}=k,\quad k\in\mathbb{R}\,, (2.10)

which is an analog of the angular momentum conservation law in the three-body problem written in the form (1.5).

The matrix B=Z˙Z1B=\dot{Z}Z^{-1} is a solution of the following Riccati equation:

B˙+B2+R=0,R=𝒜,\dot{B}+B^{2}+R=0,\quad R=-\mathcal{A}\,, (2.11)

an equation utilized in the study of the three-body problem in [4].

We introduce the matrices

I2=(1001),A~1=(m3220m2m322),A~2=(m132m10m132).I_{2}=\left(\begin{array}[]{rr}1&0\\ 0&1\end{array}\right),\quad\tilde{A}_{1}=\left(\begin{array}[]{rr}\frac{m_{32}}{2}&0\\ -m_{2}&-\frac{m_{32}}{2}\end{array}\right),\quad\tilde{A}_{2}=\left(\begin{array}[]{rr}-\frac{m_{13}}{2}&-m_{1}\\ 0&\frac{m_{13}}{2}\end{array}\right)\,. (2.12)

Then, as shown in [5, p. 6833], BB can be presented, if k=0k=0, in the following form:

B=δ˙2δI2+bm2δA~1am1δA~2,B=\frac{\dot{\delta}}{2\delta}I_{2}+\frac{b}{m_{2}\delta}\tilde{A}_{1}-\frac{a}{m_{1}\delta}\tilde{A}_{2}\,, (2.13)

where

β=12(m3m1+1),γ=12(m3m2+1),\beta=\frac{1}{2}\left(\frac{m_{3}}{m_{1}}+1\right),\quad\gamma=\frac{1}{2}\left(\frac{m_{3}}{m_{2}}+1\right)\,, (2.14)

and

δ=|Z11Z12Z21Z22|,a=|Z11Z12Z˙11Z˙12|,b=|Z21Z22Z˙21Z˙22|.\delta=\begin{vmatrix}Z_{11}&Z_{12}\\ {Z}_{21}&{Z}_{22}\end{vmatrix},\quad a=\begin{vmatrix}Z_{11}&Z_{12}\\ \dot{Z}_{11}&\dot{Z}_{12}\end{vmatrix},\quad b=\begin{vmatrix}Z_{21}&Z_{22}\\ \dot{Z}_{21}&\dot{Z}_{22}\end{vmatrix}\,. (2.15)

In order to apply Theorem 2.1 to equation (2.11), the solution BB and the matrix RR should be symmetric. This can be achieved by the linear transformation B~=P1BP\tilde{B}=P^{-1}BP with the invertible matrix PP defined by:

P=(m1m131m13m1m3Mm210),M=m1+m2+m3.P=\left(\begin{array}[]{cc}-\frac{m_{1}}{m_{13}}&\frac{1}{m_{13}}\sqrt{\frac{m_{1}m_{3}M}{m_{2}}}\\ 1&0\end{array}\right),\quad M=m_{1}+m_{2}+m_{3}\,. (2.16)

Indeed, it is easy to check that the matrices P1A~iPP^{-1}\tilde{A}_{i}P, i=1,2i=1,2, and R~=P1RP\tilde{R}=P^{-1}RP are symmetric. Since B~\tilde{B} is a linear combination of I2I_{2} and P1A~iPP^{-1}\tilde{A}_{i}P, it is also symmetric. Thus, equation (2.11) becomes:

B~˙+B~2+R~=0,\dot{\tilde{B}}+\tilde{B}^{2}+\tilde{R}=0\,, (2.17)

and Theorem 2.1 can be applied.

Lemma 2.1.

The matrix 𝒜~(ϕ1,ϕ2,ϕ3)=P1𝒜(ϕ1,ϕ2,ϕ3)P\tilde{\mathcal{A}}(\phi_{1},\phi_{2},\phi_{3})=P^{-1}\mathcal{A}(\phi_{1},\phi_{2},\phi_{3})P is negative semidefinite.

Proof.

Since 𝒜~\tilde{\mathcal{A}} is symmetric, it is sufficient to show that the eigenvalues of 𝒜\mathcal{A} are negative. We have:

det(𝒜)=M(m3ϕ1ϕ2+m2ϕ1ϕ3+m1ϕ2ϕ3)0,Tr(𝒜)=(m32ϕ1+m13ϕ2+m21ϕ3)0,\det(\mathcal{A})=M(m_{3}\phi_{1}\phi_{2}+m_{2}\phi_{1}\phi_{3}+m_{1}\phi_{2}\phi_{3})\geq 0,\quad\text{Tr}(\mathcal{A})=-\left(m_{32}\phi_{1}+m_{13}\phi_{2}+m_{21}\phi_{3}\right)\leq 0\,, (2.18)

since ϕi0\phi_{i}\geq 0 for i=1,2,3i=1,2,3.

Thus, 𝒜~\tilde{\mathcal{A}} is negative semidefinite. ∎

Let us consider two Riccati equations:

B~˙1+B~12+R~1=0,R~1=P1𝒜(ρ1,ρ2,ρ3)P,\dot{\tilde{B}}_{1}+\tilde{B}_{1}^{2}+\tilde{R}_{1}=0,\quad\tilde{R}_{1}=-P^{-1}\mathcal{A}(\rho_{1},\rho_{2},\rho_{3})P\,, (2.19)

and

B~˙2+B~22+R~2=0,R~2=P1𝒜(1β3,1β3,1β3)P\dot{\tilde{B}}_{2}+\tilde{B}_{2}^{2}+\tilde{R}_{2}=0,\quad\tilde{R}_{2}=-P^{-1}\mathcal{A}\left(\frac{1}{\beta^{3}},\frac{1}{\beta^{3}},\frac{1}{\beta^{3}}\right)P\, (2.20)

subject to the same initial conditions

B~1(0)=B~2(0)=P1X˙0X01P.\tilde{B}_{1}(0)=\tilde{B}_{2}(0)=P^{-1}\dot{X}_{0}X_{0}^{-1}P\,. (2.21)

By the linearity of 𝒜\mathcal{A} as a function of ϕi\phi_{i} (i=1,2,3i=1,2,3), we have

R~2R~1=P1𝒜(ρ11β3,ρ21β3,ρ31β3)P,\tilde{R}_{2}-\tilde{R}_{1}=P^{-1}\mathcal{A}\left(\rho_{1}-\frac{1}{\beta^{3}},\rho_{2}-\frac{1}{\beta^{3}},\rho_{3}-\frac{1}{\beta^{3}}\right)P,

and ρi1β30\rho_{i}-\frac{1}{\beta^{3}}\geq 0 for i=1,2,3i=1,2,3 by the definition of β\beta.

According to Lemma 2.1, it follows that R~1R~2\tilde{R}_{1}\geq\tilde{R}_{2}.

Equation (2.20) can be easily solved by setting R~2=Y˙Y1\tilde{R}_{2}=\dot{Y}Y^{-1}. This leads to the equivalent equation:

Y¨=θβ2Y,θβ=Mβ3/2.\ddot{Y}=-\theta_{\beta}^{2}Y,\quad\theta_{\beta}=\frac{\sqrt{M}}{\beta^{3/2}}\,. (2.22)

The solution to the Cauchy problem defined by (2.22) and the initial conditions Y(0)=Y0=P1X0Y(0)=Y_{0}=P^{-1}X_{0}, Y˙(0)=Y˙0=P1X˙0\dot{Y}(0)=\dot{{Y}}_{0}=P^{-1}\dot{X}_{0} is:

Y(t)=cos(θβt)Y0+1θβsin(θβt)Y˙0,t0.Y(t)=\cos(\theta_{\beta}t)Y_{0}+\frac{1}{\theta_{\beta}}\sin(\theta_{\beta}t)\dot{{Y}}_{0}\,,\quad t\geq 0\,. (2.23)

Therefore, the corresponding maximal solution of (2.20) is:

B~2(t)=Y˙(t)Y(t)1,t[0,t2),\tilde{B}_{2}(t)=\dot{Y}(t)Y(t)^{-1},\quad t\in[0,t_{2})\,, (2.24)

where t2t_{2} is the first positive zero of tdet(Y(t))t\mapsto\det(Y(t)).

One has

det(Y(t2))=0det(X˙0X01+θβcot(θβt2)I2)=0,\det(Y(t_{2}))=0\,\Longleftrightarrow\,\det(\dot{X}_{0}X_{0}^{-1}+\theta_{\beta}\cot(\theta_{\beta}t_{2})I_{2})=0\,, (2.25)

since X0X_{0} and PP are invertible matrices. As a consequence, we obtain:

θβcot(θβt2)Spec(C0),C0=X˙0X01,-\theta_{\beta}\cot(\theta_{\beta}t_{2})\in\text{Spec}(C_{0}),\quad C_{0}=\dot{X}_{0}X_{0}^{-1}\,, (2.26)

and

t2=1θβarccot(πsθβ).t_{2}=\frac{1}{\theta_{\beta}}\mathrm{arccot}\left(-\frac{\pi_{s}}{\theta_{\beta}}\right)\,. (2.27)

Let Ts=t1>0T_{s}=t_{1}>0 be the first syzygy moment for the solution of the three-body problem tzi(t)t\mapsto z_{i}(t), i=1,2,3i=1,2,3. Then, using a similar argument, we show that the maximal solution of the Cauchy problem defined by (2.19) and (2.21) is defined in the interval [0,t1)[0,t_{1}). Therefore, according to Theorem 2.1, Tst2T_{s}\leq t_{2}, and the upper bound in (2.8) is proven.

To prove the lower bound, consider the Riccati equations:

B~˙1+B~12+R~1=0,R~1=P1𝒜(1α3,1α3,1α3)P,\dot{\tilde{B}}_{1}+\tilde{B}_{1}^{2}+\tilde{R}_{1}=0,\quad\tilde{R}_{1}=-P^{-1}\mathcal{A}\left(\frac{1}{\alpha^{3}},\frac{1}{\alpha^{3}},\frac{1}{\alpha^{3}}\right)P\,, (2.28)

and

B~˙2+B~22+R~2=0,R~2=P1𝒜(ρ1,ρ2,ρ3)P,\dot{\tilde{B}}_{2}+\tilde{B}_{2}^{2}+\tilde{R}_{2}=0,\quad\tilde{R}_{2}=-P^{-1}\mathcal{A}(\rho_{1},\rho_{2},\rho_{3})P\,, (2.29)

where R~1R~2\tilde{R}_{1}\geq\tilde{R}_{2}. Applying the same arguments as before, the proof of Theorem 2.2 is complete. ∎

Remark 2.1.

The uniform bounds |zij|α|z_{ij}|\geq\alpha, i<j\forall\,i<j have a very natural astrophysical interpretation: such motion corresponds to the collision-free movement of three rigid planets, each having the same radius R=α/2R=\alpha/2. As seen from the proof of Theorem 2.2, the uniform bounds (2.6) and the absence of collisions are required only for the period of time preceding the first syzygy.

3. Conclusion: Numerical Validation and Open Questions

To numerically verify our findings, we will consider the figure-eight periodic solution with equal masses m1=m2=m3=1m_{1}=m_{2}=m_{3}=1 and zero angular momentum, as described in [3]. The initial positions and velocities of the bodies are set as follows:

z1(0)=1.08075i0.0126893,z2(0)=0.570154+i0.350807,z3(0)=z1(0)z2(0),z_{1}(0)=1.08075-i0.0126893,\quad z_{2}(0)=-0.570154+i0.350807,\quad z_{3}(0)=-z_{1}(0)-z_{2}(0), (3.1)

and

z˙1(0)=0.0193421+i0.467219,z˙2(0)=1.0852i0.174718,z˙3(0)=z˙1(0)z˙2(0).\dot{z}_{1}(0)=0.0193421+i0.467219,\quad\dot{z}_{2}(0)=1.0852-i0.174718,\quad\dot{z}_{3}(0)=-\dot{z}_{1}(0)-\dot{z}_{2}(0)\,. (3.2)

Numerically, we can find that the first syzygy occurs at Ts=0.55431T_{s}=0.55431 (see Figure 1). The constants α\alpha and β\beta are estimated to be α=0.690526\alpha=0.690526 and β=2\beta=2. Using formulas (2.7), we determine θα=3.01849\theta_{\alpha}=3.01849 and θβ=0.612372\theta_{\beta}=0.612372.

Using the initial conditions (3.1) and (3.2), we compute:

X0=(0.7345281.358410.7557910.470708),Spec(X0)={1.31082,1.047},πs=1.047,X_{0}=\begin{pmatrix}0.734528&1.35841\\ 0.755791&-0.470708\end{pmatrix},\quad\text{Spec}(X_{0})=\{1.31082,-1.047\},\quad\pi_{s}=-1.047, (3.3)

and the bounds (2.8) give us:

0.409781Ts0.864231,0.409781\leq T_{s}\leq 0.864231, (3.4)

which are quite satisfactory.

We would like to highlight several open questions. It would be interesting to find bounds analogous to (2.8) for the velocity syzygy moments using similar ideas based on the comparison of matrix Riccati equations. It is important to note that the estimates provided by our Theorem 2.2 are valid only if the initial configuration at t=0t=0 is not a syzygy. However, with some effort and by employing Theorem 2.1, one could establish bounds similar to (2.8) even in the case where the initial configuration is a syzygy.

Finally, generalising to the case of non-zero angular momentum is an intriguing and important challenge.

Akcnowledgments

I would like to express my special gratitude to Richard Montgomery for useful suggestions and valuable remarks and to Jean-Claude Sikorav for attracting my attention to work [6].

Refer to caption
Figure 1. The figure shows the trajectory of the figure-eight solution, as described in [3]. The initial positions of the bodies at t=0t=0 are represented by filled points. Additionally, the configuration of the first syzygy, which occurs at t=0.55431t=0.55431, is indicated by unfilled points.

References

  • [1] R. Montgomery, ”The zero angular momentum, three-body problem: All but one solution has syzygies,” Ergod. Th. Dynam. Sys., vol. 27, pp. 311-340, 2007.
  • [2] A. Wintner, The analytical foundations of Celestial Mechanics, Princeton University Press, 1941.
  • [3] A. Chenciner and R. Montgomery, ”A remarkable periodic solution of the three-body problem in the case of equal masses,” Annals of Mathematics, vol. 152, pp. 881-901, 2000.
  • [4] A. Tsygvintsev, ”On the existence of generalised syzygies in the planar three-body problem,” C.R. Acad. Sci. Paris, vol. 361, pp. 331-335, 2023.
  • [5] A. Tsygvintsev, ”On some collinear configurations in the planar three-body problem,” Nonlinearity, vol. 36, pp. 6827-6838, 2023.
  • [6] J.H. Eschenburg and E. Heintze, ”Comparison theory for Riccati equations,” Manuscripta Math, vol. 68, pp. 209-214, 1990.