On shape sphere and rotation of three-body motion

Let $q_{i}\in\mathbb{R}^{d}(i=1,\dots,n)$ be $n$ particles. Each particle $q_{i}$ is endowed with a mass $m_{i}>0$. Configuration space is the set of all $n$-tuples $q=(q_{1},\dots,q_{n})$ with $q_{i}\in\mathbb{R}^{d}$. A motion of the $n$-particle system in time interval $[a,b]$ is a continuous path $q\in C([a,b],\mathbb{R}^{d\times n})$. If $q$ is smooth, then the velocity of system $v=\dot{q}$ is given by the time derivative of $q$. In the following of this paper, we always assume that any motion is smooth so that we can get rid of problems about regularity. Our results extends to piecewise smooth path readily.

It’s well-known that translations can be separated from other motions. The velocity of a translation satisfies $v_{1}=v_{2}=\dots=v_{n}$ which represent the linear momentum of motion. For a conservative system, linear momentum is preserved by the motion. As in most papers, we assume that the center of mass is always at origin and the configuration space restricted to the following set

Elements in $G=SO(d)$ act on $\Sigma$ by rotations in $\mathbb{R}^{d}$. Following [1] of A.Guichardet, velocity at any configuration $q\in\Sigma$ can be decomposed into two parts: the rotation part $V_{R}$ and the internal part $V_{I}$. They are

Intuitively, the $V_{R}$ part follows from action of $SO(d)$ and $V_{I}$ part follows from scaling and changing of the similarity classes of configuration. If $d=2$ or $3$, clearly $v\in V_{I}$ if and only if the motion has angular momentum $0$ and $V_{R}$ is isomorphic to the space of admissible angular momentum at $q$.

Given a motion $q:[0,T]\rightarrow\Sigma$ such that the beginning configuration and the final configuration are similar, then they differ by a rotation(i.e. an elements in $SO(d)$). A basic question is: what can one tell about the rotation relating two boundary configurations? A.Guichardet in [1] proved that if $n>d$, then every two points of an arbitrary $G$-orbit can be joint by a smooth vibrational curve(a motion such that $v\in V_{I}$ all the time). Thus both the rotation part and the internal part will contribute to the angle of rotation.

In this paper, we will focus on $n=3$ while $d=2$ or $3$. This question has been studied in [3, 2]. For planar $3$-body problem, all the oriented similar classes of configuration form a two sphere called by ”shape sphere”. Since the beginning configuration and final configuration are similar, they differs by a rotation of the plane and the motion projects to a closed curve on shape sphere. Then the angle of rotation is given by the following formula

where $J$ is angular momentum and $I$ is moment of inertia. $D_{0}$ is the disk on shape sphere bounded by this closed curve and $\Omega_{0}$ is the standard area form on shape sphere.

For $d=3$, it not clear what is the angle of rotation between two similar oriented configuration. A natural idea is to put the two configurations on a plane by a certain way, as Montgomery did in [3]. He assumed that the angular momentum is preserved by the motion. Denote $\mathbf{e}$ the axis of angular momentum and $\mathbf{n}(t)$ is the normal vector for $q(t)$. After rotating the normal vectors $\mathbf{n}(0)$ and $\mathbf{n}(T)$ to $\mathbf{e}$ along minimizing geodesics on sphere, the angle of rotation $\Delta\theta$ can be defined as planar case. Montgomery [3] proved the following formula, which is an extension of (1.2).

where $\omega(t)$ depends on angular momentum, moment of inertia tensor and the axis of angular momentum, $\Omega$ is a two form on a ”reduced configuration space” and $D$ is a disk bounded by a closed curve in reduced configuration space defined by the motion. Readers can refer to [3] for details.

In Montgomery’s paper, he enlarged $\Sigma$ to form a set $\widetilde{\Sigma}$, the space of oriented configurations. An element in $\widetilde{\Sigma}$ is a configuration in $\Sigma$ together with a unit vector $\mathbf{n}$ which is orthogonal to the subspace spanned by vertices of $q$. This step is essential and the proof in [3] rely on the geometry of $\widetilde{\Sigma}$. He noticed that it may be possible to carry out all calculations directly on $\Sigma/SO(2)$ and asked the following question: can reconstruction formula and calculations be reformulated solely in terms of $\Sigma/SO(2)$? Here $SO(2)$ is the rotation group about axis $\mathbf{e}$.

The problem for spatial case faces some difficulties which do not appear in planar case. Firstly, there is no natural orientation for a spatial configuration. Different ways of choosing normal vector will affects the result. Secondly, if $\mathbf{n}=\mathbf{-e}$, the minimizing geodesic from $\mathbf{n}$ to $\mathbf{e}$ is not unique. Finally, given velocity at a collinear configuration, we can not deduce how the configuration rotate. For these reasons, Montgomery had to worked on the space $\widetilde{\Sigma}$.

In this paper, we extend the formula (1.2) and (1.3) to calculate the angle of rotation for any particle without requiring the conservation of angular momentum or the similarity of initial and final configuration. In fact, given the initial configuration $q(0)$, angular momentum $J(t)$, momentum of inertia $I(t)$ and the projected curve on shape sphere $\gamma(t)$, the motion is uniquely defined. One should be able to known the rotation of each particle even though two boundary configurations are not similar. The proof in this paper is totally different from that in [3]. All we need are some facts about shape sphere and fundamental calculus. The reconstruction formula considered in [3] is a special case of our result. As a corollary, we give a positive answer to the question proposed by Montgomery in [3].

If the particle we concern does not go through origin, the angle of rotation is uniquely defined by polar coordinates. If it goes across the origin, then we should take a perturbation. The second term of formula may differ by $2\pi$ depending on the perturbation. We regard two angles of rotation the same if they differ by $2k\pi$ for $k\in\mathbb{Z}$. Thus the angle of rotation only depends on initial and final configurations.

We state our result for planar case and spatial case separately. For planar case, the angle of rotation is obviously defined by polar coordinates. Our result is the following theorem.

For spatial case, let $\mathbf{e}$ be an unit vector and $X$ is the plane orthogonal to $\mathbf{e}$ in $\mathbb{R}^{3}$. Any oriented triangle configuration is defined by a triangle configuration $q$ and a normal vector $\mathbf{n}$ which gives the orientation. Whenever $\mathbf{n}\neq\mathbf{-e}$, there is a unique minimizing geodesic on unit sphere connecting $\mathbf{n}$ and $\mathbf{e}$. By moving $\mathbf{n}$ to $\mathbf{e}$ along the minimizing geodesic on the unit sphere, we get a configuration in $X$. Thus we can define a map $P:(q,\mathbf{n})\rightarrow x$ which maps a spatial configuration to a configuration in plane $X$. Then the angle of rotation is defined to be that of $x(t)$ as in planar case. This is the same as what R.Montgomery did in [3]. In that paper, the angular momentum is assumed to be preserved and $\mathbf{e}$ is chosen to be the axis of angular momentum.

To get a reconstruction formula in spatial case, we need some assumption for motion. Denote

We call a collinear configuration $q(t)$ as above is ”bad”. Our theorem for spatial case can be stated as follows.

Since velocity can be decomposed into two parts, the strategy is to study how the two parts effect the angle velocity of a particle. Because we use a different approach from [3], formula (1.7) is also difference from (1.3). In section 4, a map $\sigma$ is defined from $Lie(G)$ to angular momentum $J$. $\omega(t)$ in (1.3) is actually $\mathbf{e}\cdot\sigma^{-1}(J(t))$, while $\sigma_{\mathbf{e}}^{-1}(J(t))$ and $\sigma_{\mathbf{n}}^{-1}(J(t))$ are given by direct decomposition of $\sigma^{-1}(J(t))$ about a basis define by $\mathbf{e}$ and $\mathbf{n}$. The second term in the right of(1.3) is an integral on reduced configuration space, which is a two-sphere bundle over the shape sphere. The second term in our formula is an integral on the shape sphere.

The paper is organized as follows. In section $2$, we introduce some necessary foundations about shape sphere. In section $3$, we prove reconstruction formula for planar case. As an application, we give a characterization for the shape sphere. In section $4$, we prove the reconstruction formula for spacial case. In last section, we apply the formula to Montgomery’s case and give a positive answer to the question proposed in [3].

In this section, we give an introduction to the shape sphere for $3$ body problem. We only list some necessary results for later proof of main theorem. For more details readers can refer to [4].

For planar case, our configuration space is

By normalizing, one can greatly simplified the notations. Clearly, the normalized Jacobi coordinates $(Z_{1},Z_{2})$ is one-to-one corresponded to the centered configuration $q\in\Sigma$. The following fact can be proved by direct computation.

Let $\omega_{i}\in\mathbb{R}\,(i=1,2,3,4)$ be four real variables satisfying

Define the map $\pi$ from $\Sigma$ to $\mathbb{R}^{3}$ by

The following theorem is from [4].

By Theorem 2.2, the shape space is isomorphic to $\mathbb{R}^{3}$ and the Euclidian norm of the vector $(\omega_{1},\omega_{2},\omega_{3})$ is one half of the moment of inertia. By fixing the moment of inertia, we get a sphere in shape space.

As commented in [4], “the shape space is not isometric to $\mathbb{R}^{3}$: Shape space geometry is not Euclidean. However, the geometry does have spherical symmetry. Each sphere centered at triple collision is isometric to the standard sphere, up to a scale factor. We identify these spheres with the shape sphere.”

For our later use, we mark some points on the shape sphere.

By (2.9) and (2.10), $O_{1}$ is $(1/2,0,0)$ and $C_{1}$ is $(-1/2,0,0)$. Clearly, we have different ways to define Jacobi coordinates. Different choice of Jacobi coordinates differ by reflections or rotations of shape space. For example, by letting $\tilde{Z}_{1}=q_{2}-q_{3}$ in (2.9) the resulting shape sphere differs to original one by a reflection with respect to $\omega_{2},\omega_{3}$-plane. If we define Jacobi coordinates by $\tilde{Z}_{1}=q_{2}-q_{1}$ and $\tilde{Z}_{2}=q_{3}-\frac{m_{2}q_{2}+m_{1}q_{1}}{m_{2}+m_{1}}$, then the resulting shape sphere differs to original one by a rotation such that $O_{3}$ is $(1/2,0,0)$ and $C_{3}$ is $(-1/2,0,0)$. In the following of this paper, we will always use the original coordinate system defined by (2.9) and (2.10) unless explicitly told.

We can write $(\omega_{2},\omega_{3})$ in polar coordinates $(r,\xi)$. Then $\xi$ is well defined on $S\backslash\{O_{1},C_{1}\}$ and $\omega_{2}+\sqrt{-1}\omega_{3}=re^{i\xi}$. Let $Z_{1}=r_{1}e^{i\xi_{1}}$ and $Z_{2}=r_{2}e^{i\xi_{2}}$. By (2.10), it follows that

Hence, there exists some integer $k$, such that

Without loss of generality, we can assume $k=0$. Then $\xi=\xi_{2}-\xi_{1}$ is the angle of rotation from $Z_{1}$ to $Z_{2}$. $C_{1}$ is a singular point for $\xi_{1}$ while $O_{1}$ is a singular point for $\xi_{2}$. Thus we have

In this section, we give a prove of formula (1.4). Here we use a different approach by fundamental calculus which relies little on geometry.

In this section, the configuration space is

Proof of Theorem 1.1: For simplicity, we prove the theorem when $\xi_{1},\xi_{2}$ and $\xi$ are well defined for all $t\in[0,T]$. That is $\gamma(t)\neq C_{1}$ and $\gamma(t)\neq O_{1}$. At the end, we’ll see that this assumption can be removed.

Given any motion $q:[0,\,s]\rightarrow\Sigma$, then the corresponding curve in shape space $\gamma(t)$ is known. So there is a unique zero-angular-momentum motion $q^{\prime}(t)$ realizing $\gamma(t)$ and $q^{\prime}(0)=q(0)$. Let $Z_{1}^{\prime}=r_{1}^{\prime}e^{i\xi_{1}^{\prime}},Z_{2}^{\prime}=r_{2}^{\prime}e^{i\xi_{2}^{\prime}}$ be Jacobi coordinates of $q^{\prime}$. Clearly, we have

Thus

The angular momentum of $q(t)$ is

We have

Note that $\xi_{2}$ is just the angle of $q_{1}$ in polar coordinate. Hence

It is left to find $\xi_{2}^{\prime}(T)-\xi_{2}^{\prime}(0)$, which is the angle of rotation for the zero-angular-momentum motion $q^{\prime}:[0,T]\rightarrow\Sigma$. Now it suffice to consider motions with zero angular momentum.

For simplicity of notation, we denote $q$ to be a zero-angular-momentum motion(i.e. $q^{\prime}$ above is now replaced by $q$). Since we are interest in the angle of rotation $\xi_{2}(T)-\xi_{2}(0)$, by normalizing the momentum of inertia, it’s enough to only consider the projected curves on the shape sphere. Thus we can also assume $\gamma(t)$ is on the shape sphere.

Since

We have

Let $\Omega_{0}$ be the standard area form on the shape sphere defined by $\omega_{1}^{2}+\omega_{2}^{2}+\omega_{3}^{2}=\frac{1}{4}$. Let $S(t)$ be the area enclosed by two meridians $l_{\xi(0)}$ and $l_{\xi(t)}$(Here we should choose boundary orientation so that $S(t)$ increases as $\xi(t)$ increasing). Let $D_{\gamma}(t)$ be the disk enclosed by a closed curve on shape sphere composed by three pieces: $\gamma(0)$ to $\gamma(t)$, $C_{1}$ to $\gamma(0)$ along $l_{\xi(0)}$ and $C_{1}$ to $\gamma(1)$ along $l_{\xi(t)}$. The orientation is chosen so that it coincide with $S(t)$ on $l_{\xi(0)}$ and $l_{\xi(t)}$.

Note that $\xi$ is also the angle coordinate of $(\omega_{2},\omega_{3})$. There has

Let $R(t)=\int_{D_{\gamma}(t)}\Omega_{0}$ be the area of $D_{\gamma}(t)$. By direct computation from calculus, we have

Actually, $\dot{R}(t)$ is given by rotation of the curve connecting $C_{1}$ to $\gamma(t)$ along $l_{\xi(t)}$ while $\dot{S}(t)$ is given by rotation of the whole meridian $l_{\xi(t)}$. Thus $\frac{\dot{R}(t)}{\dot{S}(t)}$ is ratio of the area where $\omega_{1}\leq\omega_{1}(t)$ to the area of whole sphere, which is given by $\frac{1}{2}+\omega_{1}(t)$.

Summarizing three above equations, we have

Integrating (3.19) in $[0,T]$, we have

Recall that $\xi_{2}(T)-\xi_{2}(0)$ in (3.20) is actually $\xi_{2}^{\prime}(T)-\xi_{2}^{\prime}(0)$ in (3.15). Thus we have proved the reconstruction formula (1.4) for planar case under assumption that $\gamma(t)$ is away from $C_{1}$ and $O_{1}$.

If $\gamma(t)$ goes across $C_{1}$ or $O_{1}$, we can find a sequence of motion $q_{(n)}(t)$ away from $C_{1},O_{1}$ converging to $q(t)$ in $C^{1}$ topology. Then we have uniform convergence for angular momentum, momentum of inertia and the projected curve on shape space. The first term of formula (1.4) converges uniformly while the second term may differ by $2\pi$ depending on how we choose the disk $D_{\gamma}$. This cause no confusion because a rotation of $2\pi$ is identity. Hence the assumption $\gamma(t)$ is away from $C_{1}$ and $O_{1}$ can be removed.

For piecewise smooth curves, the angle of rotation is clearly the angle sum of these smooth pieces. Thus we have proved $(i)$ of Theorem 1.1. $(ii)$ of Theorem 1.1 can be proved in a similar way by applying above argument to $\xi_{1}$. Then $D_{\gamma}^{\prime}$ is the disk on shape sphere bounded by three curves: $\gamma(t)$, geodesic from $\gamma(0)$ to $O_{1}$, geodesic from $\gamma(T)$ to $O_{1}$.

∎

In the end of this section, we give some characterization of the shape sphere as an application of Theorem 1.1.

Note that we’ve marked some points on the shape sphere. Among them, $L_{1}$,  $L_{2}$,  $E_{1}$,  $E_{2}$,  $E_{3}$ are the five central configurations, which are very important for the study of the planar three-body problem. Once the coordinate system is chosen, then the positions of these marked points is given which depends on masses of particles $(m_{1},m_{2},m_{3})$. By our definition of the shape sphere, the six points $C_{1}$,$O_{2}$,$C_{3}$,$O_{1}$,$C_{2}$,$O_{3}$ arranged on the equator of the shape sphere counterclockwise and $C_{i}=-O_{i}(i=1,2,3)$. Let $(i,j,k)$ be a permutation of $(1,2,3)$, then $E_{i}$ is between $C_{j}$ and $C_{k}$. One can check that $E_{i}=O_{i}$ if and only if $m_{j}=m_{k}$, $E_{i}$ is between $C_{j}$ and $O_{i}$ if and only if $m_{j}>m_{k}$. When $m_{1}=m_{2}=m_{3}$, it’s clear that $E_{i}=O_{i}(i=1,2,3)$ and $L_{j}=P_{j}(j=1,2)$ are the two poles on the shape sphere. The six points $E_{i}$s and $C_{i}$s is evenly distributed on the equator of the shape sphere.

If the three masses are different, we can apply our result to find the relation of these points on the shape sphere. Assume $O_{1}=(1/2,0,0)$ and $C_{1}=(-1/2,0,0)$. Note that $P_{1}$ makes the area of triangle configuration largest under the condition $I=|Z_{1}|^{2}+|Z_{2}|^{2}=1$, $Z_{1}$ must be orthogonal to $Z_{2}$. For other versions of Jacobi coordinates, the corresponding two vector must also be orthogonal. Thus $P_{1}$ is characterized by the orientation (i.e. the positions of $1,2,3$ rotate counterclockwise) and the condition that the ortho-center coincides with the center of mass (i.e. the origin).

Consider $q:[0,T]\rightarrow\Sigma$ to be a special motion as in Fig 3.1. At $t=0$, $q(0)$ is a configuration corresponding to $P_{1}$. During the motion, $q_{3}$ is fixed and so is the center of mass of $q_{1}$ and $q_{2}$. $q_{1}$ and $q_{2}$ move uniformly linear towards their center of mass and collide at $t=T$. By Lemma 3.2, it’s a zero-angular-momentum motion. Its projected curve on the shape sphere is half of a meridian which goes from $P_{1}$ to $C_{3}$.

We can then compute the rotation of $q_{1}$ by Theorem 1.1. It is

Here $\Delta P_{1}C_{3}C_{1}$ is the spherical triangle by closing the curve as in the proof of Theorem 1.1. One finds that the above integral is exactly half of the angle of rotation from $OC_{1}$ to $OC_{3}$. It’s also the angle $\alpha_{2}$ in Fig.3.1. A direct computation shows that

Hence we find the location of $C_{3}$ and $O_{3}=-C_{3}$.

Similarly, the angle of rotation from $OC_{3}$ to $OC_{2}$ is $2\alpha_{1}$, the angle of rotation from $OC_{2}$ to $OC_{1}$ is $2\alpha_{3}$. They are given by

Since all $3$ body have positive masses, we have $\alpha_{i}\in(0,\pi/2)$, $\alpha_{1}+\alpha_{2}+\alpha_{3}=\pi$, $\alpha_{i}<\alpha_{j}$ if and only if $m_{i}<m_{j}$. By our definition of the shape sphere, the six points $C_{1}$,$O_{2}$,$C_{3}$,$O_{1}$,$C_{2}$,$O_{3}$ arranged on the equator of the shape sphere counterclockwise.

Let $(i,j,k)$ be a permutation of $(1,2,3)$, then $E_{i}$ is between $C_{j}$ and $C_{k}$. One can check that $E_{i}=O_{i}$ if and only if $m_{j}=m_{k}$, $E_{i}$ is between $C_{j}$ and $O_{i}$ if and only if $m_{j}>m_{k}$.

Next, we find the locations of $L_{1}$ and $L_{2}$ on the shape sphere. At the Lagrange point, the configuration is an equilateral triangle. Denote $\beta$ to be the angle of rotation from $Z_{1}$ to $Z_{2}$ of the equilateral triangle corresponding to $L_{1}$, then $\beta$ is a function of three masses. By Proposition 2.3, $L_{1}\in l_{\beta}$. Changing to another version of Jacobi coordinates, then there is a meridian $l_{\beta^{\prime}}^{\prime}$ from $C_{3}$ to $E_{3}$ such that $L_{1}\in l_{\beta^{\prime}}^{\prime}$. Thus the intersection point of $l_{\beta}$ and $l_{\beta^{\prime}}^{\prime}$ is $L_{1}$. The other Lagrangian point is given by $L_{2}=\sigma(L_{1})$.