A note on the geometric modeling of the full two body problem
aa r X i v : . [ phy s i c s . c l a ss - ph ] J a n A note on the geometric modeling of the full two body problemTanya Schmah ∗ and Cristina Stoica † January 4, 2019
Abstract
The two full body problem concerns the dynamics of two spatially extended rigid bodies (e.g.rocky asteroids) subject to mutual gravitational interaction. In this note we deduce the Euler-Poincar´e and Hamiltonian equations of motion using the geometric mechanics formalism.
Keywords :full two body problem, Euler-Poincar´e reduction, Hamiltonian, Poisson bracket
Contents
It is well known that the classical two body problem, in which the bodies are idealized as mass points,can be analysed with almost elementary methods. Once the “mass-point” assumption is dropped, oneis faced with a significantly more complex problem: a coupled, nonlinear 12 degrees of freedom systemwith a configuration space given by the product of two SO (3) Lie groups and two copies of R . Themain inconvenience in modeling resides in the lack of a global chart for SO (3); for this reason, even fora single rigid body, most classical mechanics textbooks use Euler angles or alike, leading to an intricatepresentation; see for example, [Iacob (1980)].Anticipating future developments in the aerospace industry, the full two body problem was studiedextensively in the last decades; see for instance, [Maciejewski (1995)], [Koon et al. (2004)], [Scheeres (2006)],[Bellerose and Scheeres (2008)], [Scheeres (2009)], [Hou and Xin (2018)] and references within. Themodeling of the problem within the geometric mechanics framework is developed in [Cendra and Marsden (2004)].However, this presentation uses extensively the geometric formalism at an abstract level. In this note weprovide a description of the full two body problem within the geometric mechanics framework working directly in the full two body problem phase space, and thus avoiding abstract generalizations. ∗ University of Ottawa, Email: [email protected] † Wilfrid Laurier University, Waterloo, Email: [email protected]
1e start our modeling by assuming that the reduction due to the linear translation symmetry hasalready been performed and that the centre of mass coincides with the origin of the inertial systemof coordinates. We write the Lagrangian, observe the SO (3) symmetry and state and prove the ap-propriate (Euler-Poincar´e) reduction theorem. We continue by computing the Euler equations. Next,we apply the reduced Legendre transform and deduce the Poisson structure of the reduced space, theHamiltonian, and the equations of motion. Finally, we deduce the Casimir invariant as a consequenceof the conservation of the size of the spatial angular momentum. We also include a small appendixwith some formulae concerning the potential. Consider two rigid bodies moving freely in space, with a coupling (gravitational) potential V dependingon the orientations of the bodies and the relative position r of their centres of mass. Choose a spatialcoordinate system with origin at the centre of mass of the entire system, which we assume remains fixed.Let r i be the vector from the centre of mass of the system to the centre of mass of body i , for each i . Let r = r − r . Let B and B , both subsets of R , be the reference configurations of the two rigid bodies,each equipped with a reference frame defining body coordinates , with origin at the body’s centre of mass.A configuration of the system is determined by ( R , R , r ), where R i specifies a rotation of body i from itsreference configuration, around its own centre of mass (see, for instance, [Marsden and Ratiu (1999)]).The configuration space of the system is Q := SO (3) × SO (3) × R \ { collisions } , where SO (3) denotesthe Lie group of spatial rotations.Let µ i be the mass measure for body i , for i = 1 ,
2. Then the total mass of body i is m i := Z B i dµ i . The translational kinetic energy of body i is m i k ˙ r i k . Following the centre of mass reduction, the reduced mass is m := m m m + m and the total translational kinetic energy of the system is m k ˙ r k .The coefficient of inertia matrix of body i , with respect to its own centre of mass, is J i := Z B i XX t dµ i ( X ) , where ( · ) t denotes the matrix transpose. The body angular velocities are ˆΩ i := R − i ˙ R i . The rotationalkinetic energy of body i is K i = 12 D ˙ R i , ˙ R i E i := 12 tr (cid:16) ˙ R i J i ˙ R ti (cid:17) = 12 tr (cid:18)(cid:16) R − i ˙ R i (cid:17) J i (cid:16) R − i ˙ R i (cid:17) t (cid:19) = 12 tr (cid:16) ˆΩ i J i ˆΩ ti (cid:17) =: 12 D ˆΩ i , ˆΩ i E i . The moment of inertia tensors are I i := tr ( J i ) Id − J i , (1)where Id is the 3 × so (3) with R via the hat map ˆ: R → so (3) , Ω = (Ω , Ω , Ω ) → ˆΩ = − Ω Ω Ω − Ω − Ω Ω , we can also write K i = 12 tr (cid:16) ˆΩ i J i ˆΩ ti (cid:17) = 12 Ω ti I i Ω i . , ˆΛ ∈ so (3) corresponding to the vectors Ω , Λ ∈ R ,we have h ˆΩ , ˆΛ i = Ω × Λ where [ · , · ] denotes the matrix Lie-bracket (i.e. [ A, B ] = AB − BA ) .In coordinates on the tangent bundle T ( SO (3) × SO (3) × R \ { collisions } ), the dynamics is givenby the Lagrangian L ( R , R , r , ˙ R , ˙ R , ˙ r ) = 12 D ˙ R , ˙ R E + 12 D ˙ R , ˙ R E + 12 m k ˙ r k − V ( R , R , r ) . (2)The spatial action of SO (3) on the configuration space is the diagonal left multiplication action, A · ( R , R , r ) = ( AR , AR , A r ) , A ∈ SO (3) . (3)Since L is invariant under this action, the dynamics may be retrieved from a reduced system. Indeed,describing the motion in the coordinates of one of the bodies allows us to render the equations as areduced system on a smaller dimensional phase space (the reduced space), together with the so-calledreconstruction equation that lifts the reduced dynamics back into the unreduced phase space.For future reference, we note that the infinitesimal action of so (3) to SO (3) × SO (3) × R is (see[Holm & al. (2009)]): ˆΩ SO (3) × SO (3) × R · ( R , R , r ) = ( ˆΩ R , ˆΩ R , ˆΩ r ) (4)Denote the relative orientation matrix of B with respect to body B , and the relative position ofthe centre of the mass of the system, respectively, by R := R − R and Γ := R − r . (5)We then calculate the tangent vector (velocity corresponding to the relative orientation) ˙ R ∈ T R SO (3)and the advected relative velocity (i.e. the velocity corresonding to the relative vector) ˙ Γ ˙ R = R ˆΩ − ˆΩ R and ˙ Γ = R − ˙ r − ˆΩ Γ . (6)Recalling that ˙ R i = R i ˆΩ i , i = 1 , , and using the above we calculate L ( R , R , r , ˙ R , ˙ R , ˙ r ) = L ( R − R , R − R , R − r , R − ˙ R , R − ˙ R , R − ˙ r )= L (cid:16) R − R , R − R , R − ( R Γ) , R − ( R ˆΩ ) , R − ( R ˆΩ ) , R − R ( ˙ Γ + ˆΩ Γ ) (cid:17) = L (Id , R, Γ , ˆΩ , R ˆΩ , ˙ Γ + ˆΩ Γ )from where we define the reduced lagrangian l : SO (3) × so (3) × so (3) × T (cid:0) R \ { collisions } (cid:1) → R l ( R, ˆΩ , ˆΩ , Γ , ˙ Γ ) := L (Id , R, Γ , ˆΩ , R ˆΩ , ˙ Γ + ˆΩ Γ ) (7)that takes the form l ( R, ˆΩ , ˆΩ , Γ , ˙ Γ ) = 12 D ˆΩ , ˆΩ E + 12 D ˆΩ , ˆΩ E + 12 m k ˙ Γ + ˆΩ Γ k − V ( R, Γ ) . (8)Let h· , ·i R be the usual dot product on R . Thus, for all Π = (Π , Π , Π ) ∈ R ≃ so (3) ∗ and Ω = (Ω , Ω , Ω ) ∈ R ≃ so (3) we have h Π , Ω i R = Π · Ω = Π Ω + Π Ω + Π Ω .
3e denote the pairing between so (3) ∗ and so (3) in matrix notation by h· , ·i (no subscript!), and definethe ‘breve’ map, ˘ : R → so (3) ∗ , by D ˘Π , ˆΩ E = h Π , Ω i R . It can be shown that D ˘Π , ˆΩ E = 12 tr( ˆΠ t ˆΩ) = 12 tr( ˆΠ ˆΩ t )for all ˘Π ∈ so (3) ∗ and Ω ∈ so (3). D ˆ M , R ˆΩ E = tr (cid:16) ˆ M t ( R ˆΩ) (cid:17) = tr (cid:16) ( R t ˆ M ) t ˆΩ (cid:17) = D R t ˆ M , ˆΩ E = D ( R t ˆ M ) A , ˆΩ E = D(cid:16) R t ˆ M − ˆ M t R (cid:17) , ˆΩ E (9)for all ˆΩ ∈ so (3), where a matrix subscript A denotes the anti-symmetric part of that matrix. (We usehere the fact that the trace pairing of any symmetric matrix with an antisymmetric matrix vanishes.)Similarly, D ˆ M , ˆΩ R E = tr (cid:16) ˆ M t ( ˆΩ R ) (cid:17) = tr (cid:16) ( ˆ M R t ) t ˆΩ (cid:17) = D ˆ M R t , ˆΩ E = D ( ˆ M R t ) A , ˆΩ E = D(cid:16) ˆ M R t − R ˆ M t (cid:17) , ˆΩ E . (10)We are ready now to state the main theorem. Theorem 2.1
Consider a Lagrangian L : T ( SO (3) × SO (3) × D ) → R , D ⊂ R open, L = L (cid:16) R , R , r , ˙ R , ˙ R , ˙ r (cid:17) . For any given curves ( R ( t ) , R ( t )) ∈ SO (3) × SO (3) and r ( t ) ∈ R , let R ( t ) = R − ( t ) R ( t ) , Γ ( t ) = R ( t ) r ( t ) and ˆΩ i ( t ) := R i ( t ) − ˙ R i ( t ) ∈ so (3) . Consider l ( R, ˆΩ , ˆΩ , Γ , ˙ Γ ) := L ( Id , R, Γ , ˆΩ , R ˆΩ , ˙ Γ + ˆΩ Γ ) and let R ( t ) be the solution of the non-autonomous differential equation ˙ R = R ( t ) ˆΩ ( t ) − ˆΩ ( t ) R ( t ) , R (0) = R . (11) where R = R (0) − R (0) . The following statements are equivalent:(i) ( R ( t ) , R ( t ) , r ( t )) satisfies the Euler-Lagrange equations for the Lagrangian L. (ii) The variational principle δ Z ba L (cid:16) R ( t ) , R ( t ) , r ( t ) , ˙ R ( t ) , ˙ R ( t ) , ˙ r ( t ) (cid:17) dt = 0 holds for variations with fixed endpoints.(iii) The reduced variational principle δ Z ba l (cid:16) R ( t ) , ˆΩ ( t ) , ˆΩ ( t ) , Γ ( t ) , ˙ Γ ( t ) (cid:17) dt = 0 holds using variations of the form δ ˆΩ i = ˙ˆΣ i + [ ˆΩ i , ˆΣ i ] and δ Γ = Λ − ˆΣ Γ where the ˆΣ i ( t ) are arbitrary paths in so (3) which vanish at the endpoints, i.e. ˆΣ i ( a ) = ˆΣ i ( b ) = ˆ0 , i = 1 , , and Λ ( t ) is an arbitrary path in R with Λ ( a ) = Λ ( b ) = R . iv) The (left invariant) “Euler-Poincar´e” equations hold: ddt (cid:18) δlδ ˆΩ (cid:19) = (cid:20) δlδ ˆΩ , ˆΩ (cid:21) + R (cid:18) δlδR (cid:19) t − δlδR R t ! , (12) ddt (cid:18) δlδ ˆΩ (cid:19) = (cid:20) δlδ ˆΩ , ˆΩ (cid:21) + R t δlδR − (cid:18) δlδR (cid:19) t R ! , (13) ddt (cid:18) δlδ ˙ Γ (cid:19) = δlδ Γ . (14) Proof.
The equivalence of (i) and (ii) is a restatement of Hamilton’s principle. To show that (ii)and (iii) are equivalent, we compute the variations δ ˆΩ , δ ˆΩ , and δ Γ and induced by the variations δR , δR , and δ r . Given that ˆΩ i = R − i ˙ R i and denoting ˆΣ i := R − i δR i ∈ so (3) , i = 1 , δ ˆΩ i = ( δR − i ) ˙ R i + R − i δ ˙ R i = − ( R − i δR i R − i ) ˙ R i + R − i δ ˙ R i = − ( R − i δR i )( R − i ˙ R i ) + R − i (cid:18) δ dR i dt (cid:19) = − ˆΣ ˆΩ i + R − i ddt ( δR i )= − ˆΣ i ˆΩ i + ddt (cid:0) R − i δR i (cid:1) − ˙ R i − δR i = − ˆΣ i ˆΩ i + d ˆΣ i dt + ( R − i ˙ R i R − i ) δR i = − ˆΣ i ˆΩ i + d ˆΣ i dt + ( R − i ˙ R i )( R − i ) δR i = d ˆΣ i dt − ˆΣ i ˆΩ i + ˆΩ i ˆΣ i = d ˆΣ i dt + [ ˆΣ i , ˆΩ i ] . Thus we have δ ˆΩ i = d ˆΣ i dt + [ ˆΣ i , ˆΩ i ] , i = 1 , . The variation of Γ is δ Γ = δ ( R − r ) = δ ( R − ) r + R − δ r = − R − ( δR ) R − r + R − δ r (15)Denoting Λ := R − δ r , the above reads: δ Γ = Λ − ˆΣ Γ . (16)To complete the proof we show the equivalence of (iii) and (iv). First note that since δR = δ ( R − R ) = δ ( R − ) R + R − δr = − ( R − ( δR ) R − ) R + R − R R − δR = − ( R − δR )( R − R ) + ( R − R )( R − δR ) = − ˆΣ R + R ˆΣ . we have δR = R ˆΣ − ˆΣ R. Now we calculate δ Z ba l (cid:16) R, Γ , ˆΩ , ˆΩ , ˙ Γ (cid:17) dt = Z ba (cid:28) δlδR , δR (cid:29) + (cid:28) δlδ Γ , δ Γ (cid:29) R + X i =1 (cid:28) δlδ ˆΩ i , δ ˆΩ i (cid:29) + (cid:28) δlδ ˙ Γ , δ ˙ Γ (cid:29) R dt = Z ba (cid:28) δlδR , R ˆΣ − ˆΣ R (cid:29) dt + (cid:28) δlδ Γ , δ Γ (cid:29) R + X i =1 Z ba (cid:28) δlδ ˆΩ i , ˙ˆΣ i + [ ˆΣ i , ˆΩ i ] (cid:29) dt + (cid:28) δlδ ˙ Γ , δ ˙ Γ (cid:29) R dt (17)5sing the relations (9) and (10), the first term of (17) becomes Z ba (cid:28) δlδR , R ˆΣ − ˆΣ R (cid:29) dt = Z ba (cid:28) δlδR , R ˆΣ (cid:29) dt − Z ba (cid:28) δlδR , ˆΣ R (cid:29) dt = Z ba * R t δlδR − (cid:18) δlδR (cid:19) t R ! , ˆΣ + dt − Z ba * δlδR R t − R (cid:18) δlδR (cid:19) t ! , ˆΣ + dt Using that ˆΠ ∈ so ∗ (3) we have D ˆΠ , [ ˆΣ , ˆΩ] E = D [ ˆΩ , ˆΠ] , ˆΣ E for all ˆΣ , ˆΩ ∈ so (3) . that δ ( d/dt ) = ( d/dt ) δ, integrating by parts and taking into account the boundary conditions, the third term of (17) becomes: Z ba (cid:28) − ddt (cid:18) δlδ ˆΩ (cid:19) + (cid:20) δlδ ˆΩ , ˆΩ (cid:21) , ˆΣ (cid:29) dt + Z ba (cid:28) − ddt (cid:18) δlδ ˆΩ (cid:19) + (cid:20) δlδ ˆΩ , ˆΩ (cid:21) , ˆΣ (cid:29) dt Finally, define Γ ⋄ P ∈ so ∗ (3) via D Γ ⋄ P , ˆΣ E := D P , ˆΣ Γ E = h P , Σ × Γ i R = h Γ × P , Σ i R for all P , Γ ∈ R , and ˆΣ ∈ so (3) . Substituting (16) the second and the last terms of (17) transform to Z ba (cid:28) δlδ Γ , δ Γ (cid:29) R + (cid:28) δlδ ˙ Γ , ddt δ Γ (cid:29) R = Z ba (cid:28) − ddt (cid:18) δlδ ˙ Γ (cid:19) + δlδ Γ , Λ − ˆΣ Γ (cid:29) R = Z ba (cid:28) − ddt (cid:18) δlδ ˙ Γ (cid:19) + δlδ Γ , Λ (cid:29) R − Z ba (cid:28) Γ ⋄ (cid:20) − ddt (cid:18) δlδ ˙ Γ (cid:19) + δlδ Γ (cid:21) , ˆΣ (cid:29) . Thus we obtain δ Z ba l (cid:16) R, Γ , ˆΩ , ˆΩ , ˙ Γ (cid:17) dt = Z ba * − ddt (cid:18) δlδ ˆΩ (cid:19) + (cid:20) δlδ ˆΩ , ˆΩ (cid:21) − δlδR R t − R (cid:18) δlδR (cid:19) t ! − Γ ⋄ (cid:20) − ddt (cid:18) δlδ ˙ Γ (cid:19) + δlδ Γ (cid:21) , ˆΣ + dt + Z ba * − ddt (cid:18) δlδ ˆΩ (cid:19) + (cid:20) δlδ ˆΩ , ˆΩ (cid:21) + R t δlδR − (cid:18) δlδR (cid:19) t R ! , ˆΣ + dt + Z ba (cid:28) − ddt (cid:28) δlδ ˙ Γ (cid:19) + δlδ Γ , Λ (cid:29) R dt. Since ˆΣ , ˆΣ and Λ are arbitrary, the conclusion follows.Recall that any orthogonal matrix R can be expressed as R := [ α , α , α ] with α i ∈ R , i = 1 , , , such that α i = 1 and α i · α j = 0 for i = j. Then for any function depending on R ∈ SO (3), i.e., f = f ( R , · ) → R the vector representation ofˆ T := R (cid:18) δfδR (cid:19) t − δfδR R t and ˆ T := R t δfδR − (cid:18) δfδR (cid:19) t R is T = X i =1 , , α i × δfδ α i and T = − X i =1 , , α i × δfδ α i , respectively. Note that in the above, we calculate δfδR as the matrix δfδR = (cid:20) ∂f∂ α ∂f∂ α ∂f∂ α (cid:21) α i = ( α i , α i , α i ) t we have ∂f∂ α i = (cid:18) ∂fα i , ∂fα i , ∂fα i (cid:19) t . This allows to writingthe vector form of the reduced equations of motion (14): ddt (cid:18) δlδ Ω (cid:19) = δlδ Ω × Ω + X i =1 , , α i × δlδ α i (18) ddt (cid:18) δlδ Ω (cid:19) = δlδ Ω × Ω − X i =1 , , α i × δlδ α i (19) ddt (cid:18) δlδ ˙ Γ (cid:19) = δlδ Γ . (20)This above system is completed by the relative orientation equation (11).Specializing the Lagrangian to the full two body problem, the reduced lagrangian is given by (8).In vectorial notation the reduced lagrangian is l ( R, Ω , Ω , Γ , ˙Γ) := 12 h Ω , I Ω i R + 12 h Ω , I Ω i R + m k ˙ Γ + Ω × Γ k − V ( R, Γ ) (21)and the equations of motion are ddt (cid:16) I Ω + m Γ × ( ˙ Γ + Ω × Γ ) (cid:17) = I Ω × Ω − m h ( ˙ Γ + Ω × Γ ) × Γ i × Ω + X i =1 , , α i × δVδ α i (22) I ˙ Ω = I Ω × Ω − X i =1 , , α i × δVδ α i (23) ddt (cid:16) ˙ Γ + Ω × Γ (cid:17) = (cid:16) ˙ Γ + Ω × Γ (cid:17) × Ω − m δVδ Γ (24) The Hamiltonian of the full two body problem may be obtained by applying the Legendre transformto the Lagrangian (2) and it reads: H : T ∗ SO (3) × T ∗ SO (3) × T ∗ R \ { collisions } → R H ( R , π R , R , π R , r , p ) = 12 h π R , π R i ∗ + 12 h π R , π R i ∗ + 12 m p + V ( R, r ) , (25)where the pairings h· , ·i ∗ i on T ∗ R i ( SO (3) for fixed R i , i=1,2 correspond to the kinetic terms in (2), and,as usual: π R i = ∂L∂ ˙ R i ∈ T ∗ R i SO (3) , i = 1 , p = ∂L∂ ˙ r ∈ T ∗ r R ≃ R . (26)In order to obtain the reduced Hamiltonian we use the reduced Legendre transform. First wecalculate the momenta Π = ddt (cid:18) δlδ Ω (cid:19) = I Ω + m Γ × (cid:16) ˙ Γ + Ω × Γ (cid:17) (27) Π = ddt (cid:18) δlδ Ω (cid:19) = I Ω (28) P = ddt (cid:18) δlδ ˙ Γ (cid:19) = m (cid:16) ˙ Γ + Ω × Γ (cid:17) . (29)7ext we calculate the reduced Hamiltonian via H ( R, Π , Π , Γ , P ) = h Π , Ω ( Π , Π , Γ , P ) i R + h Π , Ω ( Π , Π , Γ , P ) i R + D P , ˙ Γ ( Π , Π , Γ , P ) E R − l (cid:16) Ω ( R, Π , Π , Γ , P ) , Ω ( Π , Π , Γ , P ) , Γ , ˙ Γ ( Π , Π , Γ , P ) (cid:17) (30)and obtain the reduced Hamiltonian of the full two body problem H : SO (3) × so ∗ (3) × so ∗ (3) × T ∗ R → R , (31) H ( R, Π , Π , Γ , P ) = 12 (cid:10) Π + Γ × P , I − ( Π + Γ × P ) (cid:11) R + 12 (cid:10) Π , I − Π (cid:11) R + 12 m h P , P i R + V ( R, Γ ) . (32)The dynamics is given by the Poisson bracket { F, H } ( R, Π , Π , Γ , P ) = − (cid:28) Π , δFδ Π × δHδ Π (cid:29) R − (cid:28) Π , δFδ Π × δHδ Π (cid:29) R + (cid:18) δFδ Γ δHδ P − δHδ Γ δFδ P (cid:19) (33) − (cid:28) δFδR , δHδ Π R − R δHδ Π (cid:29) + (cid:28) δHδR , δFδ Π R − R δFδ Π (cid:29) . (34)This is deduced by considering the composition of real valued (smooth) functions F : SO (3) × so ∗ (3) × so ∗ (3) × T ∗ R → R with the Poisson map λ : T ∗ SO (3) × T ∗ SO ∗ (3) × T ∗ R → SO (3) × so ∗ (3) × so ∗ (3) × T ∗ R λ ( R , π R , R , π R , r , p ) = (cid:0) R − R , R t π R , R t π R , r , p (cid:1) ; (35)using the chain rule, the canonical bracket on T ∗ SO (3) × T ∗ SO ∗ (3) × T ∗ R becomes the Poisson bracket(34) (for details on this kind of techniques, see [Krishnaprasad and Marsden (1987)]).The equations of the reduced dynamics are:˙ Π = Π × (cid:2) I − ( Π + Γ × P ) (cid:3) + X i =1 , , α i × δVδ α i (36)˙ Π = Π × I − Π − α i × δVδ α i (37)˙ Γ = 1 m P + Γ × (cid:2) I − ( Π + Γ × P ) (cid:3) (38)˙ P = P × (cid:2) I − ( Π + Γ × P ) (cid:3) − ∂V∂ Γ (39)together with the reconstruction (orientation) equation:˙ R = R ˆΩ − ˆΩ R. (40)where R = [ α , α , α ] and ˆΩ and ˆΩ are calculated via the inverse of (27)- (29). Remark 3.1
Note that with the choice of B as reference frame, Π is the sum of the angular momen-tum I Ω of the rigid body B and the angular momentum Γ × P of the relative vector, both in the bodycoordinates of B : Π = I Ω + Γ × P . (41)8 emark 3.2 The change of variable ( Π , Π , P ) = ( Λ , Λ , P ) := ( Π − Γ × P , Π , P ) , is a Poisson map (see [Marsden (1992)], Section 3.7) and it leads to the Hamiltonian of the two fullbody problem as used by [Maciejewski (1995)] and [Cendra and Marsden (2004)]: H ( R, Λ , Λ , Γ , P ) = 12 (cid:10) Λ , I − Λ (cid:11) R + 12 (cid:10) Λ , I − Λ (cid:11) R + 12 m P + V ( R, Γ ) . (42) The equations of motion are ˙ Λ = Λ × I − Λ + X i =1 , , α i × δVδ α i + Γ × ∂V∂ Γ (43)˙ Λ = I − Λ − X i =1 , , α i × δVδ α i (44)˙ Γ = 1 m P + Γ × I − Λ (45)˙ P = P × Γ − ∂V∂ Γ . (46) Note that this equations coincide to those in [Maciejewski (1995)].
The spatial total angular momentum corresponds to the right SO (3) action on the phase space it isgiven by J : T ∗ (cid:0) SO (3) × SO (3) × R \ { collisions } (cid:1) ) → so ∗ (3) J ( R , R , r , ˘ π , ˘ π , p ) → (˘ π R t + ˘ π R t + ˆ r × p ) . (47)where we deduced the above using the cotangent bundle momentum map formula (see [Holm & al. (2009)]page 284) and the infinitesimal generator (4). Since the Hamiltonian (25) is invariant under the afore-mentioned action, by Noether’s theorem, the spatial angular momentum is conserved along any tra-jectory. Denoting Λ Λ the body angular momenta of B and B , respectively (i.e., Λ = I Ω and Λ = I Ω ) we have k ˘ π R t + ˘ π R t + ˆ r × p k = k ˆΩ t I R t + ˆΩ t I R t + ( R Γ × R P )ˆ k = k ( R I ˆΩ ) t + ( R I ˆΩ ) t + ( R ( Γ × P ))ˆ k = k R I Ω + R I Ω + R ( Γ × P ) k = k I Ω + ( R − R ) I Ω + Γ × P k = k Λ + R Λ + Γ × P k = k Π + R Π k (48)where we used that the relationship between the spatial and body rigid body angular momenta ˘ π i = ˆΩ ti I i (see [Holm & al. (2009)] Section 1.5). The composition of the spatial momentum map with the Casimir C : so ∗ (3) → R , C ( x ) = || x || leads to the Casimir C ( R, Λ , Λ , Γ , P ) = k Λ + R Λ + Γ × P k = k Π + R Π k (49)and further, any function of the form Φ( k Π + R Π k ) is a Casimir for the reduced dynamics. The authors were supported by two Discovery grants awarded by the National Science and EngineeringCouncil of Canada (NSERC). 9
Appendix
We append this note with some formulas on the interacting potential. In concordance with mostphysical situations, we may assume that the distance between the bodies is much larger than the bodiesdimensions. Thus we consider the potential truncated to the third order ([Maciejewski (1995)]): V ( R, Γ ) = − Gm m | Γ | − G | Γ | ( m Tr I + m Tr I ) + 3 G | Γ | ( m h Γ , I Γ i R + m h R Γ , I R Γ i R ) (50)where the rotation matrix R ∈ SO (3) is represented by R = [ α , α , α ] with α i (column) vectors suchthat α i = 1 and α i · α j = 0 for i = j . Next we calculate the terms α i × ∂V /∂ α i and Λ × ∂V /∂ Λ occurring in the equations of motion. Denoting I := diag( I , I , I ), we obtain: α × ∂V∂ α = 2Γ α α ( I − I ) − α α ( I − I ) α α ( I − I ) + 2 Γ ( α × I α + α × I α ) (51)and circular combinations. Further ∂V∂ Λ = (cid:20) Gm m | Γ | − G | Γ | ( m Tr I + m Tr I ) − G | Γ | ( m h Γ , I Γ i R + m h R Γ , I R Γ i R ) (cid:21) Λ | Λ | + 3 G | Γ | (cid:0) m I + m R t I R (cid:1) Λ | Λ | (52)and so Λ × ∂V∂ Λ = 3 G | Γ | (cid:2) m Λ × I Λ + m Λ × (cid:0) R t I R Λ (cid:1)(cid:3) . (53) References
Bellerose J.E., Scheeres D.J.: (2008),
Energy and stability in the full two body problem , Celest. Mech.Dyn. Astron., Vol. , 63Cendra, H. and Marsden, J.E.: (2004)
Geometric Mechanics and the Dynamics of Asteroid Pairs ,Dynamical Systems. An International Journal, Vol. , 3Doubochine, G. N.: (1984), Sur le problme des trois corps solides , Celestial Mech.. Vol. , no. 1, 31Hern´andez-Garduno, A. and Stoica, C.: (2015) Lagrangian relative equilibria in a modified three-bodyproblem with a rotationally symmetric ellipsoid , SIAM Journal on Applied Dynamical Systems Vol. (1), 221Xiyun Hou X. and Xin X.: A note on the full two-body problem and related restricted full three-bodyproblem , Astrodynamics, Vol. , Issue 1, 39Iacob, C. (1980) Theoretical mechanics , Editura Didactic u a c si Pedagogic u a, Bucure c stiJose, J.V. and Saletan, E.J.: (1998), Classical Dynamics: a contemporary approach , Cambridge Uni-versity PressKondurar V.T.: (1974)
On Lagrange solutions in the problem of three rigid bodies , Vol. , Issue 3,pp 327Krishnaprasad, P. and Marsden, J.E.: (1987), Hamiltonian Structures and Stability for Rigid Bodieswith Flexible Attachments , Arch. Rat. Mech. Anal., Vol. , no.1, 7110oon, W.-S., Marsden, J. E., Ross S. ,Lo M., , and D. J. Scheeres, D. J.: (2004), Geometric mechanicsand the dynamics of asteroid pairs, NY Acad of Sciences Vol. , 11Maciejewski, A.J.: (1995), Reduction, relative equilibria and potential in the two rigid body problem ,Celestial mechanics and dynamical astronomy, Vol. , 1Marsden, J.E.: (1992), Lectures on Mechanics , Vol.
London Mathematical Society Lecture Notes,Cambridge University PressMarsden, J.E. and Ratiu T.S.: (1999),
Introduction to Mechanics and Symmetry , Texts in AppliedMathematicsVol. , Springer-Verlag, Second edition.Moeckel, R.: (2017), Minimal energy configurations of gravitationally interacting rigid bodies , CelestialMech. Dynam. Astronom. , 3Moeckel, R.: (2018),
Counting relative equilibrium configurations of the full two-body problem , CelestialMech. Dynam. Astronom. Vol. , 17Holm D.D., Schmah T. and Stoica C.: (2009),
Geometric mechanics and symmetry. From finite toInfinite Dimensions , Oxford Texts in Applied and Engineering Mathematics , Oxford UniversityPressScheeres, D.J.: (2006), Relative equilibria for general gravity fields in the sphere-restricted full 2-bodyproblem , Celes. Mech. Dyn. Astron. Vol. , 317Scheeres, D.J.: (2009), Stability of the planar full 2-body problem , Celes. Mech. Dyn. Astron. Vol. ,103Scheeres, D.J.: (2017)
Constraints on bounded motion and mutual escape for the full 3-body problem ,Celes. Mech. Dyn. Astron. Vol. , Issue 2-3, pp 131Vera, J.A. and Vigueras A.: (2006)
Hamiltonian Dynamics of a Gyrostat in the N-Body Problem ,Celes. Mech. Dyn. Astron. Vol. , Issue 3, pp 289Zhuravlev, S.G., Petrutskii, A.A.: (1990) Current state of the problem of translational-rotational mo-tion of three-rigid bodies
Soviet Astron. Vol.34