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B.1 Spin-orbit coupling force

It is well known that the definition of a dipole moment of the star, which we equate to zero to determine the center of mass of the star, affects the appearance of the spin-orbit coupling force (see e.g. [108Jump To The Next Citation Point]). If one chooses i dA = 0 as in [94], the corresponding spin-orbit coupling force takes the usual form [46Jump To The Next Citation Point108Jump To The Next Citation Point152Jump To The Next Citation Point]
k F i |i = − ε4V- [(2m M il+ m M il)Δlk + 2 (m M lk + m M lk)Δli] 1SO dA=0 r312 1 2 2 1 1 2 2 1 4m1 [ i ] = ε r3- 6(⃗s2 × ⃗n12) ⋅ ⃗Vn 12 + 4⃗s2 × ⃗V − 6⃗s2 × ⃗n12 (⃗n12 ⋅ ⃗V ) 12 [ ] +ε4m2- 6(⃗s1 × ⃗n12) ⋅ ⃗V ni12 + 3⃗s1 × ⃗V − 3⃗s1 × ⃗n12(⃗n12 ⋅ ⃗V ) , (221 ) r312
where Δij ≡ δij − 3ni nj 12 12 and × in this section denotes the outer product for the usual Euclidean spatial three-vectors. The spin vector i s A is defined by
i 1 jk sA ≡ -εijkM A , (222 ) 2
where εijk is the totally antisymmetric symbol.
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