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Example 1: A three-body model in Cartesian coordinates

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    2245
  • [ "article:topic", "Graduate", "Franzen", "Author tag:Franzen" ]

    Example 1: A three-body model in Cartesian coordinates

     

    The three-body model is:

     

    The potential energy of the system is given by:

    2V = f12(Dr12)2 + f23(Dr23)2

    where Dr12 = Dx1 - Dx2 and Dr23 = Dx2 - Dx3 so that

    2V = f12 (Dx1 - Dx2)2 + f23(Dx2 - Dx3)2

     

    2V = f12(Dx1)2+(f12+f23)(Dx2)2+f23(Dx3)2-2f12(Dx2Dx3)-2f12(Dx1Dx2)

     

    The kinetic energy is

    . . .

    2T = m1(Dx1)2 + m2(Dx2)2 + m3(Dx3)2

     

    The equations of motion are then:

    ..

    m1(Dx1) + f12(Dx1) - f1(Dx2) = 0

    ..

    m2(Dx1) + (f12 + f23)(Dx2) - f23(Dx3) - f12(Dx1) = 0

    ..

    m3(Dx3) + f23(Dx3) - f23(Dx2) = 0

    The substitution

    Dxi = Aicos(Ölt + f)

    leads to the equations

     

    (f12 - m1l)A1 - f1 A2 = 0

    -f12 A1 + (f12 + f23 - m2l)A2 - f23 A3 = 0

    (f23 - m3l)A3 - f23 A2 = 0

     

    The equations have non-zero solutions only if the determinant of the coefficients vanishes:

     

    (f12 - m1l) -f12 0

    -f12 (f12 + f23 - m2l) -f23

    0 -f23 (f23 - m3l)

    There is one trivial solution of l = 0 and then two roots: