# Example 1: A three-body model in Cartesian coordinates

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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: