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

Dx_i = A_icos(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: