Example 1: A three-body model in Cartesian coordinates
- Page ID
- 2245
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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: