By Rudolf Dvorak, F. Freistetter, Jürgen Kurths

This e-book is meant as an creation to the sector of planetary structures on the postgraduate point. It comprises 4 vast lectures on Hamiltonian dynamics, celestial mechanics, the constitution of extrasolar planetary platforms and the formation of planets. As such, this quantity is very appropriate in case you have to comprehend the monstrous connections among those various topics.

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**Example text**

Xn (q, t)). Because ∂U ∂ q˙k (105) = 0, (103) reads d ∂ (T − U ) ∂ (T − U ) = dt ∂ q˙k ∂qk (106) We now deﬁne the Lagrange Function L L(q, q, ˙ t) = T (q, q, ˙ t) − U (q, t) 15 Here we use that d ∂xn dt ∂qk = ∂ x˙ n ∂qk which can be shown easily. (107) 28 Rudolf Dvorak and Florian Freistetter which is the diﬀerence between the kinetic and the potential energy. The ﬁnal form of the Lagrange equation of second kind is now d ∂L(q, q, ˙ t) ∂L(q, q, ˙ t) − =0 dt ∂ q˙k ∂qk (108) with k = 1, 2, . .

From equation (109) we get (110) q˙k = q˙k (q, p, t) We now deﬁne the Hamilton function16 : f q˙i (q, p, t) pi − L (q, q˙ (q, p, t) , t) H(q, p, t) = (111) i=1 We can use the Lagrange equation (108) to obtain the partial derivatives of H: ∂H = ∂qk f i=1 f ∂ q˙i ∂L ∂ q˙i ∂L pi − − ∂qk ∂qk i=1 ∂ q˙i ∂qk ∂L ∂qk d ∂L =− dt ∂ q˙k = −p˙k =− 16 (112) Note that the Hamilton function is obtained from the Lagrange function by a Legendre transformation. Stability and Chaos in Planetary Systems ∂H = ∂pk f i=1 f ∂ q˙i ∂L ∂ q˙i pi + q˙k − ∂pk ∂ q˙i ∂pk i=1 = q˙k ∂H = ∂t f (113) f i=1 =− 29 ∂L ∂ q˙i ∂L ∂ q˙i pi − − ∂t ∂ q˙i ∂t ∂t i=1 ∂L ∂t (114) Equations (112) and (113) are the canonical or Hamiltonian equations: ∂H(q, p, t) ∂qk ∂H(q, p, t) q˙k = ∂pk p˙k = − (115) (116) These equations follow directly from the Lagrangian equations and thus they are equivalent to the equations of motion given by (108).

X3N ) (91) q = (q1 , q2 , . . , qf ) x˙ = (x˙ 1 , x˙ 2 , . . , x˙ 3N ) (92) (93) q˙ = (q˙1 , q˙2 , . . , q˙f ) (94) Diﬀerentiation of (86) with respect to the time gives x˙ n = d xn (q, t) = dt f k=1 ∂xn (q, t) ∂xn (q, t) = x˙ n (q, q, q˙k + ˙ t) ∂qk ∂t (95) Thus it follows for x˙ n ∂ x˙ n (q, q, ˙ t) ∂xn (q, t) = ∂ q˙k ∂qk (96) The kinetic energy (in cartesian coordinates) can be written as 3N T = T (x) ˙ = mn 2 x˙ 2 n n=1 (97) If xn (q, t) does not explicitly depend on the time, equation (95) reduces to f x˙ n = k=1 14 ∂xn (q) q˙k ∂qk Note that xn stands for xn (q1 , q2 , .