Solutions_hw3 - Theoretical Dynamics Homework 3 Instructor...

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Unformatted text preview: Theoretical Dynamics September 24, 2010 Homework 3 Instructor: Dr. Thomas Cohen Submitted by: Vivek Saxena 1 Goldstein 8.1 1.1 Part (a) The Hamiltonian is given by H ( q i ,p i ,t ) = p i ˙ q i- L ( q i , ˙ q i ,t ) (1) where all the ˙ q i ’s on the RHS are to be expressed in terms of q i , p i and t . Now, dH = ∂H ∂q i dq i + ∂H ∂p i dp i + ∂H ∂t dt (2) From (1), dH = p i d ˙ q i + ˙ q i dp i- dL = p i d ˙ q i + ˙ q i dp i- ∂L ∂q i dq i + ∂L ∂ ˙ q i d ˙ q i + ∂L ∂t dt =- ∂L ∂q i dq i + ˙ q i dp i + p i- ∂L ∂ ˙ q i d ˙ q i- ∂L ∂t dt (3) Comparing (2) and (3) we get ∂H ∂q i =- ∂L ∂q i =- ˙ p i (2nd equality from Hamilton’s equation) (4) ˙ q i = ∂H ∂q i (also Hamilton’s equation) (5) p i- ∂L ∂ ˙ q i = 0 (H is not explicitly dependent on ˙ q i ) (6)- ∂L ∂t = ∂H ∂t (7) From (4) and (6) we have d dt ∂L ∂ ˙ q i- ∂L ∂q i = 0 , i = 1 , 2 ,...,n (8) which are the Euler-Lagrange equations. 3- 1 1.2 Part (b) L ( p, ˙ p,t ) =- ˙ p i q i- H ( q,p,t ) (9) = p i ˙ q i- H ( q,p,t )- d dt ( p i q i ) (10) = L ( q, ˙ q,t )- d dt ( p i q i ) (11) = L ( q, ˙ q,t )- ˙ p i q i- p i ˙ q i (12) So, dL = ∂L ∂p i dp i + ∂L ∂ ˙ p i d ˙ p i + ∂L ∂t dt (13) =- ˙ q i dp i- q i d ˙ p i + ∂L ∂t dt (from (9)) (14) Comparing (12) and (13) we get ˙ q i =- ∂L ∂p i (15) q i =- ∂L ∂ ˙ p i (16) Thus the equations of motion are d dt ∂L ∂ ˙ p i- ∂L ∂p i = 0 , i = 1 , 2 ,...,n (17) 2 Goldstein 8.6 Hamilton’s principle is δ Z Ldt = 0 (18) or equivalently δ Z 2 Ldt = 0 (19) We can subtract the total time derivative of a function whose variation vanishes at the end points of the path, from the integrand, without invalidating the variational principle. This is because such a function will only contribute to boundary terms involving the variation of q i and p i at the end points of the path, which vanish by assumption. Such a function is p i q i . So, the ‘modified’ Hamilton’s principle is δ Z 2 L- d dt ( p i q i ) dt = 0 (20) 3- 2 Using the Legendre transformation, this becomes δ Z (2 p i ˙ q i- 2 H- p i ˙ q i- ˙ p i q i ) dt = 0 (21) = ⇒ δ Z (2 H + ˙ p i q i- p i ˙ q i ) dt = 0 (22) now, ˙ p i q i- p i ˙ q i = q 1 ... q n | p 1 ... p n 1 × 2 n n × n 1 n × n- 1 n × n n × n 2 n × 2 n ˙ q 1 . . ˙ q n-- ˙ p 1 . . ˙ p n 2 n × 1 (23) = η T J ˙ η (24) So (22) becomes δ Z ( 2 H + η T J ˙ η ) dt = 0 (25) which is the required form of Hamilton’s principle. 3 Goldstein 8.9 The constraints can be incorporated into the Lagrangian L by defining a “constrained Lagrangian” L c , as L c ( q, ˙ q,t ) = L ( q, ˙ q,t )- X k λ k ψ k ( q,p,t ) (26) Applying Hamilton’s principle, and using the Legendre transformation for L , we get δ Z p i ˙ q i- H ( q,p,t )- X k λ k ψ k ( q,p,t ) !...
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Solutions_hw3 - Theoretical Dynamics Homework 3 Instructor...

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