Solutions_hw3

# Solutions_hw3 - Theoretical Dynamics Homework 3 Instructor...

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ) !...
View Full Document

{[ snackBarMessage ]}

### Page1 / 11

Solutions_hw3 - Theoretical Dynamics Homework 3 Instructor...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online