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Solutions_HW4

# Solutions_HW4 - Theoretical Dynamics Homework 4 Instructor...

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Theoretical Dynamics October 01, 2010 Homework 4 Instructor: Dr. Thomas Cohen Submitted by: Vivek Saxena Goldstein 9.7 Part (a) F 1 ( q, Q, t ) -→ F 2 ( q, P, t ) - P i = ∂F 1 ∂Q i (1) F 2 ( q, P, t ) = F 1 ( q, Q, t ) + P i Q i (2) F 1 ( q, Q, t ) -→ F 3 ( p, Q, t ) p i = ∂F 1 ∂q i (3) F 3 ( p, Q, t ) = F 1 ( q, Q, t ) - p i q i (4) F 1 ( q, Q, t ) -→ F 4 ( p, P, t ) p i = ∂F 1 ∂q i (5) P i = - ∂F 1 ∂Q i (6) F 4 ( p, P, t ) = F 1 ( q, Q, t ) - p i q i + P i Q i (7) F 2 ( q, P, t ) -→ F 3 ( p, Q, t ) p i = ∂F 2 ∂q i (8) Q i = ∂F 2 ∂P i (9) F 3 ( p, Q, t ) = F 2 ( q, P, t ) - p i q i - P i Q i (10) F 2 ( q, P, t ) -→ F 4 ( p, P, t ) p i = ∂F 2 ∂q i (11) F 4 ( p, P, t ) = F 2 ( q, P, t ) - p i q i (12) 4 - 1

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F 3 ( p, Q, t ) -→ F 4 ( p, P, t ) P i = - ∂F 3 ∂Q i (13) F 4 ( p, P, t ) = F 3 ( p, Q, t ) + P i Q i (14) Part (b) For an identity transformation, F 2 = q i P i and by equation (7), the type 4 generating function is F 4 ( p, P, t ) = F 2 ( q, P, t ) - p i q i (15) = q i P i - p i q i = 0 as p i = ∂F 2 ∂q i = P i For an exchange transformation, F 1 = q i Q i and by equation (4), the type 3 generating function is F 3 ( p, Q, t ) = F 1 ( q, Q, t ) - p i q i (16) = q i Q i - p i q i = 0 as p i = ∂F 1 ∂q i = Q i (17) Part (c) Consider a type 2 generating function F 2 ( q, P, t ) of the old coordinates and the new mo- menta, of the form F 2 ( q, P, t ) = f i ( q 1 , . . . , q n ; t ) P i - g ( q 1 , . . . , q n ; t ) (18) where f i ’s are a set of independent functions, and g i ’s are differentiable functions of the old coordinates and time. The new coordinates Q i are given by Q i = ∂F 2 ∂P i = f i ( q 1 , . . . , q n ; t ) (19) In particular, the function f i ( q 1 , . . . , q n ; t ) = R ij q j (20) where R ij is the ( i, j )-th element of a N × N orthogonal matrix, generates an orthogonal transformation of the coordinates. Now, p j = ∂F 2 ∂q j = ∂f i ∂q j P i - ∂g ∂q j = R ij P i - ∂g ∂q j (21) This equation can be written in matrix form, as p = f q P - ∂g q (22) 4 - 2
where p denotes the N × 1 column vector ( p 1 , . . . , p N ) T , ∂g/∂ q denotes the N × 1 column vector ( ∂g/∂q 1 , . . . , ∂g/∂q n ) T , and f q denotes the N × N matrix with entries f q ij = ∂f i ∂q j = R ij (23) From (22), the new momenta are given by P = f q - 1 p + ∂g q (24) = R - 1 p + ∂g q (25) = R - 1 ( p + q g ) (26) As R is an orthogonal matrix, RR T = R T R = I , so R - 1 = R T is also an orthogonal transformation. This gives the required result: the new momenta are given by the orthogonal transformation ( R - 1 ) of an n -dimensional vector ( p + q g ), whose components are the old momenta ( p ) plus a gradient in configuration space ( q g ). Goldstein 9.25 Part (a) The given Hamiltonian is H = 1 2 1 q 2 + p 2 q 4 (27) The equation of motion for q is ˙ q = ∂H ∂p = pq 4 (28) Part (b) Suppose we let Q 2 = 1 /q 2 and P 2 = p 2 q 4 . Then, Q = ± 1 /q and P = ± pq 2 . Now, { Q, P } = 1 /q, ± pq 2 } = { q - 1 , pq 2 } = { q - 1 , p } q 2 + p { q - 1 , q 2 } = ∂q - 1 ∂q ∂p ∂p - ∂q - 1 ∂p ∂p ∂q q 2 + p × 0 = - 1 q 2 q 2 = - 1 4 - 3

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So, the signs on both Q and P cannot be identical. We take Q = - 1 q (29) P = pq 2 (30) which is a valid canonical transformation. This gives the Hamiltonian, H ( Q, P ) = 1 2 ( P 2 + Q 2 ) (31) The equations of motion are ˙ Q = ∂H ∂P = P (32) ˙ P = - ∂H ∂Q = - Q (33) So, ¨ Q + Q
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Solutions_HW4 - Theoretical Dynamics Homework 4 Instructor...

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