410HW8 - dG = m X i =1 G u i du i + m X i =1 G u * i du * i...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
1 Homework #8 Consider the Hamiltonian system for the harmonic oscillator. The Hamiltonian is H ( x, π ) = π 2 2 m + kx 2 2 The corresponding system of differential equations is dx dt = ∂H ∂π = x m (1) dt = - ∂H ∂x = - kx (2) (3) 1. x * = x + π, π * = π . dx * dt = dx dt + dt = π m - kx = π * m - k ( x * - π * ) (4) * dt = dt = - kx = k ( π * - x * ) (5) 2. H * ( x * , π * ) = ( π * ) 2 2 m + k ( x * - π * ) 2 2 ∂H * ∂π * = π * m - k ( x * - π * ) = dx * dt (6) ∂H * ∂x * = k ( x * - π * ) = - * dt (7) Therefore, we get the same set of differential equations as in (1), so the transformation preserves the form of the Hamiltonian equation. 3. x * = e x , π * = π . dx * dt = e x dx dt = x * π m = x * π * m (8) * dt = dt = - kx = - k ln( x * ) (9) 1
Background image of page 1

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

View Full DocumentRight Arrow Icon
4. H * ( x * , π * ) = ( π * ) 2 2 m + k (ln( x * )) 2 2 ∂H * ∂π * = π * m = dx * dt (10) ∂H * ∂x * = k ln( x * ) x * = - * dt (11) which are different than the equations in (3). Therefore the transformation is not canonical. 5. x * = e x , we can set ˜ G = ˜ G ( x, π * ). By case 2 in class, we have x * = - ˜ G π * = e x (12) ˜ G = - e x π * (13) π = - ˜ G x = e x π * (14) π * = πe - x (15) Thus, with x * = e x , π * = πe - x is a canonical transformation. 6. For G we have
Background image of page 2
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: dG = m X i =1 G u i du i + m X i =1 G u * i du * i + G x dx (16) But since G x = 0 = H * = H . Thus, dG = m X i =1 * i du * i-m X i =1 i du i (17) = m X i =1 * i du * i-m X i =1 i du i + ( H *-H ) (18) which gives us the transformation is canonical. 7. u * = u cos(2 ) , * = u sin(2 ). * du *-du = ( 1 4 sin(4 )- ) du-2 u sin 2 (2 ) d (19) := F 1 du + F 2 d (20) 2 We can check that F 1 = F 2 u , therefore this is an exact dierential and we can solve for G such that dG = * du *-du . Specically, G = u 4 sin(4 )-u. Thus, by problem (6), this is a canonical transformation. 3...
View Full Document

This note was uploaded on 09/09/2009 for the course MATH 410 taught by Professor Staff during the Spring '08 term at Maryland.

Page1 / 3

410HW8 - dG = m X i =1 G u i du i + m X i =1 G u * i du * i...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online