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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 * im X i =1 i du i (17) = m X i =1 * i du * im 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 ) ) du2 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...
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This note was uploaded on 09/09/2009 for the course MATH 410 taught by Professor Staff during the Spring '08 term at Maryland.
 Spring '08
 staff
 Differential Equations, Equations

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