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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 π * i du * im X i =1 π i du i (17) = m X i =1 π * i du * im X i =1 π i du i + ( H *H ) (18) which gives us the transformation is canonical. 7. u * = √ u cos(2 π ) , π * = √ u sin(2 π ). π * du *πdu = ( 1 4 sin(4 π )π ) du2 u sin 2 (2 π ) dπ (19) := F 1 du + F 2 dπ (20) 2 We can check that ∂F 1 ∂π = ∂F 2 ∂u , therefore this is an exact diﬀerential and we can solve for G such that dG = π * du *πdu . Speciﬁcally, G = u 4 sin(4 π )uπ. Thus, by problem (6), this is a canonical transformation. 3...
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 Spring '08
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 Differential Equations, Equations, Trigraph, DT DT DT, Symplectic manifold

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