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s11f1038 - i n X i =1 ± ∂ρ ∂q i ˙ q i ∂ρ ∂p i...

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Solution for assignment 38 Liouville’s Theorem We consider motion of point particles with n degrees of freedom in phase space , which is described by a Hamiltonian H ( q 1 , . . . , q n ; p 1 , . . . , p n ) . Let ρ ( q 1 , . . . , q n ; p 1 , . . . , p n ; t ) be the density in phase space and the velocity of the density element is the vector v = ( ˙ q 1 , . . . , ˙ q n ; ˙ p 1 , . . . , ˙ p n ) . The gradient is now also defined in phase space (ˆ q i and ˆ p i are unit vectors): = n i =1 ˆ q i ∂q i + ˆ p i ∂p i . The continuity equation reads 0 = ∂ρ ∂t + ∇ · j = ∂ρ ∂t + ∇ · ( ) as j = holds. Therefore, 0 = ∂ρ ∂t + ρ ∇ · v + v · ∇ ρ = ∂ρ ∂t + ρ n i =1 ˙ q i ∂q i + ˙
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Unformatted text preview: i ! + n X i =1 ± ∂ρ ∂q i ˙ q i + ∂ρ ∂p i ˙ p i ! . Using Hamilton’s equations we have ∂ ˙ q i ∂q i = + ∂ ∂q i ∂H ∂p i and ∂ ˙ p i ∂p i =-∂ ∂p i ∂H ∂q i . Interchanging the derivative these terms cancel one another ( ∇· ~v = 0 in phase space) and we are left with Liouville’s theorem: 0 = ∂ρ ∂t + n X i =1 ± ∂ρ ∂q i ˙ q i + ∂ρ ∂p i ˙ p i ! = dρ dt . This is the motion of an incompressible fluid, but in phase space instead of coordinate space....
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