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AST242 LECTURE NOTES PART 5 Contents 1. Waves and ...

AST242 LECTURE NOTES PART 5 Contents 1. Waves and ...

AST242 LECTURE NOTES PART 5 Contents 1. Waves and ...

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20 <strong>AST242</strong> <strong>LECTURE</strong> <strong>NOTES</strong> <strong>PART</strong> 5in nature. We expect a dispersion relation that has limits of ω = c s k for acousticwaves <strong>and</strong> ω = N for internal gravity waves.Our zero-th order or equilibrium solution has density <strong>and</strong> pressure ρ 0 (z) <strong>and</strong> p 0 (z)related via hydrostatic equilibrium(119)dp 0dz = −ρ 0gwhere g is the gravitational acceleration. Mass conservation to first order(120)∂ρ 1∂t + ρ 0∇ · u 1 + ρ 1 ∇ · u 0 + u 0 · ∇ρ 1 + u 1 · ∇ρ 0 = 0We drop terms with u 0 as the equilibrium system is static <strong>and</strong> u 0 = 0. The gradient∇ρ 0 only contains a z component.(121)∂ρ 1∂t + u ∂ρ 0z1∂z + ρ 0∇ · u 1 = 0Euler’s equation for momentum conservation, again to first order(122)∂u 1∂t= ρ 1 dp 0ρ 2 0 dz ẑ − 1 ∇p 1ρ 0<strong>and</strong> we have ignored the variation of the gravitational acceleration. We have usedthe fact that the unperturbed pressure gradient is in the z direction.We have a system that in which two directions x, y differ from the other, z, wheregradients in the ambient pressure <strong>and</strong> density exist. Let our velocity perturbation(123)u 1 = (u, v, w).Let our perturbations be functions of z that are wavelike in the other two dimensions,(124) u, v, w, ρ 1 , p 1 ∝ exp i(ωt − k x x − k y y)with(125) k ⊥ ≡√k 2 x + k 2 y.The equation for conservation of mass (equation 121) becomes(126) iωρ 1 + w dρ [0dz + ρ 0 −i(k x u + k y v) + dw ]= 0dz

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