Lecture 3/4: Dynamics of the Earth’s core Celine´ Guervilly School of Mathematics, Statistics and Physics, Newcastle University, UK Navier-Stokes equation @u 1 + u · ru + 2Ω × u = - rP + νr2u + F @t ρ r · u = 0 F is a forcing term (e.g. buoyancy force, Lorentz force). Ω = Ωez with Ω the rotation rate. ρ is the density, ν the viscosity. Like in the oceans and atmosphere, the rotation is very important for the dynamics of the Earth’s core. Golf Stream in the Atlantic Ocean (NASA) Taking the curl, @r2u = -2Ω · r! @t Differentiating wrt time, @2r2u @! = -2Ω · r @t2 @t we obtain the inertial wave equation @2r2u = -4(Ω · r)2u @t2 We seek solutions in the modal form u = Refuˆ ei(k·x-!t)g, where uˆ is the (complex) amplitude, k is the (real) wavevector and ! is the (possibly complex) frequency. Inertial waves We linearise the invisid NS equation about a background state of uniform rotation, i.e. ub = 0. The equation for the vorticity, ! = r × u, is @! = 2Ω · ru @t Differentiating wrt time, @2r2u @! = -2Ω · r @t2 @t we obtain the inertial wave equation @2r2u = -4(Ω · r)2u @t2 We seek solutions in the modal form u = Refuˆ ei(k·x-!t)g, where uˆ is the (complex) amplitude, k is the (real) wavevector and ! is the (possibly complex) frequency. Inertial waves We linearise the invisid NS equation about a background state of uniform rotation, i.e. ub = 0. The equation for the vorticity, ! = r × u, is @! = 2Ω · ru @t Taking the curl, @r2u = -2Ω · r! @t we obtain the inertial wave equation @2r2u = -4(Ω · r)2u @t2 We seek solutions in the modal form u = Refuˆ ei(k·x-!t)g, where uˆ is the (complex) amplitude, k is the (real) wavevector and ! is the (possibly complex) frequency. Inertial waves We linearise the invisid NS equation about a background state of uniform rotation, i.e. ub = 0. The equation for the vorticity, ! = r × u, is @! = 2Ω · ru @t Taking the curl, @r2u = -2Ω · r! @t Differentiating wrt time, @2r2u @! = -2Ω · r @t2 @t Inertial waves We linearise the invisid NS equation about a background state of uniform rotation, i.e. ub = 0. The equation for the vorticity, ! = r × u, is @! = 2Ω · ru @t Taking the curl, @r2u = -2Ω · r! @t Differentiating wrt time, @2r2u @! = -2Ω · r @t2 @t we obtain the inertial wave equation @2r2u = -4(Ω · r)2u @t2 We seek solutions in the modal form u = Refuˆ ei(k·x-!t)g, where uˆ is the (complex) amplitude, k is the (real) wavevector and ! is the (possibly complex) frequency. For an initial disturbance with a given wavevector k, two modes propagate in opposite directions, each with the same frequency. The frequency of each mode is real and so these modes are waves (not instabilities). The frequency depends only on the angle between k and Ω, and not on the magnitude of k. Modes with k parallel to Ω have the fastest frequency, j!j = 2Ω. r · u = k · u = 0: transverse waves. Perturbations with k ? Ω experience no restoring effect, and so ! = 0. (But these modes do have a Coriolis force, but it is exactly balanced by the pressure force). The group velocity of the inertial waves is @! 2k × (Ω × k) c = = ± g @k jkj3 and so, cg · k = 0. Inertial waves The dispersion relation for inertial waves is: 2Ω · k ! = ± jkj The group velocity of the inertial waves is @! 2k × (Ω × k) c = = ± g @k jkj3 and so, cg · k = 0. Inertial waves The dispersion relation for inertial waves is: 2Ω · k ! = ± jkj For an initial disturbance with a given wavevector k, two modes propagate in opposite directions, each with the same frequency. The frequency of each mode is real and so these modes are waves (not instabilities). The frequency depends only on the angle between k and Ω, and not on the magnitude of k. Modes with k parallel to Ω have the fastest frequency, j!j = 2Ω. r · u = k · u = 0: transverse waves. Perturbations with k ? Ω experience no restoring effect, and so ! = 0. (But these modes do have a Coriolis force, but it is exactly balanced by the pressure force). Inertial waves The dispersion relation for inertial waves is: 2Ω · k ! = ± jkj For an initial disturbance with a given wavevector k, two modes propagate in opposite directions, each with the same frequency. The frequency of each mode is real and so these modes are waves (not instabilities). The frequency depends only on the angle between k and Ω, and not on the magnitude of k. Modes with k parallel to Ω have the fastest frequency, j!j = 2Ω. r · u = k · u = 0: transverse waves. Perturbations with k ? Ω experience no restoring effect, and so ! = 0. (But these modes do have a Coriolis force, but it is exactly balanced by the pressure force). The group velocity of the inertial waves is @! 2k × (Ω × k) c = = ± g @k jkj3 and so, cg · k = 0. In the Earth’s core, Ek ≈ 10-15 and Ro ≈ 10-6 Du Ro + 2e × u = -rP + Ekr2u Dt z where D @ = + u · r Dt @t Ekman number: Ek = ν/(Ωd2) Rossby number: Ro = U=(Ωd) with U a typical velocity scale, d a typical lengthscale Long-timescale dynamics Consider timescales that are long compared to the rotation period. The generation of waves usually results in a loss of energy, because waves carry energy away with them. On long timescales, flows that generate a lot of inertial waves will be strongly suppressed, and the dynamics is thus rotationally constrained. In the Earth’s core, Ek ≈ 10-15 and Ro ≈ 10-6 Long-timescale dynamics Consider timescales that are long compared to the rotation period. The generation of waves usually results in a loss of energy, because waves carry energy away with them. On long timescales, flows that generate a lot of inertial waves will be strongly suppressed, and the dynamics is thus rotationally constrained. Du Ro + 2e × u = -rP + Ekr2u Dt z where D @ = + u · r Dt @t Ekman number: Ek = ν/(Ωd2) Rossby number: Ro = U=(Ωd) with U a typical velocity scale, d a typical lengthscale Long-timescale dynamics Consider timescales that are long compared to the rotation period. The generation of waves usually results in a loss of energy, because waves carry energy away with them. On long timescales, flows that generate a lot of inertial waves will be strongly suppressed, and the dynamics is thus rotationally constrained. Du Ro + 2e × u = -rP + Ekr2u Dt z where D @ = + u · r Dt @t Ekman number: Ek = ν/(Ωd2) Rossby number: Ro = U=(Ωd) with U a typical velocity scale, d a typical lengthscale In the Earth’s core, Ek ≈ 10-15 and Ro ≈ 10-6 Taking the curl, we obtain the Proudman-Taylor theorem, @u (e · r)u = ' 0 z @z The flow must be nearly invariant along the rotation axis. For small (finite) Ek and Ro, the motions are mostly in the form of 2D columnar rolls aligned with the rotation axis. In the core, buoyancy forces and the magnetic field can induce some z-dependence. Long-timescale dynamics As a first approximation, we neglect inertia and viscosity, and so we have the geostrophic balance, 2ez × u ' -rP For small (finite) Ek and Ro, the motions are mostly in the form of 2D columnar rolls aligned with the rotation axis. In the core, buoyancy forces and the magnetic field can induce some z-dependence. Long-timescale dynamics As a first approximation, we neglect inertia and viscosity, and so we have the geostrophic balance, 2ez × u ' -rP Taking the curl, we obtain the Proudman-Taylor theorem, @u (e · r)u = ' 0 z @z The flow must be nearly invariant along the rotation axis. Long-timescale dynamics As a first approximation, we neglect inertia and viscosity, and so we have the geostrophic balance, 2ez × u ' -rP Taking the curl, we obtain the Proudman-Taylor theorem, @u (e · r)u = ' 0 z @z The flow must be nearly invariant along the rotation axis. For small (finite) Ek and Ro, the motions are mostly in the form of 2D columnar rolls aligned with the rotation axis. In the core, buoyancy forces and the magnetic field can induce some z-dependence. Rotating Rayleigh-Benard´ convection: Linear stability analysis Chandrasekhar (1961) Boussinesq approximation. Gravitational acceleration g = -gez. Rotation about the vertical axis Ω = Ωez. cold Fixed temperature boundary conditions: T = T0 at z z = 0 (lower plate), T = T0 - ∆T at z = d (top d plate). ρ is the constant density at temperature T . y x 0 0 α is the coefficient of thermal expansion (constant). hot ν and κ are the kinematic viscosity and thermal diffusivity (all constant). Stress-free and impenetrable boundary conditions. Horizontally periodic boundary conditions. 2 In the basic state, ub = 0 and r Tb = 0 ! Tb = T0 - ∆Tz=d. We will now perturb this basic state with small perturbations. Rotating Rayleigh-Benard´ convection: Linear stability analysis 0 The linearised equations for the velocity and temperature fluctuations (T = T - Tb) are 0 @u 1 2 ρ + 2Ωez × u = - rP + νr u - gez, @t ρ0 ρ0 r · u = 0, @T 0 ∆T - u = κr2T 0. @t z d 0 0 with the linear equation of state for the density fluctuation ρ = -ρ0αT . In dimensionless form (scaling lengths with d, time with d2/ν and temperature with ∆T): @u 2 Ra + e × u = -rP + r2u + T 0e , @t Ek z Pr z r · u = 0, @T 0 1 - u = r2T 0. @t z Pr ν αg∆Td3 ν Ek = , Ra = , Pr = Ωd2 νκ κ Rotating Rayleigh-Benard´ convection: Linear stability analysis The z-component of the curl of the momentum equation is @! 2 @u z = z + r2! , (1) @t Ek @z z where !z is the vertical component of the vorticity.