Geodynamics - Corner Flow

Geodynamics Lecture 26: Corner Flow Why aren’t subducted slabs curled over or vertical? What is the flow pattern beneath oceanic ridges? Does constant thickness basaltic crust make sense? Figure 26.1 Corner flow models: Batchelor – Introduction to fluid mechanics McKenzie – GJRAS 18, 1-32, 1969 Stevenson & Turner, Nature, 270, 334-336, 1977 Tovish et al, JGR 83, 5892, 1978 McAdoo, JGR, 87, 8684, 1982 Slab model: V=0 q=0 q = 0, vq = 0, vr = -v q q vq = 0, vr = vplate Assume ∞ plates uniform η ∞ Stokes equation 0 = η∇ 2 v + ρ∇U − ∇p % 0 = ∇⋅v Use vorticity ω = ∇×v % Take curl of Stokes equation (heading toward 4th order equation) 0 = η∇ ×∇ 2v + ∇ × ( ρ∇U − ∇p ) Recall ∇ × ( ∇ϕ ) = 0, ∇ ⋅ ( ∇ × A) = 0 % Assume ρ constant Recall ∇ 2 v = ∇ ( ∇ ⋅ v ) − ∇ ×∇ × v % % ∞ r 0 = −η ( ∇ × ∇ × ∇ × v ) = −η ( ∇ × ∇ × ω ) % = η∇ 2ω % Next – define stream function ψ by v = ∇ ×ψ % % % Automatically satisfies ∇ ⋅ v = 0 % Then ω = ∇ ×∇ ×ψ = −∇ 2ψ % % For our purposes, consider 2-D flow only ψ = ( 0, 0,ψ ) % r, θ , z Must solve scalar equation ∇ 2 ( ∇ 2ψ ) = 0 ∇ 4ψ = 0 ← Biharmonic equation [Recall Airy stress function] To help in guessing solution, recall 1 ∂ψ vr = r ∂θ ∂ψ vθ = − ∂r ∂v τ rr = − p + 2η r ∂r ⎛ 1 ∂vθ vr ⎞ + ⎟ τ θθ = − p + 2η ⎜ ⎝ r ∂θ r ⎠ ⎛ 1 ∂vr ∂vθ vθ ⎞ + − ⎟ τ rθ = η ⎜ r ⎠ ⎝ r ∂θ ∂r 1 ∂ ⎛ ∂ψ ⎞ 1 ∂ 2ψ ∂ 2ψ 2 ∇ψ = + ⎜r ⎟+ r ∂r ⎝ ∂r ⎠ r 2 ∂θ 2 ∂z 2 Let’s try solutions ψ = R ( r ) Θ (θ ) We will want solutions with vr independent of r Try R ( r ) = r ψ = rΘ d 4Θ d 2Θ + 2 + Θ = 0, dθ 4 dθ 2 or Θ ''''+ 2Θ ''+ Θ = 0 Plugging in ⇒ Solution: Θ = A sin θ + B cos θ + Cθ sin θ + Dθ cos θ Now – match boundary conditions vr = Θ ' = A cos θ − B sin θ + C ( sin θ + θ cos θ ) + D ( cos θ − θ sin θ ) vθ = −Θ Slab problem – Break into 2 regions Back-arc region θ =0 vθ = 0 = B Fore-arc region θ =0 vθ = 0 vr = 0 = A + D θ = θB vr = −v vθ = 0 = A sin θ B + Cθ B sin θ B + Dθ B cos θ B vr = v θ = θF = A cos θ B + C ( sin θ B + θ B cos θ B ) + D ( cos θ B − θ B sin θ B ) Need to solve 3 equations, 3 unknowns Solutions Back-arc ψ= Fore-arc ψ= rv ⎡⎣(θb − θ ) sin θb sin θ − θbθ sin (θ b − θ ) ⎤⎦ θb 2 − sin 2 θb − rv ⎡⎣(θ f − θ ) sin θ + θ sin (θ f − θ ) ⎤⎦ θ f + sin θ f Figure 26.3 Stream lines for flow within the mantle. Motion is with respect to the plate behind the island arc and is driven by the motion of the other plate and of the sinking slab. Thermal convection outside the slab is neglected, and the lithosphere is 50 km thick. Stresses ⎛ − p τ rθ ⎞ 1 ∂vθ vr ∂vr + = 0 ⇒τ = ⎜ = 0, ⎟ r ∂θ r ∂r ⎝ τ rθ − p ⎠ Back-arc 2vη ⎡⎣θb cos (θ b − θ ) − sin θb cos θ ⎤⎦ τ rθ = θb 2 − sin 2 θb r Fore-arc 2vη cos θ + cos (θ f − θ ) τ rθ = r θ f + sin θ f vθ = 0 vr = v 200 100 50 200 100 200 50 100 50 Figure 26.4 Shear stresses in bars caused by the flow in figure 26.3 Pressure – no direct equation But – equilibrium τ ij , j = 0 ∂p 1 ∂τ rθ + =0 ∂r r ∂θ ∂τ 1 ∂p 2 − + τ rθ + rθ = 0 ∂r r ∂θ r − Back-arc p=− →− 2vη θb sin θb at surface r θb 2 − sin 2 θb Fore-arc p= → 2vη ⎡⎣θb sin (θb − θ ) + sin θb sin θ ⎤⎦ <0 θb 2 − sin 2 θ b r →− 2vη sin 2 θb on slab r θb 2 − sin 2 θ b 2vη sin θ − sin (θ − θ f ) >0 θ f + sin θ f r 2vη sin θ f at surface and on slab r θ f + sin θ f ⇒ excess pressure in fore-arc suction in back-arc – tendency to lift slab (and also modify surface topography) This flow tends to lift slab – Resisted by 1) gravity 2) resistance to squeezing material out of wedge [not well posed for ∞ wedge Æ needs ∞ work] Consider gravity – torque balance θb l ∆ρ h l T flow = ∫ ( Pf − Pb ) ⋅ rdr 0 ⎡ sin θb sin 2 θb ⎤ = 2η vl ⎢ + 2 ⎥ 2 ⎣ (π − θb ) + sin θb θb − sin θb ⎦ l Tgrav = ∫ Δρ ghr cos θ b dr 0 = 1 Δρ ghl 2 cos θb 2 Unstable Flow Torque Gravity Stable o 63 0 o 90 θb Ridge problem v v θ By symmetry, at θ = Let ψ = r Θ π 2 , vθ = 0 , τ rθ = 0 η vr = Θ ' τ rθ = vθ = −Θ p=− r η r ( Θ ''+ Θ ) ( Θ '''+ Θ ') ∇ 4ψ = 0 Θ = A sin θ + B cos θ + Cθ sin θ + Dθ cos θ θ =0 vθ = 0 ⇒ B = 0 θ =π 2 τ rθ = 0 ⇒ D = 0 vr = v = A + D A = v , C = −(2 π )v vr = 2v ⎡⎛ π ⎤ ⎞ ⎜ − θ ⎟ cos θ − sin θ ⎥ ⎢ π ⎣⎝ 2 ⎠ ⎦ vθ = −2v ⎛ π ⎞ ⎜ − θ ⎟ sin θ π ⎝2 ⎠ vθ = 0 ⇒ A + (π 2 ) C = 0 τ rθ = 4η v cos θ πr p=− 4η v sin θ πr Note: at θ = π 2 , vr = − 2 π v (related to gradient in stream function) ⇒ upwelling zone wider than horizontal flow zone ⇒ zone of magma generation wider than depth from which melt segregates