Course 12.520, Geodynamics
Prof. Brad Hager
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
∞
r
q
vq = 0, vr = vplate
Assume ∞ plates
uniform η
∞
Stokes equation
Figure by MIT OCW.
2
0 = η∇ v + ρ∇U − ∇p
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0 = ∇⋅v
Use vorticity
ω = ∇×v
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Take curl of Stokes equation (heading toward 4th order equation)
0 = η∇ ×∇ 2v + ∇ × ( ρ∇U − ∇p )
Recall ∇ × ( ∇ϕ ) = 0, ∇ ⋅ ( ∇ × A) = 0
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Assume ρ constant
Recall ∇ 2 v = ∇ ( ∇ ⋅ v ) − ∇ ×∇ × v
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0 = −η ( ∇ × ∇ × ∇ × v )
= −η ( ∇ × ∇ × ω )
%
= η∇ 2ω
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Next – define stream function ψ by v = ∇ ×ψ
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Automatically satisfies ∇ ⋅ v = 0
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Then ω = ∇ ×∇ ×ψ = −∇ 2ψ
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For our purposes, consider 2-D flow only
ψ = ( 0, 0,ψ )
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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
Fore-arc region
θ =0
vθ = 0
vθ = 0 = B
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
Figure by MIT OCW.
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
Figure by MIT OCW.
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)
Figure by MIT OCW.
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
Figure by MIT OCW.
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
Figure by MIT OCW.
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