12.005 Lecture Notes 26
Growth and Decay of Boundary Undulations
Growth: Rayleigh - Taylor Instability
• salt domes
• diapirs
• continental delamination
X3
ηu, �u
λ = λ0cos
ηl, �l
2�x1
= λ0coskx
�
Figure 26.1
Figure by MIT OCW.
General problem: topography on an interface
ξ = ξ0 cos kx1
k=
2π
λ
(1) If ρu < ρl topography decays as ξ 0 e−t / τ .
(2) If ρu
> ρl topography grows.
Initially ξ = ξ 0 et / τ
. Eventually many wavelengths interact, problem is no longer simple. Characteristic time τ depends on ∆ρ, ηu , ηl , thickness of layers, …
X1
Before glaciation
Subsidence caused by glaciation
Ice Sheet
Surface after melting of the ice sheet but prior to postglacial rebound
Full rebound
Figure 26.2. Subsidence due to glaciation and the subsequent postglacial
rebound.
Figure by MIT OCW.
•
Weight of ice causes viscous flow in the mantle.
•
After melting of ice, the surface rebounds – “postglacial rebound”.
•
Different regions have different behaviors (e.g., Boston is now sinking).
X3
� = �0coskx
Air
X1
�m
Figure 26.3
Figure by MIT OCW.
Problem: how to reconcile physical boundary conditions with mathematical description?
Decay: Postglacial rebound (1/2 space, uniform η )
k=
� = �0 cos (kx1)
X3
Figure 26.4
Figure by MIT OCW.
•
Assume uniform η
•
Subtract out lithostatic pressure P = p − ρ gx3
•
Assume ρ g uniform
•
Use stream function Ψ
2�
�
X1
v1 = −
∂Ψ
∂x3
v3 =
∂Ψ
∂x1
⇒ ∇4 Ψ = 0
Solution: Ψ = ⎡⎣(A + Bkx3 )exp (−kx3 ) + (C + Dkx3 )exp (kx3 )⎤⎦ ⋅ sin kx1
Boundary conditions:
at x3 = 0 (mathematical, not physical)
σ 33 = ρ gζ
⎛ ∂v1
σ 13 = 0 = η ⎜
⎝ ∂x3
+
∂v3 ⎞
⎟
∂x1 ⎠
at x3 → ∞ , must be bounded
⇒C = D = 0
In order that σ 13 = 0 at x3 = 0 ,
∂2 Ψ ∂2 Ψ
+
=0
∂x32 ∂x12
⇒B=A
−
or Ψ = A(1+ kx3 )exp (−kx3 ) ⋅ sin kx1
Then
v1 = Ak 2 x3 exp (− kx3 ) ⋅ sin kx1
v3 = Ak(1 + kx3 )exp (− kx3 ) ⋅ cos kx1
at x3 = 0 v3 = ζ = Ak cos(kx1 )
Now
σ 33 = − p + 2ηε 33
ε 33 = 0 at x3 = 0
∂p
∂2 vi
+η
+ ρ x1 = 0
To get p, use −
∂xi
∂x j∂x j
for i = 1
⇒ −
∂2 v ∂2 v
∂p
+ η( 21 + 21 ) = 0
∂x1 ∂x3
∂x1
Substitute for v1 and integrating ⇒ p x
3 =0
= 2ηk 2 A cos kx1
But p = −ρ gζ ⇒ A = −
Or ζ 0 = −
ρ gζ 0
2 k 2η
ρ gζ 0
ρ gλζ 0
=−
2kη
4 πη
Or ζ 0 = ζ 0
where τ =
t =0
exp(−
ρ gt
t
) = ζ 0 t =0 exp(− )
2kη
τ
2kη 4 πη
=
ρ g ρ gλ
Solving for η : η =
ρ gλτ
4π
For curves shown,
τ : 5000 yr ⎫
⎬ ⇒ η : 10 21 Pa
λ : 3000 km ⎭
Note: stream function ∼ exp(−kx3 ) = exp(−
Falls off to ∼ 1 / e at x3 :
2 π x3
λ
)
λ
2 π
Senses to fairly great depth ⇒ postglacial rebound doesn’t reveal the details of mantle viscosity structure,
but only the gross structure.