7. Post Glacial Rebound. Ge 163 4/16/14-

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1 7. Post Glacial Rebound Ge 163 4/16/14-

2 Outline Overview Order of magnitude estimate of mantle viscosity Essentials of fluid mechanics Viscosity Stokes Flow Biharmonic equation Half-space model Channel Flow Deep Flow versus Channel Flow Relaxation spectra GPS constraint GRACE Constraints Latest models with variable viscosity

3 Fowler

4 Deglaciation Last deglaciation started around 18,000 years ago Addition to the oceans of approximately kg of water which effected a mean global rise of sea level of about 80 meters. Source of this meltwater: Laurentide complex of North America & Fennoscandian complex of N.W. Europe From time series of 0 16 and 0 18 obtained from deep sea sediment cores, major ice sheets existed on the surface for the 10 5 years prior to their disintegration (when ice sheets where slowly, if not monotonically, approaching their maximum extents). Obtain relative sea-level histories (RSL s) from C 14 dating of marine shells and other materials from relic beaches.

5 Geophysical observations Correlation of Δg FA with topography Correlation of Δg FA with geomorphologically inferred position of the ice sheets Direct measurements of surface displacements with GPS GRACE (time-dependent gravity)

6 Retreat of the Laurentide Ice Sheet Reproduced in Cathles [1975]

7 Retreat of the Ice Sheet in Fennoscandia Reproduced in Cathles [1975]

8 Ice History -- This is a Model (ICE-1) 18 Kyr 12 Kyr 8 Kyr Peltier [1981]

9 Porsangerfjord, northern Norway. Ridges are 1-4 m high with a 4 to 5 meter spacing. Prominent ridges are Holocene 6,000 to 7,000 yr BP. From cover of GSA Memoir 180.

10 Relative Sea Level (RSL) Curve

11 Free Air Gravity Over Hudson Bay

12 Gravity anomalies from 363 days of GRACE data (GGM02S)

13 Using Oxygen Isotopes as a proxy for Ice volume From a Peltier paper

14 Order of Magnitude Estimate of Newtonian Viscosity of the mantle L h h σ = 2η ε σ = ρgh ε h L why? η = σ 2 ε ~ ρghl 2 h

15 η ~ ρghl 2 h Remaining uplift will be approximated with the Bouguer Formula Δg = 2πρGh h Δg FA 2πρG η ~ gδg FA L 4πG h

16 η ~ gδg FA L 4πG h Laurentide Fennoscandia Dimensions Length X width Mean Free-air gravity anomaly Present rate of uplift η [Dimensional scaling] 4 x x 10 6 m 2.25 x x 10 6 m -35 mgal -16 mgal 1.7 cm/yr 0.9 cm/yr 2 x Pa-s 1 x Pa-s Using Rebound Parameters in Walcott[1973]

17 Viscosity We will need a constitutive relation, which for fluid mechanics relates strain rates to stress. Bulk viscosity shear viscosity σ ij = pδ ij + λ θδ ij + 2η ε ij ε ij = 1 2 u i x j + u j x i θ= ε kk (dilitation) dilitation is the strain rate tensor; u is velocity For an incompressible fluid (i.e. u = 0), ε kk = 0 σ ij = pδ ij + 2η ε ij

18 σ ij = pδ ij + 2η ε ij We may think of η (the shear or 'dynamic viscosity') as a constant, but is it?

19 We start with Stoke s formula Which for an incompressible fluid (i.e. u = 0) is 0 = p + η 2 u 0 = p + η 2 u + ρgŷ with buoyancy where η is the dynamic viscosity u is the fluid velocity p is the pressure and this is only valid in the limit of viscosuly dominated fluids. That is, the inertial term that appears in the Navier-Stokes equation does not appear. Is this valid for the mantle?

20 Navier-Stokes NS: p + η 2 u + ρgŷ = ρ Du Dt Viscous forces are: η 2 u or η u L 2 Inertial forces are proportional to ρ Du Dt ρ( u )u ρ u2 L Ratio of these forces inertial viscous ~ ρu2 L L 2 ηu = ρul η = ul υ Re = Reynolds Number A characteristic length scale (η /ρ = ν, the kinematic viscosity) (ρ / η = υ, the kinematic viscosity)

21 Re = ul υ L = 1000 km u = 5 cm/yr υ = m 2 / s (notice the units!) Re = m s m s effectively 0. ( 10 6 m) = 10 22

22 For 2-D, incompressible problems (with no lateral variations in viscosity), it is helpful to 'transform' the Stokes equation into another form (Stream Function form). 0 = p + η 2 u + ρgŷ (Stokes equation) where ρ, is a constant ( ρ / x = 0). Put the pressure, p, into a form so that there is a dynamic part, P, and a hydrostatic part P = p ρgy Substitute this into the Stokes equation and write out in component form 0 = P x + η 2 u x 0 = P y + η 2 u y

23 We can define a quantity called the Stream Function, ψ u x = ψ y u y = ψ x This definition automatically satisfies the 2-D, incompressible continuity equation u = 0

24 Streamline Continuous line drawn through the fluid so that it has the direction of the velocity vector at that point; the streamline is everywhere tangent to the velocity field. In steady flow, the streamline inclination is fixed at every point and is therefore fixed in space; a streamline is a particle path. However, in unsteady flow, the streamlines will shift in space from instant to instant.

25 Using the Stokes equation in component form, 0 = P x + η 2 u x, 0 = P y + η 2 u y Using the definition of the stream function u x = ψ y 0 = P x + η - 0 = P y + η 3 ψ x 3 u y = ψ x 3 ψ x 2 y + 3 ψ y ψ y 2 x then / y & ( 1) then / x 0 = η 4 ψ x + 4 ψ 4 x 2 y + 4 ψ 2 2 y 4 0 = 4 ψ This is called the Biharmonic equation

26 If we didn't assume ρ / x = 0, then η 4 ψ = g ρ x The flow is generated by lateral variations in density. The equation reduces to a balance between a gravitational buoyancy force and viscous resistance.

27 Response of a semi-infinite half-space to a periodic load (all details are given in section 6-10 in Turcotte and Schubert w ρ 2 ρ 1 g Δρ = ρ 1 ρ 2 Solve 4 ψ = 0, the biharmonic equation y w = w o cos 2π x initial condition. λ w = vertical displacement ("topography").

28 Want w as a function of time, w(t) This appears odd. We previously eliminated the element of time when we derived the Stoke s equation (upon which the the biharmonic equation is based). Will equate vertical velocity, u y, at y=0 to dw/dt Other constraints: Stream function, ψ, must remain finite with depth w<<λ u x =0 at y=w (or approximately y=0 because w<<λ) noslip condition on upper boundary, geophysically the presence of lithosphere. Equate hydrostatic head (of deformed surface) to normal stress (σ yy ) at upper boundary.

29 To solve 4 ψ = 0 use separation of variables ψ = X(x)Y (y) Guess ψ = A'sin 2π x λ 2π x + B'cos λ Y (y)...using the fact that the solution must be bounded as y and the no-slip boundary condition: u y = A 2π λ cos 2π λ x e 2π y/λ 1+ 2π λ y

30 To evaluate A Equate hydrostatic head, ρgw, to the total normal stress at the upper boundary of the fluid half-space. From the stress tensor: σ yy = p τ yy = p 2η u y y Set σ yy = ρgw (notice that w appears for the first time). Find p by setting up a horizontal balance, e.g. 0 = p + +ηη 2 u 2 x u x Then integrate one p = 2ηA 2π 2 λ cos 2π λ x ρgw(x) = p y=0 w(x) y=0 = 1 ρg 2ηA 2π 2 λ cos 2π λ x put A in terms of w

31 However dw dt is the vertical velocity on the top surface dw dt y=w = u y (y = w) Again, assume that w << λ Using our previous result for u y u y (y = 0) = A 2π λ cos 2π λ x = dw dw dt = λρg w nice ode 4πη dw w = λρg 4πη dt t = 0 w(0) = w o w = w o exp λρg 4πη t dt y=0

32 τ R = 4πη ρgλ = 4πυ gλ where υ kinematic viscosity w = w o e t /τ R From Turcotte & Schubert: t o = 10,000 years ago w o = 300 m, 30 me still to occur best fit τ R = 4400 years Angerman River data λ = 3000 km ρ = 3300 kg m -3 g = 10 m s -2 η = Pa-s

33 Why is there an inverse relation between relaxation time and wavelength? τ R = 4πη ρgλ τ R h h h h = 4πη ρgλ ρgh = 4πη h λ stress=(viscosity) (strain-rate) Practical: Have you ever poured honey back into a bowl of Honey? First there is a wide ridge, but after time there is a Sharp line that persists for a long time.

34 Cathles [1975]

35 Alternative model of Post-glacial rebound (flow confined to a channel). r w R w w Lithosphere on top H η Asthenosphere u x It can be shown for a cylindrical problem that τ R 2ηR2 ρgh 3

36 Cathles [1975]

37 Post-glacial rebound using a channel flow model η τ RρgH 3 2R 2 Fennoscandian radius ~ 500 km Angermann River τ R = 4,400 years η = Pa-s (not much different from our other estimates). Flow confined to Upper mantle

38 Cathles [1975]

39 RSLs at different Distances from the Ice Cap Peltier

40 Unloading of peripheral glacial lakes Circa 14,000 years ago

41 Crittenden [1963]: Shorelines [of Lake Bonneville], originally level are domed upward about 64 meters in the center of the basin, showing that during both loading and subsequent unloading, the earth reached a high degree of isostatic adjustment. To attain this degree of compensation the isostatic anomaly created during loading or unloading must be reduced to 1/e of its initial value in about 4,000 years Now this time-scale is about the same as Fennoscandia, but Bonneville has ~1/10 the diameter Channel Flow τ R = 2ηR2 Note the strong trade-off between H and η 3 ρgh η B = R F η F R B 2 H B = 70 km H F = 700 km H B H F 3 = 100 H B H F η F ~ Pa-s η B ~ Pa-s (Cathles: 4x10 19 Pa-s) 3

42 Cathles [1975]

43 Except for a low viscosity channel beneath the lithosphere does the mantle have a uniform, Pas-s, viscosity as Cathles [1975] argues?

44 Deep flow: τ R = 4πυ gλ Channel Flow: τ R = 2υR2 gh 3 channel Half-space Relaxation time λ wavenumber

45 Models like this can give the appropriate relaxation spectrum viscosity Weak upper mantle depth

46 Can the relaxation spectra be observed? (I.e. relaxation time versus wavelength) Several attempts over the years. Most recent, is to look at the localization of gravity (spectrogram of the gravity)

47 Spectogram of gravity field: Hudson s Bay is A unique feature of the gravity field. Simons & Hager [1997]

48 1. Gravity is highly correlated with former ice caps 2. Use a model for the time load (such as ICE-1) 3. Isostatically compensate ice-load (mantle viscosity is not needed) 4. Remove ice load instantly and calculate gravity 5. Correlate the model gravity with observed gravity versus wavelength, F l Simons and Hager [1997]

49 F l Simons and Hager [1997]

50 Simons and Hager [1997]

51 3-D Velocities from present Day GPS Milne et al. [2001]

52 Radial Inversion of GPS Data for Mantle Viscosity Horizontal Both Milne et al. [2001]

53 Secular change in gravity from GRACE (In units of µgal/yr) ) Homogeneous mantle models fit GRACE & RSL have viscosities of 1.4 to 2.3 X Pa s Paulson et al [2007]

54 Viscosity Structure Used in Spherical FEM model of post-glacial rebound Paulson, et al. [2005]

55 Difference in RSL between radial and FEM model with full (lateral and radial) variations in viscosity Paulson et al. [2005]

Gravitational constraints

Gravitational constraints Reading: Fowler p172 187 Gravity anomalies Free-air anomaly: g F = g g( λ ) + δg obs F Corrected for expected variations due to the spheroid elevation above the spheroid Bouguer