Isostasy in Geology andBasin Analysis

INCOMPLETE DRAFT

This exercise is drawn from Angevine, Heller and Paola (1990), with inspiration and essential planning by R. Dorsey. A. Martin-Barajas generously provided material used in this exercise.

Archimedes Principle: When a body is immersed in a fluid, the fluid exerts an upward force on the body that is equal to the weight of the fluid that is displaced by the body.”

This rule applies to all mountain belts and basins under conditions of local (Airy) isostatic compensation: the lithosphere has no lateral strength, and thus each lithospheric column is independent of neighboring columns (e.g. rift basins).

To work isostasy problems, we assume that the lithosphere (crust + upper mantle) is “floating” in the fluid asthenosphere. A simple, nongeologic example looks like this -

Solid

rs

1 2

h2h1Fluid (rf)

depth of equal compensation

Because fluid has no shear strength (yield stress =0),it cannot maintain lateral pressure differences. It will flow to eliminate the pressure gradient.

Solid

rs

1 2

h2h1Fluid (rf)

depth of equal compensation

Solid

rs

1 2

h2h1Fluid (rf)

depth of equal compensation

To calculate equilibrium forces, set forces of two columns equal to each other: F1 = F2 (f=ma)

m1xa = m2xa

m1 = m2

(gravitational acceleration

cancels out)

Because m=rxv (density x volume),convert to m=rh, and: rfh1=rsh2

Solid

rs

1 2

h2h1Fluid (rf)

depth of equal compensation

This equation correctly describes equilibrium isostatic balance in the diagram.

Onward to geology –

EXAMPLE 1 Estimate thickness of lithosphere:In this example, we’ve measured the depth to the moho (hc) using seismic refraction.Elevation (e) is known, and standard densities for thecrust, mantle, and asthenosphere are used:rc=2800 kg/m3

rm=3400 kg/m3

ra=3300 kg/m3

rc

elevation=3km

rm

hc=35km

hm=?

asthenosphere (rc)

Z=?

How deep to the base of the lithosphere?

Solve for Z:ra(Z) = rc(hc+e) + rm(Z-hc)

ra(Z) - rmZ = rc(hc+e) - rmhc

rc(hc+e) - rmhc

(ra-rm)

rc

elevation=3km

rm

hc=35km

hm=?

asthenosphere (ra)

Z=?

Z=

How deep to the base of the lithosphere?

Solve for Z:ra(Z) = rc(hc+e) + rm(Z-hc)

ra(Z) - rmZ = rm(Z-hc) - rmhc

rc(hc+e) - rmhc

(ra-rm)

2800(35+3) – 3400(35)(3300-

3400) -12,600

-100

rc

elevation=3km

rm

hc=35km

hm=?

asthenosphere (rc)

Z=?

Z=

=

= Z = 126 km

EXAMPLE 2 What is the effect of filling a basin with sediment?

Consider a basin 1km deep that is filled only with water. How much sediment would it take to fill the basin up to sea level?

rC=2800

rm=3400

ra= 3300

water

crust

mantlelithosphere

rs= 2300

sediment

crust

mantlelithosphere

depth of equal compensation

Let rw= 1000 kg/m3

Let rs= 2300 kg/m3 ho= 1km hc

hm

hs=?

rC=2800

rm=3400

1 2

rC=2800

rm=3400

ra= 3300

water

crust

mantlelithosphere

rs= 2300

sediment

crust

mantlelithosphere

depth of equal compensation

ho= 1km hc

hm

hs=?

rC=2800

rm=3400

Remember, force balance must be calculated for entire column down to depth of compensation (=depth below which there is no density difference between columns). Also, thickness of crust and mantle lithosphere does not change, so they cancel out on both sides of the equation.

1 2

rC=2800

rm=3400

ra= 3300

water

crust

mantlelithosphere

rs= 2300

sediment

crust

mantlelithosphere

depth of equal compensation

ho= 1km hc

hm

(hs-ho)

hs=?

rC=2800

rm=3400

rwho + rchc + rmhm + ra(hs-ho) = rshs + rchc + rmhmrwho + ra(hs-ho) = rshsrwho + rahs – raho = rshshs(ra- rs) = raho – rwhohs = ho(ra- rw)

(ra- rs)

rC=2800

rm=3400

ra= 3300

water

crust

mantlelithosphere

rs= 2300

sediment

crust

mantlelithosphere

depth of equal compensation

ho= 1km hc

hm

(hs-ho)

hs=?

rC=2800

rm=3400

hs = ho(ra- rw) (ra- rs)

= 1.0 (3.3-1.0)

(3.3-2.3)= 2.3 km

rC=2800

rm=3400

ra= 3300

water

crust

mantlelithosphere

rs= 2300

sediment

crust

mantlelithosphere

depth of equal compensation

ho= 1km hc

hm

(hs-ho)

hs=?

rC=2800

rm=3400

“rule of thumb”: the thickness of sediment needed to fill

a basin is ~ 2.3 times the depth of water that the sediment replaces

EXAMPLE 2B Sediment filling – Alarcon Basin example

Determine the maximum water depth in the Alarcon Basin from your profile or spreadsheet.

Calculate how much sediment would be needed to fill the Alarcon Basin up to sea level.

rC=2800

rm=3400

ra= 3300

water

crust

mantlelithosphere

rs= 2300

sediment

crust

mantlelithosphere

depth of equal compensation

ho= 1km hc

hm

(hs-ho)

hs=?

rC=2800

rm=3400

hs = ho(ra- rw) (ra- rs)

= 3 (3.3-1.0)

(3.3-2.3)= 6.9 km

crust

mantlelitho-

sphere

crust

mantlelitho-

sphere

EXAMPLE 3 How does crustal thinning effect the depth of sedimentary basins?

1 2

hc1=30km hc2

hm1=90km

hm2

ha

Newly-formed basin

Thin crust and mantle lithosphere to half of original.How deep a basin forms in response to thinning?

Solve for Z.

Note: ha = hc1 + hm1 - hc2 - hm2 - Z

Z

crust

mantlelitho-

sphere

crust

mantlelitho-

sphere

1 2

hc1=30km hc2

hm1=90km

hm2

ha

Note: ha = hc1 + hm1 - hc2 - hm2 - Z

Z

rc(hc1) + rm(hm1) = rw(Z) + rc(hc2) + rm(hm2) + ra(ha) 30rc + 90rm = rw(Z) + 15rc + 45rm + ra(120-15-45-Z)30rc + 90rm = Zrw + 15rc + 45rm + 60ra- Zra

Z(ra-rw) = 60ra-45rm-15rc

crust

mantlelitho-

sphere

crust

mantlelitho-

sphere

1 2

hc1=30km hc2

hm1=90km

hm2

ha

Z

Z(ra-rw) = 60ra-45rm-15rc

If the new basin is filled by water,what is its depth (Z)? -

A: Fill with water(rw = 1.01 g/cm2)

Z=60ra-45rm-15rc

(ra – rw)

crust

mantlelitho-

sphere

crust

mantlelitho-

sphere

1 2

hc1=30km hc2

hm1=90km

hm2

ha

Z

Z(ra-rw) = 60ra-45rm-15rc

If the new basin is filled by water -

A: Fill with water(rw = 1.01 g/cm2)

Z=60ra-45rm-15rc

(ra – rw)

60(3.3)- 45(3.4)-15(2.8) (3.3-1.01)=

=3.002.29

= 1.31 kmfor water

crust

mantlelitho-

sphere

crust

mantlelitho-

sphere

1 2

hc1=30km hc2

hm1=90km

hm2

ha

Z

Z(ra-rw) = 60ra-45rm-15rc

If the new basin is filled by sediment,what is its depth? -

B: Fill with sediment(rw = 2.3 g/cm2)

Z=60ra-45rm-15rc

(ra – rw)

crust

mantlelitho-

sphere

crust

mantlelitho-

sphere

1 2

hc1=30km hc2

hm1=90km

hm2

ha

Z

Z(ra-rw) = 60ra-45rm-15rc

Basin is formed:A: fill with waterB: fill with sediment

B: Fill with sediment(rs = 2.3 g/cm2)

Z=60ra-45rm-15rc

(ra – rs)

60(3.3)- 45(3.4)-15(2.8) (3.3-2.3)=

=3.001.0

= 3.0 kmfor sediment

EXAMPLE 3B Use Upper Delfin Basin sediment fill and crustal thickness to estimate amount of thinning of mantle lithosphere.

crust

mantlelitho-

sphere

crust

mantlelitho-

sphere

1 2

hc1 hc2 =

hm1 hm2=

Sedimentthickness

Mantle lithosphereremoved

Delfin Basin

Tiburon Basin

Location map, Northern Gulf of California

Low-angle detachment fault

intrusions

Interpreted seismic line, Delfin Basin and Tiburon Basin

Thickness of sediments and crust are interpreted from seismic lines

Minimum thicknesses, Delfin Basin Sediments 4 km

Metasediments 4 kmIntrusions 0.4 kmOther 1 km

from Dorsey, 2010, Table 1

EXAMPLE 3B Use Upper Delfin Basin sediment fill and crustal thickness to estimate amount of thinning of mantle lithosphere.

crust

mantlelitho-

sphere

crust

mantlelitho-

sphere

1 2

hc1=35km hc2=

~10km?

hm1=?km

hm2=?km

Sedimentthickness

Mantle lithosphereremoved

hs=4km

Notes, 9-20-13Example 3b - another G of CA example – how much mantle lithosphere has been removed? [“inversion” of the question. First solve for amt of sed fill, 2nd measure crust and sed. Thickness and solve for amt of mantle lithosphere removed]

Dorsey 2010 Geology paper – table of approximate sediment & metasediment fill in basin (Salton Trough and other northern G of CA basins)

Need 1) initial (pre-rifting crustal thickness – Martin-Barajas paper in press, or Couch et al. ‘91??) [~35 km]2) Thinned crust (either in Couch, or in an existing Martin-Barajas lead or co-author) [~ 14ish, of which 7ish is sed.s + metased.s]