Isostasy in Geology and Basin Analysis 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 ss 1 2 h2h2 h1h1 Fluid (  f ) 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 ss 1 2 h2h2 h1h1 Fluid (  f ) depth of equal compensation

Solid ss 1 2 h2h2 h1h1 Fluid (  f ) depth of equal compensation To calculate equilibrium forces, set forces of two columns equal to each other: F 1 = F 2 (f=ma) m 1 x a = m 2 x a m 1 = m 2 (gravitational acceleration cancels out)

Because m=  x v (density x volume), convert to m=  h, and:  f h 1 =  s h 2 Solid ss 1 2 h2h2 h1h1 Fluid (  f ) depth of equal compensation This equation correctly describes equilibrium isostatic balance in the diagram.

Onward to geology – Courtesy Sue Cashman

EXAMPLE 1Estimate 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 the crust, mantle, and asthenosphere are used:  c =2800 kg/m 3  m =3400 kg/m 3  a =3300 kg/m 3 cc cc elevation=3km mm mm hc=35km hm=? asthenosphere (  c ) Z=?

How deep to the base of the lithosphere? Solve for Z:  a (Z) =  c (hc+e) +  m (Z-hc)  a (Z) -  m Z =  c (hc+e) -  m hc  c (hc+e) -  m hc (  a -  m ) cc cc elevation=3km mm mm hc=35km hm=? asthenosphere (  a ) Z=? Z=Z=

How deep to the base of the lithosphere? Solve for Z:  a (Z) =  c (hc+e) +  m (Z-hc)  a (Z) -  m Z =  m (Z-hc) -  m hc  c (hc+e) -  m hc (  a -  m ) 2800(35+3) – 3400(35) ( ) -12, cc cc elevation=3km mm mm hc=35km hm=? asthenosphere (  c ) Z=? Z=Z= = = Z = 126 km

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

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?  C = 2800  C = 2800  m = 3400  m = 3400  a = 3300 water crust mantle lithosphere  s = 2200 sediment crust mantle lithosphere depth of equal compensation Let  w = 1000 kg/m 3 Let  s = 2200 kg/m 3 ho= 1km hc hm hs=?  C = 2800  C = 2800  m = 3400  m =

 C = 2800  C = 2800  m = 3400  m = 3400  a = 3300 water crust mantle lithosphere  s = 2200 sediment crust mantle lithosphere depth of equal compensation ho= 1km hc hm hs=?  C = 2800  C = 2800  m = 3400  m = 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

 C = 2800  C = 2800  m = 3400  m = 3400  a = 3300 water crust mantle lithosphere  s = 2200 sediment crust mantle lithosphere depth of equal compensation ho= 1km hc hm (hs-ho) hs=?  C = 2800  C = 2800  m = 3400  m = 3400  w ho +  c hc +  m hm +  a (hs-ho) =  s hs +  c hc +  m hm  w ho +  a (hs-ho) =  s hs  w ho +  a hs –  a ho =  s hs hs(  a -  s ) =  a ho –  w ho hs = ho(  a -  w )  a -  s )

 C = 2800  C = 2800  m = 3400  m = 3400  a = 3300 water crust mantle lithosphere  s = 2200 sediment crust mantle lithosphere depth of equal compensation ho= 1km hc hm (hs-ho) hs=?  C = 2800  C = 2800  m = 3400  m = 3400 hs = ho(  a -  w )  a -  s ) = 1.0 ( ) ( ) = 2.1 km

 C = 2800  C = 2800  m = 3400  m = 3400  a = 3300 water crust mantle lithosphere  s = 2200 sediment crust mantle lithosphere depth of equal compensation ho= 1km hc hm (hs-ho) hs=?  C = 2800  C = 2800  m = 3400  m = 3400 How much sediment would it take to fill a water- filled basin up to sea level? “rule of thumb”: the thickness of sediment needed to fill a basin is ~ 2.1 times the depth of water that the sediment replaces

EXAMPLE 2BSediment 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. (Kluesner, 2011)

 C = 2800  C = 2800  m = 3400  m = 3400  a = 3300 water crust mantle lithosphere  s = 2200 sediment crust mantle lithosphere depth of equal compensation ho= 1km hc hm (hs-ho) hs=?  C = 2800  C = 2800  m = 3400  m = 3400 hs = ho(  a -  w )  a -  s ) = 3 ( ) ( ) = 6.3 km

Calculate how much sediment would be needed to fill the Alarcon Basin up to sea level: Maximum water depth in the Alarcon Basin = 3km. Our calculation shows that ~6.3 km of sediment would be needed to fill the basin up to sea level.

crust mantle litho- sphere mantle litho- sphere crust mantle litho- sphere mantle litho- sphere EXAMPLE 3 How does crustal thinning affect 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 mantle litho- sphere mantle litho- sphere crust mantle litho- sphere mantle litho- sphere 1 2 hc1= 30km hc2 hm1= 90km hm2 ha Note: ha = hc1 + hm1 - hc2 - hm2 - Z Z  c (hc1) +  m (hm1) =  s (Z) +  c (hc2) +  m (hm2) +  a (ha) 30  c + 90  m =  w (Z) + 15  c + 45  m +  a ( Z) 30  c + 90  m = Z  w + 15  c + 45  m + 60  a - Z  a Z(  a -  w ) = 60  a -45  m -15  c

crust mantle litho- sphere mantle litho- sphere crust mantle litho- sphere mantle litho- sphere 1 2 hc1= 30km hc2 hm1= 90km hm2 ha Z Z(  a -  w ) = 60  a -45  m -15  c If the new basin is filled by water, what is its depth (Z)? - A: Fill with water (  w = 1.01 g/cm 2 ) Z= 60  a -45  m -15  c (  a –  w )

crust mantle litho- sphere mantle litho- sphere crust mantle litho- sphere mantle litho- sphere 1 2 hc1= 30km hc2 hm1= 90km hm2 ha Z Z(  a -  w ) = 60  a -45  m -15  c If the new basin is filled by water - A: Fill with water (  w = 1.01 g/cm 2 ) Z= 60  a -45  m -15  c (  a –  w ) 60(3.3)- 45(3.4)-15(2.8) ( ) = = = 1.31 km for water

If crust and mantle lithosphere are thinned to half of original thickness, how deep a basin would form in response to thinning? Our calculation shows that the newly-formed basin would be ~1.3 km deep if it was filled by water.

crust mantle litho- sphere mantle litho- sphere crust mantle litho- sphere mantle litho- sphere 1 2 hc1= 30km hc2 hm1= 90km hm2 ha Z Z(  a -  w ) = 60  a -45  m -15  c If the new basin is filled by sediment, how deep will it be? - B: Fill with sediment (  s = 2.2 g/cm 2 ) Z= 60  a -45  m -15  c (  a –  s )

crust mantle litho- sphere mantle litho- sphere crust mantle litho- sphere mantle litho- sphere 1 2 hc1= 30km hc2 hm1= 90km hm2 ha Z Z(  a -  w ) = 60  a -45  m -15  c Basin is formed: A: fill with water B: fill with sediment B: Fill with sediment (  s = 2.2 g/cm 2 ) Z= 60  a -45  m -15  c (  a –  s ) 60(3.3)- 45(3.4)-15(2.8) ( ) = = = 2.7 km for sediment

If crust and mantle lithosphere are thinned to half of original thickness, how deep a basin would form in response to thinning? Our calculation shows that the newly-formed basin would be ~2.7 km deep if it was filled by sediment.

(Martin-Barajas et al., 2013) EXAMPLE 3B What rifting parameters can produce the crustal structure observed in the Upper Delfin basin? Experiment with different sediment densities and different crust and mantle lithosphere initial thicknesses to find values that could produce the thicknesses shown on this cross-section.

This cross-section spans the Delfin and Tiburon basins. The core of the Baja California peninsula (NW end of cross-section) is un-rifted crust, and its thickness (35-40 km) is a reasonable approximation of pre-rift crustal thickness. (Martin-Barajas et al., 2013)

Delfin Basin Location map, Northern Gulf of California (Martin-Barajas et al., 2013)

From the cross-section, measure and record: hc1, initial crustal thickness (use crustal thickness at the NW end of the cross-section as an estimate) hc2, crustal thickness under Delfin basin after extension (0!) hs, thickness of sediments (and intrusions at base of sediments) in Delfin basin (Martin-Barajas et al., 2013)

crust mantle litho- sphere mantle litho- sphere 1 2 hc1= __km hc2=0 hm1= 90km hm2=0 ha Delfin basin Solve for Z. Then see how well your answer matches sediment thick- ness Z on the cross-section Note: ha = hc1 + hm1 - hc2 - hm2 - Z Z(=hs) Experiment #1 Use 90 km for a starting thickness of mantle lithosphere, and hc1 and hc2 values measured from the cross-section.

crust mantle litho- sphere mantle litho- sphere 1 2 hc1= __km hc2=0 hm1= 50km hm2=0 ha Delfin basin Solve for Z. Then see how well your answer matches sediment thick- ness Z on the cross-section Note: ha = hc1 + hm1 - hc2 - hm2 - Z Z(=hs) Experiment #2 Use 50 km for a starting thickness of mantle lithosphere, and hc1 and hc2 values measured from the cross-section.

crust mantle litho- sphere mantle litho- sphere 1 2 hc1= __km hc2=0 hm1= 20km hm2=0 ha Delfin basin Solve for Z. Then see how well your answer matches sediment thick- ness Z on the cross-section Note: ha = hc1 + hm1 - hc2 - hm2 - Z Z(=hs) Experiment #3 Use 20 km for a starting thickness of mantle lithosphere, and hc1 and hc2 values measured from the cross-section.

crust mantle litho- sphere mantle litho- sphere 1 2 hc1= __km hc2=0 hm1= 20km hm2=0 ha Delfin basin Solve for Z. Then see how well your answer matches sediment thick- ness Z on the cross-section Note: ha = hc1 + hm1 - hc2 - hm2 - Z Z(=hs) Experiment #4 Use same thickness values you used in experiment #1, but change  s to  s varies with total sediment thickness….

Experiment #1  c (hc1) +  m (hm1) =  s (Z) +  c (hc2) +  m (hm2) +  a (ha) and ha = hc1 + hm1 - hc2 - hm2 - Z Solve for Z. Plug in thickness values: hc1=35km, hm1=90km, hc2=0, hm2=0 and density values:  c =2800 kg/m 3,  m =3400 kg/m 3,  a =3300 kg/m 3,  s =2200 kg/m 3 Solution: Z = 7.7 km (too small, does not match cross- section)

Experiment #2  c (hc1) +  m (hm1) =  s (Z) +  c (hc2) +  m (hm2) +  a (ha) and ha = hc1 + hm1 - hc2 - hm2 - Z Solve for Z. Plug in thickness values: hc1=35km, hm1=50km, hc2=0, hm2=0 and density values:  c =2800 kg/m 3,  m =3400 kg/m 3,  a =3300 kg/m 3,  s =2200 kg/m 3 Solution: Z = 11.4 km (about right, matches cross-section)

Experiment #3  c (hc1) +  m (hm1) =  s (Z) +  c (hc2) +  m (hm2) +  a (ha) and ha = hc1 + hm1 - hc2 - hm2 - Z Solve for Z. Plug in thickness values: hc1=35km, hm1=20km, hc2=0, hm2=0 and density values:  c =2800 kg/m 3,  m =3400 kg/m 3,  a =3300 kg/m 3,  s =2200 kg/m 3 Solution: Z = 14.1 km (too large, does not match cross- section)

Experiment #4  c (hc1) +  m (hm1) =  s (Z) +  c (hc2) +  m (hm2) +  a (ha) and ha = hc1 + hm1 - hc2 - hm2 - Z Solve for Z. Plug in thickness values: hc1=35km, hm1=90km, hc2=0, hm2=0 and density values:  c =2800 kg/m 3,  m =3400 kg/m 3,  a =3300 kg/m 3,  s =2400 kg/m 3 Solution: Z = 9.4 km (about right)

Delfin Basin Location map, Northern Gulf of California Note: These solutions assume sufficient sediment supply to keep the basin filled. That’s a good assumption here because of the very high sediment input from the Colorado River. (Martin-Barajas et al., 2013)

Our assumption that there is sufficient sediment supply to keep the basin filled would not be a good assumption further south in the Gulf of California where basins are sediment starved. Delfin Basin Google Earth