Ch.2 Isostacy

2. Introduction A 1730s French expedition led by Pierre Bouguer to Peru revealed a rather interesting finding – that while a plumb line was deflected from the vertical direction by the mass of the Andes mountains, the degree of deflection was much less than initially estimated. In the 19th century, Sir George Everest found a similar reduced deflection near the Himalayas. Imagine if one made a map of gravity variations near these mountains, one might have expected huge variations arising from these huge masses. However in reality one would see much smaller variations than first expected. What can account for this phenomenon? In 1855 J.H. Pratt and G. Airy independently proposed two hypotheses to explain this phenomenon. These are known as Pratt’s hypothesis and Airy’s hypothesis. No surprise there! Figure 13 (Fowler Fig5.6) illustrates the two hypotheses. (Refer to Fowler Section 5.2.2. for a quantitative discussion.)

Figure 13: Airy’s and Pratt’s Hypotheses

For both the Andes and the Himalayas, it appears that a reduction in mass must exist either inside or underneath these mountains in order to explain the lower- than-expected gravitational attraction near them. If one calculates the amount of mass reduction (or mass deficit) required, one finds a mass comparable to the entire mountain! This calls to mind Archimedes Principle of hydrostatic equilibrium, which says that a floating body displaces its own weight of fluid in which it is immersed. But wait - who is talking about the Andes or the Himalayas floating on a fluid?

1 Isostacy (or isostasy) is a principle that is capable of accounting for observations that large-scale topographic features on Earth have little effect on large scale, regional and global, gravity field patterns.

2.1 Geoid We can represent the gravity field by introducing a surface that is everywhere perpendicular to it. For example, if we connected all the continental regions of the ocean by canals then the resulting water surface of the sea and connected canal would represent an equipotential surface, very close to what is termed the geoid. The geoid is such a surface, and is shown in Figure 14 (Stacey Fig9.1), where the landmasses are also depicted.

Figure 14: Geoid

It is clear that there is little relation between the anomalies of the geoid and excess mass of the continents. In order to appreciate the expectation of continents affecting the gravity field, consider the following idealization: if Earth were a uniform, non-rotating sphere, having constant density, objects near Earth's surface would fall directly towards its center with the same acceleration everywhere on the globe's surface. It would also be true that a pendulum anywhere on the surface would point to the center when at rest and have the same period, when put into motion, no matter where it was placed. A pendulum at rest points in the direction of local gravity. For our spherically symmetric Earth, the combined effect of all the mass elements of Earth is such that the gravity vector points to the centre. Now suppose a huge (mountain sized) mass is suddenly placed nearby the pendulum. One expect a certain degree of deflection, as per Figure 15 (to be filled in by hand):

2

Figure 15: Deflection due to “uncompensated” mass

One would expect the pendulum to be deflected towards the mass due to the attraction between the mass and the pendulum. Indeed this would happen in our thought experiment. Now consider the real Earth and imagine taking a pendulum near the Himalayas to check for a deflection. In fact when this was done, very much less deflection toward the mountain was seen than was expected. To understand this reconsider our thought experiment above and take the mountain away. Now instead of placing a mountain nearby, make a huge hole in the ground. What would our pendulum do? To answer this consider two regions of equal mass, equidistant on opposite sides of the pendulum in the undisturbed Earth as illustrated below:

Figure 16 (to be filled in): Effect of mass excess/deficit

Each of the masses attracts the pendulum by the same force so that the pendulum points directly to Earth centre. If we remove one of the masses, its attractive force disappears and the pendulum will then be deflected away from the void left by the mass removed as shown above. We can think of the void as a mass deficiency and the mountain previously added as a mass excess. Thus the deflection of the pendulum toward the excess mass above Earth could be balanced by a mass deficiency below Earth. The mass deficiency would not have to be a void but simply a reduced density as shown in the cartoon below: 3

Figure 17 (to be filled in): Effect of “compensated” mass

The mountain shown here has what is known as a root of lower density than the rock in which the mountain “floats”.

2.2 Isostatic Compensation

We now consider the elementary physical model of isostatic compensation, which is based on the principle of hydrostatic equilibrium and states simply that hydrostatic pressure is constant at some level in the mantle and this level is called the depth of compensation. It is clear that if this were not the case flow would occur until pressure became the same at this level. Before continuing with this principle, it is useful to review an elementary observation from hydrostatics given by L. Prandtl and illustrated below:

Figure 18 (to be filled in): Hydrostatic Equilibrium

In this picture there are different volumes (and masses) of fluid all connected at the base by a common reservoir. What is obvious from this figure is that the height of the fluid column is the same regardless of the shape and mass of the fluid above the reservoir. Put another way, the hydrostatic pressure or force per unit area of the fluid depends only on the height of the fluid column above (not on its volume or mass). 4 We can apply this principle to Earth's crust floating on the mantle as illustrated in the figure below:

Figure 19 (to be filled in): Isostatic Compensation involving two layers

We can find the depth of the root, r, from the height, h, above the surrounding crust of thickness tc under the assumption of hydrostatic equilibrium. We call the density of the crust ρc and the density of the mantle ρm. We calculate the mass of each of the two columns 1 and 2 with cross sectional area A, and then equate the pressures at the base of the columns. Mass of column 1 = Mass of Air + mass of Crust + mass of Mantle:

Force at base of column 1 is its weight is

Pressure at base of column 1 is

Similarly for column 2 Equating these two pressures gives

or Since air density << mantle density, then

or where the LHS is the mass deficit below and the RHS is the mass excess above the surface. We can express the depth, r , of the root in terms of the height h above the surface as

5 We can find the relationship between the height and root by using appropriate values for the densities. We take:

which yields

so that for each km of height there is a 5.6 km root! If we define the excess* crustal thickness associated with a mountain of height h as E=r+h, then

(Note *: “excess” here refers to the thickness above the depth of compensation.) This means that a 3km mountain is associated with an excess crustal thickness of about 20km.

Just as excess mass in a mountain requires a root, a mass deficiency at the surface, due to a basin filled with water, for example, requires excess mass at its root. This is shown in the sketch below:

Figure 20 (to be filled in): Isostatic Compensation involving a basin

Here the depth of the basin is d, filled with material of density ρf. Equalizing the pressures at depth gives:

or

6 Typical values of ρf the fill density are given in the table below:

The amount of crust that needs to be removed to make a basin of depth, d, is

It is useful to make up a table of relationships between r, d, and E for the three different fill materials air, water and sediments.

The interpretation of values in this table is as follows. Top 2 rows: Basin depth d represents only 15% of the total crustal material excavated to produce the basin if air filled, 22% of the excavation if water filled and 63% if filled with sediments. The root r makes up the difference. Lower two rows: For a given (observed) depth of basin, d, these numbers give r and E. Of particular interest is the amount of crustal material that needs to be excavated for a basin of depth d. For a sediment filled basin, E=1.6d; which means for a 10km deep sediment filled basin, 16km of crust must be removed. This is about 50% of total crustal thickness.

If instead of sediments the basin is water filled the situation is more extreme, since E=4.6d so that for d=10km we get E=46km, which is greater than mean crustal thickness.

Let's consider a shallow water filled basin that is d=100m deep. The table shows that E=460m. This means that a 460m thinning of the crust will be compensated in such a way that the resulting basin is only 100m deep. In other words the base of the crust (the Moho) below a basin must rise as a result of isostatic compensation.

Now suppose that sediments are added so that E=r+d remains unchanged, the extra weight will cause d to increase such that d= E/1.6 ~ 290m, leaving a root r=170m. Thus the weight of sediments can triple the depth of a basin.

In the example above, a large basin of depth 10km could be produced by additional (e.g. tectonic) forces other than isostatic compensation. These forces 7 would have to drag the basin downward as it is filling usually over a very long time, i.e. 107 - 108 years.

Some observations of sediment basins will help to answer the question whether or not isostatic compensation actually takes place.

Shown below in Figure 21 is a seismic profile of the Scotia Basin on the continental shelf off the east coast of Canada.

Figure 21: Seismic Profile of Scotia Basin

The ocean is off to the right in this figure and it can be seen that the crust thins in that direction. Deep sediments have accumulated on the submerged margin. Clearly the base of the crust (the Moho) rises as the crust gets thinner. Marked on the map is the Triumph oil well where sediments are about 10km thick; the crust has thinned from its normal value of 35km to about 19km and the Moho has risen about 6km. So here we have an example of a basin of depth 10km with a root of 6km.

In terms of the previous discussion d = 10km r = 6km E = 16km Exactly as predicted by simple isostacy.

8 A second example is illustrated in Figure 22, which is derived from seismic profiles across the North Sea.

Figure 22: North Sea basin illustrating isostatic compensation

Here the Moho discontinuity rises below the sedimentary basin since the mantle is higher density than the crust. At the mid-point of the basin d~8.5km, r~5.5km, giving E~14km; again in agreement with isostacy. It is clear that large values of E do exist as illustrated by the above examples. Also apparent is that any mechanism proposed to account for such basins must also account for major thinning of the crust.

2.3 Regional and Local Isostacy

We note that the Airy version of isostacy, as depicted in Figure 13(a) implies that the individual columns shown there must be free to 'slip' past one another. In general this will only be true on a large scale as on a small scale loads find support from local rock.

For example as illustrated here neither houses nor office towers are isostatically balanced:

Figure 23(a – to be filled in): Isostatic Compensation NOT expected 9 and when one digs a hole it does not float up:

Figure 23(b – to be filled in): Isostatic Compensation NOT expected

An exception, of course, is a fluid which has no resistance to shear so that:

Figure 24 (to be filled in): Isostatic Compensation IS expected

Conclusion? “Small” or “modest” structures are usually uncompensated. “Large” tend to be compensated. However the compensation process may take some time. What is termed regional isostatic compensation does exist in the case of concentrated loads, which are sufficiently large to allow this and is shown below for volcanic islands like Hawaii:

Figure 25 (to be filled in): Isostatic Compensation by Crustal Deformation 10 2.4 Glacial Isostatic Rebound

There have been glacial cycles lasting about 100,000 years consisting of approximately 90,000 years of ice load followed by 10,000 years without ice. Melting is considered to have taken place rapidly, lasting less than 1000 years. When the ice melts its load is removed from Earth's crust and the crust rebounds.

The sequence over Hudson's Bay went as follows: • During ice covered period there was about 3km ice thickness. • This weight led to a subsidence of ~500m that was isostatically compensated. • Ice melts quickly so is not compensated. • This leads to a flow from below and uplift takes place at a rate of 250m in 10,000 years, or 2.5cm/yr.

The rate at which uplift happens depends on the viscosity of the mantle. However we know the mantle can also behave elastically since it supports seismic waves. Accordingly we represent the mantle according to the time scales of the forces acting on it: • Long time scales of plate tectonics (108yr), the mantle is viscous • Short time scales of seismic waves (10-8yr or about 0.1 sec), the mantle is elastic • Intermediate time scales of glacial rebound (104yr), the mantle is visco- elastic.

In order to appreciate this behaviour it is useful to consider simple physical systems. For example a coil spring behaves elastically unless stretched too far as illustrated here:

Figure 26 (a- to be filled in): loaded beyond elastic limit

or if left for a very long time

Figure 26 (b- to be filled in): loaded for too long 11 We can model the mantle behaviour by a combination of springs and dashpots that we define as follows: • Spring: restoring force is proportional to extension that corresponds to an elastic mantle when forces act on a short time scale:

Figure 27 (a- to be filled in): spring model

• Dashpot has damping force proportional to velocity as illustrated here, which corresponds to a viscous mantle due to forces on a long time scale. This results in permanent deflection.

Figure 27 (b- to be filled in): dashpot model

• For intermediate time scales we use a combination of the two models as

Figure 27 (c- to be filled in): combined spring & dashpot model 12 We model the rate of glacial rebound in terms of viscosity of a dashpot, with low viscosity corresponding to quick rebound and high viscosity, slow rebound.

Earlier it was noted that ice loaded Hudson's Bay, as shown in Figure 28.

Figure 28: Ice Thickness over Hudson’s Bay during last Ice Age

The contours in this figure show the thickness of ice in metres, which was more than 3km over Hudson's Bay and thinned out at the periphery. When the ice melted the land rose up over several thousand years and left a series of elevated 'beaches' showing previous sea-levels. These levels were determined in time by dating creatures from remains on the beaches. The dates of the remains at different elevations are shown in Figure 29.

Figure 29: Uplift of Hudson’s Bay due to Isostatic Rebound 13 In Fennoscandia recent measurements of uplift due to the removal of ice is illustrated in Figure 30 (Stacey Fig5.19).

Figure 30: Uplift rate of Fennoscandia When ice loaded the land, the force due to its excess mass caused depression beneath the ice. Correspondingly there was a bulge created which surrounded the depressed area. When the ice melted the land rose leading to the elevated beaches that described above. Of course the former bulge surrounding the ice-laden region then sank, producing buried beaches. This is illustrated in Figure 31.

Figure 31: PostGlacial Rebound and Submergence

End Chapter 2 14