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 lowerthan-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 viscoelastic.
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
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