Stress, Balance Equations

8/12/2019 Stress, Balance Equations

1/41

Chapter 3

Stress, Balance Equations, and

Isostasy

We define stress to consider forces acting at depths, and derive the equations of mass,

linear momentum, angular momentum, and energy conservation. Very slow tectonic

movements allow us to neglect inertia force and simplify the equation of motion to a

balance equation. Isostasy is the most basic but important application of force balance

equations to tectonics.

3.1 Definition of stress

Suppose that a square pole is composed of five rectangler solids (Fig. 3.1(a)). If the pole bears a

load at the top and base, the parts may push each other across the bounding planes. Parts A and B

may push each other with that load. However, parts C and D may push each other with much less

force. Force across a plane in a solid depends on the direction of the plane.

In order to consider a force exerted on a rock mass, suppose there is a surface element S on

the mass, and a unit vector Nnormal to the surface and pointing outward indicates the attitude of

the surface element (Fig. 3.1(b)). In addition, let the vectorFbe the traction force acting on theelement. The force may increase with the areaSso that we define a vector quantity t(N) as the

force per unit area exerted there. Nis in parentheses because the force depends on the attitude of the

surface element. The quantity is called a stress vector. Mathematically, the stress vector is defined

by

t(N) = limS0

F

S. (3.1)

The stress vector and its components are expressed in the units Nm2 = pascals (Pa)1.

1The same units are used to count pressure. 1 bar = 105 Pa = 1000 hPa = 0.1 MPa. Engineering literature often uses the

units kgf/cm2 which are approximately equal to 0.98 MPa. The unit dyn cm2 is used in older literature. It is converted by

the factor 1 Pa = 10 dyn cm2.

59


2/41

60 CHAPTER 3. STRESS, BALANCE EQUATIONS, AND ISOSTASY

Figure 3.1: (a) A square pole loaded at the top and base is composed of five rectangular solids, A

through E. (b) A rock mass is pulled from the adjacent mass with the force F on an element of the

surfaceS. A unit vectorNindicates the attitude of the surface element. (c) Stress vector t(N) and

Cartesian coordinatesO-123 whose third axis is parallel to the unit normal N.

Consider a surface element S with the unit, outward normal N, to which the third axis of

Cartesian coordinatesO-123 is adjusted (Fig. 3.1(b) and (c)). Let e(i) be theith base vector of the

coordinates, then we write

t(N) =S31e(1)+ S32e

(2)+ S33e

(3), (3.2)

where S31, S32and S33are the components of the stress vector t(N), that is, t(N) =(S31, S32, S33)T.

The stress vector depends on the direction of the surface on which it acts. The first subscript 3 in-

dicates accordingly that the surface is normal to the third coordinate axis and its outward normal

points positive direction of the axis. The second subscript distinguishes the components of the stress

vector. A stress component normal to the surface on which it acts is known as the normal stresson

the surface, whereas the component tangent to the surface is known as shear stress. Generalizing

Eq. (3.2), the components of stress vectors acting on a surface elements normal to the coordinate

axes, we have

Sij =tj

e(i)

. (3.3)

We define thestress tensorfrom the components as

S =

S11 S12 S13S21 S22 S23S31 S32 S33

components of te(1) components of te(2) components of te(3)

.

(3.4)

The limit in Eq. (3.1) indicates that stress vector is defined at each point on an infinitesimally small

surface. The diagonal components are the normal stresses acting on a surface element normal to the

coordinate axes, while the other components represents shear stresses acting on the surface elements.

Sign conventions Figure 3.2(a) shows the positive direction of those components. Since we have

defined the unit vector Nas being the outward normal to the surface on which the stress vector is


3/41

3.1. DEFINITION OF STRESS 61

Figure 3.2: Sign convention in continuum mechanics (a, b) and solid earth science (c, d). Arrows

indicate the direction of the positive stress components acting on a cube whose surfaces are parallel

to the coordinate planes. (a) A stress component is positive when it acts in the positive direction

of the coordinate axis, and on a surface of the cube whose outwardnormal points in the one of the

positive coordinate directions. (b) A stress component is positive when its direction is opposite to

that of (a). and indicate vectors pointing to this and opposite sides of the page, respectively.

defined, the opposite faces of a cube have the unit vectors in the opposite directions. Accordingly,the normal stress on the faces has the opposite positive directions, that is, the normal stresses are

positive in sign if the cube is under a tensional stress (Fig. 3.2(b)). Compressive stresses have

negative normal stresses.

We have seen that the strain tensor E has positive diagonal components if the length parallel to

coordinate axes become larger. Therefore, it is natural that positive stress components correspond

to positive strain components. However, compressive stresses are common in the crust due to over-

burden pressure. It is the sign convention in solid-earth science to define the sign of stress as being

positive for compression, opposite to the sign convention of continuum mechanics (Table 3.1). Ac-

cordingly, we will use the symbol rfor the stress tensor with positive components for compression

to describe tectonic models. Positive directions of the components ofrare shown in Fig. 3.2(c) and


4/41


Table 3.1: Symbols and signs used in this book.

Stress Unit Normal Vector Strain

Continuum Mechanics S(tension = +) N(outward) E(elongation = +)Solid-Earth Science r(compression = +) n (inward) e (shortening = + )

Figure 3.3: Direction of the unit normal vectors n and Nat the surface of a rock mass.

(d). The stress tensor rshould be used with the inward normal n (Fig. 3.3), where

S =r, N=n. (3.5)

Stress components in cylindrical coordinates Cylindrical coordinatesO-rz are sometimes con-

venient (Fig. 3.4). Transformation between the cylindrical and rectangular Cartesian coordinates

O-xyz is given by the equations,

x = r cos , y =r sin , z = z

r2 =x2 + y2, =tan1(y/x).

Stress components in the coordinate systems are transformed by the two sets of equations

rr =xxcos2 + yysin

2 + 2xysin cos , (3.6)

=xxsin2 + yy cos

2 2xysin cos , (3.7)r =(yy xx) sin cos + xy (cos2 sin2 ). (3.8)

3.2 Cauchys stress formula

It is seen from Eq. (3.3) thatti

e(j)

=Sji =

kSjk e

(i)

k . Therefore, we have

te(i) =ST e(i). (3.9)


5/41

3.2. CAUCHYS STRESS FORMULA 63

Figure 3.4: Stress components in the cylindrical coordinatesO-r.

Note that the choice of frame of reference, the base vectors of which are e(i), is arbitrary. Equation

(3.9) is generalized as follows: the stress vector acting on a surface element with the unit outward

normalN is

t(N) =ST N, (3.10)where Nmay be oblique to the coordinate axes2. This is called Cauchys stress formula. For the

sign convention of solid-earth science we have the the equivalent formula

t(n) = rT n. (3.11)

Ift(N) is the traction on a body at its surface element with the normalN, the adjacent body has the

outward normal Nat the surface element. The action-reaction law requires that

t(N) =t(N) =t(n), (3.12)

which is shown in Fig. 3.3.

Normal and shear stresses Normal and shear stresses are the components oft(n) parallel and

normal ton, respectively. Thus,

Normal stress: N =t(n) n, (3.13)

Shear stress: S =|t(n)|2 2N. (3.14)Since the vectorn points inward at the surface of a rock mass (Fig. 3.3),Nis positive if the mass is

pushed inward at the surface. By contrast, the normal stress defined by the equation

SN =t(N) N (3.15)

is positive for tension. The shear stress for this sign convention is

SS =

|t(N)|2 SN2 .2Equation (3.10) is demonstrated via the force balance equation [134], which is introduced in 3.3. However, axiomatic

continuum mechanics defines a stress tensor as the linear transformation from t(N) to t(N), rather than the force per unit

area on the surface elements parallel to coordinate planes. The interested reader will find more information in [243].


6/41


The vectorN(n) =Nnhas a magnitude obviously equal to the normal stress Nand is parallel

ton. IfN is positive,N(n) points inward at the surface of the rock in question. The vector N(n)

is obtained by the orthogonal projection of t(n) onto the line parallel to n, so that the projector

P(n) =nn (Eq. (C.37)) is used to describe the vector:

N(n) =P(n) t(n) =nn

r n =n (r n)n. (3.16)

On the other hand, S(n) is defined as the vector that has a magnitude equal to the shear traction

and indicates the direction of the shear traction at the surface element of the mass. The elementary

orthogonal projectorP(n) =I nn(Eq. (C.38)) is useful:

S(n) =P(n)

t(n) = I nn t(n) = r n n (r n)n = t(n) tN(n). (3.17)

Hydrostatic pressure From Eq. (3.11), let us derive a useful equation for tectonics: the tensorial

formulation of hydrostatic state of stress. Consider the stress tensor defined by

r = p I, (3.18)

where I andpare the unit tensor and a scalar quantity. Coordinate rotation by Q transforms the stress

as r = Q r QT = Q (p I) QT = pQ QT = p I = r,indicating that the stress defined by Eq.(3.18) is isotropic, that is, independent from orientations. The state of stress is anisotropicif there

is the dependence.

In this case, the force acting on a surface element dS is t(n)dS = (r n)dS = (pdS)n, whichis perpendicular to the surface and pointing inward with a constant magnitude ofpdS. The stress

vector is parallel to n for hydrostatic stress state: shear stress always vanishes. Therefore, Eq. (3.18)

is known as the hydrostatic state of stress, and p is hydrostatic pressure. The equation S =p Iexpresses the same stress for the sign convention of continuum mechanics. Hydrostatic pressure

constricts a body: it shrinks with a similar figure with the original one if the body is homogeneous

inside. If the body is subject to an anisotropic stress, the shape may change.

3.3 Force balance

Tectonic motion to create a large-scale geologic structure is very slow. Forces acting on a rock mass

at depth may be thought of being balanced at every moment. Accordingly, we shall derive balance

equations for forces, and derive vertical stress at depth.

There are two categories of forces: surface forceper unit area of the surfacet andbody forceper

unit mass X. The former was discussed in the last section. The latter acts on all the internal elements

of a body from its exterior. Examples are gravity and electromagnetic forces. As an example, gravity

is written as the vector

X=(0, 0, g)T, (3.19)

whereg is gravitational accerelation and the third coordinate axis points vertically downward. Body

forceper unit volume is X.


7/41

3.3. FORCE BALANCE 65

3.3.1 Equation of motion

Consider a rock mass defined by a volume Vand a closed surface S. Body and surface forces,X

and t(N) act on every portion in the body and on the surface, respectively. If the forces are balanced,

we have

0 =

S

t(N) dS+

V

XdV. (3.20)

Since

Vv dVis the total linear momentum of the rock mass, the material derivative

D

Dt

V

v dV

represents the inertia force. Therefore, the equation of motionof the body is

D

Dt

V

v dV =

S

t(N) dS+

V

XdV. (3.21)

The surface integral in the left-hand side of Eq. (3.21) is transformed into a volume integral with

Gausss divergence theorem (Eq. (C.63)). In addition, combining Eq. (3.10), we obtain

D

Dt

V

v dV =

V

ST dS+

V

XdV. (3.22)

In this case the order ofD/Dtand integral is exchangable3 so that we haveV

v ST X

dV =0.

Since the choice of the volume is arbitrary, the integrand must vanish to satisfy this equation. Con-

sequently, we obtain the differential form of

v = ST + X or v = rT + X. (3.23)

If the forces are balanced, we have

0 = ST + X or 0 = rT + X. (3.24)

The indicial notation of Eq. (3.24) is

0 =

j

Sji

xj+ Xi or 0 =

j

ji

xj+ Xi. (3.25)

3IfV0 and V are the initial and instantaneous volumes that the rock mass occupies, respectively, at time t = 0 and t,

the exchangeability is demonstrated as follows. The initial volumeV0 does not depend on time, so that the operator D/Dt

can be included in the integral overV0. In addition, because ofV = J V0 whereJ is the Jacobian, the initial and temporal

densities, 0 and , are related to each other as = 0/J. The initial one does not depend on time, either. That is,

D0/Dt = D(J)/Dt = 0. Therefore,

D

Dt V v dV = D

Dt V0 vJdV0 = V0 v D(J)

Dt

+ JDv

Dt dV0 = V0 Dv

Dt

JdV0 = V Dv

Dt

dV.


8/41


Based on Eq. (3.24), we shall derive a simple but important quantity for tectonics. Suppose

that the ground surface is flat and there is no horizontal density variation. Rocks are subject to the

gravitational force shown in Eq. (3.19). Definingz-axis vertically downward from sea level, the

force balance in this direction is obtained from Eq. (3.24) as

xz

x +

yz

y +

zz

z =(z)g,

where density,(z), is the function of depth, z. Since there is no horizontal variation at all, the first

and second terms in the above equation vanish. We therefore have

zzz

=(z)g.

Integrating both sides, we obtain

zz =

zh

()g d, (3.26)

where h is altitude. This equation enables us to calculate the vertical stress zz from the density

profile(z). Ifg is constant over the depth range in question, Eq. (3.26) becomes zz = gz

h() d.

The integral in the right-hand side of this equation is the mass of the vertical column with the unit

cross-section of the height from the surface to the depth z. Therefore,zz is the stress at the base of

the column due to its own weight4. The vertical stress by the weight is called overburden stress. If

density can be assumed to be constant, the overburden at the depthz is simply

pL =gz. (3.27)

We have found how the vertical stress zz is determined. What are the other stress components?

Areas of compressional and extensional tectonics may have different stress states. Stress may not

only have spatial but also tempral variations, which are represented by components other than the

vertical one. However, it is convenient to define a reference stress state for later discussions. Al-

though choice of the reference is arbitrary, the isotropic stress

r =pL I (3.28)

is often used for the reference in the models of tectonics, where

pL =

zh

()g d (3.29)

is the overburden at depth z. This is the overburden with the same dimension with pressure. Equation

(3.28) has the same form as Eq. (3.18), so that the stress indicated by Eq. (3.28) is known as

lithostatic stress, andpLis calledlithostatic pressure.


9/41

3.4. SYMMETRY OF STRESS TENSOR 67

Figure 3.5: Force Fat the point P with a position vector x yields a moment of that force Mabout

the origin O. The magnitude ofMequals |x||F| sin , where is the angle between the vectors.

3.4 Symmetry of stress tensor

Both surface and body forces acting on a rock mass exert moments (torques) on the mass. If a forceFacts at a point P whose position vector isx (Fig. 3.5), the moment of the force about the origin is

M=x F. Likewise, the moment about the origin by surface force is

Sx t(N) dS, wherex is

the position vector of the point where the surface force t(N)dSacts on the rock mass. The moment

by body force is

Vx XdV, where xis the position vector of a particle in the mass where the

body force acts. Therefore, in case that the total moment is in equilibrium to conserve the angular

momentum, we have the balance equation of momentum as

0 =

S

x t(N) dS+

V

x XdV. (3.30)

If they are imbalanced, the residual moment causes the acceleration in the angular velocity of thebody:

D

Dt

V

(x v) dV =

V

(x X) dV +

S

(x t) dS. (3.31)

The choice of position of the origin is arbitrary. Hhowever, Eq. (3.31) holds wherever the origin is

defined5.

4In-situ stress measurements have shown that this picture gives a close approximation to observed zz [3]. However,

deviations from predictedzz from the density profile are rarely observed [30, 75].5 Let us calculate the total moment about a position with the position vector o, from which position vector x is defined.

That is,x = x + o. By this additive decomposition ofx, Eq. (3.31) becomes

Left-hand side = V

(x + o) v dV = V

x v dV + V

o v dV,

Right-hand side =

V

(x + o) XdV +

S

(x + o) tdS

=

V

x XdV +

S

x tdS+

V

o XdV +

S

o tdS.

Combining the terms in and outside the square brackets in the above equations, we have the equationsV

x v dV =

V

x XdV +

S

x tdS,

which is identical to Eq. (3.31) provided that x is replaced by x and

o V

v dV =o V

XdV + StdS . ()


10/41


It is important for discussions in the rest of this book that the stress tensor is symmetric:

S =ST, r = rT. (3.32)

These formulas hold for both static and dynamical problems unless there is neither couple stress

nor body torque. The former is the surface force acting on a body to rotate it about the axis

perpendicular to the object surface, and the latter is the distributed moment in the body6. Equation

(3.32) is derived from the conservation of angular momentum7.

Plane Stress It is a basic theorem in linear algebra that all the eigenvalues are of real for a real,

symmetric matrix. In addition, the eigenvectors are perpendicular to each other. If one and only one

of the eigenvalues is zero, a state ofplane stressis said to exist. In that case, taking coordinate axesparallel to the eigenvectors, the stress tensor is written as

S =

S1 0 00 S2 00 0 0

,where S1 and S2 are non-zero eigenvalues ofS. The physical interpretation of the stress state is

that traction always vanishes on a plane perpendicular to the eigenvector corresponding to the zero

eigenvalue. This is demonstrated by substitutingN = (0, 0, 1)T into t(N) = ST N. It should benoted that plane strain and plane stress do not go together for most cases (7.1.3).

3.5 Conservation of energy

Let us derive an equation governing temperature changes from the conservation of energy. Multi-

plying both sides of Eq. (3.23) byv , we haveV

v adV =

V

v ST

dV +

V

v XdV, (3.33)

whereV stands for the volume that the rock mass in question occupies. The left-hand side of this

equation is rewritten asV

v adV =

V

v v dV = DDt

V

v v

2 dV =

D

Dt

V

|v|22

dV = K.

The symbolKstands for the total kinetic energy, so that Kis the rate of its increase. On the other

hand, the integrand v (ST) in the first term in the right-hand side of Eq. (3.33) has the components

viSji

xj=

(viSji )

xj vi

xjSji =

(viSji )

xj (Dij + ij )Sji,

This is identical to the cross-product of the constant vector o and Eq. (3.22), so that Eq. () is valid for any o. Therefore, Eq.

(3.31) holds wherever the origin is chosen.6See advance continuum mechanics books such as [132].7The derivation needs a tricky calculation. See, for example, [134].


11/41

3.5. CONSERVATION OF ENERGY 69

whereLij andij are stretching and spin tensors, respectively. Therefore, Eq. (3.33) is rewritten as

K+

V

D: S dV =

V

(v S) dV

V

W: S dV +

V

v XdV. (3.34)

SinceW= (ij ) is antisymmetric we have W: S =0. Combining this and Eq. (3.34), we obtain

K+

V

D: S dV =

V

(v S) dV +

V

v XdV. (3.35)

Converting the first term on the right-hand side of this equation into a surface integral by Gausss

divergence theorem, we obtain the energy conservation equation

K+

V

D: S dV =

S

[v t(N)] dS+

V

v XdV. (3.36)

Energy equals the product of force and distance, so that the rate of energy change equals the product

of force and velocity. The right-hand side of Eq. (3.36) indicates the energy that the rock mass

accepts from outside. The second term on the left-hand side is the energy dissipation due to the

deformation of the mass against the stress S. The dissipated energy is transformed to thermal or

internal energy. Lete be the internal energy per unit mass, then the internal energy per unit volume

is equal toe, and V

e dV = V

D: S dV. (3.37)

Therefore, the energy conservation equation is rewritten as

D

Dt

V

v2

2 + e

dV =

S

v t(N) dS+

V

v XdV. (3.38)

The left-hand side of this equation represents the energy increase of the rock mass, and the right-hand

side indicates the work done by its outside.

Equation (3.38) was derived from the equation of motion multiplied by velocity. That is, only the

balance of kinetic energy was taken into account. However, vertical motion of a rock mass causes

cooling or heating of the mass. Temperature is an important factor in controlling the mechanical

properties of rocks. Accordingly, heat transfer is important in understanding the mechanical aspect

of tectonics.

Heat flux is defined by the amount of heat that passes through a unit area in a unit of time. The

heat flux is a vector quantity, so let us use the symbol qfor the quantity. The heat energy passing

through the surfacedSof a rock mass in a unit time equals Nq. Substituting this into the right-handside of Eq. (3.38), we have

D

Dt V v2

2 + e dV = Sv t(N) dS+ V v XdV SN qdS. (3.39)


12/41


Using Gausss divergence theorem again, the last term in this equation is converted to a volume

integral, and we combine all terms into the form V

( ) dV = 0. Since the volume V is arbitrary,this equation is satisfied only if

D

Dt

v2

2 + e

= S v q+ X v. (3.40)

This is called the energy equation. Combining Eqs. (3.40) and (3.33), we have

e = q+ (S v) v ( S).

Components of the first and second terms in the right-hand side arei,j

(Sij vj )

xi vi

Sij

xj

=

i,j

Sijvi

xj=S : (v).

Therefore, we finally obtain the equation

e = S : (v) q. (3.41)

3.6 Self gravitational, spherically symmetric body

The direct objects of geological observations occupy shallow levels in the Earth. However, those

objects sometimes provide constraints on ancient global changes including planetary differentiation.

The differentiation occurred at the beginning of planetary evolution, so that later geological processes

have erased all traces of the event in the Earth. However, satellites and terrestrial planets smaller than

the Earth have geological structures that evidence the early history of the planetary bodies.

For example, wide rift zones on the Jovian satellite Ganymede are thought to be evidence of

planetary differentiation (Fig. 3.6). An icy mantle ocupies one third of the radius, and 60% the mass

of the satellite. The surface of Ganymede has high and low albedo regions which appear as light

gray and dark terrains, respectively, in Fig. 3.6. Number density of impact craters indicates that the

dark terrains are older than the lighter ones. White spots in this photograph are ejecta blankets fromyoung impact craters. The light, young terrains with stripes are called sulci. These are thought to be

the surface manifestation of normal faults, and the sulci are wide rift zones [171]. The formation of

those normal normal faults is modeled in Section 12.4.

Ganymede was initially wrapped up in a dark icy layer, but global expansion broke the layer into

dark terrains. Young, light gray ice spread between the terrains. The global expansion was due to

core formation in the satellite; initially deep-seated ice was raised to shallow levels in the moon by

sinking rocks and metalic materials. The phase transformation of the ice by decompression caused

the global expansion.

We have seen in the previous section (3.3.1) that the vertical stress zzis determined by gravity,

where the gravitational acceleration gis assumed to be constant. However, g may vary with depth


13/41

3.6. SELF GRAVITATIONAL, SPHERICALLY SYMMETRIC BODY 71

Figure 3.6: Surface of Ganymede, the largest Jovian satellite with a radius of 2640 km. The circular

region labeled E is the palimpsest Epigeous, an old and degraded large impact basin 350 km in

diameter. Voyager image, 0370J2-001. Courtesy of NASA.

in the real Earth. Accordingly, in this section we study the depth-dependence of g and lithostatic

pressure.

Firstly, let us assume a density distribution with spherical symmetry, and M(r) be the mass


14/41


within the distancer from the center. The gravitational acceleration at the radius is

g(r) = GM(r)

r2 , (3.42)

whereG = 6.67 1011 Nm2 kg2 is the gravitational constant. If density at the radius is (r), wehave

M(r) =

r0

4r 2(r) dr. (3.43)

Combining Eqs. (3.42) and (3.43), we obtain the gravitational acceleration as a function ofr as

g(r) = Gr2r

0

4r 2(r) dr. (3.44)

Scondly, pressurep at r is obtained from the force balance equation (Eq. (3.25)). As the force

in this case has merely originated from the self-gravity of the spherically symmetric body, there is

no horizontal variation in the stress field. Equation (3.25) is rewritten for the spherical coordinates

O-r as

0 =rrr

1r

r

1

r sin

r

g, (3.45)

where both andg are the functions r and the latter is given by Eq. (3.44). The gravity termg

in Eq. (3.45) is negative in sign, because gravity points downward, i.e., in the direction ofr. Ifthe spherical body is composed of viscous fluid at rest, the state of stress in the body is hydrostatic.

In this case, shear stresses r and r vanish, and in Eq. (3.45) rr should be replaced by the

hydrostatic pressurep. Consequently, we have

dp

dr =g. (3.46)

As and gare always positive in sign, Eq. (3.46) indicates thatp should decrease withr.

Let us calculate gand p as the functions ofr for the two cases in Fig. 3.7. Density in the layers

is assumed to be constant. The former is a simple model for the Moon. Our Moon had its own

magnetic field a few billion years ago, suggesting that there was a melted metalic core at the center.

However, the core is so small that it is difficult to estimate its dimension. Accordingly, we assume

the Moon to be a rocky spherical body with a constant density = 3.3103 kg m3 to roughlyestimate the gravitational acceleration and pressure at depths8. In the case of a constant density,

the integral in Eq. (3.43) results inM(r) =4r3/3, and we obtain

g(r) = 4

3 Gr. (3.47)

8The Moon has a less dense crust than the mantle. However, the mean thickness of the crust is less than 100 km, two

orders of magnitude smaller than the radius of the Moon, which is about 1740 km. Accordingly, the crust is neglected in this

estimation.


15/41

3.6. SELF GRAVITATIONAL, SPHERICALLY SYMMETRIC BODY 73

Figure 3.7: Pressure p and gravitational acceleration gversus distancer from the center of a spher-

ically symmetric body. (a) Single-layer model Moon with a constant density with the parameters

= 3.3103 kg m3 and R =1740 km. (b) Two-layer model Earth with the parameters m =4103kg m3,c =10 103 kg m3,Rc =3490 km, andR = 6380 km.

It follows that gravitational acceleration for a constant density, spherically symmetric body is pro-

portional to the distance from the center (Fig. 3.7(a)). Substituting Eq. (3.47) into (3.46), wehave

dp

dr =4G

2

3 r.

Using the boundary conditionp = 0 atr =R, whereR stands for the radius of the Moon, we obtain

the pressure distribution

p(r) = 2

3 2G(R2 r2). (3.48)

The density distribution in the Earth is roughly simulated by a two-layer model, as about half

the radius of the Earth is occupied by a high density, metalic core at the center and the crust isnegligibly thin compared to the core and mantle. We do not take into account the tiny density

difference between the inner and outer cores. LetR and RC be the radius of the Earth and the core,

respectively, then the density is

=

C (rRC)m (RC < rR).

(3.49)

Gravitational acceleration in the core is the same as Eq. (3.47) and we have

g(r) = 43

GCr (rRC),


16/41


where Cis the density of the core. If that of the mantle is m, gravitational acceleration in the mantle

is obtained by using Eqs. (3.44) and (3.49) as

g(r) = 4G

3

mr + (C m)

R3C

r2

(RC < rR).

Figure 3.7(b) shows g(r) for a double-layered model Earth. It is seen in this figure that gis roughly

constant in the upper mantle. The models of tectonics in the following sections of this book deal

with the crust and the upper mantle, so that we will always assume that gravitational acceleration is

constant.

Pressure distribution in the model Earth is given by integrating Eq. (3.46) as

p(r) = 4

3 Gm(C m) R3C

1

r 1

R

+

2

3 G2m

R2 R2C+

2

3 G2C

R2C r2

(r < RC) ,

p(r) = 4

3 GmR

3c(C m)

1

Rc 1

R

+

2

3 G2m

R2 r2

(RC < r < R) .

The gravitational acceleration and pressure in the model Earth are calculated with the values, m =

4.0

103 kg m3, core = 10

103 kg m3, Rc = 3490 km, R =6380 km, and G = 6.7

1011

Nm2 kg2, and are shown in Fig. 3.7. It is seen that gravitational acceleration is more or less constant

in the upper mantle. Therefore, the acceleration is always assumed to be constant in the following

sections of this book.

3.7 Isostasy

Buoyancy is an important force to drive tectonics. The crust has smaller density than the underlying

mantle so that the crust is subject to buoyancy forces from the mantle and is floating on it. The

tectonic thinning and thickening of the crust cause the subsidence and uplift of the surface. We will

derive Archimedes principle from the force balance equation.

Suppose a solid body is immersed in a fluid with a density of(Fig. 3.8). The body and fluid are

assumed to stand still so that all forces acting on the body are balanced. The force balance equation

(Eq. (3.20)) in this case is

0 =

S

t(n) dS+

V

XdV, (3.50)

whereV andSare the volume and surface of the body. The first and second terms on the right-hand

side of this equation represent surface force from the fluid to the body and gravitational force (Eq.

(3.19)), respectively. The buoyancy force exerting on the body is the former. The traction at the

surface is pressure,p. That is,t(n) =pn. Using Gausss divergence theorem, the surface integral is


17/41

3.7. ISOSTASY 75

Figure 3.8: Schematic illustration to explain Archimedes principle.

converted to a volume integral. Therefore, the buoyancy term is rewritten asS

pndS=

V

p dV. (3.51)

A minus sign is placed on the right-hand side of this equation because the unit normal n is defined

inward, unlike the case of Eq. (C.63). It should be noted that the integration on the right-hand side ofEq. (3.51) is taken inside the body, althoughpis not the pressure in the body but fluid pressure at the

surface. Since the fluid is at rest, the pressure depends only on the depth z. The pressure gradient,

p, and the gravity term in Eq. (3.50) have no horizontal component. Therefore, an equation ofscalar quantities is enough to consider the vertical force balance of this system. LetFbbe buoyancy,

and be constant, then Eq. (3.50) is converted to the equation

Fb =

V

dV =V,

where the negative signs came from the definition of the force Fb that is positive upward. Thisequation stands for Archimedes principle: an object completely immersed in a fluid experiences an

upward buoyant force equal in magnitude to the weight of the fluid displaced by the object.

The concept of isostasy is derived from that principle. Suppose the structure shown in Fig. 3.9,

where a continental crust with a thickness oftc is floating on a mantle. According to Archimedes

principle, the buoyancy force exerted at the base of the crust is the weight of the mantle displaced

by the part of the crust that is immersed in the mantle. The weight is mghcS, wherem the mantle

density and S the area of the base. Other symbols are shown in Fig. 3.9. The gravitational force

that the crustal block experiences is cgtcS, which should be balanced with the buoyancy force.

Therefore, we have

cgtc =mghc. (3.52)


18/41


Figure 3.9: A simple model for a continental crust floating on a mantle.

This equation says that the lithostatic pressure at the level of the base is spatially constant. If densityvariations in the upper mantle are not negligible, this level should be taken to be deeper. The litho-

spheric plate that is composed of a crust and upper mantle is often envisaged as floating on a fluid

asthenosphere. Rocks behave as fluid in a geological timescale, if they have high temperature. On

the other hand, if the lower mantle is heated to become fluid, upper crustal blocks with smaller den-

sities float on it. The level is placed in the middle of the crust. The level under which rocks behave

as fluid and they cannot support a lithostatic pressure difference is called the depth of compensation.

A simple model of isostasy asserts that overburden is constant at a depth of compensation. The

following equation represents this concept:

dcsurface

gdz =constant, (3.53)

where dcis the depth of compensation, and the z-axis is defined downward with the sea level at z = 0.

If an offshore region is considered, the lower bound of this integral is taken to be at sea level to take

into account the load of the ocean water. The integral is spatially or temporarily constant. When it

is thought to be temporarily constant, we can calculate subsidence and uplift accompanied by the

temporal variation of density distribution at depths, i.e., vertical movements that are recognized from

stratigraphy give constraints to the variation.

Equation (3.53) represents a one-dimensional model of isostasy, where density variations only

along the z direction are taken into account. This idea is sometimes referred to as local isostasy

compared to regional isostasy that is introduced in Section 8.1. In the latter model, horizontal density

variations are explicitly included in its equation.

In the rest of this chapter we will use a simple isostatic model to calculate topography and vertical

tectonic movements.

3.8 Balance between ocean and continent

There are two tectonic classes of regions, continents and oceans. This division is reflected in the

frequency distribution of the height or water depth of the surface of the solid Earth (Fig. 3.10).


19/41

3.8. BALANCE BETWEEN OCEAN AND CONTINENT 77

Figure 3.10: Cumulative frequency distribution of the altitude of the surface of the solid Earth.

There are apparently two modes at about 1 km above and 34 km below sea level. They correspond

to the average altitude of the continents at 0.84 km and the average depth of oceans at 3.4 km.

Let us see that the two levels are in isostatic equilibrium [78]. Assuming g = 10 ms1 and

the depths and densities shown in Fig. 3.11(a) for representative continent and ocean, we have the

overburden at the Moho depth under the continent

pL =cgto =0.98 GPa. (3.54)

The overburden under the ocean basin at the same level is

pL =wgb + ogto + mg (tc h b to)1.006 GPa, (3.55)

which is about the same as that under the continent (Eq. (3.54)). Therefore, continents and oceans

are roughly in isostatic equilibrium.

Seawater pushes the oceanic crust downward by its weight, and pushes the continents upward.

What if the Earth loses oceans by extreme global warming? The average continental altitude from

the average oceanic surface,x in Fig. 3.11(b), would have been decreased. Neglecting other factors

including thermal expansion of rocks, the overburden at the level of continental Moho is

pL =cgto + mg (tc to x) .

Equating this pressure and that in Eq. (3.54), we obtainx =(tc to) (m c)/m3.3 km.The frequency distribution of global topographic height, hypsometry, is used to investigate the

tectonics of terrestrial planets and moons. If impact cratering was the most significant process for

shaping the surface of a planetary body, the resultant frequency distribution would be something

like a normal distribution. In spite of the existence of numberless impact craters, the Moon has two

modes in its distribution. They correspond to highlands and maria, suggesting that this distinction

was formed through some global processes.


20/41


Figure 3.11: (a) Representative large-scale structure of the shallow part of the Earth. (b) A structure

without seawater.

The bimodal distribution for the case of the Earth reflects the density contrast of the continen-

tal crust to the oceanic crust and to the mantle. The density distribution with spherical symmetry

is obviously the most stable structure in consideration of potential energy. However, the Earth has

the dichotomy of continents and ocean basins. Plate tectonics is an important action to keep the

dichotomy. Surface processes, including erosion and sedimentation, are also important for redis-tributing rocks and sediments. The strength of the lithosphere also affects the distribution. The

continental lithosphere has mechanical strength so that plate convergence causes continental thick-

ening with a limiting average altitude. As the average altitude exceeds it, increased gravitational

force flattens the crust to lower the surface (3.11).

3.9 Sediment load

A thick sedimentary pile pushes its basement downward by its weight. Let us consider the effect of

sediment loading, using the thick sedimentary sequence resting on the Atlantic passive continentalmargin (Fig. 3.12) as an example. Sedimentary layers that lie beneath the continental shelf ac-

cumulated in sublittoral paleoenvironments9. The total thickness of the sediments reaches several

kilometers. The paleo water depth, which is called paleobathymetry, and the thickness indicate that

the basement has subsided by the same amount as the thickness. Some old literature explained the

subsidence by the increasing sediment load. Sediments have greater densities than water, and sedi-

mentation replaces water by sediments. Therefore, sedimentation increases loading to the basement.

Namely, sedimentation created the sedimentary basin.

Now we are ready to deny this idea for the case of passive margin subsidence. Consider, for

9Sublittoral zone lies immediately below the intertidal zone and extending to a depth of about 200 m or to the edge of the

contiental shelf [2].


21/41

3.9. SEDIMENT LOAD 79

Figure 3.12: Crustal structure of the southern Baltimore Canyon Trough, Atlantic margin of North

America. LJ, Lower Jurassic; MJ, Middle Jurassic; UJ, Upper Jurassic; LK, Lower Cretaceous; UK,

Upper Cretaceous; T, Tertiary. Simplified from [101].

Figure 3.13: Sedimentary basin subsidence by sediment load.

simplicity, a constant density for sediments s. If an ocean basin with an initial depth ofb is filled

up with sediments (Fig. 3.13), the final thickness of the sedimentary pile x is given by the isostasy

equation

wgb + cgtc + mg (x b) =sgx + cgtc.

The left- and right-hand sides of this equation correspond to the left and right columns, respectively,

in Fig. 3.13. Therefore, we obtain

x =

m wm s

b. (3.56)

The density of sediment depends on the lithology. Using a representative value s = 2.3103kg m3, we have x = 2.3b. If excessive sediments are supplied from the hinterland, the sedimentswould be exported to the lower reaches. For the case of the thick shallow marine sequence on the

passive continental margin, b 0.2 km results in x 0.5 km. Therefore, the basement subsidedactively, not passively, because of the sediment load (3.14).

It is necessary to grasp subsidence history, and not only the initial water depth and final sediment


22/41


thickness, to understand the mechanism of subsidence. Quantitative stratigraphy provides a way to

reconstuct the history.

3.10 Quantitative stratigraphy

The subsidence history of a sedimentary basin is an observable function of time.Quantitative stratig-

raphyor geohistory analysisis the technique used to reconstruct the history from formations, where

isostatic balance plays an important role. The result of the analysis designates the vertical component

of the tectonic movements.

For this purpose, the necessary data are the present depth, lithology, porosity, sedimentation age,

the paleobathymetry of formations, and the eustatic sea level curve for the period covering the ages

of the formations. The paleobathymetry is estimated from sedimentary structures characteristic for a

certain depth such as hammocky cross-stratification, which indicates the depths were so shallow that

the sedimentation was affected by surface waves. Some kinds of molluscs and benthic foraminifers

live on the ocean bottom at certain water depths so that benthic fossils are another indicator of

paleobathymetry. Figure 3.14 shows the paleo topography inferred from those kinds of data around

Japan. The Japanese island arcs experienced vertical movements on the order of kilometers in the

Miocene [263, 272].

The basic idea of geohistory analysis is that the sum of the formation thickness and paleo-

bathymetry gives the depth to the basement from the present sea level, and applies several cor-rections. There are three key factors for the accumulation of thick sedimentary pile. Firstly, tectonic

subsidence is necessary to form a container of sediments. Secondly, sediment supply is impor-

tant. If the supply is scarce, a sediment-starved basin is the result. Thirdly, eustasy affects not

only paleobathymetry but also sedimentation. A eustatic rise is equivalent to tectonic subsidence

for sedimentation. We can abstract vertical tectonic movements from a sedimentary basin by taking

into account these and additional factors. The mathematical procedure employed for this purpose is

comparable to the stripping of formations from the top to lowest stratigraphic horizons (Fig. 3.15),

so that it is known as the backstripping technique.

The density of rocks is important for isostasy. Sediments lose their porosity and get a larger

density while they are buried. Pores between sedimentary particles are usually filled with forma-

tion fluids that are mostly water. For example, mudstone has a porosity of some 50% shortly after

deposition, However, increasing compactness of the particles expels the fluids during burial, i.e.,

compactionoccurs. The porosity is the volume fraction of a pore. The porosity of mudstone is

less than 10% at depths of a few kilometers. The total volume of sedimentary particles in a unit

volume is 1. It is known that porosity of sedimentary rocks decreases roughly exponentiallywith depth,

= 0eC, (3.57)

where is depth from the ocean bottom. The parameters 0and Cdepend on lithology. Accordingly,

it is assumed that the parameters are determined for rock types at depths, e.g., in boreholes.


23/41

3.10. QUANTITATIVE STRATIGRAPHY 81

Figure 3.14: Paleo topography around Japanese islands from 23 to 15 Ma inferred from sedimentaryenvironments. Benthic fossils and depositional facies are the keys for estimating the environment.

The Japan arc drifted to form the Japan Sea backarc basin in the Early Miocene. The paleo posi-

tion of the arc before 15 Ma, when the opening was completed, is controversial so that the paleo

environments are indicated on the present map of this region.

If a formation has a horizonally infinite extension, the porosity reduction results in the thinning of

the formation. Consider a formation with an initial thickness ofd0 decreasing todNby increasing

burial depth fromd0 todN. A column of sedimentary rocks with a unit sectional area has the total

volume of sedimentary particles d0+d0d0

(1

) d between the depths d0 andd0 + d0. Pore fluid

is expelled from this part of the column by burial, but the particles are not. The conservation of


24/41


Figure 3.15: Schematic illustration to explain the backstripping technique. The left column shows

the present state. The central and right columns indicate the situation just after the ith and (i + 1)

layers deposited, respectively.

particles is indicated by the equationd0+d0d0

(1 ) d=dN+dN

dN

(1 ) d.

Substituting Eq. (3.57), we have

d0 +0

C eCd0

eC d0 1=dN +

0

C eCdN

eC dN 1

. (3.58)

The parameters Cand 0are known for specific rock types. Therefore, if the thickness of a formation

dN when the depth of the formation wasdN is given, we can calculate the thickness d0 at a givendepth of d0 through Eq. (3.58). That is, this equation corrects the effect of compaction on the

thickness of a formation. In addition, it is easy for us to grasp the present depth and thickness of a

formation. The ancient depth of the formation is obtained as the total thicknesses of the overlying

formations. Therefore, the right-hand side of this equation is known for the youngest formation that

occupies the top of the sedimentary pile. Accordingly, Eq. (3.58) works as a recursive function

to calculate ancient thicknesses which are determined successively from the top to the base of the

layers. This procedure is calleddecompaction, which increases layer thickness.

The next task is the correction of sediment loading. Let us assign a number for each layer from

the top (Fig. 3.15). The mass of theith layer includes those of pore water iw and sedimentary

grains (1 i)g, wherei is the porosity of the layer and g is the average density of the grains. If


25/41

3.11. TECTONIC FORCE CAUSED BY HORIZONTAL DENSITY VARIATIONS 83

diis the thickness of this layer just after it was deposited, we obtain diusing the recursive equation

Eq. (3.58). The mass of the layer per unit basal area is

mi =

iw + (1 i) g

di. (3.59)

If the lowermost layer has the ordinal number n, the sedimentary column with a unit sectional area

has the mass of

Mi =mi + mi+1 + + mn1 + mn (3.60)

when theith layer was deposited. Consider that the Moho is displaced downward by the weight of

the ocean water and theith layer. If the displacement ishi, we have the equation

w(bi ei) + Mi =w(bi+1 ei+1) + Mi+1 + mhc (3.61)

from isostasy10, where bi is the water depth and ei is the level of the ocean surface relative to the

present one when the ith layer was deposited. That is, the sequence{en, en1, . . . , e0} representsthe eustatic sea level changes through time. They are estimated from ancient marine transgression

and regression observed around stable continents11. Combining Eqs. (3.58)(3.60), we obtain Mi.

Therefore,hc is calculated through Eq. (3.61).

We are able to estimate the ancient depth of burial for every formation from a sedimentary record

using this backstripping technique, and to quantify the history of tectonic subsidence. Quantitative

stratigraphy has a precision of100 m for a sedimentary sequence that was deposited near sea level.Not only subsidence, but also uplift can be evaluated with this method. A sedimentary pile

records the uplift as a decrease of paleo water depth or regression. We are able to quantify these

phenomena, provided that sedimentation lasted during the uplift. If vertical movement occurred

above sea level, fission track thermochronology is useful to evaluate uplifts of the order of kilo

meters. Recently, a geochemical signature of continental carbonates was used to estimate paleo

altitudes [66].

Figure 3.16 shows the subsidence history of the continental shelf offthe Grand Banks of Canada

determined by the backstripping technique. The vertical movements of the basement is illustrated

by a subsidence curve. The subsidence of passive continental margins is modeled as the surface

manifestation of post-rift lithospheric cooling (3.14). Kominz et al. [103] exhibited a eustatic sea

level curve from the subsidence curves of passive margins combined with the cooling model.

3.11 Tectonic force caused by horizontal density variations

Assuming the lithostatic state of stress, horizontal forces in the lithosphere are discussed in this

section. Under this condition, lithostatic pressure is proportional to depth z, and the constant of

10We asssume that the crustal thickness except for the sedimentary pile does not change so that the thickness does not

appear in Eq. (3.61). Local isostasy is assumed here, but a technique called flexural backstripping has been developed to

account for the strength of the lithosphere [74].11Epeirogeny (9.5) is an important but serious problem for this purpose.


26/41


Figure 3.16: Subsidence history of the continental shelf off the Grand Banks, Canada [1]. The

present sea level is the origin of the vertical axis. Ages on the right side of this graph depict the

depositional age of selected stratigraphic horizons.

Figure 3.17: Diagram showing the lithostatic pressure versus depth for high and low density rocks.

proportionality depends linearly on density (Fig. 3.17). We have an interesting consequence fromthis simple model if there is a horizontal density variation.

Continents and oceans are isostatically compensated. If a depth of compensation is placed at

the continental Moho, the depth dependence of the lithostatic pressures under a continent and an

ocean basin are illustrated in Fig. 3.18(a), which is drawn with the same structure as shown in Fig.

3.11(a). Namely, the pressure is greater under the continent than under the ocean around the sea level

because of the topographic bulge of the continent and of the density difference of the continental

crust and seawater. However, the lithostatic pressure under the ocean catches up with that under the

continent at the compensation depth. Pressure difference generally drives deformation. Accordingly,

the difference between the lithostatic pressures produces tectonic force, which horizontally extends

continents and constricts ocean basins (Fig. 3.18(c)).


27/41

3.11. TECTONIC FORCE CAUSED BY HORIZONTAL DENSITY VARIATIONS 85

Figure 3.18: (a) Lithostatic pressure under a continent and an ocean basin that are isostatically

compensated. It is assumed that there is no density variation in the crust and in the mantle. (b)Vertical profile of density contrast between the regions. (c) Tectonic flow driven by the pressure

difference.

A continent pushes an ocean basin with a force that is designated by the gap between the line

graphs of the lithostatic pressures under the regions. The gap depends on the profile of the density

difference, which is schematically shown in Fig. 3.18(b). The density profile exhibits deviations

with the opposite signs at two levels. The dipole moment is the magnitude of positive and negative

anomalies multiplied by the distance between the anomalies. Accordingly, the gray area in Fig.

3.18(a) increases with the increasing dipole moment of the density profile. As the continental crustis thickened, both the average altitude and the Moho depth increase. They expand the gap, and

further strengthen the horizontal force (Fig. 3.18(a)). Consequently, a continent pushes oceans with

a force of 1.61012 Nm1 = 1.6 TNm1 per unit length of the continent-ocean boundary, if thestructure shown in Fig. 3.11(a) is assumed. This value is obtained by the downward integration of

the gap between the pressures. The force is as great as other tectonic forces. That is, large-scale

variation of topography can drive tectonic deformation (12.2).

The same argument also applies to the horizontal force accompanied by a large-scale variation

in altitude. That is, highlands tend to extend horizontally and pushes the neighboring lowland.

The extensional force by this effect builds up with an uplift of the highland. The result is that the

altitude of an extensive highland cannot exceed a certain limit. Gravitational collapse occurs when


28/41


Figure 3.19: Viscous fluid has a level surface at rest. However, a moving belt installed on the bottom

of a container produces a slope at the surface.

the highland is uplifted above the limiting altitude.

The Tibetan Plateau has been uplifted to an altitude of5 km by NS cotinental convergence, andthe crustal thickness reaches 70 km. EW trending thrust faults are dominant along the Himalayas,but there are NS or NNESSW trending normal faults and strike-slip faults oblique to the normal

faults in the plateau [12, 240]. The crustal thickness is attenuated rather than increased in the plateau,

and the normal and strike-slip faulting transfers crustal blocks eastward, suggesting that the plateau

has reached the limit12.

The limiting altitude depends on the strength of the continental lithosphere, which has temporal

and spatial variations13. To see this, consider viscous fluid like honey in a container that has a

moving belt at its bottom. This thought experiment models a convergent orogen, where the fluid

simulates the continental crust. The belt drags the fluid to form a slope at the surface of the fluid

(Fig. 3.19). The slope is dynamically maintained by the competing factors, the shear traction and

gravitational spreading caused by the slope. The driving force of the latter results from the horizontal

density difference of the fluid and air across the slope. The inclination of the surface is enhanced

by the shear stress caused by the belt, and the shear stress increases with increasing viscosity. The

ratio of the gravitational and viscous forces is called the Argand number [56] and indicates the

tendency of gravitational spreading of large-scale positive undulations. Several factors including

lithology, thermal regime and pore pressure, control the effective viscosity of rocks (10.8). Since

these factors may have temporal and spatial variations, the limiting altitude has these variations, also.

If the lithosphere is soft or softened, topographic bulges on the lithosphere collapse.

3.12 Thermal isostasy

In this section we consider ocean-floor subsidence (Fig. 3.20) by combining the isostasy and thermal

evolution of the oceanic lithosphere.

The solid Earth discharges energy mostly through the cooling oceanic plates. Thermal conduc-

tion is the primary mechanism of heat transfer in the lithosphere 14. As rocks have low thermal

12The Himalayas have higher altitudes than the plateau. The peaks are supported not only by the buoyancy force associated

with isostatic compensation but also by the flexural strength of the lithosphere.13See Chapter 12.14The Galilean Satelite, Io, is a remarkable extra terrestrial body. The moon is as the size of the Moon, but discharges with


29/41

3.12. THERMAL ISOSTASY 87

Figure 3.20: Depthage relationship of ocean basins [228]. The vertical axis shows the moving

average with a window of 5 m.y. of the depth that is corrected for the load and thickness of sedimen-

tary cover. GDH1, PSM and HS are synthesized depth-age curves based on different models. PMS

assumes the cooling oceanic plate with the constant thickness and constant basal temperature [174].HS is the curve calculated by the cooling half-space with the thermal properties derived from PMS.

GDH1 is the compromised model that assumes a cooling half-space for the lithosphere younger than

20 m.y. and the plate model for the older one [228].

conductivity, the conduction is associated with steeper temperature gradients in the lithosphere than

in the asthenosphere where heat is transferred by convection. That is, the lithosphere is the thermal

boundary layer of the convecting mantle. Rocks lose their ductile strength with increasing tempera-

ture. Therefore, the lithosphere is the cold and hard lid of the convecting cells.

The thickness of the thermal boundary layer changes with cooling and heating. The density

of rocks depends on temperature so that a change in the temperature field affects isostatic balance.

This is known as thermal isostasy. Therefore, the thermal regime of the lithosphere affects the

vertical movement of ocean basins. The cooling lithosphere causes subsidence. To understand this

phenomenon, we firstly consider the heat conduction equation.

Conduction of heat

A large rock mass has a large thermal capacity so that temperature change takes a long time in the

mass. Consider a mass with the diameterL at a standstill, The temperature change is described by

the same great energy as the Earth and most of the energy is emitted from its hotspot volcanoes [135].


30/41


Figure 3.21: Diagram showing the model of cooling oceanic lithosphere.

theheat conduction equationT

t =2T, (3.62)

whereTis the temperature that is a function of time t and position, and is thethermal diffusivity

of the rock mass. If heat is conducted only in thex direction, this equation reduces to

T

t =

2T

x2. (3.63)

The representative value for the thermal diffusivity of rocks is 1 m2 s1. This unit designates that

/L2 has the dimension of time. Accordingly, we define the dimensionless length xand time tby

dividingL and L2/and the dimensionless temperature T by dividing by a reference temperature

T0. Namely,

x = x/L, t = t/L2, T =T /T0,

dx =Ldx, dt = L2dt/, dT =T0dT .(3.64)

Using these dx, dtdT, we rewrite Eq. (3.62) in the form

T

t =

2T

x2 (t >0, 0< x


31/41

3.12. THERMAL ISOSTASY 89

is placed atz =0, i.e.,z = . For simplicity, we assume that heat is transferred only by conduction.

To calculate temperature change, the thermal conduction equation,

2T

t2 =

2T

2

is solved with the initial and boundary conditions

T =Ta at 0, t = 0,T =T0 at =0, t >0,

T Ta as , t >0.

That is, we assume a cooling half-space with the constant temperatures T0

and Ta

at the ocean floor

and at the deep mantle, respectively. As the lithosphere ages, the ocean floor subsides and departs

fromz = 0 by a few kilometers. Therefore, = 0 is not the level of the floor, but this departure is

neglected in the above conditions because it is negligible compared to the dimension of the mantle.

After all, we have the solution of this problem,

T T0Ta T0

=Erf

2

t

, (3.65)

where Erf( ) is the error function(Fig. 3.22) defined by the equation

Erf(x) =

2

x0 e2 d (x0) . (3.66)The graph of the error function designates that the shallow and deep parts of the suboceanic mantle

have steep and gentle geothermal gradients (Fig. 3.22). The turning point gradually goes down

in the mantle with the denominator,

t, in the operand of the error function in Eq. (3.65). The

denominator has the dimensions of length so that

t is called the diffusion length. Obviously, an

isotherm descends with age as t.The error function has an inverse function. Therefore, the depth of the isotherm with a specific

temperatureTis given by Eq. (3.65), where the left-hand side is known. Substituting t = x/v and

=x into the equation, we have

T T0Ta T0

=Erf

z

2

v

x

. (3.67)

The thermal evolution has been investigated in the above argument. Now we can calculate the

subsidence. It is know that the density of rock at the temperature T, which is measured from the

reference temperatureT0, changes as

=0(1 T), (3.68)

where 0is the density at the reference temperature andis thethermal expansion coefficient. Rocks

have a representative value for the coefficient at about 3 105 K1. We usea =m0(1 Ta) for


32/41


Figure 3.22: Graphs of error function Erf(x) and complementary error function Erfc(x).

the density of the asthenosphere. Ifb is the depth of the sea floor from the level =0, the isostasy

equation gives

wb +

dc0

d=const. (3.69)

The value of this equation is evaluated at the spreading center, where the asthenosphere comes

immediately below the ridge axis. Therefore, we obtain the value of the above equation there,

adc + wbR, wherebR is the water depth of the axis. Accordingly, Eq. (3.69) is rewritten as

wb +

dc0

d=adc + wbR. (3.70)

We assume the half-space so that the compensation depth can be treated as dc =. Combining Eqs.(3.70), (3.67) and (3.68), we obtain

b(a w) =m0(Ta T0)

0 1 Erfz

2 v

x dz=m0(Ta T0)

0

Erfc

z

2

v

x

dz

Erfc(x) =1 Erf(x) is thecomplementary error function (Fig. 3.22). This function satisfies0

Erfc(x) dx = 1

,

therefore, the depth of the sea floor at a distance ofxfrom the spreading axis is given by the equation

b = 2a(Ta

T0)

a w xv + bR. (3.71)


33/41

3.13. HORIZONAL STRESS RESULTING FROM TOPOGRAPHY 91

The depth increases with the square root of the distance, as we have assumed a constant spreading

rate at the ridge center.

We are able to estimate the effective values for the thermal diffusivity and thermal expansion

coefficient of the oceanic mantle by comparing the above model with observations. However, it is

important that old sea floors are shallower than the depth predicted by the model (Fig. 3.20) and the

ocean basins have different deviations. Some factors other than the cooling are responsible for this

flattening. It is pointed out that reheating by hotspots cannot account for the deviations [198].

3.13 Horizonal stress resulting from topography

Large-scale topographic undulation is one of the sources of tectonic stress. Ocean floors show the

difference in their depth up to a few kilometers (Fig. 3.20) so that great stress may be generated.

Following Dahlen [47], let us consider how large is the stress level associated with the variation of

the depth of the ocean floor. In conclusion, theridge pushis the force resulting from the topography.

The force is important among other origins of tectonic stresses including the colliding force of plates.

The asthenosphere comes immediately below the spreading axis, therefore we assume a constant

density 0under the axis. The coordinates O-xz shown in Fig. 3.21 are used in this section. Consider

the density distribution

(x, z) =0 + (x, z),where the last term is the density anomaly from the reference density 0. According to Eq. (3.65),the temperature and the resultant density anomalies were derived in the previous section as

T(z, t) =TaErf

z

2

t

,

(z, t) =m0TaErfc

z

2

t

. (3.72)

The forceFR that a unit length of the ridge axis is determined by the dipole moment of vertical

density distribution (3.11). The moment is obtained from Eq. (3.72) so that

FR =

0

gz dz = m0gTa t.Consequently, the force is proportional to the age of the oceanic lithosphere15. The constant of

proportionality is about 0.035 TNm1/m.y. For example, the ocean floor with an age of 80 m.y.

has FR = 2.8 TNm1. The force resulting from the ocean-continent topographic contrast has a

magnitude of about 1.6 TNm1 (3.11). Therefore, the ridge push is as great as the force by the

contrast.

15The integral is taken over the range [0,

]. However, the contribution from the asthenosphere is negligible because of

Erfc(x) 1 forx >2 (Fig. 3.22).


34/41


In the above argument, we used the density profile under the spreading axis as the reference.

It is the next problem where the proper area that provides the reference is sought. If the density

distribution was under the summit of Mount Everest, the entire world would have been under a

compressional stress field. However, this is not the case. Accordingly, Coblentz and others excluded

the arbitrariness in the choice of reference by taking global topography into account [42].

Let the levels of the top and base of the lithosphere be z = h and z = L, respectively. Note

that the vertial force per unit area is zz (1 m2) and that the product of force and length has thedimensions of energy. Then the quantity

U= L

h

zzdz = L

h h

()g ddz

is the potential energy of the lithospheric column with the unit sectional area. A region with a large

U tends to be subject to extensional tectonics. That with a small U is ready to be horizontally

contracted. The global average is estimated at 2.4 TNm1 [42]. The threshold altitude for thecontinent is200 m, and the threshold depth of the ocean is about 4.3 km. Spreading centers havedepths actually shallower than this threshold. However, the topography is not the only cause of

lithospheric stress. Therefore, an extensional stress field rifts continents and eventually breaks up

the continental lithosphere.

3.14 Thermal subsidence of continental rifts

It was shown in the previous chapter that the subsidence history of a passive continental margin is

revealed from stratigraphic records. Here we consider Mckenzies simple stretching model to link

the subsidence and rifting together.

Post-rift subsidence

Passive margin subsidence after continental breakup is generally associated with very weak or no

fault activity. Extensional tectonics does not account for the subsidence. In contrast, the continental

lithosphere is thinned and horizontally extended during the rift phase. The fact is that the riftingaccounts for the post-rift subsidence.

Consider the thermally equilibrium state of the lithosphere illustrated in Figs. 3.23(a) and (c),

i.e., the temperature at the base of the lithosphere is Ta at a depth of z = tL. The temperature

distribution before and long after rifting is assumed to be in equilibrium. The continental crust is

thinned with the lithospheric mantle. The thinning of the crust results in subsidence. Let tcandtLbe

the initial thickness of the crust and the lithosphere, respectively (Fig. 3.23(a)). It is also assumed

that the temperature atz =tLis kept atTa. The top of the lithosphere was at sea level before rifting.

Here, the lithosphere is thought to be subject to homogeneous deformation in the rift phase. If

the lithosphere is horizontally extended by a factor of, the thickness of the lithosphere and crust

becometc/ andtL/, respectively. The rift-stage subsidence is affected not only by the thinning


35/41

3.14. THERMAL SUBSIDENCE OF CONTINENTAL RIFTS 93

Figure 3.23: Diagrams showing the simple stretching model [138]. (a) Structure and temperature

distribution in the lithosphere in the pre-rift stage, (b) at the end of rifting, and (c) in post-rift stage.

but also by the temporal variation of the thermal regime, because rock density changes with the

variation as

= 0(1 T), (3.73)

where0is the reference density atT =0 C. If the thinning of the lithosphere is quick compared to

the time constant of heat conduction, isotherms in the lithosphere are uplifted adiabatically16 (Fig.

3.23(b)). If we assume isostatic compensation throughout the rifting process, we obtain the syn-rift

subsidence

Si = tLa tcc (tL tc)m

a w

1 1

, (3.74)

where c, m anda are the representative densities of the crust, mantle lithosphere, and astheno-

sphere, respectively, and are given by

a =m0(1 Ta), c =c0

1 Ta2

tc

tL

, m =m0

1 Ta

2 Ta

2

tc

tL

.

16This assumption is valid for the rifting with a duration shorter than 25 m.y. [92].


36/41


Table 3.2: Parameters for the simple stretching model.

m0 =3.3 103 kg m3 = 3.5105 K1 tL = 100 kmc0 =2.8 103 kg m3 = 1.0106 m2 s1 tc = 35 kmw =1.0 103 kg m3 Ta = 1333 C

The coefficients, c0 and m0 are the reference densities of the crust and mantle. Si is called the

initial subsidencein the passive margin formation. Substituting the representative densities into Eq.

(3.74), we obtain the magnitude of the initial subsidence

Si =Si0

1 1

, (3.75)

Si0 =

(m0 c0)

1 Ta2

tc

tL

tc

tL m0

Ta

2

tL

m0(1 Ta) w. (3.76)

The latter quantity, (Si0) can have positive and negative signs depending on the initial thickness of

the crust. Since the denominator in the right-hand side of Eq. (3.76) is positive, the sign depends on

the numerator. Solving the equation, (numerator) =0, we obtain the solution,

tc

tL=

m0 c0(m0 c0)Ta

4(m0 c0)2 + 42(m0 c0)m0T2a (m0 c0)Ta

.

Using the parameters shown in Table 3.2, we obtain the ratiotc/tL =0.154 and 42.7. However, this

ratio should be less than unity so that the latter is a meaningless solution. The former indicates a

threshold initial thickness oftc 15 km. The lithosphere with a continental crust initially thickerthan15 km subsides through the lithospheric thinning, but the surface is uplifted by rifting if thecrust is initially thinner than this value. The uplift is due to a steepening of the geothermal gradient.

That is, the attenuation of the lithospheric mantle plays an opposite role to that of the continental

crust. Island arcs have volcanism so that their lithosphere is thinner than the ordinary continental

lithosphere. If an island arc has a lithosphere with a pre-rift thickness oftL = 50 km, then the archas the threshold at about 7 km. The geohistory analysis of syn-rift sediments can constrainwith

Eq. (3.75). This parameter is called thestretching factor.

The thermal regime is altered by the rifting. Isotherms between 0andTaare raised by the rifting(Fig. 3.23(b)). After the rifting, the isostherms descend to the equilibrium level in the long run (Fig.

3.23(c)). This post-rift cooling causes a gradual subsidence of the passive margin. This is called the

thermal subsidenceof the lithosphere. The simple stretching model (Fig. 3.23) allows us to estimate

the rate and amount of this subsidence [138].

In order to calculate the heat conduction after the termination of rifting, the initial subsidence is

neglected because Si

tL. Therefore, the thermal conduction equation T/t =

2T/z2 is solved

with the boundary conditionT =0 atz = 0. In addition, the temperature atz =tLis assumed to be


37/41

3.15. KINEMATIC MODEL OF SYN-RIFT SUBSIDENCE 95

kept atT = Ta. The timet is measured from the termination. The initial condition is shown by the

graph in Fig. 3.23(b), which is the temperature distribution at the end of rifting. Solving this system,

we obtain the thermal evolution

T

Ta=

1 z

tL

+

2

n=1

(1)n+1n

nsin

n

sin

nz

tLen

2t/, (3.77)

=2t2L

/. (3.78)

The parameters shown in Table 3.2 give the time constant =3.2 m.y.

Combining Eqs. (3.73) and (3.77), we obtain the evolution of the density structure (z, t) .

Isostatic compensation through time is expressed astL0

(z, t) dz + wSt =

tL0

(z, 0) dz,

whereSt is the thermal subsidence,

St

4m0TatL

2(m0 c0)

sin

1 et/

. (3.79)

Higher-order terms in Eq. (3.77) are neglected in deriving Eq. (3.79).

Equation (3.79) shows that the rate of thermal subsidence decays as the exponential function.

The subsidence curve shown in Fig. 3.16 has this shape. Fitting the curve given by Eq. (3.79), we

obtain the stretching factor for the area the curve was determined.

3.15 Kinematic model of syn-rift subsidence

Post-rift subsidence results from the cooling of the lithosphere, whereas syn-rift subsidence is con-

trolled by not only the thermal but also the mechanical properties of the lithosphere. The latter

affects the characteristics of rift zones such as their widths and depths. Decompression melting of

the asthenosphere under rift zones causes uplift through the buoyancy of magma. Therefore, the

observation of syn-rift subsidence sheds light on the deep seated factors and on the secular changesof these factors.

The simple stretching model (3.14) assumed an instantaneous thinning of the lithosphere and

neglected the details of the rifting processes. In this section, we consider a mathematical inverse

method to constrain the rift-phase strain rate of the lithosphere from an observed rift-phase subsi-

dence curve (Fig. 3.24) [259].

Consider a pure shear rift with horizontal and vertical strain rates of Exx and Ezz, respectively.

The lithosphere is extending horizontally so that we have Ezz < 0 < Exx. The pre-rift levels of the

top and base of the lithosphere is z =0 andz =a. The lithosphere is subject to a pure shear, so that

we have

zz = xx =G (t),


38/41


where the functionG(t) is defined to describe the temporal variation of the strain rates and is linked

to the rate of subsidence in the following discussion.

Consider that the horizontal velocity field is symmetric with respect to the plane x = 0. The

lithosphere is assumed to undergo homogeneous strain so that the horizontal velocity u(x, t) is pro-

portional tox. That is, we have

u(x, t) =G (t) x. (3.80)

Syn-rift subsidence reaches several kilometers, but is negligible compared to the representative

thickness of continental lithosphere. Therefore, we assume the top of the lithosphere is kept at z = 0

during the calculation of the lithospheric strain. This assumption allows us to write the descending

velocity of rocks at depthz as

v(x, t) =G(t) z. (3.81)Note that this is the velocity in the +zdirection.

Subsurface temperature distribution affects the subsidence of this rift. The thermal conduction

equation to be solved should include velocity, because heat is transported not only by conduction

but also by travelling rocks. Hence, the left-hand side of the thermal conduction equation, T/t =

2T, should be replaced by the material derivative (Eq. (2.20)). That is, the equation

T

t +

u

v

T =

2T

x2 +

2T

z2

(3.82)

governs the thermal evolution. From Eqs. (3.81) and (3.80), we have

u

v

T =G(t)x

T

x G(t)z T

z. (3.83)

If there was no horizontal variation in the initial temperature distribution, the pure shear would not

raise horizontal temperature variation. Therefore, we have

T

x =

2T

x2 =0.

Combining this and Eqs. (3.82) and (3.83), we obtain

T

t + Gz

T

z =

2T

z2 . (3.84)

The boundary conditions that we assume for the temperature evolution are T = 0 at z = 0 and

T = Ta atz = a. The initial condition and the equilibrium temperature distribution is the constant

geothermal gradient to a depth ofz = a.

The subsidence of this rift is the sum of the subsidence due to the thinning of the lithosphere

and to the thermal perturbation. The former is described by Eq. (3.75). The equation describing the

latter factor should have the form

Q(t) =Ba0T(z, t) T(z, ) dz, (3.85)


39/41

3.15. KINEMATIC MODEL OF SYN-RIFT SUBSIDENCE 97

Figure 3.24: (a) Strain rate history of a Triassic rift in northern Italy determined from a subsidence

curve of the rift (b). The 95% confidence region is shown in (a) by white [259].

whereB =m/(a w) is the factor to account for the buoyancy of the crust. Therefore, we havethe total subsidence

S(t) =A

1 1

BQ(t), (3.86)

whereA=

tc(m c)/(a w). Crustal thinning has a positive effect for the total subsidence, butthe thinning of the lithosphere increases the geothermal gradient which has a negative effect for the

subsidence. Rearranging Eq. (3.86), we have

= A

A S BQ(t) . (3.87)

Equation (2.39) links the stretching factor and G(t) as

=exp

tR0

G(t) dt

, (3.88)

wheretRis the time ellapsed since the rifting began. Accordingly, combining Eqs. (3.87) and (3.88),

we obtain the strain-rate history of the rift

G(t) = d

dt

ln

A

A S(t) BQ(t)

. (3.89)

The rift-phase strain-rate history can be evaluated from observations for a pure shear rift. The

right-hand side of Eq. (3.89) hasS(t) that is determined from the quantitative stratigraphy of the

rift. However,Q(t) depends on the unknownG(t) through Eq. (3.84). Therefore,G(t) is determined

by a mathematical inversion17 from the observed subsidence historyS(t). Figure 3.24(a) shows the

17See [259] for detail.


40/41


optimal strain-rate history for the subsidence history of a Triassic rift zone in northern Italy (Fig.

3.24(b)). The subsidence curve of the Triassic rift has an inflection point at about 220 Ma (Fig.

3.24(b)). After the point, the subsidence curve is similar to an exponential curve, suggesting thermal

subsidence. The point designates a rejuvenated rifting, and corresponding to the second peak in the

strain-rate history.

A sharp decline in the strain rate by several orders of magnitude indicates the termination of rift-

ing; the rift was aborted or the area became a passive margin. Timing of the termination determined

by the inverse method for the Triassic rift has a large uncertainty which is illustrated by the 90%

confidence region between 100 and 220 Ma in Fig. 3.24(a). The umbiguity stems from the difficulty

to detect very slow rifting if it overlaps simultaneous thermal subsidence. More definite constrains

come from geological structures. The termination of rifting is marked by the cessation or distinctdecline of fault activity. For example, the cross-section in Fig. 3.12 shows that the subsidence of

wedge shaped grabens had ceased by the early Jurassic. If rifting was successful to form a passive

continental margin, the magnetic anomaly near the continent-ocean boundary provides a constraint

for the timing. Some passive margins were remarkably uplifted when continents were broken offthe

margins. Such a unconformity is known as a breakup unconformity[59].

3.16 Exercises

3.1 Assume a state of stress with the principal stresses1=

4 MPa,2=

5 MPa and3=

8 MPa,and vertical1-axis and horizontal3-axis oriented due east. Determine the magnitude of the normal

and shear stresses acting on the block beneath a fracture plane of which trend and dip are N030 Eand 60SE, respectively. How much is the rake of the maximum shear stress on the plane?

3.2 Explain qualitatively why the functiong(r) shown in Fig. 3.7(b) is convex downward and why

the function decreases with increasingr in the lower mantle.

3.3 What is the magnitude of the lithostatic pressure at the Moho interface. Assume the thickness

and density of the crust at 35 km and 2 .8

103 kg m3, respectively.

3.4 Derive Eq. (3.46).

3.5 Estimate how much is the force per unit boundary length with which the Tibetan Plateau pushes

the Hindustan Plain due solely to the structure illustrated in Fig. 3.25.


41/41

3.16. EXERCISES 99

Figure 3.25: Lithostatic pressure under the neighboring plateau and lowland.

Stress, Balance Equations

Documents

Transcript of Stress, Balance Equations