Dynamics of the Lithosphere · years. The extensional tectonics that led to the opening of the Japan Sea backarc basin lasted 5 m.y. [263]. The diﬀerence is explained by the initial

Chapter 12

Dynamics of the Lithosphere

In this chapter we study how the lithosphere deforms in response to external forces, andhow factors such as the thermal regime and lithology affect the deformation. First, westudy several factors controlling the strength of the lithosphere. Then, the strength isrelated to the instability of the lithosphere that determines the fate of continental riftsand the creation of spatially periodic deformation zones.

12.1 Strength profile

We have studied several deformation mechanisms, including elasticity, brittle faulting, and ductileflow, that are relevant to describe the deformation of rocks. They exhibit different stresses for agiven tectonic condition with a specific depth, temperature, strain, strain rate, etc. When a rock massdeforms, it has options for the mechanisms. Then, the deformation mechanism that minimizes theresistance to the deformation is chosen. If the stress accompanied by brittle deformation is less thanthat of ductile deformation, the latter deformation mechanism is chosen. As the product of force anddistance is equal to energy, tectonic deformation proceeds with the minimum energy dissipation.

Consider large-scale deformations such as a continental breakup resulting from extensional de-formation of the lithosphere. In those cases we do not take elasticity into account because thosedeformations surpass the elastic limit. Brittle and ductile deformations are important. If the defor-mation of a rift can be treated as plane strain on the vertical section across the rift zone, the resistingstress is described by Eq. (6.21) as follows. No brittle faulting occurs if the differential stress,σh − σv, is less than the critical one1:

Δσbrittle = −[

2μf (1 − λf )

(1 + μ2f )1/2 + μf

]σv, (12.1)

where σv is the vertical stress that is usually equal to the overburden stress (Fig. 12.1). In the case

1Note that rifting occurs when σh < σv, so that σh − σv < 0.

303

304 CHAPTER 12. DYNAMICS OF THE LITHOSPHERE

of compressional tectonics, the critical stress is given by the equation

Δσbrittle = +

[2μf (1 − λf )

(1 + μ2f )1/2 − μf

]σv, (12.2)

and faulting occurs when the differential stress exceeds this. A rock mass breaks when it is subjectto stresses over this limit. Therefore, Eqs. (12.1) and (12.2) represent the brittle strength of therock mass under extensional and compressional tectonic regimes, respectively. The brittle strengthis insensitive to lithology.

The vertical stress, σv, is determined by overburden so that the critical differential stress de-scribes the magnitude of horizontal stress. If tectonic stress is gradually built up and the horizontalstress reaches the critical stress, faulting ‘immediately’ occurs in the geological sense. The brittledeformation relieves the state of stress under the critical state. Consequently, the state of stress isat or just below the strength under tectonically active regions: Eqs. (12.1) and (12.2) describe theapproximate magnitude of horizontal stress.

On the other hand, the stress due to ductile deformation depends on the lithology, strain rate, andtemperature. The deviatoric stress for the ductile deformation is given by the equation (§10.8.3),

s =

[A

− 1n ε

1n−1E exp

( Q

nRT

)]e, (12.3)

where εE is the equivalent strain rate

εE =

√12

e : e. (12.4)

Δσductile = τ1 − τ3 decreases rapidly with increasing temperature. If the lithology and strain rate donot change within a rock mass, temperature rises with depth and, consequently, Δσductile falls rapidly.

Because of the temperature increase with depth, the ductile strength decreases with it. Hence, theabscissa in Fig. 10.18 that displays temperatures can be translated into depths. Taking the horizontalx- and vertically downward z-axes, and assuming a pure shear

e =

⎛⎝εxx 0 00 0 00 0 εzz

⎞⎠ (εxx = −εzz),

we have εE =√

2 |εxx| from Eq. (12.4). If rocks are isotropic, we have Δσ = |τxx − τzz| and theductile strength

Δσ =√

2A− 1n∣∣εxx∣∣ 1

n−1exp

( Q

nRT

)∣∣εxx − εzz∣∣. (12.5)

Temperature T increases generally with depth, therefore Δσ decreases exponentially with it. Theductile strength strongly depends on the lithology, as Eq. (12.5) includes material constants A andQ. When differential stress applied to the lithosphere is smaller than the ductile strength or brittlestrength of rocks over the entire profile of the lithosphere, the lithosphere supports the stress byelasticity.

12.1. STRENGTH PROFILE 305

Figure 12.1: Schematic strength profile of the lithosphere with a uniform lithology. Brittle and duc-tile strengths are shown by solid lines. Rocks deform by brittle failure above the brittle-ductiletransition (BDT) because brittle deformation is easier than ductile deformation in that the formerneeds less differential stresses than the latter, whereas ductile deformation prevails under BDT. Thestrength of quasi-plastic deformation is shown by dotted lines. Since the brittle strength in exten-sional tectonic regimes is smaller than that in compressional tectonic regimes (§6.7), the lithosphereis more vulnerable to extensional than compressional tectonics.

The brittle strength increases and the ductile one decreases with increasing depth. Deformationsare accommodated either by a brittle or ductile mechanism that needs lower differential stress for thedeformation. Hence, there must be a transition region at which brittle deformation gives way to duc-tile deformation. This is known as the brittle-ductile transition2. It is often found that earthquakesoccur in the crust no deeper than 15–20 km. This is probably because the transition occurs at thatdepth.

The solid lines in Fig. 12.1 shows the yield strength of constituent rocks at each depth, so that thegraph exhibiting such strengths is called the strength profile of the lithosphere. There is a differencein the brittle strengths in normal and reverse faulting stress regimes (Fig. 6.14). The former strengthis smaller than the latter one so that the brittle-ductile transition in the normal faulting regime isdeeper than the other.

The oceanic lithosphere has abundant olivine minerals which control the mechanical strength ofboth the crust and the lithospheric mantle. Therefore, the oceanic lithosphere has strength profileslike those in Fig. 12.1. By contrast, the continental crust has a much more complex structure thathas been shaped in the long history of the crust. It has significant heterogeneity in lithology. Ductilestrength strongly depends on the lithology. In addition, impermeable rocks make overpressured rock

2The fact is that there is a regime where the deformation of rocks is accommodated by ductile and brittle mechanisms atthe same time, known as quasi-plasticity , around the brittle-ductile transition (Fig. 12.1) [213]. However, the quasi-plasticregime is not well understood, so that we do not refer to that regime any more.


Figure 12.2: Schematic picture showing the strength heterogeneity of the crust associated with theheterogeneity in lithology and other factors including pore fluid pressure and temperature. Strengthis designated by the gray scale.

Figure 12.3: Diagram showing the strength profile of continental lithosphere with homogeneouscrust. Dashed lines indicate brittle strengths for two stress regimes. In this figure a brittle regimeexists in the upper mantle for the case of extensional tectonics. However, the existence depends onthe temperature of the depths.

masses. Rocks with abundant radioactive elements affect the thermal regime in the crust. Therefore,the continental crust must have an intricate strength distribution, which is schematically shown inFig. 12.2. However, we assume simple structures for the following considerations. Namely, weassume that the entire crust is composed of one type of rock, or that the crust has two layers withdifferent lithology. If the crust has a single layer, the continental lithosphere has a strength profilelike those in Fig. 12.3.

12.2. STRETCHING INSTABILITY OF THE LITHOSPHERE 307

12.2 Stretching instability of the lithosphere

Infinitesimal strain until whole lithosphere failure

Consider that Δσ(z) denotes the strength profile, then the integral

Fc =∫ tL

0Δσ dz (12.6)

is the maximum force for the lithosphere to support without finite pure shear deformation, wherethe depth tL denotes the base of the lithosphere. The force is designated by the area bounded by thestrength profile.

Unfortunately, the lithosphere-asthenosphere boundary is fuzzy, i.e., the strength decreases con-tinuously as exp

[Q/nRT (z)

]in the boundary zone. Accordingly, the base of the lithosphere is

sometimes defined for convenience as the depth where Δσ � σzz is satisfied. More simply, thedepth is sometimes defined as the depth where the differential stress reaches a small specific valuein the range 10–20 MPa [191].

What if a tectonic force greater than Fc in Eq. (12.6) is applied to a continental lithospherethat was initially at rest? Consider the initial distribution of Δσ being equal to that of the Earthpressure at rest down to a certain depth that defines the base of the lithosphere (Fig. 12.4). Underthat depth, the initial state of stress is assumed to be lithostatic. Once the force is applied, rocks nearthe surface yield because of the near-surface low brittle strengths. The differential stress in the rocksnear the base of the lithosphere exceeds the ductile strength so that the rocks yield immediatelyafter the force is applied. Namely, the elastic core begins to be eroded at the top and base. Thereduced thickness of the elastic core leads to an increase of force that the core have to support. Theincrease further reduces the elastic core to form a positive feedback cycle [109] and, the entire depthof the lithosphere eventually yields. This is called whole lithosphere failure. Until this phenomenonoccurs, the existence of the elastic regime prohibits finite deformation of the lithosphere. After thefailure of the whole lithosphere, finite deformation of the lithosphere begins.

Finite strain

The history of finite deformation of a rift can be revealed from syn- and post-rift sedimentary records.Continental rifting can lead to continental breakup. However, some rifting is sometimes aborted. TheNorth Sea rift is an example of failed rifts. Intra-arc rifts have the two ends, also. Intra-arc riftingleads to back-arc spreading as the rifting in the Izu-Bonin-Mariana arc resulted in the spreadingin the Shikoku and the Mariana basins of the Philippine Sea plate. On the other hand, there arethe aborted rift basins of the Paleogene under the continental shelf behind the Ryukyu arc to thenorthwest of the plate.

The duration of intra-arc rifting is one order of magnitude smaller than that of intra-continentalrifting [263]. If rifting has several phases, each phase lasts tens of million years in continents. Theactivity of the Triassic rift in Fig. 3.24 is an example. By contrast, intra-arc rifting lasts a few million


�

Figure 12.4: Schematic picture showing the evolution of differential stress in the continental litho-sphere leading to whole lithosphere failure.

years. The extensional tectonics that led to the opening of the Japan Sea backarc basin lasted 5 m.y.[263].

The difference is explained by the initial thermal structure of the lithosphere. Intra-arc riftingoccurs in volcanic arcs because magmatism has heated up and reduced the ductile strength of thelithosphere before rifting. For example, temperature is estimated by a petrological method in theNortheast Japan volcanic arc at 850 and 1400◦ at 25 and 80 km, respectively [107, 238]. The crusthas a steep geothermal gradient there. On the other hand, temperatures at depths of 25 and 40 km areestimated to be 525 and 850◦ in the Basin and Range Province [179], indicating a lower geothermalgradient.

Based on a plausible structure of the lithosphere, the duration of rifting can be estimated bynumerical simulations [55, 236]. For this purpose, we take the present thermal structure underNortheast Japan and the Basin and Range Province as the references for volcanic arcs and continents.In addition, we assume pure-shear rifting by a constant force. The evolution of a pure-shear rift iscalculated by the following one-dimensional model. The upper and lower crust is considered tobe composed of granite and gabbro, respectively, so that their ductile strengths are determined bythose of quartz and plagioclase. That of the mantle lithosphere is determined by olivine. Thermalevolution is calculated by the one-dimensional equation

∂T

∂t+ v

∂T

∂z= κ

∂2T

∂z2, (12.7)

where T is the temperature, z is depth, t is the time since the beginning of rifting, v is the downwardvelocity and κ is thermal diffusivity. Because of the pure-shear, v = zEzz = −zExx (Eq. (3.81)).Based on the petrological estimates, the boundary conditions T

∣∣z=0 = 0◦ C and T

∣∣z=a

= 1300◦


Figure 12.5: Strength profiles for model rifts [236].

C, where a is a depth of 50 km for the intra-arc rift and 74 km for the intra-continental rift. Thetemperature distributions designated by the dotted lines in Fig. 12.5 are used for the initial conditionsfor the rifts. The strength profiles for the strain rates of 1.432 × 10−15 s−1 and 7.954 × 10−17 s−1,respectively, are shown in the same figure. The rate of the intra-continental rift is two orders ofmagnitude smaller than that of the intra-arc rift, merely because the former has a smaller geothermalgradient. Although the strength profile for the former has the maximum in the upper mantle, thestrength of the mantle part for the intra-arc rift is significantly smaller than the strength of the crust.This is the most conspicuous difference in the strength profiles, and is due to the sensitivity of theductile strength of olivine to temperature compared to those of quartz and plagioclase. It is assumedthat the asthenosphere is passively uplifted under the rift by the same amount as the thinning of thelithosphere. That is, mantle rocks are supplied into the region between the surface (z = 0) and thedepth z = a from below to compensate the thinning of the crust. We neglect the vertical movementof the surface by rifting because the amount of movement is smaller than those of the upper-lowercrustal interface and of the Moho interface.

The thinning of the lithosphere raises isotherms to steepen the geothermal gradient. The tem-perature changes are calculated with Eq. (12.7) for every time step in the numerical simulation.Consequently, the transient temperature structure is designated by a convex upward curve in Fig.12.6. If rifting proceeds very slowly, the change in the temperature structure is small. In this case, alargely isothermal thinning of the crust increases the occupation of the olivine-rich layer in the rangebetween z = 0 and a. Because the layer is stronger than quartz and plagioclase if the temperatureis the same, the increased occupation augments the strength of the lithosphere that is defined by∫ a

0 Δσ(z) dz. The applied force is assumed to be constant, so that the raised strength decelerates therifting. The slowdown of the rifting decreases the rate of temperature change to complete a negative


Figure 12.6: Schematic picture showing the temporal variation of temperature structure.

feedback cycle to stop rifting.On the other hand, if the lithosphere thins much more rapidly than thermal conduction through

the lithosphere, the thinning is substantially adiabatic. The result is a increased geothermal gradientwhich further leads to a weakening of the lithosphere. If the tectonic force is the constant throughtime, the reduced strength accelerates the rifting. Accordingly, we have a positive feedback cycle inthis case. Then, the rate of lithospheric attenuation goes to infinity. We interpret this as the breakupof the lithosphere.

Consequently, the fate of rifting, either leading to breakup or cessation, depends on the rateof thinning relative to thermal conduction through the lithosphere [55]. The lithosphere exhibitsstretching instability that determines the fate of rifting. The rate depends on several factors includingthe tectonic force, the composition of the lithosphere, geothermal gradient, and pore fluid pressure.

Figure 12.7(a) shows the strain rate versus time from the beginning of the intra-arc rift. The rateshows divergence if the force applied to the lithosphere is greater than a critical value of about 0.396TN−1 [236]. In this case, the duration of rifting is a few m.y. If the force is just under this value, therifting is aborted after several m.y. from the initiation of rifting. The duration is consistent with theones observed for intra-arc rifts in the world.

The critical force for intra-continental rifts is one order of magnitude greater than that for intra-arc rifts3. Namely, the island arc lithosphere is much weaker than the continental lithosphere, so thatinter-plate deformations are concentrated in the arcs (Fig. 12.7(b)). The critical force depends notonly on the strength of the lithosphere but also on the thermal diffusivity κ, because the transitionis determined by the speed of lithospheric thinning relative to thermal conduction through the litho-sphere. Rising magmas carry heat toward the surface to increase the effective thermal diffusivity of

3The critical force shown in Fig. 12.7(b) is valid specifically for the conditions assumed in the numerical simulation.It is suggested that diffusion creep of olivine instead of dislocation creep greatly reduces the critical force by one order ofmagnitude for intra-continental rifts [82].


��

��

��

��

��

��

��

��

��

Figure 12.7: Unstable behavior of the model rifts [236]. Thermal diffusivity κ is assumed to be3.5 × 10−6 m2s−1. (a) Temporal variation of the strain rate of the intra-arc rift for different tectonicforces indicated in the dimensions of TN−1. (b) The relationship between thermal diffusivity and thecritical force to separate the fate of intra-continental and intra-arc rifts.

the lithosphere4. The numerical simulation showed that intra-arc rifts tend to be formed in volcanicarcs. However, the critical stress is insensitive to κ for intra-arc rifts compared to intra-continentalrifts (Fig. 12.7(b)).

When the Japan Sea back-arc basin opened, the lithosphere under NE Japan was extended to forman intra-arc rift. The rapid subsidence of the rift is designated by the increase of paleobathymetryfrom ∼16 Ma to ∼15 Ma shown in Fig. 3.14. Figure 12.8 shows the temporal change in the rateof subsidence estimated from the stratigraphic records of the rift. The rifting began at around 20Ma. Since then, the subsidence accelerated to reach ∼3 km/m.y. The rapid subsidence was abruptlystopped at 15 Ma when the rifting was aborted. The duration of rifting was 3–5 m.y. Assumingthe local isostasy and initial thickness of the crust to be 35 km, which is estimated from the presentcrustal thickness of the Shikhote Alin (Fig. 2.12), the amount of subsidence can be transformed intothe reduced thickness of the crust. The total subsidence at around 2 km corresponds to the crustalthickness at 25–30 km at the end of rifting5. The total subsidence corresponds to an extensionalstrain of 20%. Therefore, the average strain rate was 2 × 10−15 s−1. This value agrees roughly withthe strain rate for the failed intra-arc rift that is estimated by the numerical simulation (Fig. 12.7).

4Numerical simulation by Honda to estimate the thermal structure under the volcanic arc uses the effective thermal dif-fusivity at 10−5 ms−2, which is one order of magnitude larger than that of rocks [81]. The intra-arc rifting in NortheastJapan was associated with massive volcanism. The representative thickness of the volcanics is about 2 km. Assuminga heat capacity of 1.0 × 103 J kg−1 and a density of 2.5 × 103 kg m−3, we have the heat capacity per unit volume at(1.0 × 103) × (2.5 × 103) = 2.5 × 106 Jm−3. In addition, assuming the temperature of magma to be 1000◦ C, the heattransported by the volcanism was 5.0×1012 Jm−2. Dividing this by the duration of volcanism (ca. 3 m.y.), we obtain the heatflux of 50 mWm−2. This is as great as the average heat flow from the solid Earth. A significant amount of intrusion may haveaccompanied the volcanism so that the increased effective thermal diffusivity by one order of magnitude seems reasonablefor the Early Miocene rifting in Northeast Japan.

5The present crustal thickness of Northeast Japan is greater than this thickness by 5–10 km, probably due to igneousunderplating since 15 Ma.


Figure 12.8: Temporal change in the subsidence rate of the Northeast Japan volcanic arc when theJapan Sea back-arc basin opened [263].

In addition, the duration is consistent with the calculated one.

12.3 Factors controlling te of the continental lithosphere

It was seen in Section 8.3 that the effective elastic thickness of the oceanic lithosphere is approxi-mated by the depth of the 400–600◦ isotherm, and that of the continental lithosphere has a positivecorrelation with tectonic age and heat flow (§8.7). Figure 12.9 indicates that the latter has a muchbetter correlation. Those observations suggest that the effective elastic thickness of the lithosphereis controlled by its thermal regime. Stable continents are stable because they generally have a lowaverage heat flow. However, the relationship between heat flow and te is complicated by the strengthprofile of the continental lithosphere which is more intricate than the oceanic lithosphere. In this sec-tion, we consider the effective elastic thickness of the continental lithosphere in terms of its strengthprofile.

We assume that the continental crust is composed of a single rock type for brevity, and that thelithosphere behaves as a elastic-perfectly-plastic body, i.e., we deal with the lithosphere that hasstrength profiles as shown in Fig. 12.10. The figure shows the vertical distribution of differentialstress in the flexed lithosphere for three cases. While the curvature is small, yielding occurs onlynear the top and base of the lithosphere to leave a thick elastic core in the lithosphere. However, asthe curvature increases, the differential stress generated by the bending exceeds the strength of thelower crust so that the elastic core is divided into two layers (Fig. 12.10(b)). If the interface betweenthe elastic layers with thicknesses of t1 and t2 is lubricated, the whole system has the effective elasticthickness te ≈ max(t1, t2) (§8.4.2). The effective elastic thickness before the separation was t1 + t2,which is larger than max(t1, t2). Therefore, te may decrease abruptly at a critical curvature [28].

12.3. FACTORS CONTROLLING TE OF THE CONTINENTAL LITHOSPHERE 313

Figure 12.9: Effective elastic thickness of the continental lithosphere te. (a) te versus tectonic age[28]. (b) te versus heat flow in Africa [72]. The error bars in this plot that are designated in the figurein [72] are omitted here for simplicity.

Figure 12.10: Strength profiles of a model continental lithosphere and the vertical distribution ofthe differential stress (gray) generated by the concave upward bending of the lithosphere, whereΔσ = σH − σv. (a) Bending with a small curvature K of the lithosphere gives rise to yielding nearthe top and base of the lithosphere. There is a thick elastic core indicated by the linear part ofthe stress distribution in the lithosphere. (b) As the curvature increases, the linear part reaches thecurve that designates the ductile strength of the lower crust (dashed line). Yielding in the lower crustdivides the elastic core into two parts, one in the crust and the other in the lithospheric mantle (gray).Namely, the crust and mantle is mechanically decoupled. (c) Vertical distribution of the differentialstress caused by the bending and simultaneous horizontal compressive force F .

This phenomenon is observed in collisional orogens [140]. Figure 12.10(c) shows a case where ahorizontal tectonic force F is applied to the bent lithosphere.

Given the flexure w(x) of the lithosphere, it is straightforward to calculate the curvature K = w′′.Then, the horizontal strain εxx is obtained via Eq. (8.12). Accordingly, if w(x) is also given, εxx


is obtained. The rate of strain is transformed to ductile strength. Therefore, we can calculate stressfrom w(x) and w(x). The bending moment is obtained from the stress as

M =∫ tL

0σxx (z − zn) dz. (12.8)

Then, using the formula M = −DK, the flexural rigidity D and the effective elastic thickness te ofthe flexed lithosphere can be estimated.

There may be depth ranges of yielding for given w(x) and w(x) (Fig. 12.10(b)). In that case,Eq. (12.8) is replaced by the piecewise integrals

M =N∑i=1

∫ith layer

σxx

[z − z(i)

n

]dz, (12.9)

where z(i)n is the depth of the ith layer in which deformation is accommodated by a brittle ductile or

elastic mechanism. Using the bending moment in Eq. (12.9), the effective elastic thickness of thisrheologically multi-layered lithosphere is estimated. It should be noted that te is different from thethickness of the elastic core.

If the horizontal force

F =∫ tL

0σxx dz (12.10)

does not vanish, the horizontal stress σxx in Eq. (12.9) should be replaced by (σxx − Fmean), whereFmean = F/tL is the mean horizontal stress in the lithosphere. Figure 12.11 shows (σxx − Fmean)for regions of compressive and extensional tectonics. Since the brittle strength for the former regionis larger than that for the latter if pore fluid pressure is the same, te of the horizontally stretchedlithosphere is smaller than that of the horizontally constricted lithosphere [141]. Whole lithospherefailure occurs more easily in extensional than in compressive tectonics.

Figure 12.12 shows the effect of the curvature, geothermal gradient and tectonic force Fmean onte for a model continental lithosphere. In addition, te is also affected by curvature, even if otherfactors are the same. Increased curvature increases differential stress to widen the yielded zones,which reduces te. Figure 12.12(a) shows that sensitivity on the geothermal gradient decreases withincreasing curvature. The radius of curvature at the right side of this graph is R = 1/K = 100 km,comparable with the thickness of the continental lithosphere. Thus, the model of a thin elastic platehas a large error at this end. Figure 12.12(b) shows te versus Fmean. The effective elastic thicknessdeclines more rapidly in a normal fault regime than in a reverse fault regime with increasing tectonicforce |Fmean|.

The continental lithosphere has complex structures, leading to an intricate strength profile. Thus,the effective elastic thickness of the lithosphere depends on various factors including lithology, cur-vature, thermal structure, horizontal tectonic force, radioactive heat generation in the crust, etc.[140]. The Canadian shield has a considerable spatial variation in te, although those factors have notso large differences. The variation suggests that geological structures such as old faults and suturezones play important roles in determining te [27, 254].

12.4. PERIODIC DEFORMATION OF THE LITHOSPHERE 315

Figure 12.11: Schematic strength profiles of the oceanic lithosphere and vertical distribution of dif-ferential stresses in regions of (a) compressive and (b) extensional tectonics. Due to the asymmetricbrittle strength for those cases, the effective elastic thickness for the former case is larger than forthe latter. The width of the gray area indicates Δσ − Fmean.

Figure 12.12: Effective elastic thickness te of a model continental lithosphere that has a dry graniticcrust [126]. The strain rate is assumed to be 10−15 s−1. (a) The effect of curvature K and geothermalgradient. (b) Effect of horizontal stress Fmean with an assumed curvature K = 10−5 km−1.

12.4 Periodic deformation of the lithosphere

Faults and folds are sometimes aligned parallel and uniform intervals in wide deformation zones.Horst and grabens in the Basin and Range Province are examples. The wide rift zone called sulcuson Jovian satellite Ganymede have grooves that are surface manifestations of normal faults (Fig.3.6). Many icy satellites have such zones. Figure 12.13 shows a rift zone separating the heavilycratered, therefore older, terrains on Enceradus.

The Fletcher-Hallet model explains those spatially periodic structures using the linear stabilityanalysis of a multi-layered system with a free boundary [63]. The model is also applied to theperiodic surface undulations of ropy lavas [62].


Figure 12.13: Saturnian satellite Enceradus (Voyager image 1715S2-001). The diameter of thismoon is about 500 km. Note the difference in the number density of craters. Smooth planes with asmaller crater density have grooves. The heavily cratered terrain is divided by a rift zone (arrow).

Consider the crust of an icy satellite instead of the Earth’s crust, because the latter has complexstructures. For example, Ganymede has a very thick icy layer (§9.3), so that the shallow part of thelayer may be brittle as grooves and other geological structures indicate, but deep-seated ice may beductile. Accordingly, we assume a surface brittle layer floating on a viscous fluid (Fig. 12.14(a)).The brittle behavior of the layer (1) is simulated by a power-law fluid with a very large power-lawexponent n(1) � 1. The layer (2) is also a power-law fluid but has the exponent n(2) ≈ 3. Planestrains on the xz-plane across the wide deformation zone are assumed, where the z-axis is definedupward with the origin O at the base of the initial brittle layer (Fig. 12.14). H is the initial thicknessof the layer (1). All the layers are assumed to be incompressible.

Following Section 9.4, horizontally extending pure shear is regarded as the mean flow for this


Figure 12.14: Schematic illustration showing the Fletcher-Hallet model. The layer (1) is a brittlelayer with a free boundary at its top and is underlain by the viscous layer (2). The layer (0) iscomposed of inviscid fluid or space. The interface between the layers (1) and (2) is the brittle-ductile transition. Dashed lines indicate the initial levels of the surface and brittle-ductile transition.

multi-layered system, and we have assumed incompressibility. Therefore, the results in Section 10.8are applicable to this problem. Namely, the differential equation in Eq. (10.89) holds for each oflayers (1) and (2), so that the growth rates of the coefficients of their separable solutions gives thestability of this system.

In this model, infinitesimal strains are assumed so that we can ignore temperature perturbations,i.e., temperature is a function of z in layer (2):

T = T0 − Γz (z < 0), (12.11)

where Γ is the geothermal gradient. Since the perturbation is very small, the effective viscosity inlayer (2) may be approximated by

η(2) = η0 exp(−γz). (12.12)

The effective viscosity of layer (1) is assumed to be constant.In the layers (1) and (2), we obtain the stress ratio Φ = 1/2 using the same argument as we did

to derive the slip line theory in Section 10.5. Substituting this stress ratio into Eq. (4.15), we haveTII = (ΔS/2)2 and the Mohr circles in Fig. 12.15. Therefore, we have

TII =14

(Sxx − Szz)2 + S2xz. (12.13)

The perturbations of the pure shear parallel to the coordinate axes are very small, therefore, (Sxx −Szz)2 � S2

xz (Fig. 12.15). The effective viscosity of the fluid η satisfies

2η =[BT

(n−1)/2II

]−1(12.14)

B = B∗ exp(−Q/RT ) . (12.15)

Layer (2) has the constitutive equation

Sxx = −p + 2ηDxx, Szz = −p + 2ηDzz, Sxz = 2ηDxz. (12.16)


Combining the incompressibility Dxx +Dzz = 0 and Eq. (12.16), we obtain

Sxx − Szz = 4ηDxx. (12.17)

The principal axes of the mean flow are parallel to the coordinate axes. Hence, we have Sxz = 0.Combining Eq. (12.15), we obtain

T II =14

(Sxx − Szz

)2= S

2, S ≡ 1

2

(Sxx − Szz

)> 0. (12.18)

Thus, the effective viscosity satisfies

2η = B−1S

(1−n)= B

− 1n D

( 1n−1)

xx . (12.19)

Substituting Eq. (12.15) into (12.19), and using the temperature in Eq. (12.11), we have

η =

[2(B∗) 1

n D(1− 1

n )xx

]−1

exp[ Q

nR(T0 − Γz)

].

The operand of the exponential function is expanded about z = 0 as

Q

nR(T0 − Γz)≈ Q

nRT0+

ΓQz

nRT 20

.

Therefore, using Eq. (12.12), we obtain

η0 =

[2(B∗) 1

n D(1− 1

n )xx

]−1

exp( Q

nRT0

), (12.20)

γ =ΓQ

nRT 20

. (12.21)

These equations give the effective viscosity of layer (2). That of layer (1) is assumed to be constant atη0, and the state of stress in this layer is determined by Eq. (12.17) and on the Mises yield criterion.

For a linear analysis of this system, we have to solve Eq. (10.89) under appropriate boundaryconditions. Unlike single layer folding, the boundary between layers (0) and (1) is free. The localrectangular coordinates are defined at the boundary to which the s- and n-axes are taken to be paralleland perpendicular (Fig. 12.16). The function h(x) denotes the height of the boundary. Then, thecomponents of S satisfy the conditions at the boundary

Ssn∣∣z=h

= 0, Snn∣∣z=H

= ρgh, (12.22)

where the normal stress is linearized as Eq. (D.31). These equations are transformed into theexpressions with the coordinates x and z by considering the force balance at the boundary. Let Nbe the unit normal to the boundary, then the force balance is expressed as

S(0) ·N = S(1) · (−N ). (12.23)


Figure 12.15: Mohr circles for the stress state with Φ = 1/2.

Let θ be the inclination of the s-axis with respect to the x-axis (Fig. 12.16). The stress componentsare transformed as(

S(i)ss S

(i)sn

S(i)ns S

(i)nn

)=

(cos θ sin θ− sin θ cos θ

)(S

(i)xx S

(i)xz

S(i)zx S

(i)zz

)(cos θ − sin θsin θ cos θ

).

Therefore,

Snn = S(i)xx sin2 θ − 2S (i)

xz sin θ cos θ + S (i)zz cos2 θ, (12.24)

Sns = −(S

(i)xx − S (i)

zz

)sin θ cos θ + S (i)

xz

(cos2 θ − sin2 θ

). (12.25)

These trigonometric functions are rewritten as sin θ = ∂h/∂s and cos θ = ∂x/∂s. Ignoring higher-order terms, we have

sin θ ≈ ∂h

∂x, cos θ ≈ 1.

Substituting these equations into Eqs. (12.24) and (12.25), and again ignoring the higher-orderterms, we obtain

Snn = −2S (i)xz∂h

∂x+ S (i)

zz , Sns = −(S

(i)xx − S (i)

zz

) ∂h∂x

+ S (i)xz . (12.26)

The stress tensor is divided into the mean and perturbation, S = S + S. Because of S ≈ O, the

higher-order terms of(∂h/∂x

)S can be neglected so that Eq. (12.26) is transformed into

Snn = −2S(i)xz

∂h

∂x+ S

(i)zz + S

(i)zz , (12.27)

Sns = −(S

(i)xx − S

(i)zz

) ∂h∂x

+ S(i)xz + S

(i)xz . (12.28)


Figure 12.16: Schematic picture showing the boundary condition between layers (1) and (2).

On the other hand, a non-slip condition and force balance are appropriate bondary conditions atthe boundary between layers (0) and (1):

v(1)x

∣∣∣z=0

= v(2)x

∣∣∣z=0, v

(1)z

∣∣∣z=0

= v(2)z

∣∣∣z=0

S(1)zz

∣∣∣z=0

= S(2)zz

∣∣∣z=0, S

(1)xz

∣∣∣z=0

= S(2)xz

∣∣∣z=0

.

Note that the layers are of the same density. If they are not, buoyancy force should be taken intoaccount.

Under these boundary conditions, the differential equation in Eq. (10.89) is solved. We use asolution of the form

vx = −1λ

dWdz

sin kx, vz = W cos kx, h −H = A(t) cos kx,

where k is the horizontal wavenumber. Consequently, we obtain

dAdt

= −DxxA + vz∣∣∣z=H

= (q − 1)DxxA, (12.29)

where q(k) denotes the growth rate6. The waves with q(k) > 1 are amplified, but those with q(k) < 1are erased. The waves with the maximum growth rate may emerge as macroscale boudins. Accord-ing to Fletcher and Hallet [63], q(k) is determined by the four dimensionless parameters, n(1), n(2),γH and ρgH/2cY. The third one is the characteristic depth over which the effective viscosity ofthe layer (2) decreases by 1/e. The fourth one designates the strength of layer (1). Figure 12.17(a)shows a graph of q(k), where kd indicates the wavenumber that maximizes the growth rate. Itis demonstrated that kd is insensitive to n(1) if this power-law exponent is greater than 102 (Fig.12.17(b)). This is convenient to simulate the brittle behavior of layer (1). Rocks and ice have avalue of n(2) about 3. Therefore, the remaining parameters γH and ρgH/2cY determine the kd. The

6See [62] for the concrete expression of q(k).


Figure 12.17: Growth rate of crustal boudins predicted by the Fletcher-Hallet model [63]. (a) Growthrate versus dimensionless wavenumber. The values n(1)=104, n(2)=3, γH=10 and ρgH/2cY=3 areassumed. (b) Relationship between the dominant wavenumber kd and the power-law exponent n(1).The parameter values n(2)=5, γH=10 and ρgH/2cY=3 are assumed here.

brittle-ductile boundary goes down with an increasing strain rate. Therefore, H depends on Dxx

(Eq. (12.20)). Likewise, the critical stress cY depends on the strain rate. On the other hand, γ isdetermined by Γ via Eq. (12.21). Consequently, Dxx and Γ control the dominant wavenumber kd.

Many icy satellites have grooves with roughly uniform intervals. Thus, the dominant wavelengthis observable there. However, we need one more line of evidence to constrain Dxx and Γ in theFletcher-Hallet model. Let qd be the growth rate at the dominant wavelength kd. Then, from Eq.(12.29), we obtain

ln(A/A0) = (qd − 1)Dxxt, (12.30)

where A0 and A are the initial amplitude and that of the time t. The left-hand side of Eq. (12.30)represents the logarithmic strain of the crust.

This equation enables an order-of-magnitude estimation of the controlling factors. The initialamplitude is unknown, but A0 ∼ 100 m may be acceptable. A is known from the present topography.For example, grooves on icy satellites have amplitudes of the order of A ∼ 102 m. The global strainof the moons is in the order of 1%. Thus, we estimate Dxxt ∼ 10−2. Using these estimates, weobtain qd ∼ 500 from Eq. (12.30). The Fletcher-Hallet model gives appropriate values of Dxx andΓ from this growth rate at the observed kd. If the values are unrealistic, the model is found to beinapplicable to the object. It was found that the model is applicable to Ganymede, Enceradus, andMiranda [44, 76]. As for Ganymede, the values Γ ≈ 20 K km−1 and Dxx ≈ 10−14 s−1 were obtained.Similar values were determined for Enceradus and Miranda. Thus, the extensional tectonics lasted1%/10−14 s−1 ∼ 104–105 years, surprisingly short for the global tectonics compared to the presenttectonics of the Earth.

The lithosphere of the Earth has a much more complex structure than that assumed in the


Fletcher-Hallet model. The density difference of the crust and mantle gives rise to buoyancy. Thestrength of the brittle layer depends on depth, unlike the strength in the model. In some activeregions, the hot and fluent lower crust mechanically separates the upper crust and the mantle litho-sphere, so that there are two strong layers there. If the upper and lower crusts have significantlydifferent rheology, such as granite and gabbro, the lithosphere may exhibit complex behavior. Mod-els have been developed to take account of those factors [16, 17, 196, 282]. Such a model wassuccessfully applied to the folded oceanic lithosphere in the northeastern Indian Ocean [281], wherethe simple elastic model failed to explain the dominant wavelength (§8.4.3). Linear stability analy-sis assumes very small perturbations. Numerical simulations are used to link the analysis and finiteamplitude deformations as the rifts discussed in Section 12.2 [25].

Exercises

12.1 In the model introduced in §12.2, the fate of rifts is determined by the competition betweenthe rate of lithospheric thinning and vertical heat transfer through it. Consider how to define thePeclet number, which is a dimensionless number indicating the representative time of motion of amass relative to the time constant of heat conduction.

12.2 Consider that pure shear rifting begins in the lithosphere with a negligible strength of thecrust compared to that of the mantle lithosphere, and that the geothermal gradient Γ is constant inthe lithosphere. Let

T = A− 1n E

1n−1 exp

(Q

nRT

)E (12.31)

be the constitutive equation of the mantle, where

E =

⎛⎝E 0 00 0 00 0 −E

⎞⎠ (12.32)

is the rate of strain tensor. Show that the approximate equation∫ a0Txx dz ≈ A− 1

n E1/nT 2MnR

ΓQexp

(Q

nRTM

)indicates the tectonic force needed for the rifting [55], where Txx is the horizontal component of thedeviatoric stress, TM is the Moho temperature.

Documents

Dynamics of the Lithosphere · years. The extensional tectonics that led to the opening of the Japan Sea backarc basin lasted 5 m.y. [263]. The diﬀerence is explained by the initial