Introduction to Tectonophysics
Patrice F. Rey Preface
i PREFACE This eBook, in a permanent state of unfinishedness, focusses mechanical coupling between the lithosphere and the astheno- on tectonic processes, from the forces that drive them to the pa- sphere. rameters that control the mechanical behavior of the Earth’s How the Earth’s lithosphere deforms depends on the relation- lithosphere. ship between applied forces and the mechanical response We start in Chapter 1 by looking at the Earth’s geotherm be- measured by the accumulated strain and the strain rate, hence cause the mechanical behavior of rocks - and therefore that of Chapter 4 tackles the rheology of rocks. This chapter intro- the lithosphere - is very sensitive to temperature and because duces the concepts of elastic, plastic and viscous strain and ex- tectonic processes can significantly bring the geotherm out of plores the influence of parameters such as temperature, pres- equilibrium forcing the cooling or heating of rocks. sure, and fluid, before presenting the important concept of lithospheric strength envelopes. The second chapter introduces the notion of isostatic equilib- rium, critical to understand the topography of the Earth’s sur- Chapter 5 introduces the new discipline of computational tec- face when its lithosphere is at rest. This chapter explains how tonics and geodynamics. Equipped with a reasonable under- isostatic equilibrium differs from the notion of mechanical equi- standing of the thermal and mechanical properties of the litho- librium. Isostasy introduces gravitational stresses because den- sphere, one can tap on the power of high-performance comput- sity interfaces are no longer parallel to gravitational equipoten- ing to explore through numerically experiments tectonic and tial surfaces. The notion of gravitational stress is important to geodynamic processes. understand how the Earth’s lithosphere can deform in the ab- Chapter 6 explores through numerical experiments a range of sence of plate boundary forces. tectonic processes at lithospheric scale, in a variety of tectonic Plate boundary forces are the focus of Chapter 3. Slab-pull, settings from continental collisions, to extensional tectonics ridge push, and drag from asthenospheric flow contribute to and transcurrent tectonics. drive the motion of lithospheric plates at the Earth’s surface, Finally, Chapter 7 focuses on the dynamics of mantle convec- and their deformation. Ridge-push can be evaluated analyti- tion. cally using the concept of the gravitational forces. The evalua- tion of the asthenospheric drag and that of the slab-pull is more difficult because it requires some understanding of the copyright © Patrice F. Rey, 2018
ii CHAPTER 1 The Earth’s Geotherm
The geotherm, i.e. the distribution of temperature with depth, is an important characteristic of the Earth's lithosphere because temperature impacts on all physical properties of rocks (e.g. density, viscosity, conductivity, elasticity, magnetism etc). In particular, temperature controls the rheology of rocks and therefore how they deform in response to applied deviatoric stresses, and how the Earth's lithosphere reacts to tectonic forces. In this chapter we derive, from first principles, a simple expression for the steady-state geotherm. We then consider the notion of transient geotherm.
3 SECTION 1 Heat Transfer in the Earth’s lithosphere
In this section
1. Heat transfer in the Earth’s Lithosphere 2. Heat energy & temperature 3. Heat conduction 4. Heat advection 5. Radiogenic heat production
The continental geotherm is a function of the i/ rate at which given depth does not change through time). In contrast, when heat is produced or consumed within the lithosphere, ii/ the the lithosphere has a net gain or a net loss of heat, the geotherm rate at which the lithosphere looses heat to the atmosphere/ is said to be transient (i.e the temperature changes through ocean system, and iii/ the rate at which the lithosphere gains time) until a new equilibrium is reached between heat lost and heat from the hot convective mantle. When the heat lost bal- gained. On a billion year time scale, the geotherm is always tran- ances the heat gained, an equilibrium is reached and the geo- sient because the primordial accretionary heat and the Earth’s therm is said to be in steady state (i.e. the temperature at any supply in radiogenic isotopes progressively decrease, however,
4 on a scale of ~100 myr, and in the absence of tectonic or magmatic The main processes able to change the amount of heat energy in activity, the geotherm can approach an equilibrium state. The litho- the lithosphere are: spheric geotherm varies laterally. The base of the lithosphere is de- •Heat conduction (transfer of kinetic energy between molecules fined by the isotherm ~1300ºC. In cratonic regions, this isotherm or atoms from a hot to a less hot region) can stand at a depth between 200 and 300 km. In other continental •Heat advection (replacement of a volume of rock at tempera- regions it can be as shallow as 50 km, and at mid-oceanic ridge it is ture T1 with an equivalent volume at temperature T2) met at a few kilometers underneath the oceanic floor. •Heat production (heat produced by radioactive isotopes, vis- Here, we first review the processes involved in heat generation and cous heating, exothermic metamorphic reactions) heat transfer, and we derive from the rate of these processes a gen- • Heat consumption (heat consumed by endothermic metamor- eral equation which describes the change in temperature with phic reactions, in particular partial melting) depth and through time. From this general equation we derive a particular solution for the so called "steady state" continental geo- The variation of temperature dT over an increment of time dt de- therm (temperature changes with depth but not with time, i.e. zero pends on the sum of heat variations dE due to each process. In net heat gain or loss). In a second part, we discuss how the steady what follows, we derive three expressions for i/ the rate of heat state continental geotherm is affected by a number of geological conduction, ii/ for the rate of heat advection, and iii/ for the rate processes including, lithospheric thinning and thickening, burial of radiogenic heating. From these, we derive the 1D conduction- via sedimentary or volcanic processes, and basal heating via the advection heat transfer equation from which an expression for the spreading of mantle plumes at the base of the Earth's lithosphere. steady state geotherm can be derived. Sounds more complicated than it really is. So bear with me ...
Temperature and Heat
The temperature (degree of hotness or coldest) of a small volume Heat conduction of rock somewhere in the lithosphere varies if heat energy (a form Conduction transports heat from hot to cold regions. The flow of of kinetic energy) is gained or lost. The relationship that gives the heat (Q) is proportional to the negative temperature gradient (dT/ variation of temperature dT as a function of a variation of heat dE dz) between the cold and the hot region, with the coefficient of pro- is: portionality being the conductivity (K). Mathematically this trans- . dT = dE / (Cp m) lates into the Fourier's law where Q is in W.m-2 and K is in W.m-1.K-1. In our geological reference frame, z increases downward with Cp the heat capacity, and m the mass.
5 (T(z+dz)>T(z)). Conduction occurs in the direction of decreasing Radiogenic heat production temperature (i.e. dEc is a gain for upward conduction) hence the z Radiogenic disintegration of radioac- sign "-" insures that Q is positive upward (dEc is positive when T tive isotopes (238U, 235U, 232Th, and increases downward). 40K) releases heat. The increment of dT A Q = − K radiogenic heat (dEr) produced in a E =- .a.dt z 1 Q(z) dz small cylinder of rock of section a Q(z) dE = A.a.dz.dt Let's consider a small cylinder of rock of sec- and length dz over an increment of r dEc = E2 - E1 2 time dt is: dE = .a.dt tion a (area in m ). If the incoming and outgo- c (dQ/dz).dz z+dz a ing heat at both ends of the cylinder are the a same, there is no net heat gain or loss, and A⋅a⋅dz⋅dt = A⋅dV⋅dt
E2=-Q(z+dz).a.dt the temperature remains unchanged. Tem- z+dz where A is the rate of radiogenic heat production. Radioactive heat perature changes when the heat E1 leaving is the main internal heat source for the earth as a whole (it is meas- Q(z+dz) the volume over an increment of time ( ured in W.m-3). E1=Q(z) ⋅ a ⋅ dt) is different to the heat E2 enter- ing it (E2=Q(z + dz)⋅a⋅dt). Heat Advection The entering heat flow Q(z + dz) can be approximated with a Tay- lor series in which only the two first terms are of significance. (n.b.: Advection of heat implies that mathematically f (x + dx) can be approximated from f (x ) and the n n a mass of material at tempera- a.uz . dt derivatives at location : , , etc: E1=C . .a.u .dt.T xn f ′(xn) f ′′(xn) ture T (in yellow on the z p z sketch) is being pushed out of dE = E - E 2 2 our cylinder and replaced by u 2 1 dQ dz d Q dE t.dT Q(z + dz) = Q(z) + dz⋅ + ⋅ + ⋅⋅⋅ an equivalent mass of material u = Cp. .a.uz .d 2 dz 2 dz at temperature T+dT. The incre- ment of heat gained or lost ( Therefore the increment of heat (dEc) gained or lost in an increment a of time dt is : dEu) over an increment of time a.u . dt dEc = E2 − E1 z E2=C dT) dt is proportional to the mass z+dz p. .a.uz .dt.(T+ dQ d2T of material displaced ( a⋅Q(z)⋅dt − a⋅Q(z + dz)⋅dt = − a⋅dz⋅ ⋅dt = K⋅ ⋅dV⋅dt 2 dz dz ρ⋅a⋅uz ⋅dt ), the heat capacity of
6 -1 -1 the material (Cp in J . kg . K : is the amount of energy required by left to right, the conduction term, the term of radiogenic heat pro- 1 kg of the material to increase its temperature by 1 K), and the tem- duction, and the advective term. In 3D, this equation becomes: perature contrast of the incoming and outgoing masses. (n.b.: be- ∂T K ∂2T ∂2T ∂2T A ∂T cause z increasing with depth, uz is negative for upward convec- = ⋅[ + + ] + − u ⋅ 2 2 2 z tion, however, dEu is a gain for upward convection hence the sign - ∂t ρ⋅Cp ∂x ∂y ∂z ρ⋅Cp ∂z insures that when uz is negative (i.e. upward motion) the advective heat gained is positive). We get that the rate of heat advection is:
−Cp⋅ρ⋅a⋅uz⋅dt ⋅dT The geotherm in the continental crust
Total heat gained or lost We derive here the equation for the steady state continental geo- therm. For a steady state geotherm (no variation of temperature Adding up the rate of conductive heat (dEc), the rate of convective through time) by definition dT/dt=0 and uz = 0 (no advection or ero- heat (dE ), and the rate of radiogenic heat (dE ) gives the total rate u r sion or sedimentation), therefore the 1D heat transfer equation sim- of heat gained or lost. This variation of heat triggers a change in d2T A temperature dT = dE/(Cp.m) therefore : plifies to: = − dz2 K and Cp.m. dT= dEr + dEu + dEc : This is a second order differential equation. It says that the gradi- ent of the temperature gradient is a constant equal to the negative 2 d T ratio between the radiogenic heat production and conductivity. Cp⋅ρ⋅dV⋅dT = A⋅dV⋅dt − Cp⋅ρ⋅a⋅uz⋅dt ⋅dT + K⋅ ⋅dV⋅dt dz2 This kind of equation is solved by integrating twice and by using two boundary conditions. For example we may know the tempera- This leads to the 1D conduction-advection heat transfer equation: ture at the surface let's say: T=To at z=0; and we may know the sur- face heat flow for instance at z=0 the heat flow is -Qo (remember dT K d2T A dT = ⋅ + − u ⋅ this is a positive number). Assuming that A is constant with depth, 2 z dt ρ⋅Cp dz ρ⋅Cp dz the first integration led to the temperature gradient dT/dz:
This equation describes the variation of temperature with depth dT A = − ⋅z + C and through time due to heat conduction, radiogenic heat and dz K 1 heat advection. This equation assumes no lateral heat flow (hence This gives the gradient of the geotherm as a function of depth. 1D). On the right end side of the equation the three terms are, from From this function we get that at the surface (i.e. z=0): dt/dz=C1
7 Because of the Fourier's law, the second boundary condition im- relationships, one for the crust (the very last equation), and one for poses that at z=0: the lithospheric mantle. dT Q If the production of radiogenic heat is zero in the mantle then the = 0 dz K 1D conduction-advection heat transfer equation simplifies to: 2 by combining the last two equations for z=0 we get C = Qo/K, we d T 1 = 0 therefore conclude that the thermal gradient at any depth z is ... dz2 dT A Q Integrating twice we get: = − ⋅z + 0 dz K K dT = C2 and T(z) = C2⋅z + C3 Integrating a second time led to: dz
With the two following boundary conditions: at z = zc (zc = Moho A 2 Q0 T(z) = − ⋅z + z + C depth): T(zc ) = Tc and Q(zc ) = -Qm (the basal heat flow), we get that: 2K K 2 Qm Qm The first boundary condition says that T=To at z=0, therefore we C2 = and that C3 = Tc − ⋅zc get that: K K The geotherm in the lithospheric mantle is therefore: A Q T(z) = − ⋅z2 + 0 z + T 2K K 0 Qm T(z) = ⋅(z − zc) + Tc This relationship is the steady state geotherm. It gives the distribu- K tion of temperature with depth in a layer with homogeneous radio- Hence, the geotherm in the lithosphere is defined by a two steps genic production A, conductivity K, with a surface temperature of function: To and a surface heat flow of Qo. A Q T(z) = − ⋅z2 + 0 z + T , for 0 < z ≤ z The geotherm in the continental lithosphere and beyond 2K K 0 c
The continental lithosphere consists in two layers with contrasted Qm T(z) = ⋅(z − z ) + T , for z > z thermal properties. In particular the radiogenic heat production in K c c c the mantle is negligible compared to that of the crust. The geo- therm in continental lithosphere is therefore best described by two
8 2 alternatively, if Qm instead of Qo is known the crustal geotherm (i.e. dT dT d T A “differential equation” links together the derivatives ( , , ) when 0 < z < zc): dt dz dz2 dT dT of a function T(z, t). Here, is the rate of temperature change; A Q A dt dz T(z) = − ⋅z2 + [ m + ⋅z ]⋅z + T , for 0 < z ≤ z 2K K K c 0 c is the gradient of temperature (i.e. the slope of the geotherm); and d2T In the asthenospheric mantle, convective motion is such that the is the gradient of the gradient of temperature. dz2 temperature shows relatively little variation as convection acts as From this differential equation, one can derive a solution for the an efficient mixing process. For z > zl (zl being the base of the litho- sphere) the geotherm follows the adiabatic gradient ~0.3 K per kilo- particular case when thermal equilibrium is reached (i.e. there is no dT meter. For instance, from the top of the convective mantle (i.e. the change of temperature through time: = 0 and necessarily there base of the lithosphere) to the core/mantle boundary (i.e. a dis- dt is no advection of heat, hence uz = 0). This corresponds to the tance of ca. 2700-2800km) temperature increases by only ~3000ºC. “steady geotherm” for which the differential equation simplifies to: This contrast with the evolution of the temperature within the litho- sphere where temperature increases from ca. 0ºC at the surface to d2T A T’’(z) = = − 1300ºC at depth of 100-200km. dz2 K
Appendix 1: Solving differential equations This tells us that the change of the temperature gradient with d dT depth ( )( ) is equal to a constant -A/K (we assume here that Very often, the rate of natural processes (i.e. rate of heat conduc- dz dz tion, rate of heat advection, the rate of radiogenic heating etc) can the radiogenic production A and conductivity K are constant with easily be calculated or measured. In the context of the Earth’s geo- depth). The negative sign implies that the rate of temperature in- therm, we have seen that, from the rate of individual heat transfer crease decreases with depth. mechanisms, we can express the overall rate of temperature change Lets plot T’’(z): through time (t) at any depth (z) via the following differential equa- T''(z) tion:
dT K d2T A dT = ⋅ + − u ⋅ 2 z dt ρ⋅Cp dz ρ⋅Cp dz z -A/K
9 d2T A The curve in blue is the geotherm, bearing in mind that this curve The differential equation = − can be solved to find out the dz2 K is only valid for 0 < z < zmoho. geotherm T(z). For this the differential equation is integrated twice: T(z) The first integration leads to the temperature gradient :
dT A T(0) T’(z) = = − ⋅z + C dz K 1 zmoho z Graphically we get that: T'(z) C is therefore the temperature 1 c We have recognized that the integration constants C1 and C2 have 1 -A/K gradient at z=0 (Earth’s surface), the following meanings: C1 is the temperature gradient (the slope and -A/K is the slope of the gradient. of the geotherm) at the surface (i.e. at z=0); and C2 is the tempera- ture at the surface. The temperature at the surface is typically in
z the range of 0 to 30ºC, and the temperature gradient at the surface -A/K can be determined by measuring the temperature at the surface T''(z) and at the bottom of a well a few 100s to a few 1000s of meter deep.
The second integration leads to the geotherm: Appendix 2: Taylor Series
A 2 According to the Taylor theorem any smooth function f(x) can be T(z) = − ⋅z + C1 ⋅z + C2 2K approximated as a polynomial involving its derivatives following: Graphically we get that: df(x) d x2 d2f(x) d x3 d3f(x) f (x + d x) = f (x) + d x⋅ + ⋅ + ⋅ + ⋅⋅⋅ 2 3 C2 is therefore the temperature T'(z) d x 2! d x 3! d x at z=0. c 1 -A/K f(x+dx) ≈ f(x)
f(x+dx) ≈ f(x) + f’(x) . dx c f’’(x) 2 T(z) f(x) f(x+dx) ≈ f(x) + f’(x) . dx + . dx2 2! z -A/K T''(z) x x+dx x
10 Appendix 3: Steady state geotherm with decreasing RHP A0 2 −z T(z) = − ⋅hc ⋅Exp( ) + C1z + C2 The steady state crustal geotherm we derived earlier assumed that K hc the radiogenic heat production is constant with depth. This is usu- To find out the two integration constants C1 and C2 we need to call ally not the case as cycles of partial melting and crustal differentia- two boundary conditions. Lets say that we know the temperature tion through upward flow of melt tend to deplete the lower crust at the Earth' s surface T0 and the heat flow Qm entering the base of and enrich the upper crust in radiogenic elements. The decrease the lithosphere. with depth of the radiogenic heat production is often modeled by A an exponential law: The first boundary condition leads to: C = T + 0 ⋅h 2 2 0 K c −z
A(z) = A0 . Exp( ) Qm A0 −zc hc ... and second to: C1 = − ⋅hc⋅Exp( ) K K hc This means that the radiogenic heat production measured at the surface ( A0 ) in divided by e (e=2.71) every hc meters (typically After substituting C1 and C2 into T(z), expanding and simplifying, 5000 m< hc <25000 m). Therefore the 1D conduction-advection one gets: dT K d2T A(z) dT A −z Q A −z heat transfer equation is: = ⋅ + − uz⋅ 0 2 m 0 c 2 T(z) = ⋅hc ⋅(1 − Exp( )) + ( − ⋅hc ⋅Exp( ))⋅z + T0 dt ρ⋅Cp dz ρ⋅Cp dz K hc K K hc In the case of a steady state geotherm, this equation becomes: If one knows the surface heat flow instead of the mantle heat flow then: d2T −A(z) = 2 A0 2 −z Q0 dz K T(z) = ⋅hc ⋅(1 − Exp( )) + ⋅z + T0 K hc K d2T A −z replacing A(z) we get that = − 0 ⋅Exp( ) 2 dz K hc Funbits: What love as to do with differential equations? Following two successive integrations one gets: https://fabiandablander.com/r/Linear-Love.html dT A0 −z = ⋅hc⋅Exp( ) + C1 and dz K hc
11 SECTION 2 Earth’s Geotherm
In this section TºC TºC TºC 1. Steady state geotherm 0 400 800 1200 1600 0 400 800 1200 1600 0 400 800 1200 1600
2. Transient geotherm 20 20 20
40 Moho 40 Moho 40 Moho to
t1 60 60 60 t5 Mechanical boundary layer t o t10 t100 Thermal boundary layer t1 t20
t5 80 80 t100 80 t10 t t20 8
t 100 100 8 100
to 120 120 120
140 t 140 140
8
Plume spreads at the top of the Plume spreads under the lithosphere thermal boundary layer Plume spreads at the Moho Depth (km) Depth (km) Depth (km)
Steady state geotherm tion). We then look at how sensitive is the geotherm to change In the previous section we have derived the 1D conduction- in mantle heat flow, conductivity and thickness of the radio- advection heat transfer equation, and we have derived an equa- genic crust. tion describing the equilibrium geotherm, also called the Distribution of radiogenic heat production “steady state geotherm” because the temperature at every depth doesn’t change through time. Here we explore how sensi- We have assumed so far that the volumetric radiogenic heat pro- tive is the geotherm to the radiogenic heat production when it is duction A was depth-independent ( A remains constant with not constant with depth (as we have assumed in the first sec- depth). However, because the upper crust is enriched in incom-
12 patible elements* the radiogenic heat production decreases with partial melting, they tend to concentrate into the melt phase. Due to the depth. A common model assumes that A is divided by e (e=2.71) buoyancy of the melt, incompatible elements concentrate over time into every h meters, h being the length scale of the exponential law. This the upper part of the crust. Through repeated cycles of melting, the con- model of distribution is given by: centration of radiogenic elements contributes to the cooling of the Earth’s geotherm. A(z) = A0⋅Exp( − z /h) Basal Heat Flow (Qm) ... in which A0 is the radiogenic heat production at z = 0 m. The graph below illustrates the sensitivity of the geotherm to the The graphs shows mantle heat flow (Qm), also -3 the geotherm for Radiogenic Heat (mW.m ) Temperature (TºC) called in some textbook ba- 2 4 6 8 500 1000 TºC A(z) decreasing sal heat flow. This graph 0 500 1000 1500 0 exponentially shows three geotherms cal- with various culated assuming same 25 25 20 length scale h (10, rate of radiogenic heat pro-
25, 50km). Ao is duction, same conductivity Moho Moho 40 adjusted so that 50 50 and same crustal thickness. the total radio- TMoho: Only the mantle heat flow 724, 60 Qm=36 Depth (km) genic heat produc- Depth (km) 935, varies. 10 km 75 75 1284 Qm=24 tion (R.H.P which 25 km is given by the in- 50 km A basal heat flow of 12 x 80 -2 10-3 W.m-2 (yellow geo- tegration of the R.H.P.: 0.078 W.m Qm=12 100 100 therm) is characteristic of 100 radiogenic heat Archaean (3.5-3 Ga) profile with cold and thick cratonic depth) is the same lithospheres, whereas a ba- -3 in all models. The larger is h (i.e. the deeper are the radiogenic ele- sal heat flow of 36 x 10
-2 Depth (km) ments) the hotter is the geotherm. This suggests that at an early W.m is representative of stage of the Earth evolution (before the extraction of the crust) the thinned lithosphere. In- Earth had a warmer geotherm. creasing the mantle heat flow from 12 to 36 x 10-3 W.m-2 increases the temperature at the * n.b.: Incompatible elements such as U, Th and K have a large radius and Moho from ~320 ºC to 720 ºC. therefore do not fit easily into crystals lattice. Hence, during episodes of
13 Thermal conductivity (K) thinning or fast thickening. During thinning the geotherm in- creases as warmer temperatures are reached at lesser depths (the The graph below illustrates the sensitivity of the geotherm to the slope of the geotherm increases). During thickening the geotherm thermal conductivity K (W.m-1.K-1). It shows three geotherms calcu- decreases as cooler conditions are met deeper in the crust (the lated assuming same rate of slope of the geotherm decreases). Second, as the thickness of the radiogenic heat production, TºC crust changes, the steady state geotherm also changes because a same same crustal thickness, 0 500 1000 1500 0 thicker crust will produce more radiogenic heat, therefore crustal and same mantle heat flow. thickening leads to warmer geotherm, whereas crustal thinning The conductivity is known to 20 leads to cooler geotherm. vary with temperature. Here however, we assume that K is Lithospheric thickening 40 Moho constant through depth. K is Thickening produces heat advection resulting in a rapid cooling of proportional to the ability of 60 k=1.75 the geotherm. Thickening also increases the thickness of the radio- material to conduct heat k=2.25 genic layer therefore increasing the production of radiogenic heat away. Therefore, the larger 80 in the lithosphere. Isostasy (cf. section Isostasy and Gravitational the thermal conductivity of k=2.75 Forces) produces uplift leading to erosion which in turn affects the rocks, the lesser the capacity 100 amount and distribution of the radiogenic heat elements in the of the crust to store heat, and crust which cools geotherm. The interplay between the mode of the the cooler is the geotherm. thickening (heterogeneous via thrusting vs homogeneous), the Increasing the thermal con-
Depth (km) thickening rate, and the erosion rate leads to contrasted thermal his- ductivity from 1.75 to 2.75 tories. W.m-1.K-1 decreases the tem- The graphs on the next page show transient geotherms (at 0, 0.5, 2, perature at the Moho from 10...Myr) following heterogeneous thickening (thickening is 750 to 500ºC. achieved by doubling the thickness of the crust via a single thrust, Thickness of the radiogenic crust which explains the temperature discontinuity at t=0), and homoge- neous thickening (the thicknesses of the crust and the lithospheric Crustal thickening or thinning modify the geotherm in two differ- mantle are doubled via pure shear deformation). Erosion is dis- ent ways. First, during deformation, heat is advected mainly up- carded here. The discontinuity in the case of heterogeneous thicken- ward (in the case of thinning) and mainly downward (in the case of ing is smoothed out in a few Ma. Transient geotherms in both cases thickening) as rocks carry their temperature with them during fast are similar for time > 10 Myr.
14 stasy leads the subsidence of the surface of the lithosphere leading 0 200 400 600 800 1000 1200 TºC 0 200 400 600 800 1000 1200 TºC 0 0 to sedimentation which in turn affects the amount and distribution of the radiogenic heat elements in the crust, which also impacts on the geotherm. The interplay between the geometry of the thinning, 10 myr the thinning rate, and the sedimentation rate lead to contrasted 50 50 50 myr 0.5 myr2 myr 10 myr 250 myr 50 myr 250 myr thermal histories.
Moho Moho The graph on the bottom left shows transient geotherms (0, 10, 50, 100, 250... myr) following homogeneous thinning (the thicknesses 100 100 of the crust and the lithospheric mantle are halved by pure shear deformation). Sedimentation is discarded here. Following the in- crease of the geothermal gradient due to extensional deformation, 150 150 thermal relaxation leads to cooling and therefore the thickening of Homogeneous thickening via pure shear the lithosphere. Slowly, the transient geotherm approaches the Heterogeneous thickening via thrusting doubling the thickness of the lithosphere and doubling the thickness of the crust steady state geotherm. Depth (km) Depth (km) Sedimentation of burial of the crust
Sedimentation and burial of the continental crust under a few kilo- Lithospheric thinning 0 200 400 600 800 1000 1200 TºC meter of volcano-sedimentary rocks have a significant long-term 0 Thinning drives heat advec- effect on the geotherm. The newly deposited layers effectively insu- late the buried heat producing layer. If the conductivity of the up- tion as rocks, and the heat at- Moho tached to them, are displaced per layer is lower or equal to that of the buried layer, then heat ac- vertically mainly upward re- cumulates increasing the geotherm. 50 sulting in a rapid warming of This effect could have played a major role in the differentiation of the geotherm (the geothermal 10 myr the continental crust in the Archaean. At that time, radiogenic heat 50 myr 250 myr production in the crust was 2 to 6 time larger that of present day gradient increases). However, 100 myr thinning also decreases the rate of radiogenic heat production. Furthermore it was a time 100 thickness of the radiogenic when 5 to 20 km thick continental flood basalts (the so called green- crust, and therefore reduces stones) where deposited at the surface of the Earth, insulating the 60% homogeneous thinning of heat producing crust. the production of radiogenic the lithosphere via pure shear heat in the lithosphere. Iso- Depth (km) The graphs show transient geotherms following the emplacement
15 thick layer with a temperature of 1700ºC. The three graphs show Archaean continental lithosphere Archaean continental lithosphere TºC TºC Greenstone 200 400 600 800 1000 1200 1400 Greenstone 200 400 600 800 1000 1200 1400 the results for various depth of emplacement.
Crust Crust Crust 25 25 TºC TºC TºC
Geotherm t 8 Geotherm to Geotherm to 0 400 800 1200 1600 0 400 800 1200 1600 0 400 800 1200 1600 t +10Ma to+50Ma o 50 50 to+20Ma
Geotherm t 8 20 20 20 to+100Ma to+30Ma
75 75 Moho to to+200Ma 40 40 Moho 40 Moho Mantle Mantle to+40Ma t1 t +400Ma o 60 60 60 t5 t +50Ma Mechanical boundary layer t 100 100 o o t10 t100 Thermal boundary layer t1 t20 to+100Ma t5 80 80 t100 80 t10 to+200Ma t 125 125 8 t20 to+1Ga t 100 100 8 100 Depth (km) Depth (km) 150 150 to 120 120 120
175 175 140 t 140 140
8
Plume spreads at the top of the Plume spreads under the lithosphere thermal boundary layer Plume spreads at the Moho Depth (km) of a 6 km (graph on the left) or 12 km thick greenstone covers Depth (km) Depth (km) (graph on the right), with no radiogenic heat production in them. Transient geotherms The temperature increase is large enough to lead to significant par- tial melting of the crust. Calculation of the steady state geotherm is relatively straightfor- ward. In contrast, the calculation of transient geotherms, such as Emplacement of a mantle plume those displayed above, requires computational tricks as the tem- perature changes in both space and time. Many analytical solutions Mantle plumes initiate at the core-mantle boundary and rise fast have been proposed for a range of problems. The well-known book through the convective mantle. Upon approaching the more rigid from Carslaw and Jagger, first published in 1946 "Conduction of lithosphere, they spread laterally under the lithosphere. They may Heat in Solids", provides hundreds of analytical solutions to many also thermally erode to base of the lithosphere and spread higher problems. Here, we give a couple of them. up. This is equivalent to put a hot layer, a few tens of kilometer thick, underneath or within the colder lithosphere. In the Ar- Progressive cooling of the oceanic lithosphere chaean, the Earth was warmer and plumes were most likely more numerous. The next graphs document the thermal impact of a man- The formation of oceanic lithosphere at mid-oceanic-ridge is a ther- tle plume assuming that the plume's head spreads into a 50 km mal problem involving the progressive cooling of the astheno- sphere. The temperature at the seafloor Ts is maintained constant,
16 so is the temperature in the asthenosphere Tm. The lithosphere be- (he assumed Tm =2000ºC) and that Earth had comes cooler and therefore thicker as cooling proceeds. Neglecting cooled by conduction to its present surface -3 -2 radiogenic heat in the basaltic crust and the mantle, and assuming heat flow (he assumed Q0= -30 10 W.m ). He no sedimentation, the 1D heat transfer equation becomes: assumed that all heat was lost at the surface by conduction (with κ = 10-6.m-2.s-1). Thus the dT d2T = κ⋅ problem is reduced to that of finding the tem- dt dz2 perature within a cooling half-space of infi- nite extent as a function of time after the half with the thermal diffusivity which can be expressed in terms of κ space is set at a specific temperature, a prob- conductivity (K), density ( ) and heat capacity (Cp): ρ κ = K⋅ρ⋅Cp lem for which the equations above apply. Kelvin found that the Earth was 50 myr old. This 2 order of magnitude off the mark, but The analytical solution of this differential equation is: still much better than the biblical 6000 yr. z T(z, t) = Ts + (Tm − Ts)⋅erf( ) Cooling of a dike 2 κ⋅t The cooling of a dike with half width w is another problem which By differentiating with respect to z one get the temperature gradi- has an analytical solution of the following form: ent ... T w − x w + x s dT T − T z2 T(x, t) = ⋅(erf( ) + erf( )) = m s ⋅Exp( − ) 2 2 κ⋅t 2 κ⋅t dz π⋅κ⋅t 4⋅κ⋅t If the dike has a width of 2 m, i.e. w=1 m, and its initial tempera- -6 2 -1 ...from which the variation of the surface heat flow through time ture was Ts = 1000 °C, and κ = 10 m s , then the temperature at can get extracted as: the center of the dike would be about 640°C after one week, 340°C after one month, and only 100 °C after one year. dT Tm − Ts = Some swarms, such as the giant McKenzie swarm in Canada, ap- dz π⋅κ⋅t pear to radiate from a point, commonly interpreted as a plume source for the magmas. The mid-Proterozoic Coppermine River In the 19th century, Lord Kelvin used this equation to find out the flood basalts were erupted at the same time near the plume head. age of the Earth. Kelvin made the assumption that the Earth had Individual dykes range from 10 to 50 m in width, with some up to formed as a molten body at the temperature at which basalt melts 200 m wide. Some dykes can be traced for up to 2000 km.
17 Numerical solutions and therefore:
Since the arrival of com- ∂T L ∂2T A ∂T puters, which allow for (1) − (1 − ) = κ⋅ + − U ∂t C ⋅(T − T ) ∂z2 ρ⋅C ∂z large numbers of calcula- p liq sol p tion to be performed Boundary conditions: T(z, 0)=0; T(0, t)=T0; dT/dz(zl, t)=-Qm/K very fast, new tech- niques based on numeri- An approximation of this differential equation can be obtained by cal algorithms have been replacing ∂t and ∂z with finite differences ∆t and h. The central designed to solve differ- tenet of this computational technique is that for a given time t=n, ential equations. These the temperature at a depth z = i (Tin) can be calculated from knowl- numerical recipes are based on the "discretization" of differential edge of the temperature at depth z = i-h and z = i+h. In a similar equations ... fashion, for a given depth z = i, temperature at time t = n can be cal- culated from knowledge or the temperature at time t = n-∆t and • Consider the 1D heat conduction-advection equation in a slab zl temperature at time t = n+ ∆t. The level a accuracy depend of the thick. The upper surface of the slab has at temperature To. The tran- size of the time and space finite differences. There are many ways sient temperature is then described by the conductive-advective to express T(z, t) as a function of T(z-h, t), T(z+h, t), T(z, t- ∆), T(z, t+ equation of heat balance. Here we consider a situation where inter- ∆). Here we present the Crank-Nicholson scheme where, equation nal radiogenic heat is creation at a rate A, heat is lost or gain by ad- (1) is rewritten as: vection at speed U (since z increases downward, if erosion U < 0, if sedimentation U > 0), and heat is lost through phase transition (2) solid-liquid (L < 0 for partial melting, L > 0 for crystallization) X T n+1 − T n L the melt fraction is a function of T and therefore z. i i − 1 − Δt ( Cp ⋅(Tliq − Tsol) ) ∂T ∂2T A ∂T ∂X L = κ⋅ + − U + ⋅ n n n n+1 n+1 n+1 n+1 n 2 1 Ti+1 − 2Ti + Ti−1 Ti+1 − 2Ti + Ti−1 T − T A ∂t ∂z ρ⋅C ∂z ∂t C = κ⋅ ⋅ + − U i i + p p 2 2 2 ( h h ) h ρ⋅Cp re-arranging we get: where: n n+1 Ti = T(zi, tn), T i = T(zi, tn+1), etc ∂T 1 ∂T ∂2T A ∂T − ( ⋅ ) = κ⋅ + − U 2 ∂t Tliq − Tsol ∂t ∂z ρ⋅Cp ∂z
18 The finite difference equation (2) is known as the Crank-Nicholson This formulation results in N equations and N+4 unknowns, the n n n+1 n+1 scheme. It is based on central differences for the spatial derivatives aver- four additional unknowns being u0 and uN+1 , u0 , uN+1 . How- aged forward in time over time steps n and n+1. ever, these four additional unknowns lie outside the computational grid (time, space). Also, from the boundary conditions the follow- Expanding (2) and re-arranging to express temperature at time n+1 ing conditions must be satisfied at the nth and n+1 time steps: as a function of the temperature at time n we get equation (3): T n = 1, T n = 0, T n+1 = 1, T n+1 = 0 −κ ⋅Δt L κ ⋅Δt UΔt −κ ⋅Δt 1 N 1 N ⋅T1+n + 1 − + + ⋅T1+n + ⋅T1+n 2 −1+i 2 i 2 1+i 2h ( Cp ⋅(Tliq − Tsol) h h ) ( 2h ) This gives the following set of N equations (5) for the temperature at the n+1 time step : κ ⋅Δt n L κ ⋅Δt UΔt n κ ⋅Δt n A⋅Δt = ⋅T + 1 − − + ⋅Ti + ⋅T + 2h2 −1+i ( C ⋅(T − T ) h2 h ) ( 2h2 1+i) ρ⋅C p liq sol p A⋅Δt a ⋅T1+n + b ⋅T1+n + c ⋅T1+n = a ⋅T n + b ⋅T n + c ⋅T n + , i = 1 replacing and substituting: 11 0 11 i 11 2 1 0 1 1 1 2 k A⋅Δt κ ⋅Δt U⋅Δt L 1+n 1+n 1+n n n n R = , V = , Lt = a1i ⋅Ti−1 + b1i ⋅Ti + c1i ⋅Ti+1 = a1 ⋅Ti−1 + bi ⋅Ti + c1 ⋅Ti+1 + , i = 2 to N − 1 d 2h2 h C (T − T ) k p liq sol A⋅Δt 1+n 1+n 1+n n n n a1N ⋅TN−1 + b1N ⋅TN + c1N ⋅TN+1 = aN ⋅TN−1 + bN ⋅TN + cN ⋅TN+1 + , i = N we get... k with (6): 1+n 1+n 1+n −Rd ⋅T−1+i + (1 + 2Rd + V − Lt)⋅Ti − Rd ⋅T1+i
A⋅Δt b11 = 1, c11 = 0, a1N = 0, b1N = 1 = R ⋅T n + (1 − 2R − Lt + V)⋅T n + (R ⋅T n ) + d −1+i d i d 1+i ρ⋅C p b1 = 1, c1 = 0, aN = 0, bN = 0 further replacement... These latter conditions ensure that the boundary conditions are al- ways satisfied. a1i = − Rd, c1i = −Rd, b1i = (1 + 2Rd + V − Lt) The N equations above can be rearranged into the following matrix and a = Rd, c = Rd, b = (1 − 2R + V − Lt) i i i d equations: Rearranging terms, we obtain the following set of implicit equa- tions for the temperature at the n+1 time step in terms of the tem- perature at the nth time step (4): A⋅Δt a ⋅T1+n + b ⋅T1+n + c ⋅T1+n = a ⋅T n + b ⋅T n + c ⋅T n + = dn 1i −1+i 1i i 1i 1+i i −1+i i i i 1+i k i
19 c .. .. Numerical solutions an example b11 11 0 0 0 n+1 n T1 d .. .. 1 500 1000 TºC a12 b12 c12 0 0 n+1 n T2 d • We illustrate here the problem .. 2 0 a13 b13 c13 0 0 n+1 n where a continental crust with initial T3 d3 0 ...... 0 ⋅ . = . (7) . . thickness zc is thickened by a factor 2 0 ...... 0 n n+1 d via the emplacement of one single zt 50 ...... a c TN−1 N−1 0 1N−1 b1N−1 1N−1 n thick thrust. z = 40 km, z = 40 km. n+1 dN c t ...... TN 0 a1N b1N Moho The figure on the right shows the in- 100 with: stantaneous and steady state geo- therm (potential geotherm reached c .. .. n b1 1 0 0 0 after infinite time). The aims is to dis- d1 1 a2 b c2 0 .. .. 0 n play transient geotherms down to a Heterogeneous thickening via thrusting n 2 Depth (km) and doubling the thickness of the crust d2 T2 depth z = zl + zt at various time inter- n 0 a3 b3 c3 0 .. 0 n d3 T3 A⋅Δt . = 0 ...... 0 ⋅ . ⋅ vals, where zl is the thickness of the . . k n 0 ...... 0 n lithosphere before thickening zl = 120 km. d TN−1 N−1 ...... n 0 aN−1 bN−1 cN−1 n dN TN • We choose a spatial finite difference h = 4000 m (spatial grid). 0 ...... aN b N Therefore the number of column in the tridiagonal matrixes will be
zl/h. The Crank Nicholson scheme imposes the maximum time step Thus the RHS of (7) is determined explicitly from the solution at of h2/(2.κ) = 253,000 yr. the nth time step. Note that the boundary conditions are satisfied by ensuring that the coefficients given in (6) are satisfied. Hence in or- • We get that Rd = (κ.∆t)/(2h2) = 0.25. We assume no erosion: v = 0; der to find the solution at the time step, one must solve N linear and we allow for partial melting: equations. Since the coefficient matrixes in (7) and (8) are tridiago- Lt=Latent_heat/(Cp.(Tliquidus-Tsolidus)). With Rd, v and Lt we can nal, one can make use of efficient algorithms (e.g. Thomas algo- determine ai1, bi1, ci1, the coefficient of the tridiagonal matrix at rithm) to find the solution. time n+1, and ai, bi, and ci the coefficient of the tridiagonal matrix at time n.
We choose to get the transient geotherm at 20 Myr interval up to 400 myr. The depth of the tridiagonal matrixes (number of row) is therefore 400/20 = 20.
20 • With this, we construct the tridiagonal coefficient matrixes (7) and (8) , the equation system (7) is solved via matrix inversion (Numpy, R, MatLab or Mathematica are well equipped to do the 0 200 400 600 800 1000 1200 TºC 0 dirty work...). Graph on the right show transient geotherms from 0.5 to 250 myr and to infinite time.
50 0.5 myr2 myr 10 myr 50 myr 250 myr
Moho (7)
100
150
Heterogeneous thickening via thrusting (8) and doubling the thickness of the crust Depth (km)
21 Review 1.1 Heat transfer & the Geotherm Review 1.2 Heat transfer & the Geotherm Review 1.3 Heat transfer & the Geotherm
The relationship between heat and Using the Taylor series and bearing in mind the Fou- List five processes consuming or re- temperature is: rier’s law: Q = -K . ( dT/dz), which of the following leasing heat into a volume of rock is/are correct: somewhere in the crust.
A. T(z+dz) = T(z) + dz x dT A. A. T = E x Cp x m
B. T(z+dz) = T(z) + dz x (dT/dz) B. B. dE = dT / (Cp x m)
C. T(z+dz) = T(z) x (dT/dz) C. C. dE = dT x (Cp x m)
D. T(z+dz) = T(z) - dz x (Q/K) D. D. dT=dE / (Cp x m) E. T(z+dz) = T(z) + (dT/dz) x (Q/K) E.
Check Answer Check Answer Check Answer
In a mathematical equation involving dimensional parameters, physical dimensions such as meter (m), second (s), kilogram (kg), Joule (J), Kel- vin (K) or degree centigrade (ºC) can be balanced to verify whether or not a proposed relationship between physical quantities is correct. For in- stance lets propose that the velocity (v) is a relationship between time (t) and distance (l) such that : v = t / l2. Knowing that the unit of velocity is m/s, it is clear that the proposed relationship is wrong since the unit of the right hand side of the proposed equation is s.m-2 . Considering that velocity is in m/s, one can easily figure out that the correct relationship must be: v = l/t. Review 1.4 Heat transfer and Geotherm Expected learning outcomes
Which of the following is/are correct: At this stage you should be able to:
• Explain why and how temperature increases with depth. This means: 1/ being able to comment on the A. The long term consequence of crustal various terms of the equation describing the steady state geotherm, and the meaning for the various pa- thinning is a hotter geotherm. rameters; and 2/ being able to explain the shape the geotherm in a temperature versus depth graph. This B. The long term consequence of crustal should involve concepts like conductivity, advection, radiogenic heat production etc. thickening is a hotter geotherm. • Explain the relationship between heat flow, the tem- C. The long term consequence of deposi- perature gradient and thermal conductivity. tion of a thick layer of sediments is a • Explain how tectonic and other processes can colder crustal geotherm. change the lithospheric geotherm i/ on the short terms through advective heat transfer, and ii/ on the D. Higher rock conductivity results in long term through heat conduction. hotter geotherm. • At a more advanced level you should be able to de- rive an expression of the steady state geotherm from E. For the same total radiogenic iso- the 1-D conduction-advection heat transfer equation topes content, a colder syeady steady and knowledge of two boundary conditions. state geotherm is reach when radio- genic isotopes concentrate in the up- per crust.
Check Answer CHAPTER 2 Isostasy & Gravitational Forces The Tibetan plateau stands 5000 m above sea level. It is the surface expression of a 75 km thick continental crust, which reduces the weight of the continental lithosphere. It imparts a significant horizontal gravitational push force on its surrounding. Isostasy is the physical process that explains the surface elevation of the lithosphere at rest. It is based on the Archimedes’ principle which relates the elevation of a floating body above the surrounding fluid i/ to the density contrast between the floating body and the supporting fluid, and ii/ the vertical length of the floating body.
24 SECTION 1 Isostasy, elevation, gravitational potential energy & gravitational force
Photo: DscobiePhotography.co.uk/creativecommon At rest, a lithosphere with a thick crust stands above equipotential surfaces, introduces gravitational forces adjacent non-thicken regions, and it is said to be in iso- driving the flow of rocks from regions of high pressure static equilibrium. The notion of isostatic equilibrium to regions of low pressure. In this section, we docu- must not be confused with mechanical equilibrium. In- ment the concept of isostasy and its relationship to sur- deed, a horizontal gravitational force (a volume force) face elevation. We then introduce the notion of gravita- appears when density interfaces are no longer parallel tional potential energy and that of horizontal gravita- to gravity equipotential surfaces. Hence, the action of tional force. isostasy, which forces density interfaces away from
25 The sketch below represents a schematic cross-section through a lithospheric plate at rest (i.e. no tectonics). Although the thicknesses of both the crust (in pink) and the sub-continental lithospheric mantle (in green) vary, isostatic equilibrium is maintained (i.e. no vertical mo- tion is observed) which explains the lateral variation of the surface topography. A B C • Isostatic equilibrium implies that at - and below - a particular depth called the compensation depth, the pressure becomes hydrostatic (i.e. the pressure shows no lateral variation). In other terms, at or be- low the compensation level the weight of columns with the same cross-sectional areas standing on the same gravitational equipotential surface are the same. The minimum depth for the compensation level is that of the base of the thickest lithospheric col- umn, in other terms it is the minimum depth as which there is no lateral change in density. Compensation level
Isostatic equilibrium=>Lithostatic pressure@z = constant for z > Compensation level
• Isostasy controls the elevation of the Earth's surface. Lithospheres with thin/thick continental crusts have lower/higher surface elevation compared to the average lithosphere. One can easily determine the surface elevation of column A and C with respect to the elevation of the undeformed column B, the surface of which is assumed here to be at sea level. When this heterogeneous lithosphere is at rest then isostatic equilibrium is reached via differential uplift and subsidence, and the pressure at the base of columns A, B and C, all standing on the com- pensation level, is the same. The pressure at the base of column B is: PB = ρcrust ⋅g⋅zc + ρsclm ⋅g⋅(zl − zc) + ρasth ⋅g⋅( fl ⋅zl + h − zl), and the pres- sure at the base of column C is: PC = ρcrust ⋅g⋅fc ⋅zc + ρsclm ⋅g⋅( fl ⋅zl − fc ⋅zc) where zc is the thickness of the crust, zl is the thickness of the lithosphere (i.e. crust plus sub-continental lithospheric mantle), fc is the verti- cal strain factor for the crust (i.e. the ratio of the thickness of the deformed crust over the thickness of the initial undeformed crust), fl is the vertical strain factor for the entire lithosphere. h the elevation of column C, and ρcrust, ρsclm and ρasth are the density of the crust, the sub- continental lithospheric mantle and the asthenosphere respectively. To calculate elevation of the high plateau h one has to solve PB = PC . To calculate the bathymetry of column A one has to solve PA = PB .
26 The sketch on the right illustrates the variation with depth of the lithostatic pressure B C zz = .g.z(MPa) (P(z) = ρ⋅g⋅z) along two lithospheric columns (B and C) in isostatic equilibrium. We assumed here that the density of the crust and that of the mantle are the same in both columns (with ρ ≤ ρ ), and that there is no density difference between the litho- gz (C) c m gz (B) spheric mantle and the asthenosphere. Because it has a thicker crust, column C has a surface elevation higher than that of column B. Consequently, the lithostatic pressure GPE (C) at any depth within the crust of column C is higher than the lithostatic pressure at the same depth in column B. However, at or below the compensation depth (here the base of crust in column C) the lithostatic pressure measured at a particular depth is GPE (B) the same in both column. This is because both column are in isostatic equilibrium.
Isostatic equilibrium does not mean mechanical equilibrium. Indeed column C has a larger Gravitational Potential Energy (GPE in Pa) than column B. On the graph the GPE of each column is equal to the area under their respective lithostatic pressure GPE (B) < GPE (C) => Horizontal force acting from C to B profile and down to the compensation level (i.e. orange area for column B, yellow area for column C which extends underneath the orange area). What is the implication of this?
The graph on the lower right shows water jets driven by the hydrostatic pressure in a leaking tank. Water tank The horizontal outreach of each jet is proportional to the hydrostatic pressure at the outlet. If we con- free surface nect two tanks of same height but different water depth, the difference in water level - and therefore Low pressure in pressure - drives an outflow has shown in the sketch on the dropping free surface left. This flow stops when the water level, and therefore the pres- sure, in both tanks equalizes. The same physics applies to our two rising free surface lithospheric columns. Hence, lateral variations of density produce horizontal gravitational forces (body forces) acting inside the
Out!ow pressure Increasing lithosphere and pushing material laterally from high pressure to low pressure regions. The flow stops when all the density inter- High pressure faces are parallel to equipotential gravitational surfaces, or when the gravitational deviatoric stresses are no longer strong enough to overcome the resistance to flow.
27 The gravitational force (Fg in N.m-1) acting between C and B is equal to the difference of their B C zz = .g.z(MPa) respective GPE (difference between the two colored surface areas on the graph, i.e. the thin
-1 gz (C) yellow wedge on the graph). The excess in GPE in column C (Fg > 0 N.m ) will drive exten- gz (B) sion, whereas a deficit (Fg < 0 N.m-1) will drive contraction. Mathematically: GPE (C) Top Top GPE = ρ (z)⋅g⋅z dz and GPE = ρ (z)⋅g⋅z dz C ∫ C B ∫ B GPE (B) Bottom Bottom
Fg = GPEC − GPEB GPE (B) < GPE (C) => Horizontal force acting from C to B
This movie below is the result of a numerical experiment documenting the gravitational collapse of an orogenic plateau. This model shows a 65 km thick crust adjacent to a 40 km thick crust (brown), with the upper mantle in green. The experiment starts at a stage when thermal relaxation has produced 20 to 25% of melt (dark pink) in the plateau’s lower crust. The only force acting is the gravitational force (i.e. the vertical sides are fixed). The flow of the orogenic plateau is powered by the same physics responsible of the spreading of camembert.
Movie 2.1 Collapse of orogenic plateaux
28 zz (MPa) THE CASE OF LITHOSPHERIC THICKENING AND DEBLOBBING Compression 50 100 Tension The figures on the right show the gravitational force (Fg) due to instantaneous homogeneous thick- ening (at time to), and following convective thinning (to + 30 myr). Convective thinning is the gravi- 50 tational process upon which the cold and dense lithospheric keel is dragged into the convective mantle. Following homogeneous thickening, the excess in gravitational potential energy in the up- Depth (km) per part of the thickened lithosphere is more or less balanced by a deficit in gravitational potential 100
energy in the deeper part of the lithosphere. Indeed the integration of σzz (i.e. the lithostatic pres- sure) with depth gives a very small gravitational force (Fg = 0.77 1012 N.m-1). 150
Following convective thinning (the cold and heavy lithospheric keel has been removed), the de- -1 to Fg=0.77xTNm formed lithosphere has a larger excess in gravitational potential energy. This excess in GPE gives a (MPa) 12 -1 zz gravitational force Fg = 6.7 10 N.m . Such an Fg could easily balance or overcome tectonic forces, -50 50 100 150 and in some circumstances could be sufficient to drive extensional collapse of the mountain belt. 20
40
60 THE CASE OF LITHOSPHERIC THINNING AND THERMAL RELAXATION 80 -1 100 Fg=6.7xTNm t +30Ma The two last sketches show the gravitational force (Fg) due to instantaneous homogeneous stretch- o zz (MPa) Compression -50 50 Tension ing (at time to) and following 200 myr of thermal relaxation. The yellow shaded areas represent the gravitational force acting on the lithosphere. Fg is given by the integration with depth of the differ-
50 ence in lithostatic pressure (σzz) between the thinned and the undeformed lithosphere. Depth (km) Following homogeneous thinning by a factor of 2 (50% reduction of the thickness of the crust and 100 -1 to Fg=0.17xTNm that the whole lithosphere), Fg is rather small (< 1012 N.m-1).