Geodynamics Lecture 10 The forces driving plate tectonics Lecturer: David Whipp! david.whipp@helsinki.fi!! 2.10.2014 Geodynamics www.helsinki.fi/yliopisto 1
Goals of this lecture Describe how thermal convection operates according to basic physical principles!! Quantify the basic plate-driving forces! Slab pull Ridge push Drag force 2
Mantle convection Compositional stratification around 660 km Secondary plumes Mid-ocean ridge Deflected slab Primary plume Enriched piles Penetrating slab Continent Slab grave yards 660 km discontinuity Core mantle boundary Tackley, 2008 Plate tectonics is ultimately driven by convection in the mantle reflecting heat produced and lost as the Earth cools! How convection operates in the mantle is somewhat controversial, but there is clear evidence of fossil plates in the upper and lower mantle 3
Layered versus whole-mantle convection Layered and whole-mantle convection are endmember models of mantle convection!! Layered convection is based on the strong seismic discontinuity at 670 km, among other lines of evidence, with the idea that this discontinuity separates upper and lower mantle circulation http://instruct.uwo.ca 4
Thermal convection A Rayleigh-Taylor instability occurs when a denser fluid overlies a less dense fluid, which is gravitationally! unstable (the denser fluid wants to sink, and the less dense fluid wants to rise due to its buoyancy) Fig. 6.21, Turcotte and Schubert, 2014 A fluid heated from beneath will undergo thermal expansion and may become less dense than the cooler overlying fluid! This scenario is gravitationally unstable! If the denser fluid is able to sink into the less dense fluid, thermal convection will occur 5
Thermal convection Fig. 6.21, Turcotte and Schubert, 2014 What is going to happen to these two fluids with time? 6
Thermal convection Time evolution of a Rayleigh-Taylor instability 7
Thermal convection Thermal convection in the Earth results from buoyancy forces owing to thermal expansion of mantle rocks! For an incompressible viscous fluid, the force balance of pressure, gravity and viscous forces in 2D is 0= @p @ 2 @y + g + v @x 2 + @2 v @y 2 where! is pressure, " is density, # is viscosity and $ is velocity in the % direction! To account for the buoyancy forces from thermal expansion, the density term must be variable = 0 + 0 where "0 is a reference density and "ʹ is a density perturbation much smaller than "0 8
Thermal convection If the variable density is substituted into the 2D force balance, and the hydrostatic pressure based on the reference density is eliminated by using ' =! - "0(% we find! 0= @P @y + 0 g + @ 2 v @x 2 + @2 v @y 2 A change in density as a result of thermal expansion is 0 = 0 v (T T 0 ) where )$ is the volumetric coefficient of thermal expansion and *0 is a reference temperature corresponding to the reference density "0! The vertical force balance including thermal buoyancy is thus 0= @P @ 2 @y + v @x 2 + @2 v @y 2 g 0 v (T T 0 ) 9
Thermal convection In order to determine the thermal buoyancy, we require a 2D equation for heat transfer via conduction and convection! Turcotte and Schubert give a detailed derivation, arriving at the following relationship for heat conduction and convection in 2D @T @t + u@t @x + v @T @ 2 @y = apple T @x 2 + @2 T @y 2 where + is the velocity in the, direction and - is the thermal diffusivity 10
Thermal convection In order to determine the thermal buoyancy, we require a 2D equation for heat transfer via conduction and convection! Turcotte and Schubert give a detailed derivation, arriving at the following relationship for heat conduction and convection in 2D @T @t + u@t @x + v @T @ 2 @y = apple T @x 2 + @2 T @y 2 Time dependence Convection Conduction where + is the velocity in the, direction and - is the thermal diffusivity 11
Onset of thermal convection For a fluid heated to temperature *1 at its base and cooled to temperature *0 at its upper surface thermal buoyancy will drive convection if the viscous resistance to fluid flow is overcome! Fig. 6.38, Turcotte and Schubert, 2014 When the temperature difference *1 - *0 is small, convection will not occur and the fluid velocities will be + = $ = 0, and temperature will not change with time (././ = 0) or along the, axis (./., = 0), reducing the heat transfer equation to d 2 T c dy 2 =0 where 0 indicates conductive heat transfer 12
Onset of thermal convection For the thermal boundary conditions * = *0 at % = -1/2 and * = *1 at % = 1/2, the solution to the heat transfer equation is T c = T 1 + T 0 2 + (T 1 T 0 ) b where 1 is the thickness of the fluid layer! Convection will begin when the temperature * just begins to exceed the conductive temperature *0! y Fig. 6.38, Turcotte and Schubert, 2014 In this case, the temperature difference *ʹ is extremely small T 0 T T c 13
Onset of thermal convection The temperature difference *ʹ can be inserted into the heat transfer and force balance equations for a 2D incompressible viscous fluid to determine the conditions under which convection will occur! Fig. 6.38, Turcotte and Schubert, 2014 From this, a dimensionless number known as the Rayleigh number can be defined Ra = 0g v (T 1 T 0 )b 3 apple 14
Onset of thermal convection Fig. 6.39, Turcotte and Schubert, 2014 Instability and convection will occur when Ra > ( 2 + 4 2 b 2 2 ) 3 4 2 b 2 2 where 2 is the wavelength of convection! Minimum at =2 p 2b The flow is thus stable when Ra is less than the right side of the equation above! The critical Rayleigh number when convection begins is thus Ra Ra cr = ( 2 + 4 2 b 2 2 ) 3 4 2 b 2 2 15
Onset of thermal convection For a fluid cooled from above, heated from within and with no heat flux across its base the Rayleigh number equation is Ra H = v 2 0gHb 5 k apple where 3 is the heat production per unit mass and 4 is the thermal conductivity! The equation above is appropriate for the Earth s mantle 16
Onset of thermal convection!! Ra H = v 2 0gHb 5 k apple The critical Rayleigh number for an internally heated fluid is Racr = 2772! What is the Rayleigh number for the upper mantle?! Should it convect?! What about the whole mantle?!! Assume )$ = 3 10-5 K -1, "0 = 4000 kg m -3, 3 = 9 10-12 W kg -1, 4 = 4 W m -1 K -1, # = 1 10 21 Pa s, and - = 1 mm 2 s -1 17
What drives tectonic plate motions? Forsyth and Uyeda, 1975 Driving forces FDF Resisting forces FDF FSP FTF FCD FRP FCR FSR 18
What drives tectonic plate motions? Forsyth and Uyeda, 1975 Driving forces FDF Resisting forces FDF FSP FTF FCD FRP FCR FSR 19
Slab (or trench) pull FSP Forsyth and Uyeda, 1975 Slab pull results from the gravitational body force acting on the dense, sinking oceanic lithosphere.! It can be divided into two components:! Fb1, the force resulting from the slab being colder than the surrounding mantle! Fb2, the force resulting from the elevation of the olivine spinel phase change! Mathematically, we can say F SP = F b1 + F b2 20
Slab (or trench) pull FSP For one half of a sinking mantle plume, Turcotte and Schubert show F b =2 0 g v b(t c T 0 ) u 1/2 0 apple v 0 2 u 0 We can simplify this relationship slightly for a sinking slab to estimate the slab pull force resulting from the relatively cold temperature of the slab " Reference mantle density * Temperature at LAB ( Gravitational acceleration + Horiz. velocity of upper layer ) Coeff. of thermal expansion $ Average vertical velocity 1 Convecting layer thickness - Thermal diffusivity Forsyth and Uyeda, 1975 * Temperature at conv. center 2 Width of 2 convection cells LAB = Lithosphere-asthenosphere boundary 21
Slab (or trench) pull FSP Forsyth and Uyeda, 1975 Here is the simplified convection cell geometry used for calculating the force acting on a sinking mantle plume!! Turcotte and Schubert, 2002 Note that areas in each triangle are equal, so mass is conserved, or v 0 2 = u 0b 22
Slab (or trench) pull FSP Forsyth and Uyeda, 1975 Because a sinking slab is rigid, we can assume that +06=6$0, so the force resulting from the slab being colder than the surrounding mantle is 1/2 apple F b1 =2 0 g v b(t c T 0 ) 2 u 0 If we use typical values for these variables, we see that 8116is6~3 10 13 N m -1! To be clear, this value is the force per meter of trench length 23
Slab (or trench) pull FSP Fig. 4.59, Turcotte and Schubert, 2014 Forsyth and Uyeda, 1975 The transition from olivine to spinel in the mantle generally occurs at 400-500 km depth, but the transition depends on both pressure and temperature! Because the descending slab is relatively cold, the olivine-spinel phase transition occurs at lower pressure (shallower depth) Turcotte and Schubert, 2002 24
Slab (or trench) pull FSP Fig. 4.59, Turcotte and Schubert, 2014 Forsyth and Uyeda, 1975 Warm slab Cold slab The transition from olivine to spinel in the mantle generally occurs at 400-500 km depth, but the transition depends on both pressure and temperature! Because the descending slab is relatively cold, the olivine-spinel phase transition occurs at lower pressure (shallower depth) Turcotte and Schubert, 2002 25
Slab (or trench) pull FSP Fig. 4.59, Turcotte and Schubert, 2014 Forsyth and Uyeda, 1975 Warm slab Cold slab The transition from olivine to spinel in the mantle generally occurs at 400-500 km depth, but the transition depends on both pressure and temperature! Because the descending slab is relatively cold, the olivine-spinel phase transition occurs at lower pressure (shallower depth) Fig. 6.43, Turcotte and Schubert, 2014 26
Slab (or trench) pull FSP Forsyth and Uyeda, 1975 Thus, the driving force from the phase transition depends on the location of the phase transition isotherm *os in the sinking slab! The driving force from the phase transition is F b2 = 2(T c T 0 ) os apple 0 2 u 0 1/2 The olivine-spinel phase transition increases density by ~270 kg m -3 @ Slope of Clapeyron curve A" Gravitational acceleration Fig. 6.43, Turcotte and Schubert, 2014 27
Slab (or trench) pull FSP Using typical values for a sinking slab, the gravitational body force due to the olivine-spinel phase transition in the slab 812 is ~1.5 10 13 N m -1, or about half of that from the difference in temperature alone! Forsyth and Uyeda, 1975! Going back to our original equation for the slab pull force, F SP = F b1 + F b2 we find a total slab pull force of 8E'6=6~4.5 10 13 N m -1 Fig. 6.43, Turcotte and Schubert, 2014 28
Ridge push FRP Forsyth and Uyeda, 1975 Ridge push results from the elevation of oceanic ridges relative to the seafloor! The difference in elevation results in a pressure head that drives the plate away from the ridge! This motion may also be viewed as gravitational sliding!! To calculate the ridge push force 8G' we must consider the forces acting on the top (81), bottom (82) and side (83) of the oceanic lithospheric plate: F RP = F 1 F 2 F 3 29
Ridge push FRP Forsyth and Uyeda, 1975 With this force balance in mind, we can see! Fig. 6.44, Turcotte and Schubert, 2014 1. The horizontal force on the base of the plate must be equal to the integrated lithostatic pressure in the mantle along RD! 2. The horizontal force on the top of the plate must be equal to the integrated hydrostatic pressure along AB! 3. The horizontal force acting on the lithospheric section BC is equal to the integrated pressure in the lithosphere! Note that this pressure should include that resulting from the overlying oceanic water 30
Ridge push FRP Forsyth and Uyeda, 1975 Fig. 6.44, Turcotte and Schubert, 2014! At a constant depth H P = gz assuming constant density of the overlying material! Integrated over a depth range H1 to H2 P int = Z z2 z 1 gzdz 31
Ridge push FRP Going back to the original force balance equation for ridge push, we see F RP = F 1 F 2 F 3 Forsyth and Uyeda, 1975 Fig. 6.44, Turcotte and Schubert, 2014 After some mathematical substitutions and integrations we find F RP = g m v (T 1 T 0 ) apple 1+ 2 m v (T 1 T 0 ) ( m w ) " Mantle density * Temperature at plate surface " Water density / Age of oceanic plate * Mantle temperature applet 32
Ridge push FRP Forsyth and Uyeda, 1975 Fig. 6.44, Turcotte and Schubert, 2014 Using typical values, we find the ridge push force is 8G'6=6~4 10 12 N m -1! Note that this is about an order of magnitude smaller than the slab pull force 33
Drag force FDF Forsyth and Uyeda, 1975 The drag force on the base of the oceanic lithosphere can both drive and resist plate tectonics, depending on the relative motion between the plate and the underlying mantle!! If we assume that the underlying mantle resists or drives plate motion by viscous flow across a fixed-thickness layer, the drag force on the plate is simply F DF = as u h L # Viscosity of asthenosphere h Thickness of viscous layer A+ Velocity difference J Length of the plate Fig. 6.2, Turcotte and Schubert, 2014 34
Drag force FDF Forsyth and Uyeda, 1975 Again using typical values, we find the drag force is 8K86=6~1 10 13 N m -1! Fig. 6.2, Turcotte and Schubert, 2014 Note that this value is similar in magnitude to the slab pull force 35
Recap Thermal convection in the mantle results from buoyancy forces as the mantle is heated and undergoes thermal expansion!! Slab pull and the drag force are similar and about an order of magnitude larger than the ridge push force 36
References Forsyth, D., & Uyeda, S. (1975). On the Relative Importance of the Driving Forces of Plate Motion*. Geophysical Journal International, 43(1), 163 200. doi:10.1111/j.1365-246x.1975.tb00631.x 37