# 4.1: The Forces Driving Plate Motions

The motion of tectonic plates is driven by one simple principle, convection. Convection is the idea that dense, cold things sink, and buoyant, warm things rise.

Let's look at this in terms of 3 forces.

Balancing these forces, we find that the velocity is a sum of

$F_{ridge-push}+F_{slab-pull}-F_{viscous-drag}=0$

Returning to our Chapter 1 discussion of strain, remember that $$\sigma=\frac{F}{A}$$. Rearranging, $$F=\sigma A$$. We want $$F$$ to have unit length, and thus we can write the equation as

$F=\sigma L \label{force}$

As seen in the figure below, $$L$$ is the length of the plate and $$F$$ is the unit length.

## Viscous Drag

One of the mechanisms for plate motion that is a result of convection is viscous drag. This concept relates to ideas we discussed earlier, like Couette Flow and viscous stress. The drag 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.

From the figure above, we can derive an equation to describe plate motion due to viscous drag.

$V=V_o(1-\frac{y}{h})$

$\dot{\varepsilon}=\frac{1}{2}\frac{dV}{dy}=-\frac{V_o}{2h}$

\begin{align*}\sigma &=2\eta\dot{\varepsilon} \\[4pt] &=2\eta(\frac{-V_o}{2h}) \\[4pt] &=\frac{-V_o\eta}{h} \end{align*}

The negative sign indicates an opposition to plate motion, i.e., the viscous drag. We now have an equation that describes the magnitude of drag:

$|\sigma|=\frac{V_o\eta}{h}$

placing this into our previous force equation (Equation \ref{force}):

$F=|\sigma|L$

Let's do an example calculating viscous drag under a plate.

Example Viscous Drag Under the Plate

We are given that h~100 km, $$\eta$$~1e18 Pa(s), Vo~5 $$\frac{cm}{yr}$$ = 0.05 $$\frac{m}{yr}$$= 50 $$\frac{mm}{yr}$$=50 $$\frac{km}{my}$$

Using these numbers, we can calculate the force on the descending plate, F=1.59x1011$$\frac{N}{m}$$.

On the sides of the slab, the above figures demonstrate the forces that are present. We can also see that the slab is sinking vertically, which is not often the case in the real world. From the right figure, $$\eta\approx 10^{21}$$. This is actually too high because of non linear viscosity, but we are ignoring other things so it works for this example.

Previously, we looked at viscous drag just under the plate, but now we will consider a case where there are stresses acting on both sides of the plate.

$\sigma=\frac{V_o\eta_{av}}{h}$

$F=\sigma L_{slab}$

$F_{vis.}=2F=2\sigma L_{slab}=\frac{2V_o\eta_{av}}{h}L_{slab}$

Example Viscous Drag on Two Sides of a Plate

As we did in the previous example, consider viscous drag on a plate, but now on both sides of the plate. We are given that Vo=Vplate~5 $$\frac{cm}{yr}$$, h~100 km, $$\eta_{av}$$=1021 Pa(s), and Lslab=560 m. In this case, F=-78x1013$$\frac{N}{m}$$. If we change $$\eta$$ to equal 1e18, F then equals 1.78x1010$$\frac{N}{m}$$.

Notice that $$\sigma \propto V$$

as V $$\uparrow$$, $$\sigma\uparrow$$, and $$\sigma \propto \eta$$. $$\eta$$ changes by x10-100, thus this change has the biggest effect of the overall force. $$\sigma \propto L$$ summing stress over the full length. These relationships will be useful for you to think about when answering this week's HW questions.

## Slab-Pull Force

Another main force a subducting plate will encounter is the slab-pull force. This is the force due to the density contrast between a slab and its surroundings, ie the mass anomaly of the slab in the mantle. We can approximate the force from a rectangular area of width w hanging down a distance d from the Earth's surface with a temperature $$\Delta T$$ different than its surroundings.

$\rho (T)=\rho_o(1+\propto\Delta T)$

$\rho_o=p_m\;density\;at\;T=T_m$

$\Delta T=T_m-T_{slab}$

$$\propto$$ is the thermal expansion coefficient and ~2x10-5$$\frac{1}{K}$$

$\Delta \rho=\propto\Delta T$

This is the density anomaly, the slab sinks because it is denser than its surroundings.

The figure above shows the asthenosphere before subduction. In the bottom figure, the temperature is not lost⇒cools surrounding mantle as slab warms up, but energy is conserved. If you know $$\Delta T(y)$$ before, you can use it to get $$\Delta \rho(y)$$⇒net mass $$\Delta m$$.

To get $$\Delta m$$→$$\Delta \rho$$→$$\Delta T$$→T(z).

$T(z, t)=(T_s-T_m)erfc(\frac{z}{2\sqrt{\kappa t}})+T_m$

$\Delta T=T_m-T(z, t)$

$\Delta T=-(T_s-T_m)erfc(\frac{z}{2\sqrt{\kappa t}})$

$\Delta T=(T_s-T_m)erfc(\frac{z}{2\sqrt{\kappa t}})$

Figure $$\PageIndex{8}$$: T vs. z

$\Delta \rho=\rho_o\propto\Delta T$

$\Delta \rho=\rho_o\propto\Delta T(z)$

$\Delta m=\int_{0}^{z\rightarrow\infty}{\Delta \rho(z)dz}$

$\Delta m=\rho_o\propto(T_m-T_s)\int_{0}^{\infty}erfc(\frac{z}{2\sqrt{\kappa t}})dz$

Now use variable substitution

Let $$q=\frac{z}{2\sqrt{\kappa t}}$$ where $$z\rightarrow\infty$$ and $$q\rightarrow\infty$$.

$\frac{dq}{dz}=\frac{1}{2\sqrt{\kappa t}}$

$dz=2\sqrt{\kappa t}dq$

Now substitute in

$\Delta m=\rho_o\propto(T_m-T_s)2\sqrt{\kappa t}\int_{0}^{\infty}erfc(q)dq$

The complementary error function integral solution can be found in any integral table and equals $$\sqrt{\frac{1}{\pi}}$$

$\Delta m=\rho_o\propto(T_m-T_s)2\sqrt{\frac{\kappa t}{\pi}}$

Mass per unit area⇒a single depth profile

Returning to our original equation calculating the slab pull force:

$\Delta M=\Delta m L_{slab}$

$F_{slab-pull}=\Delta M g$

$F_{sp}=\rho_o\propto g(T_m-T_s)2\sqrt{\frac{\kappa t}{\pi}}L_{slab}$

As we did for viscous drag, let's do a quick example calculating slab-pull force.

Example Slab-Pull Force

We are given that $$\rho_o$$=3300, $$\propto$$=2e-5$$\frac{1}{K}$$, g=9.81, Tm-Ts=1400 K, K=1e-6, t=age of slab at time of subduction, and Lslab=560 km.

We know that $$F_{sp}=\Delta m L_{slab} g$$

Thus, the force per unit length of the slab-pull is Fsp=2.88x1013$$\frac{N}{m}$$.

You might notice that this force is 100-1000x more than the viscous drag force.

## Ridge-Push Force

Let's cover a final force a subducting plate would experience, the ridge-push force. This force results from the elevation of oceanic ridges above the seafloor. This difference in height leads to pressure that 'pushes' the plate away from the ridge. Ridge push is the simplest force in some ways, as all its components can be easily examined.

Simply put, to determine ridge-push forces, we need to do two things. First, look at the isostatic balance and the depth of oceans w(t). Second, we need to look at the force balance on the plate.

$T(t)=(T_s-T_m)erfc(\frac{z}{2\sqrt{\kappa t}})+T_m$

$\rho(T)=\rho_m(1+\propto\Delta T)$

$\Delta T=T_m-T(t)$

T(t) is <Tm+$$\Delta T$$ because of more density.

Balance precursors:

P1=P2

$drp_wg+\int_{dr}^{dr+w}\rho (T)gdz+\int_{dr+w}^{z_c}\rho (T)gdz=(dr+w)\rho_wg + \int_{dr+w}^{z_c}\rho (T)gdz$

$\int_{dr}^{dr+w}\rho_mdz+\int_{dr+w}^{z_c}\rho_mdz=wp_w+\int_{dr+w}^{z_c}\rho (z)dz$

$w\rho_m-wp_w=\int_{dr+w}^{z_c}\rho (z)-\rho_mdz$

$\rho (z)=\rho (T)\rightarrow T(z)$

Like we did above, we will use the complementary error function and find its solution in a integral table.

$w(\rho_m-\rho_w)=\rho_m\propto(T_m-T_s)\int_{0}^{\infty}erfc(\frac{z'}{2\sqrt{\kappa t}})dz'$

The complementary error function integral solution is $$2\sqrt{\frac{\kappa t}{\pi}}$$

$w=\frac{z\rho_m\propto(T_m-T_o)}{\rho_m-\rho_w}\sqrt{\frac{\kappa t}{\pi}}$

An important thing to note is that the depth of the ocean basins deepen at ~$$\sqrt{t}$$ (the square root of age)

⇒balanced pressures (isostasy)

Ridge-push is very similar to the balanced pressures idea, but uses balanced forces instead.

⇒$$\Delta P$$⇒drives flow across A. No net force⇒no $$\Delta P$$.

$F\backsim\Delta P A$ This is the force per unit width.

Example Ridge-Push

The forces in this example should balance. After we write the integrals for the various forces, we should have an expression where FRP=F1-F2-F3.

$F_1=\int_{0}^{z_c}\rho_mgzdz=F_s$

Here, $$\rho_mgz$$ represents the pressure, dz is XA⇒Area⇒wXL$$\frac{area}{w}$$=L

We can do a simple integral in depth, this is the same pressure from the mantle below a plate.

Rewriting F1 0→zc, 0→w + w-zL

$F_1=\int_{0}^{w}\rho_mgzdz+\int_{w}^{w+z_c}\rho_mgzdz+\int_{0}^{2L}\rho_mg\bar{z}d\bar{z}$

Where we let $$\bar{z}=z-w$$

$F_2=\int_o^w\rho_wgz=F_4$

$$\rho_wgz$$ is P2. Here, we do not need to integrate over shape of w.

$F_3=\int_0^{z_L}P_Ld\bar{z}$

$P_L=P_wgw+\int_0^{\bar{z}}p_Lgd\bar{z}$

Here, Pwgw is water and $$\rho_Lg$$ is p(T)→T(t)

Remember our force balance equation:

$F_{RP}=F_1-F_2-F_3$

Plugging everything in and skipping some of the basic algebra we get:

$F_{RP}=g[\frac{1}{2}(\rho_m-\rho_w)w^2+\kappa t\rho_m\propto(T_m+T_s)]$

We get the term w2 becuase PL~w and we integrated PL.

From isostasy, we can plug in

$w(t)=\frac{2\rho_m\propto(T_m-T_s)}{(\rho_m-\rho_w)}\sqrt{\frac{\kappa t}{\pi}}$

$F_{RP}=g\kappa t\rho_m\propto(T_m-T_s)(1+\frac{2p_m\propto(T_m-T_s)}{\pi(\rho_m-\rho_w)})$

Notes on Ridge-Push

A few things to note: isostasy→controlling w(t)

The pressure across the lithosphere→ridge-push force

FRP$$\propto$$t age of the oldest part of plate

Now that we have our general solution for the ridge-push force, if we are given actual values, we can solve it. If g=9.81, k=1e-6$$\frac{m^2}{s}$$, $$\rho_m$$=3300$$\frac{ky}{m^3}$$, Tm-Ts=1400 K, $$\propto$$=2x10-5$$\frac{1}{K}$$, $$\rho_m$$=1000$$\frac{kg}{m^3}$$, we can finally find the force

$F_{RP}=3.24x10^{12}\frac{N}{m}$