5Thermal Models of the Lithosphere

Sail away from a mid-ocean ridge and the sea floor deepens: about $2.5\,\mathrm{km}$ at the axis, more than $5\,\mathrm{km}$ beneath the oldest ocean floor. Nothing has been added to the plate or taken away from it; it has only cooled, contracted and sunk. The ocean floor is a thermometer of the plate beneath it, and a heat-flow probe pushed into its sediment reads the same cooling from above. This chapter solves the cooling of a plate in two ways, as a half-space and as a plate of finite thickness, compares both with the depth of the real ocean floor and its heat flow, and then turns to the continents, whose heat comes partly from their own crust.

You already know

The diffusion equation for heat, $\partial T/\partial t = \kappa\nabla^2T$, with the thermal conductivity $k$ and the thermal diffusivity $\kappa = k/(\rho c)$, is the physics books’; Fourier series and the error function $\operatorname{erf}x = (2/\sqrt\pi)\int_0^xe^{-u^2}\,\mathrm du$ are in the mathematics books’ university volumes. The heat flow, the radiogenic heat production, the geotherm, the conductive time scale $L^2/\kappa$, the thermal boundary layer and the thermal lithosphere were defined in the Year 1 volume, in its chapter on the Earth’s heat budget; isostasy in its chapter on isostasy and gravity anomalies; the depth–age law $d = d_r + C\sqrt t$ was proved in its chapter on rifts, ridges and passive margins, admitting the temperature profile that this chapter derives. The stretching model of basins, post-rift subsidence and the thermal time constant are in the Year 2 volume, in its chapter on sedimentary basins.

5.1 Conduction, advection and their ratio

Heat in the lithosphere moves in two ways: it diffuses down the temperature gradient, and it is carried by moving rock, the plate leaving a ridge, an eroded mountain rising, a magma ascending. Which one wins depends on a single number.

Definition 5.1 (Thermal Péclet number)

For rock moving at the speed $u$ across a length $L$, the thermal Péclet number is

$$\mathrm{Pe} = \frac{uL}{\kappa},$$

the ratio of the conductive time $L^2/\kappa$ to the advective time $L/u$. Advection dominates where $\mathrm{Pe} \gg 1$, conduction where $\mathrm{Pe} \ll 1$.

A plate leaving a ridge at $25\,\mathrm{mm}/\mathrm{yr}$ has, over a distance of $95\,\mathrm{km}$, $\mathrm{Pe} \approx 94$ with $\kappa = 0.8 \times 10^{-6}\,\mathrm{m}^{2}\,\mathrm{s}^{-1}$: heat is carried sideways by the plate far faster than it diffuses sideways. Each vertical column of the plate therefore cools as if alone, by vertical conduction to the cold sea floor, its age $t$ standing for the time since it left the axis. That turns a problem in two dimensions into one in depth and time.

5.2 The cooling half-space

Definition 5.2 (Half-space cooling model)

The half-space cooling model of the oceanic lithosphere takes each column as a half-space $z \ge 0$ of uniform diffusivity, at the uniform temperature $T_m$ of the mantle when it leaves the ridge axis, whose surface is held from then on at the temperature $T_0$ of the sea floor, with no heat production and no bottom.

Theorem 5.3 (Error-function geotherm)

In the half-space cooling model, the temperature at the depth $z$ below the sea floor of age $t$ is

$$T(z, t) = T_0 + (T_m - T_0)\operatorname{erf}\frac{z}{2\sqrt{\kappa t}} .$$

This is the error-function geotherm.

The temperature at a given depth depends only on $z/\sqrt t$: an isotherm sinks as the square root of age, and the cooled layer at $40\,\mathrm{Myr}$ is twice as thick as at $10\,\mathrm{Myr}$. The surface heat flow follows by differentiation.

Proposition 5.4 (Heat flow–age relation)

In the half-space cooling model, the heat flow through the sea floor of age $t$ is

$$q(t) = k\,\frac{\partial T}{\partial z}\Big|_{z=0} = \frac{k(T_m - T_0)}{\sqrt{\pi\kappa t}} = \frac{C_q}{\sqrt t} .$$

With the values of the plate model below ($k = 3.1\,\mathrm{W}\,\mathrm{m}^{-1}\,\mathrm{K}^{-1}$, $T_m - T_0 = 1450\,\mathrm{K}$, $\kappa = 0.8 \times 10^{-6}\,\mathrm{m}^{2}\,\mathrm{s}^{-1}$), $C_q = 505\,\mathrm{mW}\,\mathrm{m}^{-2}\,\mathrm{Myr}^{1/2}$, close to the values of 430 to $510\,\mathrm{mW}\,\mathrm{m}^{-2}\,\mathrm{Myr}^{1/2}$ fitted to the global measurements by different studies; a recent compilation prefers $490$. Measured heat flow on young oceanic crust, however, grossly underestimates the heat that leaves it: sea water circulating through the cracked crust carries part of the heat away in hot springs, out of reach of the probes. Over old crust the measurements follow $1/\sqrt t$, and estimates of the Earth’s heat loss replace the measurements on crust younger than $66.5\,\mathrm{Myr}$ with $C_q/\sqrt t$.

The Year 1 volume integrated the same profile over depth, through isostasy, to obtain the depth–age law, $d = d_r + C\sqrt t$ with $C = 2\rho_m\alpha(T_m - T_0)\sqrt\kappa/\bigl(\sqrt\pi(\rho_m - \rho_w)\bigr)$ (figure below). With $\rho_m = 3300\,\mathrm{kg}\,\mathrm{m}^{-3}$, $\rho_w = 1030\,\mathrm{kg}\,\mathrm{m}^{-3}$ and $\alpha = 3.1 \times 10^{-5}\,\mathrm{K}^{-1}$, $C = 370\,\mathrm{m}\,\mathrm{Myr}^{-1/2}$, against the $350\,\mathrm{m}\,\mathrm{Myr}^{-1/2}$ fitted to the North Pacific and North Atlantic floors younger than $70\,\mathrm{Myr}$.

5.3 The plate model

Old ocean floor stops deepening: in the North Pacific and North Atlantic the depth departs from $\sqrt t$ between 60 and $80\,\mathrm{Myr}$, and heat flow levels off. The half-space keeps cooling for ever; something must hold the base of the plate warm.

Definition 5.5 (Plate cooling model)

The plate cooling model of the oceanic lithosphere takes each column as a layer $0 \le z \le L$ of uniform diffusivity, at the uniform temperature $T_m$ when it leaves the ridge, whose surface is held at $T_0$ and whose base is held at $T_m$ from then on.

Theorem 5.6 (Plate-model temperature series)

In the plate cooling model, with $T_0 = 0$,

$$T(z, t) = T_m\Bigl[\frac zL + \frac2\pi\sum_{n=1}^\infty\frac1n \sin\frac{n\pi z}{L}\,e^{-n^2t/\tau}\Bigr],\qquad \tau = \frac{L^2}{\pi^2\kappa} .$$

This is the plate-model temperature series.

Each mode decays with its own time, $\tau/n^2$: the higher harmonics, the fine structure of the initial step, vanish first, and after a time of order $\tau$ only the first one is left. The same separation of variables gives the heat flow and, through the isostatic balance of the figure above, the depth.

Proposition 5.7 (Depth and heat flow of the plate model)

In the plate cooling model, with the densities and expansion of the depth–age law,

$$q(t) = \frac{kT_m}{L}\Bigl[1 + 2\sum_{n=1}^\infty e^{-n^2t/\tau}\Bigr],\qquad d(t) = d_r + \frac{\rho_m\alpha T_mL}{2(\rho_m - \rho_w)} \Bigl[1 - \frac{8}{\pi^2}\sum_{n\ \text{odd}}\frac{e^{-n^2t/\tau}}{n^2}\Bigr].$$

As $t \to \infty$, the depth tends to $d_r + \rho_m\alpha T_mL/\bigl(2(\rho_m - \rho_w)\bigr)$ and the heat flow to $kT_m/L$, both exponentially, with the time constant $\tau$.

The plate model that best fits the global depths and heat flows has a plate $95\,\mathrm{km}$ thick with a base at $1450\,{}^{\circ}\mathrm{C}$, $\alpha = 3.1 \times 10^{-5}\,\mathrm{K}^{-1}$ and $k = 3.1\,\mathrm{W}\,\mathrm{m}^{-1}\,\mathrm{K}^{-1}$, thinner and hotter than the earlier $125\,\mathrm{km}$ and $1350\,{}^{\circ}\mathrm{C}$. With $\kappa = 0.8 \times 10^{-6}\,\mathrm{m}^{2}\,\mathrm{s}^{-1}$, $\tau = 36.2\,\mathrm{Myr}$; the depth levels off at $2500 + 3104 = 5604\,\mathrm{m}$ and the heat flow at $47\,\mathrm{mW}\,\mathrm{m}^{-2}$. Young floor cannot tell the two models apart: they differ by less than a metre at $2\,\mathrm{Myr}$, by $100\,\mathrm{m}$ at $44\,\mathrm{Myr}$. Seismology gives a similar answer: Pacific Rayleigh waves are best fitted by cooling plates whose asymptotic thickness is 85 to $95\,\mathrm{km}$.

The real ocean floor can be put on the same graph (figure below). Every cell of one degree of latitude and longitude whose age is known is placed at its age and its depth; the median of each five-million-year bin rises with age as the models say for the first $40\,\mathrm{Myr}$, but the observed floor is shallower than both models, and more and more so with age. Nothing has been corrected: old floor carries a thicker blanket of sediment, more than a kilometre in the older North Atlantic, which loads it and raises its top, and old floor has had time to collect seamounts, volcanic plateaus and the swells of hotspots. The published fits remove the sediment load and the obvious swells first; the depth–age law of the Year 1 volume, and the plate models above, come from such corrected depths.

Definition 5.8 (Thermal subsidence)

Thermal subsidence is the sinking of the surface of the lithosphere as it cools and contracts towards thermal equilibrium, the space above it being filled by water or sediment; isostasy converts the heat lost by a column into its sinking.

The ocean floor is the pure case. A stretched continental basin is another: a lithosphere thinned by a factor $\beta$ brings hot asthenosphere up, and the post-rift subsidence of the Year 2 volume is the same plate cooling, starting from a different profile, with the same modes and the same decay times $\tau/n^2$. Keeping only the first mode is therefore a good approximation once the higher ones have died away, and a poor one before: for the ocean floor of the plate model above, the first term alone gives the depth to within two metres after $20\,\mathrm{Myr}$, but is $84\,\mathrm{m}$ too deep at $5\,\mathrm{Myr}$.

5.4 Continental geotherms

Continental crust makes heat: granite carries uranium, thorium and potassium in concentrations that make the heat produced in the upper crust a large part of the heat flow at the surface. How the production is distributed with depth decides the temperature at depth.

Definition 5.9 (Heat-production depth scale; reduced heat flow)

When the radiogenic heat production of the crust decreases with depth as $A(z) = A_0e^{-z/h_r}$, $h_r$ is the heat-production depth scale. The reduced heat flow $q_m$ of a province is the heat flow that would remain at its surface if the crust produced no heat: the heat flow from below the radiogenic layer.

Proposition 5.10 (Exponential heat production)

In steady conduction with uniform conductivity $k$, surface temperature $0$, heat production $A_0e^{-z/h_r}$ and heat flow $q_m$ from great depth, the surface heat flow is

$$q_0 = q_m + A_0h_r,$$

and the geotherm is

$$T(z) = \frac{q_mz}{k} + \frac{A_0h_r^2}{k}\bigl(1 - e^{-z/h_r}\bigr).$$

If the plutons of a province share the same $h_r$ and the same $q_m$, their surface heat flows and surface heat productions lie on the straight line $q_0 = q_m + h_rA_0$.

The line is observed: in several provinces, the heat flows and the surface heat productions of plutons fall on straight lines, with depth scales of 7.5 to $10\,\mathrm{km}$ and a reduced heat flow of $0.8\,\mathrm{HFU}$ (about $33.5\,\mathrm{mW}\,\mathrm{m}^{-2}$) east of the Rocky Mountains. The exponential form is not the only one that gives a line, but it keeps the line through erosion: removing a slice of thickness $e$ from the top multiplies $A_0$ and the radiogenic part of $q_0$ by the same factor $e^{-e/h_r}$.

The figure below shows geotherms for surface heat flows of 40, 60 and $80\,\mathrm{mW}\,\mathrm{m}^{-2}$, the first typical of cratons and the last of back-arc regions, with $k = 2.5\,\mathrm{W}\,\mathrm{m}^{-1}\,\mathrm{K}^{-1}$, $h_r = 10\,\mathrm{km}$ and a crustal share $A_0h_r = 0.4\,q_0$. The $1300\,{}^{\circ}\mathrm{C}$ isotherm, taken as the base of the thermal lithosphere, lies at 129, 84 and $61\,\mathrm{km}$. Seismology finds cratonic mantle fast down to about $250\,\mathrm{km}$, and back-arc lithosphere 50 to $60\,\mathrm{km}$ thick: the cold, thick roots of cratons need a smaller reduced heat flow than this simple model gives them, and their conductivity changes with temperature, but the order of the contrast is right.

In the field — Measuring heat flow at sea

A marine heat-flow probe is a long steel lance with a heavy weight on top and a slender tube of temperature sensors along its side. From the stern of a research ship it is lowered on the winch cable to the sea floor and lets itself fall into the soft sediment, sensors and all. It stays still for some minutes while the frictional heat of the penetration dies away and the sensors settle to the temperature of the mud around them: their readings, metres apart, give the temperature gradient. A short pulse of heat is then sent along the tube, and the rate at which it decays gives the conductivity of the sediment around it; the product of the two is the heat flow. The lance is pulled out and set down again a few hundred metres further along the line, without coming back on deck.

5.5 Exercises

Exercise 5.1 ★

With $\kappa = 10^{-6}\,\mathrm{m}^{2}\,\mathrm{s}^{-1}$, compute the conductive time $L^2/\kappa$ for $L = 1$, 10 and $100\,\mathrm{km}$, in millions of years.

Solution

Solution of Exercise 5.1.

$L^2/\kappa = 10^{12}\,\mathrm{s}$, $10^{14}\,\mathrm{s}$ and $10^{16}\,\mathrm{s}$: 0.032, 3.2 and $317\,\mathrm{Myr}$. A kilometre forgets its temperature in tens of thousands of years, a hundred kilometres in hundreds of millions.

Exercise 5.2 ★

A mountain range is eroded at $1\,\mathrm{mm}/\mathrm{yr}$, and rock rises through the upper $30\,\mathrm{km}$ of crust to replace it. With $\kappa = 10^{-6}\,\mathrm{m}^{2}\,\mathrm{s}^{-1}$, compute the Péclet number. Can the geotherm of the range be computed by conduction alone?

Solution

Solution of Exercise 5.2.

$u = 10^{-3}\,\mathrm{m}/3.156 \times 10^{7}\,\mathrm{s} = 3.17 \times 10^{-11}\,\mathrm{m}\,\mathrm{s}^{-1}$, so $\mathrm{Pe} = 3.17\times10^{-11}\times3 \times 10^{4}/10^{-6} = 0.95$. Advection and conduction are of the same importance: the rising rock carries heat upwards and steepens the geotherm near the surface, and conduction alone would underestimate the gradient.

Exercise 5.3 ★

In the half-space model ($T_0 = 0$, $T_m = 1450\,{}^{\circ}\mathrm{C}$, $\kappa = 0.8 \times 10^{-6}\,\mathrm{m}^{2}\,\mathrm{s}^{-1}$), compute the temperature $10\,\mathrm{km}$ below sea floor $5\,\mathrm{Myr}$ old ($1\,\mathrm{Myr}$ $= 3.156 \times 10^{13}\,\mathrm{s}$; $\operatorname{erf} 0.445 = 0.471$).

Solution

Solution of Exercise 5.3.

$\sqrt{\kappa t} = \sqrt{0.8 \times 10^{-6}\times5\times3.156 \times 10^{13}} = 11\,236\,\mathrm{m}$, $\eta = 10000/(2\times11236) = 0.445$, and $T = 1450\times0.471 = 683\,{}^{\circ}\mathrm{C}$.

Exercise 5.4 ★

With $C_q = 490\,\mathrm{mW}\,\mathrm{m}^{-2}\,\mathrm{Myr}^{1/2}$, compute the heat flow through sea floor 1, 25 and $100\,\mathrm{Myr}$ old. Which of the three would a probe most likely underestimate?

Solution

Solution of Exercise 5.4.

$490/\sqrt1 = 490$, $490/5 = 98$ and $490/10 = 49\,\mathrm{mW}\,\mathrm{m}^{-2}$. The probe would most likely underestimate the first: young crust is cracked and open to circulating sea water, which carries part of the heat away by hot springs.

Exercise 5.5 ★★

In the half-space model, the temperature reaches $90\,\%$ of $T_m$ where $\operatorname{erf}\eta = 0.9$, $\eta = 1.163$. With $\kappa = 0.8 \times 10^{-6}\,\mathrm{m}^{2}\,\mathrm{s}^{-1}$, compute that depth at 10, 40 and $100\,\mathrm{Myr}$.

Solution

Solution of Exercise 5.5.

$z = 2\times1.163\sqrt{\kappa t}$: with $\sqrt{\kappa t} = 15.89$, 31.78 and $50.25\,\mathrm{km}$, $z = 37.0$, 73.9 and $116.9\,\mathrm{km}$. The cooled layer doubles from 10 to $40\,\mathrm{Myr}$; by $100\,\mathrm{Myr}$ it is already deeper than a $95\,\mathrm{km}$ plate.

Exercise 5.6 ★★

With the values of the text, the half-space model gives $C = 370.5\,\mathrm{m}\,\mathrm{Myr}^{-1/2}$. Keeping every other value, what thermal expansion coefficient would give the fitted $350\,\mathrm{m}\,\mathrm{Myr}^{-1/2}$? What other parameter could be changed instead?

Solution

Solution of Exercise 5.6.

$C$ is proportional to $\alpha$: $\alpha = 3.1 \times 10^{-5}\times350/370.5 = 2.93 \times 10^{-5}\,\mathrm{K}^{-1}$. $C$ is also proportional to $T_m - T_0$ (a mantle at $1450\times350/370.5 = 1370\,{}^{\circ}\mathrm{C}$) and to $\sqrt\kappa$ (a diffusivity $0.8\times(350/370.5)^2 = 0.71 \times 10^{-6}\,\mathrm{m}^{2}\,\mathrm{s}^{-1}$): depth alone cannot separate them, which is why heat flow is used as well.

Exercise 5.7 ★★

Compute the asymptotic heat flow $kT_m/L$ of the plate model for $L = 95\,\mathrm{km}$, $T_m = 1450\,{}^{\circ}\mathrm{C}$, and for $L = 125\,\mathrm{km}$, $T_m = 1350\,{}^{\circ}\mathrm{C}$, with $k = 3.1\,\mathrm{W}\,\mathrm{m}^{-1}\,\mathrm{K}^{-1}$ in both. Which is closer to the heat flow of old ocean floor computed from $C_q/\sqrt t$ at $150\,\mathrm{Myr}$ with $C_q = 490\,\mathrm{mW}\,\mathrm{m}^{-2}\,\mathrm{Myr}^{1/2}$?

Solution

Solution of Exercise 5.7.

$3.1\times1450/95 \times 10^{3} = 47.3\,\mathrm{mW}\,\mathrm{m}^{-2}$ and $3.1\times1350/125 \times 10^{3} = 33.5\,\mathrm{mW}\,\mathrm{m}^{-2}$. The half-space reaches them at $(490/47.3)^2 = 107\,\mathrm{Myr}$ and $(490/33.5)^2 = 214\,\mathrm{Myr}$: the thin, hot plate holds the heat flow of old floor at $47\,\mathrm{mW}\,\mathrm{m}^{-2}$ from about $100\,\mathrm{Myr}$ on, whereas the thicker, colder one would let it fall below $40\,\mathrm{mW}\,\mathrm{m}^{-2}$, the half-space value at $150\,\mathrm{Myr}$.

Exercise 5.8 ★★

A province has $q_0 = 60\,\mathrm{mW}\,\mathrm{m}^{-2}$, $h_r = 10\,\mathrm{km}$ and $A_0h_r = 0.4\,q_0$. Compute $A_0$, $q_m$ and the temperature at the Moho, $35\,\mathrm{km}$ deep, with $k = 2.5\,\mathrm{W}\,\mathrm{m}^{-1}\,\mathrm{K}^{-1}$.

Solution

Solution of Exercise 5.8.

$A_0h_r = 0.4\times60 = 24\,\mathrm{mW}\,\mathrm{m}^{-2}$, so $A_0 = 24 \times 10^{-3}/10^{4} = 2.4\,\text{µ}\mathrm{W}\,\mathrm{m}^{-3}$ and $q_m = 60 - 24 = 36\,\mathrm{mW}\,\mathrm{m}^{-2}$. At $35\,\mathrm{km}$, $T = 36 \times 10^{-3}\times35 \times 10^{3}/2.5 + 2.4 \times 10^{-6}\times10^{8}/2.5\times(1 - e^{-3.5}) = 504 + 96\times0.970 = 597\,{}^{\circ}\mathrm{C}$.

Exercise 5.9 ★★

Two granite plutons of one province give $q_0 = 66\,\mathrm{mW}\,\mathrm{m}^{-2}$ with $A_0 = 3.0\,\text{µ}\mathrm{W}\,\mathrm{m}^{-3}$, and $q_0 = 48\,\mathrm{mW}\,\mathrm{m}^{-2}$ with $A_0 = 1.2\,\text{µ}\mathrm{W}\,\mathrm{m}^{-3}$. Assuming they lie on the line of Proposition 5.10, compute $h_r$ and $q_m$.

Solution

Solution of Exercise 5.9.

$h_r = \Delta q_0/\Delta A_0 = 18 \times 10^{-3}/1.8 \times 10^{-6} = 10\,\mathrm{km}$, and $q_m = q_0 - h_rA_0 = 66 - 30 = 36\,\mathrm{mW}\,\mathrm{m}^{-2}$, since $h_rA_0 = 10^{4}\times3.0 \times 10^{-6}\,\mathrm{W}\,\mathrm{m}^{-3} = 30\,\mathrm{mW}\,\mathrm{m}^{-2}$.

Exercise 5.10 ★★★

(a) Show that the heat lost through the sea floor between ages 0 and $t$, per unit area, is $Q(t) = 2k(T_m - T_0)\sqrt{t/(\pi\kappa)}$ in the half-space model. (b) Show that the subsidence is $d - d_r = \alpha Q/\bigl(c(\rho_m - \rho_w)\bigr)$, where $c = k/(\rho_m\kappa)$ is the specific heat. (c) Check it numerically at $40\,\mathrm{Myr}$ with the values of the text.

Solution

Solution of Exercise 5.10.

(a) $Q(t) = \int_0^tk(T_m - T_0)/\sqrt{\pi\kappa t'}\,\mathrm dt' = 2k(T_m - T_0)\sqrt{t/(\pi\kappa)}$. (b) The heat lost is the heat content removed from the column, $Q = \rho_mc\int_0^\infty(T_m - T)\,\mathrm dz$, and isostasy gives $(\rho_m - \rho_w)(d - d_r) = \rho_m\alpha\int_0^\infty(T_m - T)\,\mathrm dz = \alpha Q/c$. (c) $c = 3.1/(3300\times0.8 \times 10^{-6}) = 1174\,\mathrm{J}\,\mathrm{kg}^{-1}\,\mathrm{K}^{-1}$; at $40\,\mathrm{Myr}$, $t = 1.262 \times 10^{15}\,\mathrm{s}$ and $Q = 2\times3.1\times1450\times\sqrt{1.262 \times 10^{15}/(\pi\times0.8 \times 10^{-6})} = 2.015 \times 10^{14}\,\mathrm{J}\,\mathrm{m}^{-2}$; $d - d_r = 3.1 \times 10^{-5}\times2.015 \times 10^{14}/ (1174\times2270) = 2343\,\mathrm{m}$, which is $370.5\sqrt{40}$.

Exercise 5.11 ★★★

In the plate model of the text ($\tau = 36.2\,\mathrm{Myr}$, $\rho_m\alpha T_mL/ \bigl(2(\rho_m - \rho_w)\bigr) = 3104\,\mathrm{m}$), the depth computed with the first Fourier term alone exceeds the full series mainly by the $n = 3$ term. Compute that term’s size at 1, 10 and $20\,\mathrm{Myr}$, and the age after which the first term alone is within $10\,\mathrm{m}$ of the series.

Solution

Solution of Exercise 5.11.

The $n = 3$ term is $3104\times(8/\pi^2)\times\tfrac19e^{-9t/36.2} = 280\,e^{-9t/36.2}$ metres: $218\,\mathrm{m}$ at $1\,\mathrm{Myr}$, $23\,\mathrm{m}$ at $10\,\mathrm{Myr}$ and $1.9\,\mathrm{m}$ at $20\,\mathrm{Myr}$. It falls to $10\,\mathrm{m}$ when $e^{-9t/36.2} = 10/280$, at $t = (36.2/9)\ln28 = 13.4\,\mathrm{Myr}$.

Exercise 5.12 ★★★

(a) Show that, in the geotherm of Proposition 5.10, $T(z) \approx q_mz/k + A_0h_r^2/k$ for $z \gg h_r$. (b) Deduce the depth of the $1300\,{}^{\circ}\mathrm{C}$ isotherm for $q_0 = 40\,\mathrm{mW}\,\mathrm{m}^{-2}$, with the values of exercise 8. (c) By how much must $q_m$ fall for that depth to reach $250\,\mathrm{km}$, the depth of the fast roots of cratons?

Solution

Solution of Exercise 5.12.

(a) For $z \gg h_r$, $e^{-z/h_r} \to 0$. (b) $q_m = 24\,\mathrm{mW}\,\mathrm{m}^{-2}$, $A_0h_r^2/k = 16 \times 10^{-3}\times10^{4}/2.5 = 64\,\mathrm{K}$, and $z = (1300 - 64)\times2.5/24 \times 10^{-3} = 128.8\,\mathrm{km}$. (c) $q_m = 2.5\times1236/250 \times 10^{3} = 12.4\,\mathrm{mW}\,\mathrm{m}^{-2}$: the reduced heat flow must be about halved, as under a craton whose thick, cold root lets little heat in from below.

5.6 Problem: Forty Million Years off the Ridge

Problem 5.1

Weekend problem — the half-space and the plate, the depth of an invented ocean, and the thickness of its plate

Take $T_0 = 0$, $T_m = 1450\,{}^{\circ}\mathrm{C}$, $\kappa = 0.8 \times 10^{-6}\,\mathrm{m}^{2}\,\mathrm{s}^{-1}$, $k = 3.1\,\mathrm{W}\,\mathrm{m}^{-1}\,\mathrm{K}^{-1}$, $\alpha = 3.1 \times 10^{-5}\,\mathrm{K}^{-1}$, $\rho_m = 3300\,\mathrm{kg}\,\mathrm{m}^{-3}$, $\rho_w = 1030\,\mathrm{kg}\,\mathrm{m}^{-3}$, a ridge $2500\,\mathrm{m}$ deep, and $1\,\mathrm{Myr}$ $= 3.156 \times 10^{13}\,\mathrm{s}$.

Part I — The half-space.

1. Compute $\sqrt{\kappa t}$ at 10 and $40\,\mathrm{Myr}$.
2. Compute the temperature $20\,\mathrm{km}$ below the floor at both ages ($\operatorname{erf}0.6294 = 0.6266$, $\operatorname{erf}0.3147 = 0.3437$).
3. The $1000\,{}^{\circ}\mathrm{C}$ isotherm lies where $\operatorname{erf}\eta = 0.6897$, $\eta = 0.7174$. Compute its depth at both ages, and the age at which it reaches $60\,\mathrm{km}$.
4. Compute the heat flow at both ages, and $C_q$.
5. A ridge spreads at $25\,\mathrm{mm}/\mathrm{yr}$ on each side. Compute the Péclet number of the plate over $95\,\mathrm{km}$. Why may each column be cooled alone?
6. Compute the conductive time across $95\,\mathrm{km}$. What does it say of a half-space model of floor $150\,\mathrm{Myr}$ old?

Part II — Subsidence.

7. Recall the isostatic balance that turns the temperature deficit of a column into subsidence, and why the integral may be taken to infinity in the half-space.
8. Compute $C$ of the depth–age law, and the depths at 10 and $40\,\mathrm{Myr}$.
9. The median depth of the real ocean floor between 40 and $45\,\mathrm{Myr}$ is $4475\,\mathrm{m}$, uncorrected. Compare with the model at $42.5\,\mathrm{Myr}$, and list what the median includes that the model leaves out.
10. By how much would $C$ change if $T_m$ were $1350\,{}^{\circ}\mathrm{C}$? Compare with the fitted $350\,\mathrm{m}\,\mathrm{Myr}^{-1/2}$.
11. Why do measured heat flows on young floor fall below $C_q/\sqrt t$?

Part III — The plate. Take $L = 95\,\mathrm{km}$.

12. Compute $\tau$ and the amplitude $\rho_m\alpha T_mL/\bigl(2(\rho_m - \rho_w)\bigr)$.
13. Compute the depth at $40\,\mathrm{Myr}$ with the terms $n = 1$, 3 and 5.
14. How much does the first term alone differ from that, and why so little?
15. Compute the difference between the half-space and the plate at 40 and $50\,\mathrm{Myr}$, and estimate by linear interpolation the age at which it reaches $100\,\mathrm{m}$.
16. Compute the heat flow of the plate at $40\,\mathrm{Myr}$, with the terms $n = 1$ to 3, and compare with the half-space.
17. Give the asymptotic depth and heat flow.

Part IV — An invented ocean. In an invented ocean whose ridge is $2500\,\mathrm{m}$ deep, the floor, once corrected for sediment and away from seamounts and swells, is $5700\,\mathrm{m}$ deep at $120\,\mathrm{Myr}$ and $4800\,\mathrm{m}$ at $40\,\mathrm{Myr}$. The plate’s $L$ is unknown; every other value is as above.

18. Keeping the first Fourier term only, write the depth at $120\,\mathrm{Myr}$ as a function of $L$, with $\tau$ itself depending on $L$.
19. Solve it for $L$ by iteration, starting from $L = 100\,\mathrm{km}$: compute $\tau$, then the $L$ that matches $5700\,\mathrm{m}$ with that $\tau$, and repeat.
20. With the $L$ found, compute the depth at $40\,\mathrm{Myr}$ and compare with the half-space and with the $4800\,\mathrm{m}$ measured.
21. Compute $\tau$ for that $L$ and the asymptotic depth of this ocean.
22. After what age does the floor lie within $100\,\mathrm{m}$ of its asymptotic depth?
23. Give the thickness $L$ of the plate that fits the old floor of the invented ocean.

Solution

Solution of Problem 5.1.

1. $\sqrt{0.8 \times 10^{-6}\times3.156 \times 10^{14}} = 15\,889\,\mathrm{m}$ and $\sqrt{0.8 \times 10^{-6}\times1.2624 \times 10^{15}} = 31\,778\,\mathrm{m}$.

2. $\eta = 20000/31778 = 0.6294$ and $20000/63556 = 0.3147$: $T = 1450\times0.6266 = 909\,{}^{\circ}\mathrm{C}$ at $10\,\mathrm{Myr}$ and $1450\times0.3437 = 498\,{}^{\circ}\mathrm{C}$ at $40\,\mathrm{Myr}$.

3. $z = 2\times0.7174\sqrt{\kappa t}$: $22.8\,\mathrm{km}$ and $45.6\,\mathrm{km}$. It reaches $60\,\mathrm{km}$ when $\sqrt{\kappa t} = 60/1.4348 = 41.82\,\mathrm{km}$, at $t = (41.82 \times 10^{3})^2/0.8 \times 10^{-6} = 2.186 \times 10^{15}\,\mathrm{s}$, $69.3\,\mathrm{Myr}$.

4. $q = 3.1\times1450/(\sqrt\pi\times15889) = 159.6\,\mathrm{mW}\,\mathrm{m}^{-2}$ at $10\,\mathrm{Myr}$ and $79.8\,\mathrm{mW}\,\mathrm{m}^{-2}$ at $40\,\mathrm{Myr}$; $C_q = 159.6\sqrt{10} = 505\,\mathrm{mW}\,\mathrm{m}^{-2}\,\mathrm{Myr}^{1/2}$.

5. $u = 7.92 \times 10^{-10}\,\mathrm{m}\,\mathrm{s}^{-1}$, $\mathrm{Pe} = 7.92\times10^{-10} \times95 \times 10^{3}/0.8 \times 10^{-6} = 94$. Heat travels sideways with the plate a hundred times faster than it diffuses sideways, so each column cools by vertical conduction only.

6. $(95 \times 10^{3})^2/0.8 \times 10^{-6} = 1.13 \times 10^{16}\,\mathrm{s}$, $357\,\mathrm{Myr}$. At $150\,\mathrm{Myr}$, $\sqrt{\kappa t} = 61.5\,\mathrm{km}$ and the cooled layer, a few times that, reaches far below $95\,\mathrm{km}$: a half-space model of so old a floor assumes that cold has reached depths where the mantle convects.

7. $(\rho_m - \rho_w)(d - d_r) = \rho_m\alpha\int(T_m - T)\,\mathrm dz$, the masses of the ridge column and of the old column being equal above a compensation depth. $T_m - T = T_m\operatorname{erfc}\eta$ falls below $0.5\,\%$ of $T_m$ for $\eta > 2$, so the deep part adds nothing.

8. $C = 2\times3300\times3.1 \times 10^{-5}\times1450/2270\times \sqrt{0.8 \times 10^{-6}\times3.156 \times 10^{13}/\pi} = 0.13069\times2834.8 = 370.5\,\mathrm{m}\,\mathrm{Myr}^{-1/2}$; $d = 2500 + 370.5\sqrt{10} = 3672\,\mathrm{m}$ and $2500 + 370.5\sqrt{40} = 4843\,\mathrm{m}$.

9. The model gives $2500 + 370.5\sqrt{42.5} = 4915\,\mathrm{m}$, $440\,\mathrm{m}$ deeper than the median. The median includes the load of the sediment, the seamounts and volcanic plateaus, the swells of hotspots, and cells on the flanks of fracture zones and margins, all of which make the floor shallower.

10. $C$ is proportional to $T_m$: $370.5\times1350/1450 = 344.9\,\mathrm{m}\,\mathrm{Myr}^{-1/2}$, close to the fitted 350.

11. Sea water circulating through the cracked young crust carries part of the heat out through hot springs, between the probes.

12. $\tau = (95 \times 10^{3})^2/(\pi^2\times0.8 \times 10^{-6}) = 1.143 \times 10^{15}\,\mathrm{s} = 36.2\,\mathrm{Myr}$; the amplitude is $3300\times 3.1 \times 10^{-5}\times1450\times95 \times 10^{3}/4540 = 3104\,\mathrm{m}$.

13. $e^{-40/36.2} = 0.33124$, $e^{-360/36.2}/9 = 5.3 \times 10^{-6}$ and the $n = 5$ term is smaller still; the bracket is $1 - 0.81057\times0.33125 = 0.73150$, and $d = 2500 + 3104\times0.73150 = 4771\,\mathrm{m}$.

14. By the $n = 3$ term, $3104\times0.81057\times5.3 \times 10^{-6} = 0.013\,\mathrm{m}$: at $40\,\mathrm{Myr}$, $e^{-9t/\tau}$ is $10^{-4}$ of $e^{-t/\tau}$, and only the first mode is left.

15. At $40\,\mathrm{Myr}$, $4843 - 4771 = 73\,\mathrm{m}$; at $50\,\mathrm{Myr}$, $2500 + 370.5\sqrt{50} = 5120$ against $2500 + 3104\times(1 - 0.81057\times0.25127) = 4972$: $148\,\mathrm{m}$. Interpolating, $40 + 10\times (100 - 73)/(148 - 73) = 43.6\,\mathrm{Myr}$, about $44\,\mathrm{Myr}$ (the full series gives 44.0).

16. $q = (3.1\times1450/95 \times 10^{3})[1 + 2(0.33124 + 0.01204 + 0.00005)] = 47.32\times1.68666 = 79.8\,\mathrm{mW}\,\mathrm{m}^{-2}$, the same as the half-space: at $40\,\mathrm{Myr}$ the base has not yet made itself felt at the surface.

17. $2500 + 3104 = 5604\,\mathrm{m}$ and $kT_m/L = 47.3\,\mathrm{mW}\,\mathrm{m}^{-2}$.

18. $d(120) = 2500 + aL\bigl[1 - (8/\pi^2)e^{-120/\tau}\bigr]$, with $a = \rho_m\alpha T_m/\bigl(2(\rho_m - \rho_w)\bigr) = 32.673\,\mathrm{m}/\mathrm{km}$ and $\tau = L^2/(\pi^2\kappa)$.

19. $L = 3200/\bigl(32.673\,[1 - 0.81057e^{-120/\tau}]\bigr)$. From $L = 100\,\mathrm{km}$: $\tau = 40.13\,\mathrm{Myr}$, bracket 0.95925, $L = 102.10\,\mathrm{km}$; then $\tau = 41.83$, $L = 102.67$; $\tau = 42.30$, $L = 102.82$; $\tau = 42.43$, $L = 102.87$; it settles at $L = 102.9\,\mathrm{km}$.

20. $\tau = 42.48\,\mathrm{Myr}$, $aL = 3362\,\mathrm{m}$: $d(40) = 2500 + 3362\times(1 - 0.81057e^{-40/42.48}) = 4799\,\mathrm{m}$, against $4843\,\mathrm{m}$ for the half-space and $4800\,\mathrm{m}$ measured: the plate fits both ages.

21. $\tau = 42.5\,\mathrm{Myr}$; asymptotic depth $2500 + 3362 = 5862\,\mathrm{m}$.

22. $3362\times0.81057\,e^{-t/42.48} = 100$ gives $t = 42.48\ln27.25 = 140\,\mathrm{Myr}$.

23. The plate that fits the old floor of the invented ocean is $102.9\,\mathrm{km}$ thick.