CHAPTER 3

Heat flow and energetics of the Earth

3.1 Introduction . The seismic activity observed in the lithosphere suggests that the surface of the Earthis in motion. The energy required for this motion must come from the Earth’s interior and we now ask:what are the energy sources and how is the energy transmitted to the surface? Potential sources of energyinclude: radioactive decay, global cooling, and gravitational redistribution of matter such as core formationor planetary accretion. Since energy is lost by the Earth mainly by heat flow through the surface, we willestimate this energy loss by examination of global heat flow measurements.

3.2 Global Energy Sources . The discovery of radioactivity radically changed the view of the global energybudget and consequent estimates of the age of the Earth. Kelvin required a very young age for the Earth (afew million years) in order to explain the observed heat flow. With the discovery of radioactivity at the turnof this century, conductive cooling no longer had to be the primary source of heat within the Earth. In fact,radioactivity could be a very important source of heat and calculation of the global energy budget requiresa good knowledge of the composition of the Earth.

The composition of the bulk Earth is usually assumed to be close to that of the carbonaceous chondritesas discussed in Chapter 1. If this is true, then the major heat producing elements in the present Earthare 238U , 235U , 232Th and 40K. Other radioactive elements either decay too slowly or are present in suchsmall abundances that they are relatively unimportant. The concentrations of the major heat producers formconsistent ratios in a wide range of terrestrial and meteoritic materials. The ratio of potassium to uraniumand thorium to uranium in the sun and the carbonaceous chondrites are about 104 and 4 respectively andprobably are inherited from the composition of the original solar nebula. Owing to the volatility of potassiumit is quite possible that the potassium/uranium ratio in the bulk Earth is somewhat lower, although how muchlower is not well constrained. If we assume a chondritic abundance as a starting point, we can make a roughguess at the heat production within the Earth due to radioactivity by:

Hr = [U ]

(HU +

[Th]

[U ]HTh +

[K]

[U ]HK

)

where HU =9.71× 10−5 W/kgHK =3.58× 10−9 W/kg

and HTh =2.69× 10−5 W/kg

Taking a concentration of Uranium [U] of 1.3× 10−8 kg/kg (∼13ppb) consistent with that in carbonaceouschondrites and the ratios of the other heat producing elements to be the same as in the sun, we get Hr to be∼ 3 × 10−12 W/kg or about 1 → 1.5 × 1013 W for the whole Earth (mass of mantle is ∼ 4 × 1024kg – it iscurrently unclear how much, if any, radioactivity is in the core).

Owing to the greater abundance of radioactive elements early in the history of the Earth, radioactive heatproduction was higher in the past than now. In Figure 3.1, we show the heat production from radioactivesources as a function of time. 26Al is a short lived isotope which would have been present a few millionyears after a supernova explosion. Whether or not this energy source was available depends on how quicklythe Earth formed after the last supernova. Some models for the formation of the solar system actually use asupernova explosion to initiate the collapse. In such a case, 26Al probably served as a major energy sourcefor a few million years. In any case, 3 billion years ago, the total radioactive heat production was abouttwice the present value.

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Fig. 3.1 Heat production from radioactive sourceswith chondritic abundances

One consequence of a greater radioactive heat production in the past is that the Earth has probably cooled.Comparing the geothermal gradient during the Archean with the present geotherm yields an estimate of thecooling rate of about 100◦/billion years. When a body cools, energy is released according to the equation:

Q(t) = −MCpdT

dt

where t is time, Cp is the specific heat at constant pressure and M is the mass of the body. [The definitionof the specific heat Cp is, the energy required to raise the temperature of a unit mass of material by 1K (atconstant pressure)]. Taking the appropriate values for the mantle, we calculate 1.5× 1013 W, comparable tothe radioactive sources. Cooling of the core at a similar rate would contribute another 0.5× 1013 W.

When matter redistributes to a lower energy configuration (denser material on the bottom), energy isreleased to the environment as heat. Thus, the formation of an iron core from an initially homogeneousEarth would be an energetic event of huge proportions. Computing this gravitational energy release at firstmay seem like an impossible task. Fortunately, gravity is a conservative force field which means that we donot need to know the exact mechanism of core formation but can compute the energy release by knowing thepotential energy before and after core formation and then subtracting the two. If we take into account thefact that g, ρ, P and T change as the Earth forms we get ∼ 1031J for the energy release due to core formation.If all the energy were kept inside the Earth we can estimate the temperature rise it would cause. The changein energy of the Earth by raising its temperature by ∆T is given by

∆E = MCp∆T

where M is the mass of the Earth. For Cp ' 1000J kg−1 K−1 we get ∆T ' 3500K. We conclude that coreformation could raise the temperature of the Earth by several thousand degrees, enough to completely melt itand drive off most of the more volatile materials as well. It is therefore likely that much of the gravitationalenergy release was lost during the accretion process by radiation into space.

Gravitational energy release by rearrangement of matter inside the Earth could be an important energysource today but the problem is knowing the rate of release of gravitational energy. Remember that poweris the rate of change of energy with respect to time (1 Watt = 1 Joule s−1) so the power available at anyparticular time depends upon the rate at which gravitational energy is being released. Is core formation stilloccurring? We simply do not know and it is difficult to assess the importance of this energy source for theproblems we seek to address. It is likely that gravitational energy release occurs during growth of the innercore and this may be an important power source for the geodynamo.

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Another source of energy is the heat generated during planetary accretion. Collision of planetessimalscan generate a huge amount of energy but, again, much of this energy was probably radiated away duringthe accretion process, The details of how the planets formed and how much heat was retained as “original”heat are unknown. If a Mars-size impact generated the Moon, we can be reasonably confident that the Earthstarted off hot.

We shall show in the next section that the total energy released through the surface of the Earth is about4.4×1013 W. Up to perhaps half of this can be accounted for by radioactive decay alone. Taking into accountplanetary cooling, core formation and original heat leads us to the conclusion that the possible supplyof energy is almost embarrassingly large and we should not be surprised that the Earth is a geologicallyvigorous body.

3.3 Heat Loss . Heat transport can be by radiation, conduction and convection. The latter is a very efficientmeans of transporting heat but can only operate if material can flow. The lithosphere is brittle and can notconvect; conduction must be the major form of heat transport through it. Conduction is a diffusion processdescribed by Fourier’s law. In three dimensions this is

q = −k∇T (3.1)

where

∇ =∂

∂x,∂

∂y,∂

∂z

and k is the thermal conductivity, a property of the material, q is the heat flow and T is the temperature. Theminus sign reminds us that heat flows down a temperature gradient.

For most problems of heat conduction in the Earth, horizontal heat flow is not important and we canreduce the problem to a one-dimensional one, i.e., q = −kdT/dz. Consider a slab of material:

where M is the mass of the Earth. For Cp ! 1000J kg!1 K!1 we get !T ! 3500K. We concludethat core formation could raise the temperature of the Earth by several thousand degrees, enough to

completely melt it and drive off most of the more volatile materials as well. It is therefore likely that

much of the gravitational energy release was lost during the accretion process by radiation into space.

Gravitational energy release by rearrangement of matter inside the Earth could be an important energy

source today but the problem is knowing the rate of release of gravitational energy. Remember that power

is the rate of change of energy with respect to time (1 Watt = 1 Joule s!1) so the power available at any

particular time depends upon the rate at which gravitational energy is being released. Is core formation

still occurring? We simply do not know and it is difficult to assess the importance of this energy source

for the problems we seek to address. It is likely that gravitational energy release occurs during growth

of the inner core and this may be an important power source for the geodynamo.

Another source of energy is the heat generated during planetary accretion. Collision of planetessimals

can generate a huge amount of energy but, again, much of this energy was probably radiated away during

the accretion process, The details of how the planets formed and howmuch heatwas retained as “original”

heat are unknown. If a Mars-size impact generated the Moon, we can be reasonably confident that the

Earth started off hot.

We shall show in the next section that the total energy released through the surface of the Earth is

about 4.4" 1013 W. Up to perhaps half of this can be accounted for by radioactive decay alone. Taking

into account planetary cooling, core formation and original heat leads us to the conclusion that the

possible supply of energy is almost embarrassingly large and we should not be surprised that the Earth

is a geologically vigorous body.

3.3 Heat Loss . Heat transport can be by radiation, conduction and convection. The latter is a very

efficient means of transporting heat but can only operate if material can flow. The lithosphere is brittle

and can not convect; conduction must be the major form of heat transport through it. Conduction is a

diffusion process described by Fourier’s law. In three dimensions this is

q = #k$T (3.1)

where

$ =!

!x,

!

!y,

!

!z

and k is the thermal conductivity, a property of the material, q is the heat flow and T is the temperature.The minus sign reminds us that heat flows down a temperature gradient.

For most problems of heat conduction in the Earth, horizontal heat flow is not important and we can

reduce the problem to a one-dimensional one, i.e., q = #kdT/dz. Consider a slab of material:

Fig 3.2

83

Fig 3.2

We can only access a thin region near the surface of the Earth so the temperature gradient can be assumedto be roughly linear, i.e., dT/dz ≈ −δT/h. To measure heat flow, all we have to do is measure a temperaturedifference over some depth range in the Earth, measure the thermal conductivity of the material and use theformula:

q = −kdTdz≈ −k δT

h(3.2)

This approximation is valid only for a thin shell.

77

Fig 3.3 Histograms of continental, oceanic and global heat flow

1) Land Measurements

The temperature gradient in the Earth’s crust (the geotherm) can be measured in mines or, to get a goodareal coverage, in boreholes. A borehole must be deep (≈ 300m) to avoid surface effects on the temperaturegradient. An example of such an effect would be the previous presence of an ice sheet. Conduction is a veryslow process and, if the surface temperature is changed, it takes a long time for the temperature gradient tosettle down to its new value. If we want to avoid these climate-induced variations we must drill below theaffected layer. The actual drilling of a borehole perturbs the temperature gradient and it is often necessaryto wait for ∼ 2 years to get a good measurement of the gradient. Boreholes invariably fill up with waterand the temperature gradient is measured by measuring the temperature of the water at different depths inthe hole. We have to be sure that the water is not circulating as this will mean there will be convective heattransport in the borehole.

Notwithstanding all these difficulties, thousands of measurements have been made giving a mean conti-nental heat flow of 65 mWm−2.

2) Sea Measurements

These are done by dropping a “corer” (a tube about 3m long with thermistors attached) into the oceanfloor sediments. The core sample is used to determine the thermal conductivity of the material later in thelaboratory. Ocean floor sediments are often porous or permeable and heat transport by convection of the seawater in the sediments can bias the measurement of the temperature gradient. If this happens the sedimentsare cooled by the water circulation leading to a low estimate of the temperature gradient.

78

Fig 3.4 Using a corer to do oceanic heatflow. The corer is releasedwhen the trigger wieght hits the ocean floor

The mean heat flow in the oceans is 101 mWm−2 (after correcting for water circulation in the oceaniccrust). This is very similar to but higher than the continental value which was initially puzzling as the heatgeneration by radioactive elements is much higher in continental rocks than in oceanic rocks. We shalldiscuss the reason for this later. The following table shows heatflow averages as a function of the age ofboth oceanic and continental lithosphere. Note that young ocean floor can have a significant amount ofwater circulation leading to low measured heat flows – the table has been corrected for this effect.

Clearly, there is a strong tendency for oceanic heatflow to be a function of the age of the ocean floorand we shall come back to this. The continental heat flow is a weak function of the age of the continentalcrust (particularly given the scatter in the data – see fig 3.6) but this trend has been used to extrapolate heatflow into regions where there are no measurements. In fig 3.7, we show the locations of actual heatflowmeasurements (this is a new compilation of almost 38,000 measurements worldwide) and fig 3.8 shows themeasured heat flux (uncorrected for water circulation effects) averaged into 2X2 cells. The mean values weare quoting for ocean and continental regions use the age dependence to extrapolate globally and also arecorrected for convective effects.

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Fig 3.5

Fig 3.6 Two different ways of averaging continental heat low data. The first (a) is by the age of

80

the last major tectonic event and (b) is by the radiometric date of crustal age. The width of each boxrepresents one standard deviation

Fig 3.7 Locations of heat flow measurements in both oceans and continents

Fig 3.8 Map of measured heat flow uncorrected for convective effects

After interpolating the data (using crustal age as a guide), we can compute the total power lost by theEarth by multiplying the mean oceanic and continental heat flow values by the areas of the continentsand oceans and adding. The area of the continents is about 2×108 km2 giving a continental power lossof 1.3 × 1013 watts. Similarly, the area of the oceans is about 3.1 × 108 km2 giving an oceanic powerloss of 3.1× 1013 watts. The total power loss of the Earth, Q is therefore about 4.4× 1013 watts.

3.4 Mechanisms of heat transfer . The lithosphere is by definition too strong to behave in a ductilefashion and the dominant mode of heat transfer through the lithosphere is by conduction. Becauseof the concentration of radioactive elements in the continental crust, the geophysical modelling ofoceanic and continental heat flow must be treated separately. The creation and destruction of oceanic

81

lithosphere is part of a grand convective process involving at least the upper mantle. In this section,we will discuss heat flow in continental and oceanic lithosphere.

1) Continental Heat flowThe radioactive elements Uranium, Thorium and Potassium have large ionic radii that do not fit

well into the high density phases that are stable in the mantle and so they are separated preferentiallyinto melts. These elements have therefore been concentrated into continental crust and contributesignificantly to the observed heat flow. Since radioactive heat produced within the crust is notavailable to do the work of moving plates around, this effect should be subtracted out. What we wishto know is the “reduced heat flow” which is the heat flowing from the mantle into the base of the crust.We need first to understand heat conduction in a material which also has heat sources within it.Consider a thin slab:

Fig 3.8 Map of measured heat flow uncorrected for convective effects

After interpolating the data (using crustal age as a guide), we can compute the total power lost

by the Earth by multiplying the mean oceanic and continental heat flow values by the areas of

the continents and oceans and adding. The area of the continents is about 2 ! 108 km2 giving a

continental power loss of 1.3 ! 1013 watts. Similarly, the area of the oceans is about 3.1 ! 108

km2 giving an oceanic power loss of 3.1 ! 1013 watts. The total power loss of the Earth, Q is

therefore about 4.4 ! 1013 watts.

3.4 Mechanisms of heat transfer . The lithosphere is by definition too strong to behave in a ductile

fashion and the dominant mode of heat transfer through the lithosphere is by conduction. Because

of the concentration of radioactive elements in the continental crust, the geophysical modelling

of oceanic and continental heat flow must be treated separately. The creation and destruction of

oceanic lithosphere is part of a grand convective process involving at least the upper mantle. In

this section, we will discuss heat flow in continental and oceanic lithosphere.

1) Continental Heat flow

The radioactive elements Uranium, Thorium and Potassium have large ionic radii that do not fit

well into the high density phases that are stable in themantle and so they are separated preferentially

into melts. These elements have therefore been concentrated into continental crust and contribute

significantly to the observed heat flow. Since radioactive heat produced within the crust is not

available to do the work of moving plates around, this effect should be subtracted out. What we

wish to know is the “reduced heat flow” which is the heat flowing from the mantle into the base

of the crust. We need first to understand heat conduction in a material which also has heat sources

within it.

Consider a thin slab:

Fig 3.9

88

Fig 3.9

For simplicity we shall assume that the temperature distribution in the slab stays constant with time,that is, it is steady state. The net heat flow per unit time per unit area of slab face is

q(z + δz)− q(z)

If δz is taken to be small we can expand q in a Taylor series:

q(z + δz) = q(z) + δzdq

dz+ · · ·

Thus

q(z + δz)− q(z) = δzdq

dz= −δzk d

2T

dz2

and we have assumed that k is not a function of z. If ρ is the density of the slab and H is the heatproduction per unit mass then the heat generated in the slab per unit area of slab face (per unit time) is

ρHδz

If there is a steady state we have

ρHδz = −δzk d2T

dz2

or kd2T

dz2+ ρH = 0 (3.3)

This equation can be integrated with suitable boundary conditions.

82

Now what about our problem of heat generation in the continental crust?

For simplicity we shall assume that the temperature distribution in the slab stays constant with

time, that is, it is steady state. The net heat flow per unit time per unit area of slab face is

q(z + !z) ! q(z)

If !z is taken to be small we can expand q in a Taylor series:

q(z + !z) = q(z) + !zdq

dz+ · · ·

Thus

q(z + !z) ! q(z) = !zdq

dz= !!zk

d2T

dz2

and we have assumed that k is not a function of z. If " is the density of the slab andH is the heat

production per unit mass then the heat generated in the slab per unit area of slab face (per unit

time) is

"H!z

If there is a steady state we have

"H!z = !!zkd2T

dz2

or kd2T

dz2+ "H = 0 (3.3)

This equation can be integrated with suitable boundary conditions.

Now what about our problem of heat generation in the continental crust?

Fig 3.10

We have qs = qm+ heat production in the crust; we want qm. Suppose we take a uniform

distribution of heat producing elements and use a value for H typical for granites (H = 9.6 "10!10Wkg!1). The contribution to qs from heat producing elements in the crust is "Hh = 91mWm!2 using " = 2700Kg m!3 and h = 35km (the thickness of the continental crust.) Somethingis wrong as this is higher than the surface heat flux (65 mW m2). We can conclude that the

concentration of heat producing elements is not uniform.

In order to choose a more realistic form for the distribution of heat sources, we take a clue from

the correlation of heat flow with crustal age. It is generally true that heat flow is lowest in the

oldest crust and highest in the youngest crust. Since the oldest crust was also the deepest (and has

89

Fig 3.10

We have qs = qm+ heat production in the crust; we want qm. Suppose we take a uniform distributionof heat producing elements and use a value for H typical for granites (H = 9.6× 10−10W kg−1). Thecontribution to qs from heat producing elements in the crust is ρHh = 91mW m−2 using ρ = 2700Kgm−3 and h = 35km (the thickness of the continental crust.) Something is wrong as this is higher thanthe surface heat flux (65 mW m2). We can conclude that the concentration of heat producing elementsis not uniform.

In order to choose a more realistic form for the distribution of heat sources, we take a clue fromthe correlation of heat flow with crustal age. It is generally true that heat flow is lowest in the oldestcrust and highest in the youngest crust. Since the oldest crust was also the deepest (and has since beeneroded), this observation suggests that the concentration of heat producing elements decreases withdepth. Let us suppose that it drops off exponentially. This is cast into mathematical form by:

H(z) = Hse−z/hr

This looks like:

since been eroded), this observation suggests that the concentration of heat producing elements

decreases with depth. Let us suppose that it drops off exponentially. This is cast into mathematical

form by:

H(z) = Hse!z/hr

This looks like:

Fig. 3.11

Now we can rewrite 3.3 for our problem:

kd2T

dz2+ !Hse

!z/hr = 0 (3.4)

Integrating this equation once gives us an equation for the temperature gradient which can be

recast into an equation for the heat flux using Fourier’s Law. The integration introduces a constant

which requires a boundary conditon for its evaluation. We shall use the boundary condition that

q(z) = !qs at z = 0 (the minus sign indicates that z is pointing in the opposite direction to q).Integrating gives

C = kdT

dz! !Hshre

!z/hr = !q(z) ! !Hshre!z/hr (3.5)

where C is a constant to be determined. At z = 0, this equation reduces to

C = qs ! !Hshr

so now 3.5 can be written

!q(z) = qs ! !Hshr

!1 ! e!z/hr

"(3.6)

This is an equation for the heat flow as a function of depth in the continental crust. Heat flow

must be a continuous function inside the Earth, in particular it will be the same on both sides of

the boundary separating the crust from the mantle. Hs can be measured as can qs, so if we knew

hr, we could use this equation to evaluate the heat flow into the bottom of the continental crust (at

z " 35km).To make further progress, we shall make an assumption that will be verifiable later. We

shall assume that hr is much less than the thickness of the continental crust. This means that the

exponential term in 3.6 will be much smaller than 1 at the base of the crust and so can be neglected.

Now let qm = !q(35) denote the heat flow into the base of the crust and evaluate 3.6 here giving

qs = qm + !Hshr (3.7)

90

Fig. 3.11

Now we can rewrite 3.3 for our problem:

kd2T

dz2+ ρHse

−z/hr = 0 (3.4)

Integrating this equation once gives us an equation for the temperature gradient which can be recastinto an equation for the heat flux using Fourier’s Law. The integration introduces a constant whichrequires a boundary conditon for its evaluation. We shall use the boundary condition that q(z) = −qsat z = 0 (the minus sign indicates that z is pointing in the opposite direction to q). Integrating gives

C = kdT

dz− ρHshre

−z/hr = −q(z)− ρHshre−z/hr (3.5)

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where C is a constant to be determined. At z = 0, this equation reduces to

C = qs − ρHshr

so now 3.5 can be written

−q(z) = qs − ρHshr

(1− e−z/hr

)(3.6)

This is an equation for the heat flow as a function of depth in the continental crust. Heat flow must bea continuous function inside the Earth, in particular it will be the same on both sides of the boundaryseparating the crust from the mantle. Hs can be measured as can qs, so if we knew hr, we could usethis equation to evaluate the heat flow into the bottom of the continental crust (at z ' 35km).

To make further progress, we shall make an assumption that will be verifiable later. We shall assumethat hr is much less than the thickness of the continental crust. This means that the exponential term in3.6 will be much smaller than 1 at the base of the crust and so can be neglected. Now let qm = −q(35)denote the heat flow into the base of the crust and evaluate 3.6 here giving

qs = qm + ρHshr (3.7)

Remember that we can measure ρ,Hs and qs and this equation tells us that if we plot qs as a function ofρHs, the data should lie on a straight line with intercept qm and slope hr. In fact, for many regions ofthe continental crust, this is true (Figure 3.12) and by using fits of straight lines to the data we estimatethat about 6 TW (terrawatts) of the surface heat flow (total 44 TW) are generated in the oceanic crustso the total coming from the mantle is about 38 TW.

Fig. 3.12 Dependence of surface heat flux on surface radioactive heatgeneration for two "heat flow provinces".

[If you are worried by the assertion that 35km is much greater than hr we note that

e−z/hr = e−35/10 = .03

which is certainly much smaller than 1]The actual distribution of radioactive elements in the crust is unclear and, indeed, it is not always

possible to fit a straight line to q/ρHs data though this appears to work well in regions where crustalradioactivity is high (e.g. granite batholiths). About all that can be said is that it is likely that radioactiveelements are more concentrated to the top of the continental crust.

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3.5 The oceanic lithosphere and oceanic heat flow . Modelling heat flow measurements in oceaniclithosphere requires a different approach from the one we adopted for continental lithosphere. Theobservation that heat flow depends upon the age of the crust holds for oceanic lithosphere as well as forthe continents and there is an overall similarity in average heat flow values ( 65 mW m−2 in continentsversus 101 mW m−2 in the oceans). However, the explanations for the two are quite different. Asdiscusssed above, heat flow in the continents is controlled partly by the heat supplied into the baseof the lithosphere (the reduced heat flow) and partly by radioactive sources in the crust. Oceaniclithosphere is not enriched in radioactive elements and radioactive sources within the oceanic crust canbe neglected. The key to oceanic heat flow is the process of sea-floor spreading whereby oceanic crustis formed at an ocean ridge and then spreads away until it is eventually subducted back into the mantleat a trench. The concept of seafloor spreading leads us to try and make a thermal model for the coolingof a lithospheric plate. Since old oceanic crust has had longer to cool than young oceanic crust, wecan explain the dependence of heat flow upon the age of the crust. We also might expect (and indeedwe do observe) a dependence of depth of the ocean on age (because of the principle of isostasy andthe fact that old oceanic lithosphere is colder and therefore denser than young oceanic lithosphere).

As a preliminary to developing a model for oceanic heat flow, we must first understand timedependent heat conduction. Consider a slab as in Fig 3.9. As in the last section we can find anexpression for the net heat flow out of the slab

q(z + δz)− q(z) = δzdq

dz= −δzk d

2T

dz2

where, in the last step, we have again used the 1-D form of Fourier’s Law (q = −kdT/dz) and we haveassumed that k is independent of position. This net heat flow is due to the fact that the slab is cooling(and not due to the presence of radioactive heat producing elements as for continental crust). We havealso encountered the energy release, ∆E, due to a temperature change, ∆T

∆E = −MCp∆T

(this equation is essentially the definition of the specific heat, Cp which is a property of the material).What we need is the energy flow per unit cross sectional area of the slab per unit time. The massper unit cross sectional area of the slab is ρδz where ρ is the density so the energy flow per unit crosssectional area per unit time is

−ρδzCp∂T

∂t

where ∂T/∂t is the rate of change of temperature with time.By conservation of energy

−δzk∂2T

∂z2= −δzρCp

∂T

∂t

or

∂T

∂t= κ

∂2T

∂z2where κ =

k

ρCp

κ is called the thermal diffusivity and has dimensions of length2/time. We will use the thermaldiffusivity again in Chapter 5.

Returning to the problem of the cooling of the oceanic lithosphere, we shall consider the model inFig 3.13. The oceanic plates are created at the ridge (x = 0) and move outward with a velocity u. Thesurface is at a temperature Ts because of cooling by seawater and we model the injection of magmaat the ridge crest with a high temperature vertical isotherm at Tm. As the material moves away fromthe ridge, cooling takes place and isotherms become more horizontal. We can think of this as cooling

85

as a function of distance from the ridge or cooling as a function of age of the lithosphere. For a platevelocity u, the age, t, is obviously related to the distance from the ridge, x, by t = x/u.

As a preliminary to developing a model for oceanic heat flow, we must first understand time

dependent heat conduction. Consider a slab as in Fig 3.9. As in the last section we can find an

expression for the net heat flow out of the slab

q(z + !z) ! q(z) = !zdq

dz= !!zk

d2T

dz2

where, in the last step, we have again used the 1-D form of Fourier’s Law (q = !kdT/dz) and wehave assumed that k is independent of position. This net heat flow is due to the fact that the slabis cooling (and not due to the presence of radioactive heat producing elements as for continental

crust). We have also encountered the energy release, !E, due to a temperature change, !T

!E = !MCp!T

(this equation is essentially the definition of the specific heat, Cp which is a property of the

material). What we need is the energy flow per unit cross sectional area of the slab per unit time.

The mass per unit cross sectional area of the slab is "!z where " is the density so the energy flowper unit cross sectional area per unit time is

!"!zCp#T

#t

where #T/#t is the rate of change of temperature with time.By conservation of energy

!!zk#2T

#z2= !!z"Cp

#T

#tor

#T

#t= $

#2T

#z2where $ =

k

"Cp

$ is called the thermal diffusivity and has dimensions of length2/time. We will use the thermaldiffusivity again in Chapter 5.

Returning to the problem of the cooling of the oceanic lithosphere, we shall consider the

following model:

Fig 3.13

The oceanic plates are created at the ridge (x = 0) and move outward with a velocity u. Thesurface is at a temperature Ts because of cooling by seawater and wemodel the injection of magma

at the ridge crest with a high temperature vertical isotherm at Tm. As the material moves away

92

Fig 3.13

At the ridge a column of mantle material is at a temperature Tm and its surface is suddenly broughtto a temperature Ts. As the column of material moves away from the ridge the surface temperature ismaintained at Ts and the column gradually cools. If we consider a sequence of such columns:

from the ridge, cooling takes place and isotherms become more horizontal. We can think of this as

cooling as a function of distance from the ridge or cooling as a function of age of the lithosphere.

For a plate velocity u, the age, t, is obviously related to the distance from the ridge, x, by t = x/u.At the ridge a column of mantle material is at a temperature Tm and its surface is suddenly

brought to a temperature Ts. As the column of material moves away from the ridge the surface

temperature is maintained at Ts and the column gradually cools. If we consider a sequence of

such columns:

Fig. 3.14

we note that, away from the ridge the isotherms are nearly horizontal, this implies that the tem-

perature gradient is just a function of depth so that heat conduction is basically one-dimensional.

In what follows we shall neglect the conduction of heat laterally which is a good approximation

for the thin oceanic lithosphere.

The equation governing the behavior of temperature in this model is

!"2T

"z2=

"T

"t

The solution is a function of position, z, and time, t, i.e., T = T (z, t) and must fit the boundaryconditions:

T (z, 0) = Tm and T (0, t) = Ts

(If we measure temperature in !C it turns out that Ts ! 0!C.)

3.6 Similarity solutions . The boundary conditions are simple constants and have no length or

time scales in them. In this case the solution to the equation is not a function of t and z separatelybut can be regarded as a function of a dimensionless parameter which incorporates the length and

time behavior. Bearing in mind the dimensions of !, one possible choice of dimensionless variableis

V =z2

!t

and we can write the solution

T (z, t) = T (V )

We now want to write the diffusion equation in terms of T (V ). We have

"T

"t=

dT

dV

"V

"t= " dT

dV

z2

!t2= " dT

dV

V

t

93

Fig. 3.14

we note that, away from the ridge the isotherms are nearly horizontal, this implies that the temperaturegradient is just a function of depth so that heat conduction is basically one-dimensional. In whatfollows we shall neglect the conduction of heat laterally which is a good approximation for the thinoceanic lithosphere.The equation governing the behavior of temperature in this model is

κ∂2T

∂z2=∂T

∂t

The solution is a function of position, z, and time, t, i.e., T = T (z, t) and must fit the boundaryconditions:

T (z, 0) = Tm and T (0, t) = Ts

(If we measure temperature in ◦C it turns out that Ts ' 0◦C.)

3.6 Similarity solutions . The boundary conditions are simple constants and have no length or timescales in them. In this case the solution to the equation is not a function of t and z separately but can beregarded as a function of a dimensionless parameter which incorporates the length and time behavior.Bearing in mind the dimensions of κ, one possible choice of dimensionless variable is

V =z2

κt

86

and we can write the solution

T (z, t) = T (V )

We now want to write the diffusion equation in terms of T (V ). We have

∂T

∂t=dT

dV

∂V

∂t= − dT

dV

z2

κt2= − dT

dV

V

t

and similarly

∂T

∂z=dT

dV

∂V

∂z=dT

dV

2z

κt=dT

dV

2V

zso

∂2T

∂z2=

d

dV

(dT

dV

2V

z

)∂V

∂z

=

[d2T

dV 2

2V

z+

2

z

dT

dV− 2V

z2dT

dV

dz

dV

]∂V

∂z

=d2T

dV 2

4V 2

z2+

4V

z2dT

dV− 2V

z2dT

dV

=d2T

dV 2

4V 2

z2+

2V

z2dT

dV

Writing dT/dV as T ′ and d2T/dV 2 as T ′′, our original equation:

κ∂2T

∂z2=∂T

∂t

becomes

κ4V 2

z2T ′′ + κ

2V

z2T ′ = −T ′V

t

Substituting into the expression for V and rearranging gives

2[2V T ′′ + T ′] = −T ′Vor finally

T ′′

T ′= −1

4− 1

2V=

d

dV( lnT ′)

Integrating this equation gives

lnT ′ = −V4− 1

2lnV + lnA

where lnA is an integration constant. Thus

lnT ′ = −V4− lnV 1/2 + lnA

soT ′V 1/2

A= exp [−V

4]

ordT

dV=

A

V 1/2exp [−V

4]

87

Integrating again gives

T (V ) = T (0) +

V∫

0

A

V 1/2exp [−V

4]dV

The integral cannot be done analytically but can be put in the form of a standard tabulated integral –the error function (so called because it appears in statistics). The error function is defined by

erf (η) =2√π

η∫

0

exp (−η′2)dη′

and has the following behavior:

T (V ) = T (0) +V!

0

A

V 1/2exp [!V

4]dV

The integral cannot be done analytically but can be put in the form of a standard tabulated integral

– the error function (so called because it appears in statistics). The error function is defined by

erf (!) =2""

!!

0

exp (!!!2)d!!

and has the following behavior:

Fig. 3.15. Note that erfc (z) = 1 ! erf (z)

If we make the identification ! = V 1/2/2 = z/2"

#t so that d! = 14V "1/2dV , we have

T (!) = T (0) + C

!!

0

exp (!!!2)d!! where C is a constant

and so

T (z, t) = T (!) = T (0) +C"

"

2erf

"z

2"

#t

#

The boundary conditions give

at z = 0, ! = 0, erf (!) = 0 so T (0, t) = T (0) = Ts = 0#Cat t = 0, ! = #, erf (!) = 1 so T (z, 0) = Tm and C

""/2 = Tm ! Ts.

Our final solution is

T (z, t) = Ts + (Tm ! Ts) erf"

z

2"

#t

#

We shall now interpret our solution. One question is the thermal structure, i.e., what are the

shapes of the isotherms in the lithosphere? Isotherms are lines of constant temperature and, because

T = T (!), an isotherm must correspond to lines on which ! is a constant. Thus z/2"

#t = a

constant so isotherms are parabolas i.e.,

95

Fig. 3.15. Note that erfc (z) = 1− erf (z)

If we make the identification η = V 1/2/2 = z/2√κt so that dη = 1

4V−1/2dV , we have

T (η) = T (0) + C

η∫

0

exp (−η′2)dη′ where C is a constant

and so

T (z, t) = T (η) = T (0) +C√π

2erf(

z

2√κt

)

The boundary conditions giveat z = 0, η = 0, erf (η) = 0 so T (0, t) = T (0) = Ts = 0◦Cat t = 0, η =∞, erf (η) = 1 so T (z, 0) = Tm and C

√π/2 = Tm − Ts.

Our final solution is

T (z, t) = Ts + (Tm − Ts) erf(

z

2√κt

)

We shall now interpret our solution. One question is the thermal structure, i.e., what are the shapesof the isotherms in the lithosphere? Isotherms are lines of constant temperature and, because T = T (η),an isotherm must correspond to lines on which η is a constant. Thus z/2

√κt = a constant so isotherms

are parabolas i.e.,

z ∝√t

88

z ∝√

t

Fig. 3.16

Since x = ut, where u is the spreading velocity, the isotherms are parabolas in space too.The model predicts the heat flow too which is given by

q = −kdT

dz

where dT/dz is to be evaluated at the surface (to get the surface heat flow!) at a given time, t. Ifwe fix t then η ∝ z so we can differentiate the expression for the error function with respect to z.The result is

∂T

∂z=

2(Tm − Ts)√π2

√κt

exp!

− z2

4κt

"

At the surface, z = 0 so

dT

dz=

Tm − Ts√πκt

Therefore

q = −k(Tm − Ts)√πκ

t−1/2 = −Ct−1/2

where C is a constant. Using k = 3 Wm−1 K−1, Tm = 1500K, Ts = 273K and κ = 1.2 × 10−6

m2 s−1 gives C = 330 mWm−2 (Myr)1/2.If you inspect Figure 3.17, you will find that the

√t dependence of q is well modeled; however,

the best values of C fitting the data is higher. The values of the constants we have chosen may bewrong or, more likely, there is a contribution to the surface heat flux from below the lithosphere

96

Fig. 3.16

Since x = ut, where u is the spreading velocity, the isotherms are parabolas in space too.The model predicts the heat flow too which is given by

q = −kdTdz

where dT/dz is to be evaluated at the surface (to get the surface heat flow!) at a given time, t. If we fixt then η ∝ z so we can differentiate the expression for the error function with respect to z. The result is

∂T

∂z=

2(Tm − Ts)√π2√κt

exp(− z2

4κt

)

At the surface, z = 0 so

dT

dz=Tm − Ts√

πκt

Therefore

q = −k(Tm − Ts)√πκ

t−1/2 = −Ct−1/2

where C is a constant. Using k = 3 Wm−1 K−1, Tm = 1500K, Ts = 273K and κ = 1.2 × 10−6 m2 s−1gives C = 330 mWm−2 (Myr)1/2.

Fig. 3.17. Heat flow versus age of ocean floor. The measured values lie close to the straight line,which shows that the heat flow is inversely proportional to the square root of the age of the oceanfloor. Note that both scales are logarithmic.

89

If you inspect Figure 3.17, you will find that the√t dependence of q is well modeled; however, the

best values of C fitting the data is higher. The values of the constants we have chosen may be wrongor, more likely, there is a contribution to the surface heat flux from below the lithosphere (see below)– in our model, surface heat flux arises only from cooling effects.Our solution predicts that q → ∞ as t → 0. This is clearly absurd and our assumption that heatconduction is only in the z direction is poorest when t is small, i.e., this is when the isotherms are notnearly horizontal. (A more complete analysis still shows the singular behavior of q at t = 0 which isreally an artifact of the bondary conditions we have used). Heat flow measurements tend to actuallybe depressed near t = 0 and the difference between the model and the data led people to suspect thatwater circulation might be an important form of heat transport near the ocean ridges. This has nowbeen verified by the discovery of vents of very hot water near ridges and large amounts of heat arebeing transported by this convection or hydrothermal circulation as it is called.As t gets very large, the model predicts a q tending to zero. The heat flow measured in the oldest partsof the oceans is about 50 mWm−2 (the abyssal value). Our model predicts this value at an age of about100 Myr which is consistent with most of the data because only a small fraction of ocean crust is olderthan about 120 Myr.

We can use the principle of isostasy to compute the elevation (or bathymetry) of the ocean floor asa function of age. Consider the following model shown in fig 3.18. The principle of isostasy statesthat the lithostatic pressure of various columns of rock and water will be the same at some depthof compensation. We take this depth to be very deep. An equivalent statement is that there is thesame mass/unit area between the surface and the depth of compensation for different vertical columns.Consider the two columns in the diagram and let ρ be a function of z and t i.e., ρ = ρ(z, t). We shall,for convenience, take the depth of compensation to be infinity. We have

ρwh0 +

∞∫

0

ρ(z, 0)dz + ∆h(t)ρ(z, 0) = ρwh0 + ρw∆h(t) +

∞∫

0

ρ(z, t)dz

i.e. ∆h(t)(ρ(z, 0)− ρw) =

∞∫

0

[ρ(z, t)− ρ(z, 0)]dz

!wh0 +!!

0

!(z, 0)dz + !h(t)!(z, 0) = !wh0 + !w!h(t) +!!

0

!(z, t)dz

i.e. !h(t)(!(z, 0) ! !w) =!!

0

[!(z, t) ! !(z, 0)]dz

Fig. 3.18

We assume the density differences in the lithosphere are due to the temperature differences pre-

dicted by our thermal model. They are related by the coefficient of thermal expansion, ";

#! = !"!#T

For small density differences we can write, correct to first order in #!

#! = !(T2) ! !(T1) = !"[T2 ! T1]!(T1)

We shall now assume that !(z, 0) = a constant = !0 = 3300 kg m"3 (this is consistent with our

thermal model which had a vertical isotherm at the ridge). Now

!(z, t) ! !(z, 0) " !"!0[T (z, t) ! T (z, 0)]

so

!(z, t) ! !(z, 0) = !"!0[T (z, t) ! Tm]

= "!0(Tm ! Ts)"1 ! erf

#z

2#

$t

$%

Substituting into the equation derived from the principle of isostasy gives

!h(t)(!0 ! !w) = "!0(Tm ! Ts)!!

0

"1 ! erf

#z

2#

$t

$%dz

This last integral can be done by integrating by parts and the result is

98

Fig. 3.18

We assume the density differences in the lithosphere are due to the temperature differences predictedby our thermal model. They are related by the coefficient of thermal expansion, α;

δρ = −αρδTFor small density differences we can write, correct to first order in δρ

90

δρ = ρ(T2)− ρ(T1) = −α[T2 − T1]ρ(T1)

We shall now assume that ρ(z, 0) = a constant = ρ0 = 3300 kg m−3 (this is consistent with our thermalmodel which had a vertical isotherm at the ridge). Now

ρ(z, t)− ρ(z, 0) ' −αρ0[T (z, t)− T (z, 0)]

so

ρ(z, t)− ρ(z, 0) = −αρ0[T (z, t)− Tm]

= αρ0(Tm − Ts)[1− erf

(z

2√κt

)]

Substituting into the equation derived from the principle of isostasy gives

∆h(t)(ρ0 − ρw) = αρ0(Tm − Ts)∞∫

0

[1− erf

(z

2√κt

)]dz

This last integral can be done by integrating by parts and the result is

∆h(t) =αρ0(Tm − Ts)

ρ0 − ρw2

(κt

π

)1/2

= Ct1/2

where C is another constant which can be evaluated. Using the numbers given above we get C = 360m (Myr)−1/2. The square root dependence of the difference in elevation of the sea floor with age isseen in the data and a fit to the data gives C = 350 m (Myr)−1/2 for the Pacific. The distance from theseafloor to the surface should follow the form

H = h0 + Ct1/2

and Figure 3.19 shows how well the data obey this relationship. (The illustration is plotted upsidedown as people think in terms of the depth of the seafloor from the ocean surface).

Fig 3.19. The depth of the ocean floor plotted versus the square root of age. Depth has been correctedfor the effect of sediments.

91

Our model is undoubtedly a crude one but does a remarkably good job of predicting both the heatflow/age and bathymetry/age observations. One apparent failure of the model is that bathymetry isoverpredicted for ages greater than about 100Ma. A model which does a better job includes someheating from below. This model (the "plate" model labelled GDH1 in fig 3.19) differs by fixingtemperature at some depth inside the earth chosen to best fit the observations. This depth, whichdefines the thickness of the thermal lithosphere in old oceans is about 100km.

We conclude this chapter by comparing representative temperature profiles ("geotherms") for con-tinental and oceanic regions (fig 3.20). Because continental heat flow is lower than oceanic heat flow,the geotherm tends to be colder under continents down to a depth of about 200 km. Under very coldshields (such as the Canadian Shield), there may be temperature differences down to 350km or more.Eventually, we get to a depth where the material is too warm to resist flow and the mantle experiencesconvection. Heat transport by convection is much more efficient than heat transport by conduction andall of the machinery we have developed for looking at conductive heat flow is irrelevant in a convectingregion. As we shall demonstrate in later chapters, the temperature profile in a convecting region isclose to "adiabatic" and temperature gradients are much smaller than they are in conductive boundarylayers.

Fig 3.20. Representative oceanic and continental shallow upper mantle geotherms.

92