Thermal Subsidence of Seafloor
Thermal structure of cooling (or heating) halfspace
We begin by considering the cooling of some freshly minted ocean floor, and we will
largely be paralleling discussions developed in Turcotte and Schubert. We will make the
problem quite easy by assuming that heat is lost through the top of the oceanic
lithosphere and not the sides. We start then with our simple equation for one dimensional
change in temperature for material not producing heat internally.
dT
d 2T
= 2
dt
dz
(1)
Let us specify the boundary conditions. At the start (t=0), the temperature in the
halfspace will be Ta, and at the surface (z=0) it will be T0. We’ll also specify that the
temperature at great depth never changes, so at z= we’ll force the temperature to be Ta.
Given these boundaries, we expect that temperatures will always be between T0 and Ta, so
a dimensionless variable indicating the fraction of the temperature difference will prove
helpful:
=
T Ta
T0 Ta
(2)
It can be shown that equation (1) is essentially unchanged by the substitution of variables:
d
d 2
= 2
dt
dz
(3)
except that the boundary conditions are now at the surface is 1, at time 0 it is 0, and at
infinity it is 0. The only part of the problem suggesting a length scale is the thermal
diffusivity, which is length2/time, so a practical approach is to try and make the problem
totally nondimensional by substituting the dimensionless parameter
=
z
2 t
We differentiate this with respect to t and z to get the proper terms for (3):
(4)
d d 1 =
=
t d t d 2 t d d 1
=
=
z d z d 2 t
(5)
1 d 2 1 d 2
2
=
=
z 2 2 t d 2 z 4t d 2
which, when placed back into (3) yields
d 1 d 2
=
d 2 d 2
and the boundary conditions have reduced to ()=0 and (0)=1. We first solve for
(6)
d
d
which will will term , so (6) becomes
=
1 d
2 d
(7)
which is easily integrated to find that
2 = ln lnc1
(8)
where c1 is a constant of integration. If we take the exponential of both sides we get
2
= c1e =
d
d
(9)
which can also be integrated to solve for :
2
= c1 0 e d + c 2
(10)
Applying our boundary conditions, when = 0 should be 1, and as the integral from 0
to 0 is 0, c2 must be 1. When = the integral becomes one commonly known to be
2
and so therefore c1 must be , which leads us to the solution
2
= 1
2
0
2
e d
(11)
This is a common enough integral that it has it’s own name, the error function, erf() and
so we rewrite our solution as
= 1 erf () = erfc (12)
where erfc is the complementary error function. We can substitute our original variables
back in to get
T Ta
z
= erfc
T0 Ta
2 t
(13)
What this tells us is that the depth of an isotherm increases with t . A simple
rearrangement of (13) is also handy:
1
T Ta
T T T T0
z
= 0
=
= erf
T0 Ta T0 Ta Ta T0
2 t
(14)
Subsidence of seafloor
If we assume that all variations in topography of oceanic lithosphere are due to
temperature variations (also called Pratt isostasy sometimes, after an early advocate of
supporting topography through variations in densities), we can directly use our analysis
to get at variations in topography. This is not a bad assumption, for most seafloor crust is
generated very near the ridgecrest and is not modified much over its whole history. In
fact, by removing the thermal variations you can see these other effects more clearly. So
let us assume local isostasy. The condition for local isostasy is that the pressure at some
reference depth is the same everywhere. The pressure in turn is the integral of the weight
of all the material above:
Pc =
zc
0
g dz = dg w +
z c
0
g dz (15)
where Pc is the pressure at the depth of compensation, which is zc below sea level or z’c
below the seafloor. We have d as the depth of the water. If we now set the pressure
under a column at the midocean ridge at a water depth of d equal that under a column
elsewhere at depth w+d of age t, we get
dridge g w +
d ridge +w
d ridge
a dz +
zc
d ridge
zc
d
g dz = ( dridge + w ) g w +
dz = w w +
+w a
w( a w ) =
zc
d ridge +w
zc
d ridge +w
zc
d ridge +w
( z) dz
g dz
(15)
((z) a )dz
We can use the coefficient of thermal expansion to relate our density to the temperature,
a = a (T Ta ) , and then use the temperature from (13), converting our limits of
integration to now be downward from the seafloor of the old ocean floor to the depth of
compensation under that seafloor:
z
dz
0
2 t
z
= a (Ta T0 ) erfc
dz
0
2 t
w( a w ) =
z c
a (T0 Ta ) erfc
(16)
We again allowed the integral to go to infinity because the temperatures converge below
z’c. We repeat the substitution used in (4) so that the integral is more simply dealt with:
w=
a (Ta T0 )
2 t
a w
0
erfc d
2 (Ta T0 ) t
= a
a w
(17)
where we have again used a standard result for the integral of the complementary error
function. This is the desired result that expresses the depth of the seafloor relative to the
ridge as the square root of the age of that sea floor. If spreading is constant, then the time
can be replaced by the distance from the ridge over the half spreading rate.
Forces driving plates
Ridge push
This is the simplest in some ways, as all the forces are available to fairly direct
examination. We note that we can take a chunk of oceanic plate normal to the spreading
center. We then balance the horizontal component of forces from the ridge out to the
area of interest:
Ocean ρw
litho
sphe
Asthenosphere ρa
re ρ
L (z)
Since the lithosphere is not accelerating, the net force must be zero. So we balance out
the pressure from the asthenosphere against the pressure of sea water and the lithosphere
at the right hand edge. We will assume that the density of the asthenosphere is constant
and that the ocean and the asthenosphere are sufficiently close to being a fluid that the
stresses are simply pressures. Then we find that
D
F=
gz dz a
0
w
0
w gz dz Dw
0
{
zw
0
}
L gdz + w w g dz
(18)
, where z’ is a depth below the seafloor and the net horizontal force F is the ridge push
force acting on lithosphere with a surface w below the ridgecrest and whose thickness is
D. Note here that this reduces to the difference in potential energy between the lefthand
and righthand columns, where gravitational potential energy is defined as
GPE =
zc
(z)gz dz
(19)
0
where zc is the depth of compensation. We will not use this particular formulation to
effect here but it will prove useful later.
If we split the first integral and reorganize, we get
w
F = g 0 ( a w ) z dz +
D
{(
0
a
w )w + a z z
0
L dz dz
}
(20)
We can use isostasy to make things easier:
D
(
w( a w ) =
When combined with the fact that a z =
F = g 12 ( a w ) w 2 +
D
0
z
0
0
0
a ) dz (21)
a dz, we recast (3) as
D
{ (
L
a ) dz L
z
(
0
L
a ) dz dz
}
(22)
We’ve been a bit silent about D, but now if we just leave it as a depth where the
lithosphere and asthenosphere densities are equal, then extending those integrals from D
to infinity will not change anything, so we can rewrite (22) as
F = g
12 ( a w ) w 2 +
D
0
z
{ (
L
a ) dz dz
}
(23)
We now make use of results from the subsidence of seafloor. Recall from (13) and the
relationship of temperature to density:
L a = a (Ta T ( z))
z Ta T ( z) = (Ta T0 ) erfc
2 t Using this in (23) we get
(24)
z D F = g 12 ( a w ) w 2 + 0 a (Ta T0 ) erfc
dz dz
2 t z
z = g 12 ( a w ) w 2 + a (Ta T0 ) 0 erfc
dz dz
2 t z
We convert the two integrals from z’ to our old friend =
F = g 12 ( a w ) w 2 + 4t a (Ta T0 ) 0
(25)
z
(from (4)) and we get
2 t
{ erfc d}d
(26)
= g[ ( a w ) w + t a (Ta T0 )]
1
2
2
We can see from this that the force goes with the square of the depth or linearly with
seafloor age. If we use the result in (17)
w=
2 a (Ta T0 ) t
(a w ) (17)
in (26) we then obtain a result as a function of time:
2 (T T ) a
0
FRP = gt a (Ta T0 )
1+ a
( a w ) (27)
Using values of w=1000 kg/m3, g=10 m/s2, a=3300 kg/m3, (Ta-T0)= 1200 K, =3x10-5
K-1, =1 mm2/s, we get 3.9x1012 N/m force for 100 million year old lithosphere.
This force is a net force: it is a combination of the normal stresses in the end of
the lithosphere and any basal shear stresses applied to the lithosphere by the
asthenosphere. It is possible to balance the topography with basal shear stresses in
theory, but this would suggest that the shear forces increase with age, which seems
unreasonable. If there are shear stresses being applied, it is more likely that they are, to
first order, nearly constant and so their contribution to the net force is linear with distance
from the ridgecrest. This would predict variations in the stress in oceanic lithosphere that
should differ between fast spreading and slow spreading ridges; unfortunately, in situ
measurements of stress in oceanic lithosphere are essentially nonexistent, and estimates
from earthquake focal mechanisms (thrust faulting vs. normal faulting) are also pretty
rare. However, stress measurements on continents seem to suggest that basal shear
tractions are fairly small, so usually we do not consider basal shear to be a large player in
driving plate tectonics.
Slab pull
The other main force is from the density contrast between a slab and its
surroundings. Most of this is thermal (some is from the way the slab encounters phase
transitions). We can approximate the force by a rectangular area of width w hanging
down a distance d from the Earth’s surface with a temperature T different that its
surroundings. The buoyancy force is, just as when we derived the Rayleigh number,
gV, where g is the acceleration of gravity and V is the volume of the body. In this case
we can ignore pressure changes in the density of the slab as these effect the surroundings
by the same amount and just consider the thermal term, which yields = T0, where
is the volumetric coefficient of thermal expansion. We then obtain the force per unit
length of trench to be gdwT0. Let us set the temperature difference to the mean
temperature of the slab relative to the asthenosphere, which using the numbers at the end
of the previous section would be 600°C. If the width is about 100 km (roughly the
thickness of old oceanic lithosphere) and the slab descends about 700 km (the deepest
seismicity is 700 km), then using the numbers from before we estimate the slab pull force
to be 4.1 x 1013 N/m, about ten times the ridge push force.
Of course, we have left out changes to the thermal structure of the slab, which are
apt to be considerable. One solution is to use the thermal structure of a steady-state
convective system. This is done in Turcotte and Schubert and the result is that the slab
pull force is
FSP = 2 0 gd (Ta T0 )
2u0
(28)
which yields an estimate of about 3.3 x 1013 N/m for convection cells of wavelength =
4000 km and plate velocity u0 = 50 mm/yr.
Why did we get such close agreement between such an obviously bogus
idealization as a huge rectangular body with constant temperature and a full blown steady
state convection solution? The main answer is that, unlike at the surface, the transfer of
heat at depth is to some degree conservative. As the slab warms, it is cooling down
surrounding material, thus making it somewhat more dense than its surroundings just as
the slab is becoming somewhat less dense. In fact, the thermal structure of the
downgoing slab isn’t trivial to determine. In one sense, the slab experiences boundary
conditions only slightly different from when it was cooling: the base of the slab remains
at mantle adiabatic temperatures, but the top now experiences higher temperatures. There
are two potential contributions to the higher temperatures: one is the creation of heat by
viscous shear as the slab tries to pull along the mantle material above it. The second is
the juxtaposition of hot materials from the mantle wedge against the old ocean floor.
These are somewhat complementary: as the wedge material moves more with the slab,
there is less frictional heating but the efficiency of movement guarantees that the slab will
first be exposed to fresh hot material rather than colder material that has already been
cooled by older slab material.
If you approximate the new thermal regime as emplacement of material of
temperature Ta against the top of the slab, all you are doing is running the original cooling
equation backwards, and to first order you would expect the slab to return to mantle
temperatures over roughly a time equal to the age of the slab at subduction. Young,
warm slabs might not penetrate as deeply as old, cold slabs. Of course, this
approximation requires that the asthenospheric material move away from the slab. The
other extreme is to simply create a new thermal structure that starts with a slab geotherm
in the slab and adiabatic temperatures above and just let the cold diffuse out.
Forces opposing plate motion
T
ax
is
Of course, the plates are not accelerating, so the forces above must be balanced by
something. The most likely candidates are mostly viscous forces, either a basal drag on a
plate at the surface, drag on a descending slab, or the forces needed to move material out
of the way to make room for a descending slab. We can get some understanding of
where these forces are likely to be operating from considering the stresses in the plate. If
you are pulling on something, then the thing being pulled on will be in tension; if you
push, it should be in compression. A body hanging from above will be in tension (try
hanging from a bar to feel that) while a body supported from below will be in
compression (just standing on the ground).
The main means of estimating the stress comes from seismology. Earthquakes,
including deep earthquakes, are shear failures: one side moves past the other. In order to
get the proper sense of shear, the maximum compressive stress has to lie within two of
four quadrants defined by the fault plane and its normal (the “auxiliary plane” to
seismologists). The center of those quadrants is called the P axis; it gets confused with
the maximum compressive stress when it is called the compressional axis or pressure
axis. (As an aside, the P axis is in the center of the quadrant where the first arrival from
an earthquake is dilatational—the ground moves down).
P
is
ax
Figure: Map view of a strike-slip earthquake on a vertical fault (thick line). Material in the
quadrants with the open arrows moves towards the epicenter (concentric circles), while material
in the other two quadrants moves away from the epicenter. P and T axes are in the center of these
quadrants. Note that a left-lateral rupture on the auxiliary plane will produce the same pattern
and so have the same P and T axes.
Even though the deep earthquakes in a slab are potentially related to very different
processes than at the surface, they appear to retain this relationship to overall stress fields.
When the P and T axes of these earthquakes are plotted relative to the slab, we find the P
axes tend to point downward along the slab for earthquakes below about 300 km, while
earthquakes from 70 to 300 km tend to have the T axis pointing downward along the slab.
This suggests that the slab is simultaneously pushing down against something below
(presumably trying to move other mantle material out of the way) while pulling against
something above. This is rather like hanging from a bar while supporting some weight
on your feet.
Given the large magnitude of the slab pull force, we might expect that the pulling
part of the top of the slab is in fact pulling the oceanic lithosphere away from the ridge.
This would suggest that the entire ocean floor is in deviatoric tension, much as pulling on
a spring stretches it out. This does not appear to be the case: earthquakes in ocean basins
away from subduction zones show compression normal to the subduction zone, not
tension. This is what we would expect from ridge push, not slab pull. So what is going
on?
Nearly all of the slab pull force seems to be expended against forces at shallow
depths in the subduction zone. These include frictional forces in the near surface, viscous
forces near the top of the subduction zone (likely including the large forces needed to
drive flow through the mantle wedge), and forces to deform the plate through the
subduction zone. It appears that these forces reduce the effect of slab pull on
unsubducted oceanic lithosphere to less than the magnitude of ridge push. Thus although
the magnitude of slab pull can be much greater than ridge push, the surface lithospheric
stress field seems to be more determined by ridge push.