# SEISMOLOGY Solution

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```							Exercises. Selection with solutions.                                                      20/03/01

SEISMOLOGY

Exercise number 3, February 18, 2000

a) Find the expression for the travel-time SDR when the shot is on the surface and the
receiver is in a borehole. Assume the origin, O, is at (0,0) and express the travel-time t as
a function of x, z and h, where x=OS, z=OR and h is the depth to the reflector. Assume
the medium has constant velocity, v.

b) Plot the travel-time curve in x2-t2 space and show how the velocity of the media can
determined and the depth to the reflector if the reflector is below the bottom of the
borehole.

Solution

a) Draw a horizontal line from R crossing the line DS at point E. Draw a vertical line
from D crossing line RE at point F and line OS at point G. Draw a vertical line from
E crossing OS at point H.
Then x1 = RF = FE and x 2 = HS . The total distance
OGS= x = 2 x1 + x 2 ,
and since DES makes up a straight line it follows that
x2      z
=       .
x1 h − z
The total travel distance RD+DE+DS can be expressed as
2 x12 + ( h − z ) 2 + x 2 + z 2 .
2

After several reductions it turns out that the total travel distance, s can be expressed as
s = x 2 + (2h − z ) 2

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Exercises. Selection with solutions.                                                  20/03/01

from which it follows that the total travel time, t can be found from

1
t=     x 2 + ( 2h − z ) 2
v

b) When plotting Y=t2 as a function of X=x2 we find
1
Y = 2 ( X + ( 2h − z ) 2 )
v
1
which is a straight line with a slope given by 2 and an intercept on the Y-axis of
v
( 2h − z ) 2
. From these two numbers the two unknowns, namely the velocity of the
v2
medium and the depth to the reflector can be determined since the depth to the receiver
point, z is known a priori.

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Exercises. Selection with solutions.                                                    20/03/01

Exercise number 4, February 18, 2000

A simplified view of a transform fault is shown in the figure below.

a) Draw fault-plane solutions for earthquakes occurring within the active zone of the
transform fault shown in the figure.
b) Discuss the relative frequency of earthquakes occurring between the two ridge
segments and outside this zone.
c) Draw fault-plane solutions for two earthquakes occurring on the fault at the 50 Ma
isochron to the west and east of the active zone. (Be careful here. Remember that
cooling of the oceanic plates is more effective when the plate is young)

50            40           30       20     10        0          10     20             30

Age of upper part [Ma]

Age of lower part
Age of lower part [Ma]

30          20            10          0    10      20          30          40          50         60

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Solution

a) The active zone of the transform fault lies between the two ridge segments where the
two plates move horizontally with respect to each other. The left plate moves to the
left with respect to the right plate, i.e. the motion is left lateral (or sinistral). The
polarity of first motions then follows the pattern shown below

+   -

-   +

b) Most earthquakes happen in the active part of the transform fault, but due to thermal
subsidence effects a small number of earthquakes also happens outside the this zone.
The reason for the latter earthquakes is that the plate, even though it belongs to the
same unit has different ages across the inactive part of the transform fault. This age
difference leads to stress accumulations (and eventually rupturing occurs) because
the subsidence is faster the younger the plate.
c) To the west of the active part of the transform fault the upper part of the plate is older
than the lower part of the plate. Hence the upper part subsides slower than the lower
part. The situation is reverse to the east of the active part. Faulting takes place as
normal faults, and the polarity of first motions are as follows:

+                                                -
-                                                +

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Exercises. Selection with solutions.                                                      20/03/01

Exercise number 3, February 19, 2001

Consider an earthquake taking place exactly at the interface between the crust and the
mantle.
Assume that the crust has a P-wave velocity of 6.5 km/s and the upper mantle a P-wave
velocity of 8 km/s. Assume further that Poisson's ratio is equal to 0.25 for crust and upper
mantle and that the crustal thickness is equal to 40 km.
a) Find the travel times for first P- and S-waves at the epicenter and at a distance of 200
km away from the epicenter.
b) Assume that the earthquake is a strike-slip earthquake with a N-S strike with the
eastern block moving from south to north with respect to the western block. Describe
how the polarity of the P-wave varies as a function of azimuth (angle between the
north direction and the vector going through the epicenter and the observer).

Solution

Poisson’s ratio is equal to 0.25. This means that the ratio between P and S wave
velocities is given by
vP
= 3
vS

a) The travel distance of the waves is equal to 200 2 + 40 2 km . Thus the travel times
for P and S waves can found as
1
t S = 3t P = 3        200 2 + 40 2 s=54,3 s
6 .5
t P = 31,4s

b) The polarity of the P-wave as it arrives to observers located at different azimuths
follows the standard “beach ball” solution for a strike slip earthquake. The nodal
planes are found at azimuths 0, 90, 180, and 270 degrees. The polarity of the P-
wave is undefined there, since there is no motion on the nodal planes. Otherwise the
polarity follows the rules:

Azimuth range[degrees]    Polarity
0-90                +
90-180               -
180-270               +
270-360               -

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Exercises. Selection with solutions.                                                         20/03/01

Exercise number 4, February 19, 2001

Consider the wave equation

∂u
ρ          = (λ + µ )∇(∇ ⋅ u) + µ∇ 2 u
∂t 2

Assume plane wave propagation in the positive z-direction in a homogeneous full-space.

a) Find the wave-speed when particle motion is in the z-direction.
b) Find the wave-speed when particle motion is in the x-direction.
c) Two observers, separated 100 km apart along the z-axis, measures the difference
between the arrival times of the P- and S-waves from a hypothetical earthquake.
The first observer, closest to the epicenter, measures on his seismogram a
travel time difference of ∆t SP (1) =90 s.
The second observer measures on his seismogram a travel time difference
of ∆t SP (2) =100 s.
♦ Find the distance from the epicenter to the first observer
♦ Find the P wave velocity assuming that Poisson's ratio is
0.25.

Solution

⎧u ⎫
⎪ ⎪
Write the displacement vector u = ⎨v ⎬ . We are dealing with a plane wave in the z-
⎪ w⎪
⎩ ⎭
direction which means that all derivatives in the x- and y-directions must be zero.

a) Particle motion is only in the z-direction. Thus

∂w           ∂ ∂      ∂2
ρ 2 = (λ + µ ) ( w) + µ 2 w ,
∂t           ∂z ∂z    ∂z

which simplifies to

∂w λ + 2µ ∂ 2 w    2 ∂ w
2
λ + 2µ
=          = vP      ; vP =
∂t 2
ρ ∂z   2
∂z 2
ρ

b) Particle motion is only in the x-direction. Hence

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Exercises. Selection with solutions.                                                   20/03/01

∂u           ∂ ∂       ∂2      ∂2
ρ 2 = (λ + µ ) ( u ) + µ 2 u = µ 2 u ,
∂t           ∂x ∂x     ∂z      ∂z

which simplifies to

∂u µ ∂ 2            2 ∂
2
µ
=        u = v S 2 u; v S =
∂t 2
ρ ∂z 2
∂z         ρ

c)
Denote the unknown epicentral distance by x and the unknown slownesses by
s S and s P (remembering that slowness is the inverse of velocity). Then we can write
∆t SP (1) = x( s S − s P ) and ∆t SP (2) = ( x + 100)( s S − s P ) .
Then
∆t SP (2) − ∆t SP (1) = 10 = 100( s S − s P ) .
A Poisson’s ratio of 0.25 means that the two slownesses are related as
sS = 3 sP
whereby velocities can be found to

vP = 7,3 km / s and vS = 4,2 km/s.
The epicenter distance x can be found from the first equation

∆t SP (1) = 90 = x( s S − s P ) = 0.1x   as

x = 900 km

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Exercises. Selection with solutions.                                                                                20/03/01

HEAT

Exercise number 5, February 18, 2000

It is well known that heat production in upper crustal rocks is much higher than the heat
production in lower crustal and mantle rocks. A popular simplified model is one where
the heat production of the crust, A is given by an exponential decay:

A = A( z ) = A0 exp(− z / z e )
where z is the depth variable and ze and A0 are constants that characterise a given small
province. The parameter zl is supposed to be much smaller than the total thickness of the
crust.

a) Suppose that part of the province has been eroded by an amount ∆z e . Calculate the
heat production as a function of depth for the eroded part of the province. What is the
surface value, and what is the value at the depth 0.5 z e ?
b) Suppose that the temperature does not vary with time and that the heat-flow from the
mantle is constant for the whole province and given by q m .
c1) Set up the differential equation for the heat-flow in the crust
c2) Find the heat flow measured at the surface of the earth in the eroded and
non-eroded part of the province.
c3) Describe how you would find the mantle heat flow from many
observations of surface heat flow and surface heat production over the same
province.

Solution

Remember that the decay depth z e is much smaller than the typical crustal thickness of
40 km. This means that at depths corresponding to the lower crust beyond 30 km depth
the heat production is practically equal to zero.

a) The heat production per unit mass is denoted by A = A( z ) = A0 exp(− z / z e ) . If part of
the province is eroded by an amount ∆z it means that the new surface is found at
z = ∆z . Or if we introduce the new z co-ordinate z’ as z ' = z − ∆z we may describe
the heat production as a function of the new depth as

A' ( z ' ) = A( z '+ ∆z ) = A0 exp( −( z '+ ∆z ) / z e ) = A0 exp( − ∆z / z e ) exp( − z ' / z e ) = A0 exp( − z ' / z e )
'

where it can be noticed that after erosion the heat production still decays exponentially
with depth with the same depth variation but with a smaller surface heat production.
The surface value is given by

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Exercises. Selection with solutions.                                                    20/03/01

A0 exp( − ∆z / z e )

and the value at 0.5 z e is given by

A0 exp( − ∆z / z e ) exp( −0.5) = 0.61 A0 exp( − ∆z / z e )

c)

c1) Under steady state conditions the temperature in the crust satisfies the stationary
heat equations
d 2T     ρ
2
= − A( z )
dz      k
where density ρ and thermal conductivity are assumed to be constant, independent of
depth, z in the crust.
c2) The differential equation can be solved under the given boundary conditions. Since
we are only interested in heat flow we can rewrite the above equation in terms of heat
flow, q using the constitutive equation

dT
q = −k      .
dz
Hence

dq
= ρA(z )
dz
which can be solved by noting that at the crust/mantle interface at depth h the heat flow
is constant and equal to (- q m ). Hence the above equation may be integrated from zero to
infinity in depth (remembering that heat flows in the negative z-direction)

− qm      ∞

∫ dq = ∫ ρA
− qs     0
0   exp(− z / z e )dz

After integration we find

(−q m + q s ) = ρA0 z e

or the surface heat flow can be expressed as the sum of the heat flow from the mantle
plus the heat production in the crust, given by the product of density, surface heat
production and depth scale:

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Exercises. Selection with solutions.                                                      20/03/01

q s = q m + ρA0 z e
This means that the total heat flow is linearly related to the surface heat flow
independent of erosion!

c3) Under the assumptions set up above it is now possible to estimate the heat flow from
the mantle in a given “heat flow province”, i.e. an area for which the mantle heat flow
does not vary laterally. We must measure the pair (surface heat production, surface heat
flow) for a lot of sites. Then plot the pair against each other and find the intercept on the
vertical axis. This intercept can be interpreted as the heat flow from the mantle.

10
Exercises. Selection with solutions.                                                         20/03/01

Exercise number 6, February 18, 2000

In order to model the effect of the sudden temperature increase that took place by the end
of the last ice age, ca 10000 years ago on the temperature distribution in the earth, we
will assume that the temperature increase was 10 degrees Celcius. Assume further that
the upper part of the crust can be taken to be homogeneous with a constant diffusivity of
10 −6 m 2 / s .

You are simply asked to find the depth at which the temperature perturbation due to
sudden temperature increase by the end of the last glaciation will be less than 0.1 degrees
Celcius.

Solution

In order to answer the question we need to find the temperature perturbation in the earth
due to the sudden temperature increase after the last glaciation. For a 1-dimensional
earth the general solution is given by

z
T = T ( z , t ) = ∆Terfc(   )
2κt
where z is depth, t is time, κ is diffusivity, and ∆T is the surface temperature increase.
10000 years corresponds to 31,5 ⋅ 1010 s . We need to find the depth, z such that

z
0 . 1 = 10 erfc (                      )
2 ⋅ 31 ,5 ⋅ 10 4
or
z = 63 ⋅ 10 4 ⋅ erfc −1 (0.01) ≈ 1453 m !

Thus a pretty deep bore-hole is needed in order to ensure that the temperature is
undisturbed by long term surface temperature variations.

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Exercise number 6, February 19, 2001

In order to study how long time it takes for a thin dyke of thickness 10 meters to solidify
when it is intruded into a solid rock we will study the following simplified model
(neglecting that a real dike solidifies from its surface inwards) :

The length and                               10 m          Initial temperature of
depth extent of the                                        surrounding rock
dike is taken to be                                        is taken to zero degrees.
infinite.                                    Dike

The thermal diffusivity                                    During the solidification
is 10-6 m-2 s-1 and the                                     process the temperature
thermal conductivity is                                    of the entire dike is taken
2 W/m/0C                                                   to be 1200 degrees.

The latent heat of fusion is assumed to be 300 kJoule/kg or 750 MJoule/m3. When the
dike has given away as much heat as the latent heat will have solidified.
♦ How long time is required before it is completely solidified?

Hint: When you solve the problem first set up a heat balance and then remember how the
heat flow of a half-space behaves as a function of time for the appropriate boundary
condition.

Solution

The problem as formulated here is very similar to the cooling of the oceanic lithosphere
using the half-space cooling model. The difference is that here the surface of the half-
space is kept at a constant temperature of T0 while the surrounding half-space is slowly
being heated by the solidifying dike. The temperature in the half-space is given by

z
T = T ( z , t ) = T0 erfc(         )
2κt

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Exercises. Selection with solutions.                                                                              20/03/01

where z is the distance away from the solidifying dike, t is time and κ is thermal
diffusivity of the surrounding rock.

∂T
. At z=0 this
The heat flow, q, i.e. the heat per unit time and unit area is given by k
∂z
becomes particularly simple (see Fowler page 239 and change sign) and is given by:

kT0
q = q (0, t ) =          .
πκt

We can now set up a heat balance such that after a certain time, t all latent heat in the
dike has been allowed to diffuse out into the surrounding rock. First we notice that the
half of the latent heat diffuses towards the left and the other half to the right. Secondly,
since we are dealing with a two-dimensional problem, we can choose to study the
conditions only for a block of cross-sectional area 1 square meter. The total latent heat
in this block of height 5 meters is then

QL = 5 ⋅ 750 MJoules = 3,75 GJoules

The total heat flowing out into the surrounding rock through a cross-sectional area of 1
square meter is given by the integral of the heat flow over time:

t                       t
kT0       1                2kT0
Qd = ∫ q(0, t ' )dt ' =            ∫           dt ' =          t
0                  πκ   0      t'             πκ

By equalising the two terms the unknown time t can be recovered as

πκ                 π ⋅ 1o −6
t=                  QL =
2
(3,75 ⋅ 10 9 ) 2 = 7,7 ⋅ 10 8 s ≈ 24 years
(2kT0 )   2
(2 ⋅ 2 ⋅ 1200)      2

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Exercises. Selection with solutions.                                                    20/03/01

AGE DATING, ISOTOPES

Exercise number 7, February 18, 2000

Explain in a qualitative way how we can distinguish between rocks that were derived
from melting in the upper mantle from rocks that were derived from melting in the lower
crust.

Solution

Generally speaking the so-called Large Ion Lithophile (LIL) elements become enriched
in magmas derived from melting of the mantle and making up major parts of the earth
crust. Conversely that part of the mantle which is the source of the melts becomes
depleted in the same elements as more and more crust is produced. This is the reason
why the upper mantle is referred to as the depleted mantle.

Rubidium is an LIL element. Its concentration is much larger in crustal rocks than in
87
Sr
mantle rocks. The initial 86    ratio in a rock can be directly related to its age if it is
Sr
derived directly from the mantle, because the mantle growth curve for this ratio is known
87
Sr
to vary very slowly. Therefore, if the 86 ratio is significantly higher than that it means
Sr
that that rock must have been formed by remelting of crustal material.

For the Samarium Neodymium system the situation is the reverse. When mantle material
melts it becomes enriched in Neodymium relative to the melt that eventually contributes
to crustal formation. The ε Nd parameter gives a direct indication of magma source. For
example mid-ocean ridge basalts and continental flood basalts have ε Nd values of 10 and
0, respectively, indicating the former basalts were derived from sources rich in
Neodymium and the latter basalts probably were mixed with continental material before
they came to the surface.

14
Exercises. Selection with solutions.                                                            20/03/01

Exercise number 5, February 19, 2001

Four samples from a meteorite have neodymium and strontium isotope ratios as follows:

143
Nd / 144Nd   147
Sm / 144Nd
0.5105          0.12
0.5122          0.18
0.5141          0.24
0.5153          0.28

a) Find the age of the meteorite.
b) Find the initial 143 Nd / 144Nd ratio for this meteorite.
c) Discuss the relevance of the initial 143 Nd / 144Nd ratio of meteorites.

Solution

First it is noticed that when plotting the 143 Nd / 144Nd ratio versus the 147 Sm / 144Nd ratio it
turns out that they fall nicely on a straight line. The slope, s of the line can be estimated
well from the first and last points in the table.

0.5153 − 0.5105 0.0048
s≈                 =        = 0.0343
0.28 − 0.14       0.14
a) The age, t of the meteorite can be found from the slope as

s ≈ λ Xm− Nd t
or
34,3
t ≈ s / λ Xm − Nd = 0.043 /(6.54 ⋅10 −12 ) a =        Ga = 5,24 Ga
6,54

b) The initial 143 Nd / 144Nd ratio is the intercept of the line above. The intercept can be
calculated from the linear equation

y = sx + β .

With y=0,5105, x=0,12 and s=0,0343 the intercept (or initial ratio) β is found to be
equal to

β = 0,5064

c) A meteorite can be considered as a closed system since its formation some 4.5 Ga
ago therefore its initial ratio can be considered as ground truth to which all
measurements on the Earth should be related.

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Exercises. Selection with solutions.                                                    20/03/01

The relevance of the initial ratio for meteorites is that it is believed to represent the
conditions in the lower mantle to which we have no direct access. By relating
measurements of initial ratios on real rocks to those of meteorites adjusted to the same
ages as the real rocks it becomes possible to estimate from what source depth the magma
came from. Upper mantle rocks are richer in Nd than crustal rocks and so are the melts
generated from these rocks.

16

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