Transcript Seismology

Seismology and the Internal
Structure of the Earth
Seismology is all about how vibrations travel through the
Earth. Some applications:
1.
Imaging: Earth structure, natural resources,
geotechnical, etc.
2.
Source Properties: earthquakes, explosions
3.
Responses to excitation: seismic waves, tsunamis
Seismology provides the highest resolution images of the subsurface
of any technique.
It is probably the most quantitative of any type of geophysical
analysis.
The good part is that we can derive everything we need to know
from a few fundamentals.
The difficult part (depending on taste) is that there can be a lot of
math.
In this lecture, I include a lot of derivation so you can see all of the
reasoning behind the techniques, BUT you really don’t need to
know or understand most of it to use it, and it’s really a lot for an
intro course.
So, if you need to know it, look for BLUE backgounds, like this.
If it is “extra” material, it will be on a GREEN
background, like this.
SEISMOLOGY
PART I:
THEORETICAL BACKGROUND
In a general sense, seismology is all about how vibrations travel
through the Earth.
A vibration is a displacement that changes with time. In order to
discuss vibrations this quantitatively, we have to have a way to talk
about displacements and the forces that cause them.
In intro level physics, we learn about forces and displacements that act
on infinitesimally small bodies (points) or talk about situations where a
finite mass can be thought of at as a point (e.g., through its center of
mass).
In a big, continuous body like the Earth, we have to be a bit more
sophisticated. We use the same classical mechanics, but
Instead of Forces, we talk about Stresses
Instead of Displacements, we talk about Strains
This is part of the general discipline of Continuum Mechanics
Stress and Strain
Continuum Mechanics – Force and displacement in a finite body.
Force -> Body forces (force per unit volume) and Stress (force per
unit area)
Displacement -> Rigid body displacement and rotation and Strain
(internal deformation).
Body Forces: Gravity and EM, for example
Fg = gV; Fg /V = force per unit volume
Stress
Generally, we begin by defining a traction T acting on a surface S as:
F
dS 0 dS
T ( nˆ )  lim
where n is a unit normal to the surface. The total force acting on the
surface is then

F   T ( nˆ )dS

In general, T is not parallel to n. If we consider a simple element
like a tetrahedra with 3 of the 4 sides defined by a cartesian
frame, then consider the traction on face i (i = 1,2,3) in direction
j (j = 1,2,3) as:
i
T j  (T1i , T2i , T3i )

We define the stress element ij in terms of this traction:
ij  Tj i
and group these terms into a matrix:

11 12 13  T11 T21 T31


 
 ij   21  22  23 T12 T22 T32
3
3
3 

T
T
T
 31  32  33
 
 1
2
3 
This is called the Stress Tensor.
What’s the difference between a Matrix and a Tensor?
We can use this tensor to describe the stress along any surface in a
body.
Consider an arbitrary surface cut
though a body. We characterize
this surface by its unit normal n,
and say it has an area dS.
Imagine this as a triangular
surface of a tetrahedra with the
other three surfaces defined by
the cartesian axes. The unit
normals to these other three
surfaces are simply (-xi).
The areas of the other three
surfaces are a projection of dS:
dSj  dS( nˆ  xˆj )  nj dS
For the body to be in equilibrium, it must be the the force applied to dS
in any direction i is balanced by tractions on the other three sides:
3
3
j 1
j 1
Ti dS   ij dSj   ij nj dS
This tells us that we can calculate the traction on dS from the stress
tensor:

3
Ti   ij nj   ij nj
j 1

using the implicit summation convention (meaning that any index
on the right that doesn’t appear on the left is summed. This is
sometimes called the Einstein notation).
Some important properties of the stress tensor:
1. Positive directions are defined by the directions of the outward
normals to the surfaces.
2. 11, 22, 33 are called normal stresses, the others are shear
stresses.
3. ii > 0 is compression, < 0 is tension (this is different from
engineers, and kind of contradicts “1”, but tension is very rare in the
Earth except as a deviatoric stress).
4. ij = ji because of no net torque (ij *dx)dy = (ji *dy)dx.
How to rotate a tensor:
First, remember that we can calculate a vector in a rotated frame by
applying the rotation matrix A:
'

T  AT
and recall that the inverse of A is equal to its transpose:
1
A A
T
also

nˆ '  Anˆ
nˆ  A T nˆ '
So
T  nˆ
T '   ' nˆ ' AT  Anˆ  AA T nˆ '
Thus

 ' AA T
We can use the above relationship to show that the trace of the

stress tensor is invariant.
In other words:
3

j 1
3
ii
  ii'
j 1
Principal Stresses
Instead of specifying 6 stresses, it is often more convenient to specify
3 stresses and 3 directions. In these directions, the shear stresses are
zero, and the remaining normal stresses are called the Principal
Stresses.
Because the Stress Tensor is symmetric, we can always do this.
The problem of finding the principal stresses can be stated as finding
those surfaces where the tractions are parallel to the normal vector
(i.e. there is no shear stress). In this case, we look for n such that
T  nˆ
or
Ti   ij n j  n i
Here’s how you do it. First, recognize that
Ti   ij n j  n i
is just an eigenvalue problem. The eignvalues are the 's, which
become the magnitudes of the principal stresses, and the
corresponding eigenvectors
are the directions of principal stress.

We solve
 ij  ij nj  0
Which has a nontrivial solution when

det  ij   ij  0
Solve the above for 1, 2, and 3, and then find the n that goes with
each . Nothingto it!
An example:
3 1 0 


  1 2 1

0 1 3 

The determinant det  ij  ij  0 gives:

3  82 19  12  0
1  1
2  3
3  4

and the corresponding eigenvectors are:
1
1 
1 
1  
1  
1  
n1 
2
;n

0
;n

1
2
3





6
2
3





1
1
 
 
1 


Stress in the Earth: Lithostatic Pressure
Consider a block with uniform density , thickness h and area A. The
volume is hA, so the total force at the bottom of the block is
Fg = mg = (hA)g
and the stress is
Fg /A = gh = yy
This simple formula allows us to calculate overburden pressure and
lithostatic stress in the Earth.
Example: Isostacy - Archimedes principle in the Earth.
The mantle behaves like a fluid over Geologic time. A consequence is
that the crust “floats” in the mantle, and there is an isopiestic depth
(equal pressure) below which the vertical stress is uniform.
If g does not change significantly with depth, then the only thing that
matters in determining pressure is mass. We solve these problems by
integrating mass over all depths over which there are lateral differences
in density.
The simplest example is continental crustal root:
Let h be the crustal thickness of density c above mantle density m.
Ignoring the density of the atmosphere, equal mass gives hc = bm ,
where b is the thickness of the crustal root.
Note that there is a net lateral stress needed to maintain the
continental crust; it should be tensional to keep it from falling
apart. In the air, this is pretty obvious, but there will be a stress
gradient below the surface as well.
This is calculated by integrating the stress over the depth to the
isopiestic level.
The force exerted at a given depth by the mantle is
Fm = mgz
Thus, integrating over z (from 0 to b) gives the total force exerted by
the neighboring mantle on the crust as
Fm = (1/2) gb2m
(technically this is a force per unit length).
Do the same for the crust – integrating this time from 0 to h;
Fc = cgz
Fc = (1/2) gh2c
In order to balance these two, there needs to be an additional
stress xx applied within the crust over the distance h. Thus:
(1/2) gh2c + xxh = (1/2) gb2m
xx = (1/2h) g(b2m - h2c)
From isostacy:
hc = bm
xx = (1/2h) g(h2c2 m2m - h2c)
xx = (ghc /2) (c m- )
This is a negative tensile stress holding the crust together.
Example: Pressure in the Interior of a Planet
We already have shown that the stress at the bottom of a column of
constant density  is
Fg /A = gh = p = yy
If  is not a constant, then we have to consider a P as
P = rgrr
Remember from gravity that for a radially symmetric body:
GM(r)
g(r) 
r2
where M(r) is the Mass within radius r:
M(r) 

r
0
4 r'2 (r')dr'
Thus, in general:

P(r) 
a
a
r
r
 (r )g(r )dr  
r
G
(r ) 2 dr  0 4 r'2 (r')dr'
r
Note that density will depend on pressure in general, so we need to
know that before going further, but we can consider the simple case
of
a constant density body (like the moon), of radius a:
a
a
r
dr
4
2
2
2
2
P(r)  4 G  2  0 r' dr '  G  rdr  G 2 a 2  r 2 
3
3
r r
r
Which has a maximum at the center (r = 0)
Strain
Displacement -> Rigid body displacement
and rotation and Strain (internal
deformation).
Imagine a length x shortened by a small
amount dx. We define the normal strain as
the relative displacement
exx = dx/x
If we have a rectangular parallelpiped and shorten each axis without
changing it's shape, then if the original volume is
The new volume is
Vo = xyz
V = (x+dx)(y+dy)(z+dz) = xyz(1+exx) (1+eyy) (1+ezz)
and the fractional change in volume is
(ignoring products of small terms)
(V-Vo)/Vo = V/Vo - 1
= (1+exx) (1+eyy) (1+ezz) - 1
= exx+eyy+ezz= Dilatation ()
More generally, let's consider two points in a body: x and x+dx and
see what happens to them on deformation. Suppose that
x -> x + u
x + dx -> x+dx + u + du
u represents rigid body
translation, while
du represents rigid body
rotation and deformation
We can come up with an expression for du by expanding u and
keeping only the first order terms.
du 3
du1
du 2
u  du  u 
x 
y 
z  O(l ) 2
dx
dy
dz
The total derivative du is related to the partials of u by (for example):
du 1
 u u u 
x   1  2  3 x
dx
x 
 x x
or
dui
du j 
x j
dx j
(summation over i = 1-3 implied)
Let's define two matrices:
so that:
About wij:
1. wii = 0


1 dui duj 
eij   

2 dxj dxi 
1 dui duj 
wij   

2 dxj dxi 
duj  eij wij xj

2. wij = -wji (i.e., it's antisymmetric, which means there are only 3
independent terms)
Now, suppose we have two axes: dx1 and dx2 - that are at right angles
to each other. Suppose that we deform the body so that an angle 2 is
between dx1' and dx1 and an angle 1 is between dx2' and dx2. Note
that:
du2
dx1
du
dx1
tan 1 
 2  1
dx1
dx1
du1
dx2
du1
dx2
tan  2 

 2
dx2
dx2
The amount of rigid body rotation
is
1   2 1 du2 du1 
 

 w21
2
2  dx1 dx2 
Thus, the w matrix isolates the rigid body rotation part of the strain.
In general, the total rigid body rotation r about an axis pointed in
the dx direction can be calculated from
r   dx
where

 k   ijkwij
and
of i, j, and k are the same
 ijk  0 if any 
 ijk  1 if i, j, and k are in cyclical order (1,2,3; 2,3,1; 3,1,2)
 ijk  1 if i, j, and k out of order (i.e. otherwise).
In the absence of any solid body rotation (1 = 2 and w = 0), the
resulting shear is called "pure shear".
If one of du2/dx1 or du1/dx2 is zero, we have a combination of rotation
and shear strain called "simple shear".
Simple shear is important on transform (strike-slip) faults (or on faults
in general). In this case
1 dui
eij 
2 dxj

1.
2.
3.
4.
5.
Uniaxial Extension
Uniaxial Compression
Simple Shear
Pure Shear (when du1/dx2 = du2/dx1)
Rigid Rotation (when du1/dx2 = -du2/dx1)
As with stress, there are principal strain directions. The techniques
for finding the directions and magnitudes are the same as for stress
(both are 3x3 symmetric tensors).
Measuring Tectonic Strain
Strain accumulates in the Earth for a variety of reasons, the most
important of which is plate motions. These are slow, so the
measurement techniques often have to be rather sophisticated.
Eg. San Andreas fault rupture in 1906: 4 meters of displacement. If
the distance perpendicular to the fault over which that strain
accumulated is 40 km (approximately) then
1 dui 1 2
eij 

 2.5105
2 dxj 2 40000
Note we use 2 meters to allow half the slip on either side of the
fault.

The strain-rate can be determined by dividing the above by an
expected recurrence interval; say 100 yrs:
2.5 10 5
eij 
 0.25 10 5 / yr
100
The preferred way to do this is through geodesy (GPS or
traditional triangulation).
The Dynamic Equation of Motion
Let’s suppose that there is a normal stress imbalance across a unit
volume in the i=1 direction, such that the stress on one side is 11 and
on the other side it is 11 + 11/x1 dx1.
The force on either side is the stress times the area = dx2dx3. The net
force is then
2 u1
2 u1 11
11
F1  m 2  v 2 
x1x 2x 2 
v
t
t
x1
x1
or
2 u1 11
 2 
t
x1

If we generalize to include (1) shear stresses and (2) all three directions
and (3) body forces (fi) we get the three dynamic equations of motion for
a continuum

 ij
2 ui
 2  fi 
t
x j
Thus, the net accelerations experienced by a body are proportional to
the stress gradient.
The body force fi will come in handy when we introduce seismic
sources (like earthquakes). For an ambient wave, we typically (but not
always!) can ignore gravity, so it is enough to solve the homogeneous
equation of motion (f = 0):
2 ui  ij
 2 
t
xj

In order to be able to solve this equation of motion, we need a
relation between stress and strain (constitutive equation) . Often
these relations are complicated (subject of Rock Mechanics).
For small, high strain rate behavior, rocks are linearly elastic, and
we can use a form of Hooke's Law:
 ij  Cijkl ekl
4 subscripts => 3 x 3 x 3 x 3 = 81 elements (elastic moduli) of C!

But because of symmetry:
Cijkl  C jikl
Cijkl  Cijlk
Cijkl  Cklij
(stress symmetry)
(strain symmetry)
(strain energy density symmetry)
So only 21 elements are independent. A little better.

We now make a BIG ASSUMPTION: ISOTROPY.
In an isotropic medium, there are only 2 independent elastic
moduli:  and . These are the Lamé parameters.
In this case:
Cijkl  ijkl   (ikjl iljk )
If we substitute this into our linear elastic constitutive equation:

 ij  ekkij  2eij  ij  2eij
Now, we put this in our equation of motion. For sake of derivation
simplicity, consider the case of i = 1:

2 u1 1 j 11 12 13
 2 



t
x j
x1
x 2
x 3
Using or constitutive equation, we write ij in terms of u:
11   2e11
u1 u2 u3 
u1
  


2


x1
x1 x2 x3 

12  2e12


13  2e13


u1 u2 
   

x2 x1 
u1 u3 
   

x3 x1 
Insert these relations back into the equation of motion and we get
2 u1
  u1 u2 u3 
u1 
 2 

 
 2 
t
x1  x1 x2 x3 
x1 
  u1 u2    u1 u3 

  
  


x2  x2 x1  x3  x3 x1 


In general, this is a difficult system of equations to solve, so we
make our next BIG ASSUMPTION: HOMOGENEITY!
2 u1 2 u2
2 u1
2 u3 
2 u1
 2   2 

 2 2
t
x1
x1 x1x2 x1x3 


2 u1 2 u2  2 u1 2 u3 
  2 
   2 

x2 x1x2  x3 x1x3 
2 u1
 u1 u2 u3 
 2    


t
x1 x1 x2 x3 

 u1 u2 u3  2 u1 2 u1 2 u1 


 
   2  2  2 
x1 x1 x2 x3  x1 x2 x3 
2 u1
 u1 u2 u3  2 u1 2 u1 2 u1 
 2       

   2  2  2 
t
x1 x1 x2 x3  x1 x2 x3 
And similarly for the other directions:
  2u 2  2u 2  2u 2 
 2u 2
  u1 u2 u3 

    2  2  2 
 2     


t
x2  x1 x2 x3 
x2
x3 
 x1
  2u3  2u3  2u3 
 2u3
  u1 u2 u3 
    2  2  2 
 2      


t
x3  x1 x2 x3 
x2
x3 
 x1
Remember that the gradient operator is:
    
   ,
, 
x1 x2 x3 
So we can summarize the above 3 equations as:

       u    2u
u
Recalling the vector identity
2u   u    u
       u     u      u
u

    2   u      u
u
Well, this still looks pretty complicated. Here's a
clever trick to make this more tractable:
The HELMHOLTZ theorem
Helmholtz showed that any vector field (like u)
can be represented as sum of the gradient of a
scalar (potential) field and the curl of a vector
(potential) field:
u      
Hermann Ludwig Ferdinand
von Helmholtz
1821-1894
Why "potential"? Because, for example, potential energy is
analogously related to force by a gradient (F = U/x). Substitute this
into the wave equation:


     2           
    
    2                   
     
  2       
Now we make use of some vector identities:
       0
2









     0
And get

    2   2     2 
     
or

  0
   2  2       2   



 





Now, since this has to be zero everywhere, it must be that the terms
in the brackets are zero everywhere, and so we now have two
independent equations:
  2 2    0
  0
 2   
or
  2 2





    2 

And these we can easily solve!
What is the solution general solution to the wave equation?
In one dimension:
2
2 u

u
2

v
t 2
x 2
Easy! It’s just
u = f(x - vt) + g(x + vt)

This is the beauty of waves: f and g are arbitrary! All that is important
is the argument to the function, not the function itself.
x±vt is called the phase of the wave; the translating functional shape is
called the wavefront.
A particularly useful realization of f and g is the sine and and cosine
solution, because we can fabricate any arbitrary function as a weighted
sum of the sines and cosines. For example, we can write:
f(x - vt) = sin(k(x - (/k) t)) = sin(kx - t)
and we identify v = /k.
The reason we do this is because k and  have simple and useful
physical meanings:
= 2f = angular frequency
f = 1/T; T = period
k = 2/ = wave number
(number of waves in a given distance).
 = wavelength
we can also write the 1D solution as:
u = A ei(kx±t)
where ei = cos() + i sin()
The extension to 3D is straightforward. For the scalar potential (),
we have
 (x,t)  Ae i (kxt )
Where k = (kx, ky, kz) is called the wavenumber vector and the
wavespeed is


  2


|k|

k points in the direction of propagation of the (in this case) plane
wave.
We can do the same trick with the vector wave equation by simply
solving for each component as above and then combining the result to
give:
(x,t)  Be i (kxt )
In this case



|k|



Note that because the two equations are independent, it is not the
case the values of k need be the same for both.

Remember that these solutions are not displacements, but
displacement potentials. To recover displacement, we need to put
the above formulas into:
u      
For the scalar part:

  Ae i (kxt )  ikAe i (kxt )
This tells us that the displacement u is
parallel to the direction of propagation
(defined by k).
For the vector part:
      Be
i (kxt )
3 2 
 

xˆ 1
x2 x3 
1 3 
 
xˆ 2
x3 x1 
2 1 


xˆ 3
 x1 x2 
let's presume the wave is
For simplicity,
propagating in the x1 direction, which means
that the terms involving d/dx2 and d/dx3 will
 k·x = kx ). In this case:
be zero (because
1
 3 
2 
      Be
 
xˆ 2  
xˆ 3
 x1 
 x1 
or, in other words, all the displacement is in the (x2, x3) plane,
perpendicular to x1.
i (kxt )
The total wavefield is the sum of the two
waves:
 3 2 
u   

xˆ 1
x1 x2 x3 
 1 3 
 

xˆ 2
x2 x3 x1 

 2 1 
 

xˆ 3
x3 x1 x2 

Remember that the change in volume (the
dilatation ) is equal to the divergence of u;
i.e.,

   u        2
Thus, the wave produced by the 
potential does not involve a change in
volume, but the one associated with the 

potential does, and in fact    2
Note that ()2 = (lBecause 
and  are always positive,  is always greater than
, so the  waves are always faster than the 
waves. In fact, for a lot of common rocks,  ~ , so
()2 ~ 3. (The case  =  of is called a Poisson
Solid).
For these reasons, the waves associated with  are
called dilatational, compressional, primary, or just P
waves, while those associated with  are rotational
or shear or secondary or just S waves.
Simon Denis Poisson
1781-1840
Also, because  = 0 in gasses and inviscid liquids, there are no shear
waves in those media, and the P waves are called acoustic (sound)
waves.
In a radially symmetric Earth, it will be convenient to define SV and
SH waves. The reason why will become evident later.
Some notes on elastic moduli:
Of the two the Lamé parameters,  has a simple physical meaning as the
ratio of shear stress to shear strain, but  does not. However, there are
several other moduli that can be easily measured that can be related to :
1. Bulk modulus or incompressibility (k) = Ratio of
pressure to dilitation:
P
2
k
  

3
2. Young's Modulus (E) = Stress/Strain ratio in uniaxial stress:
11  (3  2 )

E

e11
 
3. Poisson's Ratio () = Strain ratio in uniaxial stress:
e22



e11
2(    )
