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:

€

rT ( ˆ n ) = lim

dS −>0

v F

dS

€

vF =

r T ( ˆ n )dS∫

where n is a unit normal to the surface. The total force acting on the surface is then

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:

),,( 321iiii

j TTTT =r

We define the stress element ij in terms of this traction:

€

ij = Tji

and group these terms into a matrix:

€

ij =

σ 11 σ 12 σ 13

σ 21 σ 22 σ 23

σ 31 σ 32 σ 33

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥=

T11 T2

1 T31

T12 T2

2 T32

T13 T2

3 T33

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

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.

€

dSj = dS( ˆ n ⋅ ˆ x j ) = njdS

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:

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:

€

TidS = σ ijj =1

3

∑ dSj = σ ijj =1

3

∑ njdS

This tells us that we can calculate the traction on dS from the stress tensor:

€

Ti = σ ijj =1

3

∑ nj = σ ijnj

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:

TATrr

='

and recall that the inverse of A is equal to its transpose:

€

A−1 = AT

also

nAn ˆˆ ' =

€

ˆ n = AT ˆ n '

€

'= AσAT

Thus

We can use the above relationship to show that the trace of the stress tensor is invariant. In other words:

€

iij =1

3

∑ = σ ii'

j =1

3

∑

€

rT = σ ˆ n r T ' = σ ' ˆ n '= AT = Aσ ˆ n = AσAT ˆ n '

So

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

€

rT = λ ˆ n

€

Ti = σ ijn j = λnior

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 − λδ ij( )nj = 0

Which has a nontrivial solution when

€

detσ ij − λδ ij = 0

Solve the above for 1, 2, and 3, and then find the n that goes with each . Nothing to it!

€

Ti = σ ijn j = λni

Here’s how you do it. First, recognize that

An example:

€

=3 −1 0

−1 2 −1

0 −1 3

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

The determinant gives:

€

detσ ij − λδ ij = 0

€

3 − 8λ2 +19λ −12 = 0

λ1 = 1

λ2 = 3

λ3 = 4

and the corresponding eigenvectors are:

€

n1 =1

6

1

2

1

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥;n2 =

1

2

1

0

−1

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥;n3 =

1

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 = mgz

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 = cgz

Fc = (1/2) gh2 c

In order to balance these two, there needs to be an additionalstress 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 m

2 m - h2 c)

xx = (ghc /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 = rgrr

Remember from gravity that for a radially symmetric body:

€

g(r ) =GM(r )

r 2

where M(r) is the Mass within radius r:

€

M(r ) = 4πr '2 ρ (0

r

∫ r ')dr '

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:€

P(r ) = ρ (r )g(r )drr

a

∫ = ρ (r )G

r 2dr 4πr '2 ρ (

0

r

∫ r ')dr 'r

a

∫

Thus, in general:

€

P(r ) = 4πGρ 2 dr

r 2r '2

0

r

∫ dr 'r

a

∫ =4

3πGρ 2 rdr

r

a

∫ =2

3πGρ 2 a2 − r 2

( )

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

Vo = xyzThe new volume is

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 + ux + 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.

The total derivative du is related to the partials of u by (for example):

xx

u

x

u

x

ux ⎟

⎠

⎞⎜⎝

⎛∂∂

+∂∂

+∂∂

= 32

dxud

jj

ij x

dx

duu =d

(summation over i = 1-3 implied)

or

2321 )(dz

ud

dy

ud

dx

udd lOzyx ++++=+ uuu

Let's define two matrices:

€

eij =1

2

dui

dxj

+duj

dxi

⎛

⎝ ⎜

⎞

⎠ ⎟

€

wij =1

2

dui

dxj

−duj

dxi

⎛

⎝ ⎜

⎞

⎠ ⎟

€

duj = eij +wij( )Δxj

so that:

About wij:

1. wii = 0

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:

€

tanδ1 =

du2

dx1

dx1

dx1

=du2

dx1

≈δ1

€

tanδ2 =

du1

dx2

dx2

dx2

=du1

dx2

≈δ2

The amount of rigid body rotation is

€

1 −δ2

2=

1

2

du2

dx1

−du1

dx2

⎛

⎝ ⎜

⎞

⎠ ⎟= w21

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

€

rr = −

r ω × d

r x

€

ωk = ε ijkwij

where

and

if any of i, j, and k are the same if i, j, and k are in cyclical order (1,2,3; 2,3,1; 3,1,2) if i, j, and k out of order (i.e. otherwise).

€

ε ijk = 0

€

ε ijk = 1

€

ε ijk = −1

In the absence of any solid body rotation (1 = 2 and w = 0), the resulting shear is called "pure shear".

€

eij =1

2

dui

dxj

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. Uniaxial Extension2. Uniaxial Compression3. Simple Shear4. Pure Shear (when du1/dx2 = du2/dx1)5. 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

€

eij =1

2

dui

dxj

=1

2

2

40000= 2.5×10−5

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:

yreij /1025.0100

105.2 55

−−

×=×

=&

The preferred way to do this is through geodesy (GPS or traditional triangulation).

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

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

€

F1 = Δm∂2u1

∂t 2= ρΔv

∂2u1

∂t 2=

∂σ 11

∂x1

∂x1∂x2∂x2 =∂σ 11

∂x1

∂v

or

€

∂2u1

∂t 2=

∂σ 11

∂x1

€

∂2ui

∂t 2= f i +

∂σ ij

∂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):

€

∂2ui

∂t2=

∂σ ij

∂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 = Cijklekl

4 subscripts => 3 x 3 x 3 x 3 = 81 elements (elastic moduli) of C!

But because of symmetry:

(stress symmetry)(strain symmetry)(strain energy density symmetry)

So only 21 elements are independent. A little better.

jiklijkl CC =

€

Cijkl = Cijlk

€

Cijkl = Cklij

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 = λδ ijδ kl + μ (δ ikδ jl +δ ilδ jk )

If we substitute this into our linear elastic constitutive equation:

€

ij = λekkδ ij + 2μeij = λθδ ij + 2μeij

Now, we put this in our equation of motion. For sake of derivation simplicity, consider the case of i = 1:

€

∂2u1

∂t 2=

∂σ 1 j

∂x j

=∂σ 11

∂x1

+∂σ 12

∂x2

+∂σ 13

∂x3

€

11 = λθ + 2μe11

€

= ∂u1

∂x1

+∂u2

∂x2

+∂u3

∂x3

⎛

⎝ ⎜

⎞

⎠ ⎟+ 2μ

∂u1

∂x1

€

12 = 2μe12

€

= ∂u1

∂x2

+∂u2

∂x1

⎛

⎝ ⎜

⎞

⎠ ⎟

€

13 = 2μe13

€

= ∂u1

∂x3

+∂u3

∂x1

⎛

⎝ ⎜

⎞

⎠ ⎟

Using or constitutive equation, we write ij in terms of u:

€

∂2u1

∂t2=

∂

∂x1

λ∂u1

∂x1

+∂u2

∂x2

+∂u3

∂x3

⎛

⎝ ⎜

⎞

⎠ ⎟+ 2μ

∂u1

∂x1

⎡

⎣ ⎢

⎤

⎦ ⎥

€

+∂∂x2

μ∂u1

∂x2

+∂u2

∂x1

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥+

∂

∂x3

μ∂u1

∂x3

+∂u3

∂x1

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥

Insert these relations back into the equation of motion and we get

In general, this is a difficult system of equations to solve, so we make our next BIG ASSUMPTION: HOMOGENEITY!

€

∂2u1

∂t2= λ

∂2u1

∂x12

+∂2u2

∂x1∂x2

+∂2u3

∂x1∂x3

⎛

⎝ ⎜

⎞

⎠ ⎟+ 2μ

∂2u1

∂x12

€

+ ∂2u1

∂x22

+∂2u2

∂x1∂x2

⎛

⎝ ⎜

⎞

⎠ ⎟+ μ

∂2u1

∂x32

+∂2u3

∂x1∂x3

⎛

⎝ ⎜

⎞

⎠ ⎟

€

∂2u1

∂t2= λ

∂

∂x1

∂u1

∂x1

+∂u2

∂x2

+∂u3

∂x3

⎛

⎝ ⎜

⎞

⎠ ⎟

€

+ ∂∂x1

∂u1

∂x1

+∂u2

∂x2

+∂u3

∂x3

⎛

⎝ ⎜

⎞

⎠ ⎟+ μ

∂2u1

∂x12

+∂2u1

∂x22

+∂2u1

∂x32

⎛

⎝ ⎜

⎞

⎠ ⎟

€

∂2u1

∂t2= λ + μ( )

∂

∂x1

∂u1

∂x1

+∂u2

∂x2

+∂u3

∂x3

⎛

⎝ ⎜

⎞

⎠ ⎟+ μ

∂2u1

∂x12

+∂2u1

∂x22

+∂2u1

∂x32

⎛

⎝ ⎜

⎞

⎠ ⎟

And similarly for the other directions:

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

+∂∂

+∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

+∂∂

+∂∂

∂∂

+=∂∂

23

22

22

22

21

22

3

3

2

2

1

1

222

2

x

u

x

u

x

u

x

u

x

u

x

u

xt

uμμλρ

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

+∂∂

+∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

+∂∂

+∂∂

∂∂

+=∂∂

23

32

22

32

21

32

3

3

2

2

1

1

323

2

x

u

x

u

x

u

x

u

x

u

x

u

xt

uμμλρ

Remember that the gradient operator is:

€

∇=∂∂x1

,∂

∂x2

,∂

∂x3

⎛

⎝ ⎜

⎞

⎠ ⎟

So we can summarize the above 3 equations as:

( ) ( ) uuu 2∇+⋅∇∇+=ρ μμλ&&

Recalling the vector identity

€

∇2u = ∇ ∇⋅u( ) − ∇×∇× u

( ) ( ) ( )( )uuuu ×∇×∇−⋅∇∇+⋅∇∇+=ρ μμλ&&

( ) ( ) uuu ×∇×∇−⋅∇∇+=ρ μμλ 2&&

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=∇φ+∇×ΨHermann 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:

( ) ( ) ( )( ) ( )ΨΨΨ ×∇+∇×∇×∇−×∇+∇⋅∇∇+=×∇+∇ φφφ 2&&&&

( ) ( ) ( )( ) ( )φφφ ∇×∇×∇−∇⋅∇∇+=×∇+∇ 2Ψ&&&&

€

+ + 2μ( )∇ ∇⋅ ∇× Ψ( )( ) − μ∇×∇× ∇× Ψ( )

Now we make use of some vector identities:

€

∇⋅∇×Ψ =0

€

∇× ∇×Ψ( ) =−∇2Ψ

€

∇× ∇φ( ) =0

And get ( ) ( ) ( ) ( )ΨΨ 222 ∇×∇+∇∇+=×∇+∇ρ μφμλφ &&&&

or

( )[ ] [ ] 02 22 =ρ−∇×∇+ρ−∇+∇ ΨΨ &&&& μφφμλ

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:

( ) 02 2 =ρ−∇+ φφμλ &&

02 =−∇ ΨΨ &&μor

φφ 22∇

ρ

+=&&

ΨΨ 2∇ρ

=μ&&

And these we can easily solve!

What is the solution general solution to the wave equation?

In one dimension:

€

∂2u

∂t2= v2 ∂2u

∂x2

Easy! It’s justu = 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 - (ω/k) t)) = sin(kx - ωt)

and we identify v = ω/k.

The reason we do this is because k and ω have simple and useful physical meanings:

ω= 2f = angular frequencyf = 1/T; T = periodk = 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

€

φ(x,t) = Ae i (k⋅x±ωt )

Where k = (kx, ky, kz) is called the wavenumber vector and the wavespeed is

€

α=ω

| k |=

λ + 2μ

ρ

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:

€

Ψ(x,t) = Be i (k⋅x±ωt )

In this case

€

β =ω

| k |=

μ

ρ

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=∇φ+∇×Ψ

For the scalar part:

€

∇φ =∇Ae i (k⋅x±ωt ) = ikAe i (k⋅x±ωt )

This tells us that the displacement u is parallel to the direction of propagation (defined by k).

For the vector part:

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∇×Ψ =∇×Be i (k⋅x±ωt ) =∂Ψ3

∂x2

−∂Ψ2

∂x3

⎛

⎝ ⎜

⎞

⎠ ⎟ˆ x 1

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+∂Ψ1

∂x3

−∂Ψ3

∂x1

⎛

⎝ ⎜

⎞

⎠ ⎟ˆ x 2

€

+∂Ψ2

∂x1

−∂Ψ1

∂x2

⎛

⎝ ⎜

⎞

⎠ ⎟ˆ x 3

For simplicity, let's presume the wave is propagating in the x1 direction, which means that the terms involving d/dx2 and d/dx3 will be zero (because k·x = kx1). In this case:

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∇×Ψ =∇×Be i (k⋅x±ωt ) = −∂Ψ3

∂x1

⎛

⎝ ⎜

⎞

⎠ ⎟ˆ x 2 +

∂Ψ2

∂x1

⎛

⎝ ⎜

⎞

⎠ ⎟ˆ x 3

or, in other words, all the displacement is in the (x2, x3) plane, perpendicular to x1.

The total wavefield is the sum of the two waves:

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u=∂φ∂x1

+∂Ψ3

∂x2

−∂Ψ2

∂x3

⎛

⎝ ⎜

⎞

⎠ ⎟ˆ x 1

€

+∂φ∂x2

+∂Ψ1

∂x3

−∂Ψ3

∂x1

⎛

⎝ ⎜

⎞

⎠ ⎟ˆ x 2

€

+∂φ∂x3

+∂Ψ2

∂x1

−∂Ψ1

∂x2

⎛

⎝ ⎜

⎞

⎠ ⎟ˆ x 3

Remember that the change in volume (the dilatation ) is equal to the divergence of u; i.e.,

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=∇⋅u=∇⋅∇φ+∇×Ψ( ) =∇2φ

Thus, the wave produced by the Ψ potential does not involve a change in volume, but the one associated with the potential does, and in fact αφ 2=&&

Note that (αβ)2 = (l + 2 = 2 + Because and are always positive, α is always greater than β, so the waves are always faster than the Ψ waves. In fact, for a lot of common rocks, ~ , so (αβ)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 Ψ are rotational or shear or secondary or just S waves.

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.

Simon Denis Poisson

1781-1840

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 :

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k=−P

θ= λ +

2

3μ

2. Young's Modulus (E) = Stress/Strain ratio in uniaxial stress:

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E=11

e11

=μ (3λ + 2μ )

λ + μ

3. Poisson's Ratio () = Strain ratio in uniaxial stress:

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=−e22

e11

=λ

2(λ + μ )

1. Bulk modulus or incompressibility (k) = Ratio of pressure to dilitation: