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ANISOTROPY
So far we have assumed isotropy i.e. wavespeeds do not depend on the direction of wave
propagation. We solved the wave equation assuming plane waves:
Figure 1
P wave
S wave
x
sin i sin j
α 〉 β→ i 〉 j
z V p =α
VSH = VSV =β
Anisotropy however, cannot be ignored as it is the focus of increasing research in
seismology.
Figure 2a
In an isotropic medium wavefronts are raypath concentric circles with radius dependent on the velocity (Vp or Vs) as shown in Fig. 2a. The raypaths are perp. to wavefronts and the slowness vector is perp. to the wavefront. The energy goes with group velocity and the group velocity is perp. to wavefronts.
In anisotropic media the (^) Figure 2b
wavefronts are distorted (Fig. 2b). s ~ phase The raypaths are not perp. to the
wavefronts, therefore the direction
velocity
of the group velocity is not the
same as the direction of the phase g^ group
velocity. velocity
In a homogeneous media there are
still three solutions but now they
are called quasi P (q P), quasi SH
(q SH) and quasi Sv (q Sv).
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Now we look at some basic theory to get insights into the problem and concepts. A full
treatment is beyond the scope of this class.
We have used the Generalized Hooke’s Law:
τ ij = cijkl ε kl
τ ij =τ ji → cijkl = c jikl
ε kl =ε lk → cijkl = c jilk
And from thermodynamics
cijkl = cklij
These relationships reduce the number of independent elements from 81 to 21 elements.
In the isotropic case:
cijkl = λδ ij δ kl + μ( δ ik δ jl + δ il δ jk )
Where only the two independent parameters are the Lamé constants, λ and μ.
c 1111 = c 2222 = c 3333 =λ + 2 μ
c 1122 = c 1133 = c 2233 =λ
c 1212 = c 1313 = c 2323 =μ
Else (^) = 0
An anisotropic medium is a more complex system and uses symmetries such as:
Orthorhombic- (e.g. olivine) with 9 elements
Hexagonal- with 5 elements
Cubic- (e.g. MgO) with 3 elements
Anisotropy is not due to individual minerals but the whole medium or lattice. Lattice
preferred orientation (LPO) is the deformation of olivine by plate motion.
Hexagonal symmetry is very useful for seismology where there is rotational symmetry
around a symmetry axis (not necessarily vertical but often is).
z (^) For example a layered system with a vertical Figure 3
x,y
symmetry axis which is the most useful for Earth (Fig. 3). In the transverse direction (i.e. direction perpendicular to the symmetry axis) you have isotropy. This is also known as Vertical Transverse Isotropy (VTI).
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The exploration industry convention is similar but used Thomsen parameters ε,γ,δ
Looking at wave propagation:
τ ij = cijkl ε kl
kl =^
(∂ u + ∂ u )
k l l k
ij =^ cijkl ∂^ luk
The Equation of Motion for a homogeneous medium:
ρ u &&^ i = (∇. σ ) i = ∂ j σ ij = ∂( c ijkl ∂ l u k )^ = cijkl ∂ j ∂ luk
Where i is the only free index.
The displacement: −. u = ge
i ω ( t sx )
Where:
g =polarization vector (in the direction of particle motion)
s =slowness vector or 1/phase velocity (direction is
k ˆ^ ) c
x =position vector
s is perpendicular to the wavefront since s. x is unchanged for dx perpendicular to s.
If we substitute the displacement into the equation of motion:
ρ gi = g ck ijkls j sl or ( c ijkl s j sl − ρδ ik ) g k = 0
Where i is the only free parameter.
Now, we introduce the density- normalized elastic tensor:
cijkl ijkl
And re-write the Equation of Motion as:
2
(Γ ijkl s ˆ^ js ˆ l − c δ ik ) g k = 0
The Christoffel Matrix:
M (^) ik = Γ ijkl s ˆ (^) js ˆ l
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The Christoffel Equation:
( M ik −
2
c δ ik ) g k = 0
s 1 Where c =phase velocity and s ˆ^ = , s = s c
The Christoffel Equation is simply an eigenvalue equation. For each direction of s there
are three solutions that correspond to the three eigenvalues, c 1 ,c 2 ,c (^) 3, and corresponding
eigenvectors, g =(g 1 ,g 2 ,g 3 ), of the Christoffel matrix.
Now we will look at 2 specific cases.
CASE 1
A plane wave in a VTI medium where propagation is in the direction of the symmetry
axis. The x 3 direction is the axis of symmetry and propagation direction.
The displacement is:
sin ω
x 3 ⎞ ⎟ ⎠
ui = Ai − t c
Christoffel Matrix:
Christoffel Equation:
L 0 0
0 L 0
0 0 C
M (^) ik =
⎠ ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢
L 2 ⎤
− c 0 0
⎡ g 1 ⎤ L −
2 0 c 0 g 2 = 0
C g^3 0 0 − c^2
The three solutions are:
1 =^ ,^2 =^ , 3 =
c c
L
c
L
c
g 1 = 1 , 0 , 0 )
g 2 = )
g 3 =
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The wavespeeds in all three directions are different therefore t here is shear-wave
splitting (quasi P,SH, and Sv) where three polarization directions will propagate at three
different speeds.
A
quasi P α '= ( α =
λ +
ρ
2 μ , c 1111 =λ + 2 μ = C (^) 11 A ) ρ
The quasi P wave is the fastest wave, polarized parallel to the direction of propagation
(x 1 ).
First quasi S β '= ρ
N
( C 66 = c 1212 = N = μ β=
μ
ρ
If N>L the first quasi S wave is associated with the x 2 direction.
L λ Second quasi S β ' '= ( C (^) 44 = c 2323 =λ = L c 3 = ρ
ρ
The second quasi S is the slowest wave and polarized parallel to the x 3 direction.
