54 Basic Seismological Theory
2 Although Ψ
Ψ Ψ
Ψ Ψ is often used for the vector potential, we use ϒ
ϒ ϒ
ϒ ϒ (upsilon) to avoid confusion with the SV potential in the text section.
3 This decomposition into scalar and vector potentials, known as Helmholtz decomposition, can be done for any vector field.
This is the equation of motion for an isotropic elastic medium
written entirely in terms of the displacements, with the dependence on position and time explicitly written to remind us that
we seek a solution that varies in this way. Equation 10 can be
rewritten using the vector identity (Eqn A.6.23)
∇
∇ ∇
∇ ∇ 2 u = ∇
∇ ∇
∇ ∇(∇ ∇ ∇
∇ ∇ · u) − ∇
∇ ∇
∇ ∇ × (∇ ∇ ∇
∇ ∇ × u)
(11)
to obtain
(λ + 2µ)∇ ∇ ∇
∇ ∇(∇ ∇ ∇
∇ ∇ · u(x, t)) − µ∇ ∇ ∇
∇ ∇ × (∇ ∇ ∇
∇ ∇ × u(x, t)) =
( , ) .
ρ
∂
∂
2
2
u x t
t
(12)
Rather than solve Eqn 12 directly, we express the displacement field in terms of two other functions, φ and ϒ
ϒ ϒ
ϒ ϒ, which are
known as potentials;
u(x, t) = ∇
∇ ∇
∇ ∇φ(x, t) + ∇
∇ ∇
∇ ∇ × ϒ
ϒ ϒ
ϒ ϒ(x, t).
(13)
In this representation, the displacement is the sum of the gradient of a scalar potential, φ(x, t), and the curl of a vector potential, 2 ϒ
ϒ ϒ
ϒ ϒ(x, t), both of which are functions of space and time.
Although this decomposition appears to introduce complexity,
it actually clarifies the problem, because the vector identities
(Section A.6.4)
∇
∇ ∇
∇ ∇ × (∇ ∇ ∇
∇ ∇φ) = 0 ∇
∇ ∇
∇ ∇ · (∇ ∇ ∇
∇ ∇ × ϒ
ϒ ϒ
ϒ ϒ) = 0
(14)
separate the displacement field into two parts. The part associated with the scalar potential has no curl or rotation and gives
rise to compressional waves. Conversely, the part associated
with the vector potential has zero divergence, causes no volume
change, and corresponds to shear waves. Because taking the
curl discards any part of the vector potential that would give a
nonzero divergence, we require that the vector potential satisfy
∇
∇ ∇
∇ ∇ · ϒ
ϒ ϒ
ϒ ϒ(x, t) = 0. 3
Substituting the potentials into Eqn 12 and rearranging
terms using Eqn 14 yields
(λ + 2µ)∇ ∇ ∇
∇ ∇(∇ 2 φ) − µ∇ ∇ ∇
∇ ∇ × ∇
∇ ∇
∇ ∇ × (∇ ∇ ∇
∇ ∇ × ϒ
ϒ ϒ
ϒ ϒ) = ρ
∂
∂
2
2
t
(∇ ∇ ∇
∇ ∇φ + ∇
∇ ∇
∇ ∇ × ϒ
ϒ ϒ
ϒ ϒ).
(15)
Using Eqn 11, the second term of Eqn 15 simplifies to
∇
∇ ∇
∇ ∇ × ∇
∇ ∇
∇ ∇ × (∇ ∇ ∇
∇ ∇ × ϒ
ϒ ϒ
ϒ ϒ) = −∇ ∇ ∇
∇ ∇ 2 (∇ ∇ ∇
∇ ∇ × ϒ
ϒ ϒ
ϒ ϒ) + ∇
∇ ∇
∇ ∇(∇ ∇ ∇
∇ ∇ · (∇ ∇ ∇
∇ ∇ × ϒ
ϒ ϒ
ϒ ϒ))
= −∇ ∇ ∇
∇ ∇ 2 (∇ ∇ ∇
∇ ∇ × ϒ
ϒ ϒ
ϒ ϒ),
(16)
because the divergence of the curl is zero. After this substitution, the terms in Eqn 15 can be regrouped to give
∇ ∇ (
)
( , )
( , )
λ
µ φ
ρ
φ
+
∇
−
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
2
2 x
x
t
t
t
∂
∂
2
2
= − ×
−
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
( , )
( , ) ,
∇ ∇
∇ ∇ ϒ ϒ
ϒ ϒ
µ
ρ
2 x
x
t
t
t
∂
∂
2
2
(17)
because the elastic constants do not vary with position, and
the order of differentiation has no effect.
One solution of the equation can be found if both terms in
brackets are zero. In this case, we have two wave equations,
one for each potential. The scalar potential satisfies
∇
=
2
2
1
φ
α
φ
( , )
( , ) ,
x
x
t
t
t
∂
∂
2
2
(18)
with the velocity
α = [(λ + 2µ)/ρ] 1/2 .
(19)
As we will see shortly, this solution corresponds to P, or compressional, waves. Similarly, the vector potential satisfies
∇ ∇ ϒ ϒ
ϒ ϒ
2
2
1
( , )
( , ) ,
x
x
t
t
t
= β
∂
∂
2
2
(20)
with velocity
β = (µ/ρ) 1/2 ,
(21)
and corresponds to S, or shear, waves.
Equations 18 and 20 are wave equations that are slightly
different from those that we have previously encountered.
Waves on a string (Section 2.2) satisfied the wave equation
∂
∂
∂
∂
2
2
2
2
u x t
x
v
u x t
t
( , )
( , ) ,
=
1
2
(22)
describing the propagation of a scalar quantity in one space
dimension. The scalar potential satisfies a similar scalar wave
equation, with the difference that the space variable x is in
three dimensions. The vector potential, a vector quantity, satisfies the analogous vector wave equation in three dimensions.
The wave equations in Eqns 18 and 20 are strictly valid only
for a homogeneous medium because they were derived assuming that all derivatives of the elastic constants were zero. Although these equations were derived in Cartesian coordinates,
they are valid in any coordinate system. We next discuss solutions of the wave equation, and then return to these two types
of waves.
2.4.2 Plane waves
The scalar wave equation in three dimensions,
∇
=
2
2
1
φ
φ
( , )
( , ) ,
x
x
t
v
t
t
∂
∂
2
2
(23)
2 Although Ψ
Ψ Ψ
Ψ Ψ is often used for the vector potential, we use ϒ
ϒ ϒ
ϒ ϒ (upsilon) to avoid confusion with the SV potential in the text section.
3 This decomposition into scalar and vector potentials, known as Helmholtz decomposition, can be done for any vector field.
This is the equation of motion for an isotropic elastic medium
written entirely in terms of the displacements, with the dependence on position and time explicitly written to remind us that
we seek a solution that varies in this way. Equation 10 can be
rewritten using the vector identity (Eqn A.6.23)
∇
∇ ∇
∇ ∇ 2 u = ∇
∇ ∇
∇ ∇(∇ ∇ ∇
∇ ∇ · u) − ∇
∇ ∇
∇ ∇ × (∇ ∇ ∇
∇ ∇ × u)
(11)
to obtain
(λ + 2µ)∇ ∇ ∇
∇ ∇(∇ ∇ ∇
∇ ∇ · u(x, t)) − µ∇ ∇ ∇
∇ ∇ × (∇ ∇ ∇
∇ ∇ × u(x, t)) =
( , ) .
ρ
∂
∂
2
2
u x t
t
(12)
Rather than solve Eqn 12 directly, we express the displacement field in terms of two other functions, φ and ϒ
ϒ ϒ
ϒ ϒ, which are
known as potentials;
u(x, t) = ∇
∇ ∇
∇ ∇φ(x, t) + ∇
∇ ∇
∇ ∇ × ϒ
ϒ ϒ
ϒ ϒ(x, t).
(13)
In this representation, the displacement is the sum of the gradient of a scalar potential, φ(x, t), and the curl of a vector potential, 2 ϒ
ϒ ϒ
ϒ ϒ(x, t), both of which are functions of space and time.
Although this decomposition appears to introduce complexity,
it actually clarifies the problem, because the vector identities
(Section A.6.4)
∇
∇ ∇
∇ ∇ × (∇ ∇ ∇
∇ ∇φ) = 0 ∇
∇ ∇
∇ ∇ · (∇ ∇ ∇
∇ ∇ × ϒ
ϒ ϒ
ϒ ϒ) = 0
(14)
separate the displacement field into two parts. The part associated with the scalar potential has no curl or rotation and gives
rise to compressional waves. Conversely, the part associated
with the vector potential has zero divergence, causes no volume
change, and corresponds to shear waves. Because taking the
curl discards any part of the vector potential that would give a
nonzero divergence, we require that the vector potential satisfy
∇
∇ ∇
∇ ∇ · ϒ
ϒ ϒ
ϒ ϒ(x, t) = 0. 3
Substituting the potentials into Eqn 12 and rearranging
terms using Eqn 14 yields
(λ + 2µ)∇ ∇ ∇
∇ ∇(∇ 2 φ) − µ∇ ∇ ∇
∇ ∇ × ∇
∇ ∇
∇ ∇ × (∇ ∇ ∇
∇ ∇ × ϒ
ϒ ϒ
ϒ ϒ) = ρ
∂
∂
2
2
t
(∇ ∇ ∇
∇ ∇φ + ∇
∇ ∇
∇ ∇ × ϒ
ϒ ϒ
ϒ ϒ).
(15)
Using Eqn 11, the second term of Eqn 15 simplifies to
∇
∇ ∇
∇ ∇ × ∇
∇ ∇
∇ ∇ × (∇ ∇ ∇
∇ ∇ × ϒ
ϒ ϒ
ϒ ϒ) = −∇ ∇ ∇
∇ ∇ 2 (∇ ∇ ∇
∇ ∇ × ϒ
ϒ ϒ
ϒ ϒ) + ∇
∇ ∇
∇ ∇(∇ ∇ ∇
∇ ∇ · (∇ ∇ ∇
∇ ∇ × ϒ
ϒ ϒ
ϒ ϒ))
= −∇ ∇ ∇
∇ ∇ 2 (∇ ∇ ∇
∇ ∇ × ϒ
ϒ ϒ
ϒ ϒ),
(16)
because the divergence of the curl is zero. After this substitution, the terms in Eqn 15 can be regrouped to give
∇ ∇ (
)
( , )
( , )
λ
µ φ
ρ
φ
+
∇
−
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
2
2 x
x
t
t
t
∂
∂
2
2
= − ×
−
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
( , )
( , ) ,
∇ ∇
∇ ∇ ϒ ϒ
ϒ ϒ
µ
ρ
2 x
x
t
t
t
∂
∂
2
2
(17)
because the elastic constants do not vary with position, and
the order of differentiation has no effect.
One solution of the equation can be found if both terms in
brackets are zero. In this case, we have two wave equations,
one for each potential. The scalar potential satisfies
∇
=
2
2
1
φ
α
φ
( , )
( , ) ,
x
x
t
t
t
∂
∂
2
2
(18)
with the velocity
α = [(λ + 2µ)/ρ] 1/2 .
(19)
As we will see shortly, this solution corresponds to P, or compressional, waves. Similarly, the vector potential satisfies
∇ ∇ ϒ ϒ
ϒ ϒ
2
2
1
( , )
( , ) ,
x
x
t
t
t
= β
∂
∂
2
2
(20)
with velocity
β = (µ/ρ) 1/2 ,
(21)
and corresponds to S, or shear, waves.
Equations 18 and 20 are wave equations that are slightly
different from those that we have previously encountered.
Waves on a string (Section 2.2) satisfied the wave equation
∂
∂
∂
∂
2
2
2
2
u x t
x
v
u x t
t
( , )
( , ) ,
=
1
2
(22)
describing the propagation of a scalar quantity in one space
dimension. The scalar potential satisfies a similar scalar wave
equation, with the difference that the space variable x is in
three dimensions. The vector potential, a vector quantity, satisfies the analogous vector wave equation in three dimensions.
The wave equations in Eqns 18 and 20 are strictly valid only
for a homogeneous medium because they were derived assuming that all derivatives of the elastic constants were zero. Although these equations were derived in Cartesian coordinates,
they are valid in any coordinate system. We next discuss solutions of the wave equation, and then return to these two types
of waves.
2.4.2 Plane waves
The scalar wave equation in three dimensions,
∇
=
2
2
1
φ
φ
( , )
( , ) ,
x
x
t
v
t
t
∂
∂
2
2
(23)