Sismik Prospeksiyon ( ingilizce )
Introduction to Seismology  11
12.510 Introduction to Seismology
Surface Waves (Ground Roll)
April 2, 2008
Today we will look at the interaction of an acoustic wave (ground roll)
with a layer over half space. We will consider travel time curves for acoustic
waves and describe them using higher modes of surface wave propagation.
We will also use a propagation matrix as a re?ec tivity method to calculate
synthetic seismograms.
Case 1: Layer over Source in a Half Space
Lets look at what happens when we have 2 interfaces. We are working with
acoustic waves, so we will take P = pressure ?eld. Note that the methods
used here can be used for SH waves as well. The same principles can be
applied to PSV waves, but the algebra becomes more complicated.
Figure 1: Diagram of a source in a half space under a layer
1
The wave equation for the acoustic case is
P
¨ = k? · (
? 1
? P ) (1)
with the displacement given by
1
u =
?? 2
? P (2)
This case becomes a bit more complicated than the simple re? ection
previously discussed. The analytical methods used to decribe the simple
re?ection begin to break down when multiple layers are introduced.
´ `
We will renormalize the incoming wave to P
2
= 1 and de?ne P
2
= R.
Using the plane wave description,
´ `
P = Pe
i(kxx kz z ? t)
+ Pe
i(kxx kz z ? t)
Notice the positive and negative k
z
z terms to describe the vertical slow
´ `
ness, as well as the di?eren t amplitudes P and P . We want to know what
is happening with respect to crossing the interface, so we will ignore the
xdirection, leaving the e
ikz z
terms and giving
´ `
P = Pe
 i? z
+ Pe
 i? z
where the vertical wave number is given by
? z
= k
z
=
?
=
? cos i
= ? ?
cz c
´ `
P = Pe
 i? z
+ Pe
 i? z
can be set up for each layer. We can solve for the
pressure by taking the gradient in the zdirection:
? P 1
Pe
i? z i?
Pe
i? z
Pe
i? z
) Pe
i? z
) = (
` ´
u
z
=
? z
=
?? 2
(i?
`
 i?
´
?? 2

To solve this system we will follow the steps:
1. Look at Potentials (Pressure Field)
2. Kinematic and Dynamic Boundary Conditions
3. Zoeppritz equations
2 4. Solve for R and T
At z=0, the welded interface, the stress is continuous so
´ ` ´ `
P
2
+ P
2
= 1 + R = P
1
+ P
1
The displacement at z = 0 is given by
i?
u(0) =
?? 2
(P
`
1
 P
´
1
)
which implies that at the free surface:
P
`
2
 P
´
2
=
? 2
? 1
(P
`
1
 P
´
1
) =
? 2
(R  1)
? 1
? 2
? 2
At the free surface z =  H
P
´
1
e
 i?
1
H
+ P
`
1
e
i?
1
H
= 0 where ? 1
is the vertical wave number k
z
=
c
?
z
.
? ? The three equations above give the Zoeppritz matrix and we solve for R
1 1  1
?
 ? 2
?
?
?
? ?
´
P
1
1
? ?
? ?
? ? ? ?
? ?
`
 ? 1
? 1
? 1
 1
0
P
1
=
? 1
? 2
e
 i?
1
H
e
i?
1
H
0 R
(1 ? )+(1+? )e
2i?
1
H
R =
(1+? )+(1 ? )e
2i?
1
H
where ? =
? 1
? 2
? 2
? 1
Some remarks on the re?ection at the free surface:
 The re?ection coe?cien t is a complex number. In the cases we studied in
previous lectures, R became complex when i = i
c
.
R = 1, which means that the energy is not stored in the upper layer
but is eventually all — re?ected back. E.g., if the input is a single spike,
the output is a series of reverberations (see homework 2, problem 3). This
occurs because of conservation of energy; if there is no source in the upper
layer, there can be no residual trapped energy in the upper layer. The en
ergy is re?ected from z =?H and both re?ected and partially transmitted
from z = 0 until it fully dissipates back into the half space.
 There is a frequency?dependent phase shift
? ?
in the waveform, ie. R =
?
e
2i?
e
i(kx ? t+
?
)
.
Case 2: Source Within a Layer Over a Half Space
The source within the layer is analagous to a source in a weathered layer,
e.g. an induced source in exploration seismology or surface waves in earth
quake seismology. This scenario is similar to the Love waves case discussed
in previous lectures. In the case of angles such that i¿ic (evanescent waves),
3
no energy will be transmitted into the half space.
Figure 2: Diagram of a source in a layer over a half space
To solve this system we will follow the steps as before:
1. Look at Potentials (Pressure Field)
2. Kinematic and Dynamic Boundary Conditions
3. Zoeppritz equations
4. Solve for R and T
Because stress must be continuous,
´ ` ´ ` ´ `
P = P
1
+ P
1
= P
2
+ P
2
= 0 + T ? P
1
+ P
1
 T = 0
The total displacement must be continuous at the interface, so
?? 2
? P
??
(Pe
`
i? z
+
´
u =
1
=
i?
Pe
 i? z
)
? 1
?
? 1
(P
`
1
 P
´
1
) =
? ? 2
2
T
? (P
`
1
 P
´
1
) =
? ? 2
1
? ? 1
2
T = ? T
? P
`
1
 P
´
1
 ? T = 0 where ? =
? ? 2
1
? ? 1
2
=
? ? 2
1
c
c
2
1
=
z
z
2
1
and ? 1
c
1
=acoustic
impedance.
At z =  H,
P
`
1
+ e
2i?
1
H
P
´
1
= 0
Notice that the T in these equations is not the conventional T as before
because it is evanescent. In contrast to previous cases, we can now form a
homogeneous set of Zoeppritz equations:
? ? ? ? ? ? `
1 1  1
P
1
0
? ?
? ?
? ?
? ? ? ?
? ?
´
1  1
2i?
1
?
0
 ?
0 T 0
P
1
=
1
e
4
? ?
?
The solution of this system depends on the layer thickness (H), the ver
tical wave number (? ), the direction of the rays (
? cos i
), and the impedence
c
contrast (? =
z
z
2
1
). To obtain the nontrivial solution to this system, we must
set the determinant of the 3x3 term to zero, leading to
? ?
1 1  1
det ? ?
1  1  ? ? ?
= 0
1 e
2i?
1
H
0
? ? e
2i?
1
H
 e
2i?
1
H
 ?  1 = 0
? (?  1)e
2i?
1
H
= ? + 1
2i?
1
H ? +1
e = ?
?  1
We know that in the complex plane, generally, e
 2i?
=
a b
, giving
a+b
tan ? =
 a
ib
. Using this, we can rewrite our nontrivial solution as tan (? 1
H) =
1
=
? 2
? 1
i? i?
1
? 2
tan(? 1
H) =
1
=
? 2
? 1
?
i? i?
1
? 2
where ? 1
= ? (
c
1
1
)
2
 p
2
and ? 2
= ? (
c
1
2
)
2
 p
2
.
The dispersion relationship for ground roll (which is the same as
the dispersion relationship for Love waves) is thus given by
tan
?
? H
?
(
c
1
1
)
2
 p
2
?
=
? 2
?
(
c
1
1
)
2
 p
2
? 1
(
1
)
2
 p
2
c
2
For a given frequency and layer thickness H, for certain directions given
by p =
sin
c
1
i
1
=
sin
c
2
i
2
and known c
1
,c
2
, we can solve this system.
Given ?xed ? , ie. i
1
= 90
o
solves the dispersion equation, ie. the direct
wave is a solution to the dispersion equation. See Figure 3.
5
Figure 3: Diagram of (1) a direct wave and (2) a critical wave
Figure 4: Graphical representation of the dispersion relationship for ground
roll.
6
For all postcritical angles i, or
c
1
2