seismic sources. solution for the displacement potential properties of the disp potential 1.an...
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SEISMIC SOURCES
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SEISMIC SOURCES
2
22 2
14 , (8.1)
:P wave displacement potential
F: The effective force function applied at the elastic radius.
r eF t rt
(8.3) ,
/11,,
2
rtF
r
rtF
rr
trtru
(8.2) ,
/,
r
rtFtr
Solution for the displacement potential
Properties of the disp potential1.An outgoing D’Alembert solution in the spherical coordinate
.2.Functional shape is the same as the force-time history.
Displacement field
Near-field term Far-field term
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(8.5) , lim0
VV
fF
8.3 Elastostatics
• Purpose – to determine the static displacement u in an isotropic, infinite, homogeneous elastic medium due to a point source at point O.
O F
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V dVr
rr
(8.6) .1
0 0
8.3 Elastostatics
(8.7) ,1
4
1 2
rr
Three-dimensional delta function (in spherical polar coordinate):
With the use of Gauss’ theorem:
14
1
theorem)(Gauss' 1
4
1
1
4
1
1
4
1
2
2
dVrdSr
dSr
dVr
dVr
n
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8.3.1 Static displacement field due to a single force
(8.8) 0.-2 uuF
• Elastic equation without time-dependent displacement term (utt=0)
• The point force is balanced by the stresses/strain
(8.9) , 4 4
4 2
rrF
rFrF
aa
aafF
• Consider a point force of magnitude F at the origin
,
(2.51) :identityvector 2 uuu
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(8.10) .2
4 4 42
uu
aaaF
rr
Fr
8.3.1 Static displacement field due to a single force
The equation of equilibrium becomes:
Note that the representations of force and displacement field are similar!
(8.11)
0
0
where
2
2
SSS
PPP
SP
AAA
AAA
AAu
The solution form (Helmholtz’s theorem):
,
(2.51) :identityvector 2 AAA
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8.3.1 Static displacement field due to a single force
(8.13) . 4
4
2 22
r
F
r
FSP
aA
aA
(8.12) ,0 4
2 4
2
2
S
P
r
F
r
F
Aa
Aa
Which can be satisfied by having:
fields) "orthogonal" lly twomathmatica are and (
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(8.14) . 4
A 24
A 22
r
F
r
FSP
If we put Ap=Apa and As=Asa, we obtain two Poisson’s equations:
8.3.1 Static displacement field due to a single force
(8.15) . 8
A
28
A
Fr
Fr
S
P
The solutions (of potentials) are
rr /2 Since 2
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2 2
In indical notaion:
ˆ ˆ
1ˆ8 2
1 1ˆ ˆ ˆ8 8
P S
ji i P s i
i Pi j
s i s s iji ii j
u A j A j
rA j
x x
rA j A j A j r
x x
u A A
8.3.1 Static displacement field due to a single force
Plugging in the potentials Ap and As and expressing the vector operations with indicial notation, and yield the ith component of displacemen
t for a unit force (F=1) in the jth direction, uij:
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2
22
, ,
1 1 1
8 2 8 8
1
8 2
1or (8.16)
8
ji ij
i j i j
iji j
ji ij kk ij
r ru r
x x x x
rr
x x
u r r
(8.17) .2
Where
8.3.1 Static displacement field due to a single force
Somigliana tensor
Symmetry: uij= uj
i
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(8.18)
.12
8
8
12
8
8
8
12
8
3
233
33323
2
3
222
23313
1
3212
13
211
1
r
x
rr
Fu
r
xxFu
r
x
rr
Fu
r
xxFu
r
xxFu
r
x
rr
Fu
8.3.1 Static displacement field due to a single force
Thanks to the symmetry, there are only 6 independent permutations:
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8.3.1 Static displacement field due to a single force
Find the displacements given in polar coordinates due to a point force applied in the x1 direction (Figure 8.12), using the Jacobian coordinate transformation matrix (8.19)
jie
rjrir
kjie
rkrjrir
kjie
rkji
r
krjrir
r
ˆ cosˆ sin ˆ
sin ,ˆ cossinˆ sinsin
ˆ sinˆ sincosˆ coscos ˆ
,ˆ sinˆ sincosˆ coscos
ˆ cosˆ sinsinˆ cossin ˆ
1 ,ˆ cosˆ sinsinˆ cossin
ˆ cosˆ sinsinˆ cossin
rr
rr
rr
r
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(8.19) .
0cossin
sinsincoscoscos
cossinsincossin
13
12
11
u
u
u
u
u
ur
(8.20) .cos2
1 4
sin cos
sin 4
cos sin
13
11
13
11
r
Fuuu
r
Fuuur
(typo in textbook!)
In the x1x3 plane, φ=0:
8.3.1 Static displacement field due to a single force
The (x1x3 plane) static radial deformation and shear deformation due to a single force (Fig 8.12)
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8.3.2 Static displacement field due to a force couple
Figure 8.13 A force couple acting at O’ parallel to the x1x2 plane.
(8.21) .
,,:,,,,:,,
222
2
1
3213221
211
3213221
211
dOdu
xxxduxxxdu
i
ii
The displacement at P is the sum of the displacements from two individual forces
),,( 321
),,( 3221
21 dF
),,( 3221
21 dF
O’
P(x1,x2,x3)
x1
x2
x3
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(8.23) .222
2
1
dOdx
ui
(8.22) therefore
. that see we
233
222
211
2
k
ji
k
ji
ii
x
uu
x
rr
xxxr
k
ji
k
ji
k
ji
k
ji
x
u
x
r
r
u
r
r
uu
The displacement due to the force couple is given by:
8.3.2 Static displacement field due to a force couple
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Let dξ2→0 and F →∞ so that Fdξ2→ M, a finite moment.
Let
• Force couple acting at the origin (o’=(0,0,0))
• Offset in the x2 direction. Moment = M
With (8.18) and (8.23) the static displacement fiel
d ui
8.3.2 Static displacement field due to a force couple
2211 2 2 1 21
1 1,2 21 3 3 53
1 2 1 22 1 3
2 2
1 3 1 33 1 3
22 13
88
1 1 8
(8.18)
8
x x x xMxFu u du
r r rr r r
x xF ru ur x r r r xx xF
u ur
2 223
2
1 1 1 23 3 5
2 2
( 0)
3
x xr
r x r
x x x xr
x r r r x r
x1
x2x3
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(8.24) .38
38
32
8
5321
3
5
221
31
2
52
21
32
32
1
r
xxxMu
r
xx
r
xMu
r
xx
r
x
r
xMu
In the same way, we may derive u2 and u3 for the single couple
8.3.2 Static displacement field due to a force couple
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8.3.2 Static displacement field due to a force couple
(8.25) .38
32
8
38
5321
3
51
22
31
31
2
52
21
32
1
r
xxxMu
r
xx
r
x
r
xMu
r
xx
r
xMu
x1
x2
x3Similarly, for single couple oriented along the x2 axis with offset arm along the x1 direction
In generaljkii uu ,
j : direction of force
k : direction of offset arm
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8.3.3 Static displacement field due to a double couple
(8.26) 21,
12, iii uuu
x1
x2x3
A double couple in the x1x2 plane
Principles of superposition:
The displacement is the sum of the
displacements from two individual couples
(+/- of M)
(8.27) .34
68
311
4
3
8
22
8
311
4
3
8
22
8
3321
25321
3
2
221
251
22
31
31
2
2
212
25
212
32
32
1
r
xxx
r
M
r
xxxMu
r
x
r
x
r
M
r
xx
r
xM
r
xMu
r
x
r
x
r
M
r
xx
r
xM
r
xMu
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8.3.3 Static displacement field due to a double couple
(8.28) .2cossin1 4
2sin2sin22
1
4
2sinsin2
1 4
2
2
22
r
Mu
r
Mu
r
Mur
Convert to spherical polar coordinates with (8.19)
1 sin 22
1 cos2 . (8.29)
ru
u
On the x1x2 plane, θ=π/2, uθ=0, and
Static deformations decay rapidly.
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8.3.3 Static displacement field due to a double couple
(8.29) .2cos1 2sin2
1
uur
On the x1x2 plane, θ=π/2, uθ=0, and
uur
Figure 8.14 Azimuthal pattern of ur and uθ
on the x1x2 plane
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The displacement field due to a shear dislocation can be given by the displacement field due to a distribution of equivalent double couples that are placed in a medium without any dislocation.
Since static deformations decay rapidly with distance from the source, ground deformations are usually near the fault a point-source approximation is never valid finite fault numerically discretized distribution of double couples.
8.3.3 Static displacement field due to a double couple
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Going beyond the simple faulting model in geodetic modeling
Incorporating viscoelastic effects of the deeper crust.
Adding layering and elastic parameter heterogeneity in the Earth model.
Variable slip function or changing fault mechanism.
Curved fault plane
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8.4 Elastodynamics
2 . (2. 2 5 ) f u uu
Elastodynamic equations:
2 -4
. (8.30)4 4
F t F tr
F tr r
af r a
a a
Consider a time-dependent body force:
Following the same basic procedure as used in elastostatic problem.
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8.4 Elastodynamics
2
2
2
2
2
2
24 (8.32
.
)
4
P
S
P
S
F t
rF t
t
tr
A a
A a
A
A
0
where (8.31)0.
PP S
S
t
Au A A
A
Again, we seek a solution of the form:
Compare it to (8.14)
2
2
A 4 2
(8.14)
A . 4
P
S
F
r
F
r
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8.4 Elastodynamics
22
2 2
22
2 2
1
4 2
2 where velocity
1
4
where velocity . (8.33)
PP
SS
F t AA
r t
P
F t AA
r t
S
Putting Ap=Apa. As=Asa, we obtain twos scalar equations:
22
2
22
2
24
.4
PP
SS
F t
r tF t
r t
AA a
AA a
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8.4 Elastodynamics: The solution to an inhomogeneous wave equation
(8.34) .,,,,,,1
,,, 3213212
2
23212 txxxgtxxx
tctxxx
8.35 ., tttg rxx
The solution is: (Buy it, for now)
(8.36) .
/
4
1,
r
crtt
x
An inhomogeneous wave equation:
Where g is a “point” source both in space and time:
(Box 2.5)
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8.4 Elastodynamics
(8.37) ,11
01
22
c
rtg
rc
rtf
rc
(8.38) ,
/
4
1,
12
2
22
ξ-x
ξ-xr
ξ-x
ctt
ttc
(8.39) .
/
4
1
12
2
22
ξ-x
ξ-x
ξ-x
ctf
tftc
Standard type of D’Alembert-type solution:
For a point force at x=(ξ1, ξ2,ξ3) applied at t=τ.
For a time-dependent point force f(t) applied at x=(ξ1, ξ2,ξ3)
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8.4 Elastodynamics
(8.40) .
/,
4
1,
,1
2
2
22
dVct
t
ttc
V
ξ-x
ξ-xξΦx
xΦ
If the source is extended through a volume V, as well as in time
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8.4 Elastodynamics
/1
4 4 2
(8.41)
/1.
4 4
P
V
S
V
F tA dV
r
F tA dV
r
x - ξ
x - ξ
x - ξ
x - ξ
22
2 2
22
2 2
1
4 2 (8.33)
1
4
PP
SS
F t AA
r t
F t AA
r t
The solutions to (8.33)
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So far, so good ?
Sorry, it’s getting messy …
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How to deal with
this integration ?
2
/1
( . ) 4
8 414P
V
F tA dV
r
x - ξ
x - ξ
8.4 Elastodynamics
Integrating over V via the system of concentric spherical shells …
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8.4 Elastodynamics
2
/1
( . ) 4
8 414P
V
F tA dV
r
x - ξ
x - ξ
If is the radius of a typical shell S
, the shell thickness is , and r d dV d dS
x - ξ
2 0S
1 2
1
011S
Let the body force applied in the direction at the origin:
1 1
(4 )
1 1ˆ
(4 )
x
P
P
P
S
F tA dS dτ
r
F tA x dS dτ
r
Au A
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1 2 0 01 1S
1 03 2
1 1 1 1ˆ
(4 ) 4
Similarly,
1 1 1ˆ 0, ,
4
r
P
r
s
F tA x dS dτ F t d
r x r
A x F t dx r x r
1x-
2 2
1
Evaluation of the surface integral (AR box 4.3)
1( , ) ( )
( , ) 0 for /
1( , ) 4 for /
h dS
h
hx x
ξ
x ξξ
x x
x x
8.4 Elastodynamics
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8.4 Elastodynamics (8.43) )( .AA iSPiiu
rtF
x
rrtF
r
xdtF
x
rtF
x
r
x
r
rdtF
rxx
dtFxx
r
rdtF
rxx
dtFrxx
A
i
r
i
r
i
i
r
i
r
i
r
i
r
iPi
20
120
1
2
01
201
2
01
1
)(
(8.44.p) 4
11
4
1
4
11
4
1
1
4
1 )(
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8.4 Elastodynamics
2
/
/1
21
121
1 1,
4
1
4
1 . (8.44)
4
r
i ri
i
ii
u t F t dx x r
r r rF t
r x x
r r rF t
r x x
x
Similar procedure for ( )
1S
r r
A
Typo in textbook!
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8.4 Elastodynamics
2
/
/
2
3
3
2
1, 3
4
, (8.45)
Where is direction cosine;
1 13
1 1
1
4
1
1
4
r
i i j ij r
ii i
i
i j iji
i j ijj
j
i
r
rF
u t F t d
x r
r x
x x r r
rF t tr r
x
Stokes solution (for point force in the j direction, located at the origin):
• Near-field term Far-field term
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8.4 Elastodynamics
(8.46) 1
4
12
r
tFr
u jiPi
Properties of the far-field P-wave
1. It attenuates as 1/r
2. Arrival time=r/αwith velocity α
3. waveform is proportional to the applied force at the retarded time.
4. The displacement is parallel to the direction from the source (up×γ=0) (longitudinal wave)
5. |up| is proportional to γj
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Properties of the far-field P-wave
(8.46) 1
4
12
r
tFr
u jiPi
4. The displacement is parallel to the direction from the source (up×γ=0) (longitudinal wave)
5. |up| is proportional to γj
)0,sin,(cos
)0,sin,(cos
)0sin(cos i.e.,
circle 1ron point any for cosinedirection the
plane, x- xOn the 21
θθ
θθu
θ,θ,
j
jjjiPi
i
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8.4 Elastodynamics
Properties of the far-field S-wave
1. It attenuates as 1/r
2. Arrival time=r/β with velocityβ
3. waveform is proportional to the applied force at the retarded time.
4. The displacement is perpendicular to the direction from the source (us .γ=0) (transverse wave)
(8.47) 1
4
12
r
tFr
u jiijSi
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The displacement field for single couples and double couples can be obtained by differentiating the single-force results w.r.t appropriate coordinates. (The same as we did for the static fields.)
Only far-field displacements are discussed from now on.
8.4 Elastodynamics
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8.4 Elastodynamics – The displacement field due to a single couple
(8.48) .
/
4
1
/
4
1
2
21212222
r
rthu
r
rthu FF
iFii
Fi
(8.49) .
//
4
1
2
2112
c 22
r
rth
r
rthu FF
iii
FF2
r rrr 2
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x1
x2
11
22
2 2 21 2 3
x
rx
r
r x x x
p
In the x1-x2 plane
Direction cosines for F
x1
x2
2
2
11
2
22
2
22 22 1 2 2 3
F
F
x
r
x
r
r x x x x
p
In the x1-x2 plane
Direction cosines for F2
2x
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2
2
11
2
22
2
22 22 1 2 2 3
F
F
x
r
x
r
r x x x x
Direction cosines for F2
x1
x2
p
In the x1-x2 plane
2x
2r1r
2 2 2
1 1 12
2
2
22
2
2
1 12
1 1
2 22
2
22
(for far field)
cos
cos( )
(8.50)
ii
F F F
F
F
F
F
Fi i i
r
x
r
x x x
r r r
or
x x x
r r
x
r
r r r
r
?
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2c 1 222
2
//. (8.52)
4i i i i
h t rh t r xu
r r r
For r rrr 2
2
0 0 0
/ / / . (8.53)
/ /
/ / / (8.54)
h t r h t r r
hh t h t t t t
t
h t r r
h t r r th r
• Temporal differentiation of the source time history
8.4 Elastodynamics – The displacement field due to a single couple
Expand this term around (t-r2/α) in Taylor series.
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8.4 Elastodynamics – The displacement field due to a single couple
2 / / / +.. (8 4)/
.5h t r
h t r h t r rt
• Temporal differentiation of the source time history
c 1 222
12
22
/ 1/ /
4
/ /
4
/
/
/
/
i i i i
i i
j
h t r xu h t r r
r r r
h t r h t r r
r r
h t rx r
r
h t r
t
h t r
r
h t r
rr
2 2c 12 2
. (8.55)4
/ ji i
r x r r ru h t h t
r
r
r
h t
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(8.55) .
/
4 2
22
21c
r
thrr
thr
x
r
rthru j
ii
Near-field terms (Decays as 1/r2)
8.4 Elastodynamics – The displacement field due to a single couple
Far-field displacement (Decays as 1/r)
(8.56) .
/
4 31c
r
rthru ii
The far-field displacement is sensitive to particle velocity at the source rather than to particle displacements. (rev: Eq 8.3)
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(8.58) .4
(8.57) r) (replace
(8.56) ./
4
23
21c
2222
31c
rth
r
xu
xxx
rr
r
rthru
ii
ii
2
2
20
20
lim
lim . (8.59)
xh
xh
r rM t x h t
r rt tM hx
Introduce the moment representation: M
8.4 Elastodynamics – The displacement field due to a single couple
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(8.60) .
/
4 321c
r
rtMu ii
8.4 Elastodynamics – The displacement field due to a single couple/double coupleSolution for the single couple is
x1
Δx2
x2
x1
Δx1
x2
Symmetry in force direction and offset direction
Solution to a double couple
Dc 1 23
/2
4i
i
M t ru
r
x1
x2x3
A double couple in the x1x2 plane
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Elastodynamic Green function
ij
ijjijii
GF
rtF
r
rtF
rtu
*
1
4
11
4
1 ,
8.45)(equation solution Stokes field-far The
22
x
Gij : elastodynamic Green function – The displacement field from the simplest source – namely, the unidirectional unit impulse, which is localized precisely in both space and time
Notaion – Gij : ith response to impulse force acting on jth direction.
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8.4 Elastodynamics – general form of the far-field displacement for a couple
Using the notation of Green function, the general form of the far-field displacement field (P and S) for a couple in the pq plane is given by
c, 3
3
/
4
/( ) , (8.62/63)
4
pqn p qn pq np q
pqn p np q
M t ru M G
r
M t r-
r
P
S
Mpq : seismic moment tensor (9 couples/dipoles)
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function rateMoment :)(
radiation waveS ˆsincos ,ˆ cos2cos ,0
radiation waveP 0 ,0 ,ˆ cos2sin
(8.64) ,/
4
1
/
4
1,
03
03
tM
r
r
rtM
r
rtMt
FS
FP
FS
FP
A
A
A
Axu
8.4 Elastodynamics – Radiation pattern of the far-field displacement for double couple
Convert to spherical coordinate system (for p=1, q=3)
areafault slip
)()(0
tDtAM
Time-dependent moment function
The far-field displacements are proportional to the moment rate function.
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0 ,0 ,ˆ cos2sin rFPA
8.4 Elastodynamics – Radiation pattern of the far-field P displacement for double couple
,0 plane, x xIn the 31
1
1
ˆ sin 2 (x 0, 0)
ˆ sin 2 (x 0, )
r
r
FPA
x3
x1
Θ=180°
Θ=90° Θ=90°
Θ=0°
There are two nodal lines (fault plane and auxiliary plane)
x1
-x3
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ˆsincos ,ˆ cos2cos ,0 FSA
8.4 Elastodynamics – Radiation pattern of the far-field S displacement for double couple
x1
-x3
,0 plane, x xIn the 31
)0(x ˆ 2cos
)0(x ˆ 2cos
1
1
FSA
x3
x1
Θ=180°
Θ=90°
Θ=0°
Θ=90°
TP T
PP: pressure axis T: Tension axis
There are 6 nodal points. (There is no nodal line.)
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8.4 Elastodynamics – Radiation pattern of the far-field S displacement for double couple
11 3
1
ˆcos2 (x 0)In the x x plane, 0,
ˆcos2 (x 0)
FSA
32 ( , )
2 2ˆcos2 0 -
θ
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8.4 Elastodynamics – Example of point-source
Comparison of observed and synthetic ground motions for June 13, 1980 eruption of Mt.St. Helens (A vertical point force at the source). The comparison can be used to estimate the strength of the eruption.(Kanamori & Given, 1983)
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8.4 Elastodynamics – Example of point-source
Observed and interpretations of the source mechanism for the 1975 Kalapana, Hawaii event.(Eissler and Kanamori, 1987)
Double coupleObserved Single force
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8.4 Elastodynamics – the nature of moment rate function – step/delta function
ttM
tHtM
tDtAt
tMtDtAtM
(8.74)
(8.73)
)()( ),()(
t
M(t)
●
t
M(t)
Step function
δ function
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8.4 Elastodynamics – the nature of moment rate function – ramp/boxcar function
tBtM
tRtM
(8.74)
(8.73)t
M(t)
●
t
M(t)
Ramp function
Boxcar function
(8.77) .0MMdttMA
τ
In the case of boxcar function, the area under the boxcar is equal to M0
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8.5 The Seismic Moment Tensor P342
)78.8( .
333231
232221
131211
MMM
MMM
MMM
M
The DC solution given by (8.61) has the corresponding moment tensor:
)79.8( .
000
00
00
21
12
0
M
M
MM
Where M0 is the scalar factor.
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8.5 The Seismic Moment Tensor – Mij in terms of fault parameters
From Seth Stein & Michael Wysession “An introduction to seismology, earthquakes, and Earth structure.
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8.5 The Seismic Moment Tensor – Mij in terms of fault parameters
φf
δ
λ
(8.80) kjjkjk DDAM vv D : slip vector
V : fault normal
v D
Fig8.20 The geographic coordinate system (ray coord.)
Double couple M in the geographic frame
N
φs
E
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8.5 The Seismic Moment Tensor – Mij in terms of fault parameters
(8.80) kjjkjk DDAM vv
)82.8( .ˆ cosˆcossinˆsinsin 32f 1f xxxv
(8.81) ,ˆsinsin
ˆcossincossincos
ˆsinsincoscoscos
3
2ff
1ff
x
x
xD
D
D
D
Express D and v in terms of fault parameters (φf, δ, λ)
f2
f011 sinsin2sin2sincossin MM
Double couple
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(8.83) .cossin2cossincoscos
sinsin2coscoscoscos
2sinsin2sin2coscossin
sin2sin
cossin2sin2sincossin
sinsin2sin2sincossin
ff023
ff013
f21
f012
2211033
f2
f022
f2
f011
MM
MM
MM
MMMM
MM
MM
In the same way:
8.5 The Seismic Moment Tensor – double couple Mij in terms of fault parameters
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8.5 The Seismic Moment Tensor – moment weighted Green’s function
elements.sor moment ten respective theofeach
toingcorrespond functions sGreen' theare :
),,,,(m
(8.84) ,,
2313122211
5
1
in
iinin
G
MMMMM
Gmtu
x
It’s possible to construct the P or S motion for a moment tensor by summing the moment weighted Green’s function
The basis for many synthetic seismogram programs and waveform inversions.
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8.5 The Seismic Moment Tensor – rotation of moment tensor
(8.85) ,
000
00
00
0
0
M
M
M)79.8( .
000
00
00
21
12
0
M
M
MM
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8.5 The Seismic Moment Tensor – decomposition of Moment tensor
11 1
12 2
13 3
1 2 3
1
0 0 tr 0 0 0 01
0 0 0 tr 0 0 0 , (8.86)3
0 0 0 0 t
trace
deviatoric eigenva
r 0 0
tr(
lue
) M M M of
: os f i
M M
M M
M M
M
M
M M
M
M M
M
In general, a seismic moment tensor need not corresponding to a pure double couple, but the symmetric tensor can still be diagonalized into three orthogonal dipoles.
Moment tensors for faulting events are often determined with the constraint tr(M)=0
Isotropic component a volume change, when tr(M) ≠0
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(8.88) ,
00
000
00
3
1
00
00
000
3
1
00
00
00
3
1
tr00
0tr0
00tr
3
1
00
00
00
31
31
32
32
3
21
21
3
2
1
MM
MM
MM
MM
M
MM
MM
M
M
M
M
M
M
8.5 The Seismic Moment Tensor – decomposition of Moment tensor
An isotropic part and three double couples.
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(8.89) .
200
00
00
3
1
00
020
00
3
1
00
00
002
3
1
tr00
0tr0
00tr
3
1
00
00
00
3
3
3
2
2
2
1
1
1
3
2
1
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
8.5 The Seismic Moment Tensor – decomposition of Moment tensor
An isotropic part and three CLVDs
CLVD: componsated linear vector dipoles
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(8.90) ,
00
00
000
000
00
00
tr00
0tr0
00tr
3
1
00
00
00
13
13
11
11
3
2
1
M
MM
M
M
M
M
M
M
M
8.5 The Seismic Moment Tensor – decomposition of Moment tensor
13
12
11 MMM If
An isotropic part, a major double couple and a minor double couple.
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(8.91) ,
200
00
00
00
00
000
21
tr00
0tr0
00tr
3
1
00
00
00
3
3
3
3
3
3
2
1
M
M
M
M
M
M
M
M
M
M
M
8.5 The Seismic Moment Tensor – decomposition of Moment tensor
Where ε is a measure of the size of the CLVD component relative to the double couple. For a pure double couple, ε=0.
have then weand ,M
M- compute we,MMMFor 13
121
312
11
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Box 8.4 A non-double-couple source
Significant non-double components found using waves with different frequencies
Comparison of observed P and predictionsComparison of observed P and predictions
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Focal sphere – Beach Ball
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Focal sphere – relation between fault planes and stress axes
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Example of the determination of a complex rupture for the 1976 Guatemala earthquake.
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x1
x2
baseball
eyeball
CLVD
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A quasi-vertical inflating magma dike ~ a crack opening under tension
0 0 1 0 0 -1 0 02 2
~ 0 0 0 1 0 0 -1 03 3
0 0 2 0 0 1 0 0 2
M I
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The END
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Focal sphere – Beach Ball
From Seth Stein & Michael Wysession “An introduction to seismology, earthquakes, and Earth structure.
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Focal sphere – Beach Ball
From Seth Stein & Michael Wysession “An introduction to seismology, earthquakes, and Earth structure.
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Focal sphere – Beach Ball
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Focal sphere – Beach Ball
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8.4 Elastodynamics p.334
(8.49) .
//
4
1
2
2112
c 22
r
rth
r
rthu FF
iii
(8.50) .
2for
2for
22
22
2
22
2
jiFi
iFi
FiF
i
ii
r
x
jr
xx
jr
x
r
x
rr
x
r
x
For r rrr 2
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8.4 Elastodynamics p.335
(8.51) .
coscos
cos
cos
2
2
1
11
1
iF
F
i
rr
xr
x
(8.52) .
//
4 2
22
22
1c
r
rth
r
x
r
rthu jiii
(8.53) .///2 rrthrth
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8.4 Elastodynamics p.335-336
(8.54) ///
/////
//
000
rthrrth
rtrrtrthrth
rrth
tttt
hthth
(8.55) .
/
4
//
///
4
2
22
21c
22
21c
rth
rrth
r
x
r
rthru
r
rthr
r
rth
r
x
r
rthr
r
rth
r
rthu
jii
j
iii
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8.4 Elastodynamics p.340
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8.4 Elastodynamics p.339-340
(8.65) ,2sinsin2sinsincoscos2cossin
2cossinsincossincoscoscos
sin12sin2sinsin2sin2sinsincos
cos2coscoscossin2sin2cossin
sin2sin2cossinsinsincos2sinsin
cos2sincoscos2sinsinsincos
4
1,
4
1,
4
1,
21
221
21
222
2
3
3
3
hh
hhSH
hh
hhSV
hhh
hhP
SHSH
SVSV
PP
ii
iiR
ii
iiR
iii
iiR
rtMR
rtrU
rtMR
rtrU
rtMR
rtrU
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8.4 Elastodynamics p.340-341
(8.66) ,cossin
sin
cossincos
sin
02
0
02
2
03
00
00
d
di
ir
iET
iri
viEE hhh
(8.67) .ˆ 4
1,ˆ
3
MR
rru P
r
(8.68) .ˆ4
1,ˆ
2
32222
0
MRruE P
hhhhrhhh
(8.69) ˆ22
00 ruE
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8.4 Elastodynamics p.341
(8.70) ,ˆ,
4
1ˆ
03
MRr
hgu P
hhr
(8.71) .cos
1
sin
sin,
000
d
di
i
ihg hhhh
(8.72) ./,
4
1,ˆ
03
rtMRr
hgtu P
hhr
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8.6 Determination of Faulting Orientation P347
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8.6 Determination of Faulting Orientation P347
h2
1
h21
isin2OA' area Equal
itanOA' hicStereograp
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8.6 Determination of Faulting Orientation P348
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8.6 Determination of Faulting Orientation P348
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8.6 Determination of Faulting Orientation P349
1.
2.
3.
4.
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8.6 Determination of Faulting Orientation P350
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8.6 Determination of Faulting Orientation P351
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8.6 Determination of Faulting Orientation P350-351
(8.92) , 2
1
detx ti
(8.93) ,
sin
1
1LL
1LL
2// 4L
QiqPp
eee QUacaii
(8.94) ,
sin
1
1RR
1RR
1RR
2// 4/R
QiqPpSs
eee QUacaii
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8.6 Determination of Faulting Orientation P352
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8.6 Determination of Faulting Orientation P352
(8.95) ,2coscossinsin2sinsincos
coscoscossin2cossin
cossinsin
cos2cossinsincoscos
2cossincos2sinsinsincos
R
R
R
L
L
p
q
s
q
p
(8.96) .
21LL
21LLL
21RR
21RR
21RRR
QqPpA
QqPpSsA
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8.6 Determination of Faulting Orientation P353
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8.6 Determination of Faulting Orientation P353
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Finite fault ~ Discretized distributions of double couples (Figure 8.15)
Vertical strike slip Fault parallel motions
Fault perpendicular motions Vertical motions (Chinnery 1961)
+
+
-
-