SEISMIC SOURCES. Solution for the displacement potential Properties of the disp potential 1.An outgoing D ’ Alembert solution in the spherical coordinate

SEISMIC SOURCES

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

(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

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

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

(8.10) .2

4 4 42

uu

aaaF

rr

Fr


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


(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 (

(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:


(8.15) . 8

A

28

A

Fr

Fr

S

P

The solutions (of potentials) are

rr /2 Since 2

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


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:

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


Somigliana tensor

Symmetry: uij= uj

i

(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


Thanks to the symmetry, there are only 6 independent permutations:


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

(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:


The (x1x3 plane) static radial deformation and shear deformation due to a single force (Fig 8.12)

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

(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:


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


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

(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



(8.25) .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

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


(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.


(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

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.


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

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.

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

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

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)

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)

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

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)

So far, so good ?

Sorry, it’s getting messy …

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 …

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

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

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 )(

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!

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

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

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

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

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

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

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

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

?

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


Expand this term around (t-r2/α) in Taylor series.


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

(8.55) .

/

4 2

22

21c

r

thrr

thr

x

r

rthru j

ii

Near-field terms (Decays as 1/r2)


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)

(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


(8.60) .

/

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

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.

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)

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.

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

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.)

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

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)

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

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

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

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.

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.


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


(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

(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

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.

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

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

(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


An isotropic part and three double couples.

(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


An isotropic part and three CLVDs

CLVD: componsated linear vector dipoles

(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


13

12

11 MMM If

An isotropic part, a major double couple and a minor double couple.

(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


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

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

Focal sphere – Beach Ball

Focal sphere – relation between fault planes and stress axes

Example of the determination of a complex rupture for the 1976 Guatemala earthquake.

x1

x2

baseball

eyeball

CLVD

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

The END







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


(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

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



(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


(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


(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

8.6 Determination of Faulting Orientation P347


h2

1

h21

isin2OA' area Equal

itanOA' hicStereograp




1.

2.

3.

4.



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



(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



Finite fault ～ Discretized distributions of double couples (Figure 8.15)

Vertical strike slip Fault parallel motions

Fault perpendicular motions Vertical motions (Chinnery 1961)

+

+

-

-

Documents

SEISMIC SOURCES. Solution for the displacement potential Properties of the disp potential 1.An outgoing D ’ Alembert solution in the spherical coordinate