Transcript No Slide Title

Source Moment Tensor Solutions from Waveform
Inversion
Case study of Jabalpur event
- G.Karunakar, S.N. Bhattacharya & J.R. Kayal
Vp=4.55 km/s
Vs=2.55 km/s
density= 2.6 g/cm3
Vp=4.55 km/s
Vs=2.55 km/s
density = 3.2 g/cm3
Seismogram
Source Depth (d)
Epicentral Distance ()
Source

A Seismogram is a result of Source, propagation path and
seismograph response

It is possible to generate this mathematically (Synthetic
Seismogram)
A seismogram or a ground motion at a site can be represented as u(t)
1 
u(t) 
U() exp( it ) d


2

u ( t ) exp( it ) dt

(1)
U()  
(2)
U() = M() E(,)
(3)
Response
Fourier Transform of source time function m(t)
or moment function
U() = M() E(,) H(w)
(4)
Instrument Response
For a shear dislocation in an earthquake is a linear combination of 3
basic fundamental mechanisms for vertical and radial components
and a linear combination of 2 basic fundamental mechanisms for
Transverse component
uV(,t) = M0 [ a uVDD(,t) + b uVDS(,t) + c uVSS(,t) ]
uR(,t) = M0 [ a uRDD(,t) + b uRDS(,t) + c uRSS(,t) ]
uT(,t) = M0 [ a uTDS(,t) + b uTSS(,t) ]
Nonlinear
functions
of ,  & 
0    2,
(5)
a = sin  sin 2
b = - cos  cos  cos  + sin  cos 2 sin 
c = 0.5 sin  sin 2 cos 2 + cos  sin  sin 2
a = sin  cos 2 cos  + cos  cos  sin 
b = cos  sin  cos 2 - 0.5 sin  sin 2 sin 2
0   /2,
-    
(6)
m(t) = M0 s(t)
(7)
M0 = seismic moment = s A D,
s = modulus of rigidity at the source
A = area of the fault,
D = average slip (displacement) of the rock
above the fault
M() = M0 S()
(8)
 A general source is represented by a moment tensor [M]
with six independent elements
 However, in most tectonic earthquakes, there is no volume
change at the source and thus we have only five independent
elements of [M]
Moment tensor
We define a force couple Mij in a Cartesian co-ordinate system
as a pair opposite forces (fi) pointing in the i direction,
separated by a distance dj in the j direction. The magnitude of
Mij is given by
M ij  lim
(
f
d
)
i
j
f 
i
d j 0
There are 9 elements of Mij. However, the condition that
angular momentum be conserved requires that Mij = Mji. Thus
the moment tensor [M] is symmetric and has 6 independent
elements.
Mxx Mxy Mxz 
[M] =  Mxy Myy Myz 


 Mxz Myz Mzz
q
The earthquake source is generally deviatoric so that
Mzz = - (Mxx + Myy)
and we have 5 distinct moment tensor components
uV(,t) = Mxx GxxV + Myy GyyV + Mxy GxyV + Mxz GxzV + Myz GyzV
uR(,t) = Mxx GxxR + Myy GyyR + Mxy GxyR + Mxz GxzR + Myz GyzR
uT(,t) = Mxx GxxT Myy GyyT + Mxy GxyT + Mxz GxzT + Myz GyzT
d(t 1 )   G1 (t 1 ) G 2 (t 1 )
5
d(t )  G (t ) G (t )
1 2
2 2
2 


G i ( t q )m i 
 ...  
i 1

 
d(t n ) G1 (t n ) G 2 (t n )
)
= [G] [m]
 m1 
G 5 (t 1 )   
m

G 5 (t 2 )  2 
m 3 
 
 m 4 
G 5 ( t n )
m 5 
Green’s functions
Green’s functions Gij are the ground motion or synthetic seismogram
for the moment tensor element Mij.
G Vxx  u VDD  0.5u VSS cos2,
G Vyy  u VDD  0.5u VSS cos2
G Vxy  u VSS sin 2, G Vxz  u VDS cos, G Vyz  u VDS sin 
Same equation for Radial component
G Txx  0.5u TSS sin 2,
G Txy  u TSS cos 2,
Vertical
Radial
G Tyy  0.5u TSS sin 2,
G Txz  u TDS sin ,
G Tyz  u TDS cos Transv.
Waveform Inversion

Comparision of Synthetic and Observed Seismogram

Differences between them are minimised by adjusting
earth structure and/or source representation
Suppose the observed displacement component at the station = d(t)
the synthetic component computed = u(t)
d(t) = u(t)
[d] =[ G] [m]
This equation is solved by least square inversion method or
by Single Value Decomposition
[G]-g = [V] []-1 [UT]
where singular value decomposition of [G] is
[G] = [U] [] [VT]
Case study: Jabalpur 2003 event
Date
OT
UTC
P Arrival
S Arrival
Lat
21/05/2003 10:36:27.6 10:36:35
10:36:40
22.96
Jabalpur station coordinates
Long
Mag
80.10
2.8
Depth
??
23.8760 79.8760
Instrument Parameters of broadband digital system at Jabalpur
(A) SENSOR
Model
Transducer Type
Component
Guralp CMG 40T
Force Feedback
Three (Ver, NS & EW)
Dynamic Range
At Least 145dB
Natural period
30 seconds
(B) DATA ACQUISITION
SYSTEM
Model
Bit Resolution
72A- 07
24 Bit
Recording
Format
REFTEK
Filters:
FIR filters, Pass band=85%Nyquist
and Stopband 30dB.
( Kayal, 2000 JGS )
BA Map
40km UP Map
Depth
0.000
2.500
MULTILAYERED MODEL PARAMETERS
2.400
4.5500
2.6300
400.0
200.0
2.600
5.9400
3.4300
320.0
320.0
3.010
6.6500
3.8400
3.300
8.1900
4.7300
160.0
080.0
8.5000
4.9090
100.0
050.0
20.500
240.0
120.0
39.000
70.000
3.500
Vertical
Radial
Transverse
2003 event
Processing of Observed data
V
T
R
Offset correction (Removal of mean)
Rotation for angle of radial direction
Instrument response removal (In frequency domain)
Cosine tapering (Back to Time Domain)
Filtering at 1 Hz
Observed Seismogram
Vertical
Radial
Transverse
VP
VS
RP
RS
TS
Synthetic Seismogram
Vertical
VSS
Strike Slip
VDS
Dip Slip
VDD
CLVD
Compensated Linear
Vector Dipole
RSS
Radial
RDS
RDD
TSS
Transverse
TDS
TDD
V
R
T
Observed
G1
G2
G3
G4
G5
Synthetic
Waveform Inversion
Fault Plane Solution
--------------------------------------###--------- P ---------######----------------###
########-------------------####
#########-----------------#####
############--------------#######
#
#########------------########
# T ###########---------#########
#
############------###########
##################---############
###############################
###############-----###########
###########----------########
#####----------------####
------------------------------------------ -------
Seismic moment = 1.440 x 1020 dyne cm
Percentage of double couple = 90.8316
Eigen values (1020 Dyne cm)
 1 = -1.405
 2 = -6.7500E-02
 3 = 1.472
Nodal Planes: NP1: = 82.18 = 140.30
= -150.03
NP2: = 60.33 = 45.82
= -9.00
T, P & B Axis of the lower Hemisphere of best DC:
T-Axis: Azimuth: 269.60
Plunge: 14.70
P-Axis: Azimuth: 190.12
Plunge: 30.46
B-Axis: Azimuth:
7.11
Plunge: 26.50
Conclusions

It is a powerful tool for evaluating earth structure and
kinamatic parameters of the source even with such small
magnitude events

Moment tensor inversion based on decomposition gives
strike, dip and slip angle of the best double couple solution

Taking this as starting point, a nonlinear least square
inversion can be performed to refine the double couple
solution
 These moment tensor solutions are important for
understanding the tectonic structures of the region
SPIN OFF
With the availability of BB data on continuous basis,
this inversion technique should be carried out
on all such seismically active areas as a routine
to monitor the seismicactivity
(Could be a precursor)