Download presentation
Presentation is loading. Please wait.
1
Recall the momentum equation: ∂ 2 u i /∂t 2 = ∂ j ij +f i, where f i is the body force term An earthquake source is usually considered slip on a surface (displacement discontinuity), not a body force Fortunately, it can be shown that a distribution of body forces exists, which produces the equivalent slip (equivalent body forces)
2
Helpful to define a concept that separates the source from the wave propagation: u i (x,t)=G * f = G ij (x,t;x 0,t 0 )f j (x 0,t 0 ) f = force vector G = Green’s function = response to a ‘small’ source Linear equation Displacement from any body force can be computed as the superposition of individual point sources
3
Force Couples: Forces must occur in opposing directions to conserve momentum D no net torque double couple: D net torque no net torque
4
9 Force Couples M ij (the moment tensor), 6 different (M ij =M ji ). |M|=fd M 11 M 12 M 13 Good approximation for distant M= M 21 M 22 M 23 earthquakes due to a point source M 31 M 32 M 33 Larger earthquakes can be modeled as sum of point sources
6
u i (x,t)=G * f = G ij (x,t;x 0,t 0 )f j (x 0,t 0 ) Displacement from a force couple can be computed as u i (x,t) = G ij (x,t;x 0,t 0 )f j (x 0,t 0 )-G ij (x,t;x 0 -x k,t 0 )f j (x 0,t 0 ) = [∂G ij (x,t;x 0,t 0 ) /∂x 0 ] f j d where the force vectors are separated a distance d in the x k direction u i (x,t) = [∂G ij (x,t;x 0,t 0 ) /∂x 0 ] M jk (x 0,t 0 ) ^ ^
7
Description of earthquakes using moment tensors: Parameters: strike , dip , rake Right-lateral =180 o, left-lateral =0 o, =90 reverse, =-90 normal faulting Strike, dip, rake, slip define the focal mechanism 0 M 0 0 Example: vertical right-lateral al M= M 0 0 0 M 0 = DA scalar seismic momen 0 0 0
8
Description of earthquakes using mome Parameters: strike , dip , rake Vertical fault, right-lateral =180 o Vertical fault, right-lateral =0 o Strike, dip, rake, slip define the focal m 0 M 0 0 Example: vertical right-lateral along x M= M 0 0 0 M 0 = DA scalar seismic moment (Nm) 0 0 0
9
Because of ambiguity M ij =M ji two fault planes are consistent with a double-couple model: the primary fault plane, and the auxillary fault plane (model for both generates same far-field displacements). Distinguishing between the two requires further (geological) information
10
Far-field P-wave displacement for double-couple point source: u P i (x,t)=(1/4 3 ) (x i x j x k /r 3 )-(1/r) ∂M jk (t-r/ t r 2 =x 1 2 + x 2 2 + x 3 2 For the fault in the (x 1,x 2 ) plane, motion in x 1 direction, M 13 =M 31 =M 0 and: u P i (x,t)=(1/2 3 ) (x i x 1 x 3 /r 3 )-(1/r) ∂M j (t-r/ t In spherical coordinates: x 3 /r=cos , x 1 /r=sin cos , x i /r=r i u P =(1/4 3 ) sin2 cos (1/r) ∂M 0 (t-r/ t r ^ ^
11
Far-field S-wave displacement for double-couple point source: u S i (x,t)=[( ij - i j ) k ]/(1/4 3 )(1/r) ∂M jk (t-r/ t, i = x i /r r 2 =x 1 2 + x 2 2 + x 3 2 For the fault in the (x 1,x 2 ) plane, motion in x 1 direction, M 13 =M 31 =M 0 and: u S (x,t)=(1/4 3 )(cos2 cos -cos sin )(1/r) ∂M 0 (t-r/ t ^^
14
Earthquake focal mechanism determination from first P motion (assuming double-couple model): Only vertical component instruments needed No amplitude calibration needed Initial P motion easily determined (up or down) Up: ray left the source in compressional quadrant Down: ray left source in dilatational quadrant Plotted on focal sphere (lower hemisphere) Allows division of focal sphere into compressional/dilatational quadrants Focal mechanism is then found from two orthogonal planes (projections on the focal sphere)
15
Earthquake focal mechanism determination from first P motion (assuming double-couple model): Focal sphere is shaded in compressional quadrants, generating ‘beach ball’ Normal faulting: white with black edges Reverse faulting: black with white edges Strike-slip: cross pattern
18
Far-field pulse shapes: Earthquake rupture doesn’t occur instantaneously, thus we need a time dependent moment tensor M(t) Near-field displacement is permanent Far-field displacement (proportional to ∂M/ t) is transient (no permanent displacement after the wave passes): u S i (x,t)=[( ij - i j ) k ]/(1/4 3 )-(1/r) ∂M jk (t-r/ t area=M 0 = S ave A
19
Directivity: Haskell source model Point source: amplitude will vary with azimuth but rise time is constant Larger events: integrating over point sources 0 t r M(t) ∂M(t)∂t
20
Directivity: Haskell source model (V r ~ 0.7-0.9 ) Rupture toward you at end of fault: d = - L/ + L/V r (last arrival rupture pulse L/V r - first arrival P wave, L/ Rupture away from you at end of fault: d = L/ + L/V r (last arrival L/V r (time of rupture to the end of the fault) + L/ (time of the P waves generated by the last rupture instant at L/V r ) - first arrival 0s) VrLVrL rupture
21
Far-field displacement is the convolution of two boxcar functions, one with width r and one with width d :
22
Stress Drop = average difference between stress on fault before and after the earthquake. t 2 ) t 1 dS A is fault area Assume long skinny fault (w<<L) with average displacement D ave and slip in the direction of L. Strain is then = D ave /w, and we have D ave /w In general: C D ave /L where L is a characteristic rupture dimension, C is a nin- dimensional constant that depends on rupture geometry Infinite long strike-slip fault: L=w/2, C=2/ S
23
Earthquake magnitude Most related to maximum amplitudes in seismograms. Local Magnitude (M L ): Richter, 1930ies Noticed similar decay rate of log 10 A (displacement) versus distance Defined distance-independent magnitude estimate by subtracting a log 10 A for reference event recorded on a Wood- Anderson seismograph at the same distance M L =log 10 A(in 10 -6 m)-log 10 A 0 (in 10 -6 m) =log 10 A(in 10 -6 m)+2.56log 10 dist (in km) -1.67 for 10<dist<600km only
25
Earthquake magnitude Body wave magnitude (m b ): (used for global seismology) m b =log 10 (A/T)+Q(h, ) T is dominant period of the measured waves (usually P, 1s) Q is an empirical function of distance and depth h (details versus amplitude versus range) Surface wave magnitude (M s ): (used for global seismology, typically using Rayleigh waves on vertical components) M s =log 10 (A/T)+1.66 log 10 + 3.3 = log 10 A 20 +1.66 log 10 + 2.0 (shallow events only)
26
Earthquake magnitude Saturation problem motivated the moment magnitude M w M w =2/3 log 10 M 0 -10.7 (M 0 moment in dyne-cm, 10 7 dyne cm=1Nm) = M w =2/3 log 10 M 0 -6.1 (M 0 moment in Nm) Scaling derived so M w is in agreement with M s for small events More physical property, does not saturate for large events
Similar presentations
© 2024 SlidePlayer.com. Inc.
All rights reserved.