Last time: Seismic Source Modeling• For an earthquake, the f(t) in the source term of the wave equation:
is a (tensor) moment rate of energy release:
(& moment is M = sA!)
• The moment rate function can be used to solve for amount of slip, direction of slip, and speed of rupture propagation on a fault. This in turn tells us something about history of fault slip (including recent evolution and stress relieved in past earthquakes) & frictional stability.
Read for Fri 27 Feb: S&W 75-86 (§2.6)
€
∇2φ r( )−1
α 2
∂ 2φ
∂t2= −4πδ r( ) f t( )
€
˙ M (t) = μ ˙ s (t)dAArea
∫
Source Seismology
2011 Christchurch earthquake, M6.3, after a larger M7.0 eq further west in 2010…
€
Φ r,t( ) =f t ±
r
α
⎛
⎝ ⎜
⎞
⎠ ⎟
r
Difference isproximity…
2 injuredNZ $4B
2010M7.0
2011M6.3
185 deadNZ $15B
Can use this to get other interesting pieces of information about the earthquake rupture process…
… Including our growing recognition that many large earthquakes involve complex rupture simultaneously on several faults that may have completely different dip and orientation.*
*Lesson for Utah! Ya think future Wasatch rupture won’t cross segment boundaries? Ya got another think comin’.
Hayes et al., Nat. Geosci., 2010
Crone et al.,BSSA, 2004
2010 M7.0Haiti
2002 M7.9 Denali
Seismic Wave Energy PartitioningWith Snell’s Law in our tool-belt, we’re ready to consider what happens to seismic amplitudes when an incoming wave arrives at a change in properties (and hence, conversions occur).
One obvious thing that has to happen is conservation of energy: i.e.,
reflected energy + transmitted energy = energy of the incoming wave
As you might expect, energy is related to amplitude of the wave.
incoming PA
Impedance Contrast:Thus far we’ve focused much of the discussion on conceptsrelated to velocity & travel-time, but seismic waves also haveamplitude, A, of the particle displacements:
Amplitudes of reflections & refractions are determined by energy partitioning at the boundary. A normally-incident ( = 0) P-wave with amplitude Ai produces a reflected P with amplitude:
incoming PA
€
R ≡Arfl
Ai
=ρ 2V2 − ρ1V1
ρ 2V2 + ρ1V1
≡Z2 − Z1
Z2 + Z1
(reflectioncoefficient)
and a refracted P:
where Zi = iVi is the impedance in layer i.€
T ≡Arfr
Ai
=2ρ1V1
ρ 2V2 + ρ1V1
≡2Z1
Z2 + Z1
(transmissioncoefficient)
The energy E in a wave is directly proportional to the amplitude A, and for this example, sign (i.e. propagation direction) matters. We’ll use the sign of the z-component (positive-down) of propagation. Then we have:
Displacement must be continuous at the boundary so: Ai + Arfl = Arfr
And:
1 + R = T
Note however for the P-wave depicted here, this applies onlyto the case where i = 0°… (Why?)
