1. Stress: Force per unit area across an arbitrary plane

2. Stress Defined as a Vector ^ N = unit vector normal to plane t(n) = (t x,t y,t z ) = traction vector ^ The part of t that is perpendicular to the plane is normal stress The part of t that is parallel to the plane is shear stress

3. Stress Defined as a Tensor ^ t(x)  xx  xy  xz  =  T =  yx  yy  yz  zx  zy  zz x y z ^ ^ t(z)t(y)  zx  xz No net rotation x z

4. Relation Between the Traction Vector and the Stress Tensor ^ t(x)  t x  n  xx  xy  xz n x t(n) =  n = t y ( n ) =  yx  yy  yz n y t z ( n )  zx  zy  zz n z x y z ^ ^ t(z)t(y)  zx  xz No net rotation x z ^ ^ ^ ^ ^ ^ ^^

5. Relation Between the Traction Vector and the Stress Tensor  t x  n  xx  xy  xz n x t(n) =  n = t y ( n ) =  yx  yy  yz n y t z ( n )  zx  zy  zz n z ^ ^ ^ ^ ^ ^ ^^ That is, the stress tensor is the linear operator that produces the traction vector from the normal unit vector….

6. Principal Stresses Most surfaces has both normal and tangential (shear) traction components. However, some surfaces are oriented so that the shear traction = 0. These surfaces are characterized by their normal vector, called principal stress axes The normal stress on these surfaces are called principal stresses Principal stresses are important for source mechanisms

7. Stresses in a Fluid  P     =   P     P If  1 =  2 =  3, the stress field is hydrostatic, and no shear stress exists P is the pressure

8. Pressure inside the Earth Stress has units of force per area: 1 pascal (Pa) = 1 N/m^2 1 bar = 10^5 Pa 1 kbar = 10^8 Pa = 100 MPa 1 Mbar = 10^11 Pa = 100 GPa Hydrostatic pressures in the Earth are on the order of GPa Shear stresses in the crust are on the order of 10-100 MPa

9. Pressure inside the Earth At depths > a few km, lithostatic stress is assumed, meaning that the normal stresses are equal to minus the pressure (since pressure causes compression) of the overlying material and the deviatoric stresses are 0. The weight of the overlying material can be estimated as  gz, where  is the density, g is the acceleration of gravity, and z is the height of the overlying material. For example, the pressure at a depth of 3 km beneath of rock with average density of 3,000 kg/m^3 is P = 3,000 x 9.8 x 3,000 ~ 8.82 10^7 Pa ~ 100 MPa ~ 0.9 kbar

10. Mean (M) and Deviatoric (D) Stress  xx  xy  xz   =  yx  yy  yz  zx  zy  zz M =  xx +  yy +  zz =  ii /3  xx -M  xy  xz D  =  yx  yy -M  yz  zx  zy  zz -M

11. Strain: Measure of relative changes in position (as opposed to absolute changes measured by the displacement) U(r o )=r-r o E.g., 1% extensional strain of a 100m long string Creates displacements of 0-1 m along string

12. J can be divided up into strain (e) and rotation (Ω) is the strain tensor (e ij =e ji )  u y  u x  u y  u y  u z  x  y  y  z  y  u z  u x  u z  u y  u z  x  z  y  z  z  u x  u x  u y  u x  u z  x  y  x  z  x e = ½( + )

13. J can be divided up into strain (e) and rotation (Ω)  u y  u x  u y  u z  x  y  z  y  u z  u x  u z  u y  x  z  y  z  u x  u y  u x  u z  y  x  z  x Ω= 0 ½( - ) ½( - ) -½( - ) 0 -½( - )-½( - ) 0 ½( - ) is the rotation tensor (Ω ij =-Ω ji )

14. Volume change (dilatation)  u x  u y  u z  x  y  z  = 1/3 ( + + ) = tr(e) = div(u)  > 0 means volume increase  < 0 means volume decrease uxxuxx uxxuxx >0 <0

15.  ∂ 2 u i /∂t 2 = ∂ j  ij + f i = Equation of motion = Homogeneous eom when f i =0

16. Seismic Wave Equation (one version) For (discrete) homogeneous media and ray theoretical methods, we have   ∂ 2 u i /∂t 2 = (  )  ·u-  x  x u

17. Plane Waves: Wave propagates in a single direction u(x,t) = f(t  x/c) travelling along x axis = A(  )exp[-i  (t-sx)] = A(  )exp[-i(  t-kx)] where k=  s = (  /c)s is the wave number ^

18.  s  x sin  = v  t,  t/  x = sin  /v = u sin  = p u = slowness, p = ray parameter (apparent/horizontal slowness) rays are perpendicular to wavefronts ss xx  wavefront at t+  t wavefront at t

19. p = u 1 sin  1 = u 2 sin  2 u = slowness, p = ray parameter (apparent/horizontal slowness) 22 11 v1v1 v2v2

20. p = u 1 sin  1 = u 2 sin  2 = u 3 sin  3 Fermat’s principle: travel time between 2 points is stationary (almost always minimum) 22 11 v1v1 v2v2 33 v3v3

21. Continuous Velocity Gradients p = u 0 sin  0 = u sin  = constant along a single ray path v z   =90 o, u=u tp dT/dX = p = ray parameter X X T 00

22. X(p) generally increases as p decreases -> dX/dp < 0 v z   =90 o, u=u tp Prograde traveltime curve X X T p decreasing

23. X(p) generally increases as p decreases but not always v z Prograde X X T Retrograde Prograde caustics

24. Reduced Velocity Prograde X T Retrograde Prograde caustics T-X/Vr X

25. X(p) generally increases as p decreases -> dX/dp < 0 v z X X T Shadow zone lvz p 

30. T = ∫ 1/v(s)ds = ∫u(s)ds T j = ∑ G ij u i  T j = ∑ G ij  u i i=1 d=Gm G T d=G T Gm m g =(G T G) -1 G T d i=1 Traveltime tomography j j-th ray

31.  2 = ∑ [t i -t i p ] 2 /  i 2  i expected standard deviation  2 (m best )= ∑ [t i -t i p (m best )] 2 /ndf m best is best-fitting station  2 (m) = ∑ [t i -t i p ] 2 /  2 - contour! Earthquake location uncertainty n i=1   n i=1 n i=1

32. Fast location: S-P times: D ~ 8 x S-P(s)

33. Other sources of error: Lateral velocity variations slow fast Station distribution

34. E mean = 1/2  A 2  2 (higher frequencies carry more E!) A 2 /A 1 = (  1 c 1 /  2 c 2 ) 1/2 ds 2 ds 1

35.    1 -    2 A 1 ’’=    1 +    2 2    1 A 2 ’ =    1 +    2    1 cos  1 -    2 cos  2 S’S’’=    1 cos  1 +    2 cos  2 2    1 cos  1 S’S’ =    1 cos  1 +    2 cos  2 since    ucos  cos  For vertical incidence (     1 -    2 2    1 S’S’’ vert = S’S’ vert =    1 +    2    1 +    2

36. S waves vertical incidence  P waves vertical incidence:    1 -    2 2    1 P’P’’ vert = - P’P’ vert =    1 +    2    1 +    2    1 -    2 2    1 S’S’’ vert = S’S’ vert =    1 +    2    1 +    2

37. E 1 flux = 1/2 c 1    A 1 2  2 cos  1 E 2 flux = 1/2 c 2    A 2 2  2 cos  2 A norm = [E 2 flux /E 1 flux ] 1/2 = A 2 /A 1 [c 2   cos  2 / c 1   cos  1 ] 1/2 = A raw [c 2   cos  2 / c 1   cos  1 ] 1/2    1 cos  1 -    2 cos  2 S’S’’ norm = = S’S’’ raw    1 cos  +    2 cos  2    1 cos  (c 2   cos  2 ) 1/2 S’S’ norm = x    1 cos  1 +    2 cos  2 (c 1   cos  1 ) 1/2

38. 2    1 cos  S’S’ =    1 cos  +    2 cos  What happens beyond  c ? There is no transmitted wave, and cos  = (1-p 2 c 2 ) 1/2 becomes imaginary. No energy is transmitted to the underlying layer, we have total internal reflection. The vertical slowness  =(u2-p2)1/2 becomes imaginary as well. Waves with Imaginary vertical slowness are called inhomogeneous or evanescent waves.

39. Phase changes: Vertical incidence, free surface: S waves - no change in polarity P waves - polarity change of  Vertical incidence, impedance increases: S waves - opposite polarity P waves - no change in polarity Fig 6.4 Phase advance of  /2 - Hilbert Transform

40. Attenuation: scattering and intrinsic attenuation Scattering: amplitudes reduced by scattering off small-scale objects, integrated energy remains constant Intrinsic: 1/Q(  ) = -  E/2  E E is the peak strain energy, -  e is energy loss per cycle Q is the Quality factor A(x)=A 0 exp(-  x/2cQ) X is distance along propagation distance C is velocity

41. Ray methods: t* = ∫dt/Q ( r ), A(  )=A 0 (  )exp(-  t*/2) i.e., higher frequencies are attenuated more! pulse broadening