### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# ADV & NUMRL THERMOCHRONOLOGY GEO 391

UT

GPA 3.76

### View Full Document

## 56

## 0

## Popular in Course

## Popular in Geology

This 65 page Class Notes was uploaded by Mr. Jasmin Ratke on Sunday September 6, 2015. The Class Notes belongs to GEO 391 at University of Texas at Austin taught by Staff in Fall. Since its upload, it has received 56 views. For similar materials see /class/181658/geo-391-university-of-texas-at-austin in Geology at University of Texas at Austin.

## Reviews for ADV & NUMRL THERMOCHRONOLOGY

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/06/15

LECTUREZ 12402 2 Transformation of elastic coef cient matrix on orthogonal transformations We restrict ourselves to the Cartesian system and consider the transformation of the elastic coefficient matrix on orthogonal coordinate transformation We first start with the twodimensional case Figure 21 Coordinate transformation Let x1x2 be a twodimensional system in terms of which an arbitrary vector V can be expressed in component form as VVIJE1VZJE2 21 Now rotate the coordinate system by some arbitrary angle Fig 21 to define a new coordinate system In the new system the vector V can be expressed as V vl fl vgf 22 It is straightforward to show that the components V are related by the following equation vl cosx1 x1 cosx1 x2 v1 vz cosx2 x1 cosx2 x2 v2 Vf A11 A12 V1 23 vii 14211422 V2 Thus the components in the primed system are related to those in the unprimed system through a transformation matrix A whose elements are the direction cosines of primed system to unprimed system Thus following the notations in Fig 2 l we have V AV A cosQ sin 6 24 sin 9 cosQ Since V is arbitrary this can be inverted to give V ATIV A71 cosQ sin9 AT 25 sinQ cosQ I Therefore the transformation matrix A is orthogonal and that is why such a transformation is always called orthogonal transformation Such transformation can easily be extended to three dimensions where A becomes a 3X3 matrix Now if we consider a tensor V of order two it is straightforward to show that Vi AxkAllkl 26 Similarly for a tensor B of order three we have BU AWAHA B 27 kr pqr For a tensor C of order four we have Clsz AWquAk rAlstqrs 28 and so on Thus it is quite straightforward to transform tensor components from one coordinate system to another under orthogonal coordinate transformations We now go back to the stressstrain relation in condensed form and derive expressions for transformation of the matrix C under coordinate rotations To do this we must take a look at the transformation of the stress vector t and the strain vector e In order to do that we need to examine the transformation of the stress tensor and the strain tensor under orthogonal transformations We know that I AkAjlrkl 29 Using the above relation we can relate each component of the stress vector in the primed system with that in the unprimed system resulting in the following equation 111 111 Tgl 122 7 M 7 210 123 123 Tgl 131 zquot 139 where A121 A122 A123 2 A12 A13 2 A13 A1 1 2 A1 1 A1 2 A221 A222 A223 214221423 214231421 214211422 M 14321 14322 14323 214321433 214331431 214311432 14111431 14121432 14131433 14121433 14131432 14111433 14131431 14121431 14111432 I 14311411 14321412 14331413 14121433 14131432 14131431 14111433 14111432 14121431 14111411 14121412 14131413 14121423 14131422 14131411 14111413 14111412 14121431 21 1 Similarly we obtain transformations of strain given by a AlkA ekl 212 to obtain 11 11 2 822 3 e 33 N 33 213 2433 24323 2431 24331 2432 24312 where A121 A122 23 214121413 214131411 214111412 A121 A122 23 214121413 214131411 214111412 N A321 A322 23 214321433 214331431 214311432 214211431 214221432 214231433 14221433 14231432 14211433 14231431 14221431 14211432 I 214311411 214321412 214331413 All133 14131432 14131431 14111433 14111432 AIZASI 214111421 214121422 214131423 AIZA39ZS AISA39ZZ 14131421 14111423 AIIA39ZZ AIZASI 214 Thus the transformation matrix N of strains is very similar to that of stress except for a shift of a factor of two from upper right hand comer to the lower left hand comer The matrices M and N are called Bond Stress and Strain transformation matrices respectively Now we recall the stressstrain relation t Ce 215 and apply the Bond stress transformation to obtain t39 MtMCe 216 Now we apply the Bond strain transformation to obtain eN 1ev 217 From 2 16 and 2 17 we obtain t MCN le 218 Therefore the elastic coefficient matrix under orthogonal coordinate transformation becomes C MCN 1 219 It appears that the transformation of this coefficient matrix requires that inversion of a 6X6 matrix N This however is not necessary since from 211 and 214 we can easily show that N 1 MT 220 Thus we obtain the following simple relationship for the transformation of the elastic coefficient matrix under orthogonal coordinate transformations C MCMT 221 22A general rotation in Cartesian system anal the corresponding rotational matrices for the elastic coef cient matrix We restrict ourselves to the Cartesian system and consider the transformation of the elastic coefficient matrix for coordinate rotation We will now find a way to define a general rotation and show how the matrix C can be rotated by such general rotation Applying successive rotations about different coordinate axes can perform any general rotation of rectangular coordinates A standard method of doing that is as follows First rotate that coordinates anticlockwise through an angle 1 about the x3 axis so that the new coordinates are xl x2 and x3 Figure 22 Figure 22 Rotation of the coordinate system counterclock wise about the vertical axis The transformation matrix for such a rotation is given by cos 1 sin 1 0 A sin cos 0 222 0 0 1 Thus using the formula Eq 21 1 the Bond stress transformation matrix M a is given by cos2 1 sin2 1 0 0 0 sin 2 sin2 1 cos2 1 0 0 0 sin 2 0 0 1 0 0 0 M a 223 0 0 0 cos 1 sm 1 0 0 0 0 sin 1 cos 1 0 sin22 sin22 0 0 0 cos2 Therefore the rotated coefficient matrix for the rotation shown in Fig 12 is given by T Ca M CM 224 Step 2 Now rotate the new axes xl x2 x3 anticlock wise through an angle 9 about x2 axis to go to the new axes system x1x2 x3 as shown in Figure 23 Figure 23 Coordinate transformation The transformation matrix for such rotation becomes cosQ 0 sinQ A 0 1 0 225 sin 9 0 cos 9 Thus using the formula Eq 2 11 the Bond stress transformation matrix M0 is given by cos2 9 sin2 9 0 sin 26 0 0 0 0 0 0 sin2 9 cos2 9 0 sin29 0 226 Mg sin292 sin292 0 cos29 0 0 1 0 0 0 0 cosQ 0 sing O 0 0 0 sing O cosQ Transformation C0 of C becomes C9 M9C Mg M9M CM5M3 227 Step 3 Now rotate the system xl xfmg39 by an arbitrary angle 0 about x axis Clearly in this case Mgo becomes the same as M with 1 replaced by 0 This completes the general rotation and the rotated matrix C becomes C MWCgM MWMgMngMiMEM 228 This can also be written as C MCMT 229 where M MWMBMV 230 HOMEWORK 1 Show that the scalar product of two vectors is invariant under orthogonal transformation 2 Show that the elastic coefficient matrix for the isotropic medium is invariant under orthogonal transformation 2 3 Hexagonal Symmetry We will now derive the elastic coef cient matrix C for hexagonal symmetry which is used to mean invariance of matrix to the rotation about one of the three coordinate axes When this axis is taken as x3 x3 vertical the medium is often called transversely isotropic Since the elastic coef cient matrix is symmetric it can be written as C11 C12 C13 C14 C15 C16 C12 C22 C23 C24 C25 C26 C2 C13 C23 C33 C34 C35 C36 231 C14 C24 C34 C44 C45 C46 C15 C25 C35 C45 C55 C56 C16 C26 C36 C46 C56 C66 Let this be the coef cient matrix with respect to a coordinate system x1x2x3 Now we rotate the coordinate system by an arbitrary angle 1 about x3 axis so that the new coordinate system becomes x1 x2 x3 We know that the coef cient matrix C in this rotated coordinate system is given by Eq 224 as C MCMT 232 with Eq 223 cos2 1 sin2 1 0 0 0 sin 2 sin2 1 cos2 1 0 0 0 sin 2 0 0 l 0 0 0 M 233 0 0 0 cos 1 sm 1 0 0 0 0 sin 1 cos 1 0 sin22 sin22 0 0 0 cos2 11 For the medium to be hexagonally symmetric we have C C From 232 and 233 we have cos2 1 sin2 1 sin2 1 cos2 1 0 0 C 0 0 0 0 sin22 sin22 0 0 0 C11 C12 C13 C14 C12 C22 C23 C24 C13 C23 C33 C34 C14 C24 C34 C44 C15 C25 C35 C45 C16 C26 C36 C46 This leads to 21 linear following results p p p p p p 0 0 0 sin 2 0 0 0 sin 2 l 0 0 0 0 cos 1 sin 1 0 0 sin 1 cos 1 0 cos2 5 C16 cos2 1 sin2 1 0 0 0 sin2 N2 5 C26 sin2 1 cos2 1 0 0 0 sin2 W2 5 C36 0 0 1 0 0 0 5 C46 0 0 0 cos sin 0 5 C56 0 0 0 sin cos 0 6 C66 sin 2 sin2 0 0 0 cos 234 equations which after some algebraic manipulations give the C34 C35 0 C C55 C45 0 CE C23 C 0 235 C14 C C46 C15 C25 C56 0 C11 C23 C12 C11 2C66 C16 C26 Thus the elastic coef cient matrix for the hexagonal symmetry is given by the following C11 C11 2C66 C13 0 0 0 C11 2C66 C11 C13 0 0 0 C C13 C13 C33 0 0 0 0 0 0 C44 0 0 0 0 0 0 C44 0 0 0 0 0 0 C Therefore we require only ve elastic constants C117C137C337C447C66 to describe the elastic coef cient matrix in a transversely isotropic medium 24 Positive de nite conditions Requirement of positive de nite strain energy is C e k1200i ey 20 237 W 1 Thus we have C lijElEkl CUEIEJ gt 0 238 This means that the 6X6 matrix C is a positive de nite For hexagonal symmetry this requires that C66 gt0 C44 gt 0 C33 gt 0 C11C33 gt C1235 C C gt Cf3 C C 11 33 33 66 39 239 or C33C11 C66 gt C123 240 C11 gt C66 Ifwe compare with isotropy Eq 110 we obtain gt0 12 gt0 A2yl2ygt23 241 3y 2y2 gt 0 A g kbulk mod ulus gt 0 REFLECTIVITY 4 Propagators and Wave Propagators for Our Seismic System We will work with our system of ODE 32b 1041b f for a general strati ed anisotropic medium Eigenvectors and Eigenvalues We denote the eigenvector matrix of A by D ie each column of D is an eigenvector of A Thus ADDA where A diag11A2A6 It can be shown that As are the vertical slownesses Why is D useful To see why rst de ne a vector v such that v D1b by left multiplication of the displacement stress vector b with the inverse of the eigenvector matrix and call it a wave vector for reasons which will be made clear later Recall 32b iaHb BZD v z39wDAD1Dv D132DV iwAv D1BZDV 82 V iwAv In a homogeneous region 32D 0 and we have 32v I39m1v whose solution is given by vz eimMZ ZO vzo l9 lame 20 is called the wave propagator since it or vz Qz 20 vzo where Qzzo e propagates the wave vector from z to 20 Thus we notice that each entry in v is proportional to the amplitude of an upgoing or downgoing plane wave depending on the sign of the exponent component and hence the name wave vector for V Reflectivily4 Page 1 March 28 2002 Now since bz 1M2 D 1bz vz emquot 0 vzo bz DemAz ZOD1bzo bz Pzzo bzo where PzzoDeiWAzZOD1 20 Thus in a homogeneous region the displacementstress vector b can be propagated from a depth 20 to a new depth 2 by using the propagator matrix Pzzo Note that Pzozo I How do we get a formula for Pzzo over an inhomogeneous region Divide our model up into thin homogeneous layers and use the product rule for propagators PZZOPZZNPZNZN 1PZIZO 21 Wave vectors are useful for applying radiation type boundary conditions Since the components of the wave vector v are the coefficients of solutions that propagate with different in general vertical slownesses we are justified in naming the components of v For example we may call the first component of the wave vector the Pwave amplitude Recall that in a homogeneous region we have for fixed px pya iapxxpyyA1 z 20 tv vxyt 2 6 Jay 132 This is a plane wave propagating with plane wave velocity C1 given by 1 2 2 2 2 C1 x py A1 Thus A1 quotlC12 pi pJ2 gives the vertical slowness We will use the following notations as used in Fryer and Frazer 1984 Reflectivity4 Page 2 March 28 2002 A1 q A2 31 A3 qu 22 A4qg A5qg2 A6qg3 Thus we shall assume that the lSt column of D is the eigenvector associated with upgoing P wave the 2nd column is the eigenvector of the upgoing S1 wave etc That is after we obtain D and A from A by SVD or whatever we rearrange terms in A and D It is fairly straightforward in isotropic media since up and downgoing can be identified from the sign of the eigenvalue and vp gt vs The eigenvalues for the two S waves are identical for isotropic media S V and SH But how do we do this for anisotropic media Given our choice of FT and depth of A if follows from the radiation condition that ImqDgt 0 and ImqUlt0 As in the isotropic case the elements of v may be identified with the amplitudes of upward and downward traveling plane waves T T vquotU VD U UXU D DXD 23 Summa so far The propagator matrix is needed to propagate the displacementstress vector b from one depth level to another Recall that this requires that we find the eigenvalues vertical slownesses the eigenvector matrix D and its inverse D1 In the isotropic case these are known analytically so the construction of the propagator is straightforward In the anisotropic case analytic solutions can only be found for some simple symmetries So in general solutions will be found numerically Fortunately D and its inverse are very simply related as we shall show I Wave in an 39 region iwAz zo v iwAzo We showed above that vz 6 z0 equation 19 Thus 6 is the matrix that propagates the wave vector from 20 to z and so we refer to it as the wave propagator In an inhomogeneous medium the propagator will not have such a simple form For example suppose that the medium has a discontinuity at 21 we can derive the wave propagator from 20 to 2 as follows Reflectivity4 Page 3 March 28 2002 medium 0 1 medium 1 Recall that bz Pzzobzo and b Dv Dill z D1711 z1P Z1 zo blzo Dilpz z1 PZl zo blzo 13171D1 3MW7zl 13171130 3iwuzlif0 D61bzo ia le zl Dleoeia lei zo Vz0 leazoVzo vz vz Vz Therefore QZZo eiWAZ 21D1 1Doeiwlzl ZO Wave propagators are usually called Q just as propagators are usually called P Note that we can write the propagators in terms of the wave propagators and Vice versa 2 Pza 20 W20 vz Qz 20 vzo and 112 Dzvz and 1720 DZovzo bz DZQZZ0Z0 DZQZZOD 1ZOI7ZO Reflectivity4 Page 4 March 28 2002 Thus Pzzo DzQzzoD 1zo QZZo D1ZPZa Zo DZo We know that P22 20 P22 21Pzlzo Therefore D22Q22 Zo051zo DZzQ22ZlD11 Zo11QZlazoD51 20 DZZQZZZIQZI Zo DSIZo or Q22Zo Q22521Q21Zo Where zi s are the depths to the discontinuties It is important to note that P is continuous at interfaces in the medium but Q is not That is 2 I is not an indentity matrix Reflectivity4 Page 5 March 28 2002 Lecture 11 Orth agonality of particle displacement vectors In our analysis of motion in homogenous anisotropic solids we showed that there always exist three types of motion We will now show that they are always mutually orthogonal For this we will start with the following Christoffel equation we assume that for a given propagation direction A the three solutions are u11 withslowness 1 w ul2 withslowness and 2 ul3 withslowness w m We take one of the solutions say a and write into the Christoffel equation to get R 1 WE ux L a and dot both sides of the above equation with another solution say b to obtain a un twpltuigtltuigt Repetition of the same procedure with the roles of a and b interchanged we obtain mite gtltugt Now recall that F is symmetric Hence the subscripts 139 and j in the left hand side of the above equation can be interchanged Therefore we obtain Eli Iluzmpl I Ilw Thus if the phase velocities are different the particle displacements are orthogonal When the two phase velocities are equal the solutions a and b can be combined into solutions that are invariant if they are not already so LECTURE 8 Hexagonal Symmetry Transverse Isotropy The elastic coef cient matrix for a hexagonally symmetric medium where x3 is the axis of symmetry is given by C11 C11 2C66 C13 0 0 0 C11 2C66 C11 C13 0 0 0 C C13 C13 C33 0 0 0 0 0 0 C44 0 0 0 0 0 0 C44 0 0 0 0 0 0 C The Christoffel equation for such a medium is given by p2 C11112 C66122 C44132p p2 C12 C66llll p2 C13 C44 lllS 1 p2 C12 C66llll p2C66112 i CVllll2 p2 C13 C441213 2 p2 C13 C441113 p2 C13 C44 lzls p2 C44112 C44122 C33132 p 3 where C12 C11 2C66 This gives the characteristic equation for the hexagonal symmetry as 172C1111Z C66ZZZC44Z3Z p 172C12 C66lllZ 172C13 C44lll3 72 C12 C66lllZ pZC66112 C11122C44l32p 172C13 C44lzl3 0 172C13 C44lll3 172C13 C44lzl3 172C44112C44122 C33132p For propagation in xlx2 plane we set 3 0 to obtain the following Chritoffel equaion 72 CllllZC66122 p 72 CIZC66ZIZZ 0 ul 72 CIZC66ZIZZ 72 C66112ClllZZ p 0 quot2 0 0 0 172C44112 C44 p 3 and the characteristic equation 72 C11112C66122p 72 C12C66lllz 0 72 CIZC66lllZ 72 C66112ClllZZ p 0 0 0 0 72C44112 C44lzzp Thus for propagation in the xlx2 plane in a hexagonal crystal with x3 as the aXis of symmetry the wave motion decouples into a linear part p2C44p0 or p1 Ci 44 It is clear from the above equation that it corresponds to particle the z direction or normal to the plane of propagation SH mode The determinant equation can be expanded as 172C1111Z C66122 ppz Cssliz Cnlzz ppz C4411Z C44122 p 172 C12 C66llllpz C12 C66llllpz leiz C44122 p 0 01 172C1111Z C66122 pp2 C66112C11122 pp2 C44p 172C12 C66llllpz C12 C66llllpz C44 p 0 Therefore the quadratic part is give by 172C1111Z C66122 ppZC66112 Cnlzzppz C12 C66lllzpz C12 C66lllZ 039 which reduces to the following equation picss pp2C11 p 0 Thus the quadratic part factorizes into the following L pz C66 and L P3 C To find the direction of polarization for these solutions we find from the first row of the matirX u1 p2C11 C66llll I 2 p2 C11112 C66122 p For solution 2 with p we have C66 Ci C11 C66 llll C11 C66 llll Z 2 CLSC1115C661 p C1111C6612 C66 6 S N l I Similarly for solution 3 with p we have 11 2 Cnll2 Csslzz C11 For waves propagating in xlx2 plane A along the direction of propagation is given by From Eq 267 the particle motion 11 for solution 2 is proportional to C C ll A A CS1073 11 1 66 2 66 Thus we have C11 C66llll Cllllz C66llz C66 C11C66llzll 42 f4 C1111 C66ll C66 x1 2 z u11 clzc2 Thus solution 2 is pure shear polarized in xle plane the plane of propagation Similarly we can show that solution 3 is pure longitudinal Thus for propagation in xlx2 plane in a hexagonal crystal with x3 as the axis of symmetry the motion polarizes in three orthogonal directions CL Corresponds to pure shear motion along x3 aXis normal to the 44 Cpl plane of propagation 7 a pure SH mode Ci Corresponds to pure shear motion in the plane of propagation 7 a 66 Cpl pure SV mode 0 p3 CL Corresponds to pure longitudinal motion 7 pure F wave 11 Thus in the xlx2 plane three pure modes exist and they propagate with different velocities Note also that there is no azimuthal variation of the above three wave speeds so the slowness curves are circle like an isotropic medium This is the reason to call a hexagonally symmetric medium with x3 the vertical as the aXis of symmetry transversely isotropic medium LECTURE 9 TI MEDIUM Contd Now we consider propagation in the 9cle plane We have seen that the elastic coef cient matrix for the medium under consideration remains the same for any arbitrary rotation about the x3 axis The dispersion relation that we will derive now is valid for propagation along any meridian plane containing the x3 axis For waves propagating along 9cle plane we set 12 0 so that the unit vector along the direction of propagation is given by i 11 112 3 Therefore the Christoffel equation becomes P2 C11112 C44132p 0 p2 C13 C44 llls 1 0 p2C66112C44132 p 0 Hz 0 p2 C13 C44 lils 0 p2 C44112 Csslsz p 3 and the corresponding determinant equation that we need to solve is p2 C11112 C44132p 0 p2 C13 C44 lllS 0 p2C66112C4432 p 0 0 p2 C13 C44 0 p2C44112 C3313 which decouples into a linear part pfC lzC4412 p0 661 p 1 C66 sin2 19C44 cos2 9 and a quadratic part 2 p2 C11112 C44132 pp2 C11112 C4413 p p4C13 C44 0 The rst solution corresponds to pure shear polarized normal to the plane of propagation x3 direction 7 a pure SH mode The slownesses however is dependent on azimuth single valued in the form of an ellipse and it becomes a circle only when C11 C44 To see the behavior of the second equation we need to do some algebra after which it reduces to the following form 2 32 C11 sin2 9 C33 cos2 9 C44 C11 C44sin2 9 C44 C33cos2 9 C13 C44sin2 26 2p 0 Thus we nally get the three eigenvalues for propagation in 9cle plane or any meridian plane containing 963 for a hexagonally symmetric medium with x3 as the aXis of symmetry p 1 l lt p1 C66 sin2 19C66 cos2 9 2 P2 12 C11 sinz 9 C33 cosz 9 C44 C11 C44 sinz 9 C44 C33cos2 92 Cu C sinz 29 3 pg M C11 sinz 9 C33 cosz 9 C44 C11 C44sin2 9 C44 C33cos2 92 Cu C sinz 29 12 Solution 1 is pure shear with polarization in x2 direction Solution 2 is quasi shear wave and solution 3 is quasilongitudinal wave The direction angle 9 is measured from the Z axis At 9 0 we have the following p 1p gt1 CM p 2p gt2 CM p 3p gt3 C33 At 9 we have the following p 1p gt1 C66 p 2p gt2 CM 3103 At 9 we have the following 1 p1 C66 fa 1 2 p2 4p 1 C11C33 2CM C11 C332 4C13 C442 F l 3 p3 2 4p2 1 C11 C33 2C44 C11 C33 2 4C13 C44 2 2 where l 2 and 3 correspond to SH quasiSV and quasiP respectively Thus we note the following 0 For vertical propagation SH and SV propagate with the same phase velocity Given the density the elastic constant C44 determines the vertical velocity for SH and SV while C33 determines the vertical velocity of F wave 0 For horizontal propagation P SH and SV propagate with different velocities Given the density C66 determines the SH velocity C44 determines SV velocity and C11 determines the P velocity From the positive definite conditions section 14 we have C11 gt C66 Eq 152 thus F wave travels faster than SH 0 For all other angles the velocities are angle dependent LECTURE 6 21 Fundamentals of wave propagation in a homogenous anisotropic medium We start with the constitutive relation TC CVu 21 where T is second order stress tensor C is a fourth order elastic tensor 8 is a second order strain tensor and u is the displacement vector The linearized momentum equation without the source term is given by V 139 pii 22 where p is the mass density Substituting 22 into 2 l we obtain v 139 VrCVu 23 Now we specialize to Cartesian axes system for an anisotropic homogeneous medium and rewrite 23 in the following indicial form ijlu1kx p 24 This is the general elastodynamic equation for a homogeneous medium in CA8 Our aim is to nd a solution to this equation The most common trial solution is a plane wave solution de ned as the one which depends on position and time in the following way Figure 21 For all points along the planes is a constant K is the position vector l is the unit normal vector to the plane 0 At any time assume u constant along parallel planes with a unit normal 0 In time the solution advances in i direction with a phase speed c so that uxt U3exp iwl39 X t U3expicopx t 25 c f where we have 1ntroduced the slowness vector p Recall that c for Pwaves in isotropic media d l 1 and or Swaves in isotropic media di0 We now substitute this form into our wave equation to obtain CWka 611 pd 26 or ijlka p5 d 0 or r p511 d o r pl30 27 Both the above equations are known as Christo el equation whose solutions give the eigenvector of all possible wave motions For a nontrivial solution the determinant of the Christoffel matrix F pl vanishes Also note that the Christoffel matrix F is symmetric due to symmetry of C and positive de nite Now acceptable slowness vectors satisfy CWM 95 0 28 f A If p pomts to real directions 1n space p 1 1s a real un1t vector w1th components I c then C it cz5 dl 0 29 p The CONCEPT 0f the EFFECTIVE MEDIUM Helbig 1998 The cause of seismic anisotropy is always some internal structure on a scale that is small to the resolution of the method applied This does even include the anisotropy of crystals the internal structure is the rearrangement of the ions on a 3D lattice on a scale that is small compared to optical wavelengths though not small to the wavelengths of X rays Internal structures in geological media that can lead to anisotropy are 0 Oriented cracks 0 Lamination periodic sequences of thin layers 0 Parallel fractures 0 Oriented grains 0 Clay orientation of platelike minerals In many seismic observations in seismic frequency band that does not resolve the elements of the internal structure these media appear to be transversely isotropic One often calls the anisotropy of an effective medium apparent anisotropy or quasi anisotropy to distinguish it from intrinsic anisotropy This distinction is however not necessary as the anisotropy due to crystals is also due to ordered internal structure of the ions In addition to material anisotropy there are conditions where the propagation of waves is anisotropic even if the medium is isotropic Such conditions could be moving coordinate systems such as the propagation of sound in an air or water medium wind current either at rest observed from a xed platform or in a moving medium wind ocean observed from a xed platform The propagation of electromagnetic waves in plasmas becomes anisotropic under the in uence of external magnetic eld Anisotropy due to periodic layering Geophysical media often exhibit anisotropic behavior due to alternating strata of material each stratum itself being isotropic This can occur over a range of length scales When probed with radiation of a wavelength much larger than width of the strata such stratified regions exhibit material properties that appear to be space independent but direction dependent depending on the angle with respect to the axis perpendicular to the plane of stratification Such a medium is said to be transversely isotropic and has ve independent elastic moduli Here Iwill closely follow Schoenberg 1983 to derive the equivalent TI parameters based on the quasistatic considerations as outlined in Helbig 1968 Schoenberg 1983 also derived an exact formulation for periodic media using propagator matrices but I will not discuss that here We will now follow Schoenberg 1983 to derive the elastic behavior of periodically arranged ne layers Let 2 be the axis perpendicular to layering and let the period of layering is H Assume that one period is made up of N homogeneous layers each with shear modulus ui Poisson s ratio Vi and thickness hH i123 N Let Vibe de ned as the ratio of shear speed Bito compressional wave speed 0g so that y 1 For stress and strain elds whose scale of variation is much greater than H effective transversely isotropic modulii can be derived in terms of the u 7 and hi There are ve independent elastic constants 111 C11 C12 C13 0 0 0 11 122 C12 C11 C13 0 0 0 22 13933 2 C13 C13 C33 0 0 0 33 123 0 0 0 2G 0 0 23 13931 0 0 0 2C44 0 31 112 0 0 0 0 0 2G 12 2 where C12 C11 2C66 Now we make the following arguments 0 Under quasistatic or low frequency equilibrium requirement the field is assumed to vary slowly with respect to H the spatial period of layer aggregate That is the stresses that act on a face perpendicular to the x3 aXis ie 13913 13923 and 133 are assumed constant continuous across a set of layers of width H while 1391 1 13922 and 13912 are discontinuous 337 0 The layers are in welded contact Since all the layers are constrained to have some deformation in the x1x2 plane strains that lie in a plane parallel to the layering ie 11 22 and 812 are assumed constant continuous This is because the layers are constrained to having the same inplane motion in a medium of infinite extent in the x1 and x2 directions The 33 23 and 813 are discontinuous The 33 23 and 813 components of strain over a full spatial period H can be written in terms of the strains of the individual layers For example let the displacement components be denoted by u 1 M2 and u 3 The strain 633 for the layerz39 is given u bottom 20p 3 3 Ausx hH hH x x 1 833 The average strain 833 over a full spatial period is given by g u3 H u3 1 x3 H x3 H N Au31HiZthEis lt 833 gt a 3 11 where ltgt denotes a thickness weighted average Similarly the average strains 823 and 13 across a full period can be found using displacements u 2 and u 3 respectively to be given by H N N elf 2m quot2iZAu21iZhIe lt2 gt 4 1 1 and e ulx3Hulx3 iZN1Au iZN1heJ lt 31 gt 5 1 The inplane average stresses may be derived in a similar way If the force per unit length in the Xldirection on an X2 face the stress The average stress across the full width 1 N I 1 N N X T12 Zf12 Zh1HTi2 Zh1712 ltTi2gt 39 6 H 11 H 11 11 Similarly we have 1 N I 1 N N T11Zf11 ZI 1HTi1ZI 17i1 ltTi1gta 7 H 11 H 11 11 and 1 N 1 1 N N 722 Zf22 ZhIHTlZZZhITZ ltTl22gtI 8 H 11 H 11 11 Now consider the relations between shear stress and shear strain In each layer 1 f 2uxeal 9 From the rst of Eq 9 we have 123 8123 a 2M 01 139 323 gtlty 1 gt 31lt 1 gtlt1 gt 10 712 2ltIu gt 12 Comparison between Eq 10 and Eq 2 reveals that lttt 1 gtC66ltygt Now consider the relation between normal stress 13933 and normal strains using isotropic Hooke s law In each layer K or 7331 2lt ygtltgt 143114322ltgt 1 33 Comparison between 1 lb and 2 reveal that I C33ltyygt 1C131 2ltygtltyygt 1 11a 11b Now consider nally the relation between the normal stress 13911 and the normal strains In each layer we have Til 7 i1 En 12y1 22 833 From Eq lla and 12 we have 12 7 i1 811 12yx 822 127x 733 127x 811 822 13 21u1811 21271Iu1 11 ZZ1271133 Averaging and substituting for 13933 from Eq12 into Eq 1 l we have 1112 lty gt8112lty gt 4lt m gt1 2lt gt2 lt yy gt 1 11 822 14 1 2ltgtltygt 1 33 Comparison between 14 and 2 reveals that C114ltygt 4lt tgt1 2ltygt2 ltyygt 1 Thus we were able to express all ve coef cients in terms of the elastic parameters of the constituent layers HOMEWORK Show that for the TI medium that is a longwavelength equivalent of ne periodic layers the C66 2C44 SOLUTION TO HOMEWORK PROBLEM Prove that C66 EC44 for transversely isotropic media generated by ne isotropic layering We know that C66 lt u gt and C44 where lt1ygt N N h N ltygtZhlyl and lt1ygtZ with Zh1 11 11 I 11 Thus C66 2C44 means ltu gt2 m lt1ygt ltygtlt1ygt21 1 In other words we need to prove 1 Recall that for any numbers a1a2a3a4an and b1b2b3b4bn we have 6112a22 ab12b b a1b1a2b2anb2 2 a1b2 a2b12 alb3 a3b12 quot39 20 Therefore in general 26132 42611231 2 Now let ax I h 1 and bl This is possible since ul is always positive and so is h Therefore or EFFECTIVE MEDIUM A GENERAL DERIVATION In the previous section we followed Schoenberg 198x to derive an expression for an equivalent TI parameters for periodically strati ed isotropic layers Here we will closely follow Schoenberg and Douma 1988 and derive a generalized expression for long wavelength anisotropy due to 0 Periodic anistropic layers and 0 Parallel fratures and aligned cracks Consider a strati ed medium made up of perfectly bounded homogeneous but not necessarily isotropic layers Let the x3 aXis be perpendicular to the layering and assume that there are 71 different constituent layers arranged so that in each suf ciently large interval one nds the same proportion of each medium e g a periodic sequence of layers Each anisotropic constituent has a relative thickness h il2 n so that hl h2 hn 1 a density g and an elastic modulus tensor 0 relating sress I m to qprs strain 85 In condensed notation we have T1 711 C11 C12 C13 C14 C15 C15 81 11 T2 722 C12 C22 C23 C24 C25 C25 82 22 73 733 2 C13 C23 633 634 635 635 83 833 T4 723 614 624 634 644 645 646 4 2323 75 731 C15 C25 635 645 655 655 85 2 31 ma 1 N HQ 0 Q a Q a Aquot a MG 0 Q a an N Hm N The elastic moduli of the equivalent anisotropic medium in the long wavelength quasi static limit due to a layered medium composed of anisotropic constituent layers can be expressed in terms of thicknessweighted averages of functions of the moduli of the constituents The longwavelength assumption on stress is that all stress components acting on surfaces parallel to the layering are the same in all the layers ie The long wavelength kinematic assumption is that over many layers the layers move together so that derivatives of inplane displacements with respect to in plane coordinates x1 and x2 are the same implying that all strain components lying in the plane of the layering are the same in all layers ie 1 1 1 1 1 1 811 81 El 22 82 82 2812 86 86 The other stress and strain components may vary from layer to layer In each layer such a component may be taken as its average value across the thickness of that layer Based on thee ideas we de ne the following vectors layer dependent m N 1 we Hm I No 3 layer independent 0 0 These allow the stressstain relations in any layers to be written as 1 I I I Sl M E1PE2 3a s P TE N E 3b 2 1 2 where 611 612 616 633 CS4 635 613 614 615 1 I I I 1 I I I 1 I I I M 612 622 626 N CS4 C44 C45 P 623 624 625 616 626 666 635 C45 655 636 C46 656 Note that M and N are symmetric matrices From Eq 3b we have N 1s2 N 1 P TE1E 2 lt gt lt H gt 5 1 EN1 1s2 N 71P T E From 3a and 5 we obtain sM ElP 1s2 N 1P TE1 6 Now following the previous section we let the weighted averages over all the constituent layers be denoted by ltgt From 6 we have S1ltM ltPN 1P7gt E1 ltPN 1gtS2 7a and from 5 we have s2 ltN 1gt71ltN 1P7E1ltN 1gt71ltE2 7b Finally substituting the expression for S2 from 7b into 7a we write the elastic moduli for the media equivalent to the strati ed medium in the long wavelength limit in matriX form as S1M2El P2 E2 8a s2 PZTE1NZltE2 8b with 9 N2 ltN 1gt71 P2 ltPN 1gtN2 M2 ltM ltPN 1P7ltPN 1gtNZN IPT Thus we have the desired result in Eq 9 Given a stack of homogeneous anisotropic periodic layers equivalent long wavelength anisotropic parameters can be evaluated using eq 9 SPECIAL CASE The ith constiuent is transversely isotropic with the x3 aXis as the aXis of symmetry 1 1 1 1 611 cu 2666 0 633 0 1 I x M 611 2655 611 0 N 0 644 1 0 0 cw 0 Note that when the constituent layer is isotropic Q 1 1 1 1 1 644 666 Il ll 5 11 633 l2 1 5613 13 0 03 0 P cis c 0 If all the constituent layers are transversely isotropic the equivalent homogeneous medium is transversely isotropic and from 9 the moduli are given by 033 0 0 1 N2 0 c44 0 0 llt gt 0 633 0 0 c44 0 0 lltigt 633 633 633 013 0 0 013 1 P2 6 13 0 0 0 0 0 0 0 633 633 0 0 0 611 cu 2666 0 611 cu 2 666 0 M2 611 2666 611 0 611 2ltcssgt 611 0 9 0 0 cw 0 0 cw where identical to the result of Backus 1962 Reference Backus G E 1962 Long wavelength anisotropy produced by horizontal layering Journal of Geophysical Research 66 44274440 Schoenberg M 1983 Re ection of elastic waves from periodically strati ed media with interfacial slip Geophysical Prospecting 31 265192 Schoenberg M and J Douma 1988 Elastic wave propagation in media with parallel fractures and aligned cracks Geophysical Prospecting 36 571590 Helbig K 1958 Elastische wellen in anisotropic medium Teil II Gerlanals Beitrage zur Geophysik 64 4 256288 Helbig K 1998 Layerinduced elastic anisotropy 7 part 1 Forward relations between constituent parameters and compound medium parameters ReVista Brasileria de Geo sica vol 16 23 103112 Lecture 1 12202 FUNDAMENTALS 11 Two Fundamental Equations of linear Elasticity 1 The linearized momentum equation piiVTf 11 and 2 The constitutive relation Generalized Hookes law TCISCIV11 12 where r is second order stress tensor C is a fourth order elastic tensor 2 is a second order strain tensor u is the displacement vector p is the mass density f is a body force term Substituting 11 into 12 we obtain VrTVrCVu 13 Now we specialize to Cartesian axes system for an anisotropic homogeneous medium and rewrite 13 in the following indicial form ijluUa p j39 14 This is the general elastodynamic equation for a homogeneous medium in CA8 Equations 11 and 12 can be written in indicial notations as Pill Tw 15 and 7 ijlukl39 16 12 Vectors Tensors Isotropy Anisotropy 121 Vectors In elementary physics course the geometric aspect of vectors is emphasized A vector V is rst conceived as a directed line segment or a quantity with both a magnitude and a direction such as a velocity or a force A vector is thus distinguished from a scalar a quantity that has only magnitude such as temperature entropy or mass X3 In a right handed Cartesian aXis system as shown in Fig l the vector can be written in terms of its scalar components as follows VV1fl v1f1vzfzv3f3 17 where a summation over repeated indices is assumed and the scalar components are given by v vf 18 where a represents a scalar or a dot product 122 Scalar or Dot Product of two vectors The scalar inner or dot product of two vectors V1 and V2 is the real number de ned in geometrical language by the equation VI39VZV1 V2COSQ 19 where 9 is the angle between the two vectors measured from V1 to V2 The scalar product is commutative 123 Vector or Cross Product of two vectors The vector cross product of two vectors V1 and V2 is a vector and is written by VV1gtltV2 110 and VV1V2Sln9 111 where 9 is the angle measured fromV1 to V2 in such a way that 9 Sir The vector product is anticommutative 124 Dyads Tensor Products Let 11 V w y etc be vector elds Then 11 V is a dyad 11 V W is a triad uva isannadetc Mathematicians write dyads as u V 125 T ENSOR A tensor of order n is a scalar valued function of n vector variables linear in each variable For a second order tensor we have gt T A z 1 7i1 T12 1562 T13 15634 TZI A2amp1 ZZ 562562 7quot23 5625634 112 T31 321 T32 fcjcz T33 fcjcs Thus a second order tensor is a sum of dyads A fourth order tensor can be written as CC fcfcfckz 113 125 Double Dot Product Sometimes we use to mean that if AZAlal and 323 I J ABZZA1Bjal j 114 I J More simply uvwxuwvx 115 which is a scalar 13 Isotropy and Anisotropy The terms isotropy and anisotropy are basically meant to describe tensors 0A tensor is said to be isotropic if its components with respect to some coordinate system do not vary if either the coordinate system or the medium is rotated By rotation we mean either a pure rotation where the determinant of the rotation matrix is l or re ection where the determinant of the rotation matrix is 71 Else the tensor is said to be anisotropic In order to discuss anisotropy we must rst understand isotropy For this let us rst take a look at what tensors are isotropic 131 Isotropic Tensor of order one Consider V to be a vector tensor of order one and its components v1v2v3 with respect to some Cartesian aXis system c1x2x3 with bases fcljczjg We now rotate the coordinate system to with bases Let the new components of v in this new rotated system be vi vvz v 1 Thus 116 For V to be isotropic we require that v v and we immediately see that this is possible if El 93 ie no rotation or V v 0 Thus the only isotropic tensor of order one is the zero or the null vector 132 Isotropic Tensor of order two Using the same convention for any A we write 117 AUA AA If A is 1sotrop1c AU AU and we 1mmed1ately see that it 1s poss1ble if A xli quj Therefore the isotropic tensor of order two is the identity tensor I with components 51 j 133 Isotropic Tensor of order three A tensor of order three can be expressed as combinations of tensors of order one and two and since the only isotropic tensor of order one is the null vector there can be no isotropic third order tensor except a zero third order tensor ie a third order tensor with components equal to zero 134 Isotropic Tensor of order four A tensor of order four can be regarded as a combination of tensors of order two We know that the isotropic tensor of order two is the identity tensor Since identity tensor is symmetric possible combinations of two identity tensors are 5555 U M k N 5 5 Thus three basic isotropic tensors of order four must be 11 1k aw 615k15 b1ij 6xk6 dykl 511 611539 and the general fourth order tensor is a combination of these three given by CW 616161616261k6j16361161k39 118 14 Isotropic Elastic Medium An isotropic elastic medium is the one in which the fourth order elastic coefficient tensor which relate the stress and tensors by the relation 139 C E l l 9 where T is a second order stress tensor C is a fourth order elastic coefficient tensor 8 is the second order strain tensor and represents a double dot product Now since 239 and C are symmetric C C W W C j k1 C 111k 120 Therefore we have 6151165161 6251k51 635115jk c551kc255 03 k5 121 11 111k 61515 635114311 Giglgjk using the property 5H 51k Therefore we have 02 03 Using A c1 and u CZ 03 we obtain the familiar form of the elastic tensor for an isotropic elastic medium CW 15115k1 Il l5lk5jl 6116 122 The stressstrain relation for an isotropic elastic medium thus becomes 1U CWEH Adj klekl y61k5 5115ka or 139 AVu1 uVu VuT 123 Now if we restrict ourselves to Cartesian axes system CAS in which case the superscript and subscripts lose their meanings we arrive at the following relationship between the six independent components of stress and strain 111 A 2mg11 2322 1333 122 1311 A 2y 22 2333 133 1311 2322 A 2y 33 124 13923 2y 23 13931 2 831 13912 2312 The above can be expressed in the form of a matrix equation as follows 13911 1 2 u A A 0 0 0 11 13922 1 1 2y 1 0 0 0 22 13933 2 A A A2y 0 0 0 833 125 13923 0 0 0 u 0 0 2823 13931 0 0 0 0 u 0 2831 13912 0 0 0 0 0 u 2812 The above is the stressstrain relationship for an isotropic elastic medium written in a condensed notation 15 Anisotropic Elastic Medium From now on we specialize to a linear elastic medium which is anisotropic Such a medium exhibits a constitutive relation which in indicial notation becomes 71 Cyklgkl 126 Since 239 and E are symmetric each can have six independent components and they can be derived from the above equation as follows E ij i j ik 39 7139 Cymng 711 C1111311 C1122322 C1123 C1132 923 C1133333 C1113 C1131331 C1112 C1121 312 T11C11311C12522 C13333 C142323 C152331C162512 722 C2211311 C2222 322 C2233333 C2223 C2232323 C2213 C2231331 C2212 C2221 312 722 C21311 C22 322 C23333 C24 2 323 C25 2 331 C262312 733 C3311311 C3322322 63333333 C3323 C3332323 C3313 C3331331 C3312 C3321321 733 C31311 C32322 C33333 C34 2 323 6352331 C36 2 321 723 C2311311 C2322322 C2333333 C2323 C2332323 C2313 C2331331 C2312 C2321 321 723 C41311 C42 322 C43 333 C442323 C452331 C462321 731 C3111311 C3122922 63133333 C3123 C3132323 C3113 C3131 331 C3112 C3121321 731 C51311 C52322 C53333 C54 2 323 C55 2 331 C56 2 321 712 C1211311 C1222922 C1233333 C1223 C1232323 C1213 C1231331 C1212 C1221321 712 C61311 C62322 C63E33 C642323 C652 331 C662321 Here we used the following compact notation 11 gt1 22 gt2 33 gt3 23or32 gt4 13or31 gt5 120r21 gt6 Thus we have Tl C11 C12 C13 C14 C15 C16311 I 722 I I C21 C22 C23 C24 C25 C66 I I 322 I I C31 C32 C33 C34 C35 C36 I I 333 I C41 C42 C43 C44 C45 C46 I l 2 323 I LC51 C52 C53 C54 C55 C56 L2331j C61 C62 C63 C64 C65 C66 2 312 T 33 127 REFLECTIVITY 5 Re ection and Transmission A V Suppose that we have an earth model homogeneous above 21 and below ZN and inhomogeneous between Then it is reasonable to talk about the re ection and transmission properties of such a zone and to de ne RT coef cients ZN V we haVe bZN PZNZIbZl and Notation memen Z ZN where as before P is a propagator and Q is a wave propagator We write v vuvdT vu ugou xu T and vd d 0d xd T This means that in the equation b Dv we are using a D such that its rst column is an upgoing p wave etc We now solve two re ection problems ReflectivityS Page 1 March 28 2002 Vd f vuer wa L Z1 inhomogeneous region inhomogeneous region Vd vu Vd 1 Wave incident from above 11 Wave incident from below Case I ga39r ggyw l f 0 Q1 jQ1 Vu21 vdzirv 1 I 2 I j 34 Q11 Q12 Q2RZI Q21 Q22 which gives 0 Q11quotu 11L Q12Vd21 vd Zirv Q21Vu 21 Q22vd 21 or VuZi Q111Q12 dzi 35 1 Vd ZN Q22 Q21Q11Q1239d21 Thus we are justi ed in de ning matrices of re ection and transmission coefficients 1 1 RD Q11Q12TDQ22 Q21Q11Q12 36 Case 11 ReflectivityS Page 2 March 28 2002 9421 Q11 Q12 vb 5 37 We Q21 Q22 vu Q1 lvu Val 223 Q21Vu21 or vu r Q lm iv 38 1 vd ZN Q21Vu 21 Q21Q11Vu21v Thus we are justi ed in de ning l l TuQ11RuQ21Q11 39 Since vu z Ryder It is clear that D D S rpp r p1 VPZ R D D D D 71p 11 712 D D D 72p 721 V22 rig re ection coef cient for a downgoing Pwave to an upgoing pwave and so on Similarly D D u u u u u u til 1 IE tpp tpl tpzw rpp rpl r172 u u u u u u TDtga I15 I15 Tut1p I11 I12 RuV1p r11 3912 u u u u u u U217 I21 I22 V217 I21 I22 Vzp 21 2 We now put all ofthese PUT coef cients in a big 6X6 matrix ReflectivityS Page 3 March 28 2002 RRDETu Q111Q12 Q 1 40 TD ER Q22 Q21Q111Q12 Q21Q111 Remember that the Q in this equation is downward Q ie Q 21521 From equations 36 and 39 we have RD Q1 11Q12 41 TD Q22 Q21Q111Q12 42 Tu Q 1 43 R1 Q21Q1 11 44 From equation 43 Q11 T171 45 From equations 41 and 45 Q12 Q11RD Q12 71391 1RD 46 From equation 44 Q21 RuQ11 Ruin1 47 From equations 42 and 41 TD Q22 Q21RD Q22 TD Ru Tu 1 RD Thus we have Q z1 Qliigg r 1514 T 1Rf1 i 48 Q21 1sz LR T TD RuTu RDJ Thus we have expressed the wave propagator Q in terms of the re ection coefficients Reflectivity Page 4 March 28 2002 We can now derive a similar expression for the upward propagator Q f 2X Thus for the first case instead of using equation 34 we use v z I quot13JQ zfzv 0 J 49 VdZl d ZN and in the second case instead of using equation 37 we use v u Melif ENQ 12NWZIVJ 50 After doing similar algebra we obtain RD m 91293 Qf l 39 Ql Qj1Q21 TD Ru kQ f Q5931 and Q fZEZrQ1u1 Qllz ngl Q32 l T R TD1R RDTD1l 51 u D u Q l ZNH will TD This is the most fundamental lemma of Kennett 1974 from which almost everything that s any good follows Iteration Eguation We now establish the iteration theorem Recall that for 21 S 22 23 QZla Z3 QZla 22Q22Z3 Substituting equation 51 in the above equation we have Reflectivityf Page 5 March 28 2002 113 31 31 1 13 31 31 11 112 21 21 1 12 21 21 11 Tu RD TD Ru RD TD T RD TD Ru RD TD 1 l 1 91313 T1331 1 1 91332 T501 1 llTu23 R132 T52 1RL23 R132 T132 1 1 1 21353 r3211 1 Now doing the multiplcation and solving for R31 T51 RE and 733 in terms of Rgl etc we get 52 1 31 21 12 32 12 32 21 RD D Tu RD IRu RD TD 1 31 32 12 32 21 TD TD PRu RD 1 TD 1 1233 R33 TDR2 I 122qu Tu23 1 13 12 32 12 23 Tu Tu I RDRu IT 53 These are the iteration equations Since 21 S 22 23 is arbitrary it follows that we can compute the WT coef cients for an inhomogeneous zone by dividing that zone into a number of layers and iterating them layer by layer Interpretation of iteration eguations Use the identity I A1IA AA Now 1 R31 2 Rg TulZRgZI1 RllJZREEI T31 1259 12 Tylegrgl TJZRgZELZRg T51 TJZRmPRnggZRngI ReflectivityS Page 6 March 28 2002

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "I signed up to be an Elite Notetaker with 2 of my sorority sisters this semester. We just posted our notes weekly and were each making over $600 per month. I LOVE StudySoup!"

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.