Solutions to the Exercisesin A Breviary of Seismic TomographyYue Tian and Guust Nolet300Contents1 Preface page 52 Ray theory for seismic waves 63 Ray tracing 144 Wave scattering 225 Amplitudes of body waves 306 Travel times: observations 347 Travel times: interpretation 368 Body wave amplitudes: Observation and interpretation 419 Normal modes 4310 Surface wave interpretation: ray theory 4711 Surface waves: Finite frequency theory 5012 Model parametrization 5313 Common corrections 5814 Linear inversion 6015 Resolution and error analysis 6216 Anisotropy 6341PrefaceSolvingexercisesisessentialtomasterasubjectascomplicatedasseismicto-mography: it transforms the passive activity of reading the textbook into an activelearning process, provides feedback and deepens the understanding.Ideally, exercisesaredesignedsuchthattheyaresolvableusingtheknowledgegained, with more or less effort depending on the ability of the student.However,even a strong student may wish to check whether his solution is correct, whereasa weaker student may at times give up on a difcult problem. For that reason weprovide elaborate solutions to the exercises in A Breviary of Seismic Tomography.ThenumberingofthefollowingchaptersisidenticaltothenumberingusedinBreviary.52Ray theory for seismic wavesExercise 2.1 See Fig. B2.2; let the angle of rotation be. The change ofuxalong y-direction isux = ux(x, y + dy) ux(x, y) = dy sin 0 = dy sin.The change of uy along x-direction isuy = uy(x + dx, y) uy(x, y) = dxsin 0 = dxsin.Thusuxy=limdy0uxdy= sin, (2.1)uyx=limdx0uydx= sin. (2.2)Adding (2.1) and (2.2) givesuxy+uyx= 0.By denition of strain tensor, we have

12 = xy =12_uxy+uyx_= 0.The above proof can be repeated for all permutations of i, j = 1, 2, 3. Therefore apure rotation gives rise to ij 0.When numbers for gures or equations are preceded by a B they refer to gures/equations in A Breviary forSeismic Tomography.6Ray theory for seismic waves 7f(xct)(a) At time t0x(b) At time t + dt f(xct)0xdx=cdtFig.2.1. Functionf(x ct)representsapropagatingwavewithvelocitycinthe+xdirection.Exercise 2.2 The inverse Fourier transform givesp(t) =12_P()eitd.Inserting the condition P() = P(), we getp(t) =12_P()eitd = 12_P()eitd().Let w = , thenp(t) =12_P(w)eiwtdw = p(t).Hence p(t) is real.Exercise 2.3 Applying the chain rule of differentiation to any differentiable func-tion of the form f(x ct) yieldsPt= cf

,2Pt2= c2f

,Px= f

,2Px2= f

, (2.3)where f

and f

are the rst and second order derivatives with respect to the argu-ment x ct. Inserting (2.3) into (B2.19) we get c2f

c2f

.To see that the solution describes propagating waves, imagine that at t = t0, so thatf(x ct) represents a wave as shown in Figure 2.1a. At t = t0 + dt,the argumentx ct will remain unchanged if the distance x increases by dx = cdt, and thus theshape of function f(xct) will remain the same (Figure 2.1b). Therefore, f(xct)describes a wave propagating with velocity dx/dt = c in the +x direction.8 Ray theory for seismic wavesWavefront P=Pat time t0Wavefront P=Pat time t + dt00nr0drFig. 2.2. Equation (2.4) describes a plane wave propagating with velocity c in the directionof n.Exercise 2.4 Applying the chain rule of differentiation to any differentiable func-tion of the formP(x, y, z, t) = f(nxx +nyy +nzz ct) = f( nr ct) (2.4)yieldsPt= cf

,2Pt2= c2f

,Px= nxf

,2Px2= n2xf

,Py= nyf

,2Py2= n2yf

,Pz= nzf

,2Pz2= n2zf

; (2.5)where f

and f

are the rst and second order derivatives with respect to the argu-ment nr ct. Inserting (2.5) into the 3-D wave equation (B2.12), we obtainc2(n2x +n2y +n2z)f

= c2f

.Since n2x + n2y + n2z=1 ( n is a unit vector), the above equation is always true.Thus, (2.4) always satises the 3-D wave equation (B2.12).A wavefront is a surface of constant pressure (P=P0 in Figure 2.2). A constantargument of function f nr ct = constant (2.6)therefore guarantees constant P. At time t =t0, (2.6) represents a plane with itsRay theory for seismic waves 9normal vector being n, because all the points in this plane have the same value of nr (Figure 2.2). At timet=t0 + dt, (2.6) remains the same constant if theposition change dr satises dr n=cdt, and the simplest case is dr=cdt n.Therefore, (2.4) describes a plane wave propagating with velocity dr/dt = c in thedirection of n. (It can also propagate in other directions represented by dr so longas dr n = cdt is satised.)Exercise 2.5 Applying the chain rule of differentiation to any differentiable func-tion of the form (B2.21) yieldsPr= 1r2f + 1r(1c)f

= 1r2f 1crf

,2Pr2=2r3f 1r2(1c)f

1c( 1r2)f

1cr(1c)f

=2r3f +2cr2f

+1c2rf

,Pt=1rf

,2Pt2=1rf

, (2.7)where f

and f

are the rst and second order derivatives with respect to the argu-ment t r/c. Inserting (2.7) into the left hand side of (B2.18) we obtain2r3f +2cr2f

+1c2rf

+ 2r( 1r2f 1crf

) =1c2rf

.This describes a spherical wave. Its wavefront is a sphere centred atr=0 andexpanding with velocity dr/dt =c. Energy on the wavefront must be conserved.Since the energy per unit area is proportional to [P[2, and the area of the wavefrontis 4r2, conservation of energy requires [P[2 4r2to be constant. Hence we musthave [P[ 1/r, i.e. the pressure amplitude [P[ has to diminish as 1/r to conserveenergy.Exercise 2.6 The denition of slowness vector p isp =1cdrds, (2.8)where dr is a tangent vector of length ds along the ray. Since dr = (dx, dy, dz),(2.8) becomesp =1c_dxds, dyds, dzds_=1c(1, 2, 3), (2.9)i.e. the elements of p are equal to the direction cosines of the ray path scaled by1/c. Equation (2.9) gives the physical meaning of slowness vector p: its directionis the ray path direction, and its magnitude is equal to the slowness.10 Ray theory for seismic wavesExercise 2.7 By denition, the magnitude of the cross product is[r p[ = [r[[p[ sinr, p) (2.10)where r, p) represents the angle between r and p. In a spherical earth, this is theincidence anglei. We also have [r[ =r and [p[ =1/c (see 2.9), so that (2.10)becomes[r p[ = r sini/c.Exercise 2.8 The spherical slowness is dened in (B2.33) asp =r sinic.At rst sight, therefore, it would seem that the unit for the spherical slowness is[p] =mm/s= s,which is a unit of time, not slowness (inverse velocity)! However, we are analysinga wave that travels over an angle and when angles are involved, one should alwayscheck if the result depends on the unit of the angle if so this forces us to specifythe angle in its SI unit, the radian. Angles are fractions of a circle, inherentlydimensionless, and they do not always show up in a dimension analysis as we justdid.A look at (B2.32) shows that sini originated from the ratio between rd andds, where rd has unit of length and here r must therefore be expressed in m/rad,forcing (B2.33) to give p explicitly in s/rad. A different and more direct approachuses p = dT/d, to give the same result:[p] =srad,and upon reection this must be the correct unit, since a propagation over an angleis added that provides the missing unit to make it a slowness.For an earth with surface radius r = 6371 km, 673 s/rad corresponds to a surfacedistance of r =r/p = 6371/673 10 km travelled in one second. Thus, therays apparent surface velocity is 10 km/s.Exercise 2.9 In a spherically symmetric earth, p=r sini/c. The special casep = r/c occurs when the incidence angle i = /2. This is the point where the raybottoms, and the ray path is perpendicular to the vertical direction (Figure 2.3a).A good discussion of this principle can be found in G.J. Aubrecht et al., The Radian That TroublesomeUnit, The Physics Teacher, 31, 84-87, 1993: the radian is a supplementary unit in the Syst` eme InternationaldUnit es (SI). When SI units are consistently used in a formula, no conversion factors are needed to obtain thecorrect numerical values. For example when we write sin i i for small i, we express i in radians. Degreeis not an SI unit; if we decide to express in degrees, we could express p in s/deg, but (B2.33) would thenneed a conversion factor if this numerical value of p is used in it to nd i.Ray theory for seismic waves 11i= /2 ri= /2 (a)(b)Fig. 2.3. The rays bottoming point (i =/2) in: (a) a spherically symmetric earth; (b) ahorizontally layered medium.Note: inahorizontallylayeredmedium(Figure2.3b), therayalsobottomsatthe point ofi =/2. The horizontal slowness at the bottoming point ispx=sin(/2)/c = 1/c.Exercise 2.10 From (B2.35) and the chain rule of differentiation, we haveddr=_drd_1=1r cot i=tanir,didr=didddr=_dcdrrc 1_tanir= tani_dcc dr 1r_. (2.11)Exercise 2.11 From (B2.32):dT=dsc=drc cos i. (2.12)From (B2.33):sini = pc/r,cos i =_1 sin2i =_1 p2c2/r2. (2.13)Inserting (2.13) into (2.12)dT=drc_1 p2c2/r2,12 Ray theory for seismic wavesand henceT(p) =_dT=_drc_1 p2c2/r2. (2.14)From (2.13)tani =sinicos i=pc/r_1 p2c2/r2=1_r2/p2c21. (2.15)From (2.11), d = tani dr/r. Inserting (2.15), we getd = tani dr/r =drr_r2/p2c21.Epicentral distance is the integral of d(p) =_d =_drr_r2/p2c21. (2.16)Exercise 2.12 For a ray from an earthquake at depthh to a surface station, thetravel time in (2.14) can be expressed as a sum of two integrals: one is from thebottoming point to the earthquake at a radiusr=a h; the other is from thebottoming point to the earths surface. ThereforeT(p) =_ahbdrc_1 p2c2/r2+_abdrc_1 p2c2/r2.Exercise 2.13 According to (B2.44), the direction of displacement of seismicwavesisthedirectionoftheamplitudevectorA. Ontheotherhand, (r) =constantdenesthewavefront, andsoitsgradient denestheraydirec-tion (perpendicular to the wavefront surface). Equation (B2.46) implies that Ais perpendicular to for S waves, so that the direction of displacement of a Swave is perpendicular to the ray direction. From (B2.46) the S wave velocity isVs =_/, independent of its polarization (the direction of A).Exercise 2.14 Since(r) =constant denes the wavefront,its gradient denes the ray direction. (B2.45) says that Ais parallel to for P waves, so thatthe direction of displacement of a P wave is parallel to the ray direction. There istherefore only one polarization direction for P waves: the ray direction.Exercise 2.15 In a non-viscous uid, (B2.45) and (B2.46) become:Vp =_/, Vs = 0.Ray theory for seismic waves 13The acoustic wave velocity isc= _/ (B2.13) where the bulk modulus +23. For uids, = 0, and so = , and the acoustic velocity becomesc =_/ =_/ = Vp.Therefore, acoustic waves are P waves in a non-viscous uid, and there is no Swave in a non-viscous uid (Vs = 0).Exercise 2.16 The elastic wave equation in an isotropic, homogeneous earth is2uit2=

j( +)2ujxixj+

j2uix2j. (2.17)We use (B2.10) to dene the relationship between displacement and pressure, andapplying this to the rst term on the r.h.s. of (2.17) yields2uit2= ( +)xi_P_+

j2uix2j. (2.18)Taking the divergence of (2.18) and again using (B2.10), we get 2t2_P_=

i( +)2x2i_P_+

j2x2j_P_. (2.19)Multiplying (2.19) by (constant for a homogeneous earth) and merging the twoterms on the right2Pt2=

i( + 2)2Px2i= ( + 2)2P. (2.20)Rewriting (2.20) as2Pt2= c22P, (2.21)where c = _( + 2)/. (2.21) is a 1-D wave equation in homogeneous mediawith velocity c.Thus, pressure Psatises the wave equation and propagates withthe P wave velocity: c =_( + 2)/ = Vp.Exercise 2.17 Consider a plane wave. Since a cube is attened by the passage ofa P wave, it does not maintain its shape. Nor does a spherical shell in the case of aspherical wave (it is thinned or thickened).3Ray tracingExercise 3.1 Figure 3.1 shows the graph with letter for each node. Followingthe shortest path algorithm gives the following steps:(i) Set Q = A(TA = 0), B, C, D(TB = TC = TD = ). Set T is empty.(ii) Choose nodes=A, sinceTA is the smallest in set Q. The forward starofnodeAinset QincludesnodesBandD. Updatethetravel timesandraypathsfor nodes BandD: TB=minTB, TA+ TAB =min, 0 + 3=3, ray path AB; TD=minTD, TA + TAD=min, 0 + 1=1, ray path AD. Move node A from set Q to set T.Now set Q= B (TB=3, ray path AB); C(TC= ); D (TD=1, ray pathAD), set T = A (TA = 0).(iii) Choose node s=D, since TD is the smallest in set Q. The forward starofnodeDinset QincludesnodesBandC. Updatethetravel timesandraypathsfornodes BandC: TB=minTB, TD+ TDB =33711ABCDFig. 3.1. Simple graph for the shortest path method.14Ray tracing 15subroutine shpath(is,tt,prec,h,q,np)dimension tt(np),prec (np),h(np),q(np)dimension fs(1000),fsd(1000)integer prec ,h,q,fs,fsv,u,vdata rinf/1.0e30/do i=1,np ! put all nodes in set Qprec(i)=istt(i)=rinfh(i)=-1q(i)=-1end dott(is)=0.0 ! step (i)q(1)=is ! move source into the heap (set Q with tenddoenddoclose(1)close(2)endFig. 3.3. Fortran program that calls subroutineshpath for a 1010 2D homogeneousmodel with a source at node (5, 6).pathfornodeC: TC=minTC, TB+ TBC =min8, 2 + 3 =5, ray path ADBC. Move node B from set Q to set T. Now set Q=C (TC= 5, ray path ADBC), set T= A (TA= 0); B (TB= 2,ray pathADB); D (TD = 1, ray pathAD).(v) Choose nodes=C,sinceTCis the smallest (the only node) in set Q.There is no forward star of nodeCand no travel time to update. Movenode C from set Q to set T. Now set Q is empty, set T= A (TA=0); B(TB=2, ray pathADB); C(TC=5, ray pathADBC);D (TD = 1, ray pathAD).(vi) Since set Q is empty, stop. The nal set Tgives the travel times and raypaths from node A to the other three nodes.Exercise 3.2 Inshpath, the heaph contains alliq nodes with a provisionaltravel time less than the initial (innite) time; the top of the heap has the node withthe shortest time and is removed from the heap (into set P, which is not explicitlyidentied in shpath). The steps are identied in Figure 3.2.Ray tracing 17Fig. 3.4. Ray paths from the source node to all other nodes for a 1010 2D homogeneousmodel with a source at node (5, 6).1 2 3 4 5 6 7 8 9 10123456789100.010.020.030.040.050.060.070.08Fig. 3.5. Contour of the relative travel time errors for rays in Figure 3.4.Exercise 3.3 Figure 3.3 gives the Fortran program that calls subroutine shpath fora 10 10 2D homogeneous model with a source at node (5, 6).Figure 3.4 shows the ray paths fromthe source node to all other nodes in this model.The theoretical ray paths are straight lines but this is only correctly computed fornodes in four directions (taking the source node as the origin): N-S, E-W, NW-ES,and NE-SW. For other nodes, the ray paths are longer than the theoretical straightline paths, and so yield longer travel times.Figure 3.5 shows the contour of the relative travel time errors. Similar to the raysin Figure 3.4, there are 4 directions of zero travel time error: N-S, E-W, NW-ES,18 Ray tracingFig. 3.6. The same as Figure 3.4, except that subroutine forstar is extended to includeone more row/column on each side of the source node.and NE-SW. Between these 4 directions,the relative travel time error graduallyincreases, and the maximum error is reached in the middle between two adjacentzero error directions. The maximum relative error is about 8%.Exercise 3.4 To extend the forstar subroutine to include one more row/columnon each side of the source node, we only need to change lines 14-15 in subroutineforstar to:do jj = max(1,j-1), min(n,j+1)do ii = max(1,i-1), min(n,i+1)Figure3.6showstheraypathsfromthesourcenodetoallothernodesinthismodel. Compared with the rays in Figure 3.4 (computed with a smaller forwardstar), there are more straight line paths and there are more than one shortest path tosome nodes in Figure 3.6.Figure 3.7 shows the contour of the relative travel time errors. Again we take thesource node (5, 6) as the origin and compare it with Figure 3.5. It is clear that theoverall error level drops because the forward star nowincludes more nodes and thusprovides shorter paths, with the maximum relative error of only 2.5%. There arealso more zero error nodes. Besides the 4 directions of zero error (N-S, E-W, NW-ES, NE-SW), the zero error region in the centre expands to the larger forward starof the origin, and there are 8 more zero error nodes at the corners (e.g. nodes (2,3)Ray tracing 191 2 3 4 5 6 7 8 9 10123456789100.0050.010.0150.02Fig. 3.7. The same as Figure 3.5, except that subroutine forstar is extended to includeone more row/column on each side of the source node.and (2,7)). The maximum error region moves towards the N-S and E-W directions.Exercise 3.5 For a velocity increasing linearly with column number i : v(i) =1 + (i 1)/9, the adapted forstar subroutine is shown in Figure 3.8.Figure3.9showstheraypathsfromthesourcenodetoallothernodesinthismodel. Its different from Figure 3.6 in that there is only one shortest time path foreach node, because the velocity is no longer homogeneous. Also the ray path is nolonger symmetric about i = 5 because the velocity is no longer symmetric abouti=5. The rays do not represent the reection or transmission behaviour whenthey encounter a velocity boundary, which suggests that the shortest path methodis not very accurate.20 Ray tracingsubroutine forstar(u,np,fs,fsd,nf)! example for square 2D medium with node spacing 1! velocity increases linearly with i (column number)! from v=1 at i=1 to v=2 at i=10! input: node u, dimension n=sqrt(np)! output: fsd(i) is he travel time between u and fs(i), i=1,...,nfdimension fs(1000),fsd(1000)integer u,fsn=sqrt(0.001+np)j=(u-1)/n+1 ! row number of node u in the square gridi=u-n*(j-1) ! column numbervu = 1. + (i-1.)/9. ! velocity at node unf=0do jj=max(1,j-2),min(n,j+2)do ii=max(1,i-2),min(n,i+2)ds=sqrt(0.+(ii-i)**2+(jj-j)**2) ! distance (ii,jj) to uif(ds.gt.0.) thenvii = 1. + (ii-1.)/9.velocity=(vu+vii)/2. ! average velocity between (ii,jj) and unf=nf+1if(nf.gt.1000) stop dimension of fs exceededfs(nf)=(jj-1)*n+ii ! sequential index of node (ii,jj)fsd(nf)=ds/velocityendifenddoenddoreturnendFig. 3.8. Adapted forstar subroutine for a velocity increasing linearly with the columnnumber.Fig. 3.9. Ray paths for a velocity increasing linearly with column number. The forwardstar is the same as in Problem 3.4.Ray tracing 21Exercise 3.6 To simplify expressions, we use f=_1 c2(p21 +p22), noting thatf= 1 and if= 0 on the ray where p1 = p2 = 0.(a) iH = i(/s) = s(i= 0 since i= 0 independent of s along theray.Therefore 1H = 1(hf/c) = (h/c)f= 0 or 1(h/c) = 0.(b) h = 1 Kq1 = 1 on the ray where q1 = 0.(c) Differentiating the expression (B3.5) for H:iH =(hic cih)[1 c2(p21 +p22)] +hc2ic(p21 +p22)c2f=hic cih[1 c2(p21 +p22)]c2f(d) H/pi = (hf/c)/pi = hcpi/f, which is zero on the ray because pi = 0.(e) Since the rst derivatives of H are zero, the second order Taylor expansion is(summing over i and j = 1, 2 is implied):H H(0, 0, 0, 0) + 122Hq2iq2i+2Hqipjqipj + 122Hp2ip2iThe derivatives in this are on the ray and assumed constant, so that for exampleH/pk= (2H/pkqj)qj + (2H/p2k)pk which is zero on the ray where qiand pi are zero.(f) From (c): 1H = (h1ccf21h)/c2f. On the ray, where f= 1 and 1f= 0:2Hq21=h11c c11hc2h1c c1hc321cThe second term is zero because of the result proven in (a). Since h = 1 Kq1,the second derivative 11h = 0. Therefore 11H = (11c/c2.Exercise 3.7 This follows from a rst-order Taylor expansion:=_z2+x2+y2c=zc_1 +x2+y2z2zc_1 +x2+y22z2_From this:H11 = 2/x2= 1/zc, and H22 = 1/zc. Close to the axis z s sothat H 1csI.Exercise 3.8 dP/ds=(dH/ds)Q + H(dQ/ds) =(c1V cH2)Q +cH2Q = c2V4Wave scatteringExercise 4.1 From (B2.48) follows that the diameter of Fresnel zone is dmax=L. The wavenumber k equals 2/. ThenD =4Lka2=24La2=2La2=2d2maxa2.Exercise4.2 Sincescatteredwavesareaddedtoanincomingwaveeldthatremains unaffected,energy increases in the Born approximation and is not con-served.Exercise 4.3 Even though there is no multiple scattering in this case, energy isnot conserved so the result is not exact.Exercise 4.4 In equation (B4.15), the last two terms on the right hand side repre-sent the far-eld response, because the amplitude decreases as r1. If the sourcetime functionMjk(t) = (t), then the Fourier transform_+Mjk(t r/VP)eitdt = eir/VP= eikPr,_+Mjk(t r/VS)eitdt = eir/VS= eikSr,so that the Fourier spectrum of the far-eld response iskGjki(r, r

, ) =

jk_ijk4V3PreikPrijij4V3SrkeikPr_.22Wave scattering 23IfM is a function, M itself is the Heaviside function h(t) =_(tt

)dt

, witha spectrum i/ = 1/i (see B2.66), which explains the extra factor i.Exercise 4.5 The rst term is the dot product of (B4.10) with a forcef. Since

G is the gradient Gji/x

k we recognize the contraction (B4.13) summed overall moment tensor elements Mjk in the second term.Exercise 4.6 Since the far eld term contains the time derivative of M it is weakerfor signals in which the low frequencies are dominant (differentiation in the timedomain is equivalent to multiplication with i in the frequency domain). In thenear eld term, the Heaviside function will give a contribution_r/VSr/VPd=r22_1V2S1V2P_and displacements only damp asr2rather than ther4which is in the denom-inator of (B4.15). Note also that both the near- and intermediate terms lead to apermanent offset, whereas the far eld signal is transient. Whether or not one canignore the near eld terms is therefore dependent on the frequency of interest.Exercise 4.7 A unit force in the z-direction can be expressed with a Kroneckerdelta as fj(t) = f(t)j3. Dotting (B4.9) with this gives

jGji(r, t)fj =

j14r3(3ijij)j3_r/VSr/VPf(t )d+14V2P rijj3f(t r/VP) 14V2S r(ijij)j3f(t r/VS)=14r3(3i3i3)_r/VSr/VPf(t )d+14V2P ri3f(t r/VP) 14V2S r(i3i3)f(t r/VS)For the component in the r-direction we must dot this with the unit vector (1, 2, 3) =(cos sin, sinsin, cos ) in the r-direction to get:ur =

ijGji(r, t)fji =14r3(333)_r/VSr/VPf(t )d+14V2P r3f(t r/VP) 14V2S r(33)f(t r/VS)=cos 4r_2r2_r/VSr/VPfz(t )d +1V2Pfz(t r/VP)_24 Wave scatteringAt arbitrary , the direction is given by a vector (cos cos , sincos , sin).We can ease the algebra by recognizing that the expression for u must be indepen-dent of because of spherical symmetry. Choosing = 0, such that (1, 2, 3) =(sin, 0, cos ), we must dot (4.7) with the unit vector (cos , 0, sin):ur =14r3(3 sin cos 3 sin cos + sin)_r/VSr/VPf(t )d+14V2P r(sin cos sin cos )f(t r/VP)14V2S r(sin cos sin cos + sin)f(t r/VS)=sin4r_1r2_r/VSr/VPfz(t )d 1V2Sfz(t r/VS)_.For fz(t) = (t):_r/VSr/VPfz(t )d=_r/VSr/VP(t )d=_t if r/VP t r/VS0 otherwise.Inserting the above equations into the expressions for the displacement:ur(r, t) =_cos 4r_2tr2 +1V2P(t r/VP)_if r/VP t r/VS0 otherwise;u(r, t) =_sin 4r_tr2 1V2S(t r/VS)_if r/VP t r/VS0 otherwise.Figure4.1sketchesthebehaviourof uranduwithtime, withtheirratesofamplitude decrease denoted.When fz(t) = H(t)fz(t r/VP) = H(t r/VP) =_1 if t r/VP0 if t < r/VP,fz(t r/VS) = H(t r/VS) =_1 if t r/VS0 if t < r/VS,Wave scattering 25uru3r3rr1r/VPr/VPr/VSr1r/VS00ttFig. 4.1. Diagram of ur and u of a pulse source fz(t) = (t)._r/VSr/VPfz(t )d=_r/VSr/VPH(t )d=___0 if t < r/VP_tr/VPd=12_t2r2V2P_if r/VP t r/VS_r/VSr/VPd=r22_1V2S1V2P_if t > r/VS.Inserting this gives:ur(r, t) =___0 if t < r/VPt2cos 4r3if r/VP t r/VScos 4rV2Sif t > r/VS;u(r, t) =___0 if t < r/VPsin 8r3_t2r2V2P_if r/VP t r/VSsin 8r_1V2S+1V2P_if t > r/VS.Figure4.2sketchesthebehaviourof uranduwithtime, withtheirratesofamplitude decrease denoted. The Heaviside function kicks in after t = r/VP for ur(H(t r/VP)), which corresponds to far eld P wave and again after t = r/VS foru (H(t r/VS)), which corresponds to far eld S wave. The parabolic behaviourof both ur and u corresponds to near eld wave. It does not have a distinct pulsearrival, but extends between times t/VPand t/VS, and it cannot be identied withpure P or S wave motion.26 Wave scatteringuurr/VPr/VPr/VSr/VSr1r100ttFig. 4.2. Diagram of ur and u of a Heaviside source fz(t) = H(t).Exercise 4.8 The i-th element of Gu is

jGjivxj=

j_xj(vGji) vGjixj_.Using vector notation the above equation can be written asG v = (vG) v G.Therefore_G v d3r

=_ (vG) d3r

_v Gd3r

.Exercise 4.9 The component of scattered S wave in the direction is (see FigureB4.2)uPS= uPSxsin +uPSycos . (4.1)From the expression of uPSi:uPSx=k2seiksr4r_ (xzxz) 2VsVP(xzzx2z)_;uPSy=k2seiksr4r_ (yzyz) 2VsVP(yzzy2z)_.Notice that x = sin cos , y = sin sin, z = cos , thenuPSx=k2seiksr4r_sin cos cos + 2VsVPsin cos2 cos _; (4.2)Wave scattering 27uPSy=k2seiksr4r_sin cos sin + 2VsVPsin cos2 sin_. (4.3)Inserting (4.2) and (4.3) into (4.1)uPS= 0.Exercise 4.10(i) For scatterer , uPPr cos , uPS sin, so thatuPPr= 0 at = /2; [uPPr[ is maximum at = 0, ;uPS= 0 at = 0, ; [uPS[ is maximum at = /2.(ii) For scatterer, uPPr1, uPS0. The amplitude of the scatteredP wave is the same for all scattering angles. There is no scattered S wave(Vs =_/, so that S wave cannot see the scatterer ).(iii) For scatterer , uPPr cos2, uPS sin2, so thatuPPr= 0 at = /2; [uPPr[ is maximum at = 0, ;uPS= 0 at = 0, /2, ; [uPS[ is maximum at = /4, 3/4.Exercise 4.11(a) The incoming S wave is of the form uy= eikSz, ux=uz= 0. The far-eldGreens function of S wave excited by a force in the j-direction at r

follows from(B4.10):Gki (r, r

, ) = ikik4V2SreikSr,where r = [r r

[. Using

jr = j, we nd:(u)kj = uk/xj = kxjzikSeikSr, u =

kuk/xk = 0,(G)ikj = Gki/xj ikik4V2SrikSjeikSr,( G)i =

kGki/xk

kikik4V2SrikSkeikSr= 0,28 Wave scatteringwhere the indicates that we ignore terms of O(r2).(b) From (B4.25), ignoring the integration since we deal with point scatterers:ui =

k2Gkiukwhere uk is the incoming S wave. To derive the expression for uSSrwe must usethe Greens function for a S wave, and to obtain the r-component we dot ui withi to nd:uSSr= 21 14V2SreikSr= 0(notethatthefactoreikSz

=1sincez

=0atthescattererbychoiceofthecoordinate origin at the scatterer). Using the Greens function for a scattered Pwave and the same procedure we obtainuSPr= 2y4V2PreikPrand use y = sinsin.(c) The term in (B4.25) with is:uSSi=

kj

juj

kGkiwhich are 0 because of the zero divergence established in (a).(d) The term in (B4.25) with is:uSSi=

jk

jGki [

juk +

kuj]Substituting the Greens function for the scattered S wave one nds:

jGki [

juk +

kuj] = k2S4V2Sr(ikik)(kyz +kzy) =k2S4V2Sr(iyz +izyiyzizy)and substitution of (x, y, z) = (cos sin, sinsin, cos ) yields the requestedresult. The unit vectors needed to derive the individual components are: er = (cos sin, sinsin, cos ) e = (sin, cos , 0) e = (cos cos , sincos , sin)Wave scattering 29(e) As in (b) and (d), but now using the expression of the Greens function for thescattered P wave.(f) The expressions follow after dotting the results with the same unit vectors asgiven in problem (d).5Amplitudes of body wavesExercise 5.1 Let the solution of (B5.3) be x(t) = est, then x(t) = sest, x(t) =s2est. Inserting them into (B5.3), we gets22s +20 = 0.Thesolutionofthisquadraticequationiss=12_2 _42420_= _220. Usually 0, so s = i1, where 1= _202> 0. Thesolution isx(t) = exp[( i1)t] = exp(i1t t).Exercise 5.2 Figure5.1comparesthetrue [X()[andtheapproximationby(B5.5) for different , with 1 = 10 rad/s. Here is between 0.3 rad/s and 30 rad/s,corresponding to a period between 0.2 s and 20 s. It is clear that the approximationis equal to the true value at =1, and becomes worse as moves further awayfrom1, the natural frequency of the oscillator. The approximation is better at highfrequency than at low frequency. As (damping) increases, the height of the peakdecreases.Exercise 5.3 The amplitude of both waves decrease as exp(t/2Q), and theyhave the sameQ, so the high-frequency wave damps out faster with time. Thisimplies that with increasing travel time (distance), the seismograms are more andmore dominated by low-frequency waves, and the high-frequency waves becomemore and more difcult to observe.Exercise 5.4 Dimensionless30Amplitudes of body waves 310 10 20 3000.20.40.60.81 (rad/s)|X()|=0True valueApproximation0 10 20 3000.20.40.60.81 (rad/s)|X()|=0.2True valueApproximation0 10 20 3000.20.40.60.81 (rad/s)|X()|=0.4True valueApproximation0 10 20 3000.20.40.60.81 (rad/s)|X()|=0.6True valueApproximation0 10 20 3000.20.40.60.81 (rad/s)|X()|=0.8True valueApproximation0 10 20 3000.20.40.60.81 (rad/s)|X()|=1True valueApproximationFig. 5.1. Comparison of the true and approximate spectrum for different .Exercise 5.5 The wave amplitude damps away as A exp(kx/2QX). Sincek = 2/, the shorter-wavelength wave (higher-frequency wave, f=c/) has alarger wave number k, and thus damps away faster with distance x.Exercise 5.6 If a wave has an amplitudeA exp(kx/2QX), its energy de-creases as E A2 exp(kx/QX). Then dE/dx = kE/QX, and the energyloss in one wavelength (dx = = 2/k) is E = [ kEdx/QX[ = 2E/QX.Thus another denition of QX is QX = 2E/E, and Q1Xdescribes the relativeenergy loss per wavelength.Exercise 5.7For P waveskP=VP= _() +43()_12=1/2[(Re() +43Re()) + i(Im() +43Im())]1/2.When [Im()[ [Re()[ and [Im()[ [Re()[:kP=1/2[Re() +43Re()]1/2_1 + i(Im() +43Im())Re() +43Re()_121/2[Re() +43Re()]1/2_1 i2Im() +43Im()Re() +43Re()_.32 Amplitudes of body wavesThen from (B5.11)QPX =Re(kP)2Im(kP) Re() +43Re()Im() +43Im().For S waveskS =VS= _()_12=1/2[Re() + iIm()]1/2.When [Im()[ [Re()[kS =1/2[Re()]1/2_1 + iIm()Re()_121/2[Re()]1/2_1 i2Im()Re()_.Then from (B5.11)QSX =Re(kS)2Im(kS) Re()Im().In the derivation above, we assume that [Im()[ [Re()[ and [Im()[ [Re()[,which indicates that QPX 1 and QSX 1. Since QX=2E/E (see prob-lem 5.6), where E is the energy loss in one wavelength, QX 1 simply meansE E. Therefore the assumption we have made is consistent with the generalcondition E E.Exercise 5.8 From (B5.12) and (B5.13) and the assumption that Im() 0QPXQSX=34Re() +43Re()Re(). (5.1)BecauseVPVS=[(Re() +43Re())/]1/2[Re()/]1/2=_Re() +43Re()Re()_1/2,here VP, VS represent the real part of the velocity, which describes the wave prop-agation velocity and have nothing to do with intrinsic attenuation (imaginary partof the velocity), so (5.1) becomesQPXQSX=34V2PV2S.HenceQPX =34V2PV2SQSX.Amplitudes of body waves 33For a Poisson solid, where VP=3VSQPXQSX=34V2PV2S=34 31=94.Exercise 5.9 The complex wave number of a P wave iskP=VP=Re(VP) + i Im(VP)=[Re(VP) i Im(VP)][Re(VP)]2+ [Im(VP)]2.Inserting the above equation into (B5.11)QPX =Re(k)2 Im(k)=Re(VP)2 Im(VP)= Re(VP)2 Im(VP).Following the same procedure, we ndQSX = Re(VS)2Im(VS).6Travel times: observationsExercise 6.1 With 24 bit recording, the dynamic range is 20 log10 224= 144.5 dB.For paper recording we nd 20 log10(50/0.05)=60 dB. The largest recordablecount is 224= 1.68 107corresponding to a displacement of 1.68 107/5000 =3355 m (3.3 mm).Exercise 6.2 A period of 24 hours corresponds to 1.16 105Hz. If the seismicspectrum extends to, say, 5 Hz, the spectrum spans more than 18 octaves sincelog2(5/1.16 105) = 18.7. The dynamic range for the displacement u is of theorder of 20 log10(0.5/5109) = 160 dB. Since velocity u u and u 2ufor very different frequency, the dynamic range for velocity and accelerationmust differ from that of displacement.Exercise 6.3 The bandwidth of the lter is f=0.15 0.05=0.1 Hz. Thewindowlengthofcross-correlationisTw=60s. ThesignaltonoiseratioisSNR=4. Then from (B6.8), the Cram er-Rao lower bound estimation is2CRLB =3(1 + 2SNR)82SNR2f3Tw=3(1 + 2 4)82420.1360= 0.36 s2.The standard deviation is CRLB = 0.60 s.Exercise 6.4 Fromthedenition(B6.9)ofCPE, wehavew=0.1, Tc=10 s, 2CRLB = 0.36 s2, so2CPE = wT2c /3 + (1 w)2CRLB = 0.1 102/3 + (1 0.1) 0.36 = 3.7 s2.The standard deviation is CPE = 1.9 s.34Travel times: observations 35Exercise 6.5 If SNR=20, the Cram er-Rao lower bound estimation is2CRLB =3(1 + 2SNR)82SNR2f3Tw=3(1 + 2 20)822020.1360= 0.065 s2.The variance reduces by a factor of 5.5 (and the standard deviation, now 0.25 s, bya factor of 2.3) if the SNR increases by 4 times.Exercise 6.6 If f= 0.2 Hz, the Cram er-Rao lower bound estimation is2CRLB =3(1 + 2SNR)82SNR2f3Tw=3(1 + 2 4)82420.2360= 0.045 s2.The variance reduces by a factor of 8 if the bandwidth of the lter is doubled.Exercise 6.7 In VanDecar and Crossons method, the arrival difference Tij is anit estimate, obtained fromlocating the time of the maximumin the cross-correlationbetween the seismograms at station i and station j. Because each such estimate hasa different error, Tik is not exactly equal to TijTkj.7Travel times: interpretationExercise 7.1 With four variables, line plots of the delay ratio as a function ofonly one of the four (xing the other three) are not very informative. Better is tomake contour plots of the ratio as a function of two variables, or set the ratio at axed level and plot a surface in 3D as a function of three variables. Here we givea solution with contour plots. It is logical to x like variables, i.e. either R and S,ora and. One would xR andS when one has an idea where the anomaly islocated with respect to station and source and one wishes to study its resolvabilityas a function of its size and the seismic wavelength. Here we x the size a and thewavelength and study how resolvability depends on the distances.Of course, whenever the ratio between the nite-frequency delay and the ray the-oretical delay as computed from (B7.2) becomes negative, we are way beyond thevalidity of ray theory, the delay has completely healed and (B7.2) is not valid. Sowe set negative ratios to 0 and obtain the following plots:0100020003000400050006000Receiver distance (km)0 1000 2000 3000 4000 5000 6000Source distance(km)a=1000 =200 km0.80100020003000400050006000Receiver distance (km)0 1000 2000 3000 4000 5000 6000Source distance(km)a=500 =200 km0.20.40.60.80100020003000400050006000Receiver distance (km)0 1000 2000 3000 4000 5000 6000Source distance(km)a=200 =20 km0.40.60.8Fig. 7.1. Contour plots of eq 7.2 for different a and as a function of R and S.36Travel times: interpretation 37Exercise 7.2 It complicates the argument: the surface area under the accelerationwavelet is zero. Thus, the energy in the crosscorrelation is divided over both earlierand later times. To resolve the question one may generate a synthetic wavelet u(t)and crosscorrelate it with both u(t) + u(t ) and u(t) u(t ). This is whatwe did to create Figure 7.2: Indeed, the negative reection coefcient () givesFig.7.2. Fromlefttoright: u(t), u(t), u(t)andthecrosscorrelationwithaperturbedwaveeld from a negative (solid line) and positive (dashed line) velocity anomaly.a shift towards earlier time (dashed line). To create this effect we had to choose very high. This merely shows that the effects of innitely small point scattererscannot be well resolved.Exercise 7.3 For a pulse in a narrow frequency band with dominant frequency0, we can approximate [ m()[ ( 0), and inserting this into (B7.21) weobtainKP 0 sin(0T).Then the kernel has a maximum at0T= /2. (7.1)Since 0 = VPT0 = VP2/0,0 = 2VP/0. (7.2)Tin a medium that is homogeneous everywhere except at x (Figure 7.3) is:T=rxr +rxsrrsVP=2_q2+L2/4 LVP. (7.3)s rLxLl lql LlFig. 7.3. A scatterer at point x near the central ray sr in a medium that is homogeneouseverywhere except at x.38 Travel times: interpretationSince q L (the scatterer is near the central ray),_q2+L24=L2_1 +4q2L2L2_1 +124q2L2_=L2+q2L , and (7.3) becomesT 2(L2+q2L ) LVP=2q2LVP. (7.4)Inserting (7.2) and (7.4) into (7.1)2VP02q2LVP= /2,soq =0L22.Since the radius of the rst Fresnel zone is h =0L/2 (see B2.43), we haveq = h/2,where q is the distance from the geometrical ray to the point of maximum kernelvalue.Exercise 7.4 A scattered SH wave coming in from the North will be visible onthe EW component of the seismogram and, if it comes in early enough, will affectthe reading of the SV arrival which is the one appearing on this component. Fromthe right column of Figure B7.10, we see that such a wave can only originate froman incoming SH wave. If the Earth is isotropic and SV and SH have the samearrival time,but only the SV component is affected,this could lead to apparentanisotropy.Exercise 7.5 For a Ricker wavelet, (B7.31) shows that the kernel is zero whenT= 0. This does not say, however, that the doughnut hole exists for all possiblewaveforms. The arrival time of the maximum is not causal, and a scattered wavewith zero time delay could move the maximum to a different time if its waveshapeis not symmetrical around zero (Exercise 7.2).Picked onset arrival times are sensitive to scatterers on the geometrical ray. Thisfollows from the fact that the wave u from a scatterer on the geometrical ray doeschange the amplitude of the wave, which means it changes the time for the signalto rise above the noise level (Figure B7.12). Note, however, that this is an effect onthe error in the observation, not on the arrival time itself.Travel times: interpretation 39Exercise 7.6 DifferentiateQP Q Pw.r.t. sd( QP Q P)ds=d Qds P+QdPds dQdsP Qd Pds.Both P, QandP,Qsatisfy the Hamiltonian systemof (B3.10). Inserting (B3.10)into the above equation givesd( QP Q P)ds= c PP+Q(1c2V Q) cPP Q(1c2VQ) = 0I. (7.5)The last step uses the fact thatP, Q,P,Q, Vare diagonal for a sphericallysymmetric earth, and soPP= PP,QV Q = QVQ.By denition of P, Q,P,Q, there isQP Q P=_Q1P1Q1 P100Q2P2Q2 P2_. (7.6)(7.5) implies thatQP Q P has elements that are independent of s, and the valuesof these elements can be obtained from the initial conditions: P1(0)=P2(0)=Q1(0) =Q2(0) = 1,P1(0) =P2(0) = Q1(0) = Q2(0) = 0. Now (7.6) becomesQP Q P=_Q1(0)P1(0) Q1(0) P1(0) 00Q2(0)P2(0) Q2(0) P2(0)_=_1 00 1_= I. (7.7)Thus, we have proved the rst equality in (B7.32). The proof of the second equalityin (B7.32) is very similar.Exercise 7.7 In a homogeneous medium, for a pointx=(l, q1, q2) in the raycoordinate system, the ray length is [x[ =_l2+q21 +q22. Then the travel time Tsatises cT= [x[. Since q21 +q22 l2, we havecT= [x[ =_l2+q21 +q22 = l_1 +q21 +q22l2 l +q21 +q222l. (7.8)In order to use (B7.33), we need to know every parameter on the left hand side ofthis equation. The central ray and scattered rays in a homogeneous medium areshown in Figure 7.3. We have vr = c, rs = L. From (7.8), we havexs = l +q22l, xr = (L l) +q22(L l),40 Travel times: interpretationwhere q2= q21 +q22. Then_[det(Hxs +Hxr)[ =rsvrxsxr=Lc[l +q22l][L l +q22(Ll)].Ignoring the terms of q2and higher order in the denominator, we get_[det(Hxs +Hxr)[ =1c Ll(L l).Exercise 7.8 The Maslov index is equal to the number of dimensions lost by thebundle of rays. In a 3-Dmedium, the maximumpossible number of lost dimensionsis 2, i.e. when a 3-Dbundle of rays focuses in a single point. Therefore, the possiblevalues of the Maslov index is: 0 (e.g. P, pP, no caustic); 1 (e.g. PP, SS, at the caustic,which forms a line for different azimuths); 2 (e.g. PP, PKP, and multiple surfacereected waves that focus to a point at the antipode of the source). The increaseof the Maslov index M can accumulate to a large number as a multiply reectedray bundle passes through successive caustics. While this may seem at odds withthe fact thatHis only a 22 matrix so that sig(H) can never exceed 2, theharmonic nature of its effect, e.g. in (B7.28), makes that an index equal to 3 hasthe same effect as an index equal to -1, so that sig(H) can still be used even in thecase of multiple caustics.8Body wave amplitudes: Observation and interpretationExercise 8.1A+A = 1Tp_Tp0[u(t) +u(t)]2dt= 1Tp_Tp0u2(t)[1 +u(t)u(t) ]2dt 1Tp_Tp0u2(t)[1 + 2u(t)u(t)]dt= 1Tp_Tp0u2(t)dt +1Tp_Tp02u(t)u(t)]dt= 1Tp_Tp0u2(t)dt_1 +1Tp_Tp02u(t) u(t)dt1Tp_Tp0u2(t)dt 1Tp_Tp0u(t)2dt_1 +_Tp0u(t)u(t)dt_Tp0u(t)2dt_, (8.1)Exercise 8.2 The attenuation kernel does have a doughnut hole. This is easytoseefromtheexpressionoftheattenuationkernel(B8.16), whichhasasinedependence on the detour time Tas the travel time kernel does. For a scattereron the geometrical ray, T= 0 (assume = 0), so KQX= 0, which gives thedoughnut hole of zero sensitivity.The focusing kernel does not have a doughnut hole. From the expression of the4142 Body wave amplitudes: Observation and interpretationfocusing kernel (B8.9), KAX has a cosine dependence on the detour time T. For ascatterer on the geometrical ray, T= 0 (assume = 0), so KQX is maximum,which means that the largest focusing effect on amplitudes is from scatterers on thegeometrical ray.Of course, if the Maslov index is equal to 1, the situation is the reverse.9Normal modesExercise 9.1 To save on notation we adopt the Einstein summation notation, inwhich a summation is implied over every index that occurs twice in one term, i.e.aibi means

iaibi. In Cartesian coordinates:(Lu, v) =_V(Lu)ivi d3r =_Vxj_cijklulxk_vi d3r=_Vxj_cijklulxkvi_d3r _Vcijklulxkvixjd3r. (9.1)Apply Gausss theorem to the rst term on the RHS of (9.1)_Vxj_cijklulxkvi_d3r =_Scijklulxkvinjd2r, (9.2)where Vis the Earth and S its surface, and n is a unit normal vector of S. Noticethat ij = cijklulxk, and the traction Fon S is Fi = ijnj, so (9.2) becomes_Vxj_cijklulxkvi_d3r =_Sijvinjd2r =_SFivi d2r = 0,because S is a free surface on which F= 0. Substituting this into (9.1), we get(Lu, v) = _Vcijklulxkvixjd3r. (9.3)Similarly,(u, Lv) =_Vui(Lv)id3r =_Vuixj_cijklvlxk_d3r=_Vxj_cijkluivlxk_d3r _Vcijkluixjvlxkd3r. (9.4)4344 Normal modesApply Gausss theorem to the rst term on the RHS of (9.4)_Vxj_cijkluivlxk_d3r =_Scijkluivlxknjd2r (9.5)Again because of the zero traction on the boundary S:_Vxj_cijkluivlxk_d3r =_Suiijnjd2r =_SuiFi d2r = 0 .Substituting into (9.4), we get(u, Lv) = _Vcijkluixjvlxkd3r = _Vclkjiuixjvlxkd3r, (9.6)the last step uses the symmetry property of the elastic tensor: cijkl = cklij = clkji.Becausei, j, k, l in (9.6) are dummy subscripts, we can rearrange them (l i, k j, j k, i l), and (9.6) becomes(u, Lv) = _Vcijklulxkvixjd3r. (9.7)Compare (9.3) and (9.7), we get(Lu, v) = (u, Lv),which establishes that L is Hermitian.Exercise 9.2 From (B9.11): (up, Luq)=(Lup, uq). Using (B9.1) without aforce (such that it is a free oscillation and is an eigenfrequency):(2q)(up, uq) = 2p(up, uq)Since is real, (up, uq) = (up, uq) =pq.From this we see that (2q) =2q,so that 2must be real.Exercise 9.3 Working out the variations and collecting terms with VP, VS andNormal modes 45 gives:K +K +K

= K[(V2P 43V2S ) 83VSVS + 2VPVP]+K[2VSVS +V2S]+K

= VPKVP+ 2VS(K 43K)VS+ [(V2P 43V2S )K +V2SK +K

]which equals the conversion factors in Table B9.2 because (V2P 43V2S ) = /.Exercise 9.4 Since the angular order s = 0 gives the case for spherical symmetry(remember that the azimuthal order [m[ s so that m 0), one expects thatthe kernels reduce to the partial derivatives known from the spherically symmetriccase.Exercise 9.5 Starting from the right hand side:(eitH)m

m = (I itH + 12(it)2H2...)m

m= (m

mitHm

m + 12(it)2

kHm

kHkm...We may replace m

m by

j Zm

jZmjbecause of the orthogonality (B9.22). In-serting the expression for the elements of H:=

j[Zm

jZmjitZm

jjZmj + 12(it)2

ikZm

jjZkjZkiiZmi...]=

j[Zm

jZmjitZm

jjZmj + 12(it)2

iZm

jjijiZmi...]=

jZm

j1 itj + 12(it)22j ...]Zmi]=

jZmjeijtZm

j46 Normal modesExercise 9.6 With P= exp[itH] and using a Taylor expansion:PZ = Z itHZ + (it)2H2Z ...= Z itZ+ (it)2Z2...= Z[I it+ (it)22...]= Z exp[it]10Surface wave interpretation: ray theoryExercise 10.1 The phase velocityCn() =/kn() =cn. Takingkas theindependent variable, we get n(k) =cnk. Then the group velocity |n(k) =dn(k)/dk=cn. Hence, if there is no dispersion, the group velocity and thephase velocity are the same, and both are independent of frequency.Exercise 10.2 If the phase velocity satisesCn() =kn()= c0 +a, (10.1)thendkn()d=2c0 +a c02(c0 +a)2=(c0 + 2a)(c0 +a)2.Taking k as the independent variable, we get the group velocity|n =ddk=_dkd_1=(c0 +a)2(c0 + 2a).In order to compare the phase velocity and the group velocity, we take their ratio|nCn=c0 +ac0 + 2a= 1 ac0 + 2a. (10.2)Since a > 0 (see the slope in Figure B10.3), (10.2) indicates that |n/Cn< 1, i.e.|n< Cn.InFigureB10.3, forthersthighermodenearaperiodTof50s, Cn5.0km/s, theslopedCn/dT0.015km/s2, theinterceptisabout4.3km/s, and = 2/T 0.1257 rad/s. Then in (10.1):a =dCnd(1)= 2dCndT= 2 0.015 0.094 km/s, and c0 = 4.3 km/s.4748 Surface wave interpretation: ray theorySubstituting all these numbers into (10.2), we get|nCn= 1 0.0944.3 0.1257 + 2 0.094 0.871,and the group velocity Un 0.871Cn 0.871 5.0 4.4 km/s.Exercise 10.3 For body waves, the ray parameterp=r sini/v. For an SSSSwave, p1=r1/v1, if point 1 is the bottoming point (i1= 90) of the S leg in themantle. For an ScS wave, p2=r2 sini2/v2, taking point 2 the reecting point atthe core-mantle boundary. Its obvious thatr1>r2, v1i2,so p1>p2. The equivalent phase velocity Cn of these body waves is related to pthrough p =a/Cn, where a is the radius of the earth. Thus we have Cn18) remainoscillatory even at a period of 10 s. From Figure B6.8 this is a mostly a low noiseregion on Earth.Exercise 10.5 Expanding gi(r) in terms of the basis gk(r) for k = 1, ..., M: gi(r) =M

k=1aikgk(r). (10.3)In order to determine the coefcients aik, substitute (10.3) into the orthogonalitySurface wave interpretation: ray theory 49condition_a0 gi(r)gj(r)dr = ij,and switch the order of integral and sum, we getM

k=1aik_a0gk(r)gj(r)dr = ij.This can be written asM

k=1aikGkj = ij, (10.4)where Gkj =_a0gk(r)gj(r)dr. For each xed i, (10.4) can be written asGa = d, (10.5)where G = Gjk = Gkj,a = (ai1, ai2, , aiM)T, d = (i1, i2, , iM)T.Solving (10.5) for eachi gives the coefcientsaik, and by (10.3), the dual basis gi(r).11Surface waves: Finite frequency theoryExercise 11.1 Let u0 be the unperturbed wave zero phase, and the correspondingperturbation is u0. If the unperturbed wave has a nonzero phase 0, we can viewthis as a time shift, which is the same for both; we therefore write u = u0ei0andu = u0ei0. Insert u and u into (B11.2) and (B11.3): = Im(uu ) = Im(uei0uei0) = Im(u0u0). lnA = Re(uu ) = Re(uei0uei0) = Re(u0u0).Therefore, (B11.2) and (B11.3) remain correct for u with nonzero phase.Exercise 11.2 The boxcar windoww(t) =_1 if T/2 t < T/20 otherwise.Its spectrum isW() =_+w(t)eitdt =_T/2T/2eitdt =1i_eit/2eit/2_=2 sin T2= Tsinc_T2_, (11.1)where sinc(x) = sin(x)/x. Its power spectrum isP() = [W()[2= T2sinc2_T2_.The rst column of 11.1 plots w(t), W() and P() of a boxcar window.50Surface waves: Finite frequency theory 51T/2 0 T/201tboxcar00Tspectrum2/T 4/T 6/T00power spectrumT22/T 4/T 6/TT/2 0 T/201tcosine taper00T/2spectrum2/T 4/T 6/T00power spectrumT2/42/T 4/T 6/TFig. 11.1. Time series, spectrum, and power spectrum of a boxcar window (left) and acosine taper (right).Exercise 11.3 A cosine taper between T/2 t < T/2 is given byw(t) =_12_1 cos2(t+T/2)T_if T/2 t < T/20 otherwise.Its spectrum isW() =_+w(t)eitdt =_T/2T/212_1 cos 2(t +T/2)T_eitdt=12_T/2T/2eitdt 12_T/2T/2cos 2(t +T/2)Teitdt=T2 sinc(T2) 12I. (11.2)In the last step, we use (11.1) to get the rst term (11.2). We compute the second52 Surface waves: Finite frequency theoryintegral in (11.2)I =_T/2T/2cos 2(t +T/2)Teitdt=1i_T/2T/2cos 2(t +T/2)Td(eit)=1i_cos 2(t +T/2)TeitT/2T/2 2T_T/2T/2sin 2(t +T/2)Teitdt_=1i_eiT2eiT22iT_T/2T/2sin 2(t +T/2)Td(eit)_=2 sin(T/2) +2T2_sin 2(t +T/2)TeitT/2T/2+2T_T/2T/2cos 2(t +T/2)Teitdt_= Tsinc_T2_+_2T_2I.Solving the above equation for the integral I givesI =2T32T242sinc_T2_.Inserting this into (11.2), we get the spectrum of the cosine taperW() =T2 sinc_T2_122T32T242sinc_T2_=22T422T2sinc_T2_.The power spectrum of the cosine taper isP() = [W()[2=44T2422T2sinc_T2_.The second column of 11.1 plots w(t), W() and P() of the cosine taper.Comparison of the spectra of the two windows shows that the boxcar window hasanarrowerpeak(moredelta-functionshaped), butlargersidelobesspreadingabroader range of frequencies. On the other hand, the cosine tapered window hasvery small sidelobes at the expense of a somewhat broader central peak.12Model parametrizationExercise 12.1 We know that the radial functions are orthogonal_a0fk(r)fk (r)r2dr = kk , (12.1)and the angular functions are orthogonal_0Xml()Xm

l () sind =12ll . (12.2)The orthogonal condition of basis functions hklm(r, , ) isI =_a0_0_20hklm(r, , )hk

l

m (r, , )r2sindrdd = kk ll mm .(12.3)Rewrite (12.2) in the textbook ashklm(r, , ) = fk(r)X|m|l()m(), (12.4)wherem() =___2 cos m l m < 01 m = 02 sinm 0 < m l. (12.5)Insert (12.4) into the expression of I in (12.3) and apply (12.1)-(12.2)I =_a0_0_20fk(r)X|m|l()m()fk (r)X|m

|l

()m ()r2sindrdd=_a0fk(r)fk (r)r2dr_0X|m|l()X|m

|l

() sind_20m()m ()d= kk ll

12_20m()m ()d (12.6)5354 Model parametrizationComparison of (12.6) and (12.3) shows that we only need to proveI1 =12_20m()m ()d = mm . (12.7)Since (12.5) shows that m() has different expressions depending on the value ofm, we shall prove (12.7) for each combination separately. Also notice that m andm

are symmetric in (12.7).(i) m = m

= 0I1 =12_201 1 d = 1 = mm .(ii) m = 0, l m

< 0 (or l m < 0, m

= 0)I1 =12_202 cos m

d = 0 = mm .(iii) m = 0, 0 < m

l (or 0 < m l, m

= 0)I1 =12_202 sinm

d = 0 = mm .(iv) l m < 0, l m

< 0I1=12_202 cos m2 cos m

d=12_20[cos(m+m

) + cos(mm

)]d=12_20cos(mm

)d.If m = m

I1 =12_20d = 1 = mm .If m ,= m

I1 =12(mm

) sin(mm

)20= 0 = mm .Model parametrization 55(v) l m < 0, 0 < m

l (or 0 < m l, l m

< 0 )I1=12_202 cos m2 sinm

d=12_20[sin(m+m

) sin(mm

)]d=12_1m+m

cos(m+m

) +1mm

cos(mm

)_20(since m ,= m

)= 0 = mm .(vi) 0 < m l, 0 < m

lI1=12_202 sinm2 sinm

d=12_20[cos(m+m

) + cos(mm

)]d=12_20cos(mm

)d= mm .The last step uses the result in 4.Thus, we have proved (12.7). Inserting (12.7) into (12.6) gives (12.3), i.e. the basisdened by (B12.2) is orthogonal.Exercise 12.2 The basis hk(r) is orthogonal means_Vhk(r)hl(r)d3r = kl, (12.8)where Vis the volume over which the basis is dened. Expand the model m(r)into the basis hk(r)m(r) =

kakhk(r). (12.9)Multiply both sides of (12.9) by hl(r) and integrate over V_Vm(r)hl(r)d3r =_V_

kakhk(r)_hl(r)d3r =

kak_Vhk(r)hl(r)d3r.Using the orthogonal condition (12.8), we get_Vm(r)hl(r)d3r =

kakkl = al.56 Model parametrizationExercise 12.3 For an arbitrary corner of the rectangular box (xl, ym, zn), (B12.5)becomesm(xl, ym, zn) =2

i,j,k=1m(xl, ym, zn)_1 [xlxi[x2x1__1 [ymyj[y2y1__1 [znzk[z2z1_.(12.10)Whenever indices in the numerators are unequal, both numerator and denominatorare equal to the length of the box. Therefore, the only nonzero term in this sumoccurs when i = l, j = m, k = n, so thatm(xl, ym, zn) = m(xl, ym, zn) 1 1 1 = m(xl, ym, zn).Therefore, the interpolated value of any corner equals the assigned value of thatcorner.Exercise 12.4 The penalty function is (B12.7):E =N

i=1

jNi(Lijlij)2l2ij,where Lij =_(xixj)2+ (yiyj)2+ (zizj)2and the lij are given. E sumsover all natural neighbours of node i, on the other hand, node i is the natural neigh-bour for any j ^i. Therefore, every node pair (i, j) appears twice in the sum:Exk= 2xk

jNk(Lkjlkj)2l2kj= 4

jNkLkjlkjl2kjLkjxk= 4

jNkLkjlkjl2kj

122(xkxj)Lkj=

jNk4_1 lkjLkj_xkxjl2kj.Exercise 12.5 The excess topography is given byte() = 23P02(cos ). (12.11)We recognize:P02(cos ) =_54_12X02()Model parametrization 57Since te() has only one nonzero harmonic term, with P02(cos ), the only nonzerocoefcient is a20. We use the expression in section B12.4 and the orthogonality ofXm

:a20 = 223

_54_12_0X02X02 sind = 23

_54_12.13Common correctionsExercise 13.1 If p() is the slowness of the wave arriving at distance :() = T() p(), (13.1)withp() =dT()d. (13.2)Differentiate (13.1) w.r.t. dd=dTd p() dpd = dpd,where the last step uses (13.2). Thenddp=ddddp= dpd ddp= .Since > 0, the intercept time is a monotonically decreasing function of p.Exercise 13.2 In local problems, we use the Cartesian coordinates. Thenp=sini/c and the epicentral distance is measured byX(dimension of length). InFigure 13.5 in the textbook, sea level r = a corresponds to z = 0 in the Cartesiancoordinates, and r = a+h corresponds to z = h. Similar to the spherical case, wehavet(p=constant) = 2_h0dzccos i,X = 2_h0tani dz = 2_h0sinicos i dz.58Common corrections 59Then= t(=constant) = t(p=constant)pX = 2_h0_1c cos i p sinicos i_dz.Using p = sini/c, we get= 2_h0_1c sin2ic_dzcos i= 2_h0cos icdz = 2_h01c_1 p2c2dz,Exercise 13.3 In a spherical earth, p=r sini/c cos i =_1 sin2i =_1 p2c2/r2; then (B13.5) can be written as= 2_a+hacos icdr =_a+hacos i1c1dr +_a+hacos i2c2dr,where the rst integral is for the incoming wave, and the second integral is for thereected wave. Straight rays imply that the velocities c1, c2 and the incident anglesi1, i2 are constants, so=cos i1c1_a+hadr + cos i2c2_a+hadr = h_cos i1c1+ cos i2c2_. (13.3)This topographic correction conforms to (B13.3). Because a surface reection isalways an underside reection, there is only a positive sign in (13.3), though h canbe either positive or negative.14Linear inversionExercise 14.1 With univariant data (i = 1), the expression for 2is2(m) =N

i=1__M

j=1Aijmjdi__2. (14.1)To minimize (14.1), all derivatives with respect tomk(k=1, ..., M) should bezero:2mk=mkN

i=1__M

j=1Aijmjdi___M

l=1Ailmldi_=N

i=12__M

j=1Aijmjdi__Aik= 2N

i=1M

j=1ATkiAijmj2N

i=1ATkidi = 0.The matrix form of the last equation is (after dividing by 2):ATAmATd = 0.Exercise 14.2 Because no inverse for ATexists, the linear equations ATx = 0have non-zero solutions.Exercise 14.3 From (B14.12):VTAT= UT60Linear inversion 61Multiply on the right by U:VTATU= UTUApply (B14.11) and the orthonormality of V :VTV = = UTUFrom which it immediately follows that the ui are orthonormal if i ,= 0. For zeroeigenvalues the normalization of ui is arbitrary, we can set the length to 1 withoutaffecting (B14.11) in any way.Exercise 14.4 Taking the transpose of (B14.14):AT= VKKUTK ,and inserting this into (B14.7): m = VK2KVTKATd = VK2KVTKVKKUTKd = VK2KKUTKd= VK1KUTKd.Exercise 14.5 The normal equations for (B14.15) are found by multiplying leftand right with the transpose of the matrix:_AT

nI__A

nI_m =_AT

nI__d0_Multiplying the matrices:(ATA+2nI)m = ATd.Exercise 14.6 In the L-curve (Fig. B14.2), the rightmost point has the smallest2and the largest [m[, so this is where n = 0. The intercept with the vertical axishas the largest 2and the smallest [m[, so this is where n .15Resolution and error analysisExercise 15.1 Because such an arbitrarily scaled model does not satisfy the linearsystemAm = d.The fact that we increase the estimate of the model variance does not make upfor the fact that we do not guarantee that the scaled model satises the data at anacceptable 2level.6216AnisotropyExercise 16.1 Eq. (B16.1) gives the Voight matrix for isotropic solids:_________ + 2 0 0 0 + 2 0 0 0 + 2 0 0 00 0 0 0 00 0 0 0 00 0 0 0 0 _________, (16.1)The rst three rows/columns involve the combinations xx,yy and zz. Exchangeof x and y, for example, leaves the diagonal and off-diagonal elements the same,and so for other exchanges. For the pairs yz,xz and xy in columns 4,5,6: exchangeresultsinexchangeofdiagonalelementsthatareallequalto, sothereisnochange in the Voight matrix.Exercise 16.2 For hexagonal symmetry we have:_________A A2N F 0 0 0A2N A F 0 0 0F F C 0 0 00 0 0 L 0 00 0 0 0 L 00 0 0 0 0 N_________.As long as we leave the z-axis untouched, the third and sixth row and column donot change. The symmetry in the other rows/columns is as in the previous exercise,so again the matrix is invariant.63