Sismik Prospeksiyon ( ingilizce ) Introduction to Seismology - 11 12.510 Introduction to Seismology Surface Waves (Ground Roll) April 2, 2008 Today we will look at the interaction of an acoustic wave (ground roll) with a layer over half space. We will consider travel time curves for acoustic waves and describe them using higher modes of surface wave propagation. We will also use a propagation matrix as a re?ec tivity method to calculate synthetic seismograms. Case 1: Layer over Source in a Half Space Lets look at what happens when we have 2 interfaces. We are working with acoustic waves, so we will take P = pressure ?eld. Note that the methods used here can be used for SH waves as well. The same principles can be applied to P-SV waves, but the algebra becomes more complicated. Figure 1: Diagram of a source in a half space under a layer 1 The wave equation for the acoustic case is P ¨ = k? · ( ? 1 ? P ) (1) with the displacement given by 1 u = ?? 2 ? P (2) This case becomes a bit more complicated than the simple re? ection previously discussed. The analytical methods used to decribe the simple re?ection begin to break down when multiple layers are introduced. ´ ` We will renormalize the incoming wave to P 2 = 1 and de?ne P 2 = R. Using the plane wave description, ´ ` P = Pe i(kxx- kz z- ? t) + Pe i(kxx- kz z- ? t) Notice the positive and negative k z z terms to describe the vertical slow­ ´ ` ness, as well as the di?eren t amplitudes P and P . We want to know what is happening with respect to crossing the interface, so we will ignore the x-direction, leaving the e ikz z terms and giving ´ ` P = Pe - i? z + Pe - i? z where the vertical wave number is given by ? z = k z = ? = ? cos i = ? ? cz c ´ ` P = Pe - i? z + Pe - i? z can be set up for each layer. We can solve for the pressure by taking the gradient in the z-direction: ? P 1 Pe i? z i? Pe i? z Pe i? z ) Pe i? z ) = ( ` ´ u z = ? z = ?? 2 (i? ` - i? ´ ?? 2 - To solve this system we will follow the steps: 1. Look at Potentials (Pressure Field) 2. Kinematic and Dynamic Boundary Conditions 3. Zoeppritz equations 2 4. Solve for R and T At z=0, the welded interface, the stress is continuous so ´ ` ´ ` P 2 + P 2 = 1 + R = P 1 + P 1 The displacement at z = 0 is given by i? u(0) = ?? 2 (P ` 1 - P ´ 1 ) which implies that at the free surface: P ` 2 - P ´ 2 = ? 2 ? 1 (P ` 1 - P ´ 1 ) = ? 2 (R - 1) ? 1 ? 2 ? 2 At the free surface z = - H P ´ 1 e - i? 1 H + P ` 1 e i? 1 H = 0 where ? 1 is the vertical wave number k z = c ? z . ? ? The three equations above give the Zoeppritz matrix and we solve for R 1 1 - 1 ? - ? 2 ? ? ? ? ? ´ P 1 1 ? ? ? ? ? ? ? ? ? ? ` - ? 1 ? 1 ? 1 - 1 0 P 1 = ? 1 ? 2 e - i? 1 H e i? 1 H 0 R (1- ? )+(1+? )e 2i? 1 H R = (1+? )+(1- ? )e 2i? 1 H where ? = ? 1 ? 2 ? 2 ? 1 Some remarks on the re?ection at the free surface: - The re?ection coe?cien t is a complex number. In the cases we studied in previous lectures, R became complex when i = i c . -|R| = 1, which means that the energy is not stored in the upper layer but is eventually all — re?ected back. E.g., if the input is a single spike, the output is a series of reverberations (see homework 2, problem 3). This occurs because of conservation of energy; if there is no source in the upper layer, there can be no residual trapped energy in the upper layer. The en­ ergy is re?ected from z =?H and both re?ected and partially transmitted from z = 0 until it fully dissipates back into the half space. - There is a frequency?dependent phase shift ? ? in the waveform, ie. R = ? e 2i? e i(kx- ? t+ ? ) . Case 2: Source Within a Layer Over a Half Space The source within the layer is analagous to a source in a weathered layer, e.g. an induced source in exploration seismology or surface waves in earth­ quake seismology. This scenario is similar to the Love waves case discussed in previous lectures. In the case of angles such that i¿ic (evanescent waves), 3 no energy will be transmitted into the half space. Figure 2: Diagram of a source in a layer over a half space To solve this system we will follow the steps as before: 1. Look at Potentials (Pressure Field) 2. Kinematic and Dynamic Boundary Conditions 3. Zoeppritz equations 4. Solve for R and T Because stress must be continuous, ´ ` ´ ` ´ ` P = P 1 + P 1 = P 2 + P 2 = 0 + T ? P 1 + P 1 - T = 0 The total displacement must be continuous at the interface, so ?? 2 ? P ?? (Pe ` i? z + ´ u = 1 = i? Pe - i? z ) ? 1 ? ? 1 (P ` 1 - P ´ 1 ) = ? ? 2 2 T ? (P ` 1 - P ´ 1 ) = ? ? 2 1 ? ? 1 2 T = ? T ? P ` 1 - P ´ 1 - ? T = 0 where ? = ? ? 2 1 ? ? 1 2 = ? ? 2 1 c c 2 1 = z z 2 1 and ? 1 c 1 =acoustic impedance. At z = - H, P ` 1 + e 2i? 1 H P ´ 1 = 0 Notice that the T in these equations is not the conventional T as before because it is evanescent. In contrast to previous cases, we can now form a homogeneous set of Zoeppritz equations: ? ? ? ? ? ? ` 1 1 - 1 P 1 0 ? ? ? ? ? ? ? ? ? ? ? ? ´ 1 - 1 2i? 1 ? 0 - ? 0 T 0 P 1 = 1 e 4 ? ? ? The solution of this system depends on the layer thickness (H), the ver­ tical wave number (? ), the direction of the rays ( ? cos i ), and the impedence c contrast (? = z z 2 1 ). To obtain the nontrivial solution to this system, we must set the determinant of the 3x3 term to zero, leading to ? ? 1 1 - 1 det ? ? 1 - 1 - ? ? ? = 0 1 e 2i? 1 H 0 ? ? e 2i? 1 H - e 2i? 1 H - ? - 1 = 0 ? (? - 1)e 2i? 1 H = ? + 1 2i? 1 H ? +1 e = ? ? - 1 We know that in the complex plane, generally, e - 2i? = a- b , giving a+b tan ? = - a ib . Using this, we can rewrite our nontrivial solution as tan (? 1 H) = 1 = ? 2 ? 1 i? i? 1 ? 2 tan(? 1 H) = 1 = ? 2 ? 1 ? i? i? 1 ? 2 where ? 1 = ? ( c 1 1 ) 2 - p 2 and ? 2 = ? ( c 1 2 ) 2 - p 2 . The dispersion relationship for ground roll (which is the same as the dispersion relationship for Love waves) is thus given by tan ? ? H ? ( c 1 1 ) 2 - p 2 ? = ? 2 ? ( c 1 1 ) 2 - p 2 ? 1 ( 1 ) 2 - p 2 c 2 For a given frequency and layer thickness H, for certain directions given by p = sin c 1 i 1 = sin c 2 i 2 and known c 1 ,c 2 , we can solve this system. Given ?xed ? , ie. i 1 = 90 o solves the dispersion equation, ie. the direct wave is a solution to the dispersion equation. See Figure 3. 5 Figure 3: Diagram of (1) a direct wave and (2) a critical wave Figure 4: Graphical representation of the dispersion relationship for ground roll. 6 For all postcritical angles i, or c 1 2