Introduction to Seismology - Snell's Law

SNELL’S LAW REVISITED General approach 1. Potentials 2. BCs 3. Solve the system of equations 4. Reflection and transmission coefficients Recall Helmholtz representation: u = ∇φ + ∇ × ψ , ∇ ⋅ ψ = 0 , SV Æ solve for φ and ψ Æ reconstruct u SH Æ uy satisfies wave equation. Therefore, SH part doesn’t require potentials to find solutions. But you could analyze the SH part through potentials also by using the following Helmholtz presentation of the shear wave potential: ψ = ∇ψ SV + ∇ × ψ SH , ∇ ⋅ ψ SH = 0 so u=∇ φ + ∇ × ∇ψ SV + ∇ × ∇ × ψ SH , ∇ ⋅ ψ SH = 0 See Stein and Wysession for a good { 1424 3 14243 development. P SV SH But, customarily, we do not solve the problem this expression. In general, P-SV and SH waves are not decoupled. They are only if the medium is isotropic ) (heterogeneity is permissible) and if the normal, n , at an interface is in the plane of propagation. j2 Therefore, if the interface dips left or right with respect to the page, the P-SV and SH systems are still decoupled, but if the interface dips transverse to the page then the P-SV and SH systems are ) no longer decoupled - n no longer lies within the plane of propagation. ■ Another way to approach Snell’s Law … consider: Free surface BC requires that Tz = (σ xz , σ yz , σ zz ) = 0 , Figure 1 / P i i* \ P j \ SV and the stress components contain expressions of the form: ⎞ ⎞ ⎛ sin j ⎛ sin i ⎞ ⎛ sin i * x − t ⎟, ω ⎜ x − t ⎟, ω ⎜⎜ x − t ⎟. ω⎜ ⎟ ⎠ ⎝ β ⎝ α ⎠ ⎝ α ⎠ As the stress components vanish, they must all be equal, hence: sin i sin i * sin j = = ≡ p, ray parameter Æ i = i*. α α 1 β Now consider the interface between two layers of differing elastic properties … / S1 \ P1 / j1 i1 P1 i1* i2* Figure 2 \ P2 j2 \ S2 As before, @ z = 0, the displacement and traction elements contain the following expressions: ⎞ ⎞ ⎛ sin i2* ⎛ sin i1 ⎞ ⎛ sin j1 ⎞ ⎛ sin i1* ⎞ ⎛ sin j1 ⎜⎜ ⎟, ⎜ ⎟, ⎜ ⎜⎜ ⎜⎜ − t ⎟, − t − t ⎟, − t − t ⎟⎟ , ⎟ ⎟ ⎟ ⎜ ⎟ ⎜ ⎠ ⎠ ⎝ β2 ⎠ ⎝ α2 ⎠ ⎝ β1 ⎠ ⎝ α1 ⎝ α1 all of which must be equivalent to satisfy the kinematic and dynamic boundary conditions. Hence, sin i1 α1 = sin i1* α1 = sin i2* α2 = sin j1 β1 = sin j2 β2 ≡p The ray parameter is constant for the entire wave field! ■ From a geometric point of view, consider plane wave propagation across several interfaces, as below. The wave vector is different in each layer, but the horizontal distance (∆x) traveled along any interface (over an interval of time, ∆t) must be the same between interfaces, because the BCs require that wave fronts must propagation direction wave fronts be continuous across boundaries. Hence, the apparent horizontal velocity ∆x (cx) is everywhere constant. And since cx is constant, it follows that p = 1/cx (the ray ∆x Figure 3 parameter) is constant for the entire wave field. 2 This concept can be extended to a spherical body, but you will find that a singularity exists at the point where a downgoing raypath becomes horizontal, as it inflects from its downgoing to its upgoing branch. ■ 1/ 2 1/ 2 ⎛κ + 4 µ ⎞ 3 ⎟ > β = ⎛⎜ µ ⎞⎟ Æ 1/α < 1/β. And since sin i = sin j , it follows that Normally, α = ⎜ ⎜ρ⎟ ⎜ ⎟ ρ α β ⎝ ⎠ ⎝ ⎠ j < i. This means that converted or reflected S-waves will usually travel raypaths bent more toward the normal than transmitted or reflected P-waves. Another way of seeing this is by the Dispersion Relation. The dispersion relation states that kx ω k = (k x2 + k z2 ) = . From this fact, the shear and c compressional components of the wave vector are given by: kz, α ω) S-wave: k = k β kα β kz, β ω) P-wave: k = k α kβ α But β < α, so k β > k α , and since kx is a constant, it follows that kz,β > kz, α, meaning that the S-wave propagation direction is typically nearer to the normal than is the P-wave propagation direction. ■ Figure 4 CRITICAL ANGLES 1. Free Surface (P-SV system) The incident angle at which an incoming S-wave is first critically reflected is the angle jc that causes i = 90º. Figure 5 sin j c / j S j* i \ S Recall that p 2 + η α2 = p= sin jc β = 1 α 1 α 2 \ P β = sin i α = ⎛β ⎞ Æ jc = sin −1 ⎜ ⎟ α ⎝α ⎠ 1 The converted P-wave is critically reflected, meaning ( p ,0 ) . that kz,α = 0. So, s = ( p,ηα ) ⎯⎯⎯⎯⎯→ critical reflection . But ηα = 0 → p 2 = 1 α 2 or ; The ray parameter is exactly 1/ α at the point of critical reflection Sub-critical: j < jc ↔ ηα > 0 3 Critical: j = j ↔ η α = 0 c Post-critical: j > j c ↔ ηα is complex (evanescent waves!) 2. Welded interface (SH case) / \ S1 S1 j1 / \ \ \ S2 j2 S1 jc S1 S2 j2 Figure 6b β1 > β2 Æ j2 > j1 @ jc, j2 Æ 90º 3. Welded interface (SV case) / \ P1 Figure 7a / / S1 P1 i S1 \ i1c P1 \ \ S2 Figure 7b P2 \ S2 / P1 \ P2 / S1 / \ P1 P1 i2c \ \ P2 , S 2 Figure 7c Figure 7a: Reference case Figure 7b: 1st critical reflection; i1c is such that downgoing P2 is critically reflected at the interface. Figure 7c: 2nd critical reflection; i2c is such that downgoing S2 is also critically reflected at the interface. Note that downgoing P2 is now post-critical. i2c > i1c. Note that the transmission and refection coefficients will be real-valued for sub-critical phenomena but become complex-valued for post-critical. ■ 4 FERMAT’S PRINCIPLE Hamiltonian variational principle · “Least action” Consider the kinetic and potential energy along an arbitrary path between two points, A and B. Stationary points of the integrated difference between KE and PE over all possible paths specify paths of “least action”. Optical domain There are an infinite number of paths from A to B, but there is only one ‘correct’ path: the one with the shortest travel time. Both the shortest and the longest paths are stationary points. B A Short arc Long arc Casting the problem mathematically … P d a i x O j b e Q Figure 8 c TP−Q = d e + c1 c 2 (a = 2 + x2 c1 ) 1/ 2 (b + 2 + (c − x ) c2 ) 2 1/ 2 dT x c−x sin i sin j − = = = 0 (optimum!) + 1/ 2 1/ 2 2 2 2 2 dx c1 a + x c1 c2 ( ) c b + c − x { { 2 144244 3 14442444 3 ( ) ( ) 5 sin i Hence, Law. = sin j c1 (Snell’s Law). So Fermat’s Principle implies Snell’s c2 ■ The travel time curve, plotted as a function of offset, is typically a hyperbolic function. Near stationary points, the curve is usually fairly ‘flat’, which implies that, near optimum or stationary points, the travel time is locally insensitive to slight variations in offset. Consequently, close to a stationary point, small deviations in the ray path can be treated as second order effects. P Slight perturbations in the ray path play a second order role in the observed travel time. T(x) Figure 9 x Interval over which offset perturbations can be treated as secondary effects. Q For example … B A • Fast Body • • parameters) is first order. The presence of a fast body embedded in a homogeneous medium causes the reference (optimal) ray path to deflect from its original path between A and B. But, as a consequence of Fermat’s Principle, the change in ray path produces a second order effect on the arrival time. The effect on travel time of changes in wave speed (elastic Implication: We can generate reference ray paths assuming a homogeneous medium and use this reference model to simplify subsequent analysis for heterogeneous media. Æ Important for travel time tomography. ■ REFLECTION AND TRANSMISSION COEFFICIENTS Reflection coefficient: R ≡ Areflected Aincoming Transmission coefficient: T ≡ Atransmitted Aincoming 6 SH Case Introduction to Seismology: Lecture Notes 28 February 2005 Consider an SH-wave incident at an interface between two layers of different elastic properties. Because SH-waves are decoupled from P and SV waves, the transmitted and reflected waves will also be SH waves. The displacement of the downgoing wave in the Layer 1 (see Figure 10) can be expressed as \ \ u1 ( x, t ) = A1 g ( p,η1 , x, t ) where, for notational convenience g ( p,η1 , x, t ) ≡ exp[iω ( px + η1 z − t )] η1 ≡ η1,β x ≡ ( x, z ) and the reflected, upgoing wave in Layer 1 can be expressed as / / u1 (x, t ) = A1 g ( p,−η1 , x, t ) . Now, setting the interface to z = 0, the total displacement in Layer 1 is / \ / \ u1 + u1 = A1 g ( p,η1 , x, t ) + A1 g ( p,−η1 , x, t ) . The displacement in Layer 2 is given by \ \ u2 = A2 g ( p,η 2 , x, t ) . The kinematic boundary condition requires that the displacement be continuous across the interface at z = 0. Therefore, setting z = 0 and equating the displacements in Layers 1 and 2, we arrive at the following statement: \ / \ \ / \ \ / S1 u1 + u1 = u2 ⇔ A1 + A1 = A2 (1) \ u1 and 1 / R= A1 / u1 Figure 10 z=0 \ A1 2 \ T= S1 A2 \ A1 The dynamic boundary conditions require that 7 \ u2 \ S2 ⎛ ∂ui σ ij = λδ ij ∆ + 2 µε ij = λδ ij ∆ + µ ⎜⎜ ⎝ ∂x j + ∂u j ⎞ ⎟, ∂xi ⎠⎟ and recalling , for SH waves, u = (0, u y ,0) , we have: σ xz = 0, σ yz = µ ∂u y ∂z , σ zz = 0 \ \ ∂u σ yz ,1 = µ1 y ,1 = iωµ1η1 A1 g ( p,η1 , x, t ) ∂z \ / / ∂u σ yz ,1 = µ1 y ,1 = −iωµ1η1 A1 g ( p,−η1 , x, t ) ∂z / \ \ σ yz , 2 \ ∂u = µ1 y , 2 = iωµ2η2 A2 g ( p,η2 , x, t ) ∂z hence, @ z = 0, σ 1 = σ 1 + σ 1 = iωµ1η1 ⎛⎜ A1 − A1 ⎞⎟ exp[iω ( px − t) )] / \ \ / ⎝ \ ⎠ . \ σ 2 = σ 2 = iωµ2η2 A2 exp[iω ( px − t) )] The preceding facts imply that µ1η1 ⎛⎜ A1 − A1 ⎟⎞ = µ 2η 2 A2 …(2) \ / ⎝ \ ⎠ ■ The reflection and transmission coefficients are expressed as / \ / Rss = S S = A1 \ A1 \ \ \ Tss = S S = A2 \ A1 ■ If we re-arrange Equ. 2. / \ \ A1 − A1 = − A 2 µ 2η 2 µ1η1 Combining equs. 1 and 2 give the Zoeppritz Equation 8 1 ⎞⎛ / ⎞ −1 ⎛1⎞ ⎛⎜ ⎟⎜ A1 ⎟ µ A1 ⎜⎜ ⎟⎟ = ⎜ 1 2η 2 ⎟⎜ \ ⎟ ⎝1⎠ ⎝ µ1η1 ⎠⎝ A 2 ⎠ \ Where the left hand side (LHS) is the incoming wave and the RHS are the resulting waves. Again, if we re-arrange the Zoeppritz Equ: \ / ⎞ ⎛/ ⎛\ ⎞ ⎜ A1 − A1 ⎟ µ1η1 = −⎜ A1 + A1 ⎟ µ 2η 2 ⎠ ⎝ ⎝ ⎠ / \ A1 (µ1η1 + µ 2η 2 ) = A1 (µ1η1 − µ 2 µ 2 ) s Then the reflection coefficient is: / R≡ A1 \ = A1 µ1η1 − µ 2η 2 µ1η1 + µ 2η 2 And to calculate the transmission coefficient: ⎛ µ η − µ 2η 2 A 2 = A1 + A1 = A⎜⎜1 + 1 1 ⎝ µ1η1 + µ 2η 2 \ \ / \ \ T= A2 \ = A1 \ ⎞ \ ⎛ 2 µ1η1 ⎟⎟ = A1 ⎜⎜ ⎠ ⎝ µ1η1 + µ 2η 2 ⎞ ⎟⎟ ⎠ 2 µ1η1 µ1η1 + µ 2η 2 The dependence on the incidence angle, i1 , is illustrated by cos i1 cos i 2 η1 = ,η 2 = β1 β2 And β2 = µ ρ Using these variables to re-write the reflection and transmission coefficients: ρ β cos i1 − ρ 2 β 2 cos i2 ρ1 β1 − ρ 2 β 2 R= 1 1 = ρ1 β 1 cos i1 + ρ 2 β 2 cos i2 ρ1 β 1 + ρ 2 β 2 T= 2 ρ1 β 1 cos i1 2 ρ1 β 1 = ρ1 β 1 cos i1 + ρ 2 β 2 cos i2 ρ1 β1 + ρ 2 β 2 The reflection and transmission coefficients are equal to the RHS of the equations above for the specific case of normal incidence, when i1=i2=0. 9 / / \ For a welded interface we can also have A2 → A1 and A2 / A1 Figure 11 / A2 i1 i2 \ A2 Thus we can define a scatter matrix S: ⎛/ / S = ⎜ S/ S\ ⎜S S ⎝ \ / ⎞ S S⎟ \ \ S S ⎟⎠ (Rob showed figure on overhead of R and T versus angle of incidence for pre-critical and critical angles. The reflection coefficient lies below 0 for i1=0. At the intramission angle the reflection coefficient is 0 and increases to a value of 1 and the critical angle, ic. The transmission coefficient is less than 1 at i1=0 and increases to a value of 2 at the critical angle, ic.) 10