Reflection+ Transmission Coefficients for 2-Layer Problem

Outline 1.Two-Layer Problem: Reflection+Transmission Coefficients, Snell’s Law, Refractions, Critical angle, Post-critical Waves.

D = e i(k x - k z - wt) x z (2) U = R e i(k x + k z - wt) x z xz (2) increase x, z { - increase t { D = Te i(k x - k’ z - wt) x z k = w – k xz 222c k’ = w – k xz 222c’  c’ c  ’ 2-Layer Medium: Find R & T

(2) xz  c’ c  ’ 2-Layer Medium: Find R & T Te Te i(k x - k’ z - wt) x z P = - e i(k x - k z - wt) x z P = U + D = +e i(k x + k z - wt) x z + R Two unknowns R & T, need two equations of constraint

B.C. #1: Pressure Must Match at z=0 ik x Te Te x e ik x x e x + R = +- P = Te Te i(k x - wt) x e x e x + R Divide by e i( wt) Implies k = k xx sin O = sin O’ cc’ ik x x O O  c’ c  ’ Snell’s Law

Pressure must match at z=0 +- P = OO ik x Te Te x e ik x x e x + R = ik x + - w = w.... Recall w = -1 dP/dz .. ik x -ik Te -ik Te x -ik e ik x x e x + ik R = z=0  c’ c  ’ B.C. #2: Particle Accel. Must Match at z=0  Vertical acceleration z=0 so w+ = z=0

Pressure & accelration must match at z=0 OO ik x Te Te x e ik x x e x + R = ik x -ik Te -ik Te x -ik e ik x x e x + ik R = c’ cos O - c cos O c cos O + c’ cos O , ,  R =  c’ c  ’ c’ cos O c cos O + c’ cos O , , T = 2 

Special Cases: Bright Spots c’ cos O - c cos O c cos O + c’ cos O , ,  R = O = 0: c’ - c , R = c’ + c ,  Positive Refle. Coeff when Model is low over hi velocity Negative Refle. Coeff when Model is hi over low velocity BRIGHT SPOT Gas sand has lower impedance than wet sand. Works well in GOM and young sedimentary basins

Special Cases: Critical Angle c’ cos O - c cos O c cos O + c’ cos O , ,  R = O = critical angle: O O  c’ c  ’ sin O = sin O sin O = sin O’ c c’ 1 < c c’

Special Cases: Post Critical Reflections c’ cos O - c cos O c cos O + c’ cos O , ,  R = O = critical angle: O O  c’ c  ’ sin O = sin O sin O = sin O’ c c’ 1 < c c’

c’ cos O - c cos O c cos O + c’ cos O , ,  R = O = critical angle: O O  c’ c  ’ sin O = sin O sin O = sin O’ c c’ 1 < c c’ Special Cases: Post Critical Reflections

c’ cos O - c cos O c cos O + c’ cos O , ,  R = O = critical angle: O O  c’ c  ’ sin O = 1 c c’ 1 < c c’ Special Cases: Post Critical Reflections

c’ cos O - c cos O c cos O + c’ cos O , ,  R = O = post critical angle: O O  c’ c  ’ sin O > 1 c c’ 1 < c c’ Total Energy Reflection Post Critical Phase Shift Special Cases: Post Critical Reflections

Special Cases: Post Critical Reflection Coeff. c’ cos O - c cos O c cos O + c’ cos O , ,  R = sin = c’sinO O c = 1 – 2 c’sinO c’sinOc { { c > c’sinO c’sinOc { { = i c’sinO c’sinOc { { = e i  /2 e Ocos = 1 – sin O 2 A B e i  /2

Special Cases: Post Critical Reflection Coeff. c’ cos O - c cos O , ,  R = c’ A - c e i  /2 ,  B c’ A + c e i  /2 ,  B Reflection coefficient is now a complex number U = R e i(k x + k z - wt) x z e i c’ cos O c cos O c’sinO c’sinO c { { = e i  /2 B e cos O = cos O = A Phase change in upgoing post critical reflections Havoc AVO and stacking!

Layered Medium & Critical Angle 3 km/s 60 X (km) Z (km) Time (s) 0 Post-critical reflection ray Post-critical reflections Sea floor km/s CSGModel

Special Cases: Pressure+Marine Free Surface c’ cos O - c cos O c cos O + c’ cos O , ,  R = Free Surface  ’c’=0: R = - c cos O c cos O   = -1 P = 1 + R = 1 – 1 = 0 1 ghosts

c’ cos O + c cos O c cos O + c’ cos O , ,  R = Free Surface  ’c’=0: R = + c cos O c cos O   = 1 w = 1 + R = = ghosts Special Cases: Part. Vel. + Land Free Surface Part.

Special Cases: Part. Vel. + Land Free Surface Why is particle velocity R opposite polarity to pressure R? ik x -ik Te -ik Te x -ik e ik x x e x + ik R = so w+ = z=0 Recall w = -1 dP/dz .. - D D - U R = zz z Downgoing particle-z velocity Down particle-z vel. Up particle-z vel. R =-R part.press.

Special Cases: Impedance P = e ik z z Recall w = -1 dP/dz .. ik z z ik z dP/dz = e where ik z = P dP/dz = P Therefore w ..ik z = - P dP/dz = - P But w = -i w w.... Therefore w  ik z = P dP/dz = P iwiwiwiw or P/w =  c.Impedance

Special Cases: Conservation of Energy Flow cccc Previously we found that the Energy Density in deforming a cube was given by P 2 2 Therefore the rate at which energy is flowing across a flat interface by a propagating plane wave is by P cccc 2 Energy/area/time = = c 2 P cccc 2 Conservation of energy demands energy flow of incident wave is same as the transmitted and reflected wave: cccc 21 = R + T cccc  c’ 22 ’ Note: > 1 + cccc  c’ 2 ’ ’

Special Cases P = sinO1/v1 = sinO2/v2=sinO3/v3=sinO4/v4 v1 v2 v3 v4 Slowness = inverse apparent Vx

Time v1 v2 v3 v4 Slope = v5 v5

D = e i(k x - k z - wt) x z (2) U = R e i(k x + k z - wt) x z xz D = Te i(k x + k’ z - wt) x z  c’ c  ’ 2-Layer Medium: Find R & T P = U + D + P = D -

Special Cases: Post Critical Reflection Coeff. Post-critical reflection ray