General Reflection (and some Refraction) Theory Andrew Goodliffe University of Alabama Socorro, NM, Tuesday May 27.

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Presentation transcript:

General Reflection (and some Refraction) Theory Andrew Goodliffe University of Alabama Socorro, NM, Tuesday May 27

Seismic Reflection Surveying The most widely used and well known geophysical technique. A seismic section looks similar to a geologic cross- section – a trap for the unwary Only by understanding how the reflection method is used and seismic sections are created can geologists make informed interpretations.

Seismic Velocities Velocity depends on two main things: 1.Restoring force (analogous to the strength of a spring) As the restoring force increases, the velocity increases. 2.Mass (analogous to the mass of the spring) As the mass (density) increases, this will slow the spring, reducing the velocity S-waves: Passage involves a shear force resulting in a change in shape Size of the force depends on the shear, or rigidity modulus, μ. P-waves: Additionally involves a change in size Compressibility modulus κ is also involved. Where ρ = density. What is μ in a liquid? From Mussestt and Khan, 2000

Velocities How can we measure velocity? Refraction Velocity analysis (conventional, PSDM) Boreholes Vertical Seismic Profiles In situ logging – measuring the travel time of a high frequency acoustic pulse. Hand samples Travel time of a high frequency acoustic pulse Anisotropy Importance of confining pressure – P-wave velocity increases with confining pressure What are some typical seismic velocities?

Velocities i=1 i=2 i=3 Each layer is characterized by an interval velocity. If z 1 is the thickness of layer i and t i is the one-way travel time through it then the interval velocity of that layer is: The root-mean-square (rms) velocity of the section down to the nth interface can be approximated by: If a layer is 2000 meters thick and the two-way time (TWT) for a wave to travel through the layer is 500 ms, what is the interval velocity of that layer? What type of rock might this be?

Velocities v 1 =1500 m s -1 v 2 =2000 m s -1 v 3 =2345 m s -1 t 1 =2.14 s t 2 =1.21 s t 3 =1.13 s What is v rms at the base of layer 3?

Reflection and Transmission The proportions of the incident wave energy that are either transmitted or reflected at an interface are determined by the acoustic impedance (product of density, ρ, and velocity, v) From Kearey, Brooks, and Hill, 2002 Generally speaking, the “harder” the rock the greater its acoustic impedance. Maximum transmission of seismic energy requires a matching of acoustic impedances.

Reflection and Transmission Reflection coefficient R is a numerical measure of the effect of an interface on wave propagation. It is the ratio of the amplitude A 1 of the reflected ray to the amplitude A 0 of the incident ray: Expanding, this becomes: From Kearey, Brooks, and Hill, 2002 A negative value of R indicates a 180 o phase change in the reflected ray. What might cause this? If R = 0, all the incident energy is transmitted. Though Z = 0, ρ, and v may still be different. The transmission coefficient T is the ratio of the amplitude A 2 of the transmitted ray to the amplitude A 0 of the incident ray:

Reflection and Transmission From Mussestt and Khan, 2000

Reflection and Transmission From Goodliffe et al., 2001

Attenuation The energy E transmitted outwards from a source becomes distributed over a spherical shell From Kearey et al., 2002 If the radius of the wavefront is r, the amount of energy contained within a unit area of the shell is E/4πr 2. With increasing distance along a ray path, the energy contained in the ray falls of as r -2 due to geometrical spreading of the energy. Wave amplitude, which is proportional to the square root of the wave energy, falls of as r -1.

1.λ=5000/10 = 500 m. Attenuation = 1000/500 * 0.2 = 0.4 dB 2.λ=5000/231 = m. Attenuation = 1000/21.65 * 0.2 = 9.24 dB Attenuation The ground is imperfectly elastic – energy is gradually absorbed by internal frictional losses Absorption coefficient: proportion of energy lost during transmission through a distance equivalent to a complete wavelength – (dB λ -1 ) Higher frequency waves attenuate more rapidly than lower frequency waves as a function of time or distance 1.A 10 Hz seismic wave traveling at 5 km s -1 propagates for 1000 m through a medium with an absorption coefficient of 0.2 dB λ -1. What is the wave attenuation in dB due solely to absorption? 2.Repeat the above exercise for a 231 Hz seismic wave. 3.Comment on the differences.

What does resolution mean? What does detection mean? Dependant on seismic wavelength Individual reflectors clearly resolved when separated by > /4 v=f If v = 2000 m/s, and f = 30 Hz –Resolution = (66.67 m)/4 = m If v = 8000 m/s and f = 20 Hz –Resolution = (400 m)/4 = 100 m If v = 2000 m/s and f = 3500 Hz –Resolution = ( m)/4 = m Reflectors thicker than /10 can generally be detected. Vertical Resolution

Partly determined by distance between traces Also dependant on wavelength Parts of a reflector separated by less than the width of the Fresnel zone will not be resolved W f  (2z ) 1/2 z = depth If depth = 2000 m, = 60 m –W f  490 m If depth = 100 m, = 1 m –W f  14 m Horizontal Resolution

Snell’s Law The Earth is not a uniform sphere. Broadly speaking, it is made up of layers. When wave fronts cross from one rock type into another with a higher velocity they turn. The time between successive wave fronts remains unchanged, so the wavelength must increase in the second rock in proportion to the increase in velocity. Trigonometry tells us that: Rearranging gives: From Mussestt and Khan, 2000

Snell’s Law As BB’ and AA’ are in proportion to the velocities v 1 and v 2, the equation can be rearranged to Snell’s Law So i 2 = 48.8 o Answer the following question: A ray traveling in a rock with a seismic velocity of 3 km/s encounters an interface with a rock of 4 km/s at an angle of 45 o. At what angle from the normal does it leave the interface? From Mussestt and Khan, 2000

Snell’s Law – Multiple Flat Horizons As i’ 1 = i 1, I’ 2 = i 2, and so on From Mussestt and Khan, 2000 The ratio (sin i/v) thus remains unchanged From Mussestt and Khan, 2000

Snell’s Law - Refraction When a critical angle is reached the ray will travel along the interface between the two layers at velocity v 2 Beyond that critical angle, total internal reflection will occur A refraction survey is typically set up differently to a reflection survey – the former has much larger offsets between the source and receiver Note characteristic travel time curve: refractions form straight lines (slope = 1/v); reflections form hyperbolae From Mussestt and Khan, 2000

Seismic Sources

High Pressure Air Sources: The Air Gun ReadyFire!Fired Lower chamber has a top diameter that's smaller the bottom diameter - air pressure forces the piston down and sealing the upper, firing chamber. High pressure air is filling the firing chamber through the T- shaped passage, and the firing, or actuating air passage is blocked (solid black) by a solenoid valve. Full pressure has built up in the upper chamber. The Solenoid has been triggered, releasing high- pressure air into the active air passage, which is now yellow. The air fills the area directly below the piston, overcoming the sealing effect of the air in the lower, control chamber. The piston moves upwards, releasing the air in the upper chamber into the water. A large bubble of compressed air is expanding into the surrounding water. The air in the lower control chamber has been compressed. The triggered air, released into the space below the piston, is fully expanded, and can now exhaust at a controlled rate through the vent ports. As this takes place, the piston rapidly but gently moves downward, re-sealing the chamber, and readying the sound source for refilling. From:

Airguns suspended from stowed booms Single Air gun – note air ports Air Guns Other source?

An ideal pulse convolved with the seafloor creates a simple seismogram The Ideal Shot What do we want? We want a seismic section that looks like a geological cross section Difficult to do for a number of reasons…..

The output seismogram is a convolution of the source signal and the earth (the seafloor) Sharp seafloor signal becomes “ringy” for a number of reasons Why else might the seismic section not look like a geologic section? Reality

A single airgun creates a “ringy” signal Tuning An Air Gun Array From

Summing the signal of multiple guns creates a more desirable signal Note the relative scales of the left and right plots Tuning An Air Gun Array From

Listening Hydrophone –Piezoelectric material –Pressure changes in the water generate small currents which are amplified Geophone –Mechanical –Motion of coil relative to magnet generates a small current which is then amplified From Kearey, Brooks, and Hill, 2002