General Reflection (and some Refraction) Theory

General Reflection (and some Refraction) Theory

Andrew GoodliffeUniversity 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

mass eappropriat

force restoring wavesofvelocity =

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.

ρμ

ρ

μκ

=

+=

s

p

v

v 3

4Where ρ = density. What is μ in a liquid?

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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

ii t

zv =

i=1

i=2

i=3

•Each layer is characterized by an interval velocity.

•If z1 is the thickness of layer i and ti 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:

21

11

2, ⎥

⎦

⎤⎢⎣

⎡= ∑∑

==

n

iii

n

iinrms ttvv

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

21

11

2, ⎥

⎦

⎤⎢⎣

⎡= ∑∑

==

n

iii

n

iinrms ttvv

v1=1500 m s-1

v2=2000 m s-1

v3=2345 m s-1

t1=2.14 s

t2=1.21 s

t3=1.13 s

What is vrms at the base of layer 3?

1-

21

222

s m 064.1882

13.121.114.2

)13.12345()21.12000()14.21500(

=

⎥⎦

⎤⎢⎣

⎡

++×+×+×

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)

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ρvZ =

•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

0

1

A

AR =

12

12

1122

1122

ZZ

ZZ

vv

vvR

+−

=+−

=ρρρρ

•Reflection coefficient R is a numerical measure of the effect of an interface on wave propagation. It is the ratio of the amplitude A1 of the reflected ray to the amplitude A0 of the incident ray:

Expanding, this becomes:

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•A negative value of R indicates a 180o 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 A2 of the transmitted ray to the amplitude A0 of the incident ray:

12

1

0

2 2 becomes ray thisincident normally afor

ZZ

ZT

A

AT

+==

Reflection and TransmissionF

rom M

ussestt and Khan, 2000

Reflection and TransmissionFro

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Attenuation•The energy E transmitted outwards from a source becomes distributed over a spherical shell

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200

2

•If the radius of the wavefront is r, the amount of energy contained within a unit area of the shell is E/4πr2.

•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 = 21.65 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 = 16.67 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 = (0.5714 m)/4 =

0.1428 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

• Wf (2z )1/2 z = depth

• If depth = 2000 m, = 60 m

– Wf 490 m

• If depth = 100 m, = 1 m

– Wf 14 m

Horizontal Resolution

Snell’s Law

'

'

2

1

2

1

AA

BB

v

v

'

'sin

'

'sin

2

1

AB

AAi

AB

BBi

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: 21 sin

'

sin

''

i

AA

i

BBAB Rearrangin

g gives:

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Snell’s Law

2

2

1

1 sinsin

v

i

v

i=

As BB’ and AA’ are in proportion to the velocities v1 and v2, the equation can be rearranged to

Snell’s Law

o

o

i

i

37sin4

5sin

5

sin

4

37sin

2

2

So i2 = 48.8o

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 45o. At what angle from the normal does it leave the interface?

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Snell’s Law – Multiple Flat Horizons

3

3

2

'2

2

'2

1

'1

sinsin

sinsin

v

i

v

i

v

i

v

i

constantsinsinsin

3

3

2

2

1

1 v

i

v

i

v

i

As i’1 = i1, I’2 = i2, and so on

From

Mussestt and K

han, 2000

The ratio (sin i/v) thus remains unchanged

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Snell’s Law - Refraction

• When a critical angle is reached the ray will travel along the interface between the two layers at velocity v2

• 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

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Seismic Sources

High Pressure Air Sources: The Air Gun

Ready Fire! FiredLower 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.

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•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 ShotWhat 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

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•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

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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

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