Environmental and Exploration Geophysics II
Energy Loss and Partitioning
tom.h.wilson
wilson@geo.wvu.edu
Department of Geology and Geography
West Virginia University
Morgantown, WV
Be sure to do the following for all exercises
1) label all plotted curves,
2) label all relevant points and ….
3) Where appropriate,
provide a paragraph or so
of discussion regarding
the significance and
origins of the
interrelationships
portrayed in the resultant
time-distance plots.
In exercise II for example, how do you account for the
differences in the two reflection hyperbola? How is their
appearance explained by the equations derived in class for
the reflection time-distance relationship.
Re: 3) Discussion - In Exercise III, explain the differences
observed in the arrival times of the reflection and diffraction
observed in the shot record. Why does the diffraction event
drop below the reflection?
As noted earlier - accurately portray the arrival
times at different offsets or surface locations.
Your plots should serve as an accurate
representation of the phenomena in question.
Let’s come back to ray-tracing concepts for a minute
and consider the diffraction problem in a little more
detail. How would you set up the mathematical
representation of diffraction travel time as a function
of source receiver distance?
We’ve taken a few shortcuts here, but we need go no further. We
have defined the basic features of the diffraction and we can see
how in subtle ways it will differ from that of the reflection.
t
x2
V12

4h12
V12
Although the time intercept remains the
same - how do the shapes of the reflection
hyperbola differ in these two cases?
Can you pick out the
direct arrival?
Can you pick out the
refraction arrivals?
How many critical
refractions are there?
How would you
determine the
refraction velocity?
How would you
determine the air
wave or direct
arrival velocity?
To calculate the speed of sound go to http://www.measure.demon.co.uk/Acoustics_Software/speed.html
Discussion of Chapter 3:
What happens to the seismic energy generated by the
source as it propagates through the subsurface?
Basic ConceptsEnergy - The ability to do work. It comes in two forms potential and kinetic
Work expended (W) equals the applied force times the
distance over which an object is moved.
Power is the rate at which work is performed.
As a mechanical disturbance or wavefield propagates
through the subsurface it moves tiny particles back and
forth along it path. Particle displacements are continually
changing. Hence, it is more appropriate for us to consider
the power, or the rate at which energy is being consumed
at any one point along the propagating wavefront.
W (the work)  Fx (force times displacement)
dW
Power 
dt
The rate of change of work in unit time.
dW dFx 
dx

F
 Fv
dt
dt
dt
This force is the force exerted by the seismic wave at
specific points along the propagating wavefront.
and since force is pressure (p) x area (A), we have
dW
Power 
 Fv  pvA
dt
The power generated by the source is PS. This
power is distributed over the total area of the
wavefront A so that
PS  pvA
We are more interested to find out what is going
on in a local or small part of the wavefield and
so we would like to know * Note we are trying to quantify the effect of wavefront
spreading at this point and are ignoring heat losses.
PS
pv 
A
Over what surface area is the energy
generated by the source distributed?
Area of a hemisphere is 2R2 hence -
pv 
PS
2R 2
This is the power per unit area being dissipated along
the wavefront at a distance R from the source. (Recall
p is pressure, v is the particle velocity, PS magnitude of
the pressure disturbance generated at the source and R
is the radius of the wavefront at any given time.
In the derivation of the acoustic wave equation (we’ll
spare you that) we obtain a quantity Z which is called the
acoustic impedance. Z = V, where  is the density of the
medium and V is the interval velocity or velocity of the
seismic wave in that medium.
Z is a fundamental quantity that describes reflective
properties of the medium.
We also find that the pressure exerted at a point along the
wavefront equals Zv or Vv, where v is the particle
velocity - the velocity that individual particles in the
disturbed medium move back and forth about their
equilibrium position.
pv   Vv  v
pv  Zv 2
Combining some of these
ideas, we find that v, the
particle displacements vary
inversely with the distance
traveled by the wavefield R.
2
Zv 
v
Ps
2R 2
Ps
Z 2R 2
1
Ps
v
R Z 2
•We are interested in the particle velocity variation with
distance since the response of the geophone is
proportional to particle (or in this case) ground
displacement. So we have basically characterized how
the geophone response will vary as a function of distance
from the source.
•The energy created by the source is distributed over an
ever expanding wavefront, so that the amount of energy
available at any one point continually decreases with
distance traveled.
•This effect is referred to as spherical divergence. But in
fact, the divergence is geometrical rather than spherical
since the wavefront will be refracted along its path and its
overall geometry at great distances will not be spherical in
shape.
The effect can differ with wave type. For example, a
refracted wave will be confined largely to a cylindrical
volume as the energy spreads out in all directions
along the interface between two intervals.
z
R
The surface area along the leading edge of the
wavefront is just 2Rz.
Hence (see earlier discussion for the hemisphere)
the rate at which source energy is being expended
(power) per unit area on the wavefront is
Ps
Zv 
A
2
Remember Zv2 is just
Ps
Zv 
2Rz
2
Following similar lines of reasoning as before,
we see that particle velocity
Ps
v
R
Ps
v
R
The dissipation of energy in the wavefront
decreases much less rapidly with distance traveled
than does the hemispherical wavefront.
This effect is relevant to the propagation of waves in
coal seams and other relatively low-velocity intervals
where the waves are trapped or confined. This effect
also helps answer the question of why whales are
able to communicate over such large distances
using trapped waves in the ocean SOFAR channel.
Visit
http://www.beyonddiscovery.org/content/view.page.a
sp?I=224 for some info on the SOFAR channel.
See also -
http://www.womenoceanographers.org/doc/MTolstoy/Les
son/MayaLesson.htm
To orient ourselves, you can think of the particle
velocity as the amplitude of a seismic wave recorded
by the geophone - i.e. the amplitude of one of the
wiggles observed in our seismic records.
The amplitude of the seismic wave will decrease, and unless
we correct for it, it will quickly disappear from our records.
Since amplitude (geophone response) is proportional to
the square root of the pressure, we can rewrite our
divergence expression as
Ps
Ar 
r
As
Ar 
r
Thus the amplitude at distance r from the source (Ar)
equals the amplitude at the source (AS) divided by the
distance travelled (r).
This is not the only process that acts to decrease the
amplitude of the seismic wave.
Absorption
When we set a spring in motion, the spring oscillations
gradually diminish over time and the weight will cease to move.
In the same manner, we expect that as s seismic wave
propagates through the subsurface, energy will be consumed
through the process of friction and there will be conversion of
mechanical energy to heat energy.
We guess the following - there will be a certain loss of amplitude
dA as the wave travels a distance dr and that loss will be
proportional to the initial amplitude A.
i.e.
dA  AS dr
How many of you remember how to solve such
an equation?
dA(r )   AS dr
dA(r )
 dr
AS
 is a constant referred to
as the attenuation factor
In order to solve for A as a function of distance traveled (r)
we will have to integrate this expression In the following
discussion,let
A0  AS
A dA
r
A0 A  0dr
ln A  ln A0  r
ln A  ln A0  r
A
ln
 r
A0
ln
e
A
A0
 er
A
 er
A0
A  A0er
Mathematical Relationship
A(r )  A0er
Graphical Representation
A(r )  A0er
 - the attenuation factor is also a function of additional terms  is wavelength, and Q is the absorption constant


Q
1/Q is the amount of energy dissipated in one
wavelength () - that is the amount of mechanical
energy lost to friction or heat.


Q
 is also a function of interval velocity, period and frequency


r
A(r )  A0e Q


A(r )  A0e QV 
r
 is just the reciprocal of the frequency so we
can also write this relationship as

f
A(r )  A0e QV
r
Smaller Q translates into higher energy loss or
amplitude decay.
A(r )  A0e

f
QV
r
increase f and decrease A
Higher frequencies are attenuated to a much
greater degree than are lower frequencies.
When we combine divergence and absorption we get the
following amplitude decay relationship
A0 r
A(r ) 
e
r
The combined effect is rapid amplitude decay as the
seismic wavefront propagates into the surrounding
medium.
We begin to appreciate the requirement for high
source amplitude and good source-ground coupling to
successfully image distant reflective intervals.
But we are not through - energy continues to be
dissipated through partitioning - i.e. only some of the
energy (or amplitude) incident on a reflecting surface
will be reflected back to the surface, the rest of it
continues downward is search of other reflectors.
The fraction of the incident amplitude of the
seismic waves that is reflected back to the surface
from any given interface is defined by the reflection
coefficient (R) across the boundary between layers
of differing velocity and density.
Z 2  Z1
R

Ainc Z1  Z 2
Arefl
Z 2  Z1
R

Ainc Z1  Z 2
Arefl
Z1 and Z2 are the impedances of the bounding layers.
 2V2  1V1

1V1   2V2
Z 2  Z1
R
Z1  Z 2
The transmitted wave amplitude T is
T 1  R
Z1  Z 2  Z 2  Z1 

T
 
Z1  Z 2  Z1  Z 2 
2 Z1
T
Z1  Z 2
At a distance of 100 m from a
source, the amplitude of a P-wave is
0.1000 mm, and at a distance of 150
m the amplitude diminishes to
0.0665 mm. What is the absorption
coefficient of the rock through
which the wave is traveling?
(From Robinson and Coruh,
1988)
Any Questions?