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Lecture 2. Wave Equation for P and S Waves in an Elastic Isotropic Solid. Absorption
Assume that a volume element dxdydz inside the medium is not in equilibrium. On
face EFGH the stresses are  xx ,  yx ,  zx , but on the front face ABCD we have
 yx


 
stresses  xx  dx xx ,  yx  dx
,  zx  dx zx  , that is there arise unbalanced
x
x
x 

 yx
 
 
stress components  dx xx , dx
, dx zx  . But Stress=Force/Area, that is we
x
x
x 

   yx  zx 
dxdydz and similar
also have an unbalanced force vector  xx ,
,
x
x 
 x
expressions are obtained for the other faces. The total force acting in the direction of
 yx  zx 
 
dxdydz . Because of this force, some material will
the x axis is  xx 

y
z 
 x
be displaced a distance u along the direction x, the total amount of mass moving is
m
  dxdydz
  

mass
density
volume
Newtons 2nd Law of Motion (F=ma) gives
  xx  yx  zx 
 2u

  dxdydz    dxdydz 


 t 2
x
y
z 


mass


acceleration
force
1
stress
stress
stress




2



 xx
 xz
 u
xy
 



, and we get similar equations
x
y
z
t 2
acceleration
density
that is
for the other two components of motion, v and w. Next, we are going to express

stress in terms of strain, using Hooke's Law; and

strain in terms of the displacement components, using the definition of strain:

 xy
 xx

 2 u  xx  xy  xz







2



  xz ,
2
x
y
z
x
x
y
z
t



HOOKE 's LAW
where    xx   yy   zz .
Remembering the definition of the strain components, we relate them to the changes
in displacements:  xx 
 2u
u v
;  xy 
 ; . We get the system of equations
2
x y
x

 2u

 (   )
  2 u
2
x
t
(1)

 2v

 (   )
  2 v
2
y
t
(2)
2w

 (   )
  2 w
2
z
t
(3)

Taking the partial derivative of eq. (1) with respect to x, of eq. (2) with respect to y,
and of eq. (3) with respect to z, and adding together the three equations, we get

2
t 2
 u v w 
 u v w 
 
    2   2  



x y z 
x y z 





(4)

1 2
 2
2
2
 t
or
where
2
2 
  2

(5)
(6)
Generally, the PDF (partial differential equation)
1  2  ( x, y, z, t )  2  ( x, y, z, t )  2  ( x, y, z, t )  2  ( x, y, z, t )



v2
t 2
x 2
y 2
z 2
(7)
is called wave equation. The constant v is called propagation velocity (we shall see,
why). In 1-dimension the wave equation becomes
1  2  ( x, t )  2  ( x, t )

v2
t 2
x 2
(8)
Theorem (D'Alambert): Any two-times differentiable function of the argument x-vt,
  f ( x  vt) satisfies the wave equation (8).

 vf ' x  vt 
t
2
Proof.
 v 2 f ' ' x  vt 
t 2
1  2
2
 f ' ' ( x  vt)  2 f ( x  vt)
v 2 t 2
x
QED
Why do we call v "velocity"?
At
x0 ,t 0 
the solution is
x  x0 

f x0  vt0  . At  x1 , t 0  1
 the solution is
v 


x  x0  

  f x1  vt0  x1  x0   f x0  vt0  .
f  x1  v t 0  1
v  


Wave equation for shear waves The case of wave propagation in elastic solids has not
been finished yet, as we know from observations (seismology) that there is an other,
slower wave type, that can also propagate in an elastic solid. To derive a wave
equation with a different velocity, we proceed as follows. By subtracting the
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derivative of eq. (2) with respect to z from the derivative of eq. (3) with respect to y
2
we get:  2
t
 w v 
 w v 

    2 
  , that is
 y z 
 y z 
1  2 x
  2 x
2
2
 t
2 
where
(9)

,

(10)
which is a wave equation for the propagation of (one component of) the local shear
disturbance. We get similar equations for the other two shear components  y and  z .
Because the elastic constants are always positive, comparing eqs. (6) and (10) shows
that v P (  )  v S (  ) and it is easy to prove that
1

2 2
,

 2 1
Exercise
where σ is Poisson's ratio,  

2(   )
.
In many books you find the statement that "for fluids  is zero, and because of this 
is also zero, therefore no shear waves can propagate in a liquid". But this is "sheer"
nonsense because we have derived the wave equation for P- and S-waves on the basis
of Hooke's Law, which does not hold for a fluid or a gas.
Wave Equation in fluids It can be derived similarly as we did for solids, without
Hooke's Law, of course. The shear stress components are zero,  xy   yz   zx  0,
the diagonal ones are the negatives of the same pressure P:  xx   yy   zz   P.
Newton's Law and the condition of continuity leads to a wave equation for P-waves
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only, their velocity being  
K

where K is bulk modulus,  density. In sea-water
at ~ 20  C temperature the sound velocity is about 1450 m/s.
Wave Equation in gas In case of an ideal gas assume that the wave propagation is
adiabetic (i.e. there is no heat transfer during the wave's passage). Then pressure and
volume satisfies the Gas Law PV   const , where   c P cV (= ratio of specific
heats at constant pressures and volumes)  1.4 for air. Using Newton's Law and the
Gas Law in deriving the WE we find again that only P-waves can propagate in a gas
and their velocity is  
P
.

Comments: (i) A sedimentary rock contains all three phases (solid rock matrix, fluid
and gas) (Biot's Theory). (ii) Also, in the present derivation of the Wave Equation
(WE) we have taken partial derivatives with respect to the spatial coordinates x,y,z ,
and interchanged the order of the derivatives with respect to time and with respect to
spatial coordinates. These steps are valid only if the elastic constants are timeinvariant, and continuously changing in space, that is the WE does not hold at the
boundary between layers where the elastic properties are changing abruptly.
Considerations of the continuity of the wave-field across boundaries will yield the
laws of reflection and transmission of elastic waves.
Plane waves, Attenuation
As we have seen, both the longitudinal (P) and shear (S) waves satisfy the WE:
1  2
 2 .
2
2
V t
(1)
By D'Alambert Theorem, in 1-dimension for any twice differentiable function f , the
function   f ( x  Vt )
(eq. 2a) is a solution of WE (1). If g is an other twice
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differentiable function, then   f ( x  Vt ) (eq. 2b) also solves the WE, it is a wave
travelling in the opposite direction. A general solution of the WE can be written as
  f ( x  Vt )  g ( x  Vt ) (Eq. 3). The quantity x-Vt or x+Vt is called the phase. The
surface on which the wave has the same value that is, where the phase is constant, is
the wavefront. In the present case the wavefront is perpendicular to the x axis, the xaxis plays the role of the raypath. The wave is called a plane wave.
In the more general case when the plane wave propagates along a straight line
inclined at an angle to each of the axes, having directional cosines l , m, n , the
general form of the plane wave becomes
  f (lx  my  nz  Vt )  g (lx  my  nz  Vt )
A well-known example is the harmonic plane wave
(4)
  Ae j lx my  nz Vt 
(5)
In the spherical wave the wavefronts are concentric spherical surfaces. We rewrite eq.
(1) to spherical coordinates r,,   as
1  2  1    2  
1  
 
1  2 
 2  r

 sin 


  sin 2   2 
V 2 t 2
r  r  r  sin   
(6)
In the isotropic case, when the wave motion is independent of  and  , the equation
simplifies to
1  2  1   2  
 2
r
 (7) and a general solution progressing out
V 2 t 2
r r  r 
from a central point is  
1
1
f (r  Vt )  g (r  Vt ) (8). If r1, the spherical wave
r
r
approximates a plane wave. Observe that the wave amplitude of the spherical wave
decreases as 1 r , this is called spherical spreading. This is one kind of amplitude
attenuation during wave propagation, out of the many, such as:

Spherical spreading

Reflection-transmission losses
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
Losses due to wave conversion (between P- and S modes)

Energy scattered out of the beam because of less-than-wavelength
inhomogeneities and defects (Rayleigh scattering)

Inelastic absorption.
Reflection & Transmission Coefficients At the boundary between two different media
the elastic properties abrubtly change, in a non-differentiable manner, that is at the
boundary the WE does not hold. We must solve the equation above the boundary, and
below the boundary in the general form. If both solutions contain a sufficient number
of free parameters, we can select them in a way to ensure that the normal and
tangential components of stress would change continuously through the boundary.
(For example, if the normal component were discontinuous, there would occur an
empty space inside the medium, or one medium would penetrate each other.)
The easiest case is a P wave normally incident at the boundary, when there is no wave
conversion but only a reflected and a transmitted
P wave. The reflection and
transmission coefficients (giving the fraction of incident amplitude reflected, resp.
transmitted) can be found in terms of of the acoustic impedances Z1  V1  1 and
Z 2  V2  2 as r 
Z 2  Z1
Z 2  Z1
and t 
2Z1
(Eqs. 9.a & b). The fractions of energy
Z 2  Z1
2
 Z  Z1 
4Z1 Z 2
Z
  r 2 and ET 
 2 t2
reflected and transmitted are E r   2
2
Z 2  Z1  Z 1
 Z 2  Z1 
(Eqs. 10 a & b).
Exercise. Prove that ER  ET  1 (conservation of energy).
Remember from Mechanics that the energy density (energy per unit volume in the
neighborhood of a point as the wave passes through it) for a harmonic wave
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A cos(t  ) is: E 
1
 2 A 2 (eq. 11). Exercise: What is the physical dimension of
2
"A" in expression 11?
Absorption
If strain energy is not totally convertible to stress-energy, we have
anelastic wave propagation where wave energy is gradually dissipated in course of
propagation. The phenomenon is called absorption or internal friction. There are four
main types of mechanisms causing absorption:

Thermal losses (conduction of heat during the compressions and expansions of
the solid).

Scattering losses on the coarse grains, microcracks and other defects,
especially when the wavelength approaches the grain/pore size.

Ferromagnetic or piezoelectric losses (not important in sedimentary rocks).

Frictional losses due to one surface sliding past another when the wave passes
by, or a viscous fluid is oscillating asynchronously with the rock particles.
Probably the most important source of absorption in rocks.
Measures of absorption The most important measures of absorption are:

1 Q = specific dissipation constant: the tangent of the phase angle by which
the strain lags behind the applied stress in case of mono-frequency sinusoidal
excitation. For rocks it is essentially independent of frequency up to 10 6 Hz
("Constant Q mechanism").

 = logarithmic decrement: natural logarithm of the ratio of the amplitudes of
two successive maxima (or minima) in an exponentially decaying free
vibration.

a = damping coefficient in the expression of free vibration e  at sin 2ft .
8

 = absorption coefficient in the expression of plane harmonic waves
x

propagating in the medium: e x sin 2  t   (where v is wave velocity).
 v

f
= relative bandwidth of the resonance curve between the half-power (or
f
0.707 amplitude, i.e. -3dB amplitude) points for a solid undergoing forced
vibration.

E
= fraction of strain energy lost per stress cycle.
E
We have the following relation between these parameters:
E 2
2a 2v
f

 2 

 2
E
Q
f
f
f
(12)
Note that in a constant Q solid the absorption coefficient is linearly proportional to
frequency.
The decibel (dB) unit We mentioned dB (decibel): this is a dimensionless logarithmic
unit of amplitude attenuation in a system, defined as 20  log 10
Examples
In a system for which
Aout
.
Ain
Ain  30V ; Aout  15V , the attenuation is
20  log 10 15 30  6dB . A 100  amplification is 40dB increase; a 10 3  reduction
is a 60dB loss, etc.
Using this important concept, the absorption loss can be also expressed in dB / m
units (let us call it  ). Suppose the absorption coefficient for a certain frequency is
 , then the loss over 1m propagation path is, in logarithmic units:
  20  log 10 e    20  log 10 e  8.686 dB / m .
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 1 1

Exercise: Suppose in a rock   3.1  10 5  f 
, the frequency is 150kHz. (a)
 Hz m 
After how long distance travelled will the amplitude fall to 50% of its initial value?
(b) How many dB will be the amplitude loss after a path of 10 inches. (1 inch =
0.0254 meter). (c) Assume that the propagation velocity is 3500 m/s (typical
sandstone P-wave velocity),  & f are as above. Calculate all the different measures
of absorption mentioned in this Lecture.
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