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

      xx yx zx  stresses  xx  dx , yx  dx , zx  dx  , that is there arise unbalanced  x x x 

  xx  yx  zx  stress components dx ,dx ,dx  . But Stress=Force/Area, that is we  x x x 

  xx  yx  zx  also have an unbalanced force vector  , , dxdydz and similar  x x x  expressions are obtained for the other faces. The total force acting in the direction of

  xx  yx  zx  the x axis is    dxdydz . Because of this force, some material will  x y z  be displaced a distance u along the direction x, the total amount of mass moving is m    dxdydz      mass density volume

1 Newtons 2nd Law of Motion (F=ma) gives

      2u  xx  yx  zx   dxdydz    dxdydz   x y z      t 2   mass               acceleration force

acceleration stress stress stress density     2u    that is    xx  xy  xz , and we get similar equations t 2 x y z 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:

 2u           xx  xy  xz    2 xx   xy   xz t 2 x y z x x y z ,               HOOKE's LAW

where    xx   yy   zz .

Remembering the definition of the strain components, we relate them to the changes

 2u u v in displacements:   ;   ;⋯ . We get the system of equations xx x 2 xy x y

 2u    (  )   2u (1) t 2 x

 2v    (  )   2v (2) t 2 y

 2 w    (  )   2 w (3) t 2 z

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  2  u v w   u v w          2 2     t 2  x y z   x y z  (4)                  

1  2 or   2 (5)  2 t 2

  2 where  2  (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)    (7) v 2 t 2 x 2 y 2 z 2 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)  (8) v 2 t 2 x 2

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)  f (x  vt) QED v 2 t 2 x 2

Why do we call v "velocity"?

3  x1  x0  At x0 ,t0  the solution is f x0  vt0 . At  x1 ,t0   the solution is  v 

  x  x   1 0  f  x1  vt0    f x1  vt0  x1  x0   f x0  vt0 .   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 derivative of eq. (2) with respect to z from the derivative of eq. (3) with respect to y

 2  w v   w v        2    we get: 2     , that is t  y z   y z 

1  2 x   2 (9)  2 t 2 x

 where  2  , (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 vP ( )  vS (  ) and it is easy to prove that

1  Exercise  2 ,  2  2 1

 where σ is Poisson's ratio,   . 2(  )

4 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

K only, their velocity being   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   cP 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

P and their velocity is   . 

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 time-

5 invariant, 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  . (1) V 2 t 2

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 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 x- axis 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) (4)

A well-known example is the harmonic plane wave   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

2 2 1   1    2   1     1    2 2  2  r   sin   2 2  (6) V t r r  r  sin     sin   

6 In the isotropic case, when the wave motion is independent of  and  , the equation

1  2 1     simplifies to  r 2  (7) and a general solution progressing out V 2 t 2 r 2 r  r 

1 1 from a central point is   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

 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

7 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  Z1 2Z1 Z 2  V2  2 as r  and t  (Eqs. 9.a & b). The fractions of energy Z 2  Z1 Z 2  Z1

2 4Z Z Z  Z 2  Z1  2 1 2 2 2 reflected and transmitted are E     r and ET   t r   2 Z  Z 2  Z1  Z 2  Z1  1

(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

1 Acos(t  ) is: E   2 A2 (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).

8  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 106 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 .

  = 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 (12) E Q f f f

9 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

Aout unit of amplitude attenuation in a system, defined as 20  log10 . Ain

Examples In a system for which Ain  30V; Aout  15V , the attenuation is

3 20  log10 15 30  6dB . A 100 amplification is 40dB increase; a 10  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  log10 e   20  log10 e  8.686 dB / m .

 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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