Seismic attenuation in geophysics explained?

What is Seismic Attenuation?

Seismic Attenuation can be defined as the decrease in amplitude of seismic waves as the distance increases away from the source.

Why do seismic waves attenuate ?

The main causes of seismic energy loss through materials (rocks) medium during propagation are 

1. Geometrical Spreading 

This is the most important cause of seismic energy loss as it spreads from source and takes a spherical shaped path.

Consider the figure 1, as shown below

Figure 1: Seismic spread from the explosive source in spherical path 

Amplitude density = Energy/Surface area of wavefront.

A = E/4πr2

A is inversely to radius (distance) for surface waves, Since surface waves take circular path as 2πr

A is inversely to radius (distance) square for body waves, as body waves take spherical wavefront. See figure above.

2. Frictional Dissipation

As frictional heating due to imperfect elastic of materials (rocks).

This is also known as intrinsic attenuation or absorption.

Also Anelastic damping of the vibration of particles of materials, example passage of waves in the asthenosphere.

since, Amplitude = Ae-xr 

Where: x is the absorption coefficient. (It depends on wavelength ).

It will sound interesting if we think of seismic attenuation as damping oscillation, as amplitude decreases with increase in time. The decrease in amplitude is associated with frictional dissipative force.

If we defined retarding force, as R = bv, the restoring force F = kx

Then from Net force, from Newton's second law of motion, Fx = - F - R as they act in opposite to motion.

But Fx = max

Then, max = - kx - bv

m(d2x/dt2) = - kx - b(dx/dt)...............(1)

Equation 1 is Second Order Linear Differential equation., So we have to find the general solution of this equation as follows,

Rearrange equation 1 in the form of a(d2y/dx2)+b(dy/dx)+cy = 0..............(2)

m(d2x/dt2)+b(dx/dt)+kx = 0..................(3)

If we let D = d/dt, then D2= d2/dt2, then in D - Operator form

(mD2+ bD + k)x = 0

If we Put p = D,

Then (mp2+ bp + k)x = 0..............(4)

Since equation 4 in brackets is Auxiliary equation in form of quadratic equation, a = m, b = b, c = k

Solve the equation for p, mp2+ bp + k = 0

From general quadratic formula,

p = [- b +- √(b2 - 4mk)]/2m

Since b is small, b2 < 4mk

The solution of p will be in complex form such as

p = - b/2m +- j√(b2 - 4mk)/2m

Since the general solution for equation 2 will be in form of

y = erx (acoszx + bsinzx).............(5)

it's complex form of quadratic solution is p = r +- jz, compare with solution p above,

r = - b/2m and z = √(b2 - 4mk)/2m

Recall trigonometric identities,  we can write , Asin (zx+q) = acoszx + bsinzx

While A = √(a2 + b2), and q = tan-1 (a/b)

Then equation 5 will be y = erxAsin(zx+q)..............(6)

For If we compare our equation 2 with equation 3 we would saw that, x = y, and x = t

Put that condition above in equation 6 above then

x = ertAcos(zt+q)

Substitute for r and z, 

Then our final general solution for equation 1, will be

 x = Ae-bt/2mcos(zt + q)..................(7)

Where A - Amplitude, x - displacement, b - Damping constant, t - time taken, q - phasor angle, z - angular frequency.

z = √[k/m - (b/2m)2], as b2< 4mk

The equation 7 above shows that displacement (Amplitude) decay exponentially with time, as shown in the graph below,

As high frequency waves decay first while leaving low frequency signals remaining.

It means that surface waves attenuate less than body waves.

This describes why the surface wave train on the seismogram is more prominent than of body waves.

Thank you for your time!

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