Electromagnetic Effects Resulted from Natural Disasters - Ultra and Extremely Low Frequency Electromagnetic Fields - page 377

where w D C t =C l and h u r i stands for the mean radial displacement caused by
the acoustic emission of stochastic crack ensemble. The characteristic length of the
acoustic wave attenuation, L, is estimated as follows (Surkov et al. 2003 )

QC l
2V c

L D ǛR; Ǜ ; (10.16)

where V c is the velocity of crack growth and Q is the quality/energy-factor, which is
approximately constant in the frequency range from 10 4 to 100 Hz depending on
the variety of materials (Aki and Richards 2002 ). Taking the typical parameters
C l D 5 km/s, V c D 1:5 km/s, Q D 100 we can estimate the coefficient of
proportionality in Eq. ( 10.16 )asǛ 50. As it follows from Eqs. ( 10.15 ) and
( 10.16 ), the small cracks are of little importance in the sense that their contribution
to the net displacement field appears to be exponentially small, while the large
cracks make a main contribution to the mean displacement and therefore to the
crack-generated magnetic perturbations.
In the analysis that follows, we first seek for the solution of Eq. ( 10.14 )inthe
case of cracks with fixed radius R and then extend the solution to the case of crack
size distribution. Since the mean displacement has been obtained to be spherically
symmetric, the spherical coordinates r;;' are needed. We shall use a coordinate
system in which the z axis is positive parallel to the vector of geomagnetic field, B 0 .
For illustrative purposes, the reference frame and the cracked zone are sketched in
Fig. 10.5 b. In the case study all the values are independent of the azimuthal angle ',
and only the azimuthal component of a is nonzero. Equation ( 10.14 ) is thus reduced
to the following form

r 2 @ r r 2 @ r a ' C @ @ .a sin /

m

D B 0 h u r .r/ i sin ; (10.17)

sin

where @ r and @ denote the partial derivatives with respect to r and , accordingly,
and is the polar angle between the vectors B 0 and r . We seek for the solution of
Eq. ( 10.17 ) in the form a ' D g.r/ sin . Substituting this function and Eq. ( 10.15 )
for h u r i into Eq. ( 10.17 ), we obtain

exp

;

d r r 2 d r g 2g D

u 0 B 0
m

r
L

(10.18)

where d r D d=dr. This differential equation has a straightforward analytical
solution

1
x 2 C x 1 exp . x/ C xE 1 .x/ ; (10.19)

c 2
x 2 C

u 0 B 0
3 m

1

g.x/ D c 1 x C

where c 1 and c 2 are arbitrary constants, x D r=Land E 1 .x/ denotes the exponential
integral, that is