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Abstract and Figures

As shown by Backus (1962), the average of a stack of isotropic layers results in a transversely isotropic medium. Herein, we consider a stack of layers consisting of a randomly oriented anisotropic elasticity tensor, which-one might expect-would result in an isotropic medium. However, we show-by means of a fundamental symmetry of the Backus average-that the corresponding Backus average is only transversely isotropic and not, in general, isotropic. In the process, we formulate, and use, a relationship between the Backus and Gazis et al. (1963) averages.
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On the Backus average of layers with randomly oriented
elasticity tensors
Len Bos
, Michael A. Slawinski
, Theodore Stanoev
Abstract
As shown by Backus (1962), the average of a stack of isotropic layers results in a transversely
isotropic medium. Herein, we consider a stack of layers consisting of a randomly oriented
anisotropic elasticity tensor, which—one might expect—would result in an isotropic medium.
However, we show—by means of a fundamental symmetry of the Backus average—that the
corresponding Backus average is only transversely isotropic and not, in general, isotropic. In
the process, we formulate, and use, a relationship between the Backus and Gazis et al. (1963)
averages.
1 Introduction
In this paper, we investigate the Backus (1962) average of a stack of anisotropic layers, wherein the
tensors are oriented randomly. In spite of a conceptual relation between randomness and isotropy,
herein, the Backus average results in a medium, whose anisotropy, even though weak, is irreducible
to isotropy, regardless of increasing randomness.
Each layer is expressed by Hooke’s law,
σij =
3
X
k=1
3
X
`=1
cijk` εk` , i, j = 1,2,3,
where the stress tensor, σij , is linearly related to the strain tensor,
εk` := 1
2∂uk
∂x`
+∂u`
∂xk, k, ` = 1,2,3,
where uand xare the displacement and position vectors, respectively, and
cijk` =cj ik` =ck`ij
is the elasticity tensor, which has to be positive-definite. Under the index symmetries, this tensor
has twenty-one linearly independent components, and can be written as (e.g., B´ona et al., 2008,
Dipartimento di Informatica, Universit`a di Verona, Italy, leonardpeter.bos@univr.it
Department of Earth Sciences, Memorial University of Newfoundland, mslawins@mac.com
Department of Earth Sciences, Memorial University of Newfoundland, theodore.stanoev@gmail.com
1
arXiv:1804.06891v1 [physics.geo-ph] 18 Apr 2018
expression (2.1))
C=
c1111 c1122 c1133 2c1123 2c1113 2c1112
c1122 c2222 c2233 2c2223 2c2213 2c2212
c1133 c2233 c3333 2c3323 2c3313 2c3312
2c1123 2c2223 2c3323 2c2323 2c2313 2c2312
2c1113 2c2213 2c3313 2c2313 2c1313 2c1312
2c1112 2c2212 2c3312 2c2312 2c1312 2c1212
.(1)
Any elasticity tensor of this form is also positive-definite (e.g., B´ona et al., 2007).
A rotation of cijk` , which is in R3, expressed in terms of quaternions, is
A=A(q) =
a2+b2c2d22a d + 2 b c 2a c + 2 b d
2a d + 2 b c a2b2+c2d22a b + 2 c d
2a c + 2 b d 2a b + 2 c d a2b2c2+d2
,
where q= [ a , b , c , d ] is a unit quaternion. The corresponding rotation of tensor (1) is (B´ona et al.,
2008, diagram (3.1))
e
C=e
A C e
AT,
where ˜
Ais expression (A.1) in Appendix A.
2 The Backus and Gazis et al. averages
To examine the elasticity tensors, CR6×6, which are positive-definite, let us consider the space
of all matrices M:= R6×6. Its subspace of isotropic matrices is
Miso := {M∈ M :e
Q M e
QT=M , QSO(3)}.
Miso is a linear space, since, as is easy to verify, if M1, M2∈ Miso , then αM1+βM2∈ Miso , for
all α, β R. Let us endow Mwith an inner product,
hM1, M2iF:= tr(M1MT
2) =
6
X
i,j=1
(M1)ij (M2)ij ,
and the corresponding Frobenius norm,
kMk:= phM1, M2iF.
In such a context, Gazis et al. (1963) prove the following theorem.
Theorem 1. The closest element—with respect to the Frobenius norm—to M∈M⊂Miso is
uniquely given by
Miso := Z
SO(3) e
Q M e
QTdσ(Q),
where dσ(Q)represents the Haar probability measure on SO(3) .
2
Proof. It suffices to prove that
MMiso ⊥ Miso .
To do so, we let N∈ Miso be arbitrary. Then, for any ASO(3) ,
hMMiso , N iF= tr (MMiso )NT
= tr
MZ
SO(3) e
Q M e
QTdσ(Q)
NT
= tr
e
A
MZ
SO(3) e
Q M e
QTdσ(Q)
NT
e
AT
(as e
Ais orthogonal)
= tr
e
A M NTe
ATe
A
Z
SO(3) e
Q M e
QTdσ(Q)
e
ATe
A N Te
AT
.
But
e
A
Z
SO(3) e
Q M e
QTdσ(Q)
e
AT=Z
SO(3) e
Ae
Q M e
QTe
ATdσ(Q) (by linearity)
=Z
SO(3) e
Ae
QMe
Ae
QT
dσ(Q)
=Z
SO(3) g
A Q Mg
A Q T
dσ(Q) (by the properties of the tilde operation)
=Z
SO(3) e
Q M e
QTdσ(Q) (by the invariance of the measure)
=Miso .
Hence,
hMMiso , N iF= tr e
A M N Te
ATMiso e
A N Te
AT
= tr e
A M e
ATe
A N Te
ATMiso e
A N Te
AT
= tr e
A M e
ATNTMiso NT
= tr e
A M e
ATNTtr Miso NT,
as by assumption, N∈ Miso .
Finally, integrating over ASO(3) , we obtain
hMMiso , N iF= tr
Z
SO(3) e
A M e
ATdσ(A)
NT
tr(Miso NT)
= tr(Miso NT)tr(Miso NT)
= 0 ,
3
as required.
Since any elasticity tensor, CR6×6, is positive-definite, it follows that
Ciso =Z
SO(3) e
Q C e
QTdσ(Q)
is both isotropic and positive-definite, since it is the sum of positive-definite matrices e
Q C e
QT.
Hence, Ciso is the closest isotropic tensor to C, measured in the Frobenius norm.
If QiSO(3) , i= 1 , . . . , n , is a sequence of random samples from SO(3) , then the sample means
converge almost surely to the true mean,
lim
n→∞
1
n
n
X
i=1 f
QiCf
Qi
T=Z
SO(3) e
Q C e
QTdσ(Q) = Ciso ,(2)
which—in accordance with Theorem 1—is the Gazis et al. average of C.
This paper relies on replacing the arithmetic average in expression (2) by the Backus average, which
provides a single, homogeneous model that is long-wave-equivalent to a thinly layered medium.
According to Backus (1962), the average of the function f(x3) of “width” `0is the moving average
given by
f(x3) :=
Z
−∞
w(ζx3)f(ζ) dζ ,
where the weight function, w(x3), acts like the Dirac delta centred at x3= 0 , and exhibits the
following properties.
w(x3)>0, w(±∞)=0,
Z
−∞
w(x3) dx3= 1 ,
Z
−∞
x3w(x3) dx3= 0 ,
Z
−∞
x2
3w(x3) dx3= (`0)2.
These properties define w(x3) as a probability-density function with mean zero and standard devi-
ation `0, thus explaining the term “width” for `0.
3 The block structure of Ce
A C e
AT
The action Ce
A C e
AThas a simple block structure that is exploited in Section 4. To see this,
we consider q= [ a , 0,0, d ] , with a:= cos(θ/2) , d:= sin(θ/2) ; thus, in accordance with expres-
sion (1),
A=A(q) =
cos θsin θ0
sin θcos θ0
0 0 1
(3)
4
and, in accordance with expression (A.1),
e
A=
cos2θsin2θ0 0 0 1
2sin (2 θ)
sin2θcos2θ0 0 0 1
2sin (2 θ)
0 0 1 0 0 0
0 0 0 cos θsin θ0
0 0 0 sin θcos θ0
1
2sin (2 θ)1
2sin (2 θ) 0 0 0 cos (2 θ)
.
For q= [ 0 , b , c , 0 ] , with b:= cos(θ/2) and c:= sin(θ/2) ,
A=A(q) =
cos θsin θ0
sin θcos θ0
0 0 1
(4)
and
e
A=
cos2θsin2θ0 0 0 1
2sin (2 θ)
sin2θcos2θ0 0 0 1
2sin (2 θ)
0 0 1 0 0 0
0 0 0 cos θsin θ0
0 0 0 sin θcos θ0
1
2sin (2 θ)1
2sin (2 θ) 0 0 0 cos (2 θ)
.
In both cases permuting the rows and columns to the order ( 3 ,4,5,1,2,6 ) results in a diagonal
block structure for e
A. For expression (3), we have
e
A"e
A10
0e
A2#,
where
e
A1=
1 0 0
0 cos θsin θ
0sin θcos θ
and e
A2=
cos2θsin21
2sin (2 θ)
sin2θcos21
2sin (2 θ)
1
2sin (2 θ)1
2sin (2 θ) cos (2 θ)
.
Both e
A1,e
A2R3×3are rotation matrices. Similarly, for expression (4),
e
A"e
A10
0e
A2#;
herein, e
A1,e
A2R3×3are reflection matrices. Thus, in both cases, e
A1and e
A2are orthogonal
matrices.
In either case, the following lemma holds.
5
Lemma 2. Suppose that the rows and columns of Care permuted to the order ( 3 ,4,5,1,2,6 )
to have the block structure
CM B
K J ,
with M , B , K , J R3×3, and that the rows and columns of e
Aare also so permuted. Then,
e
A C e
AT"e
A1Me
AT
1e
A1Be
AT
2
e
A2Ke
AT
1e
A2Je
AT
2#.
Proof. Let PR6×6be the matrix obtained by permuting the rows of the identity to the or-
der ( 3 ,4,5,1,2,6 ) . Our assumption is that
P C P T=M B
K J .
Then,
Pe
A C e
ATPT=Pe
A P TP C P TPe
ATPT
="e
A10
0e
A2#M B
K J "e
AT
10
0e
AT
2#
="e
A1Me
AT
1e
A1Be
AT
2
e
A2Ke
AT
1e
A2Je
AT
2#,
as required.
4 The fundamental symmetry of the Backus average
Let us examine properties of the Backus average, which—for elasticity tensors, Ci—we denote by
(C1, . . . , Cn).
Theorem 3. For AR3×3of the form of expression (3) or (4) and any elasticity tensor, C1, . .. , CN
R6×6,
e
A(C1, . . . , Cn)e
AT=e
A C1e
AT, . . . , e
A Cne
AT,(5)
which is a symmetry condition. Conversely, if for an orthogonal matrix, AR3×3, we have equal-
ity (5), for any collection of elasticity tensors, C1, .. . , CnR6×6, then Amust be of the form of
expression (3) or (4).
Proof. As in Lemma 2, we permute the rows and columns of CR6×6to the order ( 3 ,4,5,1,2,6 ) .
Thus, we have the block structure
C=M B
K J , M , B , K , J R3×3;
herein, we use the notation of equations (5)–(9) of Bos et al. (2017). Also, e
Ahas the block structure
of
e
A="e
A10
0e
A2#,e
A1,e
A2R3×3,
6
and is orthogonal.
Let
e
C=e
A C e
AT="e
A10
0e
A2#M B
K J "e
A1T0
0e
A2T#="f
Me
B
e
Ke
J#,
where, by Lemma 2,
f
M=e
A1Me
A1T,e
B=e
A1Be
A2T,e
K=e
A2Ke
A1T,e
J=e
A2Je
A2T.
In particular,
f
M1=e
A1Me
A1T1(6)
=e
A1M1e
A1T.
The Backus-average equations are given by (Bos et al., 2017)
CBA =MBA BB A
KBA JBA ,(7)
where
MBA =M11,
BBA =M11M1B ,
KBA =K M 1M11,
JBA =JK M 1B+K M 1M11M1B,
and where denotes the arithmetic average of the expression ; for example,
M1=1
n
n
X
i=1
M1
i.
Let MBA ,BBA ,KB A and JBA denote the associated sub-blocks of the Backus average of the e
A Cje
AT.
Then,
e
A1MBA e
A1T=e
A1M11e
A1T
=e
A1M1e
A1T1
=e
A1M1e
A1T1
(by linearity)
=e
A1Me
A1T11
(by equation (6))
=f
M11
=f
MBA ,
7
e
A1BBA e
A2T=e
A1M11M1Be
A2T
=e
A1M11e
A1Te
A1M1Be
A2T
=f
MBA e
A1M1Be
A2T(by the previous result)
=f
MBA e
A1M1Be
A2T(by linearity)
=f
MBA e
A1M1e
A1Te
A1Be
A2T
=f
MBA f
M1e
B
=e
BBA ,
e
A2KBA e
A1T=e
A2K M1M11e
A1T
=e
A2K M1e
A1Te
A1M11e
A1T
=e
A2K M1e
A1Tf
MBA
=e
Kf
M1f
MBA
=e
KBA ,
and
e
A2JBA e
A2T=e
A2JK M1B+K M1M11M1Be
A2T
=e
A2Je
A2Te
A2K M1Be
A2T+e
A2K M1M11M1Be
A2T
=e
A2(J)e
A2Te
A2(K M1B)e
A2T+e
A2K M1e
A1Te
A1M11e
A1Te
A1M1Be
A2T
=e
Je
A2Ke
A1Te
A1M1e
A1Te
A1Be
A2T+
e
A2K M1e
A1Te
A1M11e
A1Te
A1M1e
A1Te
A1Be
A2T
=e
Je
Kf
M1e
B+e
Kf
M1f
MBA f
M1e
B
=e
JBA ,
which completes the proof of equality (5).
To show the converse claimed in the statement of Theorem 3, let us consider C1=Iand C2= 2 I.
Their Backus average is
B:=
3
20 0 0 0 0
03
20 0 0 0
0 0 4
30 0 0
0 0 0 4
30 0
0 0 0 0 4
30
0 0 0 0 0 3
2
.(8)
Following rotation, e
B:= e
A B e
AT,(9)
8
where e
Ais given in expression (A.1). It can be shown by direct calculation that the (3,3) entry of e
B
is
e
B33 =4
31+2b2+c22a2+d22.(10)
Since C1and C2are multiples of the identity,
e
A C1e
AT=C1and e
A C2e
AT=C2,
and the Backus average of e
A C1e
ATand e
A C2e
ATequals the Backus average of C1and C2, which is
matrix (8) . Hence,
e
A(C1, C2)e
AT=e
A C1e
AT,e
A C2e
AT
implies that, for expression (10),
4
31+2b2+c22a2+d22=4
3,
which results in
22
3b2+c2a2+d2= 0 .
Thus, either b=c= 0 or a=d= 0 . This is a necessary condition for symmetry (5) to hold, as
claimed.
Remark 4. Theorem 3 is formulated for general positive-definite matrices CR6×6, not all of
which represent elasticity tensors. However, expression (5) is continuous in the Ciand hence is
true in general only if it is also true for Ci,such as diagonal matrices, which are limits of elasticity
tensors.
5 The Backus average of randomly oriented tensors
In this section, we study the Backus average for a random orientations of a given tensor. As discussed
in Section 2, the arithmetic average of such orientations results in the Gazis et al. average, which
is the closest isotropic tensor with respect to the Frobenius norm. We see that—for the Backus
average—the result is, perhaps surprisingly, different.
Given an elasticity tensor, CR6×6, let us consider a sequence of its random rotations given by
Cj:= e
QjCe
QT
j, j = 1 , . . . , n ,
where QjR3×3are random matrices sampled from SO(3) .
The Cjare samples from some distribution and, hence, almost surely,
C:= lim
n→∞
1
n
n
X
j=1
Cj=µ(C),
the true mean,
µ(C) = Z
SO(3) e
Q C e
QTdσ(Q),
where dσ(Q) is Haar measure on SO(3) . Note that µ(C) is just the Gazis et al. average of C.
9
Similarly, for any expression X(C) of submatrices of C, which appear in the Backus-average formu-
las,
X:= lim
n→∞
1
n
n
X
j=1
Xj=µ(X).
Hence, almost surely limn→∞ (C1, . .. , Cn) equals the Backus average formula with each expres-
sion Xreplaced by
µ(X) = Z
SO(3)
Xf
QjCf
QjTdσ(Q).
Theorem 5. The limn→∞ (C1, . .. , Cn)exists almost surely, in which case it is transversely
isotropic. It is not, in general, isotropic.
Proof. Let AR3×3be an orthogonal matrix of type (3) or (4). Then
e
Cj:= e
A Cje
AT, j = 1 , . . . , n ,
=e
Af
QjCjf
QjTe
AT
=e
Af
QjCje
Af
QjT
=^
(AQj)Cj^
(AQj)
T
by the properties of the tilde operation, are also random samples from the same distribution. Hence,
almost surely,
lim
n→∞ e
C1, . . . , e
Cn= lim
n→∞ (C1, . . . , Cn) = B , say .
But by the symmetry property of the Backus average, Theorem 3,
e
C1, . . . , e
Cn=e
A(C1, . . . , Cn)e
AT.
Thus
B=e
A B e
AT,
which means that Bis invariant under a rotation of space by A. Consequently, Bis a transversely
isotropic tensor.
In general, the limit tensor is not isotropic, as illustrated by the following example. Let
C= diag [ 1 ,1,1,1,0,0 ] ,
which, as stated in Remark 4, represents a limiting case of an elasticity tensor. Numerical evidence
strongly suggests that
B=
1
2
1
4
1
40 0 0
1
4
1
2
1
40 0 0
1
4
1
4
1
20 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 1
4
,
which is not isotropic.
10
Although this is rather an artificial example, it could—with some computational effort—be “pro-
moted” to a legal proof. The conclusion is readily confirmed by the numerical examples presented
in Section 6.
In fact, it is easy to identify the limiting matrix B; it is just the Backus average expression (7), with
an expression X(C) replaced by the true mean
µ(X(C)) = Z
SO(3)
Xe
Q C e
QTdσ(Q).(11)
This limiting transversely isotropic tensor is of natural interest in its own right. It plays the role of
the Gazis et al. average in the context of the Backus average, and is the subject of a forthcoming
work.
6 Numerical example
Let us consider the elasticity tensor obtained by Dewangan and Grechka (2003); its components are
estimated from seismic measurements in New Mexico,
C=
7.8195 3.4495 2.5667 2 (0.1374) 2 (0.0558) 2 (0.1239)
3.4495 8.1284 2.3589 2 (0.0812) 2 (0.0735) 2 (0.1692)
2.5667 2.3589 7.0908 2 (0.0092) 2 (0.0286) 2 (0.1655)
2 (0.1374) 2 (0.0812) 2 (0.0092) 2 (1.6636) 2 (0.0787) 2 (0.1053)
2 (0.0558) 2 (0.0735) 2 (0.0286) 2 (0.0787) 2 (2.0660) 2 (0.1517)
2 (0.1239) 2 (0.1692) 2 (0.1655) 2 (0.1053) 2 (0.1517) 2 (2.4270)
.
(12)
Using tensor (12), let us demonstrate two methods to obtain Band their mutual convergence in the
limit.
The first method to obtain Brequires a stack of layers, whose elasticity tensors are C. We rotate
each C, using a random unit quaternion, and perform the Backus average of the resulting stack of
layers. Using 107layers, the Backus average is
B=
7.3008 2.9373 2.9379 2 (0.0000) 2 (0.0000) 2 (0.0000)
2.9373 7.3010 2.9381 2 (0.0000) 2 (0.0000) 2 (0.0000)
2.9379 2.9381 7.2689 2 (0.0000) 2 (0.0001) 2 (0.0000)
2 (0.0000) 2 (0.0000) 2 (0.0000) 2 (2.1710) 2 (0.0000) 2 (0.0000)
2 (0.0000) 2 (0.0000) 2 (0.0001) 2 (0.0000) 2 (2.1710) 2 (0.0000)
2 (0.0000) 2 (0.0000) 2 (0.0000) 2 (0.0000) 2 (0.0000) 2 (2.1819)
.
(13)
For an explicit formulation of the Backus average of generally anisotropic media, see Bos et al. (2017,
expressions (5)–(9)).
The second method requires integrals in place of arithmetic averages. Similarly to the first method,
we use a random unit quaternion, which is tantamount to a point on a 3-sphere. We approximate
the triple integral using Simpson’s and trapezoidal rules. Effectively, the triple integral is replaced
by a weighted sum of the integrand evaluated at discrete points. The sums that approximate the
integrals are accumulated and are used in expressions (7).
11
Figure 1: Difference between tensors (13) and (14)
Using the Simpson’s and trapezoidal rules, with a sufficient number of subintervals, the Backus
average is
B
RRR =
7.3010 2.9373 2.9380 2 (0.0000) 2 (0.0000) 2 (0.0000)
2.9373 7.3010 2.9380 2 (0.0000) 2 (0.0000) 2 (0.0000)
2.9380 2.9380 7.2687 2 (0.0000) 2 (0.0000) 2 (0.0000)
2 (0.0000) 2 (0.0000) 2 (0.0000) 2 (2.1711) 2 (0.0000) 2 (0.0000)
2 (0.0000) 2 (0.0000) 2 (0.0001) 2 (0.0000) 2 (2.1711) 2 (0.0000)
2 (0.0000) 2 (0.0000) 2 (0.0000) 2 (0.0000) 2 (0.0000) 2 (2.1818)
.
(14)
In the limit, the components of expressions (13) and (14) are the same; their similarity is illustrated
in Figure 1, where the horizontal axis is the number of layers and the vertical axis is the maximum
componentwise difference between the two tensors.
Expression (14) is transversely isotropic, as expected from Theorem 5, and in accordance with B´ona
et al. (2007, Section 4.3), since its four distinct eigenvalues are
λ1= 13.1658 , λ2= 4.3412 , λ3= 4.3636 , λ4= 4.3421 ,(15)
with multiplicities of m1=m2= 1 and m3=m4= 2 . The eigenvalues of expression (13) are in
agreement—up to 103—with eigenvalues (15) and their multiplicities. Furthermore, in accordance
with Theorem 5, in the limit, the distance to the closest isotropic tensor for expression (14) is
0.0326 6= 0; thus the distance does not reduce to zero.
Expressions (13) and (14) are transversely isotropic, which is the main conclusion of this work, even
though, for numerical modelling, one might view them as isotropic. This is indicated by Thomsen
(1986) parameters, which for tensor (14) are
γ= 2.4768 ×103, δ = 1.5816 ×103,  = 2.2219 ×103;
values much less than unity indicate very weak anisotropy.
7 Conclusions and future work
Examining the Backus average of a stack of layers consisting of randomly oriented anisotropic elastic-
ity tensors, we show that—in the limit—this average results in a homogeneous transversely isotropic
medium, as stated by Theorems 3 and 5. In other words, the randomness within layers does not
result in a medium lacking a directional pattern. Both the isotropic layers, as shown by Backus
(1962), and randomly oriented anisotropic layers, as shown herein, result in the average that is
12
transversely isotropic, as a consequence of inhomogeneity among parallel layers. This property is
discussed by Adamus et al. (2018), and herein it is illustrated in Appendix B.
In the limit, the transversely isotropic tensor is the Backus counterpart of the Gazis et al. average.
Indeed, the arithmetic average of randomized layers of an elasticity tensor produces the Gazis et
al. average and is its closest isotropic tensor, according to the Frobenius norm. On the other hand,
the Backus average of the layers resulting from a randomization of the same tensor produces the
transversely isotropic tensor given in expression (11). This tensor and its properties are the subject
of a forthcoming paper.
Acknowledgments
We wish to acknowledge discussions with Michael G. Rochester, proofreading of David R. Dalton,
as well as the graphic support of Elena Patarini. This research was performed in the context of The
Geomechanics Project supported by Husky Energy. Also, this research was partially supported by
the Natural Sciences and Engineering Research Council of Canada, grant 238416-2013.
References
Adamus, F. P., Slawinski, M. A., and Stanoev, T. (2018). On effects of inhomogeneity on anisotropy
in Backus average. arXiv, physics.geo-ph(1802.04075).
Backus, G. E. (1962). Long-wave elastic anisotropy produced by horizontal layering. Journal of
Geophysical Research, 67(11):4427–4440.
ona, A., Bucataru, I., and Slawinski, M. A. (2007). Coordinate-free characterization of the sym-
metry classes of elasticity tensors. Journal of Elasticity, 87(2–3):109–132.
ona, A., Bucataru, I., and Slawinski, M. A. (2008). Space of SO(3)-orbits of elasticity tensors.
Archives of Mechanics, 60(2):123–138.
Bos, L., Dalton, D. R., Slawinski, M. A., and Stanoev, T. (2017). On Backus average for generally
anisotropic layers. Journal of Elasticity, 127(2):179–196.
Dewangan, P. and Grechka, V. (2003). Inversion of multicomponent, multiazimuth, walkaway VSP
data for the stiffness tensor. Geophysics, 68(3):1022–1031.
Gazis, D. C., Tadjbakhsh, I., and Toupin, R. A. (1963). The elastic tensor of given symmetry nearest
to an anisotropic elastic tensor. Acta Crystallographica, 16(9):917–922.
Slawinski, M. A. (2018). Waves and rays in seismology: Answers to unasked questions. World
Scientific, 2 edition.
Thomsen, L. (1986). Weak elastic aniostropy. Geophysics, 51(10):1954–1966.
13
A Rotations by unit quaternions
The R6equivalent for ASO(3) of cijk` , which is the rotation of tensor (1), is (e.g., Slawinski,
2018, equation (3.42))
˜
A=
A2
11 A2
12 A2
13 2A12A13 2A11 A13 2A11A12
A2
21 A2
22 A2
23 2A22A23 2A21 A23 2A21A22
A2
31 A2
32 A2
33 2A32A33 2A31 A33 2A31A32
2A21A31 2A22 A32 2A23A33 A23 A32 +A22A33 A23 A31 +A21A33 A22 A31 +A21A32
2A11A31 2A12 A32 2A13A33 A13 A32 +A12A33 A13 A31 +A11A33 A12 A31 +A11A32
2A11A21 2A12 A22 2A13A23 A13 A22 +A12A23 A13 A21 +A11A23 A12 A21 +A11A22
.
In quaternions, this expression is
e
A=
a2+b2
c2
d22(2 b c 2a d)2
(2 b c + 2 a d)2a2
b2+c2
d22
(2 b d 2a c)2(2 a b + 2 c d)2
2 (2 b c + 2 a d) (2 b d 2a c)2 (2 a b + 2 c d)a2
b2+c2
d2
2 (2 b d 2a c)a2+b2
c2
d22 (2 b c 2a d) (2 a b + 2 c d)
2 (2 b c + 2 a d)a2+b2
c2
d22 (2 b c 2a d)a2
b2+c2
d2
(2 a c + 2 b d)22 (2 b c 2a d) (2 a c + 2 b d)
(2 c d 2a b)22 (2 c d 2a b)a2
b2+c2
d2
a2
b2
c2+d222 (2 a b + 2 c d)a2
b2
c2+d2
2 (2 c d 2a b)a2
b2
c2+d2(2 c d 2a b) (2 a b + 2 c d) + a2
b2+c2
d2 a2
b2
c2+d2
2 (2 a c + 2 b d)a2
b2
c2+d2(2 a c + 2 b d) (2 a b + 2 c d) + (2 b c 2a d)a2
b2
c2+d2
2 (2 a c + 2 b d) (2 c d 2a b) (2 c d 2a b) (2 b c 2a d) + (2 a c + 2 b d)a2
b2+c2
d2
2 (2 a c + 2 b d)a2+b2
c2
d2
2 (2 b c + 2 a d) (2 c d 2a b)
2 (2 b d 2a c)a2
b2
c2+d2
(2 b d 2a c) (2 c d 2a b) + (2 b c + 2 a d)a2
b2
c2+d2
(2 b d 2a c) (2 a c + 2 b d) + a2+b2
c2
d2 a2
b2
c2+d2
(2 b c + 2 a d) (2 a c + 2 b d) + (2 c d 2a b)a2+b2
c2
d2
2 (2 b c 2a d)a2+b2
c2
d2
2 (2 b c + 2 a d)a2
b2+c2
d2
2 (2 b d 2a c) (2 a b + 2 c d)
(2 b c + 2 a d) (2 a b + 2 c d) + (2 b d 2a c)a2
b2+c2
d2
(2 b d 2a c) (2 b c 2a d) + (2 a b + 2 c d)a2+b2
c2
d2
(2 b c + 2 a d) (2 b c 2a d) + a2+b2
c2
d2 a2
b2+c2
d2
.
(A.1)
14
B Alternating layers
Consider a randomly-generated elasticity tensor,
C=
14.5739 6.3696 2.9020 2 (9.4209) 2 (3.8313) 2 (3.5851)
6.3696 10.7276 6.2052 2 (4.0375) 2 (5.1333) 2 (6.0745)
2.9020 6.2052 11.4284 2 (1.9261) 2 (9.8216) 2 (1.3827)
2 (9.4209) 2 (4.0375) 2 (1.9261) 2 (13.9034) 2 (0.2395) 2 (2.0118)
2 (3.8313) 2 (5.1333) 2 (9.8216) 2 (0.2395) 2 (10.7353) 2 (0.0414)
2 (3.5851) 2 (6.0745) 2 (1.3827) 2 (2.0118) 2 (0.0414) 2 (9.0713)
,
(B.2)
whose eigenvalues are
λ1= 34.0318 , λ2= 18.1961 , λ3= 10.4521 , λ4= 4.8941 , λ5= 2.2737 , λ6= 0.5921 .
The Backus average of 107alternating layers composed of randomly oriented tensors (12) and (B.2)
is
B
RRR =
8.4711 1.1917 1.2572 2 (0.0000) 2 (0.0000) 2 (0.0000)
1.1917 8.4710 1.2570 2 (0.0000) 2 (0.0000) 2 (0.0000)
1.2572 1.2570 6.6648 2 (0.0001) 2 (0.0000) 2 (0.0000)
2 (0.0000) 2 (0.0000) 2 (0.0001) 2 (2.8440) 2 (0.0000) 2 (0.0000)
2 (0.0000) 2 (0.0000) 2 (0.0000) 2 (0.0000) 2 (2.8440) 2 (0.0000)
2 (0.0000) 2 (0.0000) 2 (0.0000) 2 (0.0000) 2 (0.0000) 2 (3.6340)
.
Its eigenvalues show that this is a transversely isotropic tensor,
λ1= 10.4892 , λ2= 5.8384 , λ3= 7.2794 , λ4= 7.2793 , λ5= 5.6880 , λ6= 5.6878 .
Its Thomsen parameters,
γ= 0.1400 , δ = 0.0433 ,  = 0.1353 ,
indicate greater anisotropy than for tensor (14), as expected. In other words, an emphasis of a
pattern of inhomogeneity results in an increase of anisotropy.
15
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