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=
(10) + (11)
b
+ (12)
b
2
+ 3(30)
l
2
+O
3
x
.
(11)
l
+ 2(12)
lb
+O
3
x
(20.135)
(01) + 2(02)
b
1 + 3(03)
b
2
+O
3
y
2(20)
l
+O
3
y
The matrix [
c
KL
] is given in Box
20.9
. Let us canonically compare the two positive-definite,
symmetric matrices left metric matrix G
l
=
{
G
KL
}
and left Cauchy-Green matrix C
l
=
{
c
KL
}
,
namely by the
simultaneous diagonalization
(
20.136
).
F
l
C
l
F
l
=diag(
Λ
1
,Λ
1
) susF
l
G
l
F
l
=I
2
.
(20.136)
By means of the left Frobenius matrix F
l
, the left Cauchy-Green matrix C
l
is transformed into
a diagonal matrix ((
Λ
1
,Λ
2
) are the eigenvalues). In contrast, the left metric matrix G
l
is trans-
formed into a unit matrix I
2
. The problem to simultaneously diagonalize the two symmetric
matrices
,whereG
l
is positive-definite, is equivalent to the
general eigenvalue-eigenvector
problem
. Here, we are left with
eigenvalues
{
C
l
,
G
l
}
(typical for (Riemann) polar/normal
coordinates), which are given by Box
20.10
. Finally, we compute the
maximum angular distor-
tion
(
20.137
)inBox
20.11
.
{
Λ
1
,Λ
2
=1
}
ω
:= 2 arcsin
Λ
1
−
Λ
2
Λ
1
+
Λ
2
.
(20.137)
Box 20.9 (Left Cauchy-Green deformation tensor, Riemann coordinates).
c
11
=
V
0
b
+
3
t
0
−
2
η
0
−
9
η
0
t
0
cos
2
B
0
t
0
3
2
t
0
2
−
b
2
bl
2
=1
−
−
3
V
0
η
0
(5 + 12
t
0
)
η
0
(7 + 12
t
0
))
+
t
0
(2
−
−
b
3
+
3
V
0
+
V
0
t
0
cos
4
B
0
12
(14
−
60
t
0
+
η
0
(28
−
63
t
0
)+
η
0
(14
−
3
t
0
)) cos
2
B
0
90
V
0
l
4
b
2
l
2
+
−
b
4
60
V
0
[4
η
0
(24
292
t
0
+60
t
0
)
η
0
(60
369
t
0
−
240
t
0
)
+
−
−
−
−
−η
0
(32
−
77
t
0
−
1140
t
0
)]
×
×
N
0
cos
2
B
0
,
c
12
=
c
21
=
3
V
0
bl
+
t
(2
1
−
3
η
0
−
5
η
0
)
b
2
l
+
cos
2
B
0
t
0
6
l
3
+
=
−
6
V
0
2
η
0
(17
81
t
0
)
η
0
(80
39
t
0
)
η
0
(42 + 123
t
0
)
+
4
−
−
−
−
−
b
3
l
+
(20.138)
90
V
0
bl
3
10
t
0
+
η
0
(8 + 17
t
0
)+
η
0
(4 + 27
t
0
)) cos
2
B
0
90
V
0
+
(4
−
×
N
0
cos
2
B
0
,
c
22
=
×
V
0
+
6
η
0
t
0
t
0
+
η
0
+7
η
0
t
0
)
V
0
b
2
+
cos
2
B
0
3
V
0
1
b
+
3(1
−
l
2
=
V
0
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