24
1 From Riemann manifolds to Riemann manifolds
We pose two problems. (i) Derive the right Cauchy–Green deformation tensor. (ii) Solve the right
general eigenvalue–eigenvector problem.
Solution (the first problem).
By means of detailed derivations given in Boxes 1.13 and 1.14, we aim at an analytical analysis of
the right Cauchy–Green deformation tensor in Cartesian coordinates {x, y} and in polar coordinates
{α, r}, which cover the projection plane P
2
O . The right mapping equations {Λ(x, y), Φ(x, y)} and
{Λ(α), Φ(r)} are given first. They are constituted from the identities x = X = R cos Φ cos Λ and
y = Y = R cos Φ sin Λ, where {Λ, Φ} are the spherical coordinates. {Λ, Φ} or {longitude, latitude}
label a point in S
2
R + . We use the symbol + in order to allow only positive values Z ∈ R
+ , which are
points in the northern hemisphere. Second, we compute the right Jacobi matrices J r (x, y) and J r (α, r)
in Cartesian coordinates {x, y} and in polar coordinates {α, r}. While J r (x, y) is a fully occupied
matrix, J r (α, r) is diagonal. Third, this difference continues when we are going to compute the right
Cauchy–Green matrices C r (x, y) and C r (α, r). Again, C r (x, y) is a fully occupied symmetric matrix,
while C r (α, r) is diagonal. Fourth, in Box 1.13, we represent the right Cauchy–Green deformation
tensor as a tensor of second order in the Cartesian two-basis e µ ⊗ e ν for all {µ, ν} = {1, 2}. Note that
R
2 = span{e 1 , e 2 }, where {e 1 , e 2 O} is an orthonormal two-leg at O. Remarkably, C r (x, y) includes
the components e 1 ⊗ e 2 , e 1 ⊗ e 2 + e 2 ⊗ e 1 , and e 2 ⊗ e 2 . In contrast, the algebra of the right Cauchy–
Green deformation tensor in Box 1.14, represented in polar coordinates, is slightly more complicated.
A placement vector x(α, r) ∈ P
2
O is locally described by the tangent space T x M
2
r spanned by the
tangent vectors g 1 = D α x and g 1 = D r x. In polar coordinates {α, r}, the matrix of the right metric
is given by G r = diag(r
2 , 1), a diagonal matrix. The first differential invariant of M
2
r ∼ P
2
O is given
by (ds)
2 = f (dα, dr). The basis {g
1 , g
2
}, which is dual to {g 1 , g 2 }, also called co-frame, is computed
next, namely by G
−1 (α, r). Due to the orthogonality of the two-leg {g 1 , g 2 p}, the co-frame amounts
to g
1 = g 1 /g 11 and g
2 = g 2 /g 22 , respectively. Question: “Why did we bother you with the notation of
the co-frame {g 1 , g 2 p}?” Answer: “Often the moving frame {g 1 (α, r), g 2 (α, r)} is called covariant,
accordingly its dual {g
1 (α, r), g
2 (α, r)} is called contravariant. The properly posed question can be
answered immediately. The second-order tensor C r (α, r) is represented in the contravariant or twoco-basis {g
1
⊗ g
1 , g
1
⊗ g
2 , g
2
⊗ g
1 , g
2
⊗ g
2
}, in general. Due to the diagonal structure of the right
deformation tensor C r (r), contains only components g
1
⊗ g
1 and g
2
⊗ g
2 , or g 1 ⊗ g 1 and g 2 ⊗ g 2 ,
respectively.”
End of Solution (the first problem).
Solution (the second problem).
The results on the right eigenspace analysis of the matrix pair {C r , G r } are collected in Box 1.15 and
Box 1.16, exclusively. In particular, we aim at computing the right eigenvalues, eigencolumns, and
eigenvectors, namely in Box 1.15 in Cartesian coordinates {x, y} along the fixed orthonormal frame
{e 1 , e 2 } and in Box 1.16 in polar coordinates {α, r} along the moving orthogonal frame {g 1 , g 2 p}.
First, we solve the right general eigenvalue problem, both in Cartesian representation {λ 1 (x, y), λ 2 = 1}
and in polar representation {λ 1 (r), λ 2 = 1}. The characteristic equation
C r − λ
2 G r
= 0 is solved in
Box 1.16 if both C r and G r are diagonal. The determinantal identity is factorized directly into the right
eigenvalues λ 1 and λ 2 , a result we take advantage from in a following section. Second, we derive the
simple structure of the eigencolumns {f 11 (x, y), f 21 (x, y)} and {f 12 (x, y), f 22 (x, y)} in case of Cartesian
coordinates as well as of the eigencolumns {f 11 (r), f 21 (r)} and {f 12 (r), f 22 (r)} in polar coordinates.
Third, let us derive the right eigenvectors. In Box 1.15, we succeed to represent the orthonormal right
eigenvectors in the Cartesian basis {e 1 , e 2 p}. In contrast, in Box 1.16, we are able to compute the
first right eigenvector as a tangent vector of the image of the parallel circle, while the second right
eigenvector “radial” as a tangent vector of the image (straight line) of the meridian. Such a beautiful
result is illustrated by Fig. 1.11.
End of Solution (the second problem).
1 From Riemann manifolds to Riemann manifolds
We pose two problems. (i) Derive the right Cauchy–Green deformation tensor. (ii) Solve the right
general eigenvalue–eigenvector problem.
Solution (the first problem).
By means of detailed derivations given in Boxes 1.13 and 1.14, we aim at an analytical analysis of
the right Cauchy–Green deformation tensor in Cartesian coordinates {x, y} and in polar coordinates
{α, r}, which cover the projection plane P
2
O . The right mapping equations {Λ(x, y), Φ(x, y)} and
{Λ(α), Φ(r)} are given first. They are constituted from the identities x = X = R cos Φ cos Λ and
y = Y = R cos Φ sin Λ, where {Λ, Φ} are the spherical coordinates. {Λ, Φ} or {longitude, latitude}
label a point in S
2
R + . We use the symbol + in order to allow only positive values Z ∈ R
+ , which are
points in the northern hemisphere. Second, we compute the right Jacobi matrices J r (x, y) and J r (α, r)
in Cartesian coordinates {x, y} and in polar coordinates {α, r}. While J r (x, y) is a fully occupied
matrix, J r (α, r) is diagonal. Third, this difference continues when we are going to compute the right
Cauchy–Green matrices C r (x, y) and C r (α, r). Again, C r (x, y) is a fully occupied symmetric matrix,
while C r (α, r) is diagonal. Fourth, in Box 1.13, we represent the right Cauchy–Green deformation
tensor as a tensor of second order in the Cartesian two-basis e µ ⊗ e ν for all {µ, ν} = {1, 2}. Note that
R
2 = span{e 1 , e 2 }, where {e 1 , e 2 O} is an orthonormal two-leg at O. Remarkably, C r (x, y) includes
the components e 1 ⊗ e 2 , e 1 ⊗ e 2 + e 2 ⊗ e 1 , and e 2 ⊗ e 2 . In contrast, the algebra of the right Cauchy–
Green deformation tensor in Box 1.14, represented in polar coordinates, is slightly more complicated.
A placement vector x(α, r) ∈ P
2
O is locally described by the tangent space T x M
2
r spanned by the
tangent vectors g 1 = D α x and g 1 = D r x. In polar coordinates {α, r}, the matrix of the right metric
is given by G r = diag(r
2 , 1), a diagonal matrix. The first differential invariant of M
2
r ∼ P
2
O is given
by (ds)
2 = f (dα, dr). The basis {g
1 , g
2
}, which is dual to {g 1 , g 2 }, also called co-frame, is computed
next, namely by G
−1 (α, r). Due to the orthogonality of the two-leg {g 1 , g 2 p}, the co-frame amounts
to g
1 = g 1 /g 11 and g
2 = g 2 /g 22 , respectively. Question: “Why did we bother you with the notation of
the co-frame {g 1 , g 2 p}?” Answer: “Often the moving frame {g 1 (α, r), g 2 (α, r)} is called covariant,
accordingly its dual {g
1 (α, r), g
2 (α, r)} is called contravariant. The properly posed question can be
answered immediately. The second-order tensor C r (α, r) is represented in the contravariant or twoco-basis {g
1
⊗ g
1 , g
1
⊗ g
2 , g
2
⊗ g
1 , g
2
⊗ g
2
}, in general. Due to the diagonal structure of the right
deformation tensor C r (r), contains only components g
1
⊗ g
1 and g
2
⊗ g
2 , or g 1 ⊗ g 1 and g 2 ⊗ g 2 ,
respectively.”
End of Solution (the first problem).
Solution (the second problem).
The results on the right eigenspace analysis of the matrix pair {C r , G r } are collected in Box 1.15 and
Box 1.16, exclusively. In particular, we aim at computing the right eigenvalues, eigencolumns, and
eigenvectors, namely in Box 1.15 in Cartesian coordinates {x, y} along the fixed orthonormal frame
{e 1 , e 2 } and in Box 1.16 in polar coordinates {α, r} along the moving orthogonal frame {g 1 , g 2 p}.
First, we solve the right general eigenvalue problem, both in Cartesian representation {λ 1 (x, y), λ 2 = 1}
and in polar representation {λ 1 (r), λ 2 = 1}. The characteristic equation
C r − λ
2 G r
= 0 is solved in
Box 1.16 if both C r and G r are diagonal. The determinantal identity is factorized directly into the right
eigenvalues λ 1 and λ 2 , a result we take advantage from in a following section. Second, we derive the
simple structure of the eigencolumns {f 11 (x, y), f 21 (x, y)} and {f 12 (x, y), f 22 (x, y)} in case of Cartesian
coordinates as well as of the eigencolumns {f 11 (r), f 21 (r)} and {f 12 (r), f 22 (r)} in polar coordinates.
Third, let us derive the right eigenvectors. In Box 1.15, we succeed to represent the orthonormal right
eigenvectors in the Cartesian basis {e 1 , e 2 p}. In contrast, in Box 1.16, we are able to compute the
first right eigenvector as a tangent vector of the image of the parallel circle, while the second right
eigenvector “radial” as a tangent vector of the image (straight line) of the meridian. Such a beautiful
result is illustrated by Fig. 1.11.
End of Solution (the second problem).