20
1 From Riemann manifolds to Riemann manifolds
Solution (the first problem).
Let us rewrite the mapping equations in a more systematic form by introducing the two constants
c 1 := 2R/
√
6π and c 2 := R
2π/3 in Box 1.9 in order to analyze the graticule of “Eckert II”. First,
the geometrical shape of the image of the meridians is determined by removing the root
4 − 3 sin |Φ|
from the second equation by substituting the root from the first equation. For Λ = constant, we are
led to the straight line L
1 (Λ = constant). Second, the parallel circles are immediately fixed in shape
by Φ = constant. x is a homogeneous linear form of longitude Λ and y is a constant. In summary, the
meridians are tilted straights and the parallel circles are parallel straights. Third, let us compute the
length of the circular equator x(Λ = +π, Φ = 0) − x(Λ = −π, Φ = 0) = 8Rπ/
√
6π = 4πc 1 , the length
of the central meridian x(Λ = 0, Φ = +π/2) − x(Λ = 0, Φ = −π/2) = 4Rπ/
√
6π = 2πc 1 , and the
length of the image of the pole x(Λ = +π, |Φ| = π/2) − x(Λ = −π, |Φ| = π/2) = 4Rπ/
√
6π = 2πc 1 .
Obviously, the length of image of the circular equator is twice the length of image of the central
meridian or the pole.
End of Solution (the first problem).
Solution (the second problem).
In order to derive the left Cauchy–Green deformation tensor, according to Box 1.10, we depart from
computing the left Jacobi matrix J l . First, the partial derivatives D Λ x, D Φ x, D Λ y, and D Φ y build up
the left Jacobi matrix. Second, by means of the matrix product C l = J
∗
l G r J l , we are able to compute
the left Cauchy–Green matrix for the right matrix of the metric G r = I 2 . Indeed, the chart {x, y}
is covered by Cartesian coordinates whose metric is simply given by ds
2 = dx
2 + dy
2 . Though the
special left Cauchy–Green matrix C l = J
∗
l J l looks simple, but is complicated in detail. The elements
{c 11 , c 12 = c 21 , c 22 } document these features.
End of Solution (the second problem).
Solution (the third problem).
Box 1.11 outlines the solution of the third problem, namely the laborious analytical computation of
the left eigenvalues and the left eigencolumns. First, we refer to G l as the matrix of the metric of
the sphere S
2
R of radius R, and to C l as the matrix of the left Cauchy–Green tensor, as computed
in Box 1.10. The characteristic equation of the left general eigenvalue problem leads to the solution
Λ
2
1,2 = Λ
2
+,− as functions of the two fundamental invariants (i) tr
C l G
−1
l
and (ii) det
C l G
−1
l
. While
the elements of the matrix C l G
−1
l
evoke simple, its trace is complicated. In contrast, det
C l G
−1
l
= 1.
Second, it is a straightforward proof that the product of eigenvalues squared is identical to the second
invariant, i. e. Λ
2
1 Λ
2
2 = det
C l G
−1
l
. As proven, det
C l G
−1
l
= 1 (in consequence Λ 1 Λ 2 = 1) can be
interpreted as the condition for an equiareal mapping. A detailed computation of the left eigenvalues
Λ 1 , Λ 2
=
Λ + , Λ −
is not useful due to the lengthy forms involved. Third, the same argument
holds for the computed first eigencolumn, which is associated to Λ 1 =
Λ 2
1 ∈ R
+ and for the second
eigencolumn, which is associated to Λ 2 =
Λ 2
2 ∈ R
+ , and these are very lengthy. For practical use, a
computation in a {Λ, Φ} lattice (for instance, 1
◦
× 1
◦ ) is recommended.
End of Solution (the third problem).
Solution (the fourth problem).
Box 1.12 collects the details of the proof that the “Eckert II mapping” of the point {Λ, Φ} = {0, 0} is
not an isometry. For an isometry, Λ 1 = Λ 2 = 1 is the postulate. If Λ 1 = Λ 2 , then it holds that
(tr[C l G
−1
l ])
2 = 4det[C l G
−1
l ]. Since (tr[C l G
−1
l (Λ = 0, Φ = 0)])
2 = [(64 + 9π
2 )/24π]
2
= 4 due to
det[C l G
−1
l ] = 1, it follows that Λ 1 = Λ 2 .
End of Solution (the fourth problem).
1 From Riemann manifolds to Riemann manifolds
Solution (the first problem).
Let us rewrite the mapping equations in a more systematic form by introducing the two constants
c 1 := 2R/
√
6π and c 2 := R
2π/3 in Box 1.9 in order to analyze the graticule of “Eckert II”. First,
the geometrical shape of the image of the meridians is determined by removing the root
4 − 3 sin |Φ|
from the second equation by substituting the root from the first equation. For Λ = constant, we are
led to the straight line L
1 (Λ = constant). Second, the parallel circles are immediately fixed in shape
by Φ = constant. x is a homogeneous linear form of longitude Λ and y is a constant. In summary, the
meridians are tilted straights and the parallel circles are parallel straights. Third, let us compute the
length of the circular equator x(Λ = +π, Φ = 0) − x(Λ = −π, Φ = 0) = 8Rπ/
√
6π = 4πc 1 , the length
of the central meridian x(Λ = 0, Φ = +π/2) − x(Λ = 0, Φ = −π/2) = 4Rπ/
√
6π = 2πc 1 , and the
length of the image of the pole x(Λ = +π, |Φ| = π/2) − x(Λ = −π, |Φ| = π/2) = 4Rπ/
√
6π = 2πc 1 .
Obviously, the length of image of the circular equator is twice the length of image of the central
meridian or the pole.
End of Solution (the first problem).
Solution (the second problem).
In order to derive the left Cauchy–Green deformation tensor, according to Box 1.10, we depart from
computing the left Jacobi matrix J l . First, the partial derivatives D Λ x, D Φ x, D Λ y, and D Φ y build up
the left Jacobi matrix. Second, by means of the matrix product C l = J
∗
l G r J l , we are able to compute
the left Cauchy–Green matrix for the right matrix of the metric G r = I 2 . Indeed, the chart {x, y}
is covered by Cartesian coordinates whose metric is simply given by ds
2 = dx
2 + dy
2 . Though the
special left Cauchy–Green matrix C l = J
∗
l J l looks simple, but is complicated in detail. The elements
{c 11 , c 12 = c 21 , c 22 } document these features.
End of Solution (the second problem).
Solution (the third problem).
Box 1.11 outlines the solution of the third problem, namely the laborious analytical computation of
the left eigenvalues and the left eigencolumns. First, we refer to G l as the matrix of the metric of
the sphere S
2
R of radius R, and to C l as the matrix of the left Cauchy–Green tensor, as computed
in Box 1.10. The characteristic equation of the left general eigenvalue problem leads to the solution
Λ
2
1,2 = Λ
2
+,− as functions of the two fundamental invariants (i) tr
C l G
−1
l
and (ii) det
C l G
−1
l
. While
the elements of the matrix C l G
−1
l
evoke simple, its trace is complicated. In contrast, det
C l G
−1
l
= 1.
Second, it is a straightforward proof that the product of eigenvalues squared is identical to the second
invariant, i. e. Λ
2
1 Λ
2
2 = det
C l G
−1
l
. As proven, det
C l G
−1
l
= 1 (in consequence Λ 1 Λ 2 = 1) can be
interpreted as the condition for an equiareal mapping. A detailed computation of the left eigenvalues
Λ 1 , Λ 2
=
Λ + , Λ −
is not useful due to the lengthy forms involved. Third, the same argument
holds for the computed first eigencolumn, which is associated to Λ 1 =
Λ 2
1 ∈ R
+ and for the second
eigencolumn, which is associated to Λ 2 =
Λ 2
2 ∈ R
+ , and these are very lengthy. For practical use, a
computation in a {Λ, Φ} lattice (for instance, 1
◦
× 1
◦ ) is recommended.
End of Solution (the third problem).
Solution (the fourth problem).
Box 1.12 collects the details of the proof that the “Eckert II mapping” of the point {Λ, Φ} = {0, 0} is
not an isometry. For an isometry, Λ 1 = Λ 2 = 1 is the postulate. If Λ 1 = Λ 2 , then it holds that
(tr[C l G
−1
l ])
2 = 4det[C l G
−1
l ]. Since (tr[C l G
−1
l (Λ = 0, Φ = 0)])
2 = [(64 + 9π
2 )/24π]
2
= 4 due to
det[C l G
−1
l ] = 1, it follows that Λ 1 = Λ 2 .
End of Solution (the fourth problem).