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The sum of the oriented angles of a hyperbolic triangle is minus two right
circular angles. To prove this theorem, draw a line parallel to a side and apply the Z-
theorem (figure 13.4): they will form
an angle between opposite directions,
Figure 13.4
which has the value ’π e12. The minus
sign is caused by the fact that,
following the positive orientation of
angles, a right angle is subtracted each
time an asymptote is trespassed.
Look at the figure 13.5 where
an angle inscribed in an equilateral
hyperbola is drawn. A diameter
divides the inscribed angle into the
angles ± and β. Then we draw two
radius from the origin to both extremes of the angle. All the radius have the same
length, so that the upper and lowest triangle are isosceles and have two equal angles.
Since the sum of the angles of each hyperbolic triangle is ’π e12, the third angle is
’π e12’2± and ’π e12’2β respectively. So the supplementary angles are 2± and 2β ,
and hence the theorem follows: the
inscribed angle on a hyperbola x2 ’ y2 =
r2 is the half of the central angle
(intercepted arc) so that it is constant
and independent of the location of its
vertex.
For example, take the points (5,
3) and (5, ’3) as extremes of an
inscribed angle with the vertex placed
at the negative branch, for example at
(’4, 0) or (’5, ’3). Even you can take
the vertex at the positive branch, e. g. at Figure 13.5
(4, 0). Anyway its value is log 2. Now
calculate the central angle or
intercepted arc (with vertex at the origin) and see that its value is log 4, twice the
inscribed angle.


Distance between points

The distance between two points on the hyperbolic plane is defined as the
modulus of the vector going from one point to the another:

(x ’ x P ) ’ (yQ ’ y P )
2 2
d ( P, Q ) = PQ = Q
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 161


In the hyperbolic plane, the distance is a real or imaginary positive number (depending
on the sector), which has the following properties:

1) d (P, Q ) = d (Q , P )
2) If the vectors PQ, QR and PR lie on the positive half sector then they fulfil
the triangular inequality:

d (P, R ) ≥ d (P, Q ) + d (Q, R )

The prove is obtained by means of the inner product:

PR 2 = (PQ + QR ) = PQ 2 + QR 2 + 2 PQ · QR
2




= PQ 2 + QR 2 + 2 PQ QR coshψ ≥ PQ 2 + QR 2 + 2 PQ QR

The extraction of the square root yields the triangular inequality:

PR ≥ PQ + QR

So the sum of two sides of a triangle is smaller than the third side whenever they are
taken in the positive half sector. For example, let us apply this statement to the triangle
with vertices A = (5, 3), B = (1, 0) and C = (10, 1). Taking the modulus of the sides
positive, we have:

BC = 80 BA = 7 AC = 21

’ 80 ≥ 7 + 21
BC ≥ BA + AC

This result must be commented with more detail. Firstly, we see that the straight
path has the highest length among the possible paths between two given points. In the
Minkowski™s space-time this fact causes the twin paradox. Since the pseudo-Euclidean
length of the path is the proper time for each person, the brother who followed the
straight path -going in an inertial frame- has aged more than the brother who followed
another path -subjected to accelerations-.


Area on the hyperbolic plane

Now I give a general geometric definition of area valid for both Euclidean and
hyperbolic planes. If a parallelogram has orthogonal sides, then the modulus of its area
is equal to the product of the lengths of the orthogonal sides. In Euclidean geometry a
parallelogram with orthogonal sides is called a rectangle. In the hyperbolic plane, two
sides are orthogonal if their directions are seen by us as symmetric with respect to the
direction of the quadrant bisector. So, I have preferred to avoid the word rectangle in
this case, while the term parallelogram is still valid within this context.
162 RAMON GONZALEZ CALVET


Which is the suitable algebraic expression to calculate the area of any
parallelogram? Suppose that it is the outer product. If a and b are the sides of the
parallelogram then:

A = a § b = a b sinhψ (a , b )

Let us see the consistence of this expression. First at all, recall that the angle
between orthogonal hyperbolic vectors is a right circular angle:

π
ψ ⊥ = ± e12
2

so that the area of a parallelogram with orthogonal sides is the product of their lengths
as expected:

«π  π
A = a b sinh¬ ± e12 · = a b sin = a b
2 2


For any parallelogram not having orthogonal sides, the area is the product of the
base (one side) for the altitude (the projection of the other side onto the direction
orthogonal to the base) and this product is only given by the outer product because its
anticommutativity removes the proportional projection:

A = a § b = (a || + a ⊥ ) § b = a ⊥ b

The expression for the area in Cartesian
components is equal to that for the
Euclidean plane. It means that we can
calculate areas graphically in the usual way,
fact that has allowed till now to define the
hyperbolic trigonometric functions from the
area scanned by the hyperbola radius in the
Euclidean plane, in spite of not being their
proper plane. However, as I have explained
Figure 13.6
in the footnote at the page 142, we must
perceive the pseudo-Euclidean nature of the
area in the hyperbolic plane. The figure 13.6
displays how are the radii perpendicular to the hyperbola because the bisector of the
radius and the tangent vector is parallel to the quadrant bisector. Also it may be proved
analytically by means of differentiation of the equation x2 ’ y2 = r2:

2 x dx ’ 2 y dy = 0 r · dr = 0


showing that the radius vector and its differential are orthogonal for the hyperbola with
constant radius. On the other hand, observe that the triangles drawn in the figure 13.6
are isosceles and have two equal angles approaching the right angle value at the limit of
null area.
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 163



Diameters of the hyperbola and Apollonius™ theorem

At the page 129, I postponed an interpretation of the central equation for the
hyperbola analogous to the cylindrical angles for the ellipse because it is properly of
pseudo-Euclidean nature. Here I develop this interpretation.
If P is any point on a hyperbola with major and minor half-axis OQ and OS
(figure 11.17 reproduced below) then its central equation is:

sinh χ
OQ cosh χ + OS
e2 ’ 1
OP = ±
sinh 2 χ
cosh χ ’ 2
2

e ’1

Since OQ and OS are orthogonal, the square
of this equation fulfils:

OP 2 OP 2
cosh χ + sinh 2 χ = 1
2
Figure 11.17
2 2
OQ OS

Both half-axis are related with the
eccentricity through:

OS 2 = (1 ’ e 2 ) OQ 2

Note that OS2<0 since it is the square of an ordinate on the hyperbolic plane.
Introducing the hyperbolic angle ψ in the following way:

cosh χ
coshψ = Figure 13.7
sinh 2 χ
cosh χ ’ 2
2

e ’1

now the equation of the hyperbola
becomes:

OP = ± (OQ coshψ + OS sinh ψ )

Observe in the figure 13.7 that
any hyperbola can be obtained as an
intersection of a transverse plane with the
equilateral hyperbolic prism. The plane
of the acute hyperbola (with 1 < e < 2 )
forms an angle φ with respect to the horizontal, and hence:

cosh χ cos φ
coshψ =
cosh 2 χ cos 2 φ ’ sinh 2 χ
164 RAMON GONZALEZ CALVET



So the eccentricity e of the hyperbola is related with the obliquity φ of the transverse
plane through the relationship:

cos φ = e 2 ’ 1 with 1< e < 2

The Apollonius™ conjugate diameters of any hyperbola are the intersections of
the transverse plane with a pair of orthogonal axial planes; in other words, two radii are
conjugate (figure 11.17) if their projections onto the horizontal plane are turned through
the same hyperbolic angle • :

OQ' = OQ cosh • + OS sinh•

OS' = OQ sinh• + OS cosh •

Our Euclidean eyes see the horizontal projections as symmetric lines with respect to the
quadrant bisector. However, they are actually orthogonal because:

OQ' 2 ’ OS' 2 = OQ2 ’ OS2

and, therefore, can be taken as a new system of orthogonal coordinates. Even we can
draw a new picture with the new diameters on the Cartesian axis.
The central equation of the hyperbola using the rotated axis is:

OP = ± (OQ' cosh(ψ ’ • ) + OS' sinh(ψ ’ • ))

which shows that a hyperbolic rotation of the coordinate axis has been made with respect
to the principal diameter of the hyperbola.


The law of sines and cosines

Since the modulus of the area is identical on the Euclidean and hyperbolic
planes, a parallelogram is divided by its diagonal in two triangles of equal area. This
statement is somewhat subtle since the Euclidean congruence of triangles is not valid in
the hyperbolic plane. I shall return to this question later. Now we only need to know
that the area of a hyperbolic triangle is the half of the outer product of any two sides.
Following the perimeter of a triangle, let a, b, and c be its sides respectively
opposite to the angles ±, β and γ. Then the angles formed by the oriented sides are
supplementary of the angles of the triangle and:

’ a b sinh γ = ’ b c sinh ± = ’ c a sinh β
a §b=b§c=c§a ’

a b c
= =
sinh ± sinh β sinh γ

which is the law of sines.
From a + b + c = 0 , we have:
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 165



a 2 = (’ b ’ c ) = b 2 + c 2 + 2 b · c a 2 = b 2 + c 2 ’ 2 b c cosh ±
2


which is the law of cosines. And also:

b 2 = a 2 + c 2 ’ 2 a c cosh β

Figure 13.8
c 2 = a 2 + b 2 ’ 2 a b cosh γ

When applying both theorems, we must
take care with the sides having
imaginary length and the signs of the
angles and trigonometric functions.
As an application of the law of
sines and cosines, consider the
hyperbolic triangle with vertices A=(5,
3), B=(1, 0), C=(10, 1), whose sides
belong to the real sector (figure 13.8):

(’ 4)2 ’ (’ 3)2
AB = =7
AB = B ’ A = ’4 e 2 ’ 3 e 21


BC = C ’ B = 9 e 2 + e 21 BC = 9 2 ’ 12 = 80


(’ 5)2 ’ 2 2
CA = A ’ C = ’5 e 2 + 2 e 21 CA = = 21

26
cosh ± = ’
BC 2 = CA 2 + AB 2 ’ 2 CA AB cosh ±
73

33
cosh β =
CA 2 = AB 2 + BC 2 ’ 2 AB BC cosh β
4 35

47
cosh γ =
AB 2 = BC 2 + CA 2 ’ 2 BC CA cosh γ
4 105

From where it follows that:

± = ’1.3966... ’ π e12 β = 0.8614... γ = 0.5352...

I have obtained their signs from the definition of the angles ± = BAC, β = CBA, γ = ACB
and the geometric plot (figure 13.8). Note that ± + β + γ = ’π e12 and they fulfil the law
of sines:

BC CA AB
= =
sinh ± sinh β sinh γ
166 RAMON GONZALEZ CALVET


because sinh± = sinh(’1.3966+ πe12) = ’sinh(’1.3966) = sinh 1.3966.
Consider now another triangle A = (2, 4), B = (1,0) and C = (6, 1), having sides
on real and imaginary sectors (figure 13.9):

AB = B ’ A = ’ e 2 ’ 4 e 21
Figura 13.9
(’ 1) ’ (’ 4 ) = 15 e12
2 2
AB =

BC = C ’ B = 5 e 2 + e 21

BC = 5 2 ’ 12 = 24

CA = A ’ C = ’4 e 2 + 3 e 21

(’ 4 )2 ’32
CA = =7

16 e12
cosh ± =
BC 2 = CA 2 + AB 2 ’ 2 CA AB cosh ±
105

’ e12
cosh β =
CA 2 = AB 2 + BC 2 ’ 2 AB BC cosh β
6 10

23
cosh γ =
AB 2 = BC 2 + CA 2 ’ 2 BC CA cosh γ
2 42

From the last hyperbolic cosine, which is real, we find the hyperbolic sine for γ, which
is a positive angle as shown by the plot:

19
sinh γ =
2 42

which we may use in the law of sines:

19 e12
15 e12 19 e12
24 7
sinh ± = ’ sinh β = ’
= = ’
sinh ± sinh β 19 / 2 42 105 6 10

Recalling that for ψ real:

« π « π
sinh¬ψ ± e12 · ≡ ± e12 coshψ cosh¬ψ ± e12 · ≡ ± e12 sinhψ
2 2
 

the angles ±, β and γ follow:
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 167


π π
± = ’1.2284... ’ β = 0.0527... ’ γ = 1.1757
e12 e12
2 2

Observe that the addition of the three angles is ’π e12, as expected. According to the
definition ± = BAC, β = CBA, γ = ACB, let you see the consistence with the geometric
plot. The angle γ has positive sign as shown by the figure 13.9. The bisector of ±
parallel to the quadrant bisector divides it in two angles, one real and the other complex
(with imaginary part π e12 /2). The algebraic addition of both angles is ±.. Taking into
account that the second has opposite orientation and must be subtracted and
predominates over the first, the negative value of ± is explained.


Hyperbolic similarity

Two triangles ABC and A'B'C' are said to be directly similar4 and their vertices
and sides denoted with the same letters are homologous if:

AB BC ’1 = A'B' B'C' ’1 AB’1 A'B' = BC ’1 B'C'


One can prove easily that the third quotient of homologous sides also coincides
with the other quotients:

AB’1 A'B' = BC ’1 B'C' = CA’1 C'A' = r

The similarity ratio r is defined as the quotient of every pair of homologous
sides, which is a hyperbolic number. The modulus of the similarity ratio is the size ratio
and the argument is the angle of rotation of the triangle A'B'C' with respect to the
triangle ABC.

A' B'
exp[± ( AB, A' B ' ) e1 ]
r=
AB

The definition of similarity is generalised to any pair of polygons in the
following way. The polygons ABC...Z and A'B'C'...Z' are said to be directly similar with
similarity ratio r and the sides denoted with the same letters to be homologous if:

r = AB ’1 A'B' = BC ’1 B'C' = CD ’1 C'D' = ... = YZ ’1 Y'Z' = ZA ’1 Z'A'

Here also, the modulus of r is the size ratio of both polygons and the argument is
the angle of rotation. The fact that the homologous exterior and interior angles are equal
for directly similar polygons is trivial because:
’1
= BA BC ’1 ’
B'A' B'C' angle A'B'C' = angle ABC

C'B' C'D' ’1 = CB CD ’1 ’ angle B'C'D' = angle BCD etc.

4
A direct similarity is also called a similitude and an opposite similarity sometimes an
antisimilitude.
168 RAMON GONZALEZ CALVET



The direct similarity is an equivalence relation since it has the reflexive,
symmetric and transitive properties. This means that there are classes of equivalence
with directly similar figures.
A similitude with ¦r¦=1 is a displacement, since both polygons have the same
size and orientation.
Two triangles ABC and A'B'C' are oppositely similar and the sides denoted with
the same letters are homologous if:

AB BC ’1 = ( A'B' B'C' ’1 )* = B'C' ’1 A'B'

where the asterisk denotes the hyperbolic conjugate. The former equality cannot be
arranged into quotients of pairs of homologous sides as done before. Because of this,
the similarity ratio cannot be defined for the opposite similarity but only the size ratio,
which is the quotient of the lengths of any two homologous sides. An opposite
similarity is always the composition of a reflection in any line and a direct similarity.

AB BC ’1 = v ’1 A'B' B'C' ’1 v ” BC ’1 v ’1 B'C' = AB ’1 v ’1 A'B' ”

BC ’1 ( v ’1 B'C' v ) = AB ’1 ( v ’1 A'B' ’1 v ) = r

where r is the ratio of a direct similarity whose argument is not defined but depends on
the direction vector v of the reflection axis. Notwithstanding, this expression allows to
define the opposite similarity of two polygons. So two polygons ABC...Z and A'B'C'...Z'
are oppositely similar and the sides denoted with the same letters are homologous if for
any hyperbolic vector v the following equalities are fulfilled:

AB ’1 ( v ’1 A'B' ’1 ’1
( v ’1 B'C' v ) = .... = ZA ’1 ( v ’1 Z'A' v )
v ) = BC

that is, if after a reflection one polygon is directly similar to the other. The opposite
similarity is not reflexive nor transitive and there are not classes of oppositely similar
figures.
An opposite similarity with ¦r¦=1 is called a reversal, since both polygons have
the same size but opposite orientations.
Obviously the plot of similar figures on the hyperbolic plane breaks our
Euclidean intuition about figures with the same form. The algebraic definition is very
rigorous and clear, but we must change our visual illusions. So I recommend the
exercise 13.6.
This chapter is necessarily unfinished because the geometry of the hyperbolic
plane can be developed and studied with the same extension and profundity as for the
Euclidean plane. Then, it is obvious that many theorems should follow in the same way
as Euclidean geometry. As a last example I will define the power of a point with respect
to a hyperbola.


Power of a point with respect to a hyperbola with constant radius

The locus of the points placed at a fixed distance r from a given point O is an
equilateral hyperbola centred at this point with equation:
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 169


( x ’ x O )2 ’ ( y ’ y O )2 = r 2
OP = r

The power of a point with respect to this Figure 13.10
hyperbola is the product of both oriented
distances from P to the intersections R and R'
of a line passing through P with the hyperbola.
The power of a point is constant for every line
of the pencil of lines of P. In the proof the
inscribed angle theorem is used (figure 13.10):
the angles S'RR' and S'SR' are equal so the
triangles SPR' and RPS' are oppositely similar.
Then:

PR' PS ’1 = PR ’1 PS' ’ PR PR' = PS' PS

Developing the product of distances on the line
passing through the centre of the hyperbola we find:

PS PS' = (PO + OS ) (PO + OS' ) = PO 2 + OS OS' = ( x P ’ x O ) ’ ( y P ’ y O ) ’ r 2
2 2




that is, the power of a point is obtained by substitution of the coordinates of P on the
hyperbola equation.


Exercises

13.1 Let A = (2, 2), B = (1, 0) and C = (5, 3) be the vertices of a hyperbolic triangle.
Calculate all the sides and angles and also the area.

13.2 Turn the vector 2 e2 + e21 through an angle ψ = log 2. Do a reflection in the
direction 3 e2 ’ e21 . Make also an inversion with radius 3.

13.3 Find the direction and normal vectors of the line y = 2 x + 1 and calculate the
perpendicular line passing through the point (3, 1).

13.4 Calculate the power of the point P = (’7, 3) with respect to the hyperbola x2 ’ y2 =
16 using the intersections of the lines y = 3, y = ’3x ’18 and that passing through the
centre of the hyperbola y = ’3x / 7. See that in all cases the power of P is equal to the
value found by substitution in the Cartesian equation.

13.5 In this chapter I have deduced the law of sines and cosines. Therefore a law of
tangents should be expected. Find and prove it.

13.6 Check that the triangle A = (0, 0), B = (5, 0) and C = (5, 3) is directly similar to the
triangle A' = (0, 0), B' = (25, ’15), C' = (16, 0). Find the similarity ratio and the rotation
and dilation of the corresponding homothety. Draw the triangles and you will astonish.
170 RAMON GONZALEZ CALVET


FOURTH PART: PLANE PROJECTIONS OF TRIDIMENSIONAL SPACES

The complete study of the geometric algebra of the tridimensional spaces falls
out the scope of this book. However, due to the importance of the Earth charts and of
the Lobachevsky™s geometry, the first one being more practical and the second one
more theoretical, I have written this last section. In order to make the explanations
clearer, the tridimensional geometric algebra has been reduced to the minimal concepts,
enhancing the plane projections.
The geometric quality of being Euclidean or pseudo-Euclidean is not the
signature + or ’ of a coordinate, but the fact that two coordinates have the same or
different signature, in other words, it is a characteristic of a plane. For instance, a plane
with signatures + + is equivalent, from a geometric point of view, to another with ’ ’ .
Therefore, only two kinds of three-dimensional spaces exist: the room space where all
the planes are Euclidean (signatures + + + or ’ ’ ’), and the pseudo-Euclidean space,
which has one Euclidean plane and two orthogonal pseudo-Euclidean planes (signatures
+ ’ ’ or + + ’ ).


14. SPHERICAL GEOMETRY IN THE EUCLIDEAN SPACE

The geometric algebra of the Euclidean space

A vector of the Euclidean space is an oriented segment in this space with
direction and sense, although it can represent other physical magnitudes such as forces,
velocities, etc. The set of all the segments (geometric vectors) together with their
addition (parallelogram rule) and the product by real numbers (dilation of vectors) has a
structure of vector space, symbolised with V3. Every vector in V3 is of the form:

v = v1 e1 + v 2 e 2 + v 3 e3

where ei are three unitary perpendicular vectors, which form the base of the space. If we
define an associative product (geometric or Clifford product) being a generalisation of
that defined for the plane in the first chapter of this book, we will arrive to:

ei2 = 1 ei e j = ’e j ei for i ≠ j
and

In general, the square of a vector is the square of its modulus and perpendicular vectors
anticommute whereas proportional vectors commute.
From the base vectors one deduces that the geometric algebra generated by the
space V3 has eight components:

Cl (V3 ) = Cl 3, 0 = 1, e1 , e 2 , e3 , e 23 , e31 , e12 , e123

Let us see with more detail the product of two vectors:

v w = (v1 e1 + v 2 e 2 + v 3 e 3 ) (w1 e1 + w2 e 2 + w3 e3 ) = v1 w1 + v 2 w2 + v 3 w3

+ (v 2 w3 ’ v 3 w2 ) e 23 + (v 3 w1 ’ v1 w3 ) e31 + (v1 w2 ’ v 2 w1 ) e12
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 171



The product (or quotient) of two vectors is said a quaternion1. The quaternions are the
even subalgebra of Cl3, 0 that generalise the complex numbers to the space. Splitting a
quaternion in the real and bivector parts, we obtain the inner (or scalar) product and the
outer (or exterior) product respectively:

v · w = v1 w1 + v 2 w2 + v 3 w3

v § w = (v 2 w3 ’ v 3 w2 ) e 23 + (v 3 w1 ’ v1 w3 ) e31 + (v1 w2 ’ v 2 w1 ) e12

The bivectors are oriented plane surfaces and indicate the direction of planes in the
space. Who be acquainted with the vector analysis will say that both vectors and
bivectors are the same thing. This confusion was originated by Hamilton2 himself, and
continued by the founders of vector analysis, Gibbs and Heaviside. However, vectors
and bivectors are different things just as physicists have experienced and know long
time ago. The proper vectors are called usually “polar vectors” while the pseudo-vectors
that actually are bivectors are usually called “axial vectors”. The following magnitudes
are vectors: of course a geometric segment, but also a velocity, an electric field, the
momentum, etc. On the other hand, the oriented area is, of course, a bivector, but also
the angular momentum, the angular velocity and the magnetic field. As a criterion to
distinguish both kind of magnitudes one uses the reversal of coordinates, which changes
the sense of vectors while leaves bivectors invariant.
The product of two bivectors yields a real number plus a bivector. Both parts can
be separated as the symmetric and antisymmetric product. The symmetric product is a
real number and its negative value will be denoted here with the symbol • while the
antisymmetric product is also a bivector and will be noted here with the symbol — :

1
(v w + w v ) = v 23 w23 + v 31 w31 + v12 w12
v•w=’
2

1
(v w ’ w v )
v—w=’
2

= (v 31 w12 ’ v12 w31 ) e 23 + (v12 w23 ’ v 23 w12 ) e31 + (v 23 w31 ’ v 31 w23 ) e12

v w = ’v • w ’ v — w

Let us see what happens with the outer product of three vectors. According to
the extension theory of Grassmann, the product u § v § w is the oriented volume
generated by the surface represented by the bivector u § v when it is translated
parallelly along the segment w:

Hamilton discovered the quaternions in October 16th 1843 and defined them as quotients of
1

two vectors. From this definition he deduced the properties of the product of quaternions. I
recommend you the reading of the initial chapters of the Elements of Quaternions because of its
pedagogic importance.
2
This confusion is due to the fact that vectors and bivectors are dual spaces of the algebra Cl3, 0.
However, this duality does not exist at higher dimensions, although there is also duality among
other spaces.
172 RAMON GONZALEZ CALVET



ux vx wx
u § v § w = uy vy w y e123
uz vz wz

Finally, let us see how is the product of three vectors u, v and w. The vector v
can be resolved into a component coplanar with u and v and another component
perpendicular to the plane u-v:

u v w = u v || w + u v ⊥ w

Now let us analyse the permutative property. In the plane we found u v w ’ w v u = 0. In
the space the permutative property becomes3:

u v w ’ w v u = u v ⊥ w ’ w v ⊥ u = v ⊥ (’ u w + w u )

= ’2 v ⊥ u § w = ’ 2 v § u § w = 2 u § v § w

I take the same algebraic hierarchies as in chapter 1: all the abridged products
must be operated before the geometric product, convention which is coherent with the
fact that in many algebraic situations, the abridged products must be developed in sums
of geometric products.


Spherical trigonometry

In this section the relations for the sides and angles of the spherical triangles are
deduced. I will take for convenience the sphere having unity radius, although the
trigonometric identities are equally valid for a sphere of any radius.
Let us consider any three points A,
B and C on the sphere with unity radius
centred at the origin (figure 14.1). Then
A = B = C = 1 . The angles formed by
each pair of sides will be denoted by ±, β
and γ, and the sides respectively opposite
to these angles will be symbolised by a, b
and c respectively. Then a is the arc of the
great circle passing through the points B
and C, that is, the angle between these
vectors:
Figure 14.1
sin a = B § C


3
From this result it follows that u § v § w = (u v w ’ u w v + v w u ’ v u w + w u v ’ w v u) / 6 .
In other words, the outer product is a fully antisymmetric product. However although being
beautiful, it is not so useful for geometric algebra as the permutative property u § v § w = (u v w
’ w v u)/2.
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 173



Note that in this equality the sine is positive and therefore a¤π. So, the spherical
trigonometry is deduced for strict triangles, those having a, b, c¤π.
Also, ± is the angle between the sides b and c of the triangle, that is, the angle
between the planes passing through the origin, A and B, and the origin, A and C. Since
the direction of a plane is given by its bivector, which can be obtained through the outer
product, we have:

(A § B) — (A § C )
sin ± =
A§B A§C

Now we write the products of the numerator using the geometric product:

’ ( A B ’ B A) ( A C ’ C A) + ( A C ’ C A)( A B ’ B A)
sin ± =
8 A§B A§C

’ A B A C + A B C A + B A2 C ’ B A C A + A C A B ’ A C B A ’ C A2 B + C A B A
sin ± =
8 A§B A§C

We extract the vector A as common factor at the left, but without writing it because
A = 1:

’ B A C + B C A + A B C ’ A ’1 B A C A + C A B ’ C B A ’ A C B + A ’1 C A B A
sin ± =
8 A§ B A§C

Applying the permutative property to the suitable pairs of products, we have:

6 A § B § C + 2 A ’1 A § B § C A A§ B §C
sin ± = =
8 A§ B A§C A§ B A§C

since the volume A § B § C is a pseudoscalar, which commutes with all the elements of
the algebra. Now the law of sines for spherical triangles follows:

A§ B §C
sin ± sin β sin γ
= = = (I)
A§ B B §C C § A
sin a sin b sin c

Let us see the law of cosines. Since e 23 = e31 = e12 = ’1 then:
2 2 2




(A § B) • (A § C ) = (A § B) • (A § C )
cos ± =
A§ B A§C sin c sin b

1
[( A B ’ B A) ( A C ’ C A) + ( A C ’ C A)( A B ’ B A)]
sin b sin c cos ± = ’
8
174 RAMON GONZALEZ CALVET


( A B A C ’ A B C A ’ B A 2 C + B A C A + A C A B ’ A C B A ’ C A 2 B + C A B A)
1
=’
8

Now taking into account that:

4 A · B A · C = ( A B + B A) ( A C + C A) = A B A C + A B C A + B A 2 C + B A C A

But also:

4 A · B A · C = ( A C + C A) ( A B + B A) = A C A B + A C B A + C A 2 B + C A B A

and adding the needed terms, we find:

(8 A · B A · C ’ 2 A B C A ’ 2 B A 2 C ’ 2 A C B A ’ 2 C A 2 B )
1
sin b sin c cos ± = ’
8

Extracting common factors and using A2 = B2 =1, we may write:

1
(8 A · B A · C ’ 2 A (B C + C B ) A ’ 2 (B C + C B )) = ’ A · B A · C + B · C
sin b sin c cos ± = ’
8

sin b sin c cos ± = ’ cos c cos b + cos a

cos a = cos b cos c + sin b sin c cos ± (II)

which is the law of cosines for sides. The substitution of cos c by means of the law of
cosines gives:

cos a = cos b (cos a cos b + sin a sin b cos γ ) + sin b sin c cos ±

cos a (1 ’ cos 2 b ) = cos b sin a sin b cos γ + sin b sin c cos ±

and the simplification of sin b:

cos a sin b = cos b sin a cos γ + sin c cos ±

The substitution of sin c = sin a sin γ / sin ± yields:

sin a sin γ cos ±
cos a sin b = cos b sin a cos γ +
sin ±

Dividing by sin a :

cot a sin b = cos b cos γ + sin γ cot ± (III)
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 175


The dual spherical triangle

The perpendicular to the plane containing each side of a spherical triangle cuts
the spherical surface in a point. The three points A', B' and C' obtained in this geometric
way form the dual triangle. The algebraic way to calculate them is the duality operation,
which maps bivectors into the perpendicular vectors. The dual of any element is
obtained as the product by the pseudoscalar unity ’e123, which commutes with all the
elements of the algebra:

B §C C§A A§B
A' = ’e123 B' = ’ e123 C' = ’ e123
B §C C§A A§B

The inner product of two vectors yields:

(C § A) • ( A § B ) = ’ ( A § B ) • ( A § C ) cos a' = ’ cos ±
B' · C' = ”
C § A A§ B A§B A§C

which shows that the angle a' and ± are supplementary, and so also the other angles:

a' = π ’ ± b' = π ’ β c' = π ’ γ

It is trivial that the dual of the dual triangle is the first triangle, and hence:

±' = π ’ a β' = π’b γ' = π ’ c

We may apply the laws of sines and cosines to the dual triangle. The law of sines
is self-dual and may be written in a more symmetric form:

A§ B §C
sin ± sin β sin γ
= = = (I)
A' § B' § C'
sin a sin b sin c

On the other hand, the law of cosines yields a new result when applying duality:

’ cos ±' = cos γ' cos β' ’ sin β' sin γ' cos a'

Removing the marks, since this law must be valid for any spherical triangle, we find the
law of cosines for angles:

cos ± = ’ cos γ cos β + sin β sin γ cos a (IV)

Also, applying the equality (III) to the dual triangle we find:

cot ± sin β = ’ cos β cos c + sin c cot a (V)

The five Bessel™s equalities (I to V) allow to solve every spherical triangle knowing any
three of its six elements.
176 RAMON GONZALEZ CALVET


Right spherical triangles and Napier™s rule

For the case of a right angle spherical triangle, the five Bessel™s formulas are
reduced to a simpler form, and then they may be remembered with the help of the
Napier™s pentagon rule (figure 14.2). Draw the angles and the sides of the triangle
following the order of the
perimeter removing the right
angle and writing instead of
the legs (the sides adjacent to
the right angle) the
complementary arcs. Then
follow the Pentagon rules:
1) The cosine of every
element is equal to the
product of the cotangents
Figure 14.2
of the adjacent elements.
2) The cosine of every
element is equal to the product of the sines of the nonadjacent elements.
For example:

«π  «π 
cos a = sin¬ ’ b · sin¬ ’ c · = cos b cos c
2  2 

«π «π
 
cos¬ ’ b · = cot γ cot¬ ’ c · sin b = cot γ tg c

also:
2 2
 

This rule is applied to the right side triangles in the same way: remove the right side and
write the complementary of the adjacent angles.


Area of a spherical triangle

A lune is a two-sided polygon on the sphere defined by two great circles. The
area of a lune is proportional to the angle ±
between both great circles. For an angle π/2
its area is π, therefore the area of a lune with
angle ± is 2±. Now let us consider in the
sphere shown by figure 14.3 the three lunes
having the angles ±, β and γ: Then:

± s + t = 2±

 s + u' = 2 β
 s + v = 2γ Figure 14.3


where u' is the area of the antipodal triangle
of u. Both triangles u and u' have the same angles and area and the system can be
rewritten:
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 177


±s + t = 2±

s + u = 2 β
s + v = 2γ


Adding the three equations, we find:

3 s + t + u + v = 2 (± + β + γ )

The four triangles s, t, u and v fill an hemisphere:

s+t+u+v=2π

so the area of the triangle s is the spherical excess, that is, the addition of the three
angles minus π:

s =± + β +γ ’ π


Properties of the projections of the spherical surface

No chart of the spherical surface preserving the scale of distances everywhere
exists, that is, we cannot depict any map with distances proportional to those measured
on the sphere. The distortion is the variation of the scale and the angles. Usually there is
a line of zero distortion, where the scale is constant.
Since the scale of distances is never preserved for all the points, projections with
other interesting properties have been searched in cartography. A projection is said to be
conformal if it preserves in the map the angles between great circles on the sphere. A
projection is said to be equivalent if it preserves the area, that is if the figures on the
sphere are projected into figures on the map having the same area. A projection is said
to be equidistant if the scale of distances is preserved, not everywhere but on the line
perpendicular to the line of zero distortion, or radially outwards from a point of zero
distortion.
The more general concept of projection is any one-to-one mapping of any point
(x, y, z) on the sphere into a point (u, v) on the plane, although the main types of
projections are azimuthal, cylindrical and conic. An azimuthal projection is a standard
projection into a plane, which may be considered touching the sphere. In the tangency
point there is zero distortion and the bearings or azimuths from this point are correctly
shown. A cylindrical projection is a projection into a cylindrical surface around the
sphere that will be unrolled. A conic projection is a projection into a conical surface
tangent to the sphere that will also be unrolled.
Now, I review the main and more used projections beginning with the azimuthal
projections.


The central or gnomonic projection

Let us consider the sphere with unity radius (figure 14.4) centred at the origin.
Any point on the sphere has the coordinates (x, y, z) fulfilling:
178 RAMON GONZALEZ CALVET


x2 + y2 + z2 = 1

Every point on the upper
hemisphere is projected into
another point on the plane z = 1
using as centre of projection the
centre of the sphere. Let u and v
be the Cartesian coordinates on
the projection plane. Taking
similar triangles the following
relations are found:

x y
u= v=
z z Figure 14.4

from where we obtain:

u v 1
x= y= z=
u2 + v2 + 1 u2 + v2 + 1 u2 + v2 +1

The differential of the arc length is obtained through the differentiation of the above
relationships:

ds = dx e1 + dy e 2 + dz e3

( 1 + v )du ’ 2 u v du dv + ( 1 + u 2 ) dv 2
2 2
ds = dx + dy + dz =
2 2 2 2

(1 + u + v2 )
2
2




The geodesics of the sphere are the great circles, which are the intersections with planes
passing through its centre. These planes cut the projection plane in straight lines, which
are the projections of the geodesics. In other words, any great circle is projected into a
line on the projection plane. Taking as equation of the line:

v=ku+l

with k and l constant, the substitution in ds gives:

1+ k2 + l2
ds = 2 du
u (1 + k 2 ) + 2 k l u + 1 + l 2

By integration we arrive to the following primitive:

u (1 + k 2 ) + k l
s = arctg + const
1+ k + l
2 2
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 179


The arc length between two points on this great circle is the difference of this
primitive between both points. However it is more advantageous to write it using
cosines instead of tangents by means of the trigonometric identity:

1 + tg s A tg s B
cos(s B ’ s A ) ≡
1 + tg 2 s A 1 + tg 2 s B

After removing k and l using the equation of the line, we arrive to:

1 + u A uB + v A vB
cos(s B ’ s A ) =
2 2 2 2
1 + uA + vA 1 + uB + vB

(u A e1 + v A e 2 + e3 ) · (u B e1 + v B e2 + e3 )
=
u A e1 + v A e 2 + e3 u B e1 + v B e 2 + e3

a trivial result because the arc length is the angle between the position vectors of both
points, and also of the proportional vectors going to the projection plane. However, the
interest of this result is its analogy with the result found for the hyperboloidal surface.
From this value of the cosine, we may obtain the sine of the arc:

(u A v B ’ u B v A )2 + (u A ’ v A )2 + (u B ’ v B )2
sin(s B ’ s A ) = 1 ’ cos (s B ’ s A ) =
2
2 2 2 2
1 + uA + vA 1 + uB + vB

(u A e1 + v A e2 + e3 ) § (u B e1 + v B e 2 + e3 )
=
u A e1 + v A e 2 + e3 u B e1 + v B e 2 + e3

which is also a trivial result, since the sine of the angle is proportional to the modulus of
the outer product.
Let us see the area function. The differential of the area is easily obtained taking
into account that it is a bivector and using the outer product of the differentials of the
coordinates:

du § dv
(dx § dy )2 + (dy § dz )2 + (dz § dx )2
dA = =
(1 + u + v2 )
3/ 2
2



This result shows that the central projection is not equivalent and the distortion
increases with the distance to the origin.
Let us consider a plane passing through the centre of the sphere, which cuts its
surface in the great circle determined by the equation system:

± x2 + y2 + z2 =1

 a x + b y + z = 0 a , b real

Then the angle between two great circles is the angle between both central planes:
180 RAMON GONZALEZ CALVET



(a e 23 + b e31 + e12 ) • (a' e23 + b' e31 + e12 ) a a' + b b' + 1
cos ± = =
a e 23 + b e31 + e12 a' e 23 + b' e31 + e12 a 2 + b 2 + 1 a' 2 + b' 2 + 1

The sides of a spherical triangle are great circles; therefore the central projection
of a spherical triangle is a triangle whose sides are straight lines.


Stereographic projection

In the stereographic projection the point of view is placed on the spherical
surface. As before, every point (x, y, z) on the sphere with unity radius centred at the
origin fulfils the equation:

x2 + y2 + z2 =1

We project the spherical surface into
the plane z = 0 locating the centre of
projection at the pole (0, 0, ’1) (figure
14.5). The upper hemisphere is
projected inside the circle of unity
radius while the lowest hemisphere is
projected outside. If u and v are the
Cartesian coordinates on the projection
plane, we have by similar triangles:
Figure 14.5
x y
= z +1 = z +1
v
u

Using the equation of the sphere one arrives to:

2u 2v 2
x= y= z= ’1
1+ u2 + v2 1+ u2 + v2 1 + u2 + v2

from where the differential of the arc length is obtained:

4 ( du 2 + dv 2 )
ds = dx + dy + dz =
2 2 2 2

(1 + u + v2 )
2
2



Now we see that this projection is not equidistant and the distortion increases with the
distance to the origin. The factor 4 indicates that the lengths at the origin of coordinates
on the plane are the half of the lengths on the sphere. Taking instead of the plane z = 0,
the plane z = 1 this factor becomes 1. Then, we can state correctly the scale of the chart
(for example, a polar chart).
The geodesic lines (great circles) are intersection of the sphere surface with
planes passing through the origin, which have the equation:

z=a x+b y
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 181



The substitution by the stereographic coordinates yields:
Figure 14.6
( u + a )2 + ( v + b )2 = a 2 + b 2 + 1
which is the equation of a circle centred at (’a,
’b) with radius r = a 2 + b 2 + 1 . Observe that
this radius is the hypotenuse of a right angle
triangle having as legs the distance to the
origin and the unity. That is, the great circles
on the sphere are shown in the stereographic
projection as circles that intersect the circle x2 +
y2 =1 in extremes of diameters (figure 14.6).
Usually only the projection of the upper
hemisphere is used, so that the great circles are represented as circle arcs inside the
circle x 2 + y 2 ¤ 1 . The angle ± between two of these circles, as explained in the page
89, is obtained from their radii r and r' and centres O and O' through:

r 2 + r' 2 ’ (O ’ O' )
2
a a' + b b' + 1
cos ± = =
2 r r' a 2 + b 2 + 1 a' 2 + b' 2 + 1

Just this is the angle between the planes a x + b y “ z = 0 and a' x + b' y “ z = 0, that is,
the angle between the two great circles represented by the projected circles. Therefore,
the stereographic projection is a conformal projection of the spherical surface.
If we calculate the differential of area we find:

4 du § dv
dA =
(1 + u + v2 )
2
2



Now we see that this projection is not equivalent since distortion of areas increases with
the distance from the origin (as commented above, the factor 4 becomes 1 projecting
into the plane z =1).


Orthographic projection

The orthographic projection of the sphere is a parallel projection (the point of
view is placed at the infinity). If we make a projection parallel to the z-axis upon the
plane z = 0, the Cartesian coordinates on the map are identical to x, y and we have:

x dx + y dy
z = 1 ’ x2 ’ y2 dz = ’
1 ’ x2 ’ y2

dx 2 + dy 2 + 2 x y dx dy
ds = dx + dy + dz =
2 2 2 2

1’ x2 ’ y2
182 RAMON GONZALEZ CALVET


1 + x2 + y2
dA = dx § dy
1’ x2 ’ y2

Introducing the distance r to the origin of coordinates and the angle • with respect to the
x-axis, we have:

y = r sin •
x = r cos •

dr 2 + r 2 d• 2 + 2 r dr d•
ds =
2

1’ r2

An example of orthographic projection is the Earth image that appears in many TV
news.


Spherical coordinates and cylindrical equidistant (Plate Carr©e) projection

For a sphere with unity radius, the spherical coordinates4 are related with the
Cartesian coordinates by means of:

x = sin θ cos •

y = sin θ sin •

z = cos θ

Then the differential of arc length is:

ds = (cos θ cos • dθ ’ sinθ sin• d• ) e1 + (cos θ sin • dθ + sinθ cos• d• ) e 2 ’ sinθ dθ e3

Introducing the unitary vectors eθ and e•
as:

eθ = cosθ cos • e1 + cosθ sin • e2 ’ sinθ e3

e• = ’sin • e1 + cos • e 2

the differential of arc length in spherical
coordinates becomes:

ds = dθ eθ + sin θ d• e•

ds 2 = dθ 2 + sin 2θ d• 2 Figure 14.7



For geographical coordinates, θ is the colatitude and • the longitude.
4
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 183


Note that eθ and e• are orthogonal vectors, since their inner product is zero. At θ = 0,
ds depends only on dθ and there is a pole. Then θ is the arc length from the pole to the
given point (figure 14.7), while • is the arc length over the equator (θ = π/2, sinθ = 1).
The meridians (• = constant) are geodesics, but the parallels (θ = constant) are not.
Exceptionally, the equator is also a geodesic.
In the Plate Carr©e projection u = • and v = π/ 2 ’ θ (latitude) so that the
meridians and parallels are shown in a squared graticule. This projection is equidistant
for any meridian whereas the distortion in the parallels increases, as well as for every
cylindrical projection, as we separate from the equator. Let us see the area:

dA = sin θ dθ § d• = cos v du § dv

The ratio of the real area with respect to the represented area on the chart is equal to the
cosine of the latitude and therefore the projection is not equivalent.


Mercator's projection

The Mercator's projection5 is defined as the cylindrical conformal projection. If
we wish to preserve the angles between curves, we must enlarge the meridians by the
same amount as the parallels are enlarged in a cylindrical projection, that is, by the
factor 1/sin θ (the secant of the geographical latitude):

’ dθ θ
dv = ’ v = ’ log tg
sinθ 2

The differential of arc length is:

4 exp(2 v )
ds 2 = dθ 2 + sin 2θ d• 2 = sin 2θ (du 2 + dv 2 ) = (du + dv 2 )
2

(exp(2v ) + 1) 2




2 exp( v )
ds = du 2 + dv 2
exp(2v ) + 1

where the distances are increased in an amount independent of the direction and
proportional to the inverse of the sine of θ. The differential of area is:

4 exp(2 v )
dA = sinθ dθ § d• = du § dv
(exp(2v ) + 1) 2




5
The difference between the U.T.M. (Universal Transverse Mercator) projection and the
Mercator planisphere is not geometric but geographic: in the U.T.M. the cylinder of projection
is tangent to a meridian instead of the equator. All the Earth has been divided in zones of 6º of
longitude, where the cylinder of projection is tangent to the central meridian (3º, 9º, 15º ... ).
184 RAMON GONZALEZ CALVET


Peters™ projection

The Peters™ projection is the cylindrical equivalent projection. If we wish to
preserve area, we must shorten the meridians in the same amount as the parallels are
enlarged in the cylindrical projection, what
yields:

dv = ’sin θ dθ v = cos θ



ds = dθ + sin θ d• = (1 ’ v ) du +
dv 2
2 2 2 2 2 2

1’ v2

dA = sin θ dθ § d• = du § dv

which displays clearly the equivalence of the
projection. Observe that v = cos θ means the
sphere is projected following planes
perpendicular to the cylinder of projection
(figure 14.8).

Figure 14.8
Conic projections

These projections are made into a cone surface tangent to the sphere (figure
14.9). Because the cone surface unrolled is a plane circular sector, they are often used to
display middle latitudes, while the azimuthal projections are mainly used for poles. In a
conic projection a small circle (a parallel) is shown as a circle with zero distortion. The
characteristic parameter of a conic projection is the constant of the cone n = cos θ0,
being θ0 the angle of inclination of the generatrix of the cone and also the angle from
the axis of the cone to any point of tangency with the sphere. Since the graticule of the

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