12 The Poincare´Sphere

12
The Poincare´ Sphere

12.1

INTRODUCTION

In the previous chapters we have seen that the Mueller matrix formalism and
the Jones matrix formalism enable us to treat many complex problems involving
polarized light. The use of matrices, however, only slowly made its way into physics
and optics. In fact, before the advent of quantum mechanics in 1925 matrix
algebra was rarely used. It is clear that matrix algebra greatly simplifies the treatment
of many difficult problems. In the optics of polarized light even the simplest problem
of determining the change in polarization state of a beam propagating through
several polarizing elements becomes surprisingly difficult to do without matrices.
Before the advent of matrices only direct and very tedious algebraic methods
were available. Consequently, other methods were sought to simplify these difficult
calculations.
The need for simpler ways to carry out difficult calculations began in antiquity. Around 150 BC the Greek astronomer Hipparchus was living in Alexandria,
Egypt, and working at the famous library of Alexandria. There, he compiled a
catalog of stars and also plotted the positions of these stars in terms of latitude and
longitude (in astronomy, longitude and latitude are called right ascension and
declination) on a large globe which we call the celestial sphere. In practice,
transporting a large globe for use at different locations is cumbersome.
Therefore, he devised a method for projecting a three-dimensional sphere on to
a two-dimensional plane. This type of projection is called a stereographic projection. It is still one of the most widely used projections and is particularly popular
in astronomy. It has many interesting properties, foremost of which is that the
longitudes and latitudes (right ascension and declination) continue to intersect each
other at right angles on the plane as they do on the sphere. It appears that the
stereographic projection was forgotten for many centuries and then rediscovered
during the European Renaissance when the ancient writings of classical Greece

and Rome were rediscovered. With the advent of the global exploration of
the world by the European navigators and explorers there was a need for
accurate charts, particularly charts that were mathematically correct. This

Copyright © 2003 by Marcel Dekker, Inc. All Rights Reserved.


need led not only to the rediscovery and use of the stereographic projection
but also to the invention of new types of projections, e.g., the famous Mercator
projection.
Henri Poincare´, a famous nineteenth-century French mathematician and
physicist, discovered around 1890 that the polarization ellipse could be represented on a complex plane. Further, he discovered that this plane could be
projected on to a sphere in exactly the same manner as the stereographic projection. In effect, he reversed the problem of classical antiquity, which was to project
a sphere on to a plane. The sphere that Poincare´ devised is extremely useful for
dealing with polarized light problems and, appropriately, it is called the Poincare´
sphere.
In 1892, Poincare´ introduced his sphere in his text Traite´ de Lumiere`. Before
the advent of matrices and digital computers it was extremely difficult to carry out
calculations involving polarized light. As we have seen, as soon as we go beyond the
polarization ellipse, e.g., the interaction of light with a retarder, the calculations
become difficult. Poincare´ showed that the use of his sphere enabled many of
these difficulties to be overcome. In fact, Poincare´’s sphere not only simplifies
many calculations but also provides remarkable insight into the manner in which
polarized light behaves in its interaction with polarizing elements.
While the Poincare´ sphere became reasonably well known in the optical
literature in the first half of the twentieth century, it was rarely used in the treatment
of polarized light problems. This was probably due to the considerable mathematical
effort required to understand its properties. In fact, its use outside of France appears
to have been virtually nonexistent until the 1930s. Ironically, the appreciation of
its usefulness only came after the appearance of the Jones and Mueller matrix

formalisms. The importance of the Poincare´ sphere was finally established in the
optical literature in the long review article by Ramachandran and Ramaseshan on
crystal optics in 1961.
The Poincare´ sphere is still much discussed in the literature of polarized light.
In larger part this is due to the fact that it is really surprising how simple it is to use
once it is understood. In fact, despite its introduction nearly a century ago, new
properties and applications of the Poincare´ sphere are still being published and
appearing in the optical literature. The two most interesting properties of the
Poincare´ sphere are that any point on the sphere corresponds to the three Stokes
parameters S1, S2, and S3 for elliptically polarized light, and the magnitude
of the interaction of a polarized beam with an optical polarizing element
corresponds to a rotation of the sphere; the final point describes the new set of
Stokes parameters. In view of the continued application of the Poincare´ sphere we
present a detailed discussion of it. This is followed by simple applications of the
sphere to describing the interaction of polarized light with a polarizer, retarder, and
rotator. More complicated and involved applications of the Poincare´ sphere are
listed in the references.

12.2

THEORY OF THE POINCARE´ SPHERE

Consider a Cartesian coordinate system with axes x, y, z and let the direction of
propagation of a monochromatic elliptically polarized beam of light be in the

Copyright © 2003 by Marcel Dekker, Inc. All Rights Reserved.


z direction. The equations of propagation are described by
Ex ðz, tÞ ¼ Ex exp ið!t À kzÞ

ð12-1aÞ

Ey ðz, tÞ ¼ Ey exp ið!t À kzÞ

ð12-1bÞ

where Ex and Ey are the complex amplitudes:
Ex ¼ E0x expðix Þ

ð12-2aÞ

Ey ¼ E0y expðiy Þ

ð12-2bÞ

and E0x and E0y are real quantities. We divide (12-2b) by (12-2a) and write
Ey E0y i
¼
e
Ex E0x
 
E0y
E0y
cos  þ i
sin 
¼
E0x
E0x
¼ u þ iv

ð12-3aÞ

ð12-3bÞ

where  ¼ y À x and u and v are orthogonal axes in the complex plane. On
eliminating the propagator in (12-1) and (12-2), we obtain the familiar equation of
the polarization ellipse:
E2y
Ex Ey
E2x
þ
À2
cos  ¼ sin2 
2
2
E0x E0y
E0x E0y

ð3-7aÞ

We have shown in Section 3.2 that the maximum values of Ex and Ey are E0x
and E0y, respectively. Equation (3-7a) describes an ellipse inscribed in a rectangle of
sides 2E0x and 2E0y.
This is shown in Fig. 12-1.

Figure 12-1 Parameters of the polarization ellipse having amplitude components E0x and
E0y along x and y axes, respectively. The angle  is related to E0x and E0y by tan ¼ E0y/E0x.
The major and minor axes of the ellipse are 2a and 2b, and the ellipticity is e ¼ b/a ¼ tan "; the
azimuth angle  is with respect to the x axis. (From Jerrard.)

Copyright © 2003 by Marcel Dekker, Inc. All Rights Reserved.


In general, we recall, the axes of the ellipse are not necessarily along the x and y
axes but are rotated, say, along x0 and y0 . Thus, we can write the oscillation along
x0 and y0 as
x0 ¼ a cos 

ð12-4aÞ

y0 ¼ b sin 

ð12-4bÞ

where  ¼ !t À kz. The ellipticity e, which is the ratio of the minor axis to the
major axis, is e ¼ b/a. The orientation of the ellipse is given by the azimuth angle
(0  180 ); this is the angle between the major axis and the positive x axis. From
Fig. 12-1 the angles " and  are defined by the equations:
tan " ¼

b
a

tan  ¼

E0y
E0x

90 Þ

"

ð0
ð0



90 Þ

ð12-5aÞ
ð12-5bÞ

The sense of the ellipse or the direction of rotation of the light vector depends on ; it
is designated right or left according to whether sin  is negative or positive. The sense
will be indicated by the sign of the ratio of the principal axes. Thus, tan " ¼ þb=a
or Àb/a refers to left (counterclockwise) or right (clockwise) rotation, respectively.
By using the methods presented earlier (see Section 3.4), we see that the following relations exist with respect to the parameters of the polarization ellipse, namely,
E20x þ E20y ¼ a2 þ b2

ð12-6aÞ

E20x À E20y ¼ ða2 À b2 Þ cos 2

ð12-6bÞ

E0x E0y sin  ¼ Æab

ð12-6cÞ

2E0x E0y cos  ¼ ða2 À b2 Þ sin 2

ð12-6dÞ

By adding and subtracting (12-6a) and (12-6b), we can relate E0x and E0y to a, b,
and . Thus, we find that
E20x ¼ a2 cos2  þ b2 sin2 

ð12-7aÞ

E20y ¼ a2 sin2  þ b2 cos2 

ð12-7bÞ

We see that when the polarization ellipse is not rotated, so  ¼ 0 , (12-7a) and
(12-7b) become
E0x ¼ Æa

E0y ¼ Æb

ð12-8Þ

which is to be expected, as Fig. 12-1 shows. The ellipticity is then seen to be
e¼

b E0y
¼
a E0x

when  ¼ 0 .

Copyright © 2003 by Marcel Dekker, Inc. All Rights Reserved.

ð12-9Þ


We can now obtain some interesting relations between the foregoing parameters. The first one can be obtained by dividing (12-6d) by (12-6b). Then
E0x E0y
sin 2
¼ tan 2 ¼ 2
cos 
cos 2
E0x À E20y
Substituting (12-5b) into (12-10) then yields


2 tan 
tan 2 ¼
cos 
1 À tan2 

ð12-10Þ

ð12-11Þ

The factor in parentheses is equal to tan 2. We then have
tan 2 ¼ tan 2 cos 

ð12-12Þ

The next important relationship is obtained by dividing (12-6c) by (12-6a),
whence
E0x E0y
Æab
¼ 2
sin 
2
E0x þ E20y
a þb
2

ð12-13Þ

Using both (12-5a) and (12-5b), we find that (12-13) becomes
Æ sin 2" ¼ sin 2 sin 

ð12-14Þ

Another important relation is obtained by dividing (12-6b) by (12-6a). Then
E20x À E20y a2 À b2
¼
cos 2
E20x þ E20y a2 þ b2

ð12-15Þ

Again, substituting (12-5a) and (12-5b) into (12-15), we find that
cos 2 ¼ cos 2" cos 2

ð12-16Þ

Equation (12-16) can be used to obtain still another relation. We divide (12-6d) by
(12-6a) to obtain
2E0x E0y cos  a2 À b2
¼ 2
sin 2
a þ b2
E20x þ E20y

ð12-17Þ

Next, using (12-5a) and (12-5b), we find that (12-17) can be written as
sin 2 cos  ¼ cos 2" sin 2

ð12-18Þ

Equation (12-18) can be solved for cos2" by multiplying through by sin2 so that
sin 2 sin 2 cos  ¼ cos 2" sin2 2 ¼ cos 2" À cos 2" cos2 2

ð12-19Þ

cos 2" ¼ ðcos 2" cos 2Þ cos 2 þ sin 2 sin 2 cos 

ð12-20Þ

or

We see that the term in parentheses is identical to (12-16), so (12-20) can finally be
written as
cos 2" ¼ cos 2 cos 2 þ sin 2 sin 2 cos 

Copyright © 2003 by Marcel Dekker, Inc. All Rights Reserved.

ð12-21Þ


Equation (12-21) represents the law of cosines for sides from spherical trigonometry.
Consequently, it represents our first hint or suggestion that the foregoing results
can be related to a sphere. We shall not discuss (12-21) at this time, but defer its
discussion until we have developed some further relations.
Equation (12-21) can be used to find a final relation of importance. We divide
(12-14) by (12-21):
Æ tan 2" ¼

sin 2 sin 
cos 2 cos  þ sin 2 sin 2 cos 

ð12-22Þ

Dividing the numerator and the denominator of (12-22) by sin2 cos  yields
Æ tan 2" ¼

tan 
sin 2 þ ðcos 2 cos 2Þ=ðsin 2 cos Þ

ð12-23Þ

We now observe that (12-12) can be written as
cos 2 tan 2 ¼ sin 2 cos 

ð12-24Þ

so
cos  ¼

cos 2 tan 2
sin 2

ð12-25Þ

Substituting (12-25) into the second term in the denominator of (12-23) yields the
final relation:
Æ tan 2" ¼ sin 2 tan 

ð12-26Þ

For convenience we now collect relations (12-12), (12-14), (12-16), (12-21), and
(12-26) and write them as a set of relations:
tan 2 ¼ tan 2 cos 

ð12-27aÞ

Æ sin 2" ¼ sin 2 sin 

ð12-27b)

cos 2 ¼ cos 2" cos 2

ð12-27cÞ

cos 2" ¼ cos 2v cos 2 þ sin 2 sin 2 cos 

ð12-27dÞ

Æ tan 2" ¼ sin 2 tan 

ð12-27eÞ

The equations in (12-27) have very familiar forms. Indeed, they are well-known
relations, which appear in spherical trigonometry.
Figure 12-2 shows a spherical triangle formed by three great circle arcs, AB,
BC, and CA on a sphere. At the end of this section the relations for a spherical
triangle are derived by using vector analysis. There it is shown that 10 relations exist
for a so-called right spherical triangle. For an oblique spherical triangle there exists,
analogous to plane triangles, the law of sines and the law of cosines. With respect to
the law of cosines, however, there is a law of cosines for the angles (uppercase letters)

Copyright © 2003 by Marcel Dekker, Inc. All Rights Reserved.


Figure 12-2 Spherical triangle on a sphere. The vertex angles are designated by A, B, C.
The side opposite to each angle is represented by a, b, and c, respectively.
and a law of cosines for sides (lower case letters). Of particular interest are the
following relations derived from Fig. 12-2.
cos c ¼ cos a cos b

ð12-28aÞ

sin a ¼ sin c sin A

ð12-28bÞ

tan b ¼ tan c cos A

ð12-28cÞ

cos a ¼ cos b cos c þ sin b sin c cos A

ð12-28dÞ

tan a ¼ sin b tan A

ð12-28eÞ

If we now compare (12-28a) with (12-27a), etc., we see that the equations
can be made completely compatible by constructing the right spherical triangle in
Fig. 12-3. If, for example, we equate the spherical triangles in Figs. 12-2 and 12-3,
we have
a ¼ 2" b ¼ 2

¼A

ð12-29Þ

Substituting (12-29) into, say, (12-28a) gives
cos 2 ¼ cos 2" cos 2

ð12-30Þ

which corresponds to (12-27c). In a similar manner by substituting (12-29) into the

remaining equations in (12-28), we obtain (12-27). Thus, we arrive at the very interesting result that the polarization ellipse on a plane can be transformed to a spherical
triangle on a sphere. We shall return to these equations after we have discussed some
further transformation properties of the rotated polarization ellipse in the complex
plane.
The ratio Ey/Ex in (12-3) defines the shape and orientation of the elliptical
vibration given by (3-7a). This vibration may be represented by a point m on a
plane in which the abscissa and ordinate are u and v, respectively. The diagram in
the complex plane is shown in Fig. 12-4.

Copyright © 2003 by Marcel Dekker, Inc. All Rights Reserved.


Figure 12-3

Right spherical triangle for the parameters of the polarization ellipse.

Figure 12-4 Representation of elliptically polarized light by a point m on a plane;  is the
plane difference between the components of the ellipse (From Jerrard.)
From (12-3b) we have
u¼

E0y
cos 
E0x

ð12-31aÞ

v¼

E0y

sin 
E0x

ð12-31bÞ

The point m(u, v) is described by the radius Om and the angle . The angle  is found
from (12-31) to be
tan  ¼

v
u

ð12-32aÞ

or
 ¼ tanÀ1

v
u

Copyright © 2003 by Marcel Dekker, Inc. All Rights Reserved.

ð12-32bÞ


Squaring (12-31a) and (12-31b) and adding yields
 2
E0y
2
2

u þv ¼
¼ 2
E0x

ð12-33Þ

which is the square of the distance from the origin to m. We see that we can also
write (12-33) as
  Ã
Ey Ey
E20y
u2 þ v2 ¼ ðu þ ivÞðu À ivÞ ¼
¼ Ã ¼ 2
ð12-34aÞ
Ex Ex
E0x
so
u þ iv ¼

Ey
¼
Ex

ð12-34bÞ

Thus, the radius vecor Om and the angle mOu represent the ratio Ey/Ex and the
phase difference , respectively. It is postulated that the polarization is left- or righthanded according to whether  is between 0 and  or  and 2.
We now show that (12-34a) can be expressed either in terms of the rotation
angle  or the ellipticity angle ". To do this we have from (12-33) that
 2

E0y
2
2
u þv ¼
¼ 2
ð12-35aÞ
E0x
We also have, from (12-5b)
E0y
¼ tan 
E0x

ð12-35bÞ

Squaring (12-35b) gives
E20y
¼ tan2 
E20x

ð12-35cÞ

Now,
tan 2 ¼

2 tan 
1 À tan2 

ð12-35dÞ

2 tan 

tan 2

ð12-35eÞ

so
tan2  ¼ 1 À

But, from (12-27a) we have
tan 2 ¼ tan 2 cos 

ð12-35fÞ

Substituting (12-35f) into (12-35e) gives
tan2  ¼ 1 À

2 tan 
cos 
tan 2

ð12-35gÞ

Equating (12-35g) to (12-35c) and (12-35a) we have
u2 þ v2 ¼ 1 À 2ðtan  cos = tan 2Þ
¼ 1 À 2 cot 2ðtan  cos Þ

Copyright © 2003 by Marcel Dekker, Inc. All Rights Reserved.

ð12-35hÞ

Finally, substituting (12-35b) into (12-35h) and using (12-31a), we find that
u2 þ 2 þ 2u cot 2 À 1 ¼ 0

ð12-36Þ

Thus, we have expressed u and v in terms of the rotation angle  of the
polarization ellipse. It is also possible to find a similar relation to (12-36) in terms
of the ellipticity angle " rather than . To show this we again use (12-35a), (12-35b),
and (12-35d) to form
u2 þ v2 ¼ 1 À

2 tan 
cos 2
sin 2

ð12-37aÞ

Substituting (12-27a) and (12-27b) into (12-37a) then gives
u2 þ v2 ¼ 1 Ç 2v csc 2" cos 2

ð12-37bÞ

After replacing cos 2 with its half-angle equivalent and choosing the upper sign, we
are led to
u2 þ v2 À 2v csc 2" þ 1 ¼ 0
Thus, we can describe (12-35a),
 2
E0y
u2 þ v2 ¼
¼ 2

E0x

ð12-38Þ

ð12-35aÞ

in terms of either  or ", respectively, by
u2 þ v2 þ 2u cot 2 À 1 ¼ 0

ð12-39aÞ

u2 þ v2 À 2v csc 2" þ 1 ¼ 0

ð12-39bÞ

At this point it is useful to remember that the two most important parameters
describing the polarization ellipse are the rotation angle  and the ellipticity angle ",
as shown in Fig. 12-1. Equations (12-39a) and (12-39b) describe the polarization
ellipse in terms of each of the parameters.
Equations (12-39a) and (12-39b) are recognized as the equations of a circle.
They can be rewritten in standard forms as
ðu þ cot 2Þ2 þ v2 ¼ ðcsc 2Þ2

ð12-40aÞ

u2 þ ðv À csc 2"Þ2 ¼ ðcot 2"Þ2

ð12-40bÞ

Equation (12-40a) describes, for a constant value of , a family of circles each

of radius csc 2 with centers at the point (Àcot 2, 0). Similarly (12-40b) describes, for
a constant value of ", a family of circles each of radius cot 2 and centers at the point
(0, csc 2). The circles in the two systems are orthogonal to each other. To show this
we recall that if we have a function described by a differential equation of the form
Mðx, yÞdx þ Nðx, yÞdy ¼ 0

ð12-41aÞ

then the differential equation for the orthogonal trajectory is given by
Nðx, yÞdx À Mðx, yÞdy ¼ 0

Copyright © 2003 by Marcel Dekker, Inc. All Rights Reserved.

ð12-41bÞ


We therefore consider (12-39a) and show that (12-39b) describes the orthogonal
trajectory. We first differentiate (12-39a)
u du þ v dv þ cot 2 du ¼ 0

ð12-42Þ

We eliminate the constant parameter cot2 from (12-42) by writing (12-39a) as
cot 2 ¼

1 À u2 À v 2
2u

ð12-43Þ

Substituting (12-43) into (12-42) and grouping terms, we find that
ð1 þ u2 À v2 Þ du þ 2uv dv ¼ 0

ð12-44Þ

According to (12-41a) and (12-41b), the trajectory orthogonal to (12-44) must,
therefore, be
2uv du À ð1 þ u2 À v2 Þ dv ¼ 0

ð12-45Þ

We now show that (12-39b) reduces to (12-45). We differentiate (12-39b) to obtain
u du þ v dv À csc 2" dv ¼ 0

ð12-46aÞ

Again, we eliminate the constant parameter csc 2" by solving for csc 2" in (12-39b):
csc 2" ¼

1 þ u2 þ v 2
2v

ð12-46bÞ

We now substitute (12-46b) into (12-46a), group terms, and find that
2uv du À ð1 þ u2 À v2 Þ dv ¼ 0

ð12-47Þ

Comparing (12-47) with (12-45) we see that the equations are identical so the trajectories are indeed orthogonal to each other. In Fig. 12-5 we have plotted the family

Figure 12-5 Orthogonal circles of the polarization ellipse in the uv plane.

Copyright © 2003 by Marcel Dekker, Inc. All Rights Reserved.


of circles for  ¼ 15 to 45 and for " ¼ 10 to 30 . We note that the circles intersect
at m and that at this intersection each circle has the same value of  and .
Each of the circles, (12-40a) and (12-40b), has an interesting property which we
now consider. If v ¼ 0, for example, then (12-40a) reduces to
ðu þ cot 2Þ2 ¼ ðcsc 2Þ2

ð12-48aÞ

Solving for u, we find that
u ¼ À cot 

or

u ¼ tan 

ð12-48bÞ

Referring to Fig. 12-4, these points occur at s and t and correspond to linearly
polarized light in azimuth cotÀ1u and tanÀ1u, respectively; we also note from
(12-3a) and (12-3b) that because v ¼ 0 we have  ¼ 0, so u ¼ E0y/E0x. Similarly, if
we set u ¼ 0 in (12-40a), we find that
v ¼ Æ1

ð12-48cÞ

Again, referring to (12-3b), (12-48c) corresponds to E0y =E0x ¼ 1 and  ¼ Æ/2, that
is, right- and left-circularly polarized light, respectively. These points are plotted as
P1 and P2 in Fig. 12-4. Thus, the circle describes linearly polarized light along the
u axis, circularly polarized light along the v axis, and elliptically polarized light
everywhere else in the uv plane.
From these results we can now project the point m in the complex uv plane on
to a sphere, the Poincare´ sphere. This is described in the following section.
12.2.1

Note on the Derivation of Law of Cosines and Law of
Sines in Spherical Trigonometry

In this section we have used a number of formulas that originate from spherical
trigonometry. The two most important formulas are the law of cosines and the law
of sines for spherical triangles and the formulas derived by setting one of the angles
to 90 (a right angle). We derive these formulas by recalling the following vector
identities:
A Â ðB Â CÞ ¼ ðA Á CÞB À ðA Á BÞC

ð12-N1aÞ

ðA Â BÞ Â C ¼ ðA Á CÞB À ðB Á CÞA

ð12-N1bÞ

ðA  BÞ Â ðC  DÞ ¼ ½A Á ðC  DފB À ½B Á ðC  DފA

ð12-N1cÞ

ðA Â BÞ Á ðC Â DÞ ¼ ðA Á CÞðB Á DÞ À ðA Á DÞðB Á CÞ

ð12-N1dÞ

The terms in brackets in (12-N1c) are sometimes written as
½A Á ðC  Dފ ¼ ½A, C, DŠ

ð12-N1eÞ

½B Á ðC  Dފ ¼ ½B, C, DŠ

ð12-N1fÞ

A spherical triangle is a three-sided figure drawn on the surface of a sphere as
shown in Fig. 12-N1. The sides of a spherical triangle are required to be arcs of great
circles. We recall that a great circle is obtained by intersecting the sphere with a plane
passing through its center. Two great circles always intersect at two distinct points,
and their angle of intersection is defined to be the angle between their corresponding
planes. This is equivalent to defining the angle to be equal to the plane angle between
two lines tangent to the corresponding great circles at a point of intersection.

Copyright © 2003 by Marcel Dekker, Inc. All Rights Reserved.


Figure 12-N1

Fundamental angles and arcs on a sphere.

The magnitude of a side of a spherical triangle may be measured in two ways.
Either we can take its arc length, or we can take the angle it subtends at the center of

the sphere. These two methods give the same numerical result if the radius of the
sphere is unity. We shall adopt the second of the two methods. In other words, if A,
B, and C are the vertices of a spherical triangle with opposite sides a, b, and c,
respectively, the numerical value of, say, a will be taken to be the plane angle
BOC, where O is the center of the sphere in Fig. 12-N1.
In the following derivations we assume that the sphere has a radius R ¼ 1 and
the center of the sphere is at the origin. The unit vectors extending from the center to
A, B, and C are ,
, and
, respectively; the vertices are labeled in such a way that
,
, and
are positively oriented.
We now refer to Fig. 12-N2. We introduce another set of unit vectors 0 ,
0 ,
and
0 extending from the origin and defined so that
 Â
¼
¼ sin c
0

Â
¼  ¼ sin a0
  ¼
¼ sin b
0

ð12-N2aÞ
ð12-N2bÞ

ð12-N2cÞ

In Fig. 12-N2 only 0 is shown. However, in Fig. 12-N3 all three unit vectors are
shown. The unit vectors 0 ,
0 , and
0 determine a spherical triangle A0 B0 C0 called the
polar triangle of ABC; this is shown in Fig. 12-N4. We now let the sides of the polar
triangle be a0 , b0 , and c0 , respectively. We see that B0 is a pole corresponding to the
great circle joining A and C. Also, C0 is a pole corresponding to the great circle AB. If
these great circles are extended to intersect the side B0 C0 , we see that this side is
composed of two overlapping segments B0 E and DC0 each of magnitude of 90 . Their
common overlap has a magnitude A, so we see that
a0 þ A ¼ 

ð12-N3aÞ

0

ð12-N3bÞ

0

ð12-N3cÞ

b þB¼
c þC¼

Copyright © 2003 by Marcel Dekker, Inc. All Rights Reserved.

Figure 12-N2

The construction of a spherical triangle on the surface of a sphere.

Figure 12-N3

Unit vectors within a unit sphere.

Equation (12-N3) is useful for relating the angles of a spherical triangle to the sides
of the corresponding polar triangle. We now derive the law of cosines and law of
sines for spherical trigonometry.
In the identity (12-N1d):
ðA Â BÞ Á ðC Â DÞ ¼ ðA Á CÞðB Á DÞ À ðA Á DÞðB Á CÞ

ð12-N1dÞ

we substitute  for A,
for B,  for C,
for D. Since  is a unit vector, we see that
(12-N1d) becomes
ð Â
Þ Á ð Â
Þ ¼
Á
À ð Á
Þð