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 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 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 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 cos  cos 2 E0x À E20y Substituting (12-5b) into (12-10) then yields   tan  tan 2 ¼ cos  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 sin  E0x þ E20y a þb 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 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 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   à Ey Ey E20y u2 þ v2 ¼ ðu þ ivÞðu À ivÞ ¼ ¼ à ¼ 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 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 this we have from (12-33) that  2 E0y 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 tan  À tan2  ð12-35dÞ tan  tan 2 ð12-35eÞ so tan2  ¼ À But, from (12-27a) we have tan 2 ¼ tan 2 cos  ð12-35fÞ Substituting (12-35f) into (12-35e) gives tan2  ¼ À tan  cos  tan 2 ð12-35gÞ Equating (12-35g) to (12-35c) and (12-35a) we have u2 þ v2 ¼ À 2ðtan  cos = tan 2Þ ¼ À 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 ¼ 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 À tan  cos 2 sin 2 ð12-37aÞ Substituting (12-27a) and (12-27b) into (12-37a) then gives u2 þ v2 ¼ 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 ¼ 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 ¼ ð12-39aÞ u2 þ v2 À 2v csc 2" þ ¼ 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 ð12-41aÞ then the differential equation for the orthogonal trajectory is given by Nðx, yÞdx À Mðx, yÞdy ¼ Copyright © 2003 by Marcel Dekker, Inc All Rights Reserved ð12-41bÞ ¼ sin b 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 , , and 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 ð12-N3bÞ 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  BÞ Á ðC  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 Á Þ ð12-N4Þ In Fig 12-N2 we have Á ¼ cos a,  Á ¼ cos c, and  Á ¼ cos b Hence, the righthand side of (12-N4) becomes cos a À cos b cos c Copyright © 2003 by Marcel Dekker, Inc All Rights Reserved ð12-N5Þ Figure 12-N4 The polar triangle on a sphere From (12-N2) we see that the left-hand side of (12-N4) becomes ðsin c Þ Á ðÀ sin b Þ ¼ À sin c sin bð Á Þ ð12-N6Þ Now, just as Á is equal to cosa, we see from the polar triangle in Figs 12-N3 and 12-N4 that Á ¼ cos a0 From (12-N3a) cos a0 is cos( À A), which equals Àcos A Thus, the left-hand side of (12-N4) equals sin c sin b sin A ð12-N7Þ Equating the two sides we obtain the law of cosines: cos a ¼ cos b cos c þ sin b sin c cos A ð12-N8aÞ We can, of course, imagine that Fig 12-N2 is rotated so that the roles previously played by a, b, and c, respectively, are now replaced by b, c, and a, so we can write cos b ¼ cos c cos a þ sin c sin a cos B ð12-N8bÞ cos c ¼ cos a cos b þ sin a sin b cos C ð12-N8cÞ Three other versions of the cosine law are obtained by applying the law of cosines to the polar triangle by merely changing a to a0 , b to b0 , etc., according to (12-N3), namely, cos A ¼ À cos B cos C þ sin B sin C cos a ð12-N9aÞ cos B ¼ À cos C cos A þ sin C sin A cos b ð12-N9bÞ cos C ¼ À cos A cos B þ sin A sin B cos c ð12-N9cÞ We now turn to the law of sines Here, we make use of the identity: ðA  BÞ Â ðC  DÞ ¼ ½A Á ðC  DފB À ½B Á ðC  DފA ð12-N1cÞ Replacing A by , B by [...]...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 12- 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 12- 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 12- 44Þ According to (12- 41a) and (12- 41b),... 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. .. (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 12- 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 12- 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 12- 46bÞ We now substitute (12- 46b) into (12- 46a), group terms,... 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... v2 Þ dv ¼ 0 12- 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... (12- N2) we see that the left-hand side of (12- N4) becomes ðsin c 0 Þ Á ðÀ sin b0 Þ ¼ À sin c sin bð 0 Á 0 Þ 12- N6Þ Now, just as Á is equal to cosa, we see from the polar triangle in Figs 12- N3 and 12- N4 that 0 Á 0 ¼ cos a0 From (12- N3a) cos a0 is cos( À A), which equals Àcos A Thus, the left-hand side of (12- N4) equals sin c sin b sin A 12- N7Þ Equating the two sides we obtain the law of cosines:... 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  BÞ Á ðC  DÞ ¼ ðA... , 0 , and 0 extending from the origin and defined so that   ¼ ¼ sin c 0  ¼  ¼ sin a0   ¼ ¼ sin b0 12- N2aÞ 12- N2bÞ 12- 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 ,... 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  ðB  CÞ ¼ ðA Á CÞB À ðA Á BÞC 12- N1aÞ ðA  BÞ Â C ¼ ðA Á CÞB À ðB Á CÞA 12- N1bÞ ðA  BÞ Â ðC  DÞ ¼ ½A Á ðC  DފB À ½B Á ðC  DފA 12- N1cÞ... 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 12- 48aÞ Solving for u, we find that u ¼ À cot  or u ¼ tan  12- 48bÞ Referring to Fig 12- 4, these points occur at s and t and correspond to linearly