3 DEFINITION OF ASSOCIATIVE OR DIRECT PRODUCT AND ROTATION OF VECTORS This chpter summrizes few properties of Cli ord Algebr nd describe its usefulness in e ecting vector rottions. 3.1 De nition of Associtive or Direct Product Given two vectors nd b, the resultnt is + b = c We de ne the ssocitive or direct product cc = ( + b) ( + b) = + bb + b + b For the specil cse b perpendiculr to, for which we write b?, Fig. 3.1b, this my be written cc = + b? b? +b? +b? By de nition, the direct product of vector by itself is its sclr vlue squred. Thus, cc = c 2 ; b? b? = b 2?;. From the Pythgoren theorem in Eucliden t spce, c 2 = 2 + b 2?. Therefore we must hve b? + b? = 0 (3.1)
14 3. ASSOCIATIVE OR DIRECT PRODUCT. ROTATION OF VECTORS which sys tht perpendiculr vectors nticommute. In generl, b cn be expressed s the vector sum of b k nd b?, prllel nd perpendiculr, respectively, to. b k is simply some sclr multiplier m times. Therefore, b k = m = m b k = b k (3.2) Eqs. (3.1) nd (3.2) now de ne the rules of geometric vector lgebr in 2-dimensions. 3.2 Lw of Cosines Now pply the properties de ned by Eqs. (3.1) nd (3.2) to deduce the lw of cosines. Referring to Fig. 3.1c, c = + b from which cc = + bb + b + b c b b c b θ () b (b) θ θ c b c b e 2 θ b θ c e 1 θ b θ c e 1 e 2 (c) Fig. 3.1 (d)
3.3 Lw of Sines 15 Decomposing b in the lst two terms into its components prllel nd perpendiculr to cc = + bb + b k + b? + b k + b? Using Eqs. (3.1) nd (3.2) c 2 = 2 + b 2 + 2b k c 2 = 2 + b 2 + 2b cos where the quntities re now sclr vlues. 3.3 Lw of Sines b? = e 2 b sin c c? = e 2 c sin b b? = c?! e 2 b sin c = e 2 c sin b! b sin c = c sin b Thus nd therefore b c c = sin b sin c = sin sin c ; nd b = sin sin b 3.4 Cli ord Algebr in 3-Dimensions Consider position vector x where e 1 ; e 2 ; e 3 re unit vectors long 3 orthogonl xes. x = v 1 e 1 + v 2 e 2 + v 3 e 3 Form the direct product xx = (v 1 e 1 + v 2 e 2 + v 3 e 3 ) (v 1 e 1 + v 2 e 2 + v 3 e 3 ) = v1e 2 1 e 1 + v2e 2 2 e 2 + v3e 2 3 e 3 +v 1 v 2 (e 1 e 2 + e 2 e 1 ) + v 1 v 3 (e 1 e 3 + e 3 e 1 ) + v 2 v 3 (e 2 e 3 + e 3 e 2 ) Since x 2 = v 2 1 + v 2 2 + v 2 3 which is the Pythgoren theorem in 3-dimensions, the properties of the unit vectors must be e 1 e 1 = e 2 e 2 = e 3 e 3 = 1 e 1 e 2 + e 2 e 1 = 0; etc. Tht is, e i e j = e j e i
16 3. ASSOCIATIVE OR DIRECT PRODUCT. ROTATION OF VECTORS The unit vectors re clled the genertors of the geometric lgebr nd in this cse we use Crtesin unit vectors. A convenient wy to generte ll of the independent combintions in the lgebr is to form the product (1 + e 1 ) (1 + e 2 ) (1 + e 3 ) = 1 sclr + e 1 + e 2 vector + e 3 + e 1 e 2 + e 2 e 3 + e 3 e {z } 1 + e 1 e 2 e 3 pseudo-sclr bivector The result is n 8 element lgebr. The elements consist of sclr, 3 unit vectors tht de ne lines, 3 bivectors tht de ne oriented plnes nd trivector e e 1 e 2 e 3 tht corresponds to volume. It hs the property ee = 1 nd is clled pseudo-sclr. It lso chnges sign under inversion since ( e 1 ) ( e 2 ) ( e 3 ) = e 1 e 2 e 3. Note tht the unit bivectors nticommute. For exmple, (e 1 e 2 ) (e 2 e 3 ) = (e 2 e 1 e 3 e 2 ) = (e 2 e 3 ) (e 1 e 2 ). In this sense, they behve like vectors; however, their squres re 1. For exmple, e 1 e 2 e 1 e 2 = e 1 e 1 e 2 e 2 = 1: When negtive vlues for ech of the genertors re included, the 16 elements form group. De ne, e e 1 e 2 e 3 : Note tht the ssocitive product of e with bivector is vector. For exmple: Therefore, generl bivector ee 1 e 2 = e 1 e 2 e 3 e 1 e 2 = e 2 e 3 e 2 = e 3. my be written B = e 1 e 2 B 12 + e 2 e 3 B 23 + e 3 e 1 B 31 B = e [ e (e 1 e 2 B 12 + e 2 e 3 B 23 + e 3 e 1 B 31 )] B = e (e 3 B 12 + e 1 B 23 + e 2 B 31 ) Since ee 3 = e 1 e 2 ; ee = e 2 e 3 ; ee 2 = e 3 e 1 : e 2 = 1. Thus B = ev, where v is vector. A liner combintion of the elements of the lgebr with rbitrry sclr multipliers; rel or complex, is clled multivector. Thus, spce-like multivector my be written M = S + v + ev+s 0 e 1 e 2 e 3 We now form the direct product of 2 vectors nd express the result in terms of the vector components: uv = (uv + vu) =2 + (uv vu) =2
3.4 Cli ord Algebr in 3-Dimensions 17 Express the vectors u; v in terms of e 1 ; e 2 ; e 3 vu = (v 1 e 1 + v 2 e 2 + v 3 e 3 ) (u 1 e 1 + u 2 e 2 + u 3 e 3 ) = v 1 u 1 + v 1 u 2 e 1 e 2 + v 2 u 1 e 2 e 1 + v 2 u 2 + v 1 u 3 e 1 e 3 +v 3 u 1 e 3 e 1 + v 3 u 3 + v 2 u 3 e 2 e 3 + v 3 u 2 e 3 e 2 uv = v 1 u 1 + u 2 v 1 e 2 e 1 + u 1 v 2 e 1 e 2 + v 2 u 2 + u 3 v 1 e 3 e 1 +u 1 v 3 e 1 e 3 + v 3 u 3 + u 3 v 2 e 3 e 2 + u 2 v 3 e 2 e 3 Thus, 1 2 (vu + uv) = v u = v 1u 1 + v 2 u 2 + v 3 u 3 nd 1 (vu uv) = v ^ u 2 The ltter my be written = 1 2 [v 1u 2 (e 1 e 2 e 2 e 1 ) + v 1 u 3 (e 1 e 3 e 3 e 1 ) +v 2 u 3 (e 2 e 3 e 3 e 2 ) + u 1 v 2 (e 1 e 2 e 2 e 1 ) +u 1 v 3 (e 1 e 3 e 3 e 1 ) + u 2 v 3 (e 2 e 3 e 3 e 2 )] = (v 1 u 2 u 1 v 2 ) e 1 e 2 + (v 1 u 3 u 1 v 3 ) e 1 e 3 + (v 2 u 3 u 2 v 3 ) e 2 e 3 v ^ u = e 1 e 2 e 3 [(v 1 u 2 u 1 v 2 ) e 3 + (v 3 u 1 v 1 u 3 ) e 2 + (v 2 u 3 u 2 v 3 ) e 1 ] The Gibbs cross product is Therefore, v u = e 1 (v 2 u 3 v 3 u 2 ) + e 2 (v 3 u 1 u 3 v 1 ) + e 3 (v 1 u 2 u 1 v 2 ) v ^ u = e (v u) or e (v ^ u) = v u e = e 1 e 2 e 3 ee = 1 where v u is the Gibbs vector product. Thus, the quntity e e 1 e 2 e 3 times the wedge product of 2 vectors de nes vector perpendiculr to the plne of the 2 vectors. The wedge product of 2 vectors is the sum of 3 bivectors, tht is, 3 directed plnes 1. In 4 dimensions, it would be the sum of 6 bivectors or 6 directed plnes. Thus Cli ord Algebr is more generl geometric 1 The direction of directed plne is de ned by e 1 e 2 where looking down e 1 rottes 90 CC to de ne direction perpendiculr to e 1 e 2 in the direction of right hnd screw turning from e 1 to e 2 ; the usul right hnd coordinte system.
18 3. ASSOCIATIVE OR DIRECT PRODUCT. ROTATION OF VECTORS lgebr thn vector nlysis, tht is, the Gibbs lgebr since the Gibbs cross product in 4 dimensions nd higher is not de ned. Direct Product of (vec)(biv) (v 1 e 1 + v 2 e 2 + v 3 e 3 ) (B 12 e 1 e 2 + B 31 e 3 e 1 + B 23 e 2 e 3 ) = v 1 B 12 e 2 v 1 B 31 e 3 + v 1 B 23 e 1 e 2 e 3 v 2 B 12 e 1 + v 2 B 31 e 1 e 2 e 3 + v 2 B 23 e 3 +v 3 B 12 e 1 e 2 e 3 + v 3 B 31 e 1 v 3 B 23 e 2 = (v 3 B 31 v 2 B 12 ) e 1 + (v 1 B 12 v 3 B 23 ) e 2 + (v 2 B 23 v 1 B 31 ) e 3 + (v 1 B 23 + v 2 B 31 + v 3 B 12 ) e 1 e 2 e 3 = v ^ B + v B = B v + e 1 e 2 e 3 (v B) 3.5 Rottion of Vectors Cli ord Algebr provides simple opertor for rotting vectors. To obtin this opertor we rst nd procedure for re ecting vector b in plne (mirror) contining nother vector. The mirror is perpendiculr to the plne of nd b. Fig. 3.2. The inverse of vector is de ned by 1 = 1. If is unit vector de ned by = 1, then 1 =. The re ection of b through to obtin b 0 is chieved by forming the combintion b 0 = 1 b = 1 b k 1 b? = 1 b k + 1 b? = b k + b? b? = b? by nticommuttivity of orthogonl vectors in Cli ord lgebr. To rotte vector by n ngle ' drw 2 vectors nd b seprted by ngle '=2 s shown in Fig. 3.2. nd b my be unit vectors. We wish to rotte vector v into vector v 0. Let vector v mke n ngle 1 with where 1 < '. Re ect v through ngle 1 so tht it becomes vector v 00. Re ect v 00 in b through ngle 2 = '=2 1. The net re ection of v is then through n ngle 2 1 +2 ('=2 1 ) = ': Mthemticlly, the two re ections re described by v 00 = 1 v (3.3) v 0 = b 1 v 00 b (3.4) Substituting the result of the rst re ection into the second re ection v 0 = b 1 1 vb v 0 = bvb (3.5)
3.5 Rottion of Vectors 19 v' b θ 2 θ 1 θ 2 θ 1 v" ϕ/2 ϕ/2 b () (b) Fig. 3.2 v (c) It is convenient to let = e 1 nd to express b in terms of e 1 nd the unit vector e 2 perpendiculr to e 1. Let '=2 be the ngle between e 1 nd b, where ' is the ngle through which we wish to rotte the vector v. Thus, nd therefore, which cn be written Likewise, b k = e 1 cos '=2 nd b? = e 2 sin '=2 e 1 b = e 1 (e 1 cos '=2 + e 2 sin '=2) = cos '=2 + e 1 e 2 sin '=2 (3.6) e 1 b = e e 1e 2 '=2 ; since (e 1 e 2 ) (e 1 e 2 ) = 1 be 1 = (e 1 cos '=2 + e 2 sin '=2) e 1 = cos '=2 e 1 e 2 sin '=2 = e e 1e 2 '=2 (3.7) Substituting these expressions in Eq. (3.5), we obtin v 0 = e e 1e 2 '=2 ve e 1e 2 '=2 (3.8) Eq. (3.8) speci es counter clockwise rottion of v through n ngle ' bout xis e 3 perpendiculr to the plne of e 1 nd e 2. To rotte the vector v clockwise through ', chnge ' to ' in the bove expressions. The bilterl opertors in Eq. (3.8) my be replced by the unilterl opertor v 0 = e e 1e 2 '=2 ve e 1e 2 '=2 = e e 1e 2 ' v (3.9)
20 3. ASSOCIATIVE OR DIRECT PRODUCT. ROTATION OF VECTORS when v lies in the plne of e 1 e 2. Thus, e e 1e 2 ' operting left to right on v rottes it t n ngle ' counter-clockwise bout e 3. e e 1e 2 ' operting right to left does the sme. To rotte clockwise, chnge the sign of '. To justify Eq. (3.9), use Eq. (3.8) to rotte the vector v = v x e 1 + v y e 2 into v 0 = v 0 xe 1 + v 0 ye 2 Consider v 1 e 1 e e 1e 2 '=2 = v 1 e 1 (cos '=2 + e 1 e 2 sin '=2) = v 1 e 1 [cos '=2 + e 1 e 2 e 1 (sin '=2) e 1 ] = (cos '=2 + e 1 e 1 e 2 e 1 sin '=2) v 1 e 1 = (cos '=2 e 1 e 2 sin '=2) v 1 e 1 = e e 1e 2 '=2 v 1 e 1 Likewise So v y e 2 e e 1e 2 '=2 (v x e 1 + v y e 2 ) e e 1e 2 '=2 = e e 1e 2 '=2 v y e 2 = e e 1e 2 '=2 (v x e 1 + v y e 2 ) Therefore e e 1e 2 '=2 ve e 1e 2 '=2 = e e 1e 2 '=2 e e 1e 2 '=2 (v x e 1 + v y e 2 ) = e e 1e 2 '=2 e e 1e 2 '=2 v = e e 1e 2 ' v = v 0 = v 0 xe 1 + v 0 ye 2 As n exmple, let us use Eq. (3.9) to rotte counterclockwise through n ngle ' vector x = e 1 cos + e 2 sin, mking n ngle with respect to the e 1 xis. This will yield two trigonometric ddition formuls. Thus If = 0; x 0 = e e 1e 2 ' x = e e 1e 2 ' (e 1 cos + e 2 sin ) = e 1 cos (cos ' e 1 e 2 sin ') + e 2 sin (cos ' e 1 e 2 sin ') = [e 1 (cos cos ' sin sin ') + e 2 (cos sin ' + sin cos ')] = e 1 cos ( + ') + e 2 sin ( + ') cos ( + ') = cos cos ' sin sin ' sin ( + ') = cos sin ' + sin cos ' x 0 = (e 1 cos ' + e 2 sin ') Fig. 3.3
3.6 Rottion bout n Arbitrry Axis n 21 To rotte n rbitrry vector S counterclockwise through n ngle ' nd whose end points re speci ed by R 1 nd R 2, it is necessry to rotte the two vectors de ning the end points of S through the ngle '. Thus by Fig. 3.3, one hs R 1 + S = R 2 S = R 2 R 1 R 0 1 + S 0 = R 0 2 S 0 = R 00 2 R 0 1 R 0 1 = e e 1e 2 ' R 1 R 0 2 = e e 1e 2 ' R 2 R 0 2 R 0 1 = S 0 = e e 1e 2 ' (R 2 R 1 ) = e e 1e 2 ' S S 0 = e e 1e 2 ' S To verify tht the sclr product is invrint, tht is, S 0 S 0 = S S, form Thus S 02 = S 2 S 0 S 0 = e e 1e 2 ' Se e 1e 2 ' S = e e 1e 2 ' e e 1e 2 ' S S The rottion of multivector bout e 3 is by de nition the rottion bout e 3 of the individul vectors tht constitute the multivector. To rotte multivector M bout e 3 counterclockwise through n ngle ' bout e 3, form M 0 = L 1 ML L = e e 1e 2 '=2 It is cler tht we cn lwys sndwich e e 1e 2 '=2 e e 1e 2 '=2 = 1 between the elements of multivector. For exmple, for the rottion of bivector vu e e 1e 2 '=2 vue e 1e 2 '=2 = e e 1e 2 '=2 ve e 1e 2 '=2 e e 1e 2 '=2 ue e 1e 2 '=2 = v 0 u 0 3.6 Rottion bout n Arbitrry Axis n We now show tht the rottion of vector v through n ngle ' bout n rbitrry xis speci ed by the unit vector n is given by: v 0 = e en'=2 ve en'=2 (3.10)
22 3. ASSOCIATIVE OR DIRECT PRODUCT. ROTATION OF VECTORS Fig. 3.4 To obtin this result, drw plne through the origin nd perpendiculr to n. Decompose v into component v? perpendiculr to n nd component n = v k n prllel to n. Let e 0 1 nd e 0 2 through the origin de ne two orthogonl xes in the plne. The rottion through n ngle ' of the component u perpendiculr to n is given by v 0? = e e0 1 e0 2 '=2 v? e e0 1 e0 2 '=2 (3.11) e 0 1 nd e 0 2 cn be speci ed in terms of direction cosines with respect to the xes e 1; e 2; e 3. Then e 0 1e 0 2 = (e 1 cos 1 + e 2 cos 1 + e 3 cos 1 ) (e 1 cos 2 + e 2 cos 2 + e 3 cos 2 ) = e 1 e 1 cos 1 cos 2 + e 2 e 2 cos 1 cos 2 + e 3 e 3 cos 1 cos 2 +e 1 e 2 (cos 1 cos 2 cos 2 cos 1 ) + e 1 e 3 (cos 1 cos 2 cos 2 cos 1 ) +e 2 e 3 (cos 1 cos 2 cos 2 cos 1 ) Since e 1 nd e 2 re orthogonl, the sclr product is zero: e 0 1 e 0 2 = cos 1 cos 2 + cos 1 cos 2 + cos 1 cos 2 = 0 The unit vector n is given by n = e 0 1 e 0 2 = e (e 0 1 ^ e 0 2) = e 1 (cos 1 cos 2 cos 2 cos 1 )
3.6 Rottion bout n Arbitrry Axis n 23 +e 2 (cos 1 cos 2 cos 1 cos 2 ) + e 3 (cos 1 cos 2 cos 2 cos 1 ) = e 1 cos + e 2 cos + e 3 cos where ; ; nd re the direction cosines of n. e = e 1 e 2 e 3 : The product e 0 1e 0 2 my then be written e 0 1e 0 2 = e 1 e 2 cos + e 1 e 3 cos + e 2 e 3 cos = e 1 e 2 e 3 (e 1 cos + e 2 cos + e 3 cos ) = e 1 e 2 e 3 n = en Note tht e 1 e 2 e 3 n = ne 1 e 2 e 3 Thus, e 1 e 2 e 3 n = e 1 e 2 e 3 ne 1 e 2 e 3 n = e 1 e 2 e 3 e 1 e 2 e 3 n 2 = (e 1 e 2 e 3 ) 2 n 2 = 1 Since, nn = 1 nn = (e 1 cos + e 2 cos + e 3 cos ) (e 1 cos + e 2 cos + e 3 cos ) = e 1 e 1 cos 2 + e 2 e 2 cos 2 + e 3 e 3 cos 2 + (e 1 e 2 + e 2 e 1 ) cos cos + (e 1 e 3 + e 3 e 1 ) cos + (e 2 e 3 + e 3 e 2 ) cos = cos 2 + cos 2 + cos 2 = 1 Note tht n = e e 1e 2 e 3 n'=2 ne e 1e 2 e 3 n'=2 = cos ' en sin ' n cos ' 2 2 2 = n cos ' en sin ' cos ' 2 2 2 en sin ' 2 en sin ' = n 2 Thus, for = v n = v k v 0 = e en'=2 (n + v? ) e en'=2 = e en'=2 ne en'=2 + e en'=2 v? e en'=2 v 0 = n + e en' v? = v k + e en' v? (3.12)
24 3. ASSOCIATIVE OR DIRECT PRODUCT. ROTATION OF VECTORS