DEFINITION OF ASSOCIATIVE OR DIRECT PRODUCT AND ROTATION OF VECTORS


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1 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)
2 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 2dimensions. 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 3.3 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 3Dimensions 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 v v 2 3 which is the Pythgoren theorem in 3dimensions, 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
4 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 pseudosclr 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 pseudosclr. 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, spcelike 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
5 3.4 Cli ord Algebr in 3Dimensions 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.
6 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 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 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 ) = ': 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)
7 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)
8 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 ' counterclockwise 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
9 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 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 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)
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 )
11 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 ' = n cos ' en sin ' cos ' 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)
12 24 3. ASSOCIATIVE OR DIRECT PRODUCT. ROTATION OF VECTORS
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