Orthogonal transformations

Transcription

1 Orthogoal trasformatios October 12, Defiig property The squared legth of a vector is give by takig the dot product of a vector with itself, v 2 v v g ij v i v j A orthogoal trasformatio is a liear trasformatio of a vector space that preserves legths of vectors. This defiig property may therefore be writte as a liear trasformatio, such that v Ov v v v v Write this defiitio i terms of compoets usig idex otatio. The trasformatio O must have mixed idices so that the sum is over oe idex of each type ad the free idices match, we have v i O i jv j g ij v i v j g ij v i v j g ij O i k v k O j mv m g ij v i v j gij O i ko j m v k v m g km v k v m or, brigig both terms to the same side, gij O i ko j m g km v k v m 0 Because v k is arbitrary, we ca easily make six idepedet choices so that the resultig six matrices v k v m spa the space of all symmetric matrices. The oly way for all six of these to vaish whe cotracted o gij O i k Oj m g km is if g ij O i ko j m g km 0 This is the defiig property of a orthogoal trasformatio. I Cartesia coordiates, g ij δ ij, ad this coditio may be writte as O t O 1 where O t is the traspose of O. This is equivalet to O t O 1. 1

2 Notice that othig i the preceedig argumets depeds o the dimesio of v beig three. We coclude that orthogoal trasformatios i ay dimesio must satisfy O t O 1, where O is a matrix. Cosider the determiat of the defiig coditio, det 1 det O t O 1 det O t det O ad sice the determiat of the traspose of a matrix equals the determiat of the origial matrix, we have det O ±1. Ay orthogoal trasformatio with det O 1 is just the parity operator, P 1 1 times a orthogoal trasformatio with det O +1, so if we treat parity idepedetly 1 we have the special orthogoal group, SO, of uit determiat orthogoal trasformatios. 1.1 More detail If the argumet above is ot already clear, it is easy to see what is happeig i 2-dimesios. We have v k v m v 1 v 1 v 1 v 2 v 2 v 1 v 2 v 2 which is a symmetric matrix. But g ij O i k Oj m g km v k v m 0 must hold for all choices of v i. If we make the choice v i 1, 0 the v k v m ad g ij O i k Oj m g km v k v m 0 implies g ij O i 1O j 1 g 11. Now choose v i 0, 1 so that v k v m ad we see we must also have g ij O i 2O j g 22. Fially, let v i 1, 1 so that v k v m 1 1 g ij O i 1O j 1 g 11 + g ij O i 1O j 2 g 12 + g ij O i 2O j 1 g 21 + g ij O i 2O j 2 g 22 0 but by the previous two relatios, the first ad fourth term already vaish ad we have g ij O i 1O j 2 g 12 + g ij O i 2O j 1 g 21 0 Because g ij g ji, these two terms are the same, ad we coclude g ij O i ko j m g km for all km. I 3-dimesios, six differet choices for v i will be required. 2 Ifiitesimal geerators This gives We work i Cartesia coordiates, where we ca use matrix otatio. The, for the real, 3-dimesioal represetatio of rotatios, we require O t O 1 Notice that the idetity satisfies this coditio, so we may cosider liear trasformatios ear the idetity which also satisfy the coditio. Let O 1 + ε 2

3 where ε] ij ε ij are all small, ε ij 1 for all i, j. Keepig oly terms to first order i ε ij, we have: O t 1 + ε t O 1 1 ε where we see that we have the right form for O 1 by computig OO ε 1 ε 1 ε 2 1 correct to first order i ε. Now we impose our coditio, O t O ε t 1 ε ε t ε so that the matrix ε must be atisymmetric. Next, we write the most geeral atisymmetric 3 3 matrix as a liear combiatio of a coveiet basis, ε w i J i w w 2 w 3 w 2 0 w 1 w 3 w w w where w i 1. Notice that the compoets of the three matrices Ji are eatly summarized by ] jk ε ijk where ε ijk is the totally atisymmetric Levi-Civita tesor. For example, J 1 ] ij ε 1ij The matrices are called the geerators of the trasformatios. The most geeral atisymmetric is the a liear combiatio of the three geerators. Kowig the geerators is eough to recover a arbitrary rotatio. Startig with we may apply O repeatedly, takig the it O 1 + ε O θ O 1 + ε 1 + w i Let w be the legth of the ifiitesmal vector w i, so that w i w i, where i is a uit vector. The the it is take i such a way that w θ 3

4 where θ is fiite. Usig the biomial expasio, a + b! k0 a k b k we have O 1 + w i! 1 k w i k k0 k0 k0 1 θ i k k0 exp θ i 1 2 k w i k k 1 2 k+1 θ i k We defie the expoetial of a matrix by the power series for the expoetial, applied usig powers of the matrix. This is the matrix for a rotatio through a agle θ aroud a axis i the directio of. To fid the detailed form of a geeral rotatio, we ow eed to fid powers of i. This turs out to be straightforward: i ] j k i ε j i k i 2 i ε m i k j εj k i j εi m kεj k i j εi m kε k j i j δ ij δ m δ i δ m j δ ij i j δ m + δ i i δ m j j i 3 δ m + m δ m m δ m k m k i ε k i δ m k i ε k i + m k i ε k i i ε m i + m k i ε ik i where m k i ε ik m 0. The powers come back to i with oly a sig chage, so we ca divide the series ito eve ad odd powers. For all k > 0, i 2k 1 k δ m m i 2k+1 1 k i ] For k 0 we have the idetity, i 0 m δ m. We ca ow compute the expoetial explicitly: O θ, ˆ exp θ i 4

5 k0 l0 1 + δ m + 1 θ i k 1 θ i 2l + 2l! l1 l1 1 θ i 2l 2l! 1 θ i 2l+1 2l + 1! l0 1 + θ i 2l+1 2l + 1! l0 1 l 2l! θ2l δ m m + l0 δ m + cos θ 1 δ m m + si θ i ε m i 1 l 2l + 1! θ2l+1 i where we get cos θ 1 because the l 0 term is missig from the sum. To see what this meas, let O act o a arbitrary vector v, ad write the result i ormal vector otatio, Goig fully to vector otatio, O θ, ˆ v δ m v + cos θ 1 δ m m v + si θ i ε m i v v m + cos θ 1 δ m v m v si θ i v εi v m + cos θ 1 v m v m v si θ O θ, ˆ v v + cos θ 1 v v v si θ Fially, defie the compoets of v parallel ad perpedicular to the uit vector : Therefore, v v v v v O θ, ˆ v v v v + v v cos θ si θ v v + v cos θ si θ v This expresses the rotated vector i terms of three mutually perpedicular vectors, v, v, v. The directio is the axis of the rotatio. The part of v parallel to is therefore uchaged. The rotatio takes place i the plae perpedicular to, ad this plae is spaed by v, v. The rotatio i this plae takes v ito the liear combiatio v cos θ v si θ, which is exactly what we expect for a rotatio of v through a agle θ. The rotatio O θ, ˆ is therefore a rotatio by θ aroud the axis ˆ. m 5