Chapter 6. Rotations and Tensors

Transcription

1 Vector Spaces n Physcs 8/6/5 Chapter 6. Rotatons and ensors here s a speca knd of near transformaton whch s used to transforms coordnates from one set of axes to another set of axes (wth the same orgn). Such a transformaton s caed a rotaton. Rotatons have great practca mportance, n appcatons rangng from trackng spacecraft to graphca dspay software. hey aso pay a centra roe n theoretca physcs he near transformaton correspondng to a rotaton can be descrbed by a matrx. We w descrbe the propertes of rotaton matrces, and dscuss some speca ones. A. Rotaton of Axes. In chapter we consdered a rotaton of a Cartesan coordnate system about the z-axs by an ange. A more genera case s shown n fgure 6-, where the axes are transformed by rotaton nto an arbtrary second set of axes. (We take both sets or axes to be rght handed. ransformaton to a eft-handed coordnate system s to be avoded, uness you reay know what you are dong.) A typca vector A s aso shown. Fgure 6- (a) shows the components of A n the e ˆ, eˆ, ˆ ) system, and ( e3 fgure 6- (b) shows ts coordnates n the e ˆ,' eˆ,' ˆ ') system. We see that we can wrte ( e3 A, usng components n the unprmed system, as A eˆ A ; (6-) or, usng components n the prmed frame, as A e ˆ ' A ' (6-) he standard defntons of components are cear from (6-) and (6-): A eˆ A, A ' eˆ ' A. We can, however, reate the prmed components to the unprmed, as foows: z' eˆ3' ê x' z ê 3 eˆ' eˆ' ê A x Fgure 6-. wo sets of Cartesan coordnate axes. he (x',y',z') system s obtaned from the (x,y,z) system by a rotaton of the coordnate system. y' y (6-3) 6 -

2 Vector Spaces n Physcs 8/6/5 A ' eˆ ' A j eˆ j ' ( A eˆ ) j eˆ ' eˆ A C A j j (6-4) (a) z A eˆ A 3 3 (b) z' z A eˆ A 3 3 x eˆ A eˆ A hat s to say that the reatonshp between prmed and unprmed components can be expressed as a matrx equaton, A ' CA, (6-5) where the transformaton matrx C s gven by y x ˆj ˆ j eˆ3' C C ( e ' e ). (6-6) ê x' ê 3 eˆ' eˆ' ê eˆ A eˆ A Fgure 6-. he vector A expressed n terms of ts e ˆ, eˆ, ˆ ) components (a), and n terms of ts e ˆ,' eˆ,' ˆ ') components (b). ( e3 ( e3 y' y here s an nterpretaton of the coeffcents Cj n terms of anges. Snce a of the e ˆ j and e ˆ are unt vectors, the dot product of any two s equa to the cosne of the ange between them: eˆ ' eˆ cos. (6-7) m n mn ' Exampe: Consder a rotaton by an ange n the counter-cockwse sense about the z axs, as descrbed n Chapter. Snce the z axs s the same n the prmed and unprmed systems, certan of the coeffcents C e ˆ ' e ˆ are especay smpe: j j C33, ; (6-8) C3 C3 C3 C3 he other coeffcents can be read off from fgure (-4): 6 -

3 Vector Spaces n Physcs 8/6/5 C C C î' î cos, ĵ' ĵ cos, ; (6-9) î' ĵ sn, C ĵ' î -sn. and so cos sn C sn cos R z ( ). (6-) he matrces correspondng to rotatons about the x and y axes can be cacuated n a smar way; the resuts are summarzed beow. Rx ( ) cos sn sn cos cos sn Ry ( ). (6-) sn cos cos sn Rz ( ) sn cos B. Some propertes of rotaton matrces. Orthogonaty. Consder the defnton of the transformaton matrx C : ˆ ˆ j j If we form the product of C and ts transpose, we get C C ( e e ). (6-6) CC C C C m m m m eˆ ' eˆ eˆ ' eˆ m eˆ ' eˆ eˆ eˆ ' C m m m. (6-) But, the vector e ˆ, ke any vector, can be wrtten n terms of ts components n the unprmed frame, e ˆ ˆm( ˆm ˆ e e e ). (6-3) In ths sum, ( eˆm e ˆ ) s the m-th component of the vector e ˆ n the unprmed frame. So, equaton (6-) becomes 6-3

4 Vector Spaces n Physcs 8/6/5 CC eˆ ' eˆ '. (6-4) or CC I. (6-5) he equvaent statement n ndex notaton s CC. (6-5a) A smar cacuaton for the product j j C C eads to C C I. (6-5b) CC j hs eads us to defne orthogonaty as foows: For an orthogona matrx, the transpose matrx s aso the nverse matrx. C C CC I C C C C j j (6-5c) [Note: We have not used the condton that ether e ˆ,,3 or eˆ,,3 form a rghthanded coordnate system, so orthogona rotatons ncude both rotatons and transformatons consstng of a rotaton and a refecton. More about ths dstncton ater.] Exampe: Show that the rotaton matrx of (6-) above s orthogona. Souton: CC cos sn sn cos cos sn cos sn sn cos sn cos I sn cos sn cos sn cos sn cos. (6-6) Determnant. It s easy to verfy the foowng fact: he determnant of an orthogona matrx must be equa to ether + or -. (he proof s eft to the probems.) If ts determnant s equa to +, t s sad to be a proper orthogona matrx. A rotaton matrces are proper orthogona matrces, and vce versa. Rotaton matrces must have a determnant of +. However, ths s not the ony condton for a square matrx to be a rotaton matrx. It must aso have the property that ts transpose s equa to ts nverse. 6-4

5 Vector Spaces n Physcs 8/6/5 NOE: A rotaton matrces are orthogona; but not a orthogona matrces are rotaton matrces. Can you expan why? C. he Rotaton Group Rotatons are near operators whch transform three coordnates {x} as seen n one coordnate system nto three coordnates {x'} n another system. A vectors transform n ths same way, as gven n equaton (6-4) above: A ' C A (6-4) j j We have shown that C must be orthogona, and have determnant equa to +. Mutpcaton of two rotaton matrces gves another square matrx, and t must aso be a rotaton matrx, snce carryng out two rotatons, one after another, can be represented by a snge rotaton. So, the rotaton group s defned as foows. Defnton. he "rotaton group" conssts of the set of rea orthogona 3x3 matrces wth determnant equa to +. (a) Group mutpcaton s just matrx mutpcaton accordng to the standard rues. (b) he group s cosed under mutpcaton. (c) he dentty matrx s the dentty of the group. (d) For every eement A of the group, there exsts an eement A -, the nverse of A, such that AA I. A BC AB C. (e) Mutpcaton satsfes hese propertes are easy to prove and are eft to the exercses. D. ensors What are tensors? ensors ook ke matrces; but ony certan types of matrces are tensors, as we sha see. hey must transform n a certan way under a rotaton of the coordnate system. Vectors, wth one ndex, are tensors of the frst rank. Other objects, such as the rotaton matrces themseves, may have more than one ndex. If they transform n the rght way, they are consdered to be hgher-order tensors. We w frst dscuss rotatons, then defne tensors more precsey. Under a rotaton of the coordnate system, the components of the dspacement vector x change n the foowng way: x = Rjxj. (6-7) he tensor transformaton rues represent a generazaton of the rue for the transformaton of the dspacement vector. ensors of Rank Zero. A rank-zero tensor s an object g whch s nvarant under rotaton: Rank - zero tensor : g g (6-7a) 6-5

6 Vector Spaces n Physcs 8/6/5 ensors of Rank One. Modeed on the rotaton propertes of vectors, we defne a tensor of rank one as an object wth a snge ndex, C ={ C, =,3} or just C for short, such that the components C' n a rotated coordnate system are gven by Frst - Rank ensor : C = R C. (6-8) j j ensors of Rank wo. A second-rank tensor s an object wth two ndces, D = { Dj,, j =,3} or Dj for short, such that ts components Dj' n a rotated coordnate system are gven by Second - Rank ensor: D j = Rk R j D k. (6-9) hat s to say, each of the two ndces transforms ke a snge vector ndex. ensors of Rank hree. Foowng the same pattern, a thrd-rank tensor Hjk foows the transformaton aw hrd - Rank ensor : H jk = RR jmrknh mn. (6-9) And so on. Much of the power of tensor anayss es n the ease n whch new tensors can be created, wth ther transformaton propertes under rotatons just determned by the number of tensor ndces. Here are two theorems about creaton of tensors whch we offer wthout proof. ensor Products: he drect product of a tensor of rank n and a tensor of rank m s a tensor of rank n+m. hus, for nstance, the product of two vectors, ABj, s a second-rank tensor. Contracton of Indces: If two tensor ndces are set equa (and so summed), the resutng object s a tensor of rank ower by two than that of the orgna tensor. Exampe: Let us consder the contracton of the two ndces of the tensor ABj: A B A B. (6-) he fact that ths dot product s a zero-rank tensor guarantees that t has the same vaue n a coordnate systems. Speca ensors. he Kronecker deta symbo and the Lev-Cvta symbo have ndces whch ook ke tensor ndces - but ther components are defned to be the same regardess of the coordnate system. Can they st transform ke tensors, of rank two and three, respectvey? hs s n fact the case; the proof s eft for the probems. hs means that these two tensors can be used n conjuncton wth vectors and other tensors to make new tensors. 6-6

7 Vector Spaces n Physcs 8/6/5 Exampe: Consder the foowng tensor operatons: Frst form the drect product jkabm of the Lev-Cvta tensor wth two vectors A and B. hen perform two contractons of ndces, by settng =j and m=k. hs produces the foowng object, wth one free ndex: j jk A B m m k jk AB j k (AxB ). (6-) hs s what we ca the cross product or vector product. he theorems about tensor ndces guarantee, snce t has one free ndex, that t does n fact transform ke a vector under rotatons. Exampe: he trace and determnant of a tensor are both scaars, snce a the free tensor ndces are removed by contracton. hus, the trace and determnant of a matrx are the same n a coordnate systems, provded that the matrx tsef transforms ke a tensor. E. Coordnate ransformaton of an Operator on a Vector Space. Here s an expanaton of the transformaton aw for a second-rank tensor. Consder an operator O operatng on a vector A to produce another vector B : B OA. (6-) ensors are a about ookng at somethng from another pont of vew - n a rotated coordnate system. So, f we wrte quanttes n the rotated coordnate system wth a prme, the other pont of vew s B OA. (6-3) Under rotatons, every vector s just mutped by R. If we appy ths to both sdes of equaton (6-) t becomes RB R OA. (6-4) Now nsert between O and A a factor of the dentty matrx, whch we can wrte as I R R R R, where we have used the fact that R's nverse s just ts transpose. RB ROA ROR RA ROR RA. (6-5) hs just shows the operator workng n the rotated coordnate system, as shown n (6.3), provded that the operator n the rotated coordnate system s gven by O ROR.Smarty ransformaton (6-6) hs process, of sandwchng a square matrx between R and ts nverse, s caed a smarty transformaton. But... shoudn't O transform ke a second-rank tensor? Let's wrte 6-6) n tensor notaton, 6-7

8 Vector Spaces n Physcs 8/6/5 O R O R j m mj R O R m jm R R O jm m. (6-7) hs s exacty the transformaton aw for a second-rank tensor gven n eq. (6-9). hs means that aws ke B OA. (6-8) can be used n any coordnate system, provded that vectors and operators are transformed appropratey from one coordnate system to another. hs sounds a tte ke Ensten's (reay Gaeo's) postuate about the aws of physcs beng the same n a nerta frames. And ths s no accdent. When aws are wrtten expcty n terms of tensors, they are sad to be n "covarant form." It makes for beautfu math, and sometmes beautfu physcs. F. he Conductvty ensor. here are many smpe physca aws reatng vectors. For nstance, the verson of Ohm's aw for a dstrbuted medum s z J E. (6-9) E hs aw asserts that when an eectrc fed E s apped to a regon of a conductng matera, there s a fux of charge J n the same drecton, proportona to the magntude of E. hs seems smpe and rght. But s t rght? ("A physca aw shoud be as smpe as possbe -- but no smper," Ensten.) y A more genera reaton can be obtaned, foowng the prncpe that J must transform ke a vector. We can combne J n varous ways wth tensor quanttes representng the medum, as ong as there s one free ndex. he two obvous reatons are J E J E J E J E j j scaar conductvty. (6-3) tensor conductvty he scaar reaton causes current to fow n the drecton of the eectrc fed. But the tensor reaton aows the current to fow off n a dfferent drecton. What knd of medum woud permt ths? Graphte s a matera whch s qute ansotropc, so we mght expect a reaton of more compcated drectonaty. Fgure 6-3 shows the mode we w take, of panes of carbon atoms where eectrons fow easy, the panes stacked n such a way that current fow x Fgure 6-3. Graphte, represented as ayers (of hexagonay bonded carbon) stacked one on top of another. 6-8

9 Vector Spaces n Physcs 8/6/5 from one pane to another s more dffcut. We can set up a conductvty matrx as foows. From symmetry, we woud expect that an eectrc fed n the x-y pane woud cause a current n the drecton of E, wth conductvty "n the panes." And, aso from symmetry, an eectrc fed n the z drecton shoud cause current to fow n the z drecton, but wth a much smaer conductvty. hs can be wrtten n matrx form as E J E E. (6-3) E 3 If you mutpy ths out, you see that the frst two components of E (the x and y components) are mutped by, and the thrd component, by. But what f for some reason t s better to use a rotated set of axes? hen the apped fed and the conductvty tensor both need to be transformed to the new coordnate system, and ther product w produce the current densty n the new coordnate system. Note that scaar quanttes, ke J E, are zero-th rank tensors and shoud have the same vaue n both coordnate systems. Numerca Exampe. In equaton (6-3), et's take =. * (weak current perpendcuar to the graphte panes) and have the eectrc fed n the y-z pane, at an ange of 45 from the y axs, as shown n fgure (6-3): E E,. (6-3a). hs gves for the current densty J n the orgna coordnate system, J E E E. (6-3b). he z component of the current s sma compared to the y component, and so the current vector moves coser to the pane of good conducton. Now et's see what happens n a rotated coordnate system. Let's go to the system where the eectrc fed s aong the z axs. hs requres a rotaton about the x axs of 45. Usng the form of ths matrx gven n eq. (6-) above, we have 6-9

10 Vector Spaces n Physcs 8/6/5 R Rx ( 45 ). (6-33) In ths system the eectrc fed vector s smper: E RE E E ; (6-34) And the conductvty tensor becomes R R ; (6-35) Now here s the punch ne: We can ether rotate J (cacuated n the orgna coordnate system) nto the rotated coordnate system; or we can cacuate t drecty, usng the eectrc fed and the conductvty tensor expressed n the rotated coordnate system. Here goes: 6 -

11 Vector Spaces n Physcs 8/6/5 J RJ E E.45 ; (6-36).55 And the other way: J E E E.45 ; (6-36) It works - same resut ether way. Here s the nterpretaton of ths exampe. he equaton J E produces the current densty fowng n a medum (a vector) n terms of the eectrc fed n the medum (aso a vector). But... the current does not necessary fow n the drecton of the apped eectrc fed. he most genera vector resut caused by another vector s gven by the operaton of a second-rank tensor! In a smpe physcs probem, the conductvty tensor w have a smpe form, n a coordnate system whch reates to the structure of the matera. But what f we want to know the current fowng n some other coordnate system? Just transform the apped fed and the conductvty tensor to that coordnate system, and use J E G. he Inerta ensor. In cassca mechancs, a rgd body whch s r rotatng about ts center of mass had anguar L momentum. he rate of rotaton at a partcuar nstant of tme can be descrbed by a vector,, whch s n the drecton of the axs of rotaton, wth x magntude equa to the anguar speed, n radans/sec. he anguar momentum s aso gven by a vector, L, whch s equa to the sum (or ntegra) of the ndvdua anguar momenta of the parts of the rgd body, accordng to dl dm r v. (6-37) In certan cases, such as when the rotaton s about an axs of symmetry of the object, the reaton between L and s smpe: L I. (6-38) x Fgure 6-4. Dspacement and veocty, and the resutng anguar momentum. v x 6 -

12 Vector Spaces n Physcs 8/6/5 Here I s the moment of nerta dscussed n a frst course n mechancs. However, surprsngy enough, the resutant tota anguar momentum vector L s not n genera parae to. he genera reaton nvoves an anguarmomentum tensor, I, rather than the scaar moment of nerta. he genera reaton s L I. (6-39) where now, for a body approxmated by N dscrete pont masses, N L L and (6-39) L m r v s the anguar momentum of the -th mass. he veocty of a pont n the rotatng body s gven by v r, and so dl dm r v dm r r We w evauate the doube cross product usng tensor notaton and the epson ker. dl dm r r dm r r jk j km m dm r r kj km j m dm r r dm jm m j j m r v r Fgure 6-5. A rgd body, rotatng about an axs through the orgn. (6-4) Probems he Pau matrces are speca x matrces assocated wth the spn of partces ke the eectron. hey are,, 3. (Here represents the square root of -.) Probem 6-. (a) What are the traces of the Pau matrces? (b) What are the determnants of the Pau matrces? (c) Are the Pau matrces orthogona matrces? est each one 6 -

13 Vector Spaces n Physcs 8/6/5 Probem 6-. he Pau matrces are camed to satsfy the reaton est ths reaton, for j=, k=3. j k k j jk Probem 6-3. In Probem 6- above one of the Pau matrces,, dd not turn out to be orthogona. For matrces wth compex eements a generazed reatonshp must be used: CC C C I C s untary where the "Hermetan conjugate" matrx matrx. (a) Cacuate the Hermetan conjugate C s the compex conjugate of the transpose of the second Pau matrx. (b) Check and see f t s untary. [Untary matrces become mportant n quantum mechancs, where one cannot avod workng wth compex numbers.] Probem 6-5. Usng the fact that the determnant of the product of two matrces s the product of ther determnants, show that the determnant of an orthogona matrx must be equa to ether + or -. Hnt: Start wth the orthogonaty condton A A I. Probem 6-6. For each of the three matrces beow, say whether or not t s orthogona, and whether or not t s a rotaton matrx. Justfy your answers. 4 3 A, B 3 4, C 5 5 Probem 6-7. Consder rotatons Rx() and Rz(), wth the matrces Rx and Rz as defned n the text. Here represents the ange of a rotaton about the x axs and represents the ange of a rotaton about the z axs. Frst, cacuate the matrces representng the resut of carryng out the two rotatons n dfferent orders: and A R x R B R z Rx. Comparng A and B, do you fnd that the two rotatons Rx and z z R commute? 6-3

14 Vector Spaces n Physcs 8/6/5 Next, ook at your resuts n the sma-ange approxmaton (an approxmaton where you keep ony terms up to frst order n the anges and ). Does ths change your concuson? Probem 6-8. Show agebracay that the rotaton group s cosed under mutpcaton. that s, show that, for A and B members of the rotaton group, the product matrx C = AB s orthogona and has determnant equa to +. Probem 6-. he Kronecker deta symbo s defned to have constant components, ndependent of the coordnate system. Can ths be true? hat s to say, s t true that? RR j jm m j Use the orthogonaty of the transformaton matrx Rj to show that ths s true. Probem 6-. he Lev-Cvta symbo jk, defned prevousy, has specfed constant vaues for a of ts components, n any frame of reference. How can such an object be a tensor - that s, how can these constant vaues be consstent wth the transformaton propertes of tensor ndces? We, they are! Consder the transformaton equaton jk = jm kn mn where R s a rotaton matrx, satsfyng the orthogonaty condton RjRk = jk, and jk s the Lev-Cvta symbo. he queston s, does ' have the same vaues for a of ts components as? As a parta proof (coverng of the 7 components), prove that jk = f any two of the ndces {,j,k} are equa. Note: n carryng out ths proof, set two ndces to the same vaue - but do not sum over that ndex. hs operaton fas outsde of the Ensten summaton conventon. Probem 6- ensor transformaton of the Lev-Cvta symbo. (Chaenge Probem) Prove that under transformatons of tensor ndces by an orthogona matrx R, the Lev-Cvta tensor transforms accordng to the rue jk R R = jk det( R ). Probem 6-3 Invarance of ength under rotaton. he ength of a vector s a scaar, and so shoud not change from one coordnate system to another. o check ths, use the vectors E and J from the conductvty exampe n the text. See f they have the same magntude n both coordnate systems used. Probem 6-4 Invarance of a scaar product under rotaton. he nner product of two vectors s a scaar, and so shoud not change from one coordnate system to another. o check ths, use the vectors E and J from the numerca exampe n the dscusson of tensor conductvty n ths chapter. See f the nner product E J s the same n both coordnate systems used. R, 6-4

15 Vector Spaces n Physcs 8/6/5 Probem 6-5. Return to the numerca exampe n the dscusson of tensor conductvty n ths chapter. Suppose that the eectrc fed E n the orgna system st has magntude E, but s drected aong the postve z axs; that s, E E (a) Cacuate the current densty J E n the unprmed system. (b) Now consder the same rotaton, usng R from the exampe n the text, and cacuate J E n the prmed system. (Note: has aready been cacuated n the notes.) (c) Now cacuate the current densty n the prmed system by the aternate method, J RJ. Compare wth the resut from part (b). 6-5