We can represent a vector (or higher-rank tensors) in at least three different ways:

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1 Phyc 106a, Caltech 27 November, 2018 Lecture 16: Rgd Body Rotaton, Torque Free Moton In th lecture we dcu the bac phyc of rotatng rgd bode angular velocty, knetc energy, and angular momentum ntroducng the moment of nerta tenor. We then olve a mple forcefree dynamcal ytem: the free ymmetrc top (e.g. a thrown football, or the Chandler wobble of the earth). Tenor We wll be computng the moment of nerta tenor for a rgd body. Frt, let explctly defne a tenor, n term of t properte under rotaton. A calar a rank-0 tenor; unchanged under rotaton. A vector a rank-1 tenor. Under a rotaton, a vector a tranform to the rotated vector: a a = U a, where U an orthogonal matrx: UŨ = 1, where Ũ the tranpoe of U. We can repreent a vector (or hgher-rank tenor) n at leat three dfferent way: abtract, geometrc, coordnate-free notaton a In a coordnate ytem, a vector repreented by a trplet of number (a x, a y, a z ) = (a 1, a 2, a 3 ) or a, = 1, 2, 3. In matrx language, a a column matrx wth three row. It tranpoe ã a row matrx. The nner product of two vector a calar: a b = a b = c (where we ue the Enten ummaton conventon, ummng over repeated ndce). Th alo known a a dot product or calar product. In matrx notaton, ãb = c, where c a 1 1 matrx. Under rotaton, a b a b = ãũub = ãb = c unchanged, nce ŨU = 1 the 3 3 unt matrx. A rank-2 tenor T repreented n coordnate notaton wth two ndce: T j ; n matrx repreentaton, t a 3 3 matrx. Under rotaton, t need two rotaton operator, one for each ndex: T T = U T Ũ. In component repreentaton: T j = U αt αβ Ũ βj. A rank-n tenor repreented n coordnate notaton wth N ndce. If N > 2, matrx notaton need generalzaton to more dmenon (not jut column and row). Under rotaton, one need a, orthogonal rotaton operator for each ndex: T 1 2 n = U 1 j 1 U 2 j 2... U nj n T j1 j 2 j n. (1) The nner product of a rank-n tenor wth a vector yeld a rank-(n-1) tenor. For example, the nner (dot) product of a rank-2 tenor wth a vector a vector, and the double nner product a T b = c a calar. Kp Thorne (ee Ph136a note) favor a lot notaton for tenor, that ndependent of coordnate. One can thnk of a 2nd-rank tenor a a geometrc object wth two lot: T (, ) uch that f you put a vector a n one lot: T (, a) you get a one-lot tenor, whch a vector. If you put vector n both lot: T ( a, b) you get a zero-lot tenor, whch a calar. A econd rank tenor ha two lot; an nth rank tenor ha n lot, and produce a calar from n nput 1

2 vector. Snce we know how vector tranform under rotaton (and calar do not) ether of thee relatonhp defne how the tenor change under rotaton. A tenor lnear n each of t nput lot: T (, α a + β b) = αt (, a) + βt (, b). One can form a mple rank-2 tenor by takng the outer product of two vector: T = a b, or Tj = a b j (no repeated ndce to um over). Under rotaton, T T = ( a ) ( b ) = (U a) (U b) = U() a bũ = U T Ũ. Example of tenor The moment of nerta tenor I that we wll tudy next a econd-rank tenor. A we wll ee, a rgd body that rotatng about a reference pont wth a rotaton vector ω wll have an angular momentum vector L = I ω that not, n general, n the ame drecton a ω. It wll have a (calar) knetc energy of rotaton T = 1 2 ω I ω (here, T a calar knetc energy, a zero-rank tenor). Becaue T quadratc n ω, t hould not be too urprng to ee a factor 1 2 n front of t. Another phycal example of a 2nd-rank tenor uch the electrc conductvty. In an anotropc crytal, the electrc current produced by an electrc feld not necearly along the feld. The conductvty tenor output the current vector j when we nput the electrc feld vector E. ω E L j L = I ω j = σ E T = 1 2 ω I ω P = 1 2E σ E Left: Moment of nerta tenor. Rght: Electrcal conductvty tenor. If you puh on a old body at a pont R 1 on the body, wth a force F, you wll produce a tre that decrbed by a 2nd-rank tre tenor σ kl. If the body ha ome elatcty (e, not perfectly rgd), It wll experence a tran, decrbed by a fractonal dplacement h at ome (n general, other) pont R 2, decrbed by a 2nd-rank tran tenor ɛ j. For mall tree, the lnear elatc tre-tran relaton nvolve the 4th-rank elatc complance tenor, uch that ɛ j = jkl σ kl. You wll encounter varou other mechancal, electrcal, optcal, thermodynamcal, etc tenor n your tude. 2

3 Equaton of moton for a rgd body A rgd body ha x degree of freedom correpondng to 3 tranlatonal and 3 rotatonal degree of freedom of moton. Non-rgd (elatc) bode alo have vbratonal moton; we wll conder that next week, but for now, we wll gnore t. The equaton of moton are the equaton for the total momentum and angular momentum. d P = F, d L tot = N, (2) wth P = M Rcm the total momentum gven by the ma tme the center of ma velocty, L tot the total angular momentum, F the force actng on the body and N the torque. The angular momentum and torque are calculated wth repect to ome fxed reference orgn. The angular momentum equaton wa derved n Lecture 1 for a et of partcle wth central force. In Agnment 7 you derve the reult for a rgd body ung d Alembert prncple wthout th aumpton (but aumng the contrant force do not contrbute to the vrtual work). A we wll ee, the total angular momentum, and the total knetc energy, wll eparate ncely nto the um over the tranlatonal moton of the center of ma of the entre body and the rotatonal moton relatve to the center of ma. Remember that the angular momentum defned relatve to a fxed pont and drecton, and we wll need to evaluate th for everal dfferent frame of reference: an nertal frame fxed n pace, the pace frame, wth repect to whch the body may be both tranlatng and rotatng; an nertal frame fxed on the body (another pace frame), wth repect to whch the body may be rotatng; a non-nertal frame fxed on the body and rotatng wth t (the body frame), wth repect to whch the body not movng (eg, u on the urface of the earth, whch both movng and rotatng wth repect to the un). We wll derve the equaton for rotatonal moton d L = N wth L = I ω and T rot = 1 2 ω I ω (3) 1. for pure rotaton and torque about a tatonary pont 2. for rotaton and torque about the center of ma Moment of nerta The moment of nerta tenor I calculated about the approprate pont I αβ = m (r 2 δ αβ r,α r,β ) (4) 3

4 Pure rotaton and torque about a tatonary pont For the pecal cae of a body undergong pure rotaton, and no tranlaton, about the tatonary orgn (e.g. a fxed pvot, or the ntantaneou pont of contact n a rollng problem), we have v, = ω r, wth ω the angular velocty and r, the dplacement of the th pont from the tatonary pont, and o L tot = l = r, (m v, ) = m r, ( ω r, ) = I ω, (5) wth I the moment of nerta tenor wth repect to the fxed orgn, wth unt of ma tme length 2. We can evaluate the trple cro product wth the dentty: A ( B C = B( A C) C( A B) (6) to get (omttng the ubcrpt everywhere): L x = [ m (y 2 + z 2 ] )ω x x y ω y x z ω z L y = [ m y x ω x + (z 2 + x 2 ] )ω y y z ω z L z = [ m z x ω x z y ω y + (x 2 + y 2 ] )ω z (7) (8) (9) In component notaton: L tot,α = I,αβ ω β, wth I,αβ = m (r 2,δ αβ r,,α r,,β ). (10) (orry about the cumberome ubcrpt notaton here: denote relatve to a tatonary orgn (pace frame), the partcle, and α the component). Reference pont movng wth body It often convenent to change to a reference pont movng wth the body. Ung r, = R + r, v, = V + ω r for ome reference pont K n the body wth poton wth repect to the pace frame orgn R and velocty V, gve L tot = m ( R + r ) ( V + ω r ). (11) The expreon mplfe f we take the reference pont to be the center of ma nce then m r = 0. Th gve L tot = R cm P + L, (12) where P = ( m ) V = M totv the total lnear momentum, and L the ntrnc angular momentum comng from the rotaton of the body L = m r ( ω r ) = I ω, (13) where I the moment of nerta tenor wth repect to the center of ma, wth component I αβ = m (r 2 δ αβ r,α r,β ), (14) 4

5 where we have made ome choce of the drecton of the trad of unt vector pecfyng the ba. The equaton of moton of the total angular momentum can be rewrtten d( R cm P + L) = ( R cm + r ) F. (15) Ung the equaton of moton of the total momentum, and Rcm P = 0 gve dl = r F. (16) Thu we may ue the equaton for rate of change of angular momentum and the torque calculated about a tatonary pont or about the center of ma, even f movng (but not any other movng pont). Knetc energy The knetc energy T = 1 2 choce of reference pont: m v 2 1. Reference pont the center of ma wth v = V + ω r for old body moton mplfe for two T = 1 2 M V 2 cm m ( ω r ) 2, (17) 2. Moton pure rotaton about a tatonary reference pont T = 1 m ( ω r ) 2. (18) 2 In ether cae the rotatonal knetc energy (whch from now on I wll call T ) T = 1 m ( ω r ) 2 = ω I ω = 1 2 I αβω α ω β (Enten ummaton), (19) wth I the moment of nerta ntroduced before. Moment of nerta tenor The moment of nerta about ome orgn ha component wth repect to the coordnate ba I αβ = m (r 2 δ αβ r,α r,β ), (20) wth r the poton of the th ma relatve to the orgn. You hould be able to evaluate the moment of nerta for mple ma dtrbuton and old bode (ee Hand and Fnch problem 8-3 to 8-11 for practce example). In matrx form: m (y 2 + z2 ) m x y m x z I = m x y m (z 2 + x2 ) m y z m x z (21) m y z m (x 2 + y2 ) For example, the 33 component I 33 = m (x 2 + y 2 ) Moment of nerta : number ma length 2 Vol 5 ρ(x, y, z)(x 2 + y 2 ) dx dy dz (22)

6 The dplaced ax theorem The dplaced ax theorem ay that the moment of nerta for a reference pont hfted by a from the center of ma I a,αβ = I cm,αβ + M(a 2 δ αβ a α a β ), (23) wth M the total ma. Th often ueful n the calculaton of I. Moment of nerta a tenor Under rotaton U of the coordnate axe the component of the moment of nerta tenor tranform a I µν = U µα U νβ I αβ, (24) (each ndex tranformng lke a vector) wth U the rotaton matrx relatng the bae. Thnkng of I αβ a a 3 3 matrx, th can be wrtten Prncpal component I = UIŨ or I = ŨI U. (25) Snce I a ymmetrc matrx, the theorem of lnear algebra tell u that we can fnd ome U uch that I dagonal I I = 0 I 2 0. (26) 0 0 I 3 The partcular choce of coordnate axe gvng th dagonal form are called the prncpal axe of the old body. Recallng lnear algebra, dagonalzng a matrx can be done n general by olvng the matrx egenvalue equaton (I λ1) Ω = 0, where 1 here the 3 3 unt matrx, λ the egenvalue(), and Ω the egenvector(). There wll be a oluton for arbtrary Ω ff the determnant I λ1 = 0. Th a cubc equaton n λ, yeldng three egenvalue oluton λ = I, = 1, 2, 3. The three egenvector Ω, normalzed, are orthogonal to one another, and are the prncpal axe of the body n the orgnal coordnate ytem. If needed (uually not), they can be tacked together to get U. Symmetre The moment of nerta tenor of all rgd bode can be dagonalzed n th way, even f the body not very ymmetrcally haped. If, however, there are ymmetry axe, the prncpal axe that dagonalze the moment of nerta tenor wll le along thee ymmetry axe. For example, an ellpod wth prncple axe of length a, b, c along the orthogonal drecton ˆx, ŷ, ẑ, repectvely, wth unform ma denty ρ, ha a total ma M = ρv = ρ 4 3πabc, and (dagonal, n the prncpal ax ba) moment of nerta tenor: I = M 5 (b 2 + c 2 ) (c 2 + a 2 ) 0. (27) 0 0 (a 2 + b 2 ) A prolate pherod (uch a an Amercan football) ha a = b and c > a, b o that I 1 = I 2 and I 3 < I 1,2. An oblate pherod (uch a the Earth, whoe dameter at the pole lghtly le than at the equator) ha a = b and c < a, b o that I 1 = I 2 and I 3 > I 1,2. 6

7 Example of rotatng rgd bode An Amercan football (prolate pherod). When thrown properly, t pn along t ymmetry ax. It only too eay for the pn to be a bt malgned wth the ymmetry ax. A pnnng top, wth a ymmetry ax and an arbtrary hape perpendcular to that ax (a old of revoluton). We wll conder a top wth a fxed vertex on the table, and gravty torqung t. A gyrocope. A perfectly phercal gyrocope would experence no torque and have a moment of nerta tenor that a multple of the unt matrx, along any trplet of orthogonal axe. The mot perfect man-made phere are the fued lca gyrocope made for Gravty Probe B; they are 1.5 nche n dameter and are phercal to a part n 10 7 or o. Spnnng neutron tar, wth the ma of the un but the dameter of Paadena, may be even more phercal than that; but I hope not, nce aphercty wll produce an acceleratng quadrupole moment that wll emt gravtatonal wave. The Earth, or other planet or moon, are typcally pnnng oblate pherod. They re not really rgd bode, o they wll have mode of vbraton. But they re not very elatc ether; the mode damp out dpatvely. The Earth ha a complcated tructure (lqud ocean, old contnent, molten core), o the pn ax (from the north to outh pole) n t even perfectly algned wth the ax of ymmetry; a we wll ee below, th lead to a Chandler wobble. The non-phercal dtrbuton of ma lead to torque and tdal effect, producng the preceon of the equnoxe and nutaton. Spacecraft are often bult wth ymmetry axe, and they rotate a they travel through pace. Even n the abence of torque, they experence all thee knd of rotatonal moton. Symmetrc top n free pace A a mple example conder a pnnng body wth axal ymmetry o that I 1 = I 2 I 3 wth no appled torque. Then the phyc mply that the angular momentum contant n an nertal frame, o that L = I ω = contant. (28) The ret of the oluton jut geometry. It trcky becaue the moment of nerta mot mply evaluated ung the prncpal axe of the body, but thee n turn are rotatng relatve to fxed axe (the pace frame) at the angular velocty ω. Here one way of gettng the oluton (ee fgure). Defne î, ĵ, ˆk a the prncpal axe wth the ymmetry ax along ˆk. Snce the body axally ymmetrc, at each tme we can chooe î to le n any drecton perpendcular to ˆk: chooe t to le n the plane of L (contant) and ˆk (at that tme). Then ĵ perpendcular to th plane and ω 2 = L 2 I 2 (29) (I dagonal for th ba) and th gve ω 2 = 0 nce L 2 = 0 (ĵ perpendcular to L). Hence ω alo le n the L, ˆk plane. Snce ω le n the L, ˆk plane, the moton dˆk/ of ˆk (comng from the rotaton at rate ω) perpendcular to the plane. Th mean that the angle θ between ˆk and L contant θ = 0. 7

8 L "! k Th mean that at the next tme ntant we can draw the ame pcture (wth a new choce of î). ω 1 = L n θ/i 1 and ω 3 = L co θ/i 3 are the ame, and o the ame argument apple. Thu the moton of ω, ˆk teady preceon about the drecton of L. The moton of the body ax gven by dˆk = ω ˆk = ω 1 î ˆk = L n θ î ˆk L = I ˆk. (30) 1 Snce L contant th correpond to teady preceon at the angular velocty ω P = L/I 1 whch a rotaton rate L/I 1 along L. Th preceon the wobble you ee f a non-expert throw a football wth the angular velocty and the angular momentum not algned. Th treatment dfferent from the one n Hand and Fnch. They defne the pn Ω of the body ubtractng off from ω 3 (the component of ω along the ymmetry ax) the part comng from the preceon ( 1 Ω = ω 3 ω P co θ = L 1 ) co θ. (31) I 3 I 1 Th a trange noton of the pn, nce t goe to zero for a pnnng phere I 1 = I 3. They tart off by aumng that Ω contant, wherea th hould be deduced from the analy baed on contant L (or the Euler equaton method, ee below), and they talk about the mng pece of the angular velocty, whch I fnd confung. They alo ue the conervaton of energy to argue θ contant, whch fne. Euler equaton A econd approach to evaluate the term n the torque equaton n term of tme dervatve n the body frame. Th gve the Euler equaton. For the top problem th approach partcularly convenent to undertand the moton from the perpectve of an oberver on the top. Recall that for any vector a, t tme evoluton n an nertal pace frame and n a body frame rotatng wth rotaton vector ω are related by d a = d a + ω a. (32) b I 1 8

9 In the nertal (pace) frame we have d L = N, (33) wth N the torque. Th gve n the body frame rotatng at angular velocty ω relatve to the pace frame dl = N ω L. (34) b Th Euler Equaton. Do not confue th wth the Euler-Lagrange equaton, even though t the ame Euler. Now wrte L n term of component along the prncple axe î, ĵ, ˆk o that L = I 1 ω 1 î + I 2 ω 2 ĵ + I 3 ω 3ˆk. In the body frame î, ĵ, ˆk are ndependent of tme, o that n component form (the component are wth repect to axe fxed n the body frame) I 1 dω 1 ω 2ω 3 (I 2 I 3 ) = N 1, (35) I 2 dω 2 ω 3ω 1 (I 3 I 1 ) = N 2, (36) I 3 dω 3 ω 1ω 2 (I 1 I 2 ) = N 3, (37) where the equaton are related by cyclc permutaton of the ndce. Thee are equaton of moton for the component of ω wth repect to the prncpal axe of the body, whch are themelve rotatng relatve to the pace frame wth the angular velocty ω. Tranlatng back to the pace frame only geometry, but t can be qute complex; a we wll ee, t bet, n that cae, to ue the Euler-Lagrange equaton wth carefully choen angular coordnate (the Euler angle). Note that f the pn vector ω algned wth a prncpal ax (a n a perfectly thrown football), then only one of the three component ω 1, ω 2, ω 3 non-zero. In that cae, and n the abence of torque ( N = 0), ω contant. At leat two of the component of ω mut be non-zero to get a changng ω n the body frame. Symmetrc top n free pace from the body perpectve: Chandler wobble What would we ee f we were ttng on the wobblng football? Well, we are! Ignorng the mall effect of gravtatonal force actng on the aphercal ma dtrbuton, the Earth effectvely a ymmetrc top n free pace and t oblate I 3 > I 1 = I 2. We could tranfer the reult we have calculated to the body frame (ee below), but let ntead redo the calculaton from cratch. In the body frame we can convenently ue Euler equaton, and nce the top free, N = 0. The equalty I1 = I 2 (let call th I ) gve ω 3 = contant from the thrd equaton. Then the other two equaton become dω 1 dω 2 [( ) ] I3 = 1 ω 3 ω 2, (38) I [( ) ] I3 = 1 ω 3 ω 1. (39) I The quantty nde the [ ] contant, and o thee equaton can be olved to gve ( ) I3 ω 1 = A co(ω p t + φ), ω 2 = A n(ω p t + φ) wth Ω p = 1 ω 3, (40) I 9

10 wth A, φ ntegraton contant. Thu the angular velocty vector teay prece about the ymmetry ax at the rate Ω p. Note that Ω p potve for the oblate earth, negatve for the football. The Earth pn ax almot perfectly algned wth a prncpal ax of the moment of nerta tenor, wth ω 3 = 2π/P, where P the perod of the Earth rotaton (whch one dereal day, 24*(1-1/365) = hour. Becaue the Earth ha a complcated nternal tructure (ncludng a rotatng molten core), t lghtly malgned: the prncpal ax approxmately 10 meter away from the north pole (and changng wth tme). For the Earth, (I 3 /I 1) The calculaton predct we hould ee the rotaton ax prece about the geometrc NS ax wth a perod of about 306 day. Such an effect oberved, and called Chandler wobble, but the perod about 435 day, apparently due to the dynamc deformaton of the earth due to the tde. Hand and Fnch dcu th more n For obervaton of the Chandler Wobble ee Celetal pole; Chandler Wobble: x-y plot; Chandler Wobble: x(t) plot From: (0.5 arcec 15m at the pole) Wobble a een from the body frame and a pace frame The wobble moton calculated n the body (Earth) frame the ame a dcued for the wobblng football, but the reult look rather dfferent. For example, for I I 3 the Chandler frequency Ω p get mall, but the football wobble frequency L/I jut the frequency of the pn of the football and reman large. Here one way to relate the dcuon n the two frame. In the body frame we have d ω = Ω pˆk ω, b dˆk = 0. (41) b Tranformng to the tme dependence n the pace frame n the uual way gve d ω dˆk = Ω pˆk ω, = ω ˆk, (42) (the equaton for ω unchanged of coure nce ω ω = 0). Thee equaton are hard to nterpret, nce ω rotate about ˆk, and ˆk rotate about ω. However, ntroducng ω p = Ω pˆk + ω, (43) 1 Note that ther Fg. 8.8 a lttle mleadng n that the wobble only about 10 meter, nothng lke a large a mpled by the ketch. The Z ax n ther pcture the drecton of the contant angular momentum, and o th too very cloe to the ymmetry ax and not perpendcular to the plane of the orbt, a you mght gue from the fgure and your knowledge of the tlt of the ymmetry ax. 10

11 the equaton can be wrtten d ω = ω p ω, dˆk = ω p ˆk, (44) (the extra pece gve zero on takng the cro product) o that both vector rotate at ω p. Furthermore, evaluatng the component of ω n term of L, ( 1 ω p = 1 I 1 I 3 ) ( L n θ L co θˆk + I 1 î + L co θ ) L ˆk =, (45) I 3 I 1 o that the moton of both vector teady preceon at angular frequency L/I 1 about the (contant) angular momentum n the pace frame. The algebra ω = ω p +( Ω p )ˆk decrbed pctorally n Fg. 8.7 of Hand and Fnch (cf. the decompoton of the angular velocty of the cone rollng on a plane you nvetgated n Agnment 7). Alternatvely, let tart wth the equaton for the teady preceon of ω n the pace frame d ω L = ω, (46) I 1 and go to the body frame. The tme dervatve of ω the ame n the body frame, but Eq. (46) le ueful there, nce L not contant n that frame. Intead ˆk contant and we would lke to expre the moton of ω n term of rotaton about th drecton. From L = I ω we can wrte L = I 1 ω + (I 3 I 1 )(ˆk ω) ˆk. (47) The frt term doe not contrbute to the cro product n Eq. (46) and o we get ( ) d ω I3 = 1 ω 3 ˆk ω = ΩP ˆk ω = ωp ω, (48) b I 1 gvng the unform preceon of ω about ˆk at the rate Ω p (note that Eq. (48) how that ω 3 contant). To ummarze: n the pace frame, both ˆk and ω rotate wth angular velocty ω p = Ω pˆk + ω. In the body frame, ω rotate wth angular velocty Ω pˆk or ωp (net effect the ame), whle ˆk (of coure) contant. 11

12 Free ymmetrc top: pace and body frame (a) pace frame (b) body frame L 3 k p t k j L ω p = ( ) Ωp I3 = 1 ω 3ˆk I 1 I 1 Free ymmetrc top: rollng cone pcture, wth ω p = Ω pˆk + ω Hand and Fnch Fg

13 Appendx: Torque-free moton of an aymmetrc top Ponot contructon* (Jut for fun: not for examnaton) For an aymmetrc top, the moment of nerta component wth repect to the prncpal axe are I 1, I 2, I 3, all dfferent. The torque free moton n the pace frame can be contructed a follow (ee Hand and Fnch 8.9 for more detal): L contant, and ω evolve o that the knetc energy T a contant 1 2 (I 1ω I 2 ω I 3 ω 2 3) = T. (49) Th the equaton for an ellpod n ω pace wth em major axe 2T/I 1 etc. Ue the component form L α = I αβ ω β and T = 1 2 I αβω α ω β to how ω contant T ellpod L = ω T, (50) wth ω the gradent wth repect to ω. Th mean that L perpendcular to the contant T ellpod at each tme. Snce L contant n the torque free moton, th gve a contrant on the tme evoluton of the ellpod. The component form alo how ω L = 2T = contant. (51) L plane contant L The T ellpod rotate wth angular velocty ω about the contact pont wth the plane perpendcular to L, mantanng the heght of the center contant. Thee equaton how that the moton of the prncple axe (and therefore the body) gven by rollng the contant T ellpod about the pont of contact wth the plane perpendcular to the contant vector L, mantanng the heght of the center fxed (Eq. (51)), at a rate gven by the angular velocty vector, whch the vector jonng the center of the ellpod to the contact pont. Thee argument would, of coure, be vald for the ymmetrc top, and o provde a dfferent approach to that problem too. 13