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1 COPYRT NOTCE: TRBBE: Pnceon ue o Avance Physcs s publshe by Pnceon Unvesy Pess an copyghe, (c) 996, by Pnceon Unvesy Pess. All ghs eseve. Ths ex may be use an shae n accoance wh he fa-use povsons of US copygh law, an may be achve an esbue n eleconc fom, pove ha hs ene noce s cae an pove ha Pnceon Unvesy Pess s nofe an no fee s chage fo access. Achvng, esbuon, o epublcaon of hs ex on ohe ems, n any meum, eques he consen of Pnceon Unvesy Pess. o COURSE PAC PERMSSONS, efe o eny on pevous menu. o moe nfomaon, sen e-mal o pemssons@pupess.pnceon.eu

2 .0 CASSCA MECANCS. UNDAMENTA TECNQUES.. The Val Theoem The equaon of moon of a sysem can be wen n he fom We ae neese n he quany Dffeenang hs expesson gves om Equaon. hs euces o p = 0. (.) = p. (.) = p + p = m + p. (.3) = T+. (.4) Aveagng ove a peo of me τ, we oban 0 τ τ 0 τ = τ ( ) ( ) = T +. (.5) f he moon s peoc, wh τ = peo, hen hen lef han se of Equaon.5 s eo an we see ha T =. (.6)

3 .. D'Alembe's Pncple om Equaon. follows ha he vual wok one by hs sysem s also eo, p δ = 0. (.7) The oal foce wll be a combnaon of exenally apple foces an nenal consans, so ha Equaon.8 euces o a = + f, (.8) a e p δ f δ = 0. (.9) f we esc ou aenon o g boes an ohe sysems fo whch he foces of consan o no wok hen we conclue ha he conon fo equlbum of a sysem s gven by D'Alembes pncple whch saes a e p δ = 0. (.0)..3 agange's Equaon f s a funcon of nepenen vaables q, hen δ = δ q q. (.) Doppng he supescp "a" fo convenence, he fs em fom Equaon.0 s δ = δq = Qδq, (.) q,

4 whee Q s he geneale foce. The secon em n Equaon.0 s By efnon an Smlaly p δ = m δ = m q, m = m m q q v q = q = q δ q. (.3) q, (.4). (.5) v = q q k + (.6) k k so follows ha v q =. (.7) q Usng Equaons.5 an.7 n Equaon.4 we fn ha m q = mv v q mv Equaons. an.8 ae combne o gve q mv q mv Q q δ v. (.8) q = 0, (.9)

5 o T q T q Q q δ = 0. (.0) Consequenly, T q T q Q =. (.) f V =, hen Q q V q V q = = =. (.) Equaon. s equvalen o T q q T V ( ) = 0. (.3) We efne =T V o be he agangan. f V s no a funcon of me hen q q = 0, (.4) whch s known as agange's equaon, can be use o eemne he equaons of moon. We also efne p q =, (.5) o be he canoncal, o conugae, momenum. f he agangan oes no conan a gven coonae q hen he coonae s sa o be cyclc o gnoable.

6 ..3. The Two Boy Cenal oce Poblem As an example of he applcaon of agange's equaon, conse a sysem of wo mass pons m an m subec o an neacon poenal V, whee V s any funcon of he veco beween he pacles. The knec enegy s T = m + m. (.6) The knec enegy can also be wen as he knec enegy of he cene of mass plus he knec enegy abou he cene of mass. We efne R = poson of he cene of mass, an = = veco beween m an m. The knec enegy of he cene of mass s gven by Tcm = ( m + m) R. (.7) Relave o he cene of mass, he poson of m an m ae gven by an ' ' m = m + m, (.8) m = m + m. (.9) Theefoe, he knec enegy abou he cene of mass s gven by Consequenly, he agangan s T ' = mm m m m m ' + ' = +. (.30) = mm m m R m m ( + ) + + V (,,...). (.3) We see mmeaely ha because he poenal s only a funcon of he veco beween he pacles he conugae momenum of R, (he momenum of he cene of mass), s consan. Tha s, he moon of he cene of mass has no

7 effec on he moon abou he cene of mass. Ths also mples ha hee wll be no ou of plane moon...3. The nvese Squae aw of oces When V s a funcon of only, as s he case fo gavaonal o elecosac foces, Equaon.3 may be expesse n pola coonaes as = µ e + θ V( ). (.3) Noe ha we have chosen o gnoe he em escbng he moon of he cene of mass snce has no effec on ohe paamees an we have nouce he efnon mm µ= m + m, (.33) whee µ s eme he euce mass. The equaons of moon ae foun fom agange's equaon, (Equaon.4). o he vaable q = θ we have whle fo q = we have ( µ θ) θ θ = = 0, (.34) V µ µ θ = + = 0. (.35) o many poblems of nees, such as obal mechancs, m >> m an µ m. The physcal consequence s ha he smalle pacle, m, s subece o he lages peubaon n s moon. om hs pon fowa we wll follow he usual convenon an eplace µ wh he symbol m wh he unesanng ha efes o he moon of he smalle of m an m abou he cene of mass. Equaon.34 s he saemen of consevaon of angula momenum. Tha s, l = m θ, (.36)

8 s a consan. Equaon.36 can be ewen n he fom whch mples an l = m θ, (.37) l = m l = m θ, (.38) l m θ θ. (.39) The aea swep ou by a movng boy s gven by A= bθg. (.40) follows ha A l = m θ M N O l b θg =. (.4) P Q Because angula momenum s conseve A/ s also consan. Ths s eple's Secon aw whch saes ha he planes sweep ou equal aeas n equal mes. om he efnon of l, equaon.35 becomes m l m 3 V =, (.4) o, because V s only a funcon of, m = V + l m. (.43)

9 The pacle moves n an effecve poenal gven by l Veff = V + m. (.44) Equaon.43 euces o Veff m =, (.45) hus m = m Veff Veff = =. (.46) Consequenly, m + V eff = 0, (.47) whch s he saemen ha enegy s conseve. om hs equaon, we also fn ha o smply m E V m E V l =± eff =±, (.48) m = m E V l m. (.49) Subsung he elaon beween an θ, Equaon.37, an we fn ha θ= me mv l l. (.50)

10 k Conse he case when he poenal s of he fom V = = ku. Equaon.50 becomes θ= me mk + u u l l u. (.5) negang hs expesson gves θ f θ = acsn mk u+ l 8mE mk + l l l mk = acsn. (.5) El + mk nveng hs expesson gves R mk El = M + S l M mk T N Ths s usually wen n he fom whee O P P Q U snθf θ V. (.53) W mk = { e snθ f θ }, (.54) l El e = + mk. (.55) Ths s he equaon fo a conc secon havng seveal classes of soluons as shown below..) f e >, an E > 0, he ob s a hypebola..) f e =, an E = 0, he ob s a paabola.

11 3.) f e <, an E < 0, he ob s an ellpse. mk 4.) f e = 0, an E = l, he ob s a ccle.. VARATONA TECNQUES.. The Calculus of Vaaons Conse a funcon f ( y, y, x) efne on a pah y=y(x) beween x an x whee y = y/ x. We wsh o fn a pacula pah y(x) such ha he negal x = f( yyxx,, ), (.56) x has a saonay value elave o pahs ffeng nfnesmally fom he coec funcon y(x). Snce mus have a saonay value fo he coec pah elave o any neghbong pah, he vaaon mus be eo elave o some pacula se of neghbong pahs. Such a se of pahs can be enoe by y( x, α)= y( x, 0 )+ αη( x), (.57) whee y(x, 0) s he coec pah an η(x ) η(x ) = 0. Explcly, x ( α)= f ( y( x, α), y( x, α), x) x. (.58) x A necessay conon fo a saonay pon s om Equaon.58 α α = 0 = 0. (.59)

12 x x f y f y = + x. (.60) α y α y α s easly seen ha f y f y =, (.6) y α y xα so ha x x f y x x x f y f y f y = x = x y α y α x y α x y. (.6) α x x om he bounay conons, he fs em on he gh han se vanshes an Equaon.60 euces o x x f f = α y x y x x y x. (.63) α The funamenal lemma of he calculus of vaaons saes ha f x M( x) η ( x) x = 0, (.64) x fo all η(x) connuous hough he secon evave hen M(x) mus be encally eo on he neval. Thus s saonay only f f f s a funcon of many nepenen vaables hen f f y x y = 0. (.65) f y f x y = 0. (.66)

13 .. amlon's Pncple amlon's pncple saes ha he moon, n confguaon space, of a sysem whee all non-consanng foces ae evable fom a geneale scala poenal ha may be a funcon of coonaes, veloces, an me s such ha he negal =, (.67) has a saonay value fo he coec pah of he moon. Tha s, δ = δ = 0. (.68) The negal s eme he acon, an amlon's pncple saes ha he vaaon n s eo. n ohe wos, he acon s mnme. By compason wh Equaon.65 follows ha q q = 0, (.69) whch s agange's equaon...3 agange Mulples D'Alembes pncple, an he esulng fom of agange's equaon, assume no consan foces. Conse a eamen when he equaons of consan can be pu n he fom alkqk + al = 0, (.70) k whee he a lk an a l 's may be funcons of a,. o vual splacemens follows ha

14 f hs s ue, hen mus also follow ha λ alkδ qk = 0. (.7) k a δq = 0, (.7) l lk k k whee he λ l ae uneemne coeffcens calle agange mulples. om Equaons.63,.68, an.69 s seen ha amlon's pncple s equvalen o By he same pocess, Equaon.7 s equvalen o k δqk = 0. (.73) q q k k λlalkδqk = 0. (.74) kl, We combne hese wo elaons o oban n = k q k + q k l λ a l lk δq = 0. (.75) k The δq k 's ae no necessaly nepenen, bu because he values of he λ l 's ae uneemne we may choose hem such ha q k λlalk q =. (.76) Thes equaons, ogehe wh Equaon.70, can be use o eemne he equaons of moons fo sysems wh consanng foces. k l

15 Example. Conse he case of a lae of lengh ha s nclne agans a fconless wall an floo as shown a gh. n he equaons of moon. θ The poson of he cene of mass of he lae, an s oenaon, can be escbe wh he vaables x, y, θ. The moon of he lae s consane by he wall an floo. We have he wo consans x = snθ, (.77) an y = cosθ. (.78) om Equaon.60 follows ha hese gve he consanng elaons an λ λ M N M N especvely. By nspecon, he knec enegy s O x cos θθ 0 P =, (.79) Q O y + sn θθ 0 P =, (.80) Q T = m x + y + e θ, (.8) whee = m. Smlaly, he poenal enegy s V = mgy, (.8)

16 so ha he agangan s m = m x + y e + θ mgy. (.83) 4 om Equaons.76,.79, an.80 he equaons of moon ae mx =λ, (.84) my + mg =λ, (.85) an m 4 cos θ= λ θ λ sn θ +, (.86) especvely. om Equaons.77 an.78 we see ha sn x = θθ + cos θθ, (.87) an cos y = θθ sn θθ. (.88) om Equaons.84 an.87 we see ha λ = m θθ θθ sn + cos, (.89) whle fom Equaon.85 an.88 we see ha λ = m θθ θθ cos sn + g. (.90) When hese ae combne wh Equaon.86 an smplfe we oban θ= e 6 sn θθ + cos θθ b cos θg e cos θθ sn θθ + g b sn θg, (.9)

17 whch s equvalen o 5 θ= 6 sn θ cos θθ g sn θ. (.9). RD BODY MOTON.. Roaons A g boy n space nees 6 nepenen geneale coonaes o specfy s confguaon. o example, 3 coonaes ae neee o specfy he locaon of he cene of mass elave o some exenal axes an 3 ohe coonaes ae neee o specfy he oenaon of he boy elave o a coonae sysem paallel o he exenal axes. The oenaon s specfe by sang he econ cosne of he boy axes elave o he exenal axes. Tha s, f he pme enoes boy axes hen $ $ $ $ $ $ $' = $' + $' + e$' k k, (.93) o $ $ $ $' = α + α+ α3 k, (.94) o $ $ $ $ $ $ $' = cos $', + cos $', + cos e$', k k. (.95) Smlaly, whee R = x'= Rx, (.96) α α α β β β γ γ γ (.97)

18 Because x' = x, (.98) we have RR =, (.99) o α + β + γ =, (.00) fo l =,, 3. Tha s, x'= ax, (.0) an x' x'= aakxxk = xx. (.0) Theefoe α α k =, f = k, α α k = 0 ohewse. n wo mensons R = cos φ snφ snφ cos φ. (.03)... eneal Popees of Roaons f = A an we ansfom o a new coonae sysem hen - B = BA = BAB B we say ha BAB - s he fom of A n he new coonae sysem. A'= BAB - efnes a smlay ansfomaon. n some coonae sysem cos φ snφ 0 A' = snφ cos φ 0, (.04) 0 0 so ha TA = + cos φ. Ths popey hols ue n all coonae sysems.

19 Example. n he axs of oaon an compue he angle of oaon fo R = T = can easly be vefe ha RR, so R sasfes he equemens of a oaon max. The axs of oaon can be efne as he econ ha any veco whch emans unchange by R pons. Tha s, f x = Rx, hen x s he axs of oaon. We mus solve whch gves he equaons x 6 x 3 4 = bx x x g, (.05) 3 x3 4 3x+ 6x + x3 = 4x, (.06) 6x+ x + 6x3 = 4x, (.07) x 6x + 3x3 = 4x3. (.08) These hee equaons can be solve o show ha x = x 3 an x = 0. The nomale axs of oaon s heefoe b0,, g. By examnaon, T R = + cos θ =, so ha θ = 60º.

20 ... The Eule Angles Rahe han specfy he 9 nepenen elemens of he oaon max we may escbe he oenaon n ems of 3 Eule angles. o example, oae he nal sysem of axes, xy, by an angle ψ couneclockwse abou. Ths efnes he ξηζ axes. Nex oae abou he ξ axs by an angle θ n he couneclockwse econ. Ths efnes he ξ'η'ζ' axes. nally oae couneclockwse by an angle φ abou he ζ' axes. Ths efnes he ξ'ψ'ζ' axes. n max fom whee an B = C = D = x'= Ax, (.09) A= BCD, (.0) cos ψ sn ψ 0 snψ cos ψ 0, (.) cosθ snθ, (.) 0 snθ cos θ cos φ snφ 0 snφ cos φ 0. (.3) The Cayley-len Paamees Conse a geneal lnea ansfomaon n mensonal space an u'= αu+ β v, (.4)

21 whee he ansfomaon max s v'= γu+ δ v, (.5) Q = α β γ δ. (.6) Noe ha α, β, γ, δ may be complex. f we eque fn ha β = γ* an δ = α*, ha s QQ = an Q =+ we α β Q = β* α*. (.7) Conse a max of he fom P x y = x+ y, (.8) such ha P'= QPQ. (.9) The heman popey an he ace of a max ae unaffece by smlay ansfomaons. Consequenly, P' s of he fom f we le x+ = x+ y an x = x y, hen ' x' y' P' =. (.0) x' + y' ' x P' = ' ' x ' ' = α γ + β δ x + x δ β γ α. (.) n hs way, we may efne a 9 elemen oaon max n ems of 4 Cayley- len paamees.

22 .. The Rae of Change of a Veco The ae of change of a veco as seen by an obseve n he boy sysem of axes wll ffe fom he coesponng change as seen by an obseve fxe n space. f he boy axes ae oang wh angula velocy ω he geneal soluon s We have = + ω. (.) space boy an a successve applcaon of Equaon. gves v = v +ω, (.3) s b ω as = ab +( ω vb) + ω ( ω ) +. (.4) f he angula velocy of he boy s consan Newon's law s = mas = mab + m( ω vb)+ mω ( ω ). (.5) To an obseve n he oang sysem appeas as hough he pacle s movng une an effecve foce eff = mab = m( ω vb ) mω ( ω ). (.6) The secon em on he gh han se s calle he Cools foce an he las em on he gh han se s calle he cenfugal foce.

23 Example.3 Show ha f a pacle s hown up vecally wh nal spee v o an eache a hegh h, ha wll expemen a Cools eflecon ha s oppose n econ, an fou mes geae n magnue, han he eflecon woul expeence f wee oppe a es fom he same maxmum hegh. The ae of oaon of he Eah, ω, s encal fo each scenao. The ffeence s he nal velocy an poson of he pacle. om equaon.6, he Cools foce s bω v b g. Defnng he coonae axes as shown, we have ωx = ωy =0 an ω = ω, whle v x = 0, vy = vcosθ, an v = v snθ. As a esul, x s s ω θ x b b y s y b bω v $ $ g= + bωvcos θg + 0 b g+ 0 b gk$, (.7) b As a esul, he acceleaon ue o he Cools foce s smply x = ωvcosθ. (.8) A boy fallng une he nfluence of gavy sasfes he conon v = vb0g g, (.9) whee v(0) s efne as he velocy a me = 0. Subsung hs no Equaon.8 an negang gves x = cos O ω θ vb0g g. (.30) M P N Q negang a secon me gves 3O x = ωcosθ vb0g g, (.3) M P 6 N Q

24 To solve he poblem a han, we le be he me akes fo he pacle o complee he p. The poson of he pacle a any me s gven by h b g= hb0g+ vb0g g, (.3) n one case, he pacle sas a es, v(0) = 0, a hegh h. om he efnon of he poblem, can be seen ha he me eque o complee he op o he goun s h vo = =, (.33) g g Consequenly, he eflecon when oppe fom es a hegh h s x g v 3 3 o O vo = g 6 g 3 g N M Q P = ωcosθ ωcos θ. (.34) f nsea, we le be he me akes o fall fom hegh h, we have v = v o an =. The eflecon s x 3 3 vo vo O 4 vo = vo g g 6 g 3 N M Q P =+ ωcos θ ωcos θ. (.35) g Thus, he eflecon when hown fom he goun up s oppose n econ, an fou mes geae n magnue, han hen eflecon when oppe fom es...3 The Rg Boy Equaons of Moon We have pevously seen ha he oal knec enegy of a g boy may be expesse as he sum of he knec enegy of he ene boy as f concenae a he cene of mass, plus he knec enegy of he moon abou he cene of mass. f a g boy moves wh one pon saonay, he angula momenum abou he cene of mass s

25 Snce s a fxe veco n he boy Thus = m ( v). (.36) v =ω. (.37) m m = ( ( ω ))= ( ω ( ω)). (.38) The x-componen s x x y = ω m ( x ) ω m x y ω m x, (.39) o x = xxωx + xyωy + xω. (.40) We efne k = ρ ( )( δ k x xk ) V, (.4) v o be he nea enso so ha The knec enegy of moon abou a pon s = ω. (.4) T = mv mv = ( ω ), (.43) o ω T = = ω ω. (.44)

26 The momen of nea,, abou some gven axs s elae o he momen abou a paallel axs hough he cene of mass. e he veco fom he ogn o he cene of mass be R an le he a vecos fom he ogn an he cene of mass o he h pacle be an ' especvely. Tha s a b R CoM The momen of neal abou axs a s o = R+ '. (.45) a = m( n$ ) = m [(R+ ') n$ ], (.46) a = M( R n$ ) + m ( '+ n$ ) + m( R n$ )( ' n$ ). (.47) The las em s ( R n$ ) ( n$ m '), bu m ' =0 an b = m( ' n) $ so a = b + M( R n$ ). (.48)..3. The Eule Equaons an Toque-ee Moon By efnon n he boy sysem, usng = ω, we fn N = + ω =. (.49) s ω b + ε ω ω = N. (.50) k k k

27 The Eule equaons ae ω ω ω ( ) = N, (.5) 3 3 ω ω ω ( ) = N, (.5) 3 3 ω ω ω ( ) = N. (.53) The pncple axes of a sysem ae hose whee he enso s he case, T = s agonal. f hs. (.54) Because T s consan, he elaon efnes an ellpso fxe n he boy axes. Because angula momenum s conseve mus be on a fxe sphee efne by 3 = + +. (.55) o he gven nal conons, knec enegy an angula momenum, he pah of s consane o be he nesecon of he sphee,, an he ellpso, T. f we have a boy symmecal abou he axs so ha = he Eule equaons ae, n he absence of oque's, We have ω 3 = consan, an whee ω =( ) ω ω, (.56) 3 3 ω = ( ) ω ω, (.57) 3 3 ω 3 ω 3 = 0. (.58) Ω ω, ω Ω ω, (.59) = = Ω= 3 ω 3. (.60)

28 Thus, ω = Ω ω, (.6) whch has soluons ω = Acos Ω an ω = AsnΩ. ence, he oal angula velocy s consan n magnue bu pecesses wh fequency Ω abou he ζ- axs. We may solve fo A an ω 3 by nong ha an T = A + 3ω, (.6) = A + ω. (.63)..3. The eavy Symmecal Top Conse a heavy op wh a symmey axs aken o be he -axs of he coonae sysem fxe n he boy as shown. The 3 Eule angles ae: θ = nclnaon of he -axs fom he vecal, φ = amuh of he op abou he vecal, an ψ = oaon angle of he op abou s own -axs. The knec enegy s o The poenal enegy s T = ( ω + + ω ) 3ω, (.64) 3 T = ( θ + φ sn θ) + 3( ψ + φ cos θ ). (.65) so ha he agangan s gven by V = mgl cosθ, (.66)

29 = ( θ + φ sn θ) + 3( ψ + φ cos θ) mgl cosθ. (.67) We have p ψ = = ( ψ + φ 3 cos θ) = 3ω3 = a, (.68) ψ p φ = = ( sn θ+ cos θ) φ + ψ cos θ= b 3 3, (.69) φ an E = T + V = ( θ + φ sn θ) + ω + Mgl cos θ. (.70) 3 3 Combnng hese fs wo expessons gves cos θ φ = b a, (.7) sn θ an a b acosθ ψ = cos θ. (.7) sn θ 3.3 OSCATORY MOTON.3. Oscllaons Conse a sysem ha s subece o a poenal ha s only a funcon of coonaes. f we expan he coonaes q abou he equlbum poson accong o q = q +η, (.73) o an hen pefom a Taylo sees expanson on he poenal we oban

30 V V V( q)= V( q )+ η o q + qq o + o ηη (.74) The fs em on he gh han se may be eefne o be eo by shfng he eo poenal o be he equlbum value. Smlaly, f he geneale foces ae eo he secon em s also eo. As a esul, he poenal can be appoxmae by he max elaon Smlaly, we can efne so ha he agangan s V V( q) qq o ηη. (.75) T = m q q = m ηη = T ηη, (.76) = T ηη V ηη. (.77) The equaons of moon ae gven by T η + V η = 0. (.78) ω f we y a soluon of he fom η = Ca exp we fn ha whch s equvalen o V a ω T a = 0, (.79) V ω T = 0. (.80) Example.4

31 Two pacles move n one menson a he uncon of hee spngs as shown. The spngs all have unseche lengh a an foce consans as shown. n he fequences of he nomal moes of oscllaon. m m x x We efne he coonaes x an x o escbe he splacemen of he wo blocks, elave o he lef aachmen pon. n hese coonaes T = mx + mx. (.8) Expanng abou equlbum, we efne x = a + η, x = a + η. Equaon.8 s equvalen o whee T = mη + mη = T η η, (.8) m T = 0 0 m. (.83) Smlaly, V = k x a + k x x a + k a x 3 b g b g, (.84) o V = V ηη, (.85) wh V V = xx a. (.86)

32 By examnaon V x = kbx ag 3kbx x ag, (.87) V x = k + 3k = 4k, (.88) an V x = 3kbx x ag kba x g, (.89) V x = 3k + k = 4k. (.90) kewse, he coss ems ae V x x V = = 3k, (.9) x x so ha V = 4k 3k k k. (.9) 3 4 The secula equaon s whch euces o V ω T = 4k ω m 3k 3k 4k ω m = 0, (.93) 4 e4k ω m 9k = m ω 8kmω + 7k = 0. (.94) om he quaac equaon, hs has soluons

33 ω = k m, (.95) an 7k ω = m. (.96).4 AMTON'S EQUATONS.4. egene Tansfomaons an amlon's Equaons of Moon Conse a funcon f of wo vaables such ha f = ux + vy, (.97) f whee u = f an v =. We wsh o change fom he vaables x, y o he x y vaables u, y. We efne a funcon We have g = f ux. (.98) g = f ux ux = vy xu, (.99) g so ha x = u an v g =. Ths s an example of a egene ansfomaon. y s fequenly use n hemoynamcs. Recall ha agange's equaons ae q whee = ( q, q, ), an he conugae momena ae q = 0, (.00)

34 p q q = (,,). (.0) q We can ansfom fom he vaables ( q, q,) o he vaables ( q, p, ) by he use of ( q, p, ) = q p ( q, q,), (.0) whee s calle he amlonan. We have oweve, fom he efnon q q p p = + +. (.03) p q q p q q q q = +, (.04) o smply q p q q =. (.05) Compang he wo expessons fo we oban amlon's equaons q = = p, (.06) q p = q, (.07) =. (.08) f he equaons efnng he geneale coonaes on' epen on me explcly, an f he foces ae evable fom a consevave poenal V, hen = T + V = E. (.09)

35 f we efne a column max η wh n elemens such ha η fo n, hen an = q, η n p + =, η = q, (.0) η + n = p. (.) f we efne = 0 0, hen amlon's equaons may be wen n he fom Ths s calle symplecc noaon. η =. (.) η Example.5 The agangan fo a smple spng s gven by = mx kx. n he amlonan an he equaons of moon usng he amlonan fomulaon. enfy any conseve quanes. om he efnon of p we have px = x =mx. (.3) om he efnon of he amlonan, equaon.7, we see ha = x bmx g mx + kx, (.4)

36 o smply om amlon's equaons we have m p = x + kx. (.5) q p = kx = p, (.6) x px = = m x, (.7) = 0. (.8) The fs equaon s he equaon of moon n one menson, mx + kx =0, (.9) he secon equaon s he efnon of momenum, an he las equaon s he saemen of consevaon of enegy..5. Canoncal Tansfomaons f a geneale coonae q has consan conugae momena s sa o be cyclc. f hs s he case, hen p = 0, whch ells us ha he amlonan s nepenen of ha p. f all coonaes q ae cyclc he conugae momena can be efne by p = α. Consequenly, q = = ω, (.0) α o q = ω + β. (.) A poblem s ofen ease o solve f we can fn a sysem whee he numbe of cyclc coonaes s maxmum. ow o we ansfom o hs se of

37 coonaes? We nee a new se of coonaes Q, P, whee Q = Q bq, p, g, P = P bq, p, g. We eque Q, P o be canoncal coonaes. Theefoe, some funcon = bq, P, g exss such ha Q =, (.) P an P = Q. (.3) f Q, P ae canoncal coonaes hey mus sasfy a mofe amlon's pncple ha can be pu n he fom δ PQ ( Q, P, ) = 0, (.4) because he ol coonaes sasfy δ pq ( q, p, ) g = 0. (.5) b Boh equemens can be sasfe f we eque a elaon λ pq PQ b g = +, (.6) whee λ = consan an s any funcon of he phase space coonaes connuous hough he secon evave. λ s elae o a scale ansfomaon. f λ = he elaon efnes a canoncal ansfomaon. The funcon s eme he geneang funcon. may be a funcon of q, p, Q, P, an efnes he ansfomaon..5.3 Symplecc Tansfomaons an Posson Backes

38 Recall ha amlon's equaons can be wen n he fom η =. (.7) η f we have a canoncal ansfomaon fom η ξ = ξ( η), hen ξ ξ η =. (.8) η n max fom, ξ = M η, (.9) whee M = ξ η. We have ξ= M, (.30) η also η ξ =, (.3) ξ η an = M. (.3) η ξ Consequenly, Theefoe, a ansfomaon s canoncal f ξ= MM =. (.33) ξ ξ MM =. (.34)

39 The Posson backe of a funcon s efne by We have u uv, v u PB = q p p v. (.35) q ηη, PB =, (.36) an ξξ, PB ξ ξ = = MM =. (.37) η η n ohe wos, he funamenal Posson backes ae nvaan une canoncal ansfomaons. We also have o Smlaly, u u q q u p p u = + +, (.38) u u = u, + PB. (.39) q = q,, p p, PB =. (.40) PB.6 CONTNUOUS SYSTEMS.6. The Tanson fom a Dscee o a Connuous Sysem Conse an nfnely long elasc o ha can unego small longunal vbaons. We appoxmae hs by an nfne chan of equal mass pons a sance a apa an connece by unfom massless spngs havng foce consans k. The knec enegy s

40 T = m η, (.4) whee m s he mass of each pacle an η s he locaon of he h pacle. The poenal enegy s We have o The equaons of moon ae m a V = kb η+ η g. (.4) O = m η kbη + η g, (.43) NM QP a m η = η ka a NM η + η a η η η η ka ka a + a + O. (.44) QP = 0. (.45) We have m/a = µ = mass/un lengh. ooke's law saes ha he exenson of a o/un lengh s popoonal o he foce,.e. = Yξ, (.46) η+ η whee ξ = s he exenson/un lengh. The foce necessay o a sech he sng by an amoun x s η+ η = kbη+ η g = ka a. (.47) Consequenly, ka = Y = Young's moulus. Noe ha

41 , η+ η η( x+ a) η( x) η =, (.48) a a x as a 0. Also as a 0 he summaon ove he pacles becomes an negal an Equaon.44 becomes The equaon of moon s = Y x η µη x. (.49) η η µ Y = 0. (.50) x Ths s a wave equaon wh wave popagaon velocy v.49 s sa o efne a agangan ensy η = Y η µ x Y = µ. Equaon. (.5).6. The agangan omulaon Conse a agangan ensy η η, η, x, x. (.5) = amlon's pncple s δ = δ x = 0. (.53) We choose value of η such ha η( x, ; α) = η( x, ; 0 ) + αξ( x, ), (.54)

42 whee η(x,;0) s he funcon ha sasfes amlon's pncple. We have δ α α = = =0 0, (.55) whee x x x x x α η η α η α η η α η = + + R S T U V W. (.56) Ths expesson may be smplfe as follows. s, η α η η η α R S T U V W = N M M M M O Q P P P P, (.57) plus a bounay em ha goes o eo. Also, x x x x x x x x x x η α η η η α R S T U V W = N M M M M O Q P P P P, (.58) plus anohe bounay em ha goes o eo. Theefoe, δ η η η η α α x x x x x = R S T U V W = = 0 0, (.59) an

43 ,,, η x Smlaly fo ohe agangans. + x η = 0. (.60) η.6.3 Noehe's Theoem A fomal escpon of he connecon beween nvaance o symmey popees an conseve quanes s conane n Noehe's heoem. We conse ansfomaons whee ' µ µ µ µ x x = x +δx, (.6) ' ' ηρ xµ η e ρ x µ = ηρ xµ + ηρ δxµ, (.6) ' ' ' ' ' eηρ xµ ηρ, ν xµ, xµ ' η x, η x, x e e ρ µ ρ, ν µ µ. (.63) We make hee assumpons:.) 4-space s Euclean,.) The agangan ensy has he same funconal fom afe ansfomaon, 3.) The magnue of he acon negal s nvaan une he ansfomaon. om assumpons, 3 we have Ω' ' ' ' ' ρ ρ, ν µ µ ρ ρ, ν µ µ Ω eη η, x x eη η, x x =0. (.64) x µ ' s a ummy vaable, so le x ' µ µ x o oban

44 ',, ', ',,,, ' ' eη x x, x x x, ρ µ η ρν η η, µ µ µ e ρ µ ρ, ν xµ xµ xµ =0. (.65) Ω' Ω Une he ansfomaons he fs oe ffeence beween he negals consss of wo pas, one s an negal ove Ω an he ohe s an negal ove Ω' Ω. o example, n one menson, b+ δb a+ δa b a b+ δb f ( x) + δf ( x) x f ( x) x = b a a+ δa δf ( x) x f ( x) δf ( x) x f ( x) δf ( x) x. (.66) To fs oe, he las wo ems ae b The ffeence beween negals s b b b b + δb a+ δa f ( x) x f ( x) x = δbf ( b) δaf ( a). (.67) b a b N a b O δf ( x) x+ f ( x) δx = δf ( x) + δxf ( x) g x. (.68) a M P x Consequenly, fo he agangans we have o Ω' µ µ µ µ Ω η x x η, x x Ω' To fs oe, = η x η x x + ηδx S = Ω' b g µ µ µ µ µ R xµ S η xµ η xµ + eη, x δx x T µ S a µ µ Q 0, (.69) U V= 0. (.70) W

45 ', ', η x η, x, η δη µ µ = ρ + δηρν, (.7) η ρ ρν, o η xµ η, xµ = x ν η ρ, ν δη ρ. (.7) The nvaance conon s We efne because an we have Equaon.305 euces o R xµ S δηρ + δx xν η T ρ, ν δx ν U νv W = 0. (.73) = ε X, δηρ = ε Ψ ρ, (.74) ν ' ' η e ρ x µ = ηρ xµ + δηρ xµ, (.75) δη ρ ηρ = δηρ + δx σ, (.76) x σ δη = ε e η X. (.77) ρ Ψ ρ ρ, σ σ R U ε S ηρσ, δνσ X σ xν ηρ ν Ψ ηρν ρ Vx T,, W µ = 0. (.78)

46 The esul s Noehe's heoem, whch saes ha x ν R S η T, ρ ν η δ X ρσ, νσ σ η ρν, U ρv W Ψ = 0. (.79).7 Bblogaphy ancos, C., The Vaaonal Pncples of Mechancs, 4h E., (Toono, Canaa: Unvesy of Toono Pess, 970). olsen,., Classcal Mechancs, n E., (Reang, MA: Ason-Wesley Publshng Co., 980.) Symon,. R., Mechancs, 3 E., (Reang, MA: Publshng Co., 97.) Ason-Wesley