The Principles of Dynamics

Transcription

1 The Prncples of Dynamcs Dr C. T. Whelan. 1 Lent L A TEXed by Tom Bentley

2 Techncaltes Copyrght & Dsclamer c 000, T. J. Bentley, Pembroke Collge, Cambrdge. Permsson s granted to copy and dstrbute these notes so long as the followng condtons are met: Ths page of techncaltes s dstrbuted wth the notes, n whatever form that dstrbuton may take. The notes reman free. Nothng more than the cost of prntng should be charged for ther use or dstrbuton. That nether T. J. Bentley, nor C. T. Whelan be held responsble for the consequences of any errors or omssons contaned heren. In partcular, t should be noted that these notes are a work n progress. It s antcpated that a corrected and expanded set wll be made avalable n the near future. Thanks I would lke to thank Dr. Colm Whelan for allowng me to produce these notes, and for hs nvaluable help n fndng the many errors. Any remanng errors are mne and correctons should be sent to tjb37@cam.ac.uk. Books By far the best book for ths course s Goldsten s Classcal Mechancs: It covers all the major parts of the course n a helpful and clear style. Landau and Lfshtz s Mechancs s also qute good, especally for the secton on Adabatc Invarants. Some Other Ponts to Note It s a great shame that ths course s so very avodable n the Mathematcal Trpos: The materal t teaches you les behnd vast swathes of theoretcal physcs and appled mathematcs, not least among whch are: Quantum Mechancs, General Relatvty. However the theory has applcatons n dfferental equatons, fnancal modellng, populaton modellng, and many other areas. In short, f you ntend to study these thngs later, or f you smply want to understand mechancs at t s most natural level, ths s a very good course to study.

3 Contents 0 Introducton: Sn and Death 1 1 From Newton to Lagrange Summary of Newton s Laws Momentum, Work and Forces Systems of Partcles Energy Constrants D Alembert s Prncple and Lagrange s Equatons Two prelmnary lemmas D Alembert s Prncple Lagrange s Equatons Generalsatons of Lagrange s Equatons Velocty Dependent Potentals Constrants Lagrange s Equatons for Impulsve Forces Some Defntons Hamlton s Prncple Hamlton s Prncple Conservaton Laws and Symmetres Rgd Bodes 5.1 Frame of Reference Rotatng Frames Transformng Between Frames Euler s Theorem The Moment of Inerta Propertes of the Moment of Inerta Tensor Spnnng Tops Dervng the Lagrangan Conserved Quanttes Steady Moton Stablty Investgaton Hamlton s Equatons, & Onwards to Abstracton Hamlton s Equatons An alternatve approach Cyclc Coordnates and Conservaton Theorems

4 v CONTENTS The prncple of Least Acton Canoncal Transformatons Canoncal Transformatons Generalsaton to Hgher Dmensons Posson Brackets The Sympletc Condton and CTs The Specal Case The General Case More on ICTs The Hamltonan as the generator of an ICT Symmetry and Conserved Quanttes The Hamlton-Jacob Equaton Nce coordnates The prncple Functon and the Lagrangan Integrable Systems Integrable Systems Acton-Angle Varables Adabatc Invarants

5 Chapter 0 Introducton: Sn and Death Synopss: A short hstorcal ntroducton to serve as motvaton for the course, and to put the materal n context. The underlyng Prncple of Mechancs as beng a Varatonal Prncple. Newton s Laws Durng the 17th century thnkng man was obssed wth Sn and Death. How, f God was perfect could He create a Unverse whch was not perfect? Why should there be so much sufferng and apparent waste n the world? The answer was that t was all n some way necessary to lead to the fnal pont: The salvaton of Manknd. That path was determned by the end pont. Newton had gven the world a mechancsal unverse, workng accordng to a set of smple fxed laws and the whole majestc clockwork had no need for a Dvne Hand to drve t. Some of the greatest mnds of the tme were seduced nto tryng to fnd the underlyng metaphyscal reason for Newton s Laws: Tryng, f not to fnd the hand of the creator then at least to fnd hs fnger prnts on the Cosmos. Lebntz, n partclular, was determned to prove that all was for the best n the best possble world. He felt that the world we lve n exhbts: The greatest smplcty n ts premses and the greatest wealth n t phenomena. Lebtz had 3 major problems wth Newtonan Mechancs 1. Occult Vrtues: Lebntz held that Newton had not explaned Gravty by postulatng a Gravtatonal Force - Forces are defne n terms of drectly measurable quanttes (masses and veloctes and ther rates of change), e as a property of ther moton. Lebttz felt that the underlyng mechansm had not been found: He argued that Newtonan theory was a knematcal one, that s a scence of moton. What he sought was a scence of powers. Ie. Dynamcs Lebntz recognsed that energy was conserved n certan mechancal systems and suggested that a prncple of energy conservaton mght be the underlyng one, from whch all Laws of Moton could be derved. He deduced somethng lke a potental energy functon.. Acton at a Dstance: To get round ths he postulated an ether of very fne partcles. Much of hs deas on ths subject antcpated what we would call a feld 1

6 CHAPTER 0. INTRODUCTION: SIN AND DEATH theory.. 3. Absolute Space: How could the stars be treated as an absolute frame of reference? Lebtz argued that space was not a thng n tself, just a relaton between objects n t. He clamed that all nertal frames should be as good as the next. Acton In ths ntellectual clmate Maupertus advanced an argument based on God s effcency. He clamed that the Laws of Nature were acted out n a way where the least possble acton was expended. He was unclear as to what acton was, but s had somethng to do wth mvs. The Varatonal Prncple movng from A to B as Euler lked the dea and defned the acton of a partcle B A mv ds. He postulated that for any gven partcle, the path taken was chosen so that the acton would be least. Actually, he always assumed the exstance of a potental energy functon V (r) from whch all forces were derved. Ie n our terms we are dealng wth a conservatve force. To progress further he nvented the Calculus of Varatons 1, that s, a necessary condton to extremse of the ntegral s that B A F (y, dy, t) F y d ( ) F = 0. ẏ We can generalse ths to a set of N ndependent coordnates y n : B δ F (y n, ẏ n, t) = 0 A F d ( ) F = 0, 1 n N. y n ẏ n Ths s called the Varatonal Prncple. Havng shown ths Euler was able to show that f we had a conservatve system (.e. there exsts a potental energy functon V ) then the path of a partcle as deduced from the varatonal prncple was precsely the same as Newton s Laws: Consder B mv ds = t(b) mv = t(b) T A t(a) t(a) 1 See Methods IB f you need to revew ths - t forms a crucal rôle n ths course.

7 3 where T = knetc energy = m energy functon V then (ẋ + ẏ ). If there exsts a potental T + V = constant = E L T V = T E. We want to make the ntegral t(b) t(a) d d L(x, ẋ, y, ẏ) statonary, but the Euler Lagrange equatons mply that ( ) L L ẋ x = 0 (mẋ) = V x, and smlarly for the ys. But these are Newton s Laws F = V = d (mẋ) So the Varatonal Prncple and energy conservaton mply Newton s Law. Qute clever maybe, but t does t gve us anythng new? Yes: The y s n the E-L equatons are mplctly dependent of any partcluar coordnate system. We used Cartesan coordnates, but there was no reason to do ths. Generalzed Coordnates Let us ntroduce generalzed coordnates {q 1,..., q 3N } If we have a system of N partcles (n 3 dmensons) free from constrants, t has 3N degrees of freedom, and we can choose to descrbe the moton n terms of any 3N ndependent varables {q } 3N =1. Usually these generalzed coordnates wll not form a convenent set of N vectors n R 3. Example: Planetry Moton. We have a radal force µm r the potental functon V (r) s radal. If we choose our coordnates to be (r, θ) we can defne L T V = 1 m (ṙ + r θ ) V (r) Now f δ L = 0 then the E-L equatons for θ mply: L θ d L θ = 0 d ( ) mr θ = 0 mr θ = constant = l.

8 4 CHAPTER 0. INTRODUCTION: SIN AND DEATH We ve derved the conservaton of angular momentum, you ll recall from IA Dynamcs that f m s constant ths mples Kepler s Second Law. Applyng the E-L equatons for r mples: L r d L ṙ = 0 d (mṙ) mr θ + V r = 0 But now the conservaton of angular momentum can be used to replace θ to get.e. m r l mr 3 = V r m r = d ( V + 1 l ) dr mr m rṙ d ( ) 1 mṙ = d ( V + 1 l ) dr dr mr d ( ) 1 mṙ + l mr + V (r) = 0 Ths s the conservaton of energy. The Dfferences between Analytc and Vectoral Mechancs In analytc dynamcs the equatons of moton can be deduced from a sngle unfyng prncple. In vectoral mechancs we have Newton s Laws In vectoral mechancs we look at the moton of the ndvdual partcles that make up the system. In analytc dynamcs we treat the system as a whole It frequently happens that there are constrants on the system (eg. n a rgd body we have the constrant that the dstances between partcles remans fxed) In the Newtonan pont of vew the must ascrbe forces to these constrants. In analytc mechancs we don t care about these forces, t s enough to know the constrants.

9 Chapter 1 From Newton to Lagrange Synopss: A bref recap of dynamcs, followed by the development of the Lagrangan formalsm. Lagrange s Equatons must take varous forms dependng on the nature of the forces (conservatve, dervable from a velocty dependent potental, or even more general), and Constrants (holonomc, monogenc, etc.). Hamlton s Prncple. Conserved quanttes. 1.1 Summary of Newton s Laws If r s the radus vector of a partcle wrt some orgn, then the velocty, v s v = dr The lnear momentum P s P = mv. Newton s frst two laws mply F = d (mv) (1.1) A reference frame n whch (1) holds s called nertal or Gallelan. Newton s Thrd Law: To every acton there s an equal and opposte reacton. What does ths mean? Is t always true? Suppose we have two partcles and j and suppose exerts a force F j on j. Then we can translate NIII to read F j = F j Ie the force j exerts of s equal and opposte. Ths s the weak formulaton of the law. If F j = F j and the forces act along the lne connectng the partcles, we have a central force. Ths s the strong form of the law, and t holds for many forces n Nature eg. Gravty, Electrostatcs. Consder the example of the Bot-Savart Law between movng charges: 1. If we have two charges movng wth parallel velocty vectors that are not perpendcular to the lne jonng the two partcles Then the weak form holds, but not the strong form. 5

10 6 CHAPTER 1. FROM NEWTON TO LAGRANGE. Consder two charges movng nstantaneously such that ther velocty vectors are perpendcular The nd charge exerts a non-zero force on the frst whle experencng no reacton force at all Momentum, Work and Forces If the force actng on the partcle s zero then d (mv) = 0 mv = const = P Defne the angular momentum of a partcle about O to be L = r P Defne the moment of the force, or the torque, to be N = r F = r d (mv) = d (r mv) = d L So f the total torque s zero the angular momentum s constant/conserved. The work done by an external force upon a partcle n gong from A to B s but v = ṡ so W AB = W AB = B A tb = m = m t A tb F ds, m dv v d t A v ( v A vb) work done = change n knetc energy. Defnton: If the force feld s st. the work done s the same for any path then we have a conservatve system. Ths s true ff F ds = 0 Whch mples there exsts a functon V (r) st. F = V. B A F ds = V B + V A whch mples W AB = V A V B = T B T A T A + V A = T B + V B e. energy s conserved.

11 1.1. SUMMARY OF NEWTON S LAWS Systems of Partcles Suppose we have a system of N partcles. We dstngush between the external appled force and the nternal forces between partcles. Newton s frst two laws become d P = j F j + F ext (1.) (note F = 0.) Applyng NIII n the weak form means that the F j term cancels. Defne the centre of mass by Then R = m r m = m r M ( ) d P = d mv ( ) = d m r = M d R = F ext P = MṘ (1.3) So the momentum of the system s the same as the momentum of t s centre of mass. Note that to get ths result we have only requred the weak form of NIII. Now consder the total torque of the system r Ṗ = = L = d But f we now assume the strong form of NIII ( ) r Ṗ r F ext + j r F j r F j + r F j = (r r j ) F j = 0 (1.4) So N ext = dl Ths s the conservaton of angular momentum Defne (1.5) r = r R v = v + V

12 8 CHAPTER 1. FROM NEWTON TO LAGRANGE e. Workng n the CoM frame. Now L = R mv + r m v + ( ) ( ) m r v + R d m r But the last two terms are zero. Now m r defne the radus vector of the centre of mass n a coord system wth t s orgn at the CoM. Ie. m r = 0 L = R m V + r mv or R MV + r m v The total angular momentum about a pont O s the angular momentum of the system concentrated at the CoM plus the angular momentum about the centre Energy We wsh to calculate the work done by all the forces n movng the system from an ntal confguraton A to a fnal one B. B Recall that F = 0 then W AB = = A B A F ds F ext ds + j B A F j ds W AB = = B A B A m v v ( ) 1 d mv W AB = T B T A Now we want to transform to the CoM frame: T = 1 ( ) ( ) m V + v V + v = 1 m V + 1 m v T = 1 MV + 1 m v

13 1.1. SUMMARY OF NEWTON S LAWS 9 If we can assume that the external forces can be derved from a potental energy functon then the frst term n?? can be wrtten B A F ext ds = = B A V B ( V ) ds Where we are usng V for the potental. If, now, the nternal forces are conservatve then the mutual forces on the th and jth partcles can be obtaned from a potental functon V j If the strong form of the acton-reacton law holds then V j can only be a functon of the dstance between the th and jth partcles, V j = V j ( r r j ) A F j = V j = j V j = F j (1.6) If the V j were also functons of the dfference of some other varable (e.g. velocty) then the forces would stll be equal and opposte but not necessarly le along the lne connectng the two partcles. If?? holds then V j = (r r j ) f where f s some other functon. Now when all the forces are conservatve the second term n?? becomes a sum over terms of the form B A ( V j ) ds + B A ( j V j ) ds j But, n Cartesan coordnates ds ds j = dr dr j dr j, so the term for the -jth par s ( j V j ) dr j Then the total work due to the nternal forces s 1 B,j, j A ( j V j ) dr j = 1,j, j V j B A (1.7) (note the factor of 1 s present because we re double countng n the ndces.) If both the external and the nternal forces can be derved from potentals, and nternal forces are radal, then we can defne a total potental energy V = V + 1 V j (1.8) such that the total energy s conserved. The nd term n?? s the nternal potental energy of the system. In general V j s not constant and can change as the system changes wth tme. But a specal case j

14 10 CHAPTER 1. FROM NEWTON TO LAGRANGE s... Defnton: A rgd body s a system of partcles n whch the dstances r j are constant and cannot vary wth tme. e But then r j = r r j = c j, const, j, t d (r j) = 0 r j dr j = 0 Thus f Newton s Law holds n the strong form: F j dr j = 0, and nternal forces do no work Constrants Take as examples: Rgd bodes Gas molecules n a contaner A partcle movng on a sold sphere Defnton:If the condton of constrant s such that t can be wrtten n the form f(r 1,..., r n, t) = 0 then we have a holonomc constrant. An example s the rgd body. Constrants whch cannot be wrtten ths way are called non-holonomc, eg a gas n a contaner. If the constrant contans tme explctly then t s sad to be rheonomous, f t does not t s called scleronomous, eg. bead on rgd wre s subject to the latter type of constrant, but f the bead s on a movng wre then we have the former type. Note that ths means a holonomc constrant must allow us to elmnate some varables Very often the constrant can be wrtten g (x 1,..., x n ) dx = 0 then the constrant wll be holonomc. If an ntegratng functon exsts f(x 1,..x n ) such that g = e the constrant s holonomc only f f x fg x j = f x j g + f g x j = g j g + f f x x j = fg j x (1.9) Example:The Rollng Dsc. Other examples of non-holonomc constrants are a partcle on the sphere, and all constrants dependng on hgher dervatves 1. D Alembert s Prncple and Lagrange s Equatons 1..1 Two prelmnary lemmas The cancellaton of the dots If we have a functon x = x(q, q ) then ẋ = x q q

15 1.. D ALEMBERT S PRINCIPLE AND LAGRANGE S EQUATIONS 11 So that ẋ q = x q The nterchange of the d and the ( ) d x = q j x q q j = q j ẋ q j = ẍ q So that d ( ) x q = ẋ q by the cancellaton of the dots. 1.. D Alembert s Prncple Defnton: A vrtual dsplacement of a system refers to a change n the confguraton of the system as a result of an arbtrary nfntesmal dsplacement δr consstent wth the forces and constrants at tme t. The dsplacement s called vrtual so as to dstngush t from an actual dsplacement occurrng n tme durng whch the forces and constrants can vary. Suppose the system s n equlbrum, e. the total force on each partcle s zero, F = 0. Then clearly F δr = 0 so as not to affect constrants and forces. If we decompose F as F = F ext F ext δr + f ext δr = 0 + f then We now make the assumpton that constrant forces do no work (e the second term s zero) under the vrtual dsplacement. Ie we assume we have a rgd body. Then F ext δr = 0 (1.10) Ths s the Prncple of Vrtual Work, or what some authors call D Alemberts Prncple, 1 e. The condton for he equlbrum of a system s that the vrtual work of the appled forces s zero.. Consder a system descrbed by n generalsed coordnates. Let us assume al constrants are holonomc. We remark that {q } may be less n number than the total number 3N of degrees of freedom of the system (constrants). Now the work can be done n an nfntesmal dsplacement wll be proportonal to the elements dq, dw = r Q r dq r, 1 We shall reserve ths for a later result

16 1 CHAPTER 1. FROM NEWTON TO LAGRANGE Q r s then defned as the generalsed force. Consder now a system of N partcles, let F be the force on the th partcle, let P be ts momentum. From Newton F Ṗ = 0 ( ) F Ṗ δr = 0 where δr s a vrtual dsplacement. But F = F ext + f so ( ) F ext + f Ṗ δr = 0 We make the assumpton that forces of constrant do no work, e f δr = 0 and we obtan ( ) F ext Ṗ δr = 0, (1.11) what we shall call D Alemberts Prncple - ths s the dynamc prncple of vrtual work Lagrange s Equatons contnung from above: Ṗ δr = = j = j m r δr ( ) r m r δq j ( ( d r d m ṙ m ṙ ( )) ) r δq But as we ve seen d ( ) r = v, and dv d q j = dr dq j So m r r = = j [ d ( d ( v mv q j q j ( ) v m v 1 m v ] )) ( ) 1 m v δq j Now let us make use of the fact that we have holonomc constrants - we can defne our coordnates {q } such that they form a complete set r = r (q 1,...q n, t)

17 1.. D ALEMBERT S PRINCIPLE AND LAGRANGE S EQUATIONS 13 So v = dr = k r dq k q k + r t Hence δr = j r δq j snce δr s ndept of tme. So we have now that: F ext δr = j F ext r δq j We now defne the Generalsed Forces correspondng to our generalsed coords as Q j = So usng T = 1 v KE we an wrte j F ext r [{ ( ) d T dt } ] Q j δq j = 0 q j dq j whch s just D Alemberts prncple agan! Snce ths s true for any vrtual dsplacement and the q j s are ndependent (holonomc) ( ) d T T = Q j (1.1) q j Assume we are dealng wth a conservatve system e that F ext = V then Q j = F ext r = ( V) r e Q j = V so substtuton nto equaton?? ( ) d T (T V) = 0 q j But we know that V q j = 0 so defne the Lagrangan, L by L = T V then d ( L q ) L q = 0 Note 1. Ths s set of n second order odes, the s are there only as part of the notaton The soluton of LEs wll nvolve fndng n functons eg. at t = 0 q α (0) = A α, q α = B α

18 14 CHAPTER 1. FROM NEWTON TO LAGRANGE Example:The Sphercal Pendulum a partcle of mass m whch moves under gravty s attached to a fxed pont by a rod of length a. It s a partcle constraned to move on a sphere of radus a. In terms of sphercal coords (a, θ, φ) wth θ measured upwards from the downward drecton the knetc and potental energes can be wrtten: So T = 1 ( ma θ + sn θ φ ) V = mga cos θ and L = T V L θ = ma θ, L θ = ma sn θ cos θ φ mga sn θ, L φ = maa sn θ φ L φ = 0 So by LEs and ma θ ma sn θ cos θ φ + mga sn θ = 0 ma d ( sn θ φ ) = 0 sn θ φ = const 1.3 Generalsatons of Lagrange s Equatons Velocty Dependent Potentals Suppose there exsts a potental U(q, q ) such that Q j = U + d ( ) U q j then we would be able to defne a Lagrangan L = T U and the form of LEs would be unaltered. U wll be called a generalsed or velocty dependent potental Example:Maxwell s Equatons The force on a charge q s not smply E + 1 c Ḃ = 0 D = 4πρ H 1 c Ḋ = 0 B = 0 F = qe = φ Presented here usng the auxlary felds D and H and n Gaussan unts: So for those who attended Electromagnetsm we have B = H 4πM and E = D 4πP

19 1.3. GENERALISATIONS OF LAGRANGE S EQUATIONS 15 but rather F = q (E + 1c ) v B E s not the gradent of a scalar functon: B = 0 A st. B = A. A s called the Magnetc Vector Potental We can wrte?? as E + 1 c A t ( ( A) = E + 1 t c ) A = 0 t If we now set E + 1 c = φ ths becomes { F = q φ 1 A c t + 1 } (v [ A]) c Consder: ( Ay (v ( A)) x = v y x A ) ( x Ax v z y x A ) z x A y = v y x + v A z z x + v A x x x v A x y y v A x z z v A z x x Now we note that So that da x = A x t A x + v x x + v A y y y + v A z z z v ( A) = x (v A) da x + A x t ( F x = q (φ 1c ) x A v = u x + d ( ) u ẋ 1 c ( )) d (A v) v x U = qφ q c A v So we defne L T U Suppose now that not all the forces are dervable from a potental. We can retan Lagrange s Equatons n the form ( ) d L L = q j q Q j j Where L contans the potental of the conservatve and velocty dependent forces and Q j represents those forces whch cannot be derved from a potental of ether type.

20 16 CHAPTER 1. FROM NEWTON TO LAGRANGE Suppose now that we have a frctonal force, F 3 whch s proportonal to velocty F x = k x v x Defne F = 1 ( kx vx + k y vy + k z v ) z (Ths s known as Raylegh s Functon) Then F x = F v x, or F = F, and the work done by the th partcle aganst frcton s The component of the generalsed force s dw = F dr = k v Q j = F r And LEs become ( ) d L L + F = 0 dotq j 1.3. Constrants Now let us look at a system whch may be rheonomc (tme dependent constrants), non-conservatve and non-holonomc. Consder a system of N partcles wth masses m and postons r, and acceleratons a. Let F = m a or N 1 (m a F ) = 0. for any vrtual dsplacement we have Defne N (m a F ) δr = 0 1 δw = N F δr 1 Let us consder a maxmal set {q α } α=1, where all holonomc constrants have been absorbed. Suppose the more general (non-holonomc) constrants can be wrtten n A βα (q, t) q α + A β (q, t) = 0 (1.13) where β = 1,.., m. Ie. there are n < N constrants. 4 We can wrte?? as n A αβ (q, t) dq α + A β (q, t) = 0 α=1 3 Notaton: For ths secton F s the frctonal force, s the partcle number, and x s the drecton 4?? Notaton here??

21 1.3. GENERALISATIONS OF LAGRANGE S EQUATIONS 17 But the knetc energy of the system s T = 1 N =1 m ṙ ṙ, so Defne S α by ṙ = n α=1 S α = d r q α + r q α t ( T q α ) T q α (Remember cancellaton of the dots and the nterchange of the d and the ). Now let δq α satsfy S α = N =1 m a r q α n A βα δq α = 0,, β = 1,..., m α=1 Remark 1. We are dealng wth vrtual dsplacements, e. terms n are lost. δr are vrtual dsplacements satsfyng the constrants We may wrte n S α δq α = δw α=1 δw = α Q α δq α n (S α Q α ) δq α = 0 α=1 S α = Q α because the δq α are not ndependent, but are subject to n A βα δq α = 0, β = 1,..., m. α=1 Now wrte S α Q α = B α and defne F = (B 1 λ 1 A 11 λ A 1... λ m A m1 ) δq 1 + (B λ 1 A 1 λ A... λ m A m ) δq. + (B n λ 1 A 1n λ A n... λ m A mn ) δq n (The λ are Lagrange multplers, and as such are arbtrary.)

22 18 CHAPTER 1. FROM NEWTON TO LAGRANGE Note. F = 0 δq α satsfyng n α=1 B αδq α = 0, n α=1 A βαδq α = 0 We have m arbtrary λs, we chose these so that B 1 = λ 1 A 11 + λ A λ m A m1. Then B m = λ 1 A 1m + λ A m λ m A nm F = (B m+1 λ 1 A 1,m+1 λ A m+1... λ m A m,m+1 ) δq m+1. + (B n λ 1 A 1n λ A n... λ m A mn ) δq n F = 0 snce each column s zero by the constrants. Hence the equatons of moton are: n ( ) d T A βα q α + A β = 0, T m = Q α + λ β A βα q α q α α=1 Ths wll work for holonomc constrants and for many non-holonomc constrants. Example: The Rollng Loop Consder a loop rollng wthout slppng down an nclned pane. Ths s actually a holonomc constrant, but t wll stll serve to llustrate the prncple. The constrant s Also t s clear that r dθ = dx, V = mg(l x) sn φ (1.14) β T = 1 mẋ + 1 mr θ (1.15) L = 1 mẋ + 1 mr θ mg (() l x) sn φ (1.16) One constrant one Lagrange Multpler. The constrant s of the form n A 1α q α = 0 α=1 wth A 1θ = r, A 1x = 1 So Lagrange s Equatons mẍ mg sn φ + λ = 0 (1.17) mr θ λr = 0 (1.18) r θ = ẋ (1.19) where the last equaton s the constrant. We have 3 equatons for 3 unknowns, θ, x, λ d (constrant) (1.0) r θ = ẍ mẍ = λ (1.1) ẍ = g sn φ mg sn φ and λ = and = g sn φ r (1.)

23 1.3. GENERALISATIONS OF LAGRANGE S EQUATIONS 19 hoop rolls down the plane wth half the acceleraton t would have f t slpped down mg sn φ a frctonless plane. The force constrant s λ =, and notce that the force of constrant appears va the Lagrange Multpler. Example:Atwood s Machne We have two masses and a frctonless massless pulley. There s only one ndependent coordnate, x, the poston of the second weght determned by the constrant that the strng has length l. So Notce V = m 1 gx m g(l x) T = 1 (m 1 + m ) ẋ L = T V = 1 (m 1 + m ) ẋ + m 1 gx + m g(l x) L x = (m 1 m ) g L and ẋ = (m 1 + m ) ẋ LEs (m 1 + m ) ẍ = (m 1 m ) g ( ) m1 m or ẍ = g m 1 + m The force of constrant (tenson n the strng) appears nowhere. We don t need to say anythng about t to fnd the equatons of moton We can t deduce anythng about t. e. we cannot determne the tenson. Example:Moton on the sphere A partcle of mass m movng under gravty on a smooth sphere of radus b. The constrant s x + y + z = b (ths s actually holonomc) xẋ + yẏ + zż = 0. We also have that T = m (ẋ + ẏ + ż ) Defne the generalsed forces X = 0, Ỹ = 0, Z = mg, then the equatons of moton are xẋ + yẏ + zż = 0, (1.3) mẍ = λx, (1.4) mÿ = λy, (1.5) m z = mg + λz (1.6) Lagrange s Equatons for Impulsve Forces Consder any dynamcal system whch moves accordng to Lagrange s Equatons: ( ) d T T = Q α q α q α

24 0 CHAPTER 1. FROM NEWTON TO LAGRANGE Now ntegratng w.r.t. t we have: [ T q α ] t t t 1 t 1 T t = Q α q α t 1 Thnk Drac delta-functon Let us assume that as t 1 t, Q α such that ˆQ t α = lm t1 t t 1 Q α remans fnte. Then we call the ˆQ α the generalsed mpulsve forces. In the nfntesmal nterval t 1 t we assume the generalsed coordnate don t change and the generalsed veloctes reman fnte.then we wrte [ ] T = q ˆQ α α Some Defntons Defnton:For a holonomc system ( ) d T = T + q q Q The T q are sometmes called fcttous forces due to our change of coordnates. Note 3. These are dfferent from the fcttous forces ntroduced to make a non-nertal frame appear nertal Defnton: The nstantaneous confguraton of a system can be descrbed by the n generalsed coordnates q 1,.., q n. Ths corresponds to a partcular pont n the Cartesan hyperspace where the q s form the coordnate axes. Ths n-dmensonal space s called confguraton space or coordnate space. Note 4. As tme goes on the system pont moves n confguraton space tracng out the path of the system. Confguraton space s not necessarly the same as physcal space Defnton: We say a system s monogenc f all the forces (except the forces of constrant) are dervable from a generalsed potental whch may be a functon of the (generalsed) coordnates, the (generalsed) veloctes and the tme. For such a system we have Hamlton s Prncple Hamlton s Prncple The moton of the system from tme t 1 to t s such that the ntegral Ths s where we started the course: The Varatonal Prncple. Only now we re n coordnate space, and we re seekng only a statonary value, not necessarly a mnmum. I = t t 1 has a statonary value. We wrte ths as δi = 0. L(q, q, t),

25 1.4. HAMILTON S PRINCIPLE 1 Remarks 1. then 1. We ve seen that f L = T U and we have holonomc constrants d ( L q ) L q = 0 Newton s Laws, and also Maxwell s Equatons.. Recall that δ f(y, y, x) dx = 0 d ( ) f dx y f = 0 y e: The varatonal prncple Lagrange s Equatons. (assumng the q s are ndependent). If we have non-holonomc constrants then A βα dq α + A β = 0, β = 1,..., m α And that usng Lagrange multplers we can wrte ( ) d L L = λ β A βα (1.7) q α q α β A αβ q α + A β = 0 (1.8) α e. n + m equatons for n + m varables, {q } n, {λ } m 1. Note 5. From now on we shall assume the holonomcy of the constrants 1.4. Conservaton Laws and Symmetres If L s not a functon of a gven q then ( ) d L q = 0 L q = A. That s, A s a constant of the moton, t s conserved. eg. Suppose L = 1 mṙ then L q = dl dẋ = mẋ and because L x = 0 we have mẋ = const Ie Newton s Frst Law, that momentum s conserved. We say that momentum s a conjuagte varable to poston. Defnton: It wll be convenent to defne generalzed momenta by p = L q. Note that p need not be an ordnary momentum. Suppose that the Lagrangan s ndependent of tme dl = L t + d = ( L q + q )) ( ( L q q q L q L ) L d q q (1.9) (1.30) = const (1.31)

26 CHAPTER 1. FROM NEWTON TO LAGRANGE So we defne H = P q L (n Cartesan coordnates H = m ṙ T V = T + V = energy). Consder a generalsed coordnate q j for whch a change n dq j represents a translaton of the entre system (eg. q j s the CoM). Now v s claerly dependent of the orgn of coordnates v system, e = 0 and hence T ( ) d T = ṗ j = V = Q j = q j = 0. Suppose also that we have a conservatve r F Now consderng the effect of the nfntesmal dq j (translaton of the system along some axs) r (q j ) r (q j + dq j ) r r (q j + dq j ) r (q j ) lm dq j 0 dq j (1.3) = dˆn. dq j (1.33) Where ˆn s a unt vector n the drecton of the translaton. Q j = r F = ˆn F = ˆn F Now suppose that q j does not appear n V (and hence n L). Then T = 1 and m ṙ p j = = m ṙ r q j (1.34) m ṙ r q j (1.35) = m r ˆn (1.36) ( ) = ˆn m v (1.37) Now snce q j s not n L Q j = 0 F ˆn = 0 ˆn m v = const

27 1.4. HAMILTON S PRINCIPLE 3 rned however: Never an gnorable coor- A varable n generalsed coordnates whch does not appear n the Lagrangan s sad to by cyclc or gnorable. We have seen that f the Lagrangan s ndependent of translaton n a gven drecton ˆn there s no force n ths drecton and the momentum component s conserved. Suppose q j s a cyclc coordnate, and dq j corresponds to a rotaton of the system about some axs. Now, just as before, we wll argue that a rotaton of the coordnate system T cannot affect the magntude of the veloctes e. = 0. We are assumng that q j s gnorable so L = 0, hence V = 0. d ( ) T q j = p j = V Now the dervatve has a dfferent meanng. The change dq j must correspond to an nfntesmal rotaton keepng the magntude of r fxed e So r(q j ) = r (q j + dq j ) dr = r sn θ dq j (1.38) r = r sn θ (1.39) Let ˆn ndcate a unt vector defnng the axs about whch we rotate. (snce dr r and ˆn). So r = ˆn r Q j = = F r (1.40) F ( ) n ˆ r (1.41) = ˆn (r F ) (1.4) ( ) = ˆn N, the torque on the th partcle (1.43) So Q j = 0 ˆn N = 0, where N s the total torque.. But ths p j = const = n m r v = n L We deduce that q j (rotaton about ˆn) s gnorable zero torque angular momentum s conserved. Summary 1. We have revewed IA Dynamcs, and seen how Lagrange s equatons are equvalent to Newton s Laws. It should be apparent, however, that they offer a more powerful approach to fndng and solvng the equatons of moton: The equatons themselves are easy to fnd; Conserved quanttes are mmedately apparent.

28 4 CHAPTER 1. FROM NEWTON TO LAGRANGE

29 Chapter Rgd Bodes Synopss: Ths chapter s an n-depth applcaton of the Lagrange formalsm as developed n the prevous chapter. We study rotatng frames of reference, Euleran angles, the Moment of Inerta Tensor and go on to nvestgate rgd body rotaton, n partcular the moton of a symmetrcal spnnng top..1 Frame of Reference.1.1 Rotatng Frames Let OXY represent a fxed (nertal) frame. Let Oxy be smlar. Then = j j = I I = J J = 1, and j = I J = 0. The relatonshp between the frames s So that = I cos θ + J sn θ, j = I sn θ + J cos theta, d = ( I sn θ + J cos θ) θ = j θ (.1) dj = ( I cos θ J sn θ) θ = θ (.) So some general vector r = XI + Y J = x + yj dr Now we defne ω = θk, so that = ẋ + ẏj + x + y j (.3) = ẋ + ẏj + x θj y θ (.4) dr = r t + ω Where the frst term s from ẋ+ẏj and the second from xωj yω. Now ths s clearly true of any vector so ( ) d = t + ω (.5) 5

30 6 CHAPTER. RIGID BODIES For example F = m v, where m s constant so (??) ( ) v F = m t + ω v (.6) ( ) = m t + ω (ṙ + ω r) (.7) So = m r r + ω t t + dω r + ω (ω r) (.8) m r r = F ω t t dω r ω (ω r) And you see we ve shown that the Corols Force s mv ṙ and the Centrfugal Force s mω (ω r). We nterpret ths as sayng that f we wsh to pretend a non-nertal frame s nertal, we must nvent fcttous forces..1. Transformng Between Frames A rgd body wth N partcles can have at most 3N degrees of freedom. Ths number wll be reduced by the constrant rj = c j, fxed. You need at most 6 coordnates: To establsh the poston of one partcle n the body we need three coordnates. Call ths partcle 1. Now to fx the poston of partcle we need only two coordnates (t must le on a sphere centred on partcle 1 and wth radus c j ). If we take a thrd partcle, 3, we can only rotate the axs jonng partcles 1 and, ths s the fnal degree of freedom. All other partcles are unquely fxed. The relaton between Cartesan frames can be wrtten (x 1, x, x 3 ) (x 1, x, x 3); x x = Ax. Then, for a transformaton preservng length x x = Ax Ax (.9) k a jk a k = δ j (.10) Let us assume that {x } = {x } at tme t = 0. e. AA T = A T A = I (.11) det A = ±1, (.1) Remark. det A = 1 cannot be acheved by any rgd change of coordnate axs. We wll assume det A = +1 always. We can transform from a gven Cartesan frame to another wth the same orgn by at most three rotatons. Suppose we start wth X, Y, Z. Then rotate through φ counter-clockwse about the Z axs. Then rotate c.clockwse through θ about the ξ axs:

31 .1. FRAME OF REFERENCE 7 Fnally rotate c.clockwse through ψ about ρ Ths defnes our new axes x, y, z Let us formalse ths n matrces cos ψ sn ψ 0 B = sn ψ cos ψ 0 (.13) C = cos θ sn θ (.14) 0 sn θ cos θ cos φ sn φ 0 D = sn φ cos φ 0 (.15) Then the entre rotaton has the form A = BCD, or cos ψ cos φ cos θ sn φ sn ψ cos ψ sn φ + cos θ cos φ sn ψ sn ψ sn θ A = sn ψ cos φ cos θ sn φ cos ψ sn ψ snφ + cos θ cos φ cos ψ cos ψ cos θ sn θ sn φ sn θ cos φ cos θ.1.3 Euler s Theorem The most general dsplacement of a rgd body wth one fxed pont s a rotaton about some axes. At any nstant the orentaton of such a body can be specfed by an orthogonal transformaton A(t), for smplcty we assume A(0) = I. A(t) wll be a cts. functon of tme. The transformaton wll be a rotaton f 1. The transformaton leaves one drecton unchanged. (the axs about whch t rotates).. The magntude of vectors are left unchanged Note: () follows from the orthogonalty condton. Proposton 1. The real orthogonal transformaton specfyng the physcal moton of a system wth one fxed pont always has egenvalue +1. Proof (A λi) r = 0 has a soluton ff (.16) det (A λi) = 0 (.17) (A I) A T = ( I A T) (.18)

32 8 CHAPTER. RIGID BODIES Fgure.1: Euleran Angles: Frst rotate about the z-axs, then about the (new) ξ-axs, and fnally about the z -axs.

33 .. THE MOMENT OF INERTIA 9 So det(a I) det A T = det(i A T ). But (I A) T = I A T, hence det(a I) = det(i A) usng det A T = det A. For any n n matrx we have that det( B) = ( 1) n det(b), so det(a I) = det(i A) = det(a I) det(a I) = 0. Hence the egenvalues s λ = +1.. Now we can transform A st. XAX 1 = λ λ λ 3 where the λ are the egenvalues of A. Ths det A = λ 1 λ λ 3 and λ = 1 for some =1,, 3. Suppose that λ 3 = 1 then λ 1 = λ and λ 1 = λ = 1, snce A s a rotaton. Then there are three cases: 1. λ 1 = λ λ 3 A = I, a rotaton by π.. λ 1 = λ = λ 3 rotaton through π. 3. λ 1 = e φ, λ = e φ then e φ 0 0 cos φ sn φ 0 0 e φ 0 smlar sn φ cos φ e. rotaton about the z-axs n the new frame. Ths proves Euler s Theorem. Note For e φ 0 0 A = 0 e φ we have mathrmtr(a) = 1 cos φ, and remember that the trace s the same for smlar matrces.. The sense of drecton of the rotaton s not yet well defned, snce f λ s an egenvalue, so too s λ. Ie. f x s an egenvector the Ax = λx then x s also an egenvector wth the same egenvalue. We assgn φ wth A and φ wth Ā(= A 1 ), and use a rght hand rule. Smlarly we have Chasles Theorem: The most general dsplacement of a rgd body s a translaton plus a rotaton. Ths suggests that the 6 coordnates needed could well be the 3 cartesan coordnates to fx the body n space and then the 3 Euleran angles.. The Moment of Inerta We know that the total knetc energy of the system can be wrtten = mv + T (θ, φ, ψ) Ie. the sum of translatonal and rotatonal energes. The total angular momentum about We assume here that θ etc. are not ndependent of θ, φψ.

34 30 CHAPTER. RIGID BODIES some pont O s L = R Mv + r P (agan the ang. mom. of the body concentrated at CoM plus the ang. mom. about the CoM.) The essence of rgd body moton s that all the partcles that make up the body move and rotate together. When a rgd body moves wth one pont statonary then the total ang. mom. about that pont s L = m (r v ) wth r and v gven wrt the fxed pont. Snce r s fxed relatve to the body the velocty v wrt. the space arses solely from the rotaton And e. L x L y L z v = r t + ω r L = m (r (ω r )) = ( m ωr r (r ω) ) = m (r x ) m x y m x z m x y m (r y ) m z y m x z m z y m (r z ) ω x ω y ω z The ang. mom. vector s related to the ang. mom. by the lnear transformaton I xx I xy I xz L = I yx I yy I yz ω x ω y I zx I zy I zz ω x The dagonal terms are called the moments of nerta whle the off-dagonal terms are called the products of nerta. In the case of a contnuous mass dstrbuton we would replace the sums by ntegrals n the obvous way I xy = ρ(r)xy dτ V Notaton: Sometme we wll make use of the notaton (x 1, x, x 3 ) = (x, y, z). In ths notaton I j = ρ(r)(r δ j x x j ) dτ So then V L = I ω where I s the moment of nerta tensor. A second rank tensor.

35 .. THE MOMENT OF INERTIA 31 Remark 3. Sometmes one makes use of dyads. A dyad, ab a T b ba s the outer product of two vectors, a and b: It s a nd rank tensor. Don t confuse t wth the nner product, a b ab T In ths language I = m (r δ j r r j ) L = I ω = m (r ω r (r ω)) Let n be a unt vector n the drecton of rotaton e. T = 1 m v (.19) = 1 m v (ω r ) (.0) = ω m (r v ) (.1) = ω L (.) = 1 ω I ω (.3) ω = ωn T = ω nt In = 1 Iω Then we say that I s the moment of nerta about the axs n of rotaton. Now let us consder the vector r n. It s magntude wll be the perpendcular from the axs of the rotaton. So I = T ω = m ω (v v ) (.4) = m ω (ω r ) (ω r ) (.5) = m (n r ) (n r ) (.6) I the moment of nerta about an axs s the sum over all the partcles n the body, of the product of the masses tmes ther perpendcular dstance from the axs. Let the vector from the orgn, O, to the CoM be R. Let the radus vector from O and R be r and r respectvely. Then the moment of nerta about an axs a s I a = m (r n) (.7) = m ((r + R) n) (.8) ( ) = m (R n) + m (r n) + m (R n) (r (.9) n) If we wrte M = m, the total mass of the system, and m r = 0, by the defnton of the CoM we have that I a = I b + M (R n) Ths s called the parallel axs theorem. It states that the moment of nerta about a gven axs s the same as the MoI about a parallel axs gong through the CoM + the MoI of the CoM wrt. the orgnal axs.

36 3 CHAPTER. RIGID BODIES Body Dmenson(s) Axs Moment of Inerta Sold Sphere radus a dameter 5 Ma Hollow Sphere radus a dameter 3 Ma 1 Sold Cylnder radus a, length l symmetry axs Ma 1 through centre, symmetry axs 4 Ma Ml Hollow Cylnder radus a, length l symmetry axs Ma 1 Sold Cubod sdes a, b, c through centre, sde of length c 3 M ( a + b ) 3 Sold Cone radus a, heght l symmetry axs 10 Ma Table.1: Some common moments of nerta..1 Propertes of the Moment of Inerta Tensor The Moment of Inerta Tensor has the followng propertes: 1. It s symmetrc, I xy = I yx.. All ts values are real real egenvalues. 3. Together these mply that t s self-adjont. Notaton: I s an egenvalue and I s the dentty matrx Lemma: All egenvalues of I are real and t s egenvectors are mutually orthogonal. We know that I can be put n dagonal form. The axes correspondng to ths dagonal form are known as the prncple axes and the dagonal elements I 1, I, I 3 (e. the egenvalues of the tensor) are the prncple moments of nerta. They satsfy det(i I I) = 0

37 .3. SPINNING TOPS 33 Now I xx = ( ) m y + z 0 I1, I, I 3 0. Now consder an nertal frame B whose orgn s at a fxed pont of a rgd body (on a system of space axes S wth the orgn at the centre). For an axs fxed n the body ( dl ) S = ( L t ) B + ω L = N e. N = L t + sum jk ε jk ω j L k. The angular momentum components are L = I ω, the prncple moments of nerta are tme ndependent: I dω + jk These are Euler s Equatons. In full.3 Spnnng Tops.3.1 Dervng the Lagrangan ε jk ω j ω k I k = N N 1 = I 1 ω 1 ω ω 3 (I I 3 ) (.30) N = I ω ω 3 ω 1 (I 3 I 1 ) (.31) N 3 = I 3 ω 3 ω 1 ω (I 1 I ) (.3) Fgure.: Euleran Angles as appled to the spnnng top Consder a symmetrc top wth one pont fxed. We wll used a body fxed set (x, y, z). One of the prncple axes wll be the z-axs, as fxed n the body. Snce one pont s fxed - the Euleran angles are all we need to descrbe the body. θ gves the nclnaton of the z-axs about the vertcal. φ measures the azmuth at the top of the vertcal and ψ measures the rotaton angle of the top about t s own z-axs. The general nfntesmal rotaton assocated wth ω can be consdered the result of 3 rotatons:

38 34 CHAPTER. RIGID BODIES 1. φ = ωφ about Z (space frame). θ = ωθ about ξ 3. ψ = ω ψ about z (body frame) Now ω s parallel to the space-fxed axs Z to put t n terms of the body frame we need to apply the orthogonal transformaton A = BCD, wth (ω φ ) x = φ sn θ sn ψ (.33) (ω φ ) y = φ sn θ cos ψ (.34) (ω φ ) z = φ cos θ (.35) Now the drecton of ω θ concdes wth the ξ axs. So the components of ω θ wrt. the body fxed axes s gven by applyng B: (ω θ ) x = θ cos ψ (.36) (ω θ ) y = θ sn ψ (.37) (ω θ ) z = 0 (.38) No transformaton s needed for ω ψ snce t s already about z. Addng the components of ω = ω φ + ω ψ + ω θ we get Hence the body s symmetrc. θ sn θ sn ψ + θ cos ψ ω = φ sn θ cos ψ + θ sn ψ φ cos θ + ψ T = 1 I ( 1 ω x + ωy ) 1 + I 3ωz (.39) = 1 ( I θ 1 + φ ) sn θ + 1 ( I 3 ψ + φ cos θ) (.40) V = m r g = MR g (.41) = Mgl cos θ (.4) where l s the dstance from the fxed pont to the CoM, and the angles are Euleran Angles. Hence L = 1 ( I θ 1 + φ ) sn θ + 1 ( I 3 ψ + φ cos θ) + Mgl cos θ. ( The φ and ψ are cyclc. Hence p ψ = L ψ = I 3 ψ + φ cos θ) = const. = I 3 ω 3. And p φ = L φ = ( I 1 sn θ + I 3 cos θ ) ṗh + I 3 ψ cos θ = const. = I 1 b, and we defne I 1 a = I 3 ω 3. So the two constrants of the moton p ψ and p φ can be expressed n terms of a and b.

39 .3. SPINNING TOPS Conserved Quanttes The total energy s gven by Now E = T + V = 1 ( I1 θ + φ ) sn θ + I 3 ω 3 + Mgl cos θ If we substtute n for p φ we get Then equatons (??) and (??) I 3 ψ = I 1 a I 3 φ cos θ (.43) I 1 φ sn θ + I 1 a cos θ = I 1 b (.44) and ψ = I 1a I 3 ψ = b a cos θ sn θ cos θ (b acosθ) sn θ (.45) (.46) Now ω 3 = I1a I 3 s a constant of the moton. It s (sometmes denoted n and) called the spn. Defne E = E I3 ω 3, another constant. We can wrte Or E = I 1 θ + I 1 (b a cos θ) sn + Mgl cos θ } θ {{} E = I 1 θ Ṽ (θ) + Ṽ (θ) Ths looks lke a one dmensonal problem, wth an effectve potental Ṽ (θ). Makng the change of varable u = cos θ we have: E (1 u ) = I u + I 1 (b au) + Mglu ( 1 u ) And lettng α = E I 1 and β = Mgl I 1 then u = (1 u )(α βu) (b au) Hence ɛ = u1(ɛ) u 1(ɛ) du (1 u )(α βu) (b au) Unfortunately ths ntegral s ellptc, and the solutons for θ, φ, ψ are n terms of ellptc ntegrals.

40 36 CHAPTER. RIGID BODIES If we look at the equatons of moton, as derved from Lagrange s Equatons we have that ( ) I 1 θ I1 φ sn θ cos θ + I 3 sn θ φ + φ cos θ Mgl sn θ = 0 (.47) d ( ( I 1 φ sn θ + I 3 cos θ ψ + φ )) cos θ = 0 (.48) d ( ( I 3 ψ + φ )) cos θ = 0 (.49) Ths last reveals And so we can wrte ψ + φ cos θ = const. = ω 3 = spn I 1 θ I1 φ sn θ cos θ + I 3 ω 3 φ sn θ Mgl sn θ = 0 (.50) I 1 θ I 1 φ sn θ + Mgl cos θ = const = F (.51) Ths s the conservaton of energy. And fnally the mddle equaton gves I 1 φ sn θ + I 3 ω 3 cos θ = const. = D The moton due to the change n θ s called nutaton, and the moton due to change n φ s called precesson..3.3 Steady Moton Steady moton has θ = α = const. φ = const. = Ω, say. Now provded we have θ 0 we have: I 1 Ω cos α I 3 ω 3 Ω + mgl = 0 a par of real dstnct precessonal angular veloctes Ω 1 and Ω, provded ω 3 > 4I 1 Mgl cos α. Ie. for suffcently fast spn, ω, about the axs of symmetry the top can perform steady moton, wth θ = α, wth possble precessonal veloctes, ω 1 and ω..3.4 Stablty Investgaton Let γ(θ) = I 1 φ sn θ cos θ + I 3 ω 3 φ sn θ Mgl sn θ and the stablty condton γ(α) = 0. Now puttng θ = α + ɛ where ɛ s small, we have I 1 ɛ + ɛγ (ɛ) = 0 γ(α + ɛ) γ (α) If we can now show that γ (ɛ) > 0 then ths equaton wll reduce to SHM: The steady pont s stable. Now we know that φ = (D I 3ω 3 cos θ) I 1 sn θ I 1 γ(θ) sn 3 θ = cos θ (D I 3 ω 3 cos θ) (.5) +I 3 ω 3 sn θ (D I 3 ω 3 cos θ) I 1 Mgl sn θ (.53)

41 .3. SPINNING TOPS 37 Now dfferentatng both sdes and puttng γ(α) = 0, we have I 1 γ (α) sn 3 α = sn α (D I 3 ω 3 cos α) (.54) I 3 ω 3 cos α sn α (D I 3 ω 3 cos α) (.55) +I 3 ω 3 cos α sn α (D I 3 ω 3 cos α) (.56) +I3 ω3 sn α 4I 3 Mgl sn α cos α (.57) So that Thus So that D I 3 ω 3 cos α = I 1 Ω sn α (.58) I 3 ω 3 = I 1 Ω cos α + Mgl (.59) Ω ( ) Mgl I 1 γ (α) = I1 Ω I 1 Mglcosα + Ω (.60) ɛ + γ (α)ɛ = 0 I 1????? (.61) γ { (α) = Ω I 1 ( )} Mgl Mgl (1 cosα) + I 1 Ω I 1 Therefore γ (α) > 0 α 0, and so the moton s SHM about the stable pont.

42 38 CHAPTER. RIGID BODIES

43 Chapter 3 Hamlton s Equatons, & Onwards to Abstracton Synopss: The alternatve Hamltonan formalsm s developed. Momenta and Coordnates become ndstngushable. Canoncal transformatons/posson brackets allow us to reformulate the Hamltonan: By fndng a good set of coordnates solvng the equatons of moton becomes trval (Hamlton-Jacob theory). Conserved quanttes and symmetres are related. Louvlle s Theorem provdes an mportant lnk to Statstcal/Contnuum mechancs. Notaton 1. Untl now we have used P, or P for momenta - whether or not they were generalsed momenta. However, central to the Hamltonan method of dong thngs s the concept that coordnates and momenta are vewed equally: Henceforth we shall wrte p for the momenta of a system wth coordnates q. (The transformed coordnates shall then be wrtten as P and Q.) 3.1 Hamlton s Equatons An alternatve approach So far we have formulated everythng n terms of the n ndependent varables q. We have, n effect, treated q as dstnct varables, ndependent of the q of whch they are the tme dervatve. For example we have used L q to mean the dervatve wrt. q wth q j q and q j held constant, and the symbol L q has been taken to mean the dervatve wrt. q wth all q j and all q j held constant. We could work n a space defne by q, q, t but t wll ntroduce a greater symmetry f we work wth the conjugate momenta p j = L q j. The generalsed momenta p j are sad to be conjugate to the q j and the quanttes are sad to be canoncal varables. Defnton: The Hamltonan of the system s defned by H(q, p, t) = q p L(q, q, t) So, n the Lagrangan formulaton when a varable s absent from L, ts conjugate varable s conserved. 39

44 40 CHAPTER 3. HAMILTON S EQUATIONS, & ONWARDS TO ABSTRACTION Then ( d But p = d varables we have dh = H dq + H dp + H q p t = ) L q = L q q = dh dp, ( q dp + p d q ) L dq L dq L q q t (3.1) (3.) so that dh = q dp ṗ dq L t so equatng ṗ = H q, L t = H t These are the canoncal equatons of Hamlton. They are a set of coupled partal dfferental equatons. Note 7. Features: Hamlton s Equatons are frst order. Lagrange s Equatons were second order. Hamlton s Equatons are n n varables q and p, Lagranges equatons were n the n constrants q. We now need to determne n constants. Notaton: here q s the charge, not a generalsed coordnate H t = 0 dh = H s a constant of the moton. H q dq + H p dp H H dh H = 0 q p dq p Example 1. One Dmensonal Moton Consder one dmensonal moton and suppose the exstence of a potental V (x) st F = d dx V (x). Then L = 1 mv V (x) (3.3) H = mv 1 mv + V (x) (3.4) = T + V = Total Energy (3.5) Example. Electromagnetc Feld Consder a small (non-relatvstc) partcle n an EM feld: L = T U = 1 mv qφ + q c A v (3.6) = 1 mẋ ẋ + q c A ẋ qφ, (3.7) n Cartesan coords. The generalsed momenta are gven by p = mẋ + q c A Then H = 1 ( p q ) m c A + qφ (3.8) 1 ( H = p q ) m c A + qφ (3.9)

45 3.1. HAMILTON S EQUATIONS Cyclc Coordnates and Conservaton Theorems Observe ṗ j = d ( ) L q j = H. So f a coordnate s absent from the Hamltonan the correspondng (conjugate) generalsed momentum s conserved. If the generalsed momentum p j s absent from the Hamltonan then So q j s conserved. H p j = 0 q j = The prncple of Least Acton In coordnate space we have t δ L(q, q, t) = 0, t 1 so we should be able to wrte ( ) T δ p q H(p, q, t) = 0. t 1 But let us stop here and thnk: Ths equaton s mplctly n phase space. We have to thnk about what we mean by an ndependent varaton : Snce we derved Lagrange s Equaton (n n varables) be assumng δq = 0, and by slght of hand we now have n varables (q, p ), and so by wrtng the above we are assumng δq = 0 and δp = 0. In the -varaton: The vared path over whch the ntegral s evaluated may end at dfferent tmes from the correct path There may be a varaton n the coordnates at the end ponts. Consder the famly of vared paths defned by q (t, α) = q (t, 0) +αη (t) }{{} true path