A Review of Analytical Mechanics

Transcription

1 Chapter 1 A Revew of Analytcal Mechancs 1.1 Introducton These lecture notes cover the thrd course n Classcal Mechancs, taught at MIT snce the Fall of 01 by Professor Stewart to advanced undergraduates (course 8.09) as well as to graduate students (course 8.309). In the prerequste classcal mechancs II course the students are taught both Lagrangan and Hamltonan dynamcs, ncludng Kepler bound moton and central force scatterng, and the basc deas of canoncal transformatons. Ths course brefly revews the needed concepts, but assumes some famlarty wth these deas. References used for ths course nclude Goldsten, Poole & Safko, Classcal Mechancs, 3rd edton. Landau and Lfshtz vol.6, Flud Mechancs. Symon, Mechancs for readng materal on non-vscous fluds. Strogatz, Nonlnear Dynamcs and Chaos. Revew: Landau & Lfshtz vol.1, Mechancs. (Typcally used for the prerequste Classcal Mechancs II course and hence useful here for revew) 1. Lagrangan & Hamltonan Mechancs Newtonan Mechancs In Newtonan mechancs, the dynamcs of a system of N partcles are determned by solvng for ther coordnate trajectores as a functon of tme. Ths can be done through the usual vector spatal coordnates r (t) for {1,..., N}, or wth generalzed coordnates q (t) for {1,..., 3N} n 3-dmensonal space; generalzed coordnates could be angles, et cetera. 1

2 Veloctes are represented through v ṙ for spatal coordnates, or through q for generalzed coordnates. Note that dots above a symbol wll always denote the total tme dervatve d. Momenta are lkewse ether Newtonan p = m v or generalzed p. For a fxed set of masses m Newton s nd law can be expressed n equvalent ways: 1. It can be expressed as N second-order equatons F = d (m ṙ ) wth N boundary condtons gven n r (0) and ṙ (0). The problem then becomes one of determnng the N vector varables r (t).. It can also be expressed as an equvalent set of N 1 st order equatons F = ṗ & p /m = ṙ wth N boundary condtons gven n r (0) and p (0). The problem then becomes one of determnng the N vector varables r (t) and p (t). Note that F = ma holds n nertal frames. These are frames where the moton of a partcle not subject to forces s n a straght lne wth constant velocty. The converse does not hold. Inertal frames descrbe tme and space homogeneously (nvarant to dsplacements), sotropcally (nvarant to rotatons), and n a tme ndependent manner. Nonnertal frames also genercally have fcttous forces, such as the centrfugal and Corols effects. (Inertal frames also play a key role n specal relatvty. In general relatvty the concept of nertal frames s replaced by that of geodesc moton.) The exstence of an nertal frame s a useful approxmaton for workng out the dynamcs of partcles, and non-nertal terms can often be ncluded as perturbatve correctons. Examples of approxmate nertal frames are that of a fxed Earth, or better yet, of fxed stars. We can stll test for how nonnertal we are by lookng for fcttous forces that (a) may pont back to an orgn wth no source for the force or (b) behave n a non-standard fashon n dfferent frames (.e. they transform n a strange manner when gong between dfferent frames). We wll use prmes wll denote coordnate transformatons. If r s measured n an nertal frame S, and r s measured n frame S wth relaton to S by a transformaton r = f(r, t), then S s nertal ff r = 0 r = 0. Ths s solved by the Gallean transformatons, r = r + v 0 t t = t, whch preserves the nertalty of frames, wth F = m r and F = m r mplyng each other. Gallean transformatons are the non-relatvstc lmt, v c, of Lorentz transformatons whch preserve nertal frames n specal relatvty. A few examples related to the concepts of nertal frames are: 1. In a rotatng frame, the transformaton s gven by [ ] [ ] [ ] x cos(θ) sn(θ) x y = sn(θ) cos(θ) y If θ = ωt for some constant ω, then r = 0 stll gves r 0, so the prmed frame s nonnertal.

3 . In polar coordnates, r = rrˆ, gves and thus Fgure 1.1: Frame rotated by an angle θ drˆ dθˆ = θθ, ˆ = θrˆ (1.1) r = r rˆ + ṙθθ ˆ + r θθ ˆ θ rˆ. (1.) Even f r = 0 we can stll have r 0 and θ 0, and we can not n general form a smple Newtonan force law equaton mq = F q for each of these coordnates. Ths s dfferent than the frst example, snce here we are pckng coordnates rather than changng the reference frame, so to remnd ourselves about ther behavor we wll call these non-nertal coordnates (whch we may for example decde to use n an nertal frame). In general, curvlnear coordnates are non-nertal. Lagrangan Mechancs In Lagrangan mechancs, the key functon s the Lagrangan L = L(q, q, t). (1.3) Here, q = (q 1,..., q N ) and lkewse q = (q 1,..., q N). We are now lettng N denote the number of scalar (rather than vector) varables, and wll often use the short form to denote dependence on these varables, as n Eq. (1.3). Typcally we can wrte L = T V where T s the knetc energy and V s the potental energy. In the smplest cases, T = T (q ) and V = V (q), but we also allow the more general possblty that T = T (q, q, t) and V = V (q, q, t). It turns out, as we wll dscuss later, that even ths generalzaton does not descrbe all possble classcal mechancs problems. The soluton to a gven mechancal problem s obtaned by solvng a set of N second-order dfferental equatons known as Euler-Lagrange equatons of moton, d L L = 0. (1.4) q q 3

4 These equatons nvolve q, and reproduce the Newtonan equatons F = ma. The prncple of statonary acton (Hamlton s prncple), t δs = δ L(q, q, t) = 0, (1.5) t 1 s the startng pont for dervng the Euler-Lagrange equatons. Although you have covered the Calculus of Varatons n an earler course on Classcal Mechancs, we wll revew the man deas n Secton 1.5. There are several advantages to workng wth the Lagrangan formulaton, ncludng 1. It s easer to work wth the scalars T and V rather than vectors lke F.. The same formula n equaton (1.4) holds true regardless of the choce of coordnates. To demonstrate ths, let us consder new coordnates Q = Q (q 1,..., q N, t). (1.6) Ths partcular sort of transformaton s called a pont transformaton. Defnng the new Lagrangan by L = L (Q, Q, t) = L(q, q, t), (1.7) we clam that the equatons of moton are smply d L L = 0. (1.8) Q Q Proof: (for N = 1, snce the generalzaton s straghtforward) Gven L (Q, Q, t) = L(q, q, t) wth Q = Q(q, t) then Q = d Q(q, t) = Q Q q +. (1.9) q t Therefore Q q = Q, (1.10) q a result that we wll use agan n the future. Then L L = q q L L = q q L Q L Q = +, (1.11) Q q Q q L Q L Q = =. Q q Q q 4

5 Q q L q Snce Q = 0 there s no term L q Q ( Pluggng these results nto 0 = d snce d Q q n the last lne. ) L gves q [ ] [ d L Q L d Q L Q L ] Q 0 = + + Q q Q q Q q Q q [ ] d L L Q, = (1.1) Q Q q = (q + ) Q = q t q q (q + q t q )Q = Q cancel. Fnally for non-trval transformaton where Q = q Note two thngs: so that the second and fourth terms 0 we have, as expected, d L L 0 =. (1.13) Q Q Ths mples we can freely use the Euler-Lagrange equatons for nonnertal coordnates. We can formulate L n whatever coordnates are easest, and then change to convenent varables that better descrbe the symmetry of a system (for example, Cartesan to sphercal). 3. Contnung our lst of advantages for usng L, we note that t s also easy to ncorporate constrants. Examples nclude a mass constraned to a surface or a dsk rollng wthout slppng. Often when usng L we can avod dscussng forces of constrant (for example, the force normal to the surface). Lets dscuss the last pont n more detal (we wll also contnue to dscuss t n the next secton). The method for many problems wth constrants s to smply make a good choce for the generalzed coordnates to use for the Lagrangan, pckng N k ndependent varables q for a system wth k constrants. Example: For a bead on a helx as n Fg. 1. we only need one varable, q 1 = z. Example: A mass m attached by a massless pendulum to a horzontally sldng mass m 1 as n Fg. 1.3, can be descrbed wth two varables q 1 = x and q = θ. Example: As an example usng non-nertal coordnates consder a potental V = V (r, θ) n polar coordnates for a fxed mass m at poston r = rrˆ. Snce ṙ = ṙrˆ + rθθ ˆ ( we have T = m ṙ = m ṙ + r θ ), gvng m L = ṙ + r θ V (r, θ). (1.14) 5

6 Fgure 1.: Bead on a helx Fgure 1.3: Pendulum of mass m hangng on a rgd bar of length l whose support m 1 s a frctonless horzontally sldng bead For r the Euler-Lagrange equaton s d L L 0 = ṙ r = d (mṙ) mrθ V +. (1.15) r Ths gves mr mrθ V = = F r, (1.16) r from whch we see that F r mr. For θ the Euler-Lagrange equaton s d L L d 0 = = mr θ V θ +. (1.17) θ θ Ths gves d mr V θ = = F θ, (1.18) θ whch s equvalent to the relaton between angular momentum and torque perpendcular to the plane, L z = F θ = τ z. (Recall L = r p and τ = r F.) 6

7 Fgure 1.4: Partcle on the nsde of a cone Example: Let us consder a partcle rollng due to gravty n a frctonless cone, shown n Fg. 1.4, whose openng angle α defnes an equaton for ponts on the cone tan(α) = x + y /z. There are 4 steps whch we can take to solve ths problem (whch are more general than ths example): 1. Formulate T and V by N = 3 generalzed coordnates. Here t s ( most convenent ) to choose cylndrcal coordnates denoted (r, θ, z), so that T = m ṙ + r θ + ż and V = mgz.. Reduce the problem to N k = ndependent coordnates and determne the new Lagrangan L = T V. In ths case we elmnate z = r cot(α) and ż = ṙ cot(α), so m [ ( L = 1 + cot α ) ] ṙ + r θ mgr cot α. (1.19) ( L ) 3. Fnd the Euler-Lagrange equatons. For r, 0 = d ṙ L, whch here s r d [ ( 0 = m 1 + cot α ) ṙ ] mrθ + mg cot α (1.0) gvng ( 1 + cot α ) r rθ + g cot α = 0. (1.1) For θ we have 0 = d L L, so gvng θ θ d 0 = mr θ 0, (1.) (ṙθ + rθ)r = 0. (1.3) 4. Solve the system analytcally or numercally, for example usng Mathematca. Or we mght be only nterested n determnng certan propertes or characterstcs of the moton wthout a full soluton. 7

8 Hamltonan Mechancs In Hamltonan mechancs, the canoncal momenta p L are promoted to coordnates q on equal footng wth the generalzed coordnates q. The coordnates (q, p) are canoncal varables, and the space of canoncal varables s known as phase space. The Euler-Lagrange equatons say ṗ = L. These need not equal the knematc momenta q m q f V = V (q, q ). Performng the Legendre transformaton H(q, p, t) = q p L(q, q, t) (1.4) (where for ths equaton, and henceforth, repeated ndces wll mply a sum unless otherwse specfed) yelds the Hamlton equatons of moton H q = (1.5) p H ṗ = q whch are N 1 st order equatons. We also have the result that Proof: (for N = 1) Consder H t L =. (1.6) t H H H dh = dq + dp + q p t (1.7) L L L = pdq + q dp dq q dq. q t (1.8) Snce we are free to ndependently vary dq, dp, and ths mples L H = L. t t We can nterpret the two Hamlton equatons as follows: q = p, L q = ṗ, and q = H s an nverson of p = L p q = p (q, q, t). ṗ = H provdes the Newtonan dynamcs. q However, these two equaton have an have equal footng n Hamltonan mechancs, snce the coordnates and momenta are treated on a common ground. We can use p = L to q construct H from L and then forget about L. 8

9 As an example of the manner n whch we wll usually consder transformatons between Lagrangans and Hamltonans, consder agan the varables relevant for the partcle on a cone from Fg. 1.4: z=r cot α L(r, θ, z, ṙ, θ, ż) new L(r, θ, ṙ, θ) Euler-Lagrange Eqtns. (1.9) not here H(r, θ, z, p r, p θ, p z ) = H(r, θ, p r, p θ ) Hamlton Eqtns. Here we consder transformng between L and H ether before or after removng the redundant coordnate z, but n ths course we wll only consder constrants mposed on Lagrangans and not n the Hamltonan formalsm (the step ndcated by = ). For the curous, the topc of mposng constrants on Hamltonans, ncludng even more general constrants than those we wll consder, s covered well n Drac s lttle book Lectures on Quantum Mechancs. Although Hamltonan and Lagrangan mechancs provde equvalent formalsms, there s often an advantage to usng one or the other. In the case of Hamltonan mechancs potental advantages nclude the language of phase space wth Louvlle s Theorem, Posson Brackets and the connecton to quantum mechancs, as well as the Hamlton-Jacob transformaton theory (all to be covered later on). Specal case: Let us consder a specal case that s suffcent to mply that the Hamltonan s equal to the energy, H = E T + V. If we only have quadratc dependence on veloctes n the knetc energy, T = 1 T jk (q)q jq k, and V = V (q) wth L = T V, then L q p = q = q 1 1 q T kq k + q jt j q = T. (1.30) Hence, whch s just the energy. H = q p L = T + V = E (1.31) Another Specal case: Consder a class of Lagrangans gven as 1 L(q, q, t) = L 0 + a j q j + q jt jk q k (1.3) where L 0 = L 0 (q, t), a j = a j (q, t), and T jk = T kj = T jk (q, t). We can wrte ths n shorthand as 1 L = L 0 + a q + q ˆT q. (1.33) Here the generalzed coordnates, momenta, and coeffcents have been collapsed nto vectors, lke q (rather than the boldface that we reserve for Cartesan vectors), and dot products of 9

10 vectors from the left mply transposton of that vector. Note that q s an unusual vector, snce ts components can have dfferent dmensons, eg. q = (x, θ), but nevertheless ths notaton s useful. To fnd H, L p j = = a j + T jk q k, (1.34) q j meanng p = a + T ˆ q. Invertng ths gves q = T ˆ 1 (p a), where Tˆ 1 wll exst because of the postve-defnte nature of knetc energy, whch mples that Tˆ s a postve defnte matrx. Thus, H = q p L yelds 1 H = (p a) Tˆ 1 (p a) L 0 (q, t) (1.35) as the Hamltonan. So for any Lagrangan n the form of Eq. (1.3), we can fnd Tˆ 1 and wrte down the Hamltonan as n Eq. (1.35) mmedately. Example: let us consder L = 1 mv eφ + ea v, where e s the electrc charge and SI unts are used. In Eq. (1.3), because the coordnates are Cartesan, a = ea, T ˆ = m1, and L 0 = eφ, so 1 H = (p + m ea) eφ. (1.36) As you have presumably seen n an earler course, ths Hamltonan does ndeed reproduce the Lorentz force equaton e(e + v B) = mv. A more detaled Example. Fnd L and H for the frctonless pendulum shown n Fg Ths system has two constrants, that m 1 s restrcted to le on the x-axs sldng wthout frcton, and that the rod between m 1 and m s rgd, gvng y 1 = 0, (y 1 y ) + (x 1 x ) = l. (1.37) Pror to mposng any constrants the Lagrangan s m1 m L = T V = ẋ 1 + (ẋ + ẏ ) m gy m 1 gy 1. (1.38) Lets choose to use x x 1 and the angle θ as the ndependent coordnates after mposng the constrants n Eq. (1.37). Ths allows us to elmnate y 1 = 0, x = x + l sn θ and y = l cos θ, together wth ẋ = ẋ + l cos θ θ, ẏ = l sn θ θ, ẋ 1 = ẋ. The Lagrangan wth constrants mposed s m 1 m L = ẋ + ẋ + l cos θ ẋθ + l cos θ θ + l sn θ θ + m gl cos θ. (1.39) Next we determne the Hamltonan. Frst we fnd L p x = = m 1 ẋ + m (ẋ + l cos θθ) = (m 1 + m )ẋ + m l cos θ θ, (1.40) ẋ L p θ = = m l cos θ ẋ + m l θ. θ 10

11 Note that p x s ( not ) smply pr ( oportonal ) to ẋ here (actually p x s the center-of-mass momenp x ˆ ẋ tum). Wrtng = T gves p θ θ m1 + m ˆT = m l cos θ, m l cos θ m l (1.41) 1 ẋ wth L = (ẋ θ) T ˆ + L 0 where L 0 = mgl cos θ. Computng θ ( ) ˆT 1 1 m = l m l cos θ m 1 m l + m l sn, (1.4) θ m l cos θ m 1 + m we can smply apply Eq. (1.35) to fnd the correspondng Hamltonan 1 p H = (p x p θ ) Tˆ 1 x m p gl cos θ (1.43) θ 1 [ ] = ml p + (m 1 + m )p m l cos θp x p θ m gl cos θ. m l x θ (m1 + m sn θ) Lets compute the Hamlton equatons of moton for ths system. Frst for (x, p x ) we fnd H p x cos θ p ẋ = = p x m 1 + m sn θ θ, l(m 1 + m sn (1.44) θ) H ṗ x = = 0. x As we mght expect, the CM momentum s tme ndependent. Next for (θ, p θ ): H 1 [ ] θ = = (m 1 + m )p θ m l cos θp x, (1.45) p θ m l (m 1 + m sn θ) H sn θ cos θ [ ] ṗ θ = = m l p + (m1 + m )p m l cos θp x p θ θ l 1 + m sn θ) x θ (m sn θpxpθ m gl sn θ. l(m 1 + m sn θ) These non-lnear coupled equatons are qute complcated, but could be solved n mathematca or another numercal package. To test our results for these equatons of moton analytcally, we can take the small angle lmt, approxmatng sn θ θ, cos θ 1 to obtan p x p ẋ = θ 1 [ ], ṗ x = 0, θ = ( m m 1 lm 1 m 1 m l 1 + m )p θ m lp x, θ [ ] θp x p θ ṗ θ = m l p x + (m 1 + m )p θ m l cos θp x p θ m l m glθ. (1.46) 1 lm 1 11

12 To smplfy t further we can work n the CM frame, thus settng p x = 0, and lnearze the equatons by notng that p θ θ should be small for θ to reman small, and hence θp θ s a hgher order term. For the non-trval equatons ths leaves p ẋ = θ pθ, θ =, ṗ θ = m lm 1 µl glθ, (1.47) where µ = m 1 m /(m 1 + m ) s the reduced mass for the two-body system. Thus θ = ṗ θ /(µl ) = m g θ as expected for smple harmonc moton. µ l 1.3 Symmetry and Conservaton Laws A cyclc coordnate s one whch does not appear n the Lagrangan, or equvalently n the Hamltonan. Because H(q, p, t) = q p L(q, q, t), f q j s absent n L for some partcular j, t wll be absent n H as well. The absence of that q j corresponds wth a symmetry n the dynamcs. In ths context, Noether s theorem means that a symmetry mples a cyclc coordnate, whch n turn produces a conservaton law. If q j s a cyclc coordnate for some j, then we can change that coordnate wthout changng the dynamcs gven by the Lagrangan or Hamltonan, and hence there s a symmetry. Furthermore the correspondng canoncal momentum p j s conserved, meanng t s a constant through tme. The proof s smple. If L = 0 then ṗ j = d L = L = 0, or even more smply, H q j = 0 s equvalent to ṗ j = 0, so p j s a constant n tme. Specal cases and examples of ths abound. Lets consder a few mportant ones: 1. Consder a system of N partcles where no external or nternal force acts on the center of mass (CM) coordnate R = 1 m r, where the total mass M = M m. Then the CM momentum P s conserved. Ths s because F R = R V = 0 (1.48) so V s ndependent of R. Meanwhle, T = 1 mṙ, whch when usng coordnates relatve to the center of mass, r r R, becomes 1 d T = m R +R 1 mr + 1 m ṙ 1 = MR + m ṙ. (1.49) Note that m r = 0 from the defntons of M, R, and r, so T splts nto two terms, one for the CM moton and one for relatve moton. We also observe that T s ndependent of R. Ths means that R s cyclc for the full Lagrangan L, so P = MR s a conserved quantty. In our study of rgd bodes we wll also need the forms of M and R for a con tnuous body wth mass dstrbuton ρ(r), whch for a three dmensonal body are M = d 3 r ρ(r) and R = 1 d 3 r ρ(r) r. M 1

13 Note that Ṗ = 0 s satsfed by havng no total external force, so F ext = 0, and by the nternal forces obeyng Newton s 3 rd law F j = F j. Hence, F ext = MR = F ext + F j = 0. (1.50). Let us consder a system that s nvarant wth respect to rotatons of angle φ about a symmetry axs. Ths has a conserved angular momentum. If we pck φ as a generalzed coordnate, then L = T V s ndependent of φ, so ṗ φ = L = 0 meanng p φ s constant. φ In partcular, for a system where V s ndependent of the angular velocty φ we have T ṙ p φ = = m ṙ ϕ ϕ Smplfyng further usng the results n Fg.. yelds,j r = m v. (1.51) ϕ p ϕ = m v (nˆ r ) = nˆ r m v = nˆ L total. (1.5) Fgure 1.5: Rotaton about a symmetry axs Note that L about the CM s conserved for systems wth no external torque, τ ext = r F ext = 0 and nternal forces that are all central. Defnng r j r r j and ts magntude approprately, ths means V j = V j (r j ). Ths mples that F j = V j (no sum on the repeated ndex) s parallel to r j. Hence, dl = r ṗ = r 13 F ext + r F j. (1.53),j

14 However, r F ext = 0, so dl = rj F j = 0. (1.54) <j 3. One can also consder a scalng transformaton. Suppose that under the transformaton r λr the potental s homogeneous and transforms as V λ k V for some constant k. Lettng T be quadratc n ṙ and takng tme to transform as t λ 1 k/ t then gves ṙ λ k/ ṙ k. So by constructon T λ T also, and thus the full Lagrangan L λ k L. Ths overall factor does not change the Euler-Lagrange equatons, and hence the transformaton s a symmetry of the dynamcs, only changng the overall scale or unts of the coordnate and tme varables, but not ther dynamcal relatonshp. Ths can be appled for several well known potentals: a) k = for a harmonc oscllator. Here the scalng for tme s gven by 1 k/ = 0, so t does not change wth λ. Thus, the frequency of the oscllator, whch s a tme varable, s ndependent of the ampltude. b) k = 1 for the Coulomb potental. Here 1 k/ = 3/ so there s a more ntrcate relaton between coordnates and tme. Ths power s consstent wth the behavor of bound state orbts, where the perod of the orbt T obeys T a 3, for a the sem-major axs dstance (Kepler s 3rd law). c) k = 1 for a unform gravtatonal feld. Here 1 k/ = 1/ so for a freely fallng object, the tme of free fall goes as h where h s the dstance fallen. 4. Consder the Lagrangan for a charge n electromagnetc felds, L = 1 mṙ eφ + ea ṙ. As a concrete example, let us take φ and A to be ndependent of the Cartesan coordnate x. The canoncal momentum s p = L = mṙ + ea, whch s notably dfferent ṙ from the knetc momentum. Then x beng cyclc means the canoncal momentum p x s conserved. 5. Let us consder the conservaton of energy and the relatonshp between energy and the Hamltonan. Applyng the tme dervatve gves H = H q + H ṗ + H. However, q = H and ṗ = p H. Thus q q p t There are two thngs to consder. Ḣ = H t L =. (1.55) t If H (or L) has no explct tme dependence, then H = q p L s conserved. Energy s conserved f E = 0, where energy s defned by E = T + V. If H = E then the two ponts are equvalent, but otherwse ether of the two could be true whle the other s false. 14

15 Example: Let us consder a system whch provdes an example where H = E but energy s not conserved, and where H E but H s conserved. The two stuatons wll be obtaned from the same example by explotng a coordnate choce. Consder a system consstng of a mass m attached by a sprng of constant k to a cart movng at a constant speed v 0 n one dmenson, as shown n Fg Let us call x the dsplacement Fgure 1.6: Mass attached by a sprng to a movng cart of m from the fxed wall and x s ts dsplacement from the center of the movng cart. Usng x, m k L(x, ẋ) = T V = ẋ (x v 0 t), (1.56) where the knetc term s quadratc n ẋ and the potental term s ndependent of ẋ. Ths means that H falls n the specal case consdered n Eq. (1.31) so H = E = T + V = p k + (x v 0 t), (1.57) m However H 0 so the energy s not conserved. (Of course the full energy would t be conserved, but we have not accounted for the energy needed to pull the cart at a constant velocty, treatng that nstead as external to our system. That s what led to the tme dependent H.) If we nstead choose to use the coordnate x = x v 0 t, then m L (x, ẋ m ) = ẋ + mv 0 x k + v 0 x. (1.58) Note that p = mẋ +mv 0 = mẋ = p. Ths Lagrangan fts the general form n equaton (1.3) wth a = mv 0 and L 0 = mv 0 / kx /. So 1 H (x, p ) = ẋ p L = (p m mv 0) + 15 k x m v 0, (1.59)

16 Here the last terms s a constant shft. The frst and second terms n ths expresson for H look knd of lke the energy that we would calculate f we were sttng on the cart and dd not know t was movng, whch s not the same as the energy above. Hence, H = E, but H = 0 because H = 0, so H s conserved. 1.4 Constrants and Frcton Forces t So far, we ve consdered constrants to a surface or curve that are relatonshps between coordnates. These fall n the category of holonomc constrants. Such constrants take the form f(q 1,..., q N, t) = 0 (1.60) where explct tme dependence s allowed as a possblty. An example of holonomc constran s mass n a cone (Fgure 1.4), where the constran s z r cot α = 0. Constrants that volate the form n Eq. (1.60) are non-holonomc constrants. An example of a non-holonomc constrant s a mass on the surface of a sphere. The Fgure 1.7: Mass on a sphere constrant here s an nequalty r a 0 where r s the radal coordnate and a s the radus of the sphere. Another example of a non-holonomc constrant s an object rollng on a surface wthout slppng. The pont of contact s statonary, so the constrant s actually on the veloctes. A smple example s a dsk of radus a rollng down an nclned plane wthout slppng, as shown n Fg Here the condton on veloctes, aθ = ẋ s smple enough that t can be ntegrated nto a holonomc constrant. As a more sophstcated example, consder a vertcal dsk of radus a rollng on a horzontal plane, as shown n Fg The coordnates are (x, y, θ, φ), where (x, y) s the pont of contact, φ s the rotaton angle about ts own axs, and θ s the angle of orentaton along the xy-plane. We wll assume that the flat edge of the dsk always reman parallel to z, so the 16

17 Fgure 1.8: Dsk rollng down an nclne wthout slppng Fgure 1.9: Vertcal rollng dsk on a two dmensonal plane dsk never tps over. The no-slp condton s v = aφ where v s the velocty of the center of the dsk, and v = v. Ths means ẋ = v sn(θ) = a sn(θ)φ and ẏ = v cos(θ) = a cos(θ)φ, or n dfferental notaton, dx a sn(θ)dφ = 0 and dy + a cos(θ)dφ = 0. In general, constrants of the form aj (q)dq j + a t (q) = 0 (1.61) j are not holonomc. We wll call ths a sem-holonomc constrant, followng the termnology of Goldsten. Let us consder the specal case of a holonomc constrant n dfferental form, f(q 1,..., q 3N, t) = 0. Ths means df = f f dq j + = 0, (1.6) t j 17

18 so a j = f and a t = f. The symmetry of mxed partal dervatves means t a j q = a, a t a =. (1.63) q t These condtons mply that a seemngly sem-holonomc constrant s n fact holonomc. (In math we would say that we have an exact dfferental form df for the holonomc case, but the dfferental form n Eq.(1.61) need not always be exact.) Example: To demonstrate that not all semholonomc constrants are secretly holonomc, consder the constrant n the example of the vertcal dsk. Here there s no functon h(x, y, θ, φ) that we can multply the constrant df = 0 by to make t holonomc. For the vertcal dsk from before, we could try (dx a sn(θ) dφ)h = 0 wth a x = h, a a φ = a sn(θ)h, a θ = 0, and a y = 0 all for some functon h. As we must have φ = a θ, θ φ then 0 = a cos(θ) a sn(θ) h, so h = k. That sad, ax = a θ gves h = 0 whch s a θ sn(θ) θ x θ contradcton for a non-trval h wth k 0. If the rollng s nstead constraned to a lne rather than a plane, then the constrant s holonomc. Take as an example θ = π for rollng along xˆ, then ẋ = aφ and ẏ = 0. Integratng π we have x = aϕ + x 0, y = y0, and θ =, whch together form a set of holonomc constrants. A useful concept for dscussng constrants s that of the vrtual dsplacement δr partcle. There are a few propertes to be noted of δr. of It s nfntesmal. It s consstent wth the constrants. It s carred out at a fxed tme (so tme dependent constrants do not change ts form). Example: let us consder a bead constraned to a movng wre. The wre s orented along Fgure 1.10: Bead on a movng wre the x-axs and s movng wth coordnate y = v 0 t. Here the vrtual dsplacement of the 18

19 bead δr s always parallel to xˆ (snce t s determned at a fxed tme), whereas the real dsplacement dr has a component along ŷ n a tme nterval. For a large number of constrants, the constrant force Z s perpendcular to δr, meanng Z δr = 0, so the vrtual work (n analogy to work W = F dr) of a constrant force vanshes. More generally, there s no net work from constrants, so Z δr = 0 (whch holds for the actons of surfaces, rollng constrants, and smlar thngs). The Newtonan equaton of moton s ṗ = F + Z, where F encapsulates other forces. Vanshng vrtual work gves (ṗ F ) δr = 0 (1.64) whch s the D Alembert prncple. Ths could be taken as the startng prncpal for classcal mechancs nstead of the Hamlton prncple of statonary acton. Of course Eq.(1.64) s not fully satsfactory snce we are now used to the dea of workng wth generalzed coordnates rather than the cartesan vector coordnates used there. So lets transform to generalzed coordnates through r = r (q, t), so δr = r δq j, where agan we sum over repeated ndces (lke j here). Ths means where we have defned generalzed forces r F δr = F δq j Q j δq j (1.65) r Q j F. (1.66) We can also transform the ṗ δr term usng our earler pont transformaton results as well as the fact that d r r = + r t k q k = v q k. Wrtng out the ndex sums explctly, ths gves r ṗ δr = m r δq j q,j j = ( ) d r d r mṙ m ṙ δq j q,j j = ( ) d v v m v m v δq j,j q j = ( ) d T T δq j (1.67) j q j for T = 1 m v. Together wth the D Alembert prncple, we obtan the fnal result ( ) d T T Q j δq j = 0. (1.68) q j j 19

20 We wll see momentarly that ths result s somewhat more general than the Euler-Lagrange equatons, contanng them as a specal case. We wll start by consderng systems wth only holonomc constrants, postponng other types of constrants to the next secton. Here we can fnd the ndependent coordnates q j wth j = 1,..., N k that satsfy the k constrants. Ths mples that the generalzed vrtual dsplacements δq j are ndependent, so that ther coeffcents n Eq. (1.68) must vansh, d T T Q j = 0. (1.69) q j There are several specal cases of ths result, whch we derved from the d Alembert prncple. 1. For a conservatve force F = V, then r V Q j = ( V ) = (1.70) where we assume that the potental can be expressed n the generalzed coordnates as V = V (q, t). Then usng L ( = T) V, we see that Eq. (1.69) smply reproduces the Euler-Lagrange equatons d L L = 0.. If Q j = V + d ( V q j ) q j for V = V (q, q, t), whch s the case for velocty dependent forces dervable from a potental (lke the electromagnetc Lorentz force), then the Euler-Lagrange equatons d L L = 0 are agan reproduced. q j 3. If Q j has forces obtanable from a potental as n case, as well as generalzed forces R j that cannot, then d L L = R j (1.71) q j s the generalzaton of the Euler-Lagrange equatons wth non-conservatve generalzed forces. An mportant example of a nonconservatve forces R j s gven by frcton. Statc frcton s F s F max s = µ s F N for a normal force F N. Sldng frcton s F = µfn v, so ths s a constant force that s always opposte the v drecton of moton (but vanshes when there s no moton). Rollng frcton s F = µ R F N v. v Flud frcton at a low velocty s F = bv v = v 0 bv.

21 A general form for a frcton force s F = h (v ) v (where as a remnder there s no v mplct sum on here snce we specfed on the rght-hand-sde). For ths form R j = v h rj v = v h. (1.7) v v q j Smplfyng further gves R j = h ( v ) v = h v q j q j F = q j = v v dv q j v h (v ) = q j 0 v dv h (v ) 0 (1.73) where F = v dv h (v ) (1.74) 0 s the dsspaton functon. Ths s a scalar functon lke L so t s relatvely easy to work wth. Example: Consder a sphere of radus a and mass m fallng n a vscous flud. Then T = 1 m ẏ where m < m accounts for the mass of dsplaced flud (recall Archmedes prnc- ple that the buoyant force on a body s equal to the weght of flud the body dsplaces). Also V = m gy, and L = T V. Here h ẏ, so F = 3πηaẏ, where by the constant of proportonalty s determned by the constant η, whch s the vscosty. From ths, d L L = gves the equaton of moton m ÿ + m g = 6πηaẏ. The frcton force 6πηaẏ s known as Stokes Law. (We wll derve ths equaton for the frcton force from frst prncples later on, n our dscusson of fluds.) Ths dfferental equaton can be solved by addng a partcular soluton y p (t) to a soluton of the homogeneous equaton m ÿ H + 6πηaẏ H = 0. For the tme dervatves the results are ẏ p = m g/(6πηa) and ẏ H = A exp( 6πηat/m ), where the constant A must be determned by an ntal condton. The result ẏ = ẏ H + ẏ p can be ntegrated n tme once more to obtan the full soluton y(t) for the moton. Example: f we add sldng frcton to the case of two masses on a plane connected by a sprng (consdered on problem set #1), then h = µ f m g for some frcton coeffcent µ f, and ) F = µ f g(m 1 v 1 + m v ) = µ f g (m 1 ẋ 1 + ẏ 1 + m ẋ + ẏ. (1.75) If we swtch to a sutable set of generalzed coordnates q j that smplfy the equatons of moton wthout frcton, and then compute the generalzed frcton forces R j = F, we can q j get the equatons of moton ncludng frcton. Further detals of how ths frcton complcates the equatons of moton were provded n lecture. 1 ẏ y F ẏ

22 1.5 Calculus of Varatons & Lagrange Multplers Calculus of Varatons In the calculus of varatons, we wsh to fnd a set of functons y (s) between s 1 and s that extremze the followng functonal (a functon of functons), J[y ] = s ds f(y 1 (s),..., y n (s), ẏ 1 (s),..., ẏ n (s), s), (1.76) s 1 where for ths general dscusson only we let ẏ dy rather than d. To consder the acton of ds the functonal under a varaton we consder y (s) = y (s) + η (s) where η (s 1 ) = η (s ) = 0, meanng that whle the two endponts are fxed durng the varaton δy = η, the path n between s vared. Expandng the varaton of the functonal ntegral δj = J[y ] J[y ] = 0 to 1 st order n δy we have s [ ] f f 0 = δj = ds δy + δẏ. (1.77) y ẏ s 1 Usng ntegraton by parts on the second term, and the vanshng [ of the ( varaton )] at the s f d f endponts to remove the surface term, δj vanshes when δy (s) ds = s 1 y ds 0. For ndependent varatons δy (for example, after mposng holonomc constrants), ths can only occur f f d f = 0. (1.78) y ds ẏ The scope of ths calculus of varaton result for extremzng the ntegral over f s more general than ts applcaton to classcal mechancs. Example: Hamlton s prncple states that moton q (t) extremzes the acton, so n ths case s = t, y = q, f = L, and J = S. Demandng δs = 0 then yelds the Euler-Lagrange equatons of moton from Eq. (1.78). Example: As an example outsde of classcal mechancs, consder showng that the shortest dstance between ponts on a sphere of radus a are great crcles. Ths can be seen by s mnmzng the dstance J = ds where for a sphercal surface, s 1 ds = (dx) + (dy) + (dz) = ẏ a (dθ) + a sn (θ)(dφ) (1.79) snce dr = 0. Takng s = θ and y = φ, then dφ ds = a 1 + sn (θ) dθ, (1.80) dθ

23 ( so f = 1 + sn (θ)φ. The soluton for the mnmal path s gven by solvng d dθ f ϕ ) f = 0. After some algebra these are ndeed found to be great crcles, descrbed by sn(φ α) = β cot(θ) where α, β are constants. Example: Hamlton s prncple can also be used to yeld the Hamlton equatons of moton, by consderng the varaton of a path n phase space. In ths case δj[q, p] = δ t t 1 [ ] p q H(q, p, t) ϕ = 0 (1.81) must be solved wth fxed endponts: δq (t 1 ) = δq (t ) = 0 and δp (t 1 ) = δp (t ) = 0. Here, the role of y, of s played by the N varables (q 1,..., q N, p 1,..., p N ). As f = p q H, then d f f H = 0 = ṗ =, (1.8) q q q d f f H = 0 = q =, ṗ p p gvng Hamlton s equatons as expected. Note that because f s ndependent of ṗ, the term ( f/ ṗ )δṗ = 0, and t would seem that we do not really need the condton that δp (t 1 ) = δp (t ) = 0 to remove the surface term. However, these condtons on the varatons δp are actually requred n order to put q and p on the same footng (whch we wll explot later n detal when dscussng canoncal transformatons). It s nterestng and useful to note that D Alembert s prncple ( ) d L L R j δq j = 0 (1.83) q j s a dfferental verson of the equatons that encode the classcal dynamcs, whle Hamlton s prncple t ( ) L d L δ J = δq j = 0 (1.84) q j t1 (for R j = 0 where all forces come from a potental) s an ntegrated verson. Method of Lagrange Multplers Next we wll consder the method of Lagrange multplers. For smplcty we wll assume there are no generalzed forces outsde the potental, R j = 0, untl further notce. The method of Lagrange multplers wll be useful for two stuatons that we wll encounter: 1. When we actually want to study the forces of constrant that are holonomc.. When we have sem-holonomc constrants. 3

24 Let us consder k constrants for n coordnates, wth α {1,..., k} beng the ndex runnng over the constrants. These holonomc or sem-holonomc constrants take the form g α (q, q, t) = a jα (q, t)q j + a tα (q, t) = 0 (1.85) where agan repeated ndces are summed. Thus, g α = a jα dq j + a tα = 0. For a vrtual dsplacement δq j we have = 0, so n a jα δq j = 0, (1.86) j=1 whch gves us k equatons constranng the vrtual dsplacements. For each equaton we can multply by a functon λ α (t) known as Lagrange multplers, and sum over α, and the combnaton wll stll be zero. Addng ths zero to D Alembert s prncple yelds [ ] d L L λ α a jα δq j = 0 (1.87) q j where the sums mplctly run over both α and j. Its clear that the Lagrange multpler term s zero f we sum over j frst, but now we want to consder summng frst over α for a fxed j. Our goal s make the term n square brackets zero. Only n k of the vrtual dsplacements δq j are ndependent, so for these values of j the square brackets must vansh. For the remanng k values of j we can smply choose the k Lagrange multplers λ α to force the k square bracketed equatons to be satsfed. Ths s known as the method of Lagrange multplers. Thus all square bracketed terms are zero, and we have the generalzaton of the Euler-Lagrange equatons whch ncludes terms for the constrants: d L q j L = λ α a jα. (1.88) Ths s n equatons, for the n possble values of j, and on the rght-hand-sde we sum over α for each one of these equatons. The sum λ α a jα can be nterpreted as a generalzed constrant force Q j. The Lagrange multplers λ α and generalzed coordnates q j together form n + k parameters, and equaton (1.88) n conjuncton wth g α = 0 for each α from (1.85) together form n + k equatons to be solved. There are two mportant cases to be consdered. 1. In the holonomc case, f (q, t) = 0. Here, g = f α α α = fα q j + fα, so a jα = fα. Ths t gves d L L k f α = λ α (1.89) q j α=1 for holonomc constrants. The same result can be derved from a generalzed Hamlton s prncple t J[q j, λ α ] = (L + λ α f α ) (1.90) t 1 4

25 by demandng that δj = 0. It s convenent to thnk of λ α f α as an extra potental energy that we add nto L so that a partcle does work f t leaves the surface defned by f α = 0. Recall that gven ths potental, the Force q = q ( λ α f α ) = λ α q f α, where the dervatve q f α gves a vector that s normal to the constrant surface of constant f α = 0. Ths agrees wth the form of our generalzed force above. α. In the sem-holonomc case, we just have g α = a jα (q, t)q j +a tα (q, t) = 0, wth a q j Ths gves d L L k g α = λ α (1.91) q j q j α=1 j α = g. for sem-holonomc constrants. Ths result cannot be derved from Hamlton s prncple n general, justfyng the tme we spent dscussng d Alembert s prncple, whch we have used to obtan (1.91). Recall that statc frcton mposes a no-slp constrant n the form of our equaton g α = 0. For g q, the form, g, s consstent wth the form q of generalzed force we derved from our dsspaton functon, F from our dscusson q of frcton. We end ths chapter wth several examples of the use of Lagrange multplers. Example: Consder a partcle of mass m at rest on the top of a sphere of radus a, as shown above n Fg The partcle s gven an nfntesmal dsplacement θ = θ 0 so that t sldes down. At what angle does t leave the sphere? We use the coordnates (r, θ, φ) but set φ = 0 by symmetry as t s not mportant. The constrant r a s non-holonomc, but whle the partcle s n contact wth the sphere the constrant f = r a = 0 s holonomc. To answer ( ths queston ) we wll look for the pont where the constrant force vanshes. Here T = m ṙ + r θ and V = mgz = mgr cos(θ) so ( that L = T V, then d L ) L = λ f gves whle d ( L θ ) L = λ f θ θ ṙ r mr = 0 gves d r mrθ + mg cos(θ) = λ, (1.9) mr θ mgr sn(θ) = 0. (1.93) Ths n conjuncton wth r = a gves 3 equatons for the 3 varables (r, θ, λ). Puttng them together gves ṙ = 0 so r = 0. Ths means ma θ = mga sn(θ), maθ + mg cos(θ) = λ. Multply the frst of these by θ and ntegrate over tme, knowng that θ = 0 when θ = 0, gves θ = g (1 cos(θ)). Thus, a λ = mg(3 cos(θ) ) (1.94) 5

26 s the radal constrant force. The mass leaves the sphere when λ = 0 whch s when cos(θ) = (so θ 48 o ). 3 What f we nstead mposed the constrant f = r a = 0? If we call ts Lagrange multpler λ we would get λ f = aλ when r = a, so aλ = λ s the constrant force from r before. The meanng of λ s dfferent, and t has dfferent unts, but we stll have the same constrant force. What are the equatons of moton for θ > arccos? Now we no longer have the 3 constrant so d mr mrθ + mg cos(θ) = 0 and mr θ mgr sn(θ) = 0. The ntal condtons are r 1 = a, θ 1 = arccos (, 3) ṙ1 = 0, and θ1 = g from before. Smpler 3a coordnates are x = r sn(θ) and z = r cos(θ), gvng m L = ẋ + ż mgz, (1.95) so ẍ = 0 and z = g wth ntal condtons z 1 = a, x 1 = 5a, and the ntal veloctes 3 3 smply left as ż 1 and ẋ 1 for smplcty n wrtng (though the actual values follow from ż 1 = a sn θ 1 θ 1 and ẋ 1 = a cos θ 1 θ 1 ). Ths means x(t) = ẋ 1 (t t 1 ) + x 1, (1.96) g z(t) = (t t 1 ) + ż 1 (t t 1 ) + z 1, (1.97) where t 1 s the tme when the mass leaves the sphere. That can be found from θ g 4g θ = (1 cos(θ)) = sn, (1.98) a a a 4g arccos( 3) dθ so t 1 = θ 0 sn( θ where θ0 s the small ntal angular dsplacement from the top ) of the sphere. Example: Consder a hoop of radus a and mass m rollng down an nclned plane of angle φ wthout slppng as shown n Fg. 1.11, where we defne the xˆ drecton as beng parallel to the ramp as shown. What s the frcton force of constrant, and how does the acceleraton compare to the case where the hoop s sldng rather than rollng? The no-slp constrant means aθ = ẋ, so h = aθ ẋ = a, whch can be made holonomc but whch we wll treat as sem-holonomc. Then T = T CM + T rotaton = 1 mẋ + 1 ma θ as I hoop = ma. Meanwhle, V = mg(l x) sn(φ) so that V (x = l) = 0. Ths means L = T V = m ma ẋ + θ + mg(x l) sn(φ). (1.99) 6

27 Fgure 1.11: Hoop rollng on nclned plane The equatons of moton from d ( L ẋ x ) L = λ h ẋ and d ( L θ ) L = λ h are θ θ mẍ mg sn(φ) = λ and ma θ = λa, (1.100) along wth ẋ = aθ. Takng a tme dervatve of the constrant gves ẍ = aθ, so mẍ = λ, and ẍ = g sn(φ). Ths s one-half of the acceleraton of a sldng mass. Pluggng ths back n we fnd that 1 λ = mg sn(φ) (1.101) s the frcton force n the xˆ drecton for the no-sldng constrant, and also θ = g sn(φ). Example: Consder a wedge of mass m and angle α restng on ce and movng wthout frcton. Let us also consder a mass m 1 sldng wthout frcton on the wedge and try to fnd the equatons of moton and constrant forces. The constrants are that y = 0 so the a Fgure 1.1: Wedge sldng on ce wedge s always sttng on ce, and y 1 y = tan(α) so the pont mass s always sttng on the x 1 x wedge. (We wll gnore the constrant force for no rotaton of the wedge, and only ask about these two.) The knetc energy s smply T = m 1 (ẋ 1 + ẏ1)+ m (ẋ + ẏ), whle the potental 7

28 energy s V = m 1 gy 1 + m g(y + y 0 ), where y 0 s the CM of the wedge taken from above ts bottom. Then L = T V, wth the constrants f 1 = (y 1 y ) (x 1 x ) tan(α) = 0 and f = y = 0. The equatons of moton from the Euler-Lagrange equatons wth holonomc constrants are d L L f 1 f = λ 1 + λ ẋ 1 x 1 x 1 x 1 = m 1 ẍ 1 = λ 1 tan(α), (1.10) d L L f 1 f = λ 1 + λ ẏ 1 y 1 y 1 y 1 = m 1 ÿ 1 + m 1 g = λ 1, d L L f 1 f = λ 1 + λ ẋ x x x = m ẍ = λ 1 tan(α), d L L f 1 f = λ 1 + λ ẏ y y y = mÿ + m g = λ 1 + λ, whch n conjuncton wth y 1 y = (x 1 x ) tan(α) and y = 0 s sx equatons. We number them (1) to (6). Equaton (6) gves ÿ = 0 so (4) gves m g = λ λ 1 where λ s the force of the ce on the wedge and λ 1 s the vertcal force (component) of the wedge on the pont mass. Addng (1) and (3) gves m 1 ẍ 1 + m ẍ = 0 meanng that the CM of m 1 and m has no overall force actng on t. Addtonally, as (5) mples ÿ 1 = (ẍ 1 ẍ ) tan(α), then usng (1), (), and (3) we fnd the constant force g λ 1 =. (1.103) 1 tan (α) + m 1 cos (α) Wth ths result n hand we can use t n (1), (), and (3) to solve for the trajectores. Snce m tan(α) ẍ = λ 1, (1.104) m tan(α) ẍ 1 = λ 1, m 1 ÿ 1 = λ 1 g, m 1 the acceleratons are constant. As a check on our results, f m, then ẍ = 0 so ndeed the wedge s fxed; and for ths case, ẍ 1 = g sn(α) cos(α) and ÿ 1 = g sn (α) whch both vansh as α 0 as expected (snce n that lmt the wedge dsappears, flattenng onto the cy floor below t). 8

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