Chapter 2. Lagrange s and Hamilton s Equations

Transcription

1 Chapter 2 Lagrange s and Hamlton s Equatons In ths chapter, we consder two reformulatons of Newtonan mechancs, the Lagrangan and the Hamltonan formalsm. The frst s naturally assocated wth confguraton space, extended by tme, whle the latter s the natural descrpton for workng n phase space. Lagrange developed hs approach n 1764 n a study of the lbraton of the moon, but t s best thought of as a general method of treatng dynamcs n terms of generalzed coordnates for confguraton space. It so transcends ts orgn that the Lagrangan s consdered the fundamental object whch descrbes a quantum feld theory. Hamlton s approach arose n 1835 n hs unfcaton of the language of optcs and mechancs. It too had a usefulness far beyond ts orgn, and the Hamltonan s now most famlar as the operator n quantum mechancs whch determnes the evoluton n tme of the wave functon. We begn by dervng Lagrange s equaton as a smple change of coordnates n an unconstraned system, one whch s evolvng accordng to Newton s laws wth force laws gven by some potental. Lagrangan mechancs s also and especally useful n the presence of constrants, so we wll then extend the formalsm to ths more general stuaton. 35

2 36 CHAPTER 2. LAGRANGE S AND HAMILTON S EQUATIONS 2.1 Lagrangan for unconstraned systems For a collecton of partcles wth conservatve forces descrbed by a potental, we have n nertal cartesan coordnates mẍ = F. The left hand sde of ths equaton s determned by the knetc energy functon as the tme dervatve of the momentum p = T/ ẋ, whle the rght hand sde s a dervatve of the potental energy, U/ x.ast s ndependent of x and U s ndependent of ẋ n these coordnates, we can wrte both sdes n terms of the Lagrangan L = T U, whch s then a functon of both the coordnates and ther veloctes. Thus we have establshed d =0, dt ẋ x whch, once we generalze t to arbtrary coordnates, wll be known as Lagrange s equaton. Note that we are treatng L as a functon of the 2N ndependent varables x and ẋ, so that / ẋ means vary one ẋ holdng all the other ẋ j and all the x k fxed. Makng ths partcular combnaton of T ( r) wth U( r) to get the more complcated L( r, r) seems an artfcal constructon for the nertal cartesan coordnates, but t has the advantage of preservng the form of Lagrange s equatons for any set of generalzed coordnates. As we dd n secton 1.3.3, we assume we have a set of generalzed coordnates {q j } whch parameterze all of coordnate space, so that each pont may be descrbed by the {q j } or by the {x },, j [1,N], and thus each set may be thought of as a functon of the other, and tme: q j = q j (x 1,...x N,t) x = x (q 1,...q N,t). (2.1) We may consder L as a functon 1 of the generalzed coordnates q j and q j, 1 Of course we are not sayng that L(x, ẋ, t) s the same functon of ts coordnates as L(q, q, t), but rather that these are two functons whch agree at the correspondng physcal ponts. More precsely, we are defnng a new functon L(q, q, t) =L(x(q, t), ẋ(q, q, t),t), but we are beng physcsts and neglectng the tlde. We are treatng the Lagrangan here as a scalar under coordnate transformatons, n the sense used n general relatvty, that ts value at a gven physcal pont s unchanged by changng the coordnate system used to defne that pont.

3 2.1. LAGRANGIAN FOR UNCONSTRAINED SYSTEMS 37 and ask whether the same expresson n these coordnates d dt q j q j also vanshes. The chan rule tells us ẋ j = k q k q k ẋ j + k q k q k ẋ j. (2.2) The frst term vanshes because q k depends only on the coordnates x k and t, but not on the ẋ k. From the nverse relaton to (1.10), q j = q j x ẋ + q j t, (2.3) we have Usng ths n (2.2), q j ẋ = q j x. ẋ = j q j q j x. (2.4) Lagrange s equaton nvolves the tme dervatve of ths. Here what s meant s not a partal dervatve / t, holdng the pont n confguraton space fxed, but rather the dervatve along the path whch the system takes as t moves through confguraton space. It s called the stream dervatve, a name whch comes from flud mechancs, where t gves the rate at whch some property defned throughout the flud, f( r, t), changes for a fxed element of flud as the flud as a whole flows. We wrte t as a total dervatve to ndcate that we are followng the moton rather than evaluatng the rate of change at a fxed pont n space, as the partal dervatve does. For any functon f(x, t) of extended confguraton space, ths total tme dervatve s df dt = j f x j ẋ j + f t. (2.5) Usng Lebntz rule on (2.4) and usng (2.5) n the second term, we fnd d = dt ẋ j ( ) d qj + dt q j x j ( 2 ) q j ẋ k + 2 q j. (2.6) q j k x x k x t

4 38 CHAPTER 2. LAGRANGE S AND HAMILTON S EQUATIONS On the other hand, the chan rule also tells us x = j q j q j x + j q j q j x, where the last term does not necessarly vansh, as q j n general depends on both the coordnates and veloctes. In fact, from 2.3, q j x = k 2 q j x x k ẋ k + 2 q j x t, so x = j q j q j x + j ( 2 ) q j ẋ k + 2 q j. (2.7) q j x x k x t Lagrange s equaton n cartesan coordnates says (2.6) and (2.7) are equal, and n subtractng them the second terms cancel 2,so k 0 = j ( d dt q j q j ) qj x. The matrx q j / x s nonsngular, as t has x / q j as ts nverse, so we have derved Lagrange s Equaton n generalzed coordnates: d =0. dt q j q j Thus we see that Lagrange s equatons are form nvarant under changes of the generalzed coordnates used to descrbe the confguraton of the system. It s prmarly for ths reason that ths partcular and pecular combnaton of knetc and potental energy s useful. Note that we mplcty assume the Lagrangan tself transformed lke a scalar, n that ts value at a gven physcal pont of confguraton space s ndependent of the choce of generalzed coordnates that descrbe the pont. The change of coordnates tself (2.1) s called a pont transformaton. 2 Ths s why we chose the partcular combnaton we dd for the Lagrangan, rather than L = T αu for some α 1. Had we done so, Lagrange s equaton n cartesan coordnates would have been αd(/ ẋ j )/dt / x j = 0, and n the subtracton of (2.7) from α (2.6), the terms proportonal to / q (wthout a tme dervatve) would not have cancelled.

5 2.2. LAGRANGIAN FOR CONSTRAINED SYSTEMS Lagrangan for Constraned Systems We now wsh to generalze our dscusson to nclude contrants. At the same tme we wll also consder possbly nonconservatve forces. As we mentoned n secton 1.3.2, we often have a system wth nternal forces whose effect s better understood than the forces themselves, wth whch we may not be concerned. We wll assume the constrants are holonomc, expressble as k real functons Φ α ( r 1,..., r n,t) = 0, whch are somehow enforced by constrant forces F C on the partcles {}. There may also be other forces, whch we wll call F D and wll treat as havng a dynamcal effect. These are gven by known functons of the confguraton and tme, possbly but not necessarly n terms of a potental. Ths dstncton wll seem artfcal wthout examples, so t would be well to keep these two n mnd. In each of these cases the full confguraton space s R 3, but the constrants restrct the moton to an allowed subspace of extended confguraton space. 1. In secton we dscussed a mass on a lght rgd rod, the other end of whch s fxed at the orgn. Thus the mass s constraned to have r = L, and the allowed subspace of confguraton space s the surface of a sphere, ndependent of tme. The rod exerts the constrant force to avod compresson or expanson. The natural assumpton to make s that the force s n the radal drecton, and therefore has no component n the drecton of allowed motons, the tangental drectons. That s, for all allowed dsplacements, δ r,wehave F C δ r = 0, and the constrant force does no work. 2. Consder a bead free to slde wthout frcton on the spoke of a rotatng bcycle wheel 3, rotatng about a fxed axs at fxed angular velocty ω. That s, for the polar angle θ of nertal coordnates, Φ := θ ωt =0s a constrant 4, but the r coordnate s unconstraned. Here the allowed subspace s not tme ndependent, but s a helcal sort of structure n extended confguraton space. We expect the force exerted by the spoke on the bead to be n the ê θ drecton. Ths s agan perpendcular to any vrtual dsplacement, by whch we mean an allowed change n 3 Unlke a real bcycle wheel, we are assumng here that the spoke s drectly along a radus of the crcle, pontng drectly to the axle. 4 There s also a constrant z =0.

6 40 CHAPTER 2. LAGRANGE S AND HAMILTON S EQUATIONS confguraton at a fxed tme. It s mportant to dstngush ths vrtual dsplacement from a small segment of the trajectory of the partcle. In ths case a vrtual dsplacement s a change n r wthout a change n θ, and s perpendcular to ê θ. So agan, we have the net vrtual work of the constrant forces s zero. It s mportant to note that ths does not mean that the net real work s zero. In a small tme nterval, the dsplacement r ncludes a component rω t n the tangental drecton, and the force of constrant does do work! We wll assume that the constrant forces n general satsfy ths restrcton that no net vrtual work s done by the forces of constrant for any possble vrtual dsplacement. Newton s law tells us that p = F = F C + F D. We can multply by an arbtrary vrtual dsplacement ( F D p ) δ r = F C δ r =0, where the frst equalty would be true even f δ r dd not satsfy the constrants, but the second requres δ r to be an allowed vrtual dsplacement. Thus ( F D p ) δ r =0, (2.8) whch s known as D Alembert s Prncple. Ths gves an equaton whch determnes the moton on the constraned subspace and does not nvolve the unspecfed forces of constrant F C. We drop the superscrpt D from now on. Suppose we know generalzed coordnates q 1,...,q N whch parameterze the constraned subspace, whch means r = r (q 1,...,q N,t), for =1,...,n, are known functons and the Nq s are ndependent. There are N = 3n k of these ndependent coordnates, where k s the number of holonomc constrants. Then r / q j s no longer an nvertable, or even square, matrx, but we stll have r = r q j + r j q j t t. For the velocty of the partcle, dvde ths by t, gvng v = r q j + r j q j t, (2.9) but for a vrtual dsplacement t = 0wehave δ r = j r q j δq j.

7 2.2. LAGRANGIAN FOR CONSTRAINED SYSTEMS 41 Dfferentatng (2.9) we note that, v q j = r q j, (2.10) and also v q j = k 2 r q k + 2 r q j q k q j t = d dt r, (2.11) q j where the last equalty comes from applyng (2.5), wth coordnates q j rather than x j, to f = r / q j. The frst term n the equaton (2.8) statng D Alembert s prncple s F δ r = j F r q j δq j = j Q j δq j. The generalzed force Q j has the same form as n the unconstraned case, as gven by (1.9), but there are only as many of them as there are unconstraned degrees of freedom. The second term of (2.8) nvolves p δ r = = j = j = j = j dp r δq j dt q j ( d p r ) δq j ( ) d r p δq j dt q j j dt q j ( d p v ) δq j p v δq j dt q j j q j [ d m v v m v v ] δq j dt q j q j [ d T T ] δq j, dt q j q j where we used (2.10) and (2.11) to get the thrd lne. Pluggng n the expressons we have found for the two terms n D Alembert s Prncple, [ d T T ] Q j δq j =0. dt q j q j j

8 42 CHAPTER 2. LAGRANGE S AND HAMILTON S EQUATIONS We assumed we had a holonomc system and the q s were all ndependent, so ths equaton holds for arbtrary vrtual dsplacements δq j, and therefore d T T Q j =0. (2.12) dt q j q j Now let us restrct ourselves to forces gven by a potental, wth F = U({ r},t), or Q j = r q U = Ũ({q},t). j q j t Notce that Q j depends only on the value of U on the constraned surface. Also, U s ndependent of the q s, so d T T + U =0= d (T U) (T U), dt q j q j q j dt q j q j or d =0. (2.13) dt q j q j Ths s Lagrange s equaton, whch we have now derved n the more general context of constraned systems Some examples of the use of Lagrangans Atwood s machne Atwood s machne conssts of two blocks of mass m 1 and m 2 attached by an nextensble cord whch suspends them from a pulley of moment of nerta I wth frctonless bearngs. The knetc energy s T = 1 2 m 1ẋ m 2ẋ Iω2 U = m 1 gx + m 2 g(k x) =(m 1 m 2 )gx + const where we have used the fxed length of the cord to conclude that the sum of the heghts of the masses s a constant K. We assume the cord does not slp on the pulley, so the angular velocty of the pulley s ω =ẋ/r, and L = 1 2 (m 1 + m 2 + I/r 2 )ẋ 2 +(m 2 m 1 )gx,

9 2.2. LAGRANGIAN FOR CONSTRAINED SYSTEMS 43 and Lagrange s equaton gves d dt ẋ x =0=(m 1 + m 2 + I/r 2 )ẍ (m 2 m 1 )g. Notce that we set up our system n terms of only one degree of freedom, the heght of the frst mass. Ths one degree of freedom parameterzes the lne whch s the allowed subspace of the unconstraned confguraton space, a three dmensonal space whch also has drectons correspondng to the angle of the pulley and the heght of the second mass. The constrants restrct these three varables because the strng has a fxed length and does not slp on the pulley. Note that ths formalsm has permtted us to solve the problem wthout solvng for the forces of constrant, whch n ths case are the tensons n the cord on ether sde of the pulley. Bead on spoke of wheel As a second example, reconsder the bead on the spoke of a rotatng bcycle wheel. In secton (1.3.4) we saw that the knetc energy s T = 1 2 mṙ mr2 ω 2. If there are no forces other than the constrant forces, U(r, θ) 0, and the Lagrangan s L = 1 2 mṙ mr2 ω 2. The equaton of moton for the one degree of freedom s easy enough: d dt ṙ = m r = r = mrω2, whch looks lke a harmonc oscllator wth a negatve sprng constant, so the soluton s a real exponental nstead of oscllatng, r(t) =Ae ωt + Be ωt. The velocty-ndependent term n T acts just lke a potental would, and can n fact be consdered the potental for the centrfugal force. But we see that the total energy T s not conserved but blows up as t, T mb 2 ω 2 e 2ωt. Ths s because the force of constrant, whle t does no vrtual work, does do real work.

10 44 CHAPTER 2. LAGRANGE S AND HAMILTON S EQUATIONS Mass on end of gmballed rod Fnally, let us consder the mass on the end of the gmballed rod. The allowed subspace s the surface of a sphere, whch can be parameterzed by an azmuthal angle φ and the polar angle wth the upwards drecton, θ, n terms of whch z = l cos θ, x = l sn θ cos φ, y = l sn θ sn φ, and T = 1 2 ml2 ( θ 2 + sn 2 θ φ 2 ). Wth an arbtrary potental U(θ, φ), the Lagrangan becomes L = 1 2 ml2 ( θ 2 + sn 2 θ φ 2 ) U(θ, φ). From the two ndependent varables θ, φ there are two Lagrange equatons of moton, d dt ml 2 θ = U θ sn(2θ) φ 2, (2.14) ( ml 2 sn 2 θ φ ) = U φ. (2.15) Notce that ths s a dynamcal system wth two coordnates, smlar to ordnary mechancs n two dmensons, except that the mass matrx, whle dagonal, s coordnate dependent, and the space on whch moton occurs s not an nfnte flat plane, but a curved two dmensonal surface, that of a sphere. These two dstnctons are connected the coordnates enter the mass matrx because t s mpossble to descrbe a curved space wth unconstraned cartesan coordnates. Often the potental U(θ, φ) wll not actually depend on φ, n whch case Eq tells us ml 2 sn 2 θ φ s constant n tme. We wll dscuss ths further n Secton Hamlton s Prncple The confguraton of a system at any moment s specfed by the value of the generalzed coordnates q j (t), and the space coordnatzed by these q 1,...,q N s the confguraton space. The tme evoluton of the system s gven by

11 2.3. HAMILTON S PRINCIPLE 45 the trajectory, or moton of the pont n confguraton space as a functon of tme, whch can be specfed by the functons q (t). One can magne the system takng many paths, whether they obey Newton s Laws or not. We consder only paths for whch the q (t) are dfferentable. Along any such path, we defne the acton as S = t2 t 1 L(q(t), q(t),t)dt. (2.16) The acton depends on the startng and endng ponts q(t 1 ) and q(t 2 ), but beyond that, the value of the acton depends on the path, unlke the work done by a conservatve force on a pont movng n ordnary space. In fact, t s exactly ths dependence on the path whch makes ths concept useful Hamlton s prncple states that the actual moton of the partcle from q(t 1 )=q to q(t 2 )=q f s along a path q(t) for whch the acton s statonary. That means that for any small devaton of the path from the actual one, keepng the ntal and fnal confguratons fxed, the varaton of the acton vanshes to frst order n the devaton. To fnd out where a dfferentable functon of one varable has a statonary pont, we dfferentate and solve the equaton found by settng the dervatve to zero. If we have a dfferentable functon f of several varables x, the frst-order varaton of the functon s f = (x x 0 ) f/ x x0, so unless f/ x x0 =0forall, there s some varaton of the {x } whch causes a frst order varaton of f, and then x 0 s not a statonary pont. But our acton s a functonal, a functon of functons, whch represent an nfnte number of varables, even for a path n only one dmenson. Intutvely, at each tme q(t) s a separate varable, though varyng q at only one pont makes q hard to nterpret. A rgorous mathematcan mght want to descrbe the path q(t) ont [0, 1] n terms of Fourer seres, for whch q(t) =q 0 + q 1 t + n=1 a n sn(nπt). Then the functonal S(f) gven by S = f(q(t), q(t),t)dt becomes a functon of the nfntely many varables q 0,q 1,a 1,... The endponts fx q 0 and q 1, but the statonary condton gves an nfnte number of equatons S/ a n =0. It s not really necessary to be so rgorous, however. Under a change q(t) q(t) +δq(t), the dervatve wll vary by δ q = dδq(t)/dt, and the

12 46 CHAPTER 2. LAGRANGE S AND HAMILTON S EQUATIONS functonal S wll vary by δs = ( f q = f f q δq + ) f δq + q δ q dt [ f q d dt ] f δqdt, q where we ntegrated the second term by parts. The boundary terms each have a factor of δq at the ntal or fnal pont, whch vansh because Hamlton tells us to hold the q and q f fxed, and therefore the functonal s statonary f and only f f q d f dt q =0 fort (t,t f ) (2.17) We see that f f s the Lagrangan, we get exactly Lagrange s equaton. The above dervaton s essentally unaltered f we have many degrees of freedom q nstead of just one Examples of functonal varaton In ths secton we wll work through some examples of functonal varatons both n the context of the acton and for other examples not drectly related to mechancs. The fallng partcle As a frst example of functonal varaton, consder a partcle thrown up n a unform gravtonal feld at t = 0, whch lands at the same spot at t = T. The Lagrangan s L = 1 2 m(ẋ2 +ẏ 2 +ż 2 ) mgz, and the boundary condtons are x(t) = y(t) = z(t) = 0att = 0andt = T. Elementary mechancs tells us the soluton to ths problem s x(t) =y(t) 0, z(t) =v 0 t 1 2 gt2 wth v 0 = 1 gt. Let us evaluate the acton for any other path, wrtng z(t) n 2 terms of ts devaton from the suspected soluton, z(t) = z(t)+ 1 2 gtt 1 2 gt2. We make no assumptons about ths path other than that t s dfferentable and meets the boundary condtons x = y = z =0att = 0 and at t = T.

13 2.3. HAMILTON S PRINCIPLE 47 The acton s S = { T 1 0 ( ) 2 d z 2 m ẋ 2 +ẏ g(t 2t) d z dt dt } mg z 1 2 mg2 t(t t) dt g2 (T 2t) 2 The fourth term can be ntegrated by parts, T 1 mg(t 2t)d z dt = 1 T T 0 2 dt 2 mg(t 2t) z + mg z(t) dt. 0 0 The boundary term vanshes because z = 0 where t s evaluated, and the other term cancels the sxth term n S, so T [ 1 1 S = 0 2 mg2 4 (T 2t)2 t(t t)] dt ( ) T 2 1 d z m ẋ 2 +ẏ 2 +. dt The frst ntegral s ndependent of the path, so the mnmum acton requres the second ntegral to be as small as possble. But t s an ntegral of a nonnegatve quantty, so ts mnmum s zero, requrng ẋ = ẏ = d z/dt =0. As x = y = z =0att = 0, ths tells us x = y = z = 0 at all tmes, and the path whch mnmzes the acton s the one we expect from elementary mechancs. Is the shortest path a straght lne? The calculus of varatons occurs n other contexts, some of whch are more ntutve. The classc example s to fnd the shortest path between two ponts n the plane. The length l of a path y(x) from (x 1,y 1 )to(x 2,y 2 ) s gven 5 by l = x2 x 1 ds = x2 x 1 ( ) 2 dy 1+ dx. dx 5 Here we are assumng the path s monotone n x, wthout movng somewhere to the left and somewhere to the rght. To prove that the straght lne s shorter than other paths whch mght not obey ths restrcton, do Exercse 2.2.

14 48 CHAPTER 2. LAGRANGE S AND HAMILTON S EQUATIONS We see that length l s playng the role of the acton, and x s playng the role of t. Usng ẏ to represent dy/dx, we have the ntegrand f(y, ẏ, x) = 1+ẏ 2, and f/ y = 0, so Eq gves d f dx ẏ = d ẏ dx =0, 1+ẏ 2 and the path s a straght lne. so ẏ = const. 2.4 Conserved Quanttes Ignorable Coordnates If the Lagrangan does not depend on one coordnate, say q k, then we say t s an gnorable coordnate. Of course, we stll want to solve for t, as ts dervatve may stll enter the Lagrangan and effect the evoluton of other coordnates. By Lagrange s equaton d = =0, dt q k q k so f n general we defne P k :=, q k as the generalzed momentum, then n the case that L s ndependent of q k, P k s conserved, dp k /dt =0. Lnear Momentum As a very elementary example, consder a partcle under a force gven by a potental whch depends only on y and z, but not x. Then L = 1 2 m ( ẋ 2 +ẏ 2 +ż 2) U(y, z) s ndependent of x, x s an gnorable coordnate and P x = ẋ = mẋ s conserved. Ths s no surprze, of course, because the force s F = U and F x = U/ x =0.

15 2.4. CONSERVED QUANTITIES 49 Note that, usng the defnton of the generalzed momenta P k = q k, Lagrange s equaton can be wrtten as d dt P k = q k = T q k U q k. Only the last term enters the defnton of the generalzed force, so f the knetc energy depends on the coordnates, as wll often be the case, t s not true that dp k /dt = Q k. In that sense we mght say that the generalzed momentum and the generalzed force have not been defned consstently. Angular Momentum As a second example of a system wth an gnorable coordnate, consder an axally symmetrc system descrbed wth nertal polar coordnates (r, θ, z), wth z along the symmetry axs. Extendng the form of the knetc energy we found n sec (1.3.4) to nclude the z coordnate, we have T = 1 2 mṙ mr2 θ mż2. The potental s ndependent of θ, because otherwse the system would not be symmetrc about the z-axs, so the Lagrangan L = 1 2 mṙ mr2 θ mż2 U(r, z) does not depend on θ, whch s therefore an gnorable coordnate, and P θ := θ = mr2 θ = constant. We see that the conserved momentum P θ s n fact the z-component of the angular momentum, and s conserved because the axally symmetrc potental can exert no torque n the z-drecton: τ z = ( r U ) z = r ( U ) θ = r2 U θ =0. Fnally, consder a partcle n a sphercally symmetrc potental n sphercal coordnates. In secton (3.1.2) we wll show that the knetc energy n

16 50 CHAPTER 2. LAGRANGE S AND HAMILTON S EQUATIONS sphercal coordnates s T = 1 2 mṙ mr2 θ mr2 sn 2 θ φ 2, so the Lagrangan wth a sphercally symmetrc potental s L = 1 2 mṙ mr2 θ mr2 sn 2 θ φ 2 U(r). Agan, φ s an gnorable coordnate and the conjugate momentum P φ s conserved. Note, however, that even though the potental s ndependent of θ as well, θ does appear undfferentated n the Lagrangan, and t s not an gnorable coordnate, nor s P θ conserved 6. If q j s an gnorable coordnate, not appearng undfferentated n the Lagrangan, any possble moton q j (t) s related to a dfferent trajectory q j(t) =q j (t) +c, n the sense that they have the same acton, and f one s an extremal path, so wll the other be. Thus there s a symmetry of the system under q j q j + c, a contnuous symmetry n the sense that c can take on any value. As we shall see n Secton 8.3, such symmetres generally lead to conserved quanttes. The symmetres can be less transparent than an gnorable coordnate, however, as n the case just consdered, of angular momentum for a sphercally symmetrc potental, n whch the conservaton of L z follows from an gnorable coordnate φ, but the conservaton of L x and L y follow from symmetry under rotaton about the x and y axes respectvely, and these are less apparent n the form of the Lagrangan Energy Conservaton We may ask what happens to the Lagrangan along the path of the moton. dl dt = q dq dt + d q q dt + t In the frst term the frst factor s d dt q 6 It seems curous that we are fndng straghtforwardly one of the components of the conserved momentum, but not the other two, L y and L x, whch are also conserved. The fact that not all of these emerge as conjugates to gnorable coordnates s related to the fact that the components of the angular momentum do not commute n quantum mechancs. Ths wll be dscussed further n secton (6.6.1).

17 2.4. CONSERVED QUANTITIES 51 by the equatons of moton, so dl dt = d dt ( ) q + q t. We expect energy conservaton when the potental s tme nvarant and there s not tme dependence n the constrants,.e. when / t = 0, so we rewrte ths n terms of H(q, q, t) = q L = q P L q Then for the actual moton of the system, dh dt = t. If / t =0,H s conserved. H s essentally the Hamltonan, although strctly speakng that name s reserved for the functon H(q, p, t) on extended phase space rather than the functon wth arguments (q, q, t). What s H physcally? In the case of Newtonan mechancs wth a potental functon, L s an nhomogeneous quadratc functon of the veloctes q. If we wrte the Lagrangan L = L 2 + L 1 + L 0 as a sum of peces purely quadratc, purely lnear, and ndependent of the veloctes respectvely, then q q s an operator whch multples each term by ts order n veloctes, q n q = nl n, q q =2L 2 + L 1, and H = L 2 L 0. For a system of partcles descrbed by ther cartesan coordnates, L 2 s just the knetc energy T, whle L 0 s the negatve of the potental energy L 0 = U, soh = T + U s the ordnary energy. There are, however, constraned systems, such as the bead on a spoke of Secton 2.2.1, for whch the Hamltonan s conserved but s not the ordnary energy.

18 52 CHAPTER 2. LAGRANGE S AND HAMILTON S EQUATIONS 2.5 Hamlton s Equatons We have wrtten the Lagrangan as a functon of q, q, and t, sotsa functon of N + N + 1 varables. For a free partcle we can wrte the knetc energy ether as 1 2 mẋ2 or as p 2 /2m. More generally, we can 7 reexpress the dynamcs n terms of the 2N + 1 varables q k, P k, and t. The moton of the system sweeps out a path n the space (q, q, t) ora path n (q, P, t). Along ths lne, the varaton of L s dl = ( d q k + ) dq k + k q k q k t dt = ( Pk d q k + ) P k dq k + k t dt where for the frst term we used the defnton of the generalzed momentum and n the second we have used the equatons of moton P k = / q k. Then examnng the change n the Hamltonan H = k P k q k L along ths actual moton, dh = k = k (P k d q k + q k dp k ) dl ( qk dp k P k dq k ) t dt. If we thnk of q k and H as functons of q and P, and thnk of H as a functon of q, P, and t, we see that the physcal moton obeys q k = H, P P k k = H H, = q,t q k P,t t q,p t The frst two consttute Hamlton s equatons of moton, whch are frst order equatons for the moton of the pont representng the system n phase space. Let s work out a smple example, the one dmensonal harmonc oscllator. Here the knetc energy s T = 1 2 mẋ2, the potental energy s U = 1 2 kx2,so 7 In feld theory there arse stuatons n whch the set of functons P k (q, q ) cannot be nverted to gve functons q = q (q j,p j ). Ths gves rse to local gauge nvarance, and wll be dscussed n Chapter 8, but untl then we wll assume that the phase space (q, p), or cotangent bundle, s equvalent to the tangent bundle,.e. the space of (q, q). q, q

19 2.5. HAMILTON S EQUATIONS 53 L = 1 2 mẋ2 1 2 kx2, the only generalzed momentum s P = / ẋ = mẋ, and the Hamltonan s H = P ẋ L = P 2 /m (P 2 /2m 1 2 kx2 )=P 2 /2m+ 1 2 kx2. Note ths s just the sum of the knetc and potental energes, or the total energy. Hamlton s equatons gve ẋ = H P = P x m, P = H = kx = F. x P These two equatons verfy the usual connecton of the momentum and velocty and gve Newton s second law. The dentfcaton of H wth the total energy s more general than our partcular example. If T s purely quadratc n veloctes, we can wrte T = 1 2 j M j q q j n terms of a symmetrc mass matrx M j. If n addton U s ndependent of veloctes, L = 1 M j q q j U(q) 2 j P k = = M k q q k whch as a matrx equaton n a n-dmensonal space s P = M q. Assumng M s nvertble, 8 we also have q = M 1 P,so H = P T q L ( ) 1 = P T M 1 P 2 qt M q U(q) = P T M 1 P 1 2 P T M 1 M M 1 P + U(q) = 1 2 P T M 1 P + U(q) =T + U so we see that the Hamltonan s ndeed the total energy under these crcumstances. 8 If M were not nvertble, there would be a lnear combnaton of veloctes whch does not affect the Lagrangan. The degree of freedom correspondng to ths combnaton would have a Lagrange equaton wthout tme dervatves, so t would be a constrant equaton rather than an equaton of moton. But we are assumng that the q s are a set of ndependent generalzed coordnates that have already been pruned of all constrants.

20 54 CHAPTER 2. LAGRANGE S AND HAMILTON S EQUATIONS 2.6 Don t plug Equatons of Moton nto the Lagrangan! When we have a Lagrangan wth an gnorable coordnate, say θ, and therefore a conjugate momentum P θ whch s conserved and can be consdered a constant, we are able to reduce the problem to one nvolvng one fewer degrees of freedom. That s, one can substtute nto the other dfferental equatons the value of θ n terms of Pθ and other degrees of freedom, so that θ and ts dervatves no longer appear n the equatons of moton. For example, consder the two dmensonal sotropc harmonc oscllator, L = 1 2 m ( ẋ 2 +ẏ 2) 1 2 k ( x 2 + y 2) = 1 2 m ( ṙ 2 + r 2 θ2 ) 1 2 kr2 n polar coordnates. The equatons of moton are P θ = 0, where P θ = mr 2 θ, m r = kr + mr θ / 2 = m r = kr + Pθ 2 mr 3. The last equaton s now a problem n the one degree of freedom r. One mght be tempted to substtute for θ nto the Lagrangan and then have a Lagrangan nvolvng one fewer degrees of freedom. In our example, we would get L = 1 2 mṙ2 + P θ 2 2mr kr2, whch gves the equaton of moton Ths s wrong m r = P θ 2 mr kr. 3 Notce that the last equaton has the sgn of the Pθ 2 term reversed from the correct equaton. Why dd we get the wrong answer? In dervng the Lagrange equaton whch comes from varyng r, we need d dt ṙ r,θ, θ = r. ṙ,θ, θ

21 2.7. VELOCITY-DEPENDENT FORCES 55 But we treated P θ as fxed, whch means that when we vary r on the rght hand sde, we are not holdng θ fxed, as we should be. Whle we often wrte partal dervatves wthout specfyng explctly what s beng held fxed, they are not defned wthout such a specfcaton, whch we are expected to understand mplctly. However, there are several examples n Physcs, such as thermodynamcs, where ths mplct understandng can be unclear, and the results may not be what was ntended. 2.7 Velocty-dependent forces We have concentrated thus far on Newtonan mechancs wth a potental gven as a functon of coordnates only. As the potental s a pece of the Lagrangan, whch may depend on veloctes as well, we should also entertan the possblty of velocty-dependent potentals. Only by consderng such a potental can we possbly fnd velocty-dependent forces, and one of the most mportant force laws n physcs s of that form. Ths s the Lorentz force 9 on a partcle of charge q n the presence of electromagnetc felds E( r, t) and B( r, t), ( F = q E + v ) c B. (2.18) If the moton of a charged partcle s descrbed by Lagrangan mechancs wth a potental U( r, v, t), Lagrange s equaton says 0= d = m r d U + U, dt v r dt v r so F = d U U. dt v r We want a force lnear n v and proportonal to q, so let us try Then we need to have U = q ( φ( r, t)+ v C( r, t) ). E + v c B = d dt C φ j v j Cj. (2.19) 9 We have used Gaussan unts here, but those who prefer S. I. unts (ratonalzed MKS) can smply set c =1.

22 56 CHAPTER 2. LAGRANGE S AND HAMILTON S EQUATIONS The frst term s a stream dervatve evaluated at the tme-dependent poston of the partcle, so, as n Eq. (2.5), d dt C = C t + j v j C x j. The last term looks lke the last term of (2.19), except that the ndces on the dervatve operator and on C have been reversed. Ths suggests that these two terms combne to form a cross product. Indeed, notng (A.17) that we see that (2.19) becomes v ( C ) = j v j Cj v j C x j, E + v c B = C t φ j v j Cj + j v j C x j = C t φ v ( C ). We have successfully generated the term lnear n v f we can show that there exsts a vector feld C( r, t) such that B = c C. A curl s always dvergenceless, so ths requres B = 0, but ths s ndeed one of Maxwell s equatons, and t ensures 10 there exsts a vector feld A, known as the magnetc vector potental, such that B = A. Thus wth C = A/c, we need only to fnd a φ such that Once agan, one of Maxwell s laws, E = φ 1 c A t. E + 1 B c t =0, guarantees the exstence of φ, the electrostatc potental, because after nsertng B = A, ths s a statement that E +(1/c) A/ t has no curl, and s the gradent of somethng. 10 Ths s but one of many consequences of the Poncaré lemma, dscussed n secton 6.5 (well, t should be). The partcular forms we are usng here state that f B =0and F =0nallofR 3, then there exst a scalar functon φ and a vector feld A such that B = A and F = φ.

23 2.7. VELOCITY-DEPENDENT FORCES 57 Thus we see that the Lagrangan whch descrbes the moton of a charged partcle n an electromagnetc feld s gven by a velocty-dependent potental U( r, v) =q ( φ(r, t) ( v/c) A( r, t) ). Note, however, that ths Lagrangan descrbes only the moton of the charged partcle, and not the dynamcs of the feld tself. Arbtrarness n the Lagrangan In ths dscusson of fndng the Lagrangan to descrbe the Lorentz force, we used the lemma that guaranteed that the dvergenceless magnetc feld B can be wrtten n terms of some magnetc vector potental A, wth B = A.But A s not unquely specfed by B; n fact, f a change s made, A A + λ( r, t), B s unchanged because the curl of a gradent vanshes. The electrc feld E wll be changed by (1/c) A/ t, however, unless we also make a change n the electrostatc potental, φ φ (1/c) λ/ t. If we do, we have completely unchanged electromagnetc felds, whch s where the physcs les. Ths change n the potentals, A A + λ( r, t), φ φ (1/c) λ/ t, (2.20) s known as a gauge transformaton, and the nvarance of the physcs under ths change s known as gauge nvarance. Under ths change, the potental U and the Lagrangan are not unchanged, ( L L q δφ v ) c δ A = L + q λ c t + q c v λ( r, t) =L + q dλ c dt. We have here an example whch ponts out that there s not a unque Lagrangan whch descrbes a gven physcal problem, and the ambguty s more that just the arbtrary constant we always knew was nvolved n the potental energy. Ths ambguty s qute general, not dependng on the gauge transformatons of Maxwell felds. In general, f L (2) (q j, q j,t)=l (1) (q j, q j,t)+ d dt f(q j,t) (2.21) then L (1) and L (2) gve the same equatons of moton, and therefore the same physcs, for q j (t). Whle ths can be easly checked by evaluatng the Lagrange equatons, t s best understood n terms of the varaton of the acton. For

24 58 CHAPTER 2. LAGRANGE S AND HAMILTON S EQUATIONS any path q j (t) between q ji at t = t I to q jf at t = t F, the two actons are related by ( S (2) tf = L (1) (q j, q j,t)+ d ) dt f(q j,t) dt t I = S (1) + f(q jf,t F ) f(q ji,t I ). The varaton of path that one makes to fnd the statonary acton does not change the endponts q jf and q ji, so the dfference S (2) S (1) s a constant ndependent of the trajectory, and a statonary trajectory for S (2) s clearly statonary for S (1) as well. The conjugate momenta are affected by the change n Lagrangan, however, because L (2) = L (1) + j q j f/ q j + f/ t, so p (2) j = (2) = p (1) j + f. q j q j Ths ambguty s not usually mentoned n elementary mechancs, because f we restct our attenton to Lagrangans consstng of canoncal knetc energy and potentals whch are velocty-ndependent, a change (2.21) to a Lagrangan L (1) of ths type wll produce an L (2) whch s not of ths type, unless f s ndependent of poston q and leaves the momenta unchanged. That s, the only f whch leaves U velocty ndependent s an arbtrary constant. Dsspaton Another famlar force whch s velocty dependent s frcton. Even the constant sldng frcton met wth n elementary courses depends on the drecton, f not the magntude, of the velocty. Frcton n a vscous medum s often taken to be a force proportonal to the velocty, F = α v. We saw above that a potental lnear n veloctes produces a force perpendcular to v, and a term hgher order n veloctes wll contrbute a force that depends on acceleraton. Ths stuaton cannot handled by Lagrange s equatons. More generally, a Lagrangan can produce a force Q = R j q j wth antsymmetrc R j, but not for a symmetrc matrx. An extenson to the Lagrange formalsm, nvolvng Raylegh s dsspaton functon, can handle such a case. These dsspatve forces are dscussed n Ref. [6]. Exercses

25 2.7. VELOCITY-DEPENDENT FORCES (Galelean relatvty): Sally s sttng n a ralroad car observng a system of partcles, usng a Cartesan coordnate system so that the partcles are at postons r (S) (t), and move under the nfluence of a potental U (S) ({ r (S) }). Thomas s n another ralroad car, movng wth constant velocty u wth respect to Sally, and so he descrbes the poston of each partcle as r (T ) (t) = r (S) (t) ut. Each takes the knetc energy to be of the standard form n hs system,.e. T (S) = 1 ( ) (S) 2 2 m r and T (T ) = 1 ( ) (T ) 2. 2 m r (a) Show that f Thomas assumes the potental functon U (T ) ( r (T ) ) to be the same as Sally s at the same physcal ponts, U (T ) ( r (T ) )=U (S) ( r (T ) + ut), (2.22) then the equatons of moton derved by Sally and Thomas descrbe the same physcs. That s, f r (S) (t) s a soluton of Sally s equatons, r (T ) (t) =r (S) (t) ut s a soluton of Thomas. (b) show that f U (S) ({ r }) s a functon only of the dsplacements of one partcle from another, { r r j },thenu (T ) s the same functon of ts arguments as U (S), U (T ) ({ r })=U (S) ({ r }). Ths s a dfferent statement than Eq. 2.22, whch states that they agree at the same physcal confguraton. Show t wll not generally be true f U (S) s not restrcted to depend only on the dfferences n postons. (c) If t s true that U (S) ( r) =U (T ) ( r), show that Sally and Thomas derve the same equatons of moton, whch we call form nvarance of the equatons. (d) Show that nonetheless Sally and Thomas dsagree on the energy of a partcular physcal moton, and relate the dfference to the total momentum. Whch of these quanttes are conserved? 2.2 In order to show that the shortest path n two dmensonal Eucldean space s a straght lne wthout makng the assumpton that x does not change sgn along the path, we can consder usng a parameter λ and descrbng the path by two functons x(λ) and y(λ), say wth λ [0, 1]. Then 1 l = dλ ẋ 2 (λ)+ẏ 2 (λ), 0 where ẋ means dx/dλ. Ths s of the form of a varatonal ntegral wth two varables. Show that the varatonal equatons do not determne the functons x(λ) and y(λ), but do determne that the path s a straght lne. Show that the par of functons (x(λ),y(λ)) gves the same acton as another par ( x(λ), ỹ(λ)), where x(λ) = x(t(λ)) and ỹ(λ) = y(t(λ)), where t(λ) s any monotone functon mappng [0, 1] onto tself. Explan why ths equalty of the lengths s obvous

26 60 CHAPTER 2. LAGRANGE S AND HAMILTON S EQUATIONS n terms of alternate parameterzatons of the path. [In feld theory, ths s an example of a local gauge nvarance, and plays a major role n strng theory.] 2.3 Consder a crcular hoop of radus R rotatng about a vertcal dameter at a fxed angular velocty Ω. On the hoop there s a bead of mass m, whch sldes wthout frcton on the hoop. The only external force s gravty. Derve the Lagrangan and the Lagrange equaton usng the polar angle θ as the unconstraned generalzed coordnate. Fnd a conserved quantty, and fnd the equlbrum ponts, for whch θ = 0. Fnd the condton on Ω such that there s an equlbrum pont away from the axs. 2.4 Early steam engnes had a feedback devce, called a governor, to automatcally control the speed. The engne rotated a vertcal shaft wth an angular velocty Ω proportonal to ts speed. On opposte sdes of ths shaft, two hnged rods each held a metal weght, whch was attached to another such rod hnged to a sldng collar, as shown. As the shaft rotates faster, the balls move outwards, the collar rses and uncovers a hole, releasng some steam. Assume all hnges are frctonless, the rods massless, and each ball has mass m 1 and the collar has mass m 2. (a) Wrte the Lagrangan n terms of the generalzed coordnate θ. (b) Fnd the equlbrum angle θ as a functon of the shaft angular velocty Ω. Tell whether the equlbrum s stable or not. m m 1 1 m 2 Ω L L Governor for a steam engne. 2.5 A transformer conssts of two cols of conductor each of whch has an nductance, but whch also have a couplng, or mutual nductance.

27 2.7. VELOCITY-DEPENDENT FORCES 61 If the current flowng nto the upper posts of cols A and B are I A (t) andi B (t) respectvely, the voltage dfference or EMF across each col s V A and V B respectvely, where VA I A A VB B I B di A V A = L A dt + M di B dt di B V B = L B dt + M di A dt Consder the crcut shown, two capactors coupled by a such a transformer, where the capactances are C A and C B respectvely, wth the charges q 1 (t) andq 2 (t) servng as the generalzed coordnates for ths problem. Wrte down the two second order dfferental equatons of moton for q 1 (t) and q 2 (t), and wrte a Lagrangan for ths system. q q A 0 B 0 q q A cylnder of radus R s held horzontally n a fxed poston, and a smaller unform cylndrcal dsk of radus a s placed on top of the frst cylnder, and s released from rest. There s a coeffcent of statc frcton µ s and a coeffcent of knetc frcton µ k <µ s for the contact between the cylnders. As the equlbrum at the top s unstable, the top cylnder wll begn to roll on the bottom cylnder. (a) If µ s s suffcently large, the small dsk wll roll untl t separates from the fxed cylnder. Fnd the angle θ at whch the separaton occurs, and fnd the mnmum value of µ s for whch ths stuaton holds. (b) If µ s s less than the mnmum value found above, what happens dfferently, and at what angle θ does ths dfferent behavor begn? θ R A small cylnder rollng on a fxed larger cylnder. 2.7 (a) Show that f Φ(q 1,..., q n,t) s an arbtrary dfferentable functon on extended confguraton space, and L (1) ({q }, { q j },t) and L (2) ({q }, { q j },t) are two a

28 62 CHAPTER 2. LAGRANGE S AND HAMILTON S EQUATIONS Lagrangans whch dffer by the total tme dervatve of Φ, L (1) ({q }, { q j },t)=l (2) ({q }, { q j },t)+ d dt Φ(q 1,..., q n,t), show by explct calculatons that the equatons of moton determned by L (1) are the same as the equatons of moton determned by L (2). (b) What s the relatonshp between the momenta p (1) and p (2) determned by these two Lagrangans respectvely. 2.8 A partcle of mass m 1 moves n two dmensons on a frctonless horzontal table wth a tny hole n t. An nextensble massless strng attached to m 1 goes through the hole and s connected to another partcle of mass m 2, whch moves vertcally only. Gve a full set of generalzed unconstraned coordnates and wrte the Lagrangan n terms of these. Assume the strng remans taut at all tmes and that the motons n queston never have ether partcle reachng the hole, and there s no frcton of the strng sldng at the hole. Are there gnorable coordnates? Reduce the problem to a sngle second order dfferental equaton. Show ths s equvalent to sngle partcle moton n one dmenson wth a potental V (r), and fnd V (r). 2.9 Consder a mass m on the end of a massless rgd rod of length l, the other end of whch s free to rotate about a fxed pont. Ths s a sphercal pendulum. Fnd the Lagrangan and the equatons of moton (a) Fnd a dfferental equaton for θ(φ) for the shortest path on the surface of a sphere between two arbtrary ponts on that surface, by mnmzng the length of the path, assumng t to be monotone n φ. (b) By geometrcal argument (that t must be a great crcle) argue that the path should satsfy cos(φ φ 0 )=K cot θ, and show that ths s ndeed the soluton of the dfferental equaton you derved Consder some ntellgent bugs who lve on a turntable whch, accordng to nertal observers, s spnnng at angular velocty ω about ts center. At any one tme, the nertal observer can descrbe the ponts on the turntable wth polar coordnates r, φ. If the bugs measure dstances between two objects at rest wth respect to them, at nfntesmally close ponts, they wll fnd

29 2.7. VELOCITY-DEPENDENT FORCES 63 dl 2 = dr 2 r ω 2 r 2 /c 2 dφ2, because ther meterstcks shrnk n the tangental drecton and t takes more of them to cover the dstance we thnk of as rdφ, though ther meterstcks agree wth ours when measurng radal dsplacements. The bugs wll declare a curve to be a geodesc, or the shortest path between two ponts, f dl s a mnmum. Show that ths requres that r(φ) satsfes dr dφ = ± r α 1 ω 2 r 2 /c 2 2 r 2 1, where α s a constant. Straght lnes to us and to the bugs, between the same two ponts Hamlton s Prncple tells us that the moton of a partcle s determned by the acton functonal beng statonary under small varatons of the path Γ n extended confguraton space (t, x). The unsymmetrcal treatment of t and x(t) s not sutable for relatvty, but we may stll assocate an acton wth each path, whch we can parameterze wth λ, soγsthetrajectoryλ (t(λ), x(λ)). In the general relatvstc treatment of a partcle s moton n a gravtatonal feld, the acton s gven by mc 2 τ, where τ s the elapsed proper tme, τ = dτ. But dstances and tme ntervals are measured wth a spatal varyng metrc g µν, wth µ and ν rangng from 0 to 3, wth the zeroth component referrng to tme. The four components of extended confguraton space are wrtten x µ, wth a superscrpt rather than a subscrpt, and x 0 = ct. The gravtatonal feld s descrbed by the space-tme dependence of the metrc g µν (x ρ ). In ths language, an nfntesmal element of the path of a partcle corresponds to a proper tme dτ =(1/c) µν g µνdx µ dx ν,so S = mc 2 τ = mc dλ µν g µν (x ρ ) dxµ dλ dx ν dλ. (a) Fnd the four Lagrange equatons whch follow from varyng x ρ (λ). (b) Show that f we multply these four equatons by ẋ ρ and sum on ρ, wegetan dentty rather than a dfferental equaton helpng to determne the functons

30 64 CHAPTER 2. LAGRANGE S AND HAMILTON S EQUATIONS x µ (λ). Explan ths as a consequence of the fact that any path has a length unchanged by a reparameterzaton of the path, λ σ(λ), x µ (λ) =x µ (σ(λ) (c) Usng ths freedom to choose λ to be τ, the proper tme from the start of the path to the pont n queston, show that the equatons of moton are d 2 x λ dτ 2 + ρσ Γ λ dx ρ dx σ ρσ dτ dτ =0, and fnd the expresson for Γ λ ρσ (a): Fnd the canoncal momenta for a charged partcle movng n an electromagnetc feld and also under the nfluence of a non-electromagnetc force descrbed by a potental U( r). (b): If the electromagnetc feld s a constant magnetc feld B = B 0 ê z, wth no electrc feld and wth U( r) = 0, what conserved quanttes are there?