4. Laws of Dynamics: Hamilton s Principle and Noether's Theorem

Transcription

1 4. Laws of Dynamcs: Hamlton s Prncple and Noether's Theorem Mchael Fowler Introducton: Galleo and Newton In the dscusson of calculus of varatons, we antcpated some basc dynamcs, usng the potental energy mgh for an element of the catenary, and conservaton of energy 1 mv mgh E + = for moton along the brachstochrone. Of course, we haven t actually covered those thngs yet, but you re already very famlar wth them from your undergraduate courses, and my am was to gve easly understood physcal realzatons of mnmzaton problems, and to show how to fnd the mnmal shapes usng the calculus of varatons. At ths pont, we ll begn a full study of dynamcs, startng wth the laws of moton. The text, Landau, begns (page!) by statng that the laws come from the prncple of least acton, Hamlton s prncple. Ths s certanly one possble approach, but confronted wth t for the frst tme, one mght well wonder where t came from. I prefer a gentler ntroducton, more or less followng the hstorcal order: Galleo, then Newton, then Lagrange and hs colleagues, then Hamlton. The two approaches are of course equvalent. Naturally, you ve seen most of ths earler stuff before, so here s a very bref summary. To begn, then, wth Galleo. Hs two major contrbutons to dynamcs were: 1. The realzaton, and expermental verfcaton, that fallng bodes have constant acceleraton (provded ar resstance can be gnored) and all fallng bodes accelerate at the same rate.. Gallean relatvty. As he put t hmself, f you are n a closed room below decks n a shp movng wth steady velocty, no experment on droppng or throwng somethng wll look any dfferent because of the shp s moton: you can t detect the moton. As we would put t now, the laws of physcs are the same n all nertal frames. Newton s major contrbutons were hs laws of moton, and hs law of unversal gravtatonal attracton. Hs laws of moton: 1. The law of nerta: a body movng at constant velocty wll contnue at that velocty unless acted on by a force. (Actually, Galleo essentally stated ths law, but just for a ball rollng on a horzontal plane, wth zero frctonal drag.). F = ma. 3. Acton = reacton. In terms of Newton s laws, Gallean relatvty s clear: f the shp s movng at steady velocty v relatve to the shore, than an object movng at u relatve to the shp s movng at u+ v relatve to the shore. If there s no force actng on the object, t s movng at steady velocty n both frames: both are nertal frames, defned as frames n whch Newton s frst law holds. And, snce v s constant, the acceleraton s the same n both frames, so f a force s ntroduced the second law s the same n the two frames. (Needless to say, all ths s classcal, meanng nonrelatvstc, mechancs.)

2 Any dynamcal system can be analyzed as a (possbly nfnte) collecton of parts, or partcles, havng mutual nteractons, so n prncple Newton s laws can provde a descrpton of the moton developng from an ntal confguraton of postons and veloctes. The problem s, though, that the equatons may be ntractable we can t do the mathematcs. It s evdent that n fact the Cartesan coordnate postons and veloctes mght not be the best choce of parameters to specfy the system s confguraton. For example, a smple pendulum s obvously more naturally descrbed by the angle the strng makes wth the vertcal, as opposed to the Cartesan coordnates of the bob. After Newton, a seres of French mathematcans reformulated hs laws n terms of more useful coordnates culmnatng n Lagrange s equatons. The Irsh mathematcan Hamltonan then establshed that these mproved dynamcal equatons could be derved usng the calculus of varatons to mnmze an ntegral of a functon, the Lagrangan, along a path n the system s confguraton space. Ths ntegral s called the acton, so the rule s that the system follows the path of least acton from the ntal to the fnal confguraton. Dervaton of Hamlton s Prncple from Newton s Laws n Cartesan Co-ordnates: Calculus of Varatons Done Backwards! We ve shown how, gven an ntegrand, we can fnd dfferental equatons for the path n space tme between two fxed ponts that mnmzes the correspondng path ntegral between those ponts. Now we ll do the reverse: we already know the dfferental equatons n Cartesan coordnates descrbng the path taken by a Newtonan partcle n some potental. We ll show how to use that knowledge to construct the ntegrand such that the acton ntegral s a mnmum along that path. (Ths follows Jeffreys and Jeffreys, Mathematcal Physcs.) We begn wth the smplest nontrval system, a partcle of mass m movng n one dmenson from one pont to another n a specfed tme, we ll assume t s n a tme-ndependent potental U( x ), so ( ) /. mx = du x dx Its path can be represented as a graph x( t ) aganst tme for example, for a ball thrown drectly upwards n a constant gravtatonal feld ths would be a parabola. Intal and fnal postons are gven: x( t ) = x, x( t ) 1 1 = x, and the elapsed tme s t t1 Notce we have not specfed the ntal velocty we don t have that opton. The dfferental equaton s only second order, so ts soluton s completely determned by the two (begnnng and end) boundary condtons. We re now ready to embark on the calculus of varatons n reverse. Trvally, multplyng both sdes of the equaton of moton by an arbtrary nfntesmal functon the equalty stll holds: ( / ) δ ( ) ( ) = ( ) mx δx t du x dx x t.

3 3 and n fact f ths equaton s true for arbtrary δ x( t), the orgnal equaton of moton holds throughout, because we can always choose a δ x( t) nonzero only n the neghborhood of a partcular tme t, from whch the orgnal equaton must be true at that t. By analogy wth Fermat s prncple n the precedng secton, we can pcture ths δ x( t) as a slght varaton n the path from the Newtonan trajectory, x( t) x( t) + δ x( t), and take the varaton zero at the fxed ends, δx( t ) = δx( t ) =. 1 0 In Fermat s case, the ntegrated tme elapsed along the path was mnmzed there was zero change to frst order on gong to a neghborng path. Developng the analogy, we re lookng for some dynamcal quantty that has zero change to frst order on gong to a neghborng path havng the same endponts n space and tme. We ve fxed the tme, what s left to ntegrate along the path? For such a smple system, we don t have many optons! As we ve dscussed above, the equaton of moton s equvalent to (puttng n an overall mnus sgn that wll prove convenent) t t1 ( mx( t) du( x( t) ) dx) δx( t) dt = δx( t) / 0 to leadng order, for all varatons. Integratng the frst term by parts (recallng δ ( x) = 0 at the endponts): t t t t 1 ( ) δ ( ) ( ) δ ( ) δ ( ( )) δ ( ( )) mx t x t dt = mx t x t dt = mx t dt = T x t dt, t1 t1 t1 t1 usng the standard notatont for knetc energy. The second term ntegrates trvally: t t ( ( ) / ) δ ( ) δ ( ) du x dx x t dt = U x dt t1 t1 establshng that on makng an nfntesmal varaton from the physcal path (the one that satsfes Newton s laws) there s zero frst order change n the ntegral of knetc energy mnus potental energy. The standard notaton s t t ( ) δs = δ T U dt = δ Ldt = 0. t1 t1 The ntegral S s called the acton ntegral, (also known as Hamlton s Prncpal Functon) and the ntegrand s called the Lagrangan. Ths equaton s Hamlton s Prncple. T U = L

4 4 The dervaton can be extended straghtforwardly to a partcle n three dmensons, n fact to n nteractng partcles n three dmensons. We shall assume that the forces on partcles can be derved from potentals, ncludng possbly tme-dependent potentals, but we exclude frctonal energy dsspaton n ths course. (It can be handled see for example Vujanovc and Jones, Varatonal Methods n Nonconservatve Phenomena, Academc press, 1989.) But Why? Fermat s prncple was easy to beleve once t was clear that lght was a wave. Imagnng that the wave really propagates along all paths, and for lght the phase change along a partcular path s smply the tme taken to travel that path measured n unts of the lght wave oscllaton tme. That means that f neghborng paths have the same length to frst order the lght waves along them wll add coherently, otherwse they wll nterfere and essentally cancel. So the path of least tme s heavly favored, and when we look on a scale much greater than the wavelength of the lght, we don t even see the dffracton effects caused by mperfect cancellaton, the lght rays mght as well be streams of partcles, mysterously choosng the path of least tme. So what has ths to do wth Hamlton s prncple? Everythng. A standard method n quantum mechancs these days s the so-called sum over paths, for example to fnd the probablty ampltude for an electron to go from one pont to another n a gven tme under a gven potental, you can sum over all possble paths t mght take, multplyng each path by a phase factor: and that phase factor s none other than Hamlton s acton ntegral dvded by Planck s constant, S /. So the true wave nature of all systems n quantum mechancs ensures that n the classcal lmt S the well-defned path of a dynamcal system wll be that of least acton. (Ths s covered n my notes on Quantum Mechancs.) Hstorcal footnote: Lagrange developed these methods n a classc book that Hamlton called a scentfc poem. Lagrange thought mechancs properly belonged to pure mathematcs, t was a knd of geometry n four dmensons (space and tme). Hamlton was the frst to use the prncple of least acton to derve Lagrange s equatons n the present form. He bult up the least acton formalsm drectly from Fermat s prncple, consdered n a medum where the velocty of lght vares wth poston and wth drecton of the ray. He saw mechancs as represented by geometrcal optcs n an approprate space of hgher dmensons. But t ddn t apparently occur to hm that ths mght be because t was really a wave theory! (See Arnold, Mathematcal Methods of Classcal Mechancs, for detals.) Lagrange s Equatons from Hamlton s Prncple Usng Calculus of Varatons We started wth Newton s equatons of moton, expressed n Cartesan coordnates of partcle postons. For many systems, these equatons are mathematcally ntractable. Runnng the calculus of varatons argument n reverse, we establshed Hamlton s prncple: the system moves along the path through confguraton space for whch the acton ntegral, wth ntegrand the Lagrangan L= T U, s a mnmum. We re now free to begn from Hamlton s prncple, expressng the Lagrangan n varables that more naturally descrbe the system, takng advantage of any symmetres (such as usng angle varables for rotatonally nvarant systems). Also, some forces do not need to be ncluded n the descrpton of the system: a smple pendulum s fully specfed by ts poston and velocty, we do not need to know the tenson n the strng, although that would appear n a Newtonan analyss. The greater effcency (and elegance) of the Lagrangan method, for most problems, wll become evdent on workng through actual examples.

5 We ll defne a set of generalzed coordnates q ( q q ) 5 = by requrng that they gve a complete descrpton of 1, n the confguraton of the system (where everythng s n space). The state of the system s specfed by ths set plus the correspondng veloctes q = ( q q ) some functon of the, 1 n. For example, the x -coordnate of a partcular partcle a s gven by,, f x xa = fx q a 1 qn and the correspondng velocty component x a a = q k. q q s, ( ) The Lagrangan wll depend on all these varables n general, and also possbly on tme explctly, for example f there s a tme-dependent external potental. (But usually that sn t the case.) Hamlton s prncple gves k k that s, t t1 ( ) δs = δ L q, q, t dt = 0 t t1 δq + δq dt = 0. q q Integratng by parts, t t d δs = δq δqdt 0. q + = q dt q t t 1 1 Requrng the path devaton to be zero at the endponts gves Lagrange s equatons: d 0. dt q q = Non-unqueness of the Lagrangan The Lagrangan s not unquely defned: two Lagrangans dfferng by the total dervatve wth respect to tme of some functon wll gve the same dentcal equatons on mnmzng the acton, t1 t1 t1 (, ) t t t df q t S = L ( q, q, t) dt = L( q, q, t) dt + dt = S + f ( q ( t), t) f ( q ( t1), t1), dt q t1, t1, q t, t are all fxed, the ntegral over df / varyng the path to mnmze S gves the same result as mnmzng S. Ths turns out to be mportant later t gves us a useful new tool to change the varables n the Lagrangan. and snce ( ) ( ) dt s trvally ndependent of path varatons, and

6 6 Frst Integral: Energy Conservaton and the Hamltonan Snce Lagrange s equatons are precsely a calculus of varatons result, t follows from our earler dscusson that f the Lagrangan has no explct tme dependence then: L q L = constant. q (Ths s just the frst ntegral y f / y f = constant dscussed earler, now wth n varables.) Ths constant of moton s called the energy of the system, and denoted by E. We say the energy s conserved, even n the presence of external potentals provded those potentals are tme-ndependent. (We ll just menton that the functon on the left-hand sde, q/ q L, s the Hamltonan. We don t dscuss t further at ths pont because, as we ll fnd out, t s more naturally treated n other varables.) We ll now look at a couple of smple examples of the Lagrangan approach. Example 1: One Degree of Freedom: Atwood s Machne In 1784, the Rev. George Atwood, tutor at Trnty College, Cambrdge, came up wth a great demo for fndng g. It s stll wth us. The tradtonal Newtonan soluton of ths problem s to wrte F = ma for the two masses, then elmnate the tensont. (To keep thngs smple, we ll neglect the rotatonal nerta of the top pulley.) x The Lagrangan approach s, of course, to wrte down the Lagrangan, and derve the equaton of moton. m1 m Measurng gravtatonal potental energy from the top wheel axle, the potental energy s and the Lagrangan ( ) = ( ) U x m gx m g x 1 ( ) ( ) L = T U = m + m x + m gx + m g x Lagrange s equaton:

7 7 gves the equaton of moton n just one step. d = ( m + m ) x ( m m ) g = dt x x It s usually pretty easy to fgure out the knetc energy and potental energy of a system, and thereby wrte down the Lagrangan. Ths s defntely less work than the Newtonan approach, whch nvolves constrant forces, such as the tenson n the strng. Ths force doesn t even appear n the Lagrangan approach! Other constrant forces, such as the normal force for a bead on a wre, or the normal force for a partcle movng on a surface, or the tenson n the strng of a pendulum none of these forces appear n the Lagrangan. Notce, though, that these forces never do any work. On the other hand, f you actually are nterested n the tenson n the strng (wll t break?) you use the Newtonan method, or maybe work backwards from the Lagrangan soluton. Example : Lagrangan Formulaton of the Central Force Problem A smple example of Lagrangan mechancs s provded by the central force problem, a mass m acted on by a force r ( ) / F = du r dr. To contrast the Newtonan and Lagrangan approaches, we ll frst look at the problem usng just F = ma. To take advantage of the rotatonal symmetry we ll use ( r, θ ) coordnates, and fnd the expresson for acceleraton by the standard trck of dfferentatng the complex number ( θ ) = ( ) ( θ + r θ) = 0. z = re θ twce, to get m r r du r / dr m r The second equaton ntegrates mmedately to gve mr θ =, a constant, the angular momentum. Ths can then be used to elmnateθ n the frst equaton, gvng a dfferental equaton for r( t ). The Lagrangan approach, on the other hand, s frst to wrte and put t nto the equatons 1 ( θ ) ( ) L= T U = m r + r U r

8 8 d = 0, dt r r d = 0. dt θ θ Note now that snce L doesn t depend on θ, the second equaton gves mmedately: L = constant, θ and n fact / θ = mr θ, the angular momentum, we ll call t. The frst ntegral (see above) gves another constant: r + θ L= constant. r θ Ths s just the energy. ( θ ) ( ) 1 m r + r + U r = E Angular momentum conservaton, mr θ =, then gves m r + + U( r) = E mr 1 gvng a frst-order dfferental equaton for the radal moton as a functon of tme. We ll deal wth ths n more detal later. Note that t s equvalent to a partcle movng n one dmenson n the orgnal potental plus an effectve potental from the angular momentum term: 1 ( ). E = mv + U r + mr Ths can be understood by realzng that for a fxed angular momentum, the closer the partcle approaches the center the greater ts speed n the tangental drecton must be, so, to conserve total energy, ts speed n the radal drecton has to go down, unless t s n a very strongly attractve potental (the usual gravtatonal or electrostatc potental sn t strong enough) so the radal moton s equvalent to that wth the exstng potental plus the / mr term, often termed the centrfugal barrer. Exercse: how strong must the potental be to overcome the centrfugal barrer? (Ths can happen n a black hole!)

9 9 Generalzed Momenta and Forces For the above orbtal Lagrangan, dl / dr= mr= p r, the momentum n the r -drecton, and dl / d θ = mr θ = p θ, the angular momentum assocated wth the varable θ. The generalzed momenta for a mechancal system are defned by p =. q (Warnng: these generalzed momenta are an essental part of the formalsm, but do not always drectly correspond to the physcal momentum of a partcle, an example beng a charged partcle n a magnetc feld, see my quantum notes.) Less frequently used are the generalzed forces, F =/ q, defned to make the Lagrange equatons look Newtonan, F = p. Conservaton Laws and Noether s Theorem The two ntegrals of moton for the orbtal example above can be stated as follows: Frst: f the Lagrangan does not depend on the varable θ, / θ = 0, that s, t s nvarant under rotaton, meanng t has crcular symmetry, then angular momentum s conserved. p θ = = constant, θ Second: As stated earler, f the Lagrangan s ndependent of tme, that s, t s nvarant under tme translaton, then energy s conserved. (Ths s nothng but the frst ntegral of the calculus of varatons, recall that for an f y y not explctly dependent on x, y f / y f s constant.) ntegrand functon (, ) q / q L = E, a constant. Both these results lnk symmetres of the Lagrangan nvarance under rotaton and tme translaton respectvely wth conserved quanttes. Ths connecton was frst spelled out explctly, and proved generally, by Emmy Noether, publshed n The essence of the theorem s that f the Lagrangan (whch specfes the system completely) does not change when some contnuous parameter s altered, then some functon of the q, q stays the same t s called a constant of the moton, or an ntegral of the moton. To look further at ths expresson for energy, we take a closed system of partcles nteractng wth each other, but closed means no nteracton wth the outsde world (except possbly a tme-ndependent potental).

10 10 The Lagrangan for the partcles s, n Cartesan coordnates, A set of general coordnates ( q q ) 1, n L= mv U r r. ( ) 1 1,, coordnate and velocty of a partcular partcle a are gven by, by defnton, unquely specfes the system confguraton, so the (, 1 ), f xa x. a = fx q q a n xa = qk q k k From ths t s clear that the knetc energy term (meanng every term s of degree two), so T = mv s a homogeneous quadratc functon of the q s 1 1 L= ak q qq k U q k, Ths beng of degree two n the tme dervatves means ( ) ( ) T q = q = T. q q. (If ths sn t obvous to you, check t out wth a couple of terms: q, qq.) 1 1 Therefore, recallng from the tme-ndependence of the Lagrangan that E = q/ q L s constant, we see that ths constant s smply E = T + U, whch as we shall see, s the Hamltonan, to be dscussed later. Momentum Another conservaton law follows f the Lagrangan s unchanged by dsplacng the whole system through a dstance δr = ε. Ths means, of course, that the system cannot be n some spatally varyng external feld t must be mechancally solated. It s natural to work n Cartesan coordnates to analyze ths, each partcle s moved the same dstance r r + δr = r + ε, so δl= δr = ε r where the dfferentaton by a vector notaton means dfferentatng wth respect to each component, then addng the three terms. (I m not crazy about ths notaton, but t s Landau s, so get used to t.), r

11 11 For an solated system, we must have δ L = 0 on dsplacement, movng the whole thng through empty space n any drectonε changes nothng, so t must be that the vector sum / r = 0 Lagrange equatons, wrtng r = v, d d d 0 = / r = = =, dt r dt v dt v so, takng the system to be composed of partcles of mass m and velocty v, the momentum of the system. L = mv = P = constant, v, so from the Cartesan Euler- Ths vector conservaton law s of course three separate drectonal conservaton laws, so even f there s an external feld, f t doesn t vary n a partcular drecton, the component of total momentum n that drecton wll be conserved. (nto page) O An nfntesmal rotaton s represented by a vector of length along the axs. In the Newtonan pcture, conservaton of momentum n a closed system follows from Newton s thrd law. In fact, the above Lagrangan analyss s really Newton s thrd law n dsguse. Snce we re workng n Cartesan coordnates, / r = V / r = F, the force on the th partcle, and f there are no external felds, / r = 0 just means that f you add all the forces on all the partcles, the sum s zero. For the Lagrangan of a two partcle system to be nvarant under translaton through space, the potental must have the form V ( r r ), from whch automatcally F1 = F1. 1 Center of Mass If an nertal frame of reference K s movng at constant velocty V relatve to nertal frame K, the veloctes of ndvdual partcles n the frames are related by v = v + V, so the total momenta are related by P= mv = mv + V m = P + MV, M = m. If we choose V = P/ M, then P = mv = 0, the system s at rest n the frame K. Of course, the ndvdual partcles mght be movng, what s at rest n K s the center of mass defned by

12 MR 1 = m r cm. (Check ths by dfferentatng both sdes wth respect to tme.) The energy of a mechancal system n ts rest frame s often called ts nternal energy, we ll denote t by E. nt (Ths ncludes knetc and potental energes.) The total energy of a movng system s then (Exercse: verfy ths.) Angular Momentum E = MV + E 1 nt. Conservaton of momentum followed from the nvarance of the Lagrangan on beng dsplaced n arbtrary drectons n space, the homogenety of space, angular momentum conservaton s the consequence of the sotropy of space there s no preferred drecton. So angular momentum of an solated body n space s nvarant even f the body s not symmetrc tself. The strategy s just as before, except now nstead of an nfntesmal dsplacement we make an nfntesmal rotaton, δ r = δφ r and of course the veloctes wll also be rotated: δ v = δφ v. We must have Now / v =/ r = p so the sotropy of space mples that δ L L r v 0. r δ v δ = + = / r = d / dt / r = p, by defnton, and from Lagrange s equatons ( )( ) ( p δφ r + p δφ v) = 0. Notce the second term s dentcally zero anyway, snce two of the three vectors n the trple product are parallel: dr / dt p = v mv = 0 ( ) That leaves the frst term. The equaton can be wrtten:

13 13 d δφ r p = 0, dt Integratng, we fnd that r p = L s a constant of moton, the angular momentum. The angular momentum of a system s dfferent about dfferent orgns. (Thnk of a sngle movng partcle.) The angular momentum n the rest frame s often called the ntrnsc angular momentum, the angular momentum n a frame n whch the center of mass s at poston R and movng wth velocty V s L = L + R P (Exercse: check ths.) cm frame. For a system of partcles n a fxed external central feld V ( r ), the system s nvarant wth respect to rotatons about that pont, so angular momentum about that pont s conserved. For a feld cylndrcally nvarant for rotatons about an axs, angular momentum about that axs s conserved.