12. The HamiltonJacobi Equation Michael Fowler


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1 1. The HamltonJacob Equaton Mchael Fowler Back to Confguraton Space We ve establshed that the acton, regarded as a functon of ts coordnate endponts and tme, satsfes ( ) ( ) S q, t / t+ H qpt,, = 0, and at the same tme p = S( q, t) / q, so (, ) Ths s the HamltonJacob equaton. Notce that we re now back n confguraton space! S q t obeys the frstorder dfferental equaton S S S + H q1,, qs;,, ; t = 0. t q1 qs For eample, the HamltonJacob equaton for the smple harmonc oscllator n one dmenson s (, ) S( t, ) S t t = 0. (Notce that ths has some resemblance to the Schrödnger equaton for the same system.) If the Hamltonan has no eplct tme dependence S t H( qp) so the acton has the form S S ( q) Et 0 / +, = 0 becomes just S / t = E, =, and the HamltonJacob equaton s S S H q1,, qs;,, = E. q1 qs (Ths s analogous to the tme ndependent Schrödnger equaton for energy egenstates.) The HamltonJacob equaton s therefore a thrd complete descrpton of the dynamcs, equvalent to Lagrange s equatons and to Hamlton s equatons. Snce S only appears dfferentated, f we have a soluton to the equaton, we can always add an arbtrary constant term, to gve an equally vald soluton. For the general case, there wll be a further s constants of ntegraton, so a complete soluton has the form ( ) ( α α ) S q, t = f tq,,, q;,, + A, 1 s 1 s the α 's and A beng the constants of ntegraton. We're not sayng t's easy to solve ths dfferental n general, just that we know how many constants of ntegraton there must be n a fnal soluton. Snce the acton determnes the moton of the system completely, the constants of ntegraton wll be determned by the gven ntal and fnal coordnates, or, they could equally be regarded as functons of
2 the ntal coordnates and momenta (the ntal momenta themselves beng determned by the gven ntal and fnal coordnates). The Central Role of These Constants of Integraton To descrbe the tme development of a dynamcal system n the smplest way possble, t s desrable to fnd parameters that are constant or change n a smple way. For eample, moton n a sphercally symmetrc potental s descrbed n terms of (constant) angular momentum components. Now, these constant α's are functons of the ntal coordnates and momenta. Snce they reman constant durng the moton, they are evdently among the "varables" that descrbe the dynamcal development n the smplest possble way. So, we need to construct a canoncal transformaton from our current set of varables (fnal coordnates and momenta) to a new set of varables that ncludes these constant of ntegraton "momenta". (The correspondng canoncal "postons" wll then be gven by dfferentatng the generatng functon wth respect to the "momenta".) How do we fnd the generatng functon for ths transformaton? A clue comes from one we ve already dscussed: that correspondng to development n tme, gong from the ntal set of varables to the fnal set, or back. That transformaton was generated by the acton tself, epressed n terms of the two sets of postons. That s, we allowed both ends of the acton ntegral path to vary, and wrote the acton as a 1 endpont varables and tmes: functon of the fnal ( ) and ntal ( ) ( ) ( 1) ( ) ( ) ( ) ( 1) ( 1) ( 1) ( ) = + ds q, t, q, t p dq H dt p dq H dt. 1 1 In the present secton, the fnal endpont postons are denoted smply by tq, 1,, qs these are the same ( ) ( ) 1 ( ) ( 1) ( 1) ( 1) ( ) s s as the earler t, q,, qs. Eplctly, we're wrtng ( ) S q, t, q, t S q,, q, t, q, q, t Compare ths epresson for the acton wth the formal epresson we just derved from the Hamlton Jacob equaton, ( ) ( α α ) S q,, q, t = f q,, q, t;,, + A. 1 s 1 s 1 s These two epressons for S have just the same form: the acton s epressed as a functon of the endpont poston varables, plus another s varables needed to determne the moton unquely. Ths tme, nstead of the orgnal poston varables, though, the second set of varables s these constants of ntegraton, theα 's. Now, just as we showed the acton generated the transformaton (ether way) between the ntal set of coordnates and momenta and the fnal set, t wll also generate a canoncal transformaton from the fnal set of coordnates and momenta to another canoncal set, havng the α 's as the new "momenta". We'll label the new "coordnates" (the canoncal conjugates of theα 's) β,, 1 βs.
3 3 Takng then the acton (neglectng the constant A whch does nothng) S= f( tq,,, q; α,, α ) 1 s 1 as the generatng functon, t depends on the old coordnates q and the new momentaα. Ths s the same set of varables  old coordnates and new momenta  as those of the (prevously dscussed) generatng functon Φ ( qpt),,. s Recall so here (,, ) ( ) dφ q P t = pdq + QdP + H H dt, (, α, ) β α ( ) df q t = p dq + d + H H dt, and p = f / q, β = f / α, H = H + f / t. Ths defnes the new "coordnates" β, and ensures that the transformaton s canoncal. To fnd the new Hamltonan H, we need to fnd f / t and add t to H. But ( ) ( α α ) S q, t = f tq,,, q;,, + A, 1 s 1 s where A s just a constant, so f / t = S / t. The frst equaton n ths secton was ( ) S/ t+ H qpt,, = 0, so the new Hamltonan H = H + f / t = H + S / t = 0. We have made a canoncal transformaton that has led to a ero Hamltonan! What does that mean? It means that the nether the new momenta nor the new coordnates vary n tme: [ H ] [ H ] α =, α = 0, β =, β = 0. (The fact that all momenta and coordnates are fed n ths representaton does not mean that the system doesn t move  as wll become evdent n the followng smple eample, the orgnal coordnates are functons of these new (nonvaryng!) varables and tme.)
4 4 The s equatons f / α = β can then be used to fnd the q as functons of α, β, ths works, t s necessary to work through an eample. A Smple Eample of the HamltonJacob Equaton: Moton Under Gravty The Hamltonan for moton under gravty n a vertcal plane s so the HamltonJacob equaton s 1 H = ( p + p) + mg m (,, ) (,, ) (,, ) t 1 S t S t S t + + mg + = 0 m t Frst, ths Hamltonan has no eplct tme dependence (gravty sn't changng!), so from S / t+ H q, p = 0 = S / t+ E, we can replace the last term n the equaton by E. ( ). To see how all A Smple Separaton of Varables Snce the potental energy term depends only on, the equaton s solvable usng separaton of varables. To see ths works, try ( ) ( ) ( ) S,, t = W + W Et. Puttng ths form nto the equaton, the resultng frst term depends only on the varable, the second plus thrd depend only on, the last term s just the constant E. A functon dependng only on can only equal a functon ndependent of f both are constants, smlarly for. Labelng the constants α, α, ( ) ( ) = α + mg = α E = α + α 1 dw 1 dw,,. m d m d So theseα 's are constants of the moton, they are our new "momenta" (although they have dmensons of energy). Solvng, 8 W ( ) = ± mα, W ( ) = ± ( α mg) 3/. 9mg (We could add n constants of ntegraton, but addng constants to the acton changes nothng.) So now we have ( α α ) ( α ) ( α ) ( α α ) S= S,,,, t = W, + W, + t.
5 5 Ths s our generatng functon (equvalent to Φ ( qpt,, ) ), n terms of old coordnates and these new momenta, α, α. Followng the HamltonJacob analyss, ths acton wll generate a canoncal transformaton whch reduces the Hamltonan to ero, meanng that not only these new momenta stay constant, but so do ther conjugate coordnate varables, ( α mg) S m S β = = ± t, β = = ± t α α α mg These equatons solve the problem. Rearrangng, the trajectory s α α g = ± ( β ), ( ) + t = β + t. m mg The four constants of moton α, α, β, βare unquely fed by the ntal coordnates and veloctes, and they parametere the subsequent tme evoluton of the system. Separaton of Varables for a Central Potental; Cyclc Varables Landau presents n some detals the separaton of varables method for a 1/r potental, nterestng here because t results n equatons you ve met before  those arsng n the standard quantum treatment of the hydrogen atom. How do we make any progress wth these formdable dfferental equatons? One possblty s that some coordnates are cyclc, meanng that q, 1 say, does not appear eplctly n the Hamltonan  for eample, an angle varable n a sphercally symmetrc feld. Then we have mmedately that the correspondng momentum, p1 = S / q1 = α1, a constant. The Hamltonan for a central potental s: The HamltonJacob equaton s therefore 1 p pφ θ H = pr + + V + ( r). m r r sn θ V ( r) 1 S 1 S 1 S = E. m r mr θ mr sn θ φ The frst thng to note s thatφ s cyclc (t doesn't appear n the Hamltonan), so we can mmedately replace S / φ 0 wth a constant. Then we have: p φ 1 S0 1 S p 0 φ + V ( r) + E. + = m r mr θ sn θ
6 6 Now we seek a soluton of the form ( ) ( ) ( ) S0 r, θφ, = Sr r + Sθ θ + pφφ. Substtutng n the equaton, notce that the epresson n square brackets wll become S p θ + θ sn θ φ, ndependent of r, but on multplyng the full equaton by r, and starng at the result, we see that n fact t s purely a functon of r. Ths means that t s a constant, say and then S p θ + = θ sn θ φ β, 1 S r β + V ( r) + E. = m r mr These frstorder equatons can then be solved, at least numercally (and of course eactly for some cases). Physcally, β =, beng the total angular momentum, and E s the total energy. Note: recall that n quantum mechancs, for eample n solvng the Schrödnger equaton for the hydrogen atom, the separaton of varables was acheved by wrtng the wave functon as a product of functons belongng to the dfferent varables. Here we use a sum remember that the acton corresponds closely to the phase of a quantum mechancal system, so a sum of actons s analogous to a product of wave functons.
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