Analytical classical dynamics

Transcription

1 Analytcal classcal ynamcs by Youun Hu Insttute of plasma physcs, Chnese Acaemy of Scences Emal: Abstract These notes were ntally wrtten when I rea tzpatrck s book[] an were later revse to a more contents. Generalze coornates or a ynamcal system wth egrees of freeom, the Cartesan coornates can be expresse n terms of generalze coornates (q,q 2,,q, x =x (q,q 2,.,q,t, for,, ( 2 Generalze force The work on a ynamcal system when ts Cartesan coornates change by δx s gven by δw = δx = x δq = = x δq = = = Q δq, where Q s efne by Q = x, (2 whch s calle generalze force. or conservatve system, can be wrtten n the form Usng Eq. (3, the generalze force s wrtten as = U x. (3 Q = U x x = U. (4 3 Euler-Lagrange equaton Newton s secon law s wrtten as m ẍ =. (5

2 2 Secton 3 Usng we obtan ẋ = x t + x q, (6 = ẋ q I = x, (7 where the partal ervatve on the left-han se s taken by holng q,, q constant (ths conventon s mportant for ervng the Euler-Lagrange equaton. Multplyng Eq. (7 by mẋ an summng over, we obtan ẋ m ẋ = x m ẋ. (8 q I Takng tme fferental, the above equaton s wrtten as ẋ m ẋ = x m ẋ t q I t (9 The left-han se of Eq. (9 s wrtten as ẋ m ẋ t q I = 2 t 2 m ẋ q I = t q I 2 m 2 ẋ = ( K t q I, (0 where K m 2 ẋ 2 s the knetc energy. The rght-han se of Eq. (9 s wrtten as x m ẍ + m t ( x Usng Newton s secon law, the above expresson s wrtten as x + m t ( x ẋ. ( ẋ, (2 whch can be further wrtten as, by usng the efnton of the generalze force, Q I + m t ( x ẋ. (3 In orer to smplfy the secon term n expresson (3, we try to prove that x = ẋ, (4 t where the partal ervatve wth respect to q I on the rght-han se s taken by holng q, q 2,.,q constant. [Proof: The rght-han se of the above equaton s wrtten as ẋ = x t + x q = = x t + x q, (5 =

3 Prove that Euler-Lagrange equaton s coornates nepenent 3 In obtanng the last equalty, we have use the fact that the orer of the partal ervatve of x wth respect to q, q I, an t s nterchangeable. The rght-han se of Eq. (5 can be further wrtten as t ( x + = ( x q, (6 whch s obvously the total tme ervatve of x /,..e., x, (7 t whch s exactly the left-han se of Eq. (4. Thus Eq. (4 s prove.] Usng Eq. (4, the expresson (3 s wrtten as whch can be further wrtten as.e., Thus we obtan Q I + m ẋ ẋ, Q I m ẋ, Q I + K (8 K =Q I + K. t q I (9 If the generalze force s gven by Eq. (4, equaton (9 s wrtten as K = U + K. (20 t q I Defne L=K U (2 an notng that U s nepenent of q,q, q, Eq. (20 s wrtten as L = L, (22 t q I whch s the Euler-Lagrange equaton, where, recallng the remarks below Eqs. (7 an (4, we know that the partal ervatve wth respect to q I an q I are taken by treatng q an q, respectvely, as nepenent varables. Defne the canoncal momentum then the Euler-Lagrange equaton s wrtten as p I = L q İ, (23 t (p I= L. (24 4 Prove that Euler-Lagrange equaton s coornates nepenent Try to use L = L, (25 t ẋ x where x are rectangular coornates, to prove that L = L. (26 t q

4 4 Secton 5 where q are arbtrary generalze coornates. Proof: The left-han se of Eq. (26 s wrtten as ( L = L x + t q t x q ( = 0+ L ẋ t ẋ q = L ẋ t ẋ q = [ L ẋ t ẋ q Usng Eq. (25, the above equaton s wrtten as ( L t q Usng the fact that Eq. (29 s wrtten as Usng the fact that t = ( L q t = L ẋ ẋ q + L ẋ t [ L ẋ + L x q ẋ t ( ẋ = ẋ ] ẋ q ] ẋ q (27 (28 (29. (30 ] (3 q [ L ẋ + L ẋ x q ẋ ẋ q Eq. (3 s further wrtten as L = t q.e., whch s what we want. t = x (32 [ L x + L ] ẋ x ẋ = L, (33 ( L q = L, (34 5 Usng Euler-Lagrange equaton to erve Hamlton s equaton The Euler-Lagrange equaton s gven by wth p I efne by t (p I= L, (35 p I = L q İ, (36 The Lagrangan L s efne as the fference of the knetc energy an the potental energy of the system,.e., L=K U. (37 In the Euler-Lagrange equaton, the nepenent varables are chosen to be q an q wth =, 2, where s the freeom of the system,.e., L=L(q,q 2, q,q, q Ḟ. (38 Now we use the above results to erve Hamlton s equaton. In Hamlton s equaton the nepenent varables are chosen to be q an p wth =,2,.. The Hamltonan s equal to the total energy of the system,.e. H=K+U. (39

5 Prove that 2K= p q 5 An the Hamltonan must be expresse as a functon of q an p wth =,2 (.e., t s the total energy expresse n terms of q an p that can be calle Hamltonan. Notng that 2K= p q. (40 (The proof of ths result s gven n Sec. 6, we can obtan the relaton between the Lagrangan an Hamltonan L+H= p q. (4 Snce the nepenent varables n Hamlton s equaton are q an p, thus the q shoul be vew as a functon of general coornates an general momentum,.e., q =q (q,q 2, q,p,p 2, p. (42 Unerstanng ths epenence, we can take the fferental of Eq. (4 wth respect to p I, whch gves (usng the chan rule L + L + H = p +q İ. (43 Notng that q an p are nepenent varables, thus / = 0, the above equaton s wrtten as 0+ L + H = p +q İ. (44 Notng the efnton of the general momentum, the two summaton terms cancel each other, we are left wth q İ = H, (45 whch s the frst Hamlton s equaton. Smlarly, we take the fferental of Eq. (4 wth respect to q I, whch gves L + L + H = q p. (46 q Notng the efnton of the general momentum, the two summaton terms cancel each other, we are left wth L = H. (47 Usng the Euler-Lagrange equaton (35, the above equaton s wrtten whch s the secon Hamlton s equaton. p İ = H, 6 Prove that 2K = p q To prove that 2K= p q. (48 Proof: Usng the efnton of the generalze momentum p, the rght-han se of the above equaton s wrtten as p q = L q q. (49

6 6 Secton 8 Notng that L=K U, where the potental U s nepenent of q, Eq. (49 s wrtten as p q = Usng the efnton of the knetc energy, Eq. (50 s wrtten as Usng the fact that Eq. (5 s wrtten as p q = = = p q = K q. (50 m 2 2 ẋ q 2 2 m ẋ q q ẋ m ẋ q q. (5 ẋ = x, (52 = m ẋ x q m ẋ x q (53 If the coornates transformaton x =x (q,, q,t oes not explctly epens on t,.e., x = x (q,,q, then the above equaton s wrtten as p q = m ẋ ẋ = 2K (54 Thus Eq. (48 s prove. 7 Varatonal prncple The ervaton of Lagrange s equaton gven n the last secton starts wth Newton s law. The ervaton can also starts wth a varatonal prncple. Defne the acton ntegral t 2 J = Lt, (55 Then the acton prncple says that the equaton of moton s gven by t δj =0. (56 rom Eq. (56, we can erve the Euler-Lagrange equaton L = L (57 t q I 8 el theory not fnshe Generally speakng, the Lagrangan of a ynamcal system s a functon that summarzes the ynamcs of the system. Moern formulatons of classcal fel theores ten to be expresse by usng Lagrangan. The Lagrangan s a functon that, when subecte to an acton prncple, gves rse to the fel equatons an a conservaton law for the theory.

7 Noether s theorem not fnshe 7 The Lagrangan n many classcal systems s a functon of generalze coornates q an ther veloctes q. These coornates (an veloctes are, n turn, parametrc functons of tme. In the classcal vew, tme s an nepenent varable an (q, q are epenent varables. Ths formalsm was generalze further to hanle fel theory. In fel theory, the nepenent varable s replace by an event n space-tme (x, y,z,t, or more generally by a pont s on a manfol. An the epenent varables q are replace by ϕ, the value of a fel at that pont n space-tme, so that fel equatons are obtane by means of an acton prncple, wrtten as: δs =0. (58 δϕ Lagrangan enstes n fel theory The tme ntegral of the Lagrangan s calle the acton enote by S. In fel theory, a stncton s occasonally mae between the Lagrangan L, of whch the acton s the tme ntegral: S= Lt (59 an the Lagrangan ensty L, whch one ntegrates over all space-tme to get the acton: S[ϕ]= L[ϕ(x] 4 x. (60 The Lagrangan s then the spatal ntegral of the Lagrangan ensty. L=L(ϕ, ϕ, ϕ,,x (6 9 Noether s theorem not fnshe We know that f the Lagrangan s nepenent of a coornate q (.e., q k s a gnorable coornate, the corresponng canoncal momentum p k wll be conserve. The absence of the gnorable coornate q k from the Lagrangan mples that the Lagrangan s unaffecte by a change or transformaton of q k. Ths means the Lagrangan exhbt a symmetry uner the transformaton. Ths s the see ea generalze n Noether s theorem. L=L(q,q 2,,q k, q,q,,q k,.,q,t (62 If there exsts a transformaton q = q (s q = q (s wth =,., an the Lagrangan L s nvarant uner ths transformaton, that s Then t s easy to prove that C efne n the followng s conserve, Proof C t C= L =0. (63 s p q (s s. (64 = q p (s t s = [ q (s p + q ] (s t s s t p = [ q (s p + q ] (s L t s s = [ q (s p + q ] (s L s t s = [ L q q s + q ] (s L s = L s = 0 (65

8 8 Secton QED. 0 Posson bracket not fnshe Hamlton s equatons are an p İ = H (66 q İ = H. (67 Defne the Posson brackets of two functons f(q,p an g(q,p, {f,g} ( f g f g, (68 p p then Hamlton s equatons are wrtten n symmetrcal form an q İ ={q I,H} (69 p İ ={p I,H}. (70 More generally, for any functon f =f(q,p, we obtan If f explctly epens on tme,.e., f =f(q,p,t, then f ={f,h}. (7 f ={f,h}+ f t. (72 0. Propertes of Posson bracket rom the efnton of the Posson bracket, t s easy to prove that {f,g}= {g,f}, (73 {f,g+h}={f,g}+{f,h}, (74 an {f +h,g}={f,g}+{h,g}, (75 {f,gh}={f,g}h+{f,h}g. (76 Jacob entty for the Posson bracket s (I o not check ths n entty {f,{g,h}}+{g,{h,f}}+{h,{f,g}}=0. (77 Bblography [] Analytcal Classcal Dynamcs. Rchar tzpatrck, [2] Robert G. Lttleohn. Dfferental forms an canoncal varables for rft moton n toroal geometry. Phys. lus, 28(6: , 985. [3] Shaoe Wang. Canoncal hamltonan theory of the gung-center moton n an axsymmetrc torus, wth the fferent tme scales well separate. Phys. Plasmas, 3(5:052506, 2006.