Mechancs Physcs 151 Lectue 18 Hamltonan Equatons of Moton (Chapte 8) What s Ahead We ae statng Hamltonan fomalsm Hamltonan equaton Today and 11/6 Canoncal tansfomaton 1/3, 1/5, 1/10 Close lnk to non-elatvstc QM May not cove Hamlton-Jacob theoy Cute but not vey elevant What shall we do n the last lectues? Classcal chaos? Petubaton theoy? Classcal feld theoy? Send me e-mal f you have pefeence! Hamltonan Fomalsm Newtonan Lagangan Hamltonan Descbe same physcs and poduce same esults Dffeence s n the vewponts Symmetes and nvaance moe appaent Flexblty of coodnate tansfomaton Hamltonan fomalsm lnked to the development of Hamlton-Jacob theoy Classcal petubaton theoy Quantum mechancs Statstcal mechancs 1
Lagange Hamlton Lagange s equatons fo n coodnates d L L = 0 = 1,, n nd -ode dffeental dt q q equaton of n vaables n equatons n ntal condtons q ( t = 0) q ( t = 0) Can we do wth 1 st -ode dffeental equatons? Yes, but you ll need n equatons We keep q and eplace q wth somethng smla Lq ( j, q j, t) We take the conjugate momenta p q Confguaton Space We consdeed ( q1,, q n ) as a pont n an n-dm. space Called confguaton space Moton of the system q = q() t A cuve n the confg space When we take vaatons, we consde q and q as ndependent vaables.e., we have n ndependent vaables n n-dm. space Isn t t moe natual to consde the moton n n-dm space? Phase Space Consde coodnates and momenta as ndependent State of the system s gven by ( q1,, qn, p1,, pn) Consde t a pont n the n-dmensonal phase space We ae swtchng the ndependent vaables ( q, q, t) ( q, p, t) A bt of mathematcal tck s needed to do ths q = q() t p = p () t
Legende Tansfomaton Stat fom a functon of two vaables Total devatve s f f df = dx + dy udx + vdy x y Defne g f ux f ( xy, ) and consde ts total devatve dg = df d( ux) = udx + vdy udx xdu = vdy xdu.e. g s a functon of u and y g g If f = L and ( x, y) = ( q, q) = v = x y u Lqq (, ) g( pq, ) = L pq Ths s what we need Hamltonan Opposte sgn fom Legende tansf. Defne Hamltonan: Hqpt (,, ) = qp Lqqt (,, ) Total devatve s L L L dh = pdq + q dp dq dq dt q q t L Lagange s equatons say = p q L dh = q dp p dq dt t Ths must be equvalent to Puttng them togethe gves dh = dp + dq + dt p q t Hamlton s Equatons L We fnd = q = p and = p q t t n equatons eplacng the n Lagange s equatons 1 st -ode dffeental nstead of nd -ode Symmety between q and p s appaent Thee s nothng new We just eaanged equatons Fst equaton lnks momentum to velocty Ths elaton s gven n Newtonan fomalsm Second equaton s equvalent to Newton s/lagange s equatons of moton 3
Quck Example Patcle unde Hooke s law foce F = kx m k L L= x x p = = mx x m k H = xp L= x + x p Replace x wth p k x m = + m Hamlton s equatons p x = = p = = kx p m x Usual hamonc oscllato Enegy Functon Defnton of Hamltonan s dentcal to the enegy L functon hqqt (,, ) = q Lqqt (,, ) q Dffeence s subtle: H s a functon of (q, p, t) Ths equals to the total enegy f Lagangan s L= L0( qt, ) + L1( qtq, ) + L( qtqq, ) j k Constants ae tme-ndependent Ths makes T = L ( q, t ) q j q k Foces ae consevatve See Lectue 4, o Ths makes V = L ( q) Goldsten Secton.7 0 Hamltonan and Total Enegy If the condtons make h to be total enegy, we can skp calculatng L and go dectly to H Fo the patcle unde Hooke s law foce p k H = E = T + V = + x m Ths woks often, but not always when the coodnate system s tme-dependent e.g., otatng (non-netal) coodnate system when the potental s velocty-dependent e.g., patcle n an EM feld Let s look at ths 4
Patcle n EM Feld Fo a patcle n an EM feld m L= x qφ + qax p = mx + qa We can t jump on H = E because of the last tem, but mx H = ( mx + qa) x L= + qφ Ths s n fact E We d be done f we wee calculatng h Fo H, we must ewte t usng p = mx + qa ( p qa) H( x, p) = + qφ m Patcle n EM Feld ( p qa) H( x, p) = + qφ m Hamlton s equatons ae p qa pj qaj Aj φ x = = p = = q q p m x m x x Ae they equvalent to the usual Loentz foce? Check ths by elmnatng p d Aj φ ( mx + qa ) = qx q dt x x d ( mv ) = qe + q ( v B) dt A bt of wok Consevaton of Hamltonan Consde tme-devatve of Hamltonan dh ( q, p, t) = q + p + dt q p t Hamltonan s = pq + qp + conseved f t does not t depend explctly on t H may o may not be total enegy If t s, ths means enegy consevaton Even f t sn t, H s stll a constant of moton 5
Cyclc Coodnates A cyclc coodnate does not appea n L By constucton, t does not appea n H ethe H ( q, pt, ) = qp L( q, qt, ) Hamlton s equaton says p = = 0 q Conjugate momentum of a cyclc coodnate s conseved Exactly the same as n the Lagangan fomalsm Cyclc Example Cental foce poblem n dmensons m L= ( + θ ) V( ) = m p = m θ p θ 1 p θ H = p + () + V θ s cyclc p m = const = θ l Hamlton s equatons 1 l = p + + V() p l V() m = p = 3 m m Cyclc vaable dops off by tself Gong Relatvstc Pactcal appoach Fnd a Hamltonan that woks Does t epesent the total enegy? Pust appoach Constuct covaant Hamltonan fomalsm Fo one patcle n an EM feld Don t expect macles Fundamental dffcultes eman the same 6
Pactcal Appoach Stat fom the elatvstc Lagangan that woks L mc β V = 1 ( x) L mv p = = v 1 β 4 H = h= p c + m c + V( x) It does equal to the total enegy Hamlton s equatons pc p x = = = p 4 pc + mc mγ Dd ths last tme V p = = = F x x Pactcal Appoach w/ EM Feld Consde a patcle n an EM feld L= mc 1 β qφ( x) + q( v A) Hamltonan s stll total enegy H = mγc + q φ 4 = m γ v c + m c + qφ Can be easly checked Dffeence s n the momentum p = mγ v + qa 4 H = ( p qa) c + m c + qφ Not the usual lnea momentum! Pactcal Appoach w/ EM Feld 4 H = ( p qa) c + m c + qφ Consde H qφ 4 ( H qφ ) ( p qa) c = m c constant It means that ( H qφ, pc qac) s a 4-vecto, and so s ( H, pc) Smla to 4-momentum (E/c, p) of a elatvstc patcle Remembe p hee s not the lnea momentum! Ths patcula Hamltonan + canoncal momentum tansfoms as a 4-vecto Tue only fo well-defned 4-potental such as EM feld 7
Pust Appoach Covaant Lagangan fo a fee patcle Λ= Λ p p p = = mu H = u m We know that p 0 s E/c We also know that x 0 s ct Geneally tue fo any covaant Lagangan You know the coespondng elatonshp n QM 1 mu uν ν Enegy s the conjugate momentum of tme Pust Appoach Value of Hamltonan s p p mc H = = m What s mpotant s H s dependence on p Hamlton s equatons dx p = = p m dp = = 0 x Tme components ae dct ( ) E dec ( ) = = γ c = 0 mc Ths s constant! 4-momentum consevaton Enegy defnton and consevaton Pust Appoach w/ EM Feld Wth EM feld, Lagangan becomes Λ ( x, u ) = mu u + qu A p = mu + qa 1 mu u ( p qa )( p qa ) H = = m Hamlton s equatons ae dx p qa dp ( pν qaν) A = = = = p m x m x A bt of wok can tun them nto du ν A A m = q uν = K x x ν ν 4-foce 8
EM Feld and Hamltonan In Hamltonan fomalsm, EM feld always modfy the canoncal momentum as pa = p0 + qa Wth EM feld Wthout EM feld A handy ule: Hamltonan wth EM feld s gven by eplacng p n the feld-fee Hamltonan wth p qa Often used n elatvstc QM to ntoduce EM nteacton Summay Constucted Hamltonan fomalsm Equvalent to Lagangan fomalsm Smple, but twce as many, equatons Hamltonan s conseved (unless explctly t-dependent) Equals to total enegy (unless t sn t) (duh) Cyclc coodnates dops out qute easly A few new nsghts fom elatvstc Hamltonans Conjugate of tme = enegy p qa ule fo ntoducng EM nteacton 9