11. Introduction to Liouville s Theorem Michael Fowler

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1 11 Introducton to Louvlle s Theorem Mchael Fowler Paths n Smple Phase Spaces: the SHO and Fallng Bodes Let s frst thnk further about paths n phase space For example the smple harmonc oscllator wth Hamltonan H = p /2 m+ mω q /2 descrbes crcles n phase space parameterzed wth the varables ( pm q) ω (A more usual notaton s to wrte the potental term as Queston: are these crcles the only possble paths for the oscllator to follow? kq ) Answer: yes: any other path would ntersect a crcle and at that pont wth both poston and velocty defned there s only one path forward (and back) n tme possble so the ntersecton can t happen Here s an example from Taylor of paths n phase space: four dentcal fallng bodes are released smultaneously see fgure x measures dstance vertcally down Two are released wth zero momentum from the orgn O and from a pont x 0 meters down the other two are released wth ntal momenta p 0 agan from the ponts O x 0 (Note the dfference n ntal slope) O Four phase space trajectores for fallng objects Check: convnce yourself that all these paths are parts of parabolas centered on the x -axs (Just as the smple harmonc oscllator phase space s flled wth crcular paths ths one s flled wth parabolas) Bodes released wth the same ntal velocty at the same tme wll keep the same vertcal dstance apart those released wth dfferent ntal veloctes wll keep the same velocty dfference snce all accelerate at g Therefore the area of the parallelogram formed by the four phase space ponts at a later tme wll have the same area as the ntal square Exercse: convnce yourself that all the ponts of an ntal vertcal sde of the square all stay n lne as tme goes on even though the lne does not stay vertcal The four sdes of the square deform wth tme to the four sdes of the parallelogram pont by pont Ths means that f we have a fallng stone correspondng ntally to a pont nsde the square t wll go to a pont nsde the parallelogram because f somehow ts path reached the boundary we would have two

2 2 paths n phase space ntersectng and a partcle at one pont n phase space has a unquely defned future path (and past) Followng Many Systems: a Gas n Phase Space We ve looked at four paths n phase space correspondng to four fallng bodes all begnnng at t = 0 but wth dfferent ntal co-ordnates n ( px ) Suppose now we have many fallng bodes so that at t = 0 a regon of phase space can be magned as flled wth a gas of ponts each representng one p x = 1 N fallng body ntally at The argument above about the phase space path of a pont wthn the square at t = 0 stayng nsde the square as tme goes on and the square dstorts to a parallelogram must also be true for any dynamcal system and any closed volume n phase space snce t depends on phase space paths never ntersectng: that s f at t = 0 some closed surface n phase space contans a number of ponts of the gas those same ponts reman nsde the surface as t develops n tme -- none ext or enter For the number of ponts N suffcently large the phase space tme development looks lke the flow of a flud Louvlle s Theorem: Local Gas Densty Is Constant along a Phase Space Path The fallng bodes phase space square has one more lesson for us: vsualze now a unformly dense gas of ponts nsde the ntal square Not only does the gas stay wthn the dstortng square the area t covers n phase space remans constant as dscussed above so the local gas densty stays constant as the gas flows through phase space Louvlle s theorem s that ths constancy of local densty s true for general dynamcal systems Landau s Proof Usng the Jacoban Landau gves a very elegant proof of elemental volume nvarance under a general canoncal transformaton provng the Jacoban multplcatve factor s always unty by clever use of the generatng functon of the canoncal transformaton Jacobans have wde applcablty n dfferent areas of physcs so ths s a good tme to revew ther basc propertes whch we do below as a prelmnary to gvng the proof It must be admtted that there are smpler ways of dervng Louvlle s theorem drectly from Hamlton s equatons the reader may prefer to skp the Jacoban proof at frst readng Jacoban for Tme Evoluton As we ve establshed tme development s equvalent to a canoncal coordnate transformaton ( p q) ( p q ) ( PQ ) t t t+ τ t+ τ

3 Snce we already know that the number of ponts nsde a closed volume s constant n tme Louvlle s theorem s proved f we can show that the volume enclosed by the closed surface s constant that s wth V denotng the volume V evolves to become we must prove V dq dq dp dp dq dq dp dp 1 s 1 s = 1 s 1 s? V If you re famlar wth Jacobans you know that (by defnton) dq dq dp dp = Ddq dq dp dp 1 s 1 s 1 s 1 s where the Jacoban ( Q1 Qs P1 Ps) ( q q p p ) D = 1 s 1 s Louvlle s theorem s therefore proved f we can establsh that D = 1 If you re not famlar wth Jacobans or need remndng read the next secton! Jacobans 101 You can skp ths secton f you're already famlar wth Jacobans Suppose we are ntegratng a functon over some regon of ordnary three-dmensonal space V ( ) I = f x x x dx dx dx but we want to change varables of ntegraton to a dfferent set of coordnates ( 1 2 ) example ( r θφ ) The new coordnates are of course functons of the orgnal ones ( ) q q q such as for q1 x1 x2 x etc and we assume that n the regon of ntegraton they are smooth well-behaved functons We can t smply re-express f n terms of the new varables and replace the volume dfferental dx1dx2dx by dq1dq2dq that gves the wrong answer n a plane you can t replace dxdy wth drdθ you have to use rdrdθ That extra factor r s called the Jacoban t s clear that n the plane a small element wth sdes of fxed lengths( δ r δθ ) s bgger the further t s from the orgn not all δ rδθ elements are equal so to speak Our task s to construct the Jacoban for a general change of coordnates We need to thnk carefully about the volumes n the three-dmensonal space represented by dx1dx2dx and by dq1dq2dq Of course the x s are just ordnary perpendcular Cartesan axes so the volume s just the product of the three sdes of the lttle box dx1dx2dx Imagne ths lttle box ts corner closest x x x and ts furthest pont at the other end of the body dagonal at to the orgn at ( ) ( x dx x dx x dx ) Let s take these two ponts n the q coordnates to be at ( ) q q q and

4 4 ( q dq q dq q dq ) In vsualzng ths bear n mnd that the q axes need not be perpendcular to each other (but they cannot all le n a plane that would not be well-behaved) For the x coordnate ntegraton we magne fllng the space wth lttle cubcal boxes For the q ntegraton we have a system of space fllng nfntesmal paralleleppeds n general pontng dfferent ways n dfferent regons (thnk ( r θ ) ) What we need to fnd s the volume of the ncremental parallelepped wth sdes we ll wrte as vectors n x -coordnates dq 1 dq 2 dq These three ncremental vectors are along the correspondng q coordnate axes and the three added together are x + dx x + dx x + dx q + dq q + dq q + dq the dsplacement from ( x x x ) to Hence n components q1 q1 q 1 dq1 = dx1 dx2 dx x1 x2 x Now the volume of the parallelepped wth sdes the three vectors from the orgn abc s a b c (recall b c s the area of the parallelogram then the dot product sngles out the component of a perpendcular to the plane of bc ) So the volume correspondng to the ncrements dq1 dq2 dq n q space s q1 q1 q1 x1 x2 x q q q dq dq dq = dx dx dx = Ddx dx dx x1 x2 x q q q x x x wrtng D (Landau s notaton) for the determnant whch s n fact the Jacoban often denoted by J The standard notaton for ths determnantal Jacoban s D = ( q1 q2 q) ( x x x ) So the approprate replacement for the three dmensonal ncremental volume element represented n the ntegral by dq1dq2dq s

5 5 dq dq dq ( q1 q2 q) ( x x x ) dx dx dx The nverse D = ( x x x ) ( q q q ) 1 ths s easly establshed usng the chan rule for dfferentaton Exercse: check ths! Thus the change of varables n an ntegral s accomplshed by rewrtng the ntegrand n the new varables and replacng ( x1 x2 x) ( q q q ) I = f x x x dx dx dx = f q q q dq dq dq V V The argument n hgher dmensons s just the same: on gong to dmenson n + 1 the hypervolume element s equal to that of the n dmensonal element multpled by the component of the new vector perpendcular to the n dmensonal element The determnantal form does ths automatcally snce a determnant wth two dentcal rows s zero so n addng a new vector only the component perpendcular to all the earler vectors contrbutes We ve seen that the chan rule for dfferentaton gves the nverse as just the Jacoban wth numerator and denomnator reversed t also readly yelds and ths extends trvally to n dmensons x x x q q q x x x = q q q r r r r r r It s also evdent form the determnantal form of the Jacoban that x x x x x = q q x q q dentcal varables n numerator and denomnator can be canceled Agan ths extends easly to n dmensons Jacoban proof of Louvlle s Theorem After ths rather long detour nto Jacoban theory recall we are tryng to establsh that the volume of a regon n phase space s unaffected by a canoncal transformaton we need to prove that dq dq dp dp = dq dq dp dp 1 s 1 s 1 s 1 s

6 6 and that means we need to show that the Jacoban ( Q1 Qs P1 Ps) ( q q p p ) D = = 1 1 s 1 Usng the theorems above about the nverse of a Jacoban and the chan rule product s Q1 Qs P1 Ps q1 qs p1 ps D = / q q P P q q P P 1 s 1 s 1 s 1 s Now nvokng the rule that f the same varables appear n both numerator and denomnator they can be cancelled D ( s) ( ) ( s) ( ) Q Q p p / 1 1 = q1 qs P constant 1 P P= s q= constant Up to ths pont the equatons are vald for any nonsngular transformaton but to prove the numerator and denomnator are equal n ths expresson requres that the equaton be canoncal that s be gven by a generatng functon as explaned earler Recall now the propertes of the generatng functon ( qpt ) Φ ( ) ( ) dφ q P t = d F + PQ = p dq + Q dp + H H dt from whch p = Φ qpt / q Q= Φ qpt / P H = H+ Φ qpt / t In the expresson for the Jacoban D the kelement of the numerator s Q / q k In terms of the generatng functon Φ ( qp ) ths element s / P 2 Φ qk Exactly the same procedure for the denomnator gves the kelement to be P p =Φ q P 2 / k / k In other words the two determnants are the same (rows and columns are swtched but that doesn t affect the value of a determnant) Ths means D = 1 and Louvlle s theorem s proved Smpler Proof of Louvlle s Theorem Landau s proof gven above s extremely elegant: snce phase space paths cannot ntersect pont nsde a volume stay nsde no matter how the volume contorts and snce tme development s a canoncal transformaton the total volume gven by ntegratng over volume elements dqdp stays the same snce t s an ntegral over the correspondng volume elements dqdp and we ve just shown that dqdp = dqdp

7 7 Here we ll take a slghtly dfferent pont of vew: we ll look at a small square n phase space and track how ts edges are movng to prove ts volume sn t changng (We ll stck to one dmenson but the generalzaton s straghtforward) The ponts here represent a gas of many systems n the two dmensonal ( qp ) phase space and wth a small square area q p tagged by havng all the systems on ts boundary represented by dots of a dfferent color What s the ncremental change n area of ths ntally square pece of phase space n tme dt? Begn wth the top edge: the partcles are all movng wth veloctes ( qp ) but of course the only change n area comes from the p term that s the outward movement of the boundary so the area change n dt from the movement of ths boundary wll be p qdt Meanwhle there wll be a smlar term from the bottom edge and the net contrbuton top plus bottom edges wll depend on the change n p from bottom to top that s a net area change from p / p p qdt movement of these edges Addng n the other two edges (the sdes) wth an exactly smlar argument the total area change s ( / / ) p p +q q p qdt But from Hamlton s equatons p =H / q q = H / p so and therefore = = 2 2 p/ p H/ pq q/ q H/ pq p / p+q / q= 0 establshng that the total ncremental area change as the square dstorts s zero The concluson s that the flow of the gas of systems n phase space s lke an ncompressble flud but wth one mportant qualfcaton: the densty may vary wth poston! It just doesn t vary along a dynamcal path Energy Gradent and Phase Space Velocty For a tme-ndependent Hamltonan the path n phase space ( qp ) s a constant energy lne and we can thnk of the whole phase space as delneated by many such lnes exactly analogous to contour lnes jonng ponts at the same level on a map of uneven terran energy correspondng to heght above sea level The gradent at any pont the vector pontng exactly uphll and therefore perpendcular to the constant energy path s H = H / q H / p

8 8 here H = E The velocty of a system s pont movng through phase space s v = q p = H / p H / q Ths vector s perpendcular to the gradent vector as t must be of course snce the system moves along a constant energy path But nterestngly t has the same magntude as the gradent vector! What s the sgnfcance of that? Imagne a small square sandwched between two phase space paths close together n energy and suppose the dstance between the two paths s decreasng so the square s gettng squeezed at a rate equal to the rate of change of the energy gradent But at the same tme t must be gettng stretched along the drecton of the path an amount equal to the rate of change of phase space velocty along the path and they are equal So ths s just Louvlle agan ts area doesn t change