Classcal Mechancs Symmetry an Conservaton Laws Dpan Kumar Ghosh UM-DAE Centre for Excellence n Basc Scences Kalna, Mumba 400085 September 7, 2016 1 Concept of Symmetry If the property of a system oes not change uner some efne operatons, the system s sa to have a symmetry wth respect to so efne operatons. Thought he unverse s both homogeneous an sotropc, there are local nhomogenety an lack of sotropy. Nevertheless, a close system (.e. a system whch oes not nteract wth partcles or fels outse of the system) may exhbt space homogenety an/or sotropy. 1.1 Homogenety of Space: Homogenety of space mples that the physcal laws are nvarant uner space translaton. Ths means that f we change the poston of every partcle (an fel) by the same constant vector ɛ, the Lagrangan of our system woul reman the same. Mathematcally, t mples that when r r = r + ɛ, the change n the Lagrangan of our system, δl = ɛ = ɛ (1) r r Here the summaton nex s over all the partcles of the system. arbtrary translaton mples δl = 0, whch results n Usng Euler-Lagrange equaton, ths mples ( ) = 0 v Symmetry uner r = 0 (2)
c D. K. Ghosh, IIT Bombay 2 For a close system, V epens only on postons of the partcles, We then get L = 1 m v 2 V ({ r } 2 ( ) ( ) m v = p = 0 (3).e. the total momentum of the system s constant. Thus homogenety of space leas to conservaton of lnear momentum of the system. 1.2 Isotropy of Space: If L s nepenent of the orentaton of the system, t results n the conservaton of angular momentum of the system. Conser (see fgure) the rotaton of the system about the z-axs by an angle θ so that OA OA. z C δϕ θ r r O It follows from the fgure that OA = OA + AA where the magntue of AA s gven by AA = r sn θδϕ. Defnng δϕ to be of magntue δϕ an recte along the axs of rotaton, we can wrte, AA = δϕ r Thus as r r = r + δ r, δ r = δϕ r. We then have where δ v = δϕ v. v v = v + δ v
c D. K. Ghosh, IIT Bombay 3 If the Lagrangan s nvarant uner rotaton escrbe agove, we have δl = δ r + r = ( ) δ r + ṙ = ( ) ṙ ṙ δ ṙ ṙ δ ṙ ( r δϕ) Once agan, f we conser a close system, L = 1 m v 2 V ({ r } 2 we have (4) can be wrtten as δl = Wrtng δϕ = ˆnδϕ, we can express (5) as Thus δl = 0 mples ṙ = m ṙ = p [ ṗ ( r δϕ) + p ( ṙ δϕ) ] δl = δϕˆn ( r p ) = 0 ṙ ( v δϕ) (4) (5) ( r p ) (6) so that the angular momentum of the system J = ( r p ) s constant. 1.3 Homogenety of Tme an Energy Conservaton: Close system also exhbt symmetry wth respect to tme translaton. What ths mples s that repeatng an experment uner entcal conons after a lapse of some tme wll yel entcal result. Tme homogenety s a consequence of the fact that the Lagrangan of the system has no explct tme epenence: = 0, so that L = Lq, q only. t The total tme ervatve of the Lagrangan, n such a case s gven by L = = = q + q q q q + ( ) q q q ( ) q q
c D. K. Ghosh, IIT Bombay 4 We efne the Hamltonan of the system as H = ( ) q L (7) q so that the preceng equaton gves us H = 0. For a close system, the Hamltonan also gves the total energy of the system (see below), so that the tme homogenety results n conservaton of energy. Let us conser a close system for whch the knetc energy s expressble as a blnear prouct of generalze veloctes an the potental energy s a functon of the generalze coornates alone, T = 1 C,j q q j 2,j where C,j = C j, coul be functons of generalze coornates but not of veloctes or of tme. Thus we have L = T ({ q }) +V ({q }) We then have q k = ( ) 1 q k C,j q q j q k q k 2 k k,j = [ ] 1 q k C,j δ,k q j + 1 C,j q δ j,k 2 2 k,j,j ( = 1 C k,j q j + ) C,k q q k 2 j k = j,k C j,k q j q k = 2T where n the last but one step, we have use C,k = C k, an have change the ummy summaton nex n the secon term from to j. Usng (7) for the efnton of the Hamltonan, we then have, H = ( ) q L q = 2T L = 2T (T V ) = T + V whch s the total energy. Thus for a close system, homogenety of tme leas to a conservaton of the total energy.
c D. K. Ghosh, IIT Bombay 5 2 Invarance uner Gallean Transformaton Let S an S be two nertal frames. Though the values of knematcal quanttes wll be measure fferently n two frames, by efnton of nertal frames, Newton s laws wll be val n both the frames,.e. whch gves whch can be ntegrate to yel m 2 r 2 r r = m 2 r 2 (= F ) 2 ( r r ) 2 = 0 = u 0 (t t 0 ) u 0 t, where u 0 s a constant velocty, the last relaton follows f the two frames are mae to conce at t = t 0. The two frames are thus relate by Gallean transformaton: r = r u 0 t How oes the Lagrangan transform? If the potental energy epens only on the relatve coornates of two partcles then so that n S, L = 1 2 mv 2 V r 2 r 1 = r 2 r 1 = 1 2 m( v u 0) 2 V = 1 2 mv2 V m v u 0 + 1 2 mu2 0 = L + [ ] 1 2 mu2 0t m u 0 r L + F where the functon F s efne through (8) an (9). Note that both L an L satsfy Euler-Lagrange equaton. Ths s an example of gauge symmetry of the Lagrangan. If L satsfes the Euler-Lagrange equaton an F s any fferentable functon of (q, t), then L + F also satsfes the Euler-Lagrange equaton. Ths s realy prove by observng that the extra term n the Euler-Lagrange equaton (8) (9)
c D. K. Ghosh, IIT Bombay 6 ue to F s [ ( )] F q ( ) F = q [ q ( j ) ( )] F q j + F F q j + F q j t q q j j t (10) Note that [ ] F = ( ) F = ( ) F q t t q t q snce the explct fferentaton can always be carre out at the en. Further, [ ( )] ( ) F q j = F q j = ( ) F q j q q j j q q j j q j j q snce q j s nepenent of the coornates an the nepenent coornates q an q j are nterchangeable. Usng the above two relatons, the rght han se of (10) becomes j 2 F q j + 2 F q j q t q j 2 F q j 2 F q q j q t = 0 whch proves the asserton. The reason for the exstence of the gauge symmetry can be trace back to the mnmsaton of the acton ntegral. Let, L (q, q, t) = L(q, q, t) + F (q, t) The corresponng acton ntegral woul then be gven by S [q(t)] = = tf t tf t L ( L + F ) = S + F (q, t) t f t = S + constant snce δq(t ) = δq(t f ) = 0 for the varatonal functon, ang a constant to the acton ntegral s nconsequental to the etermnaton of t. Example: Conser the Lagrangan for a charge partcle n electromagnetc fel. We know that
c D. K. Ghosh, IIT Bombay 7 the followng gauge transformatons of the scalar an vector potentals leave the fels unchange: A = A + F ( r, t) ϕ = ϕ t F We have seen earler that the electromagnetc forces can be nclue n the Lagrangan formulaton f we efne a velocty epenent potental, eϕ e v A, where e s the charge of the partcle. Thus L = 1 2 mv2 eϕ + e v A F ( r, t) Suppose we a to the Lagrangan above a term equal to e s then gven by L = 1 2 mv2 eϕ + e v A + e F = 1 [ 2 mv2 e ϕ F ] [ + v e( A t ] + e F ) The new Lagrangan where we have use whch gves F = F x v x + F y v y + F z v z + F t = F v + F t L = 1 2 mv2 eϕ + e v A 3 Symmetry Operatons an Noether s Theorem 3.1 Coornate Transformaton vrs. Symmetry Operatons Suppose we have a system wth Lagrangan L(q, q, t). If we make a coornate transformaton such that q(t) q (t). (When we wrte q, t actually stans for the set of coornates {q k } whch are requre to escrbe the system. ) Ths s smply a way to relabel the system, the form of Lagrangan woul change to L (q, q, t) though the physcal propertes of the system woul not change. Example : Suppose the Lagrangan s gven by L = 1 2 m q2 1 2 kq2. coornate system such that q q = q + a, the new Lagrangan woul be If we change the L = 1 2 m q 2 1 2 k(q a) 2 (11)
c D. K. Ghosh, IIT Bombay 8 whch s essentally sayng that L (q, q, t) = L(q(q, q, t), q(q, q, t), t) (12) Notce that L s functonal epenence on q, q, t s not the same as the functonal epenence of L on q, q, t. What we have one s to smply relabel the system coornates. Ths coul be one by ether a passve manner (whch s to leave the system as t s an shft the coornate system or by an actve transformaton n whch we keep the coornate frame the same but movng the entre system (partcles an fels) relatve to the orgnal coornates such that a pont n the system whch ha a coornate q now has the coornate q. Ths s llustrate n the fgure below. Note that mport of both ways of ong s entcal rrespectve of whether we choose actve or passve transformaton, the physcal result s the same. Passve Actve Example: (1) Suppose we ece that the coornate x alone wll become x = x+a. In the passve transformaton, we woul have shfte the orgn along the x-axs to lett keepng axes parallel so that a pont whch was at x now s at x + a. In the actve transformaton, we woul not change the coornate axs but wll move system to the rght by an amount a. (2) Suppose we rotate the coornate axes by an angle θ so that the new coornates of a pont whch was earler (x, y, z) s x, y, z where x = x cos θ y sn θ y = x sn θ + y cos θ z = z (13) In the passve case, we rotate the coornate axes about the z axs by an angle θ so that a pont whch was at x, y, z now has the coornates x, y, z gven by the relatonshp above.
c D. K. Ghosh, IIT Bombay 9 On the other han, we coul arrve at the same values for the new coornates f we an actve transformaton, n whch we not touch the coornate axs but rotate the entre physcal system by an angle θ about the z axs. These two examples (translaton an rotaton) are what are known as contnuous transformaton as n these there s a parameter (a n case of translaton an the angle θ n case of rotaton) whch coul be vares contnuously. [An example of screte transformaton s reflecton transformaton n whch x x = x]. In the followng scusson we woul be ntereste n contnuous transformaton alone where the coornates epen on a contnuously varyng parameter s, so that we express the transformaton as follows: q q = Q(s, t) (14) wth the entty transformaton beng the stuaton when s = 0,.e. Q(0, t) = q(t). Let us look at some very specal types of coornate transformaton. Instea of the Lagrangan of the new stuaton beng gven by (12) suppose we eman that L (q, q t) = L(q, q, t) (15).e. nstea of the relaton gven by (11), f we ha somehow got nto a stuaton where L = 1 2 m q 2 1 2 kq 2 (16) was val, we woul have ha what s known as a Symmetry Transformaton of the system. Of course, uner the transformatons scusse n examples (1) an (2) we cannot get (16). However, ths woul have happene f we ha consere a reflecton transformaton where q = q. (Ths woul not have been a contnuous transformaton, but that s another matter!). As an example of contnuous transformaton conser a mass connecte to a sprng such that the sprng force s raally nwar an the Lagrangan for whch s L(x, y, z, ẋ, ẏ, ż, t) = 1 2 m(ẋ2 + ẏ 2 + ż 2 ) 1 2 k(x2 + y 2 + z 2 ) If we o a rotaton transformaton (13) on the coornates, we woul have L (x, y, z ) = L(x(x, y, z ), y((x, y, z ), z(x, y, z )) = 1 2 m ( (ẋ cosθ + ẏ sn θ) 2 + ( ẋ sn θ + ẏ cos θ) 2 + ż 2) 1 2 k ( (x cosθ + y sn θ) 2 + ( x sn θ + y cos θ) 2 + ż 2) = 1 2 m(ẋ 2 + ẏ 2 + ż 2 ) 1 2 k(x 2 + y 2 + z 2 ) (17) whch s equal to L(x, y, z ) showng that the rotaton s a contnuous symmetry operaton for ths Lagrangan. However, of we consere a translaton operaton such that x = x a, y = y, z = z, the
c D. K. Ghosh, IIT Bombay 10 knetc energy term woul have the same form but the potental energy term woul be 1 2 k ((x + a) 2 + y 2 + z 2 ) whch woul not result on a symmetry operaton. Thus (15) s a test for symmetry operaton. One can have alternate but equvalent prescrpton for testng a symmetry operaton. That s to check f L(x, y, z ) = L(x, y, z) (18) Let us check ths prescrpton for the example consere above. Conser rotaton transformaton (13). We have L(x, y, z ) = 1 2 m(ẋ 2 + ẏ 2 + ż 2 ) 1 2 k(x 2 + y 2 + z 2 ) Usng (13) ths expresson can be wrtten as L(x, y, z ) = 1 ( (ẋ cos θ ẏ sn θ) 2 + (ẋ sn θ + ẏ cos θ) 2 + ż 2) 2 1 2 k [ (x cos θ y sn θ) 2 + (x sn θ + y cos θ) 2 + z 2] = 1 2 (ẋ2 + ẏ 2 + ż 2 ) 1 2 k(x2 + y 2 + z 2 ) = L(x, y, z) Ientcal concluson can be rawn for reflecton operaton but not for translaton. 3.2 Noether s Theorem: Whenever a physcal system exhbts a contnuous symmetry, there s a conserve quantty (calle Noether s charge n fel theory) assocate wth t. [Converse of ths s not necessarly true as n fel theory nstances are known of conservaton wth whch no apparent symmetry has been foun to be assocate.] Let s be a constant parameter whch can be contnuously vare an whose value tells us how far are we from the entty operaton. Example of the parameter s θ n case of rotaton, a n case of translaton (reflecton, not beng a contnuous symmetry oes not have such a parameter assocate wth t). Let Q(s, t) s a symmetry operaton such that the Lagrangan oes not epen on t: s L[Q(s, t), Q(s, t), t] = L[q(t), q(t), t] = 0 s The statement oes seem confusng as Q(s, t) has explct epenence on s, but the Lagrangan can be wrtten n such a way that t oes not have s epenence. We then have, Q Q s + Q Q s = 0
c D. K. Ghosh, IIT Bombay 11 (the /s term s zero because t an s are nepenent). Usng Euler-Lagrange equaton on the frst term of above, whch can be wrtten as ( Q ) Q s + Q Q s = 0 [ Q ] Q = 0 s As the canoncal momentum corresponng to Q s p =, we have Q Λ = p Q s s=0= constant (19) We have evaluate Q/s at s = 0 as the prouct beng constant, we are free to evaluate t at any value of s. In general, f the Lagrangan possesses a set of n contnuous symmetry operatons, there woul then be n conserve quanttes. If the Lagrangan s a functon of a set of generalze k coornates an veloctes, the conserve Noether charges are Λ j ({q k }, { q k }, t = k p k Q k s j sj =0 (20) for n parameters s j, j = 1,... n. 3.3 Cyclc coornates: A coornate s sa to be cyclc (or gnorable) f L oes not explctly epen on t. In such a case, the corresponng momentum s conserve. Ths follows trvally from the Euler-Lagrange equaton,as q = 0 = ( ) = 0 q whch gves p = 0. It also follows trvally from Noether s theorem, as f we conser a contnuous transformaton q(t) q(t) + s, q(t) q(t), we have f L s cyclc. δl = L(q + s, q, t) L(q, q, t) q s = 0
c D. K. Ghosh, IIT Bombay 12 3.4 Examples from sngle parameter famly of Transformaton: Example 1: Conser a partcle movng n two mensons n a central potental. The Lagrangan s gven by L = 1 2 m(ṙ2 + r 2 ϕ 2 ) V (r) In ths case ϕ s cyclc an hence p ϕ = mr 2 ϕ = constant. Note that r an ϕ epens on t n such a way that the prouct r 2 ϕ s nepenent of tme. In ths case t s approprate to conser the transformaton: The Noether s charge s r(s) = r; r(0) = 0 ϕ(s) = ϕ + s; Λ = r ṙ s s=0 + ϕ ϕ(0) = ϕ ϕ s s=0 = (mṙ) 0 + (mr 2 ϕ) 1 = mr 2 ϕ Example 2: Let us look at the same problem n the Cartesan coornates takng a slghtly fferent form of the contnuous symmetry. The Lagrangan s rewrtten L = 1 2 m(ẋ2 + ẏ 2 ) V ( (x 2 + y 2 )) Conser the followng contnuous transformaton whch leaves the Lagrangan unchange. x(s) = x cos s y sn s ỹ(s) = x sn s + y cos s We have Noether s charge s gven by x s = ỹ; ỹ s = x Λ = x ẋ s s=0 + ẏ ỹ s s=0 = (mẋ) ( y) + (mẏ) x = m(xẏ yẋ) = m( r ṙ) z mr 2 ϕ Example 3: Conser the moton of a partcle where the potental s a functon of stance ρ an the combnaton aϕ+z, where (ρ, ϕ, z) are the usual cylnrcal coornates. The Lagrangan s gven by L = 1 2 m[ ρ2 + ρ 2 ϕ 2 + ż 2 ] V (ρ, aϕ, z)
c D. K. Ghosh, IIT Bombay 13 Defne a one parameter famly of transformaton, ρ = ρ, ϕ = ϕ + s, z = z as Note that a ϕ + z = aϕ + z. The conserve Noether s charge s gven by Λ = ρ ρ s s=0 + ϕ ϕ s s=0 + ż z s s=0 = m ρ 2 0 + mρ 2 ϕ 1 + mż ( a) = mρ 2 ϕ maż Let us verfy whether ths quantty s nee conserve. Usng Euler-Lagrange equatons, we have, ( ) = ϕ (mρ2 ϕ) = ϕ = V ϕ = V (aϕ + z) (aϕ + z) ϕ = av ( ) = (mż) = ż z = V z = V (aϕ + z) (aϕ + z) z = V where we have wrtten V = V/ (aϕ + z). Thus (mr2 ϕ maż) = av + av = 0 Example 4: Suppose L s nvarant uner an nfntesmal translaton of the system along ˆn. We have x = x + sˆn. Noether s charge s Λ = ẋ ˆn = P ˆn where P = p s the total lnear momentum of the system. Thus contnuous translatonal symmetry leas to conservaton of lnear momentum. Example 5: If L s nvarant uner an nfntesmal rotaton about an axs ˆn. We then have x = R(s, ˆn) x = x + sˆn x + O(s 2 )
c D. K. Ghosh, IIT Bombay 14 The Noether s charge n ths case s the total angular momentum of the system about ˆn, Λ = = ˆn ẋ (ˆn x ) x ẋ = ˆn x ˆp = ˆn L where L s the total angular momentum about the axs ˆn. 4 Aonal Avance Informaton* (* not nclue as a part of the course.) In provng Noether s theorem we ha requre that L be nvarant uner a contnuous transformaton. Ths s actually a more strngent requrement than s requre. What we requre s that the acton S be nvarant, whch means that the Lagrangan can vary by an overall tme ervatve, L L = L + F We have seen that presence of ths aonal term oes not change acton except for an ave constant. If q (t) Q (s α ) = q (t) + s α δq (t) be an nfntesmal contnuous transformaton of the generalse coornates by parameters s α such that uner ths transformaton then the quanttes gven by L L = L + s α F Λ α = q α δq F are conserve. The proof s straghtforwar. The change n the Lagrangan ue to change n the coornates an the veloctes s δl = δ q + δq = ( ) δq q q q For each s α, the change s gven by s α ( ) F δq = s α q
c D. K. Ghosh, IIT Bombay 15 whch gves ( ) δq F = 0 q Let us conser a consequence of ths. Conser tme translaton,.e. the case of symmetry In such a case δq = q snce Q(s, t) = q(t + s) sδq = q(t + s) q(t) = s q t s=0= s q an we get, s L(Q(s)) s=0= L Thus n ths case F can be taken to be L tself. Thus the conserve quantty s q δq L whch s just the Hamltonan H. Normally ths also turns out to be the total energy (we wll see later that Hamltonan s not always the total energy). If L = 1 2 m q2 V (q). The Hamltonan s then gven by m q q ( 1 2 m q2 V ) = 1 2 m q2 + V = T + V = E Conser a free partcle for whch L = 1 2 m q2 whch has translatonal symmetry. We can take q s (t) = q(t) + sa In ths case δq = a an δ q = 0. Snce q s cyclc, δl = 0. We therefore have F = 0 n ths case whch leas to conservaton of momentum.