Symmetries and Conservation Laws in Classical Mechanics

Transcription

1 Symmetres nd Conservton Lws n Clsscl Mechncs Wllm Andrew Astll September 30, 0 Abstrct Ths pper wll provde detled explorton nd explnton of symmetres n clsscl mechncs nd how these symmetres relte to conservton lws. The pper wll strt out loong t how Newton s lws of moton cn be found for ny rbtrry set of coordntes through the Euler-Lgrnge equton. It wll then move on to loong t the nvrnce of physcl lws under contnuous trnsformtons nd dentfy the conserved quntty whch s ssocted wth whch symmetry. Contents Introducton Lest Acton Prncple. The Clculus of Vrton Equtons of Moton n rbtrry reference frmes Summry Symmetres nd ther Assocted Conserved Qunttes 5 3. Descrbng Trnsformtons nd Symmetres Mthemtclly Rottonl Symmetry Tme Trnslton Symmetry Lenz Vector Symmetry under Certn Potentls Concluson 5 Acnowledgements Introducton Sr Isc Newton s orgnl formulton of the equtons of motons were bsed on solvng sets of dfferentl equtons to fnd how n object velocty chnges wth tme. Whlst these re smple to solve when the moton of n object s n strght lne n crtesn coordntes nd hence usully esy to solve for dy to dy stutons they cn become qute complex qute qucly. An lterntve, but eqully vld nterpretton of the moton of prtcles, s the Hmltonn lest cton prncple whch sttes tht prtcle wll lwys follow the pth whch mnmses the totl cton[]. Whlst ths my seem more complcted n smple stutons the use of the lest prncple cton llows for ner effortless trnsformtons between coordnte systems menng problems whch cnnot be descrbed esly n crtesn coordntes cn be expressed n more pproprte coordnte system wthout cretng new dffcultes[]. The blty to be ble to esly chnge the coordntes when usng the lest cton prncple mes loctng symmetres much eser thn wth Netwon s formulton.

2 Lest Acton Prncple The lest cton prncple sttes tht prtcle wll follow pth whch mnmsed the totl cton. Ths s cheved by mnmsng the ntegrl I = t t 0 L dt.0. whch s nown s the cton ntegrl[]. The functon L s the Lgrngn functon nd s defned s L = T V = mẋ + ẏ + ż V x, y, z.0. whch the reder should recognse s the dfference between the netc nd potentl energy nd thus equton.0. s the sum of the Lgrngn t every pont long the pth ten from t 0 to t As Newton s equtons of motons nd the lest cton prncple re equvlent then the equtons of moton re the solutons whch mnmses the cton ntegrl. However unle the clculus the reder my be used to where one fnds the vlue whch gves the mnmum vlue of functon,.e. settng df dx = 0, you must nsted fnd the functon whch leves the ntegrl unchnged when vred slghtly such tht when fy, y fy, y + δfy, y.0.3 the cton ntegrl s unchnged whch mens tht I I + δi = I.0.4 δi = 0 To solve ths problem wll requre the use of new type of mthemtcs nown s the clculus of vrtons.. The Clculus of Vrton The clculus of vrtons s the study of fndng the extrem of functonl rther thn fndng the extrem of functons whch s wht one sees to do n regulr clculus. A functonl s mppng from functons to number s opposed to functon whch mps numbers onto vector spce nd for the context of ths pper wll consst of defnte ntegrls of n unnown functon note tht defnte ntegrl gets number out of functon. To see how ths wors n prncple one cn loo t the smple exmple of the shortest pth between two ponts n flt spce, whch one would ntutvely now to be strght lne. Ths exmple s ten from [] but expnded to show most of the ntermedte steps so tht the reder cn get better understndng of wht s beng done nd why. Settng ths problem up one defnes the curve to be of the form y = yx such tht y 0 = yx 0 nd y = yx. Thus short dstnce long the rope, dl, s gven by dl = dx + dy = + dy dx dx dy = + dx dx The totl length of the strng L s therefore gven by the ntegrl L = = + y dx.. L L 0 dl = x x 0 + y dx..

3 If one now consders n ntegrl of the form I = F y, y, xdx..3 nd replce yx wth yx + αηx where α s smll nd ηx s functon n sutble form to be dded to yx. The functon ηx s lso chosen such tht η = 0 nd ηb = 0 whch mens tht δy = 0 t the lmts of the ntegrl. If the vlue of I s nvrnt then di dα for ll ηx..4 α=0 Under the lterntve functon equton..3 becomes Iy, α = = F y + αη, y + αη, xdx F y, y, xdx + αη + y y αη dx..5 nd for equton..4 to hold the dfference between..3 nd..5, δi, s equl to zero δi = = y η + y ηdx + y η αdx = 0 To fnd the soluton the second ntegrl s ntegrted by prts to obtn As η = 0 nd ηb = 0 δi = = [ η y equton..7 therefore smplfes to δi = y ηdx + ] b + [ η y ] b η dx..6 y y d dx [ η ] b = 0 y y d dx y y d dx y ηdx ηdx..7 ηdx = 0..8 As the functon ηx s rbtrry for equton..8 to be zero for ny pth nd ny functon ηx the functon nsde the brcet must be equl to zero, thus obtnng the the very mportnt result of y d dx y = 0..9 whch s nown s the Euler Lgrnge equton nd wll be used mny tme throughout ths pper. Gong bc to the orgnl problem of fndng the equton of the shortest curve between two ponts n plne by pplyng equton..9 to equton... Dong so one obtns the followng results: y = 0 nd therefore from equton..9 d = 0 dx y 3

4 whch mens tht y =..0 where s constnt. The equton of the shortest curve cn now be obtned by solvng equton..0: y = y + y = y = + y y = + y y = y = or y = However only the postve root stsfes the orgnl problem nd so the negtve root s dscrded dy dx = y = x + c mx + c.. Ths result confrms tht the shortest curve between two ponts on plne s strght whch s exctly wht one would ntutvely now. Whlst ths exmple my therefore seem rther tedous nd unnecessry t shows method whch cn be used to solve mny more complcted problems.. Equtons of Moton n rbtrry reference frmes The results from the prevous secton now llow one to fnd the equtons of moton whch mnmse equton.0., the cton ntegrl, by usng equton..9 wth L replcng F, t replcng x, q replcng y nd q replcng y nd thus to mnmse the cton ntegrl the specfc form of the Euler Lgrnge equton requred s d =.. dt q q where q s n rbtrry coordnte nd q s the tme dervtve of the rbtrry coordnte. worng n crtesn coordntes nd thus the equtons of moton re L = mẋ + ẏ + ż V x, y, z d dt mx = V = F x x When whch the reder wll recognse s Newton s equtons of moton. Whlst ths s my seem long wnded wy of re-sttng the obvous ths wy of loong t how object moves mes t much eser to fnd new equtons of moton when the coordnte system chges. For exmple f we te the exmple of prbolc coordntes ξ = r + x nd η = r x then fndng the equtons of moton n ths system s reltvely smple usng the Lgrngn. Frstly rerrnge the expressons of ξ nd η to get expresson for x nd y: x = ξ η y = ξη 4

5 Second, te the tme dervtves to get expressons for ẋ nd ẏ: ẋ = ξ η nd then squre both of them to get ẏ = ξη ξη + ξ η ẋ = 4 ξ ξ η + η ẏ = 4 η ξ ξ + ξ η + ξ η η These cn now be substtuted nto the Lgrngn to get L = 8 m ξ + η + η + ξ V ξ, η ξ η from whch we cn esly get the equtons of moton d m ξ + η = m η dt 4 ξ 8 η η ξ ξ V ξ d dt m η + ξ 4 η = m ξ 8 ξ ξ η η V η These equtons of moton re much more complcted thn those for crtesn coordntes nd to try nd fnd these wthout the lest cton prncple nd Euler Lgrnge equton would be much hrder ts lthough stll doble..3 Summry Ths secton hs therefore shown tht by f n object follows trjectory whch mnmses the Lgrngn long the pth ten then ths s equvlent to Newton s equtons of moton. Furthermore ths secton hs shown tht the Lgrngn llows for the coordntes to be esly chnged llowng you to use the most pproprte coordnte system. 3 Symmetres nd ther Assocted Conserved Qunttes A symmetry n physcs s when physcl lw s unchnged by trnsformton n one or more of the coordntes. These trnsformtons cn be vrety of opertons on the coordntes, for exmple they could be trnslton n spce, rotton round n xs or trnslton n tme. Assocted wth ech contnuous symmetry s conserved quntty,.e. there s quntty ssocted wth the system whch does not chnge f the system s symmetrc under certn trnsformton. Ths result s nown s Noether s Theorem nd s nmed fter the Germn Physcst who dscovered t Emmy Noether. 3. Descrbng Trnsformtons nd Symmetres Mthemtclly Tng the Lgrngn functon L = Lq, q, t for the resons dscussed erler lthough n ths cse the Lgrngn wll not be dependent on tme s the systems delt n ths pper re closed nd then chngng the coordntes by n mount δq then the functon becomes L L + q δq + δq q 5

6 The dfference n q cn be expressed s δq = ωd = ω + j b j q j + j c j q j where ω s n nfntesml nd, b nd c re not. trnsformton cn be wrtten s L L + q ωd + Ths mens tht the new functon under the ωḋ q = L + dl dω ω 3.. If one now ssumes the functon fter the trnsformton cn lso be wrtten n the form L L + ω dk dt 3.. whch mens tht dl dω = dk dt nd tht there s conserved quntty Q whch one defned s Q = 3..3 d K 3..4 q If Q s ndeed conserved quntty then the totl tme dervtve s equl to zero,.e. the totl tme dervtve one fnds tht: dq dt = d d + d dt q q dt d dk dt = d + d dk from equton..9 q q dt = dl dω dk from equton 3.. dt = dk dt dk dt = 0 from equton 3..3 Quod Ert Demonstrndum dq dt = 0. Tng Thus f the trnsformton of L cn be expressed s totl tme dervtve of nother functon K then the functon s symmetrc under the trnsformton nd there s conserved quntty Q whch s defned by equton Rottonl Symmetry When system s rotted round one of ts xs the chnge n the coordntes δ q s gven by δ q = ω j b j q j = ωt z r when n crtesn coordntes. The symbol T z s the genertor of the rotton whch cn be obtned by loong t how the coordntes chnge when rotted round the x xs. If you rotte crtesn coordnte system round the z xs by n mount θ the rotton cn be represented by the mtrx cos θ sn θ 0 Rz = sn θ cos θ

7 nd f θ s smll then by Tylor expnson R z = +... θ θ = θ θ = θ = I + θt z +... If the θ s very smll then only the frst two terms of the Tylor expnson me ny sgnfcnt contrbuton nd the chnge n vector when rotted by n mount θ s gven by R z θ r = r + θt z r nd therefore, when n crtesn coordntes, d s gven by d = Tz r 0 0 x = 0 0 y z y = x Usng equton 3.. one cn now loo t under wht condtons the Lgrngn s spherclly symmetrc nd fnd the connected conserved quntty. L L + ωd + ωd q q L + ωy + ωx + ωẏ + x y ẋ ẏ ωẋ L + ω x y y + ωmẋẏ mẋẏ x L + ω y V x x V y When the potentl V s spherclly symmetrc,.e. V r = V x + y + z then nd therefore 3.. becomes y V x = x V y L L whch cn be expressed n the form of equton 3.. f K s ny rbtrry constnt. The conserved quntty cn now be clculted by usng equton Q = d K q = mxẏ myẋ K = xp y yp x K 3.. = L z K

8 As K cn be ny constnt nd stll stsfy equton 3.. f K s mde to equl zero then equton 3..3 becomes Q = L z 3..4 nd thus the the Lgrngn s nvrnce under rotton round the z xs corresponds to the conservton of the z-component of the ngulr moment. It s worth notng tht f K s defned to equl non-zero constnt then vlue of Q s stll conserved t s just no longer the ngulr momentum but nsted sd component mnus constnt. If ths process s repeted for rotton round the x nd y xs then one rrves t dentcl results wth the Lgrngn symmetrcl under rotton round ny xs nd tht ths symmetry corresponds to the conservton of the ngulr momentum component long the rotton xs. 3.3 Tme Trnslton Symmetry If one now consders trnslton n tme,.e. tng system nd tng t nto the pst or future, the chnge n the coordntes s gven by δq = ω q The chnge n the Lgrngn s therefore: L L + ω q + q L + ω m q q + L + ωm q ω q V q q q q 3.3. whch mes nd the conserved quntty s dk dt = m q q 3.3. Q = q K q = m r K If you then defne K s then the totl tme dervtve of K s therefore dk dt = m K = L = T r V r q q + V q q = m q q whch stsfes equton 3.3. menng tht the defnton K = L s correct. Substtutng the expresson for K nto equton gves conserved quntty of: Q = m r m r + V r = m r + V r = E where E s the totl energy. Thus the nvrnce of the Lgrngn from tme trnsformton hs the Energy s the correspondng conserved quntty to go wth the symmetry. 8

9 3.4 Lenz Vector Symmetry under Certn Potentls If one now consders n rbtrry trnsformton of the form δx = ωp x x x p s δ s r p = ωmẋ x s mẋ sx δ sm x ẋ 3.4. then d s therefore nd d s therefore d = mẋ x s mẋ sx δ sm x ẋ 3.4. d = mẍ x s + mẋ ẋ s mẋ ẋ s x ẍ s δ sm ẋ δ sm x ẍ = mẍ x s + mẋ ẋ s x ẍ s δ sm ẋ δ sm x ẍ The dervtve of L wth respect to ω, dl dω d dl dω = d + x ẋ = x ẋ V x = s therefore: mẋ x s mẋ sx m x ẋ + x s mẍ x s + mẋ ẋ s mẋ ẋ s x ẍ s m mx ẋ s mẋ x s + V x s mx ẋ + m ẋ ẍ x s m x ẋ ẍ s m x ẍ ẋ s ẋ x ẍ ẋ s If the Lgrngn s symmetrc under the trnsformton then the equton cn be wrtten s tme dervtve of K. When tryng to fnd the functon K t s eser to splt the functon up so tht K = f + g where df dt = V x mx ẋ s mẋ x s + V x s mx ẋ nd dg dt = m ẋ ẍ x s m x ẋ ẍ s m x ẍ ẋ s such tht dk dt = df dt + dg dt = dl dω From nspecton, f one defnes functon g s g = m ẋ x s x ẋ ẋ s then the totl tme dervtve s dg dt = m ẋ ẍ x s m x ẋ ẍ s m x ẍ ẋ s 9

10 whch stsfes equton All tht s left s to fnd n expresson for f whch stsfes equton Frstly usng to expnd equton to get V = dv x dr r = dv x x dr r df dt = = dv dr dv dr mx ẋ s r mx ẋ s r mx ẋ x s + dv mx ẋ x s r dr r mx ẋ x s r If the equton for expresson for f s now defned s f = m x sv then the totl tme dervtve of f s df dt = m ẋsv m x dv s dr = m ẋsv dv dr mx ẋ x s r r dx x dt For equton to be dentcl to equton then there must be potentl V whch stsfes the dfferentl equton V = x dv r dr Solvng the dfferentl equton obtns the soluton: dv V = r dr x dv V = dr r ln V = ln r + c ln V = ln r + c V = r + c Now tht both f nd g hve been found the functon K cn now be defned s K = = m ẋ x s x ẋ ẋ s m x sv m ẋ x s x ẋ ẋ s m x s r

11 nd the conserved vlue Q cn be found Q s = d K ẋ = mẋ mẋ x s mx ẋ s δ s mẋ x K =m ẋ x s x ẋ ẋ s K =m ẋ x s x ẋ ẋ s = m ẋ x s x ẋ ẋ s + m m ẋ x s x ẋ ẋ s + m x s r x s r 3.4. Combnng ll 3 components of Q s nto the vector Q one obtns the result Q = P P r P r P + m ˆr s = P r P + m ˆr s = P L + m ˆr s 3.4. All tht s left s to ensure tht the dmensons re consstent throughout equton 3.4. nd so the conserved quntty becomes: Q = P L + α m ˆr s where α s constnt wth sutble dmensons. Ths conserved quntty s nown s the Lenz vector nd s constnt n tme,.e. the drect nd mgntude do not chnge s tme progresses. The vector cn be used to defne n orbt s t s unchnged regrdless of the where n orbtng object s n ts object, whlst the drecton n whch the orbtng object s movng chnges the Lenz vector never chnges. Ths s true s long s the potentl s n nversely proportonl to the dstnce from the centre. 4 Concluson Ths pper hs explored when physcl system s nvrnt under certn trnsformtons nd how these symmetres correspond to specfc conserved quntty. We hve seen how rottonl symmetry nd the conservton of ngulr momentum re two sdes of the sme con nd tme trnslton symmetry nd energy conservton re relted. Ths pper hs lso dscussed how usng the Euler Lgrnge equton from vrtonl clculus to solve the lest cton prncple provdes wy to fnd the equtons of moton n ny rbtrry coordnte system. 5 Acnowledgements Ths pper would not hve been possble wthout the help nd gudnce of my supervsor Dr Sebstn Jäger who helped pont me n the rght drecton when ll pths seemed ded ends nd explned ny new concepts whch I struggled to grsp on my own. References [] T.W.B. Kbble. Clsscl Mechncs. McGrw-Hll, second edton, 973. [] K. F. Rley, M. P. Hobson, nd S. J. Bence. Mthemtcl Methods for Physcs nd Engneerng. Cmbrdge Unversty Press, thrd edton, 006.