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Thomas precession

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Lecture notes by G. F. Smoot (Physics Department, UC Berkeley), dated February 1998, from a special relativity course. They show with successive orthogonal Lorentz boosts how a rotation of the rest frame arises, derive the precession rate for small velocities, and apply it to the electron spin-orbit term to obtain the factor of one half. A simple polygon-orbit derivation after Purcell is also given.

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Ph ysics /1/3/9 Relativit yThomas Precession F ebruary /1/9/9/8G/. F/. SMOOTDepartmen to f P h ysics/,Univ ersit y of California/, Berk eley /, USA /9/4/7/2/0/1 Thomas PrecessionThomas Precession is a kinematic e/#0Bect disco v ered b y L/. T/. Thomas in /1/9/2/6 /#28L/.T/. Thomas Phil/. Mag/. /3 /, /1 /#28/1/9/2/7/#29/#29/. It is fairly subtle and mathematicall ysophisticated but it has great imp ortance in atomic ph ysics in connection with spin/-orbit in teraction/. Without including Thomas Precession/, the rate of spin precessionof an atomic electron is o/#0B b y a factor of /2/. Later w e will see that there is a similare/#0Bect for gra vitational /#0Celds/.The e/#0Bect is connected with the fact that t w o successiv e Loren tztransformations in di/#0Beren t directions are equiv alen t to a Loren tz transformation plusa three dimensional rotation/. This rotation of the lo cal frame of rest is the kinematic e/#0Bect that causes the Thomas precession/.F or the lecture w e will not do the full mathematical treatmen t/, since it israther in v olv ed/. Instead w e will sho wb y a simple example ho w the rotation and th usprecession comes ab out/.Mak et w o successiv e Loren tz transformations in orthogonal directions/: from Sto S /0with v elo cit y v along the x axis/, follo w ed b y a transformation from S /0to S /0/0withv elo cit y v /0along the y /0axis/, as sho wn b y the follo wing diagram/./#12 /0/0/#1A /#1A /#1A /#1A/#3E/#12/#1A /#1A /#1A /#1A /#1A /#1A /#1A /#1A /#1A /#1A /#1A /#1A /#1A /#1A/#3E O /0/0 S /0/0 y /0/0x /0 /0v /0v /6/- /- /6O /0 y /0x /0S /0 /- /6O yxS /- /6The line from the origin O of S to the origin O /0/0of S /0/0making and angle /#12 inS and an angle /#12 /0/0in S /0/0/.W e can calculate the angles in the t w o frames b y applying/1 the Loren tz transformations and ev aluating them in eac h frame/.x /0/= /#0D /#28 x /, vt /#29 x /= /#0D /#28 x /0/+ vt /0/#29t /0/= /#0D /#28 t /, v x/=c /2/#29 t /= /#0D /#28 t /0/+ vx /0/=c /2/#29y /0/= y y /= y /0y /0/0/= /#0D /0/#28 y /0/, v /0t /0/#29 y /0/= /#0D /0/#28 y /0/0/+ vt /0/0/#29x /0/0/= x /0x /0/= x /0/0where/#0D /=/1 /= q/1 /, v /2/=c /2/#0D /0/=/1 /= q/1 /, /#28 v /0/=c /#29 /2 /#28/1/#29Com bing these equations one /#0Cnds/:y /0/0/= /#0D /0/#5B y /, v /0/#0D /#28 x /, vt /#29/#5Dx /0/0/= /#0D /#28 x /, vt /#29 /#28/2/#29No ww e can calculate the angle /#12 made b y the line b et w een origins/. F or aGalilean transform one w ould ha v etan/#12 /= yx /= v /0tvt /= v /0v /#28/3/#29but Sp ecial Relativit y sho ws us that /3/-D v elo cities do not transform lik e /3/-D v ectors/.So w em ust calculate carefully /.tan/#12 /= yx /= y /0vt /= /#0D /0/#28 y /0/0/+ v /0t /0/0/#29vt jy /0/0/=/0 /= /#0D /0v /0t /0/0vt /#28/4/#29t /= /#0D /#28 t /0/+ vx /0/=c /2/#29 jx /0/0/= x /0/=/0 /= /#0D/#0D /0/#28 t /0/0/+ v /0y /0/0/=c /2/#29 jy /0/0/=/0 /= /#0D/#0D /0t /0/0/#28/5/#29so thattan/#12 /= /#0D /0v /0t /0/0v/#0D /#0D /0t /0/0 /= v /0/#0Dv /#28/6/#29Note that this answ er is v ery near the Galilean result but with the factor of/1/#2F /#0D whic h reminds us of ab erration/.No ww e calculate /#12 /0/0/:tan/#12 /0/0/= y /0/0x /0/0 /= /#0D /0/#5B y /0/, v /0t /0/#5Dx /0 /#28/7/#29where x /0/0and y /0/0are the co ordinates of the origin O of system S in the S /0/0system/.Th ustan/#12 /0/0/= /#0D /0/#5B y /, v /0/#5Dx /0 jy /=/0 /= /, /#0D /0v /0t /0x /0 /= /, /#0D /0v /0t /0/#0D /#28 x /, vt /#29 jx /=/0 /= /, /#0D /0v /0t /0/, /#0Dv t /#28/8/#29t /0/= /#0D /#28 t /, v x/=c /2/#29 jx /=/0 /= /#0Dt /; /#28/9/#29tan/#12 /0/0/= /#0D /0v /0v /#28/1/0/#29/2 This lo oks again similar to the Galilean angle except for the extra factor of /#0D /0/.No w consider a particle on a curv ed path/#0E v /6 yx X X /#08 /#08/#20 /#20 /#08 CCCCCCCCCCCW/?A t a certain time it is at the origin O of our system S/. Put the x axis parallelto the path/, and y axis to w ard the cen ter of curv ature/. A t t /= /0/, the rest frame S /0ismo ving in the x direction with v elo cit y v /.A t a sligh tly later time its rest frame S /0/0ismo ving p erp endicular to x /0in the y direction with v elo cit y v /0/= /#0Ev /.De/#0Cne/#0E/#12 /= /#12 /0/0/, /#12 /= tan /, /1 /#20v /0/#0D /0v /!/, tan /, /1 /#20v /0/#0Dv /!/#28/1/1/#29F or a v ery short time in terv al the motion is circular/. That is /#0Ct the lo cal curv e witha tangen t circle with appropriate radius of curv ature/.vx /= /! Rcos/#1E vy /= /! Rsin/#1Evx /= v vy /= /#0Ev /= v /0 /#28/1/2/#29sotan/#1E /= v /0v/#0E/#12 /= /#12 /0/0/, /#12 /= tan /, /1/#28 /#0D /0tan/#1E /#29 /, tan /, /1 /#20tan/#1E/#0D /!/#28/1/3/#29Cho ose /#1E to b e v ery small/;/#1E /= /#0ESR /= v/#0EtRThen/#0E/#12 /#19 v/#0EtR /#20/#0D /0/, /1/#0D /!/#28/1/4/#29/!T /= /#0E/#12/#0Et /#19 vR /#20/#0D /0/, /1/#0D /!In a circle the acceleration isa /= v /2R so that vR /= av/3 giving/!T /= av /#20/#0D /0/, /1/#0D /!Supp ose w e are in a non/-relativistic region v/#3C /#3C c /, lik e an electron in an atom/:/#0D /0/, /1/#0D /= /1q/1 /, /#28 v /0/=c /#29 /2 /, q/1 /, /#28 v/= c /#29 /2/#19 /1/+ /1/2 /#28 v /0c /#29 /2/, /1/+ /1/2 /#28 vc /#29 /2/#18 /1/2 /#28 vc /#29 /2since tan/#1E /= v /0/=v /#3C/#3C /1/. Putting this bac ki n to the expression for /!T/!T /#19 av v /2/2 c /2 /= va/2 c /2Th us /#12 /0/0/#3E/#12 /,t h us a coun ter/-clo c kwise rotation/, implying/~/!T /= /~ v /#02 /~ a/2 c /2 /#28/1/5/#29The rigorous result is/~/!T /= /#0D /2/#0D /+/1 /~ v /#02 /~ a/2 c /2 /#28/1/6/#29/2 Spin/-Orbit In teraction of Electron withNucleus in an A tomNo ww e are set to apply this kinematic e/#0Bect to spin precession in an atom/. In itso wn rest frame the electron /#5Csees/" the n ucleus /#0Dying b y /.The electron/'s magnetic momen t/, /~/#16 /, and spin angular momen tum /, /~S /, arerelated b y/~ mu /= emc c /~S /#28/1/7/#29The torque on the magnetic momen ti s/~ /#1C /= d /~Sdt /= /~/#16 /#02 /~B /0/#28/1/8/#29where /~B /0is the magnetic /#0Celd in the e /,frame/./~B /0/= /#0D /#20/~B /, /~ vec /#02 /~E /!/#28/1/9/#29Where /~B is the magnetic /#0Celd and /~E is the electric /#0Celd in the n ucleus rest frame/.v/= c /#3C /#3C /1 so that /#0D /#19 /1/,d /~Sdt /= /~/#16 /#02 /#20/~B /, /~ vec /#02 /~E /!/#28/2/0/#29/4 arises from the in teraction energyU /0/= /, /~/#16 /#01 /#20/~B /, /~ vec /#02 /~E /!/#28/2/1/#29If /~E is due to a spherically symmetri cal c harge distribution /#7B as for a one/-electron atom or one outside a closed shell /#7B thene /~E /= /, /~r V /#28 r /#29 /, /~ rr dVdr /: /#28/2/2/#29ThenU /0/= /, eme c /~S /#01 /~B /+ eme c /2 /~S /#01 /~ v /#02 /#20/, /~ rr dVdr /!/#28/2/3/#29/~S /#01 /~ v /#02 /~/#28 /, r /#29/=/+ /~S /#01 /~f /#02 /~ vU /0/= /, eme c /~S /#01 /~B /+ em /2e c /2 /~S /#01 /#28 /~ r /#02 /~ v /#29 /1r dVdr/= /, eme c /~S /#01 /~B /+ eme c /2 /~S /#01 /~L /1r dVdr /#28/2/4/#29since m /~ r /#02 /~ v /= /~L /#11 angular momen tum /. This second term is the spin/-orbitin teraction/.No w/, if the electron rest frame is rotating /#7B Thomas angular v elo cit y /~/! /,d /~S /=dt /6/= /~/#16 /#02 /~B /0/. The general kinematic result from classical ph ysics is/:/@/@t jrotation co ordinates /= /@/@t jinertial co ordinate s /, /~/! /#02 /#28/2/5/#29as an op erator on an y v ector/. So/@ /~S/@t jrotation co ordinates /= /@ /~S/@t jinertial co ordinate s /, /~/! /#02 /~S /#28/2/6/#29With this expression the in teration energy is c hanged to/:U /= U /0/, /~S /#01 /~/!T /#28/2/7/#29where /~/!T is prop ortional to the cen trip etal acceleration due to Er /./~/!T /#19 /1/2 c /2 /~ v /#02 /~ a /= /1/2 c /2 /~ v /#02 /0/@ e /~Eme /1A/= /1/2 mc /2 /~ v /#02 /#20/, /~ rr dVdr /!/5 /= /1/2 me c /2 /#28 /~ r /#02 /~ v /#29 /1r dVdr /= /~L/2 m /2e c /2 /1r dVdr /#28/2/8/#29Th usU /= U /0/, /1/2 m /2e c /2 /~S /#01 /~L /1r dVdr/= /, eme c /~S /#01 /~B /+/#28 /1 /, /1/2 /#29 /1m /2e c /2 /~S /#01 /~L /1r dVdr /#28/2/9/#29The /-/1/#2F/2 is the famous one half/. Including it/, the observ ed /#0Cne/-structure spacings inatomic sp ectra/, due to electron spin/, are correctly predicted/.This sc hematic giv es a heuristic indication of ho w the torque arises/./~/#16/#01 /#01 /#01 /#01/#15/`/#01/#01/#01/#01/#01/#0B /#01 /#01 /#01 /#01 /#01 /#01/#15 /+q/-q /#12 /#01/#01/#01/#01/#01/#01/#01/#01/#01/#01/#01 s /~E/6The force on eac hc harge /#28p ositiv e and negativ e/#29 is F /= qE /. The magneticmomen ti s /#16 /= g/` /. The net torque is/#1C /= q E /`sin/#12 /= /#16E sin/#12The energy relativ et o /#12 /= /#19 is/#01 E /= /, /2 qE /`/2 cos/#12 /= /, /~/#16 /#01 /~E/6 /3 A Simple Deriv ation of the Thomas PrecessionThe follw oing deriv ation is based up on a suggestion b y E/.M/. Purcell/.Imagine an aricraft /#0Dying in a large circular orbit/. Appro ximate the orbit b yap olygon of N sides/, with N av ery large n um b er/. As the aircraft tra v erses eac ho ft h eN sides/, it c hanges its angle of /#0Digh tb y the angle /#12 /=/2 /#19/= N as sho wn in the /#0Cgure/.polygon side of /#12B B B B B B B B B B B BM B B BMZ Z Z Z/#7DWL /#12/6After the aircraft has /#0Do wn N segmen ts/, it is bac k at its starting p oin t/. IN thelab oratory frame/, the aircraft has rotated through an angle of /2 /#19 radians/. Ho w ev er inthe aircraft/'s instan taneous rest frame/, the triangles sho wn ha v e a Loren tz/-con tractionalong the direction it is /#0Dying but not transv ersely /.T h us at the end of eac h segmen t/, inthe aircraft frame/, the aircraft turns b y a larger angle than the lab oratory /#12 /=/2 /#19/= N /,but b y an angle /#12 /0/= /#0D/#12 /= W/= /#28 L/=/#0D /#29/= /2 /#19/#0D /= N /. After all N segemen ts in the aircraftinstan teous rest frame the total angle of rotation is /2 /#19/#0D /.The di/#0Berence in the reference frame is/#01 /#12 /=/2 /#19 /#28 /#0D /, /1/#29Since N has dropp ed out of the form ula for the angle and angle di/#0Berence/, one can letit go to in/#0Cnit y and the motion is circular and the form ula is for the rate of precession/./!P/! /= /#01 /#12/= T/2 /#19/= T /= /#0D /, /1This equation/, dispite the simplicit y of the deriv ation/, is the exact expressionfor the Thomas precession /. The equation do es not include the oscillationg termb ecause the deriv ation neglected the fact that the fron t and rear of the inertial barsare not accelerated sim ultaneously /./7