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