Physics 139 Relativity Thomas Precession February 1998 G. F. SMOOT DepartmentofPhysics, University of California, Berkeley, USA 94720 1 Thomas Precession Thomas Precession is a kinematic e ect discovered by L. T. Thomas in 1926 L. T. Thomas Phil. Mag. 3, 1 1927. It is fairly subtle and mathematically sophisticated but it has great imp ortance in atomic physics in connection with spin- orbit interaction. Without including Thomas Precession, the rate of spin precession of an atomic electron is o by a factor of 2. Later we will see that there is a similar e ect for gravitational elds. The e ect is connected with the fact that two successive Lorentz transformations in di erent directions are equivalent to a Lorentz transformation plus a three dimensional rotation. This rotation of the lo cal frame of rest is the kinematic e ect that causes the Thomas precession. For the lecture we will not do the full mathematical treatment, since it is rather involved. Instead we will showby a simple example how the rotation and thus precession comes ab out. Maketwo successive Lorentz transformations in orthogonal directions: from S 0 0 00 to S with velo city v along the x axis, followed by a transformation from S to S with 0 0 velo city v along the y axis, as shown by the following diagram. 00 y 6 00 S > 00 00 O 0 0 - x > 0 6 y v 6 0 6 y 0 - - - 0 x x O O 0 v S S 00 00 The line from the origin O of S to the origin O of S making and angle in 00 00 S and an angle in S .We can calculate the angles in the two frames by applying 1 the Lorentz transformations and evaluating them in each frame. 0 0 0 x = x vt x = x + vt 0 2 0 0 2 t = t v x=c t = t + vx =c 0 0 y = y y = y 1 00 0 0 0 0 0 0 00 00 y = y v t y = y + vt 00 0 0 00 x = x x = x where q q 0 2 2 0 2 =1= 1v =c =1= 1v=c Combing these equations one nds: 00 0 0 y = [y v x vt] 00 x = x vt 2 Nowwe can calculate the angle made by the line b etween origins. For a Galilean transform one would have 0 0 y v t v tan = = = 3 x vt v but Sp ecial Relativity shows us that 3-D velo cities do not transform like 3-D vectors. So wemust calculate carefully. 0 0 00 0 00 0 0 00 y y + v t v t y 00 = = j = 4 tan = y =0 x vt vt vt 0 0 2 0 00 0 00 2 0 00 00 0 00 t = t + vx =c j = t + v y =c j = t 5 x =x =0 y =0 so that 0 0 00 0 v t v tan = = 6 0 00 v t v Note that this answer is very near the Galilean result but with the factor of 1/ which reminds us of ab erration. 00 Nowwe calculate : 0 0 0 0 00 [y v t ] y 00 = 7 tan = 00 0 x x 00 00 00 where x and y are the co ordinates of the origin O of system S in the S system. Thus 0 0 0 0 0 0 0 0 0 0 0 [y v ] v t v t v t 00 tan = j = = j = 8 y =0 x=0 0 0 x x x vt vt 0 2 t = t v x=c j = t; 9 x=0 0 0 v 00 tan = 10 v 2 0 This lo oks again similar to the Galilean angle except for the extra factor of . Now consider a particle on a curved path C C C y C C C C C C C CW X X ? x 6 v At a certain time it is at the origin O of our system S. Put the x axis parallel 0 to the path, and y axis toward the center of curvature. At t = 0, the rest frame S is 00 moving in the x direction with velo city v .At a slightly later time its rest frame S is 0 0 moving p erp endicular to x in the y direction with velo city v = v. De ne ! ! 0 0 0 v v 00 1 1 = = tan tan 11 v v For a very short time interval the motion is circular. That is t the lo cal curve with a tangent circle with appropriate radius of curvature. v = ! Rcos v = ! Rsin x y 12 0 v = v v = v = v x y so 0 v tan = v ! tan 00 1 0 1 13 = = tan tan tan Cho ose to b e very small; S vt = = R R Then ! vt 1 0 14 R ! v 1 0 ! = T t R In a circle the acceleration is 2 v v a a = so that = R R v 3 giving ! a 1 0 ! = T v Supp ose we are in a non-relativistic region v<<c, like an electron in an atom: q 0 1 1 1 v 1 v 1 v 0 2 2 2 2 q = 1 v=c 1+ 1+ 0 2 2 c 2 c 2 c 1 v =c 0 since tan = v =v << 1. Putting this backinto the expression for ! T 2 a v va ! = T 2 2 v 2c 2c 00 Thus > ,thus a counter-clo ckwise rotation, implying ~v ~a ~! = 15 T 2 2c The rigorous result is 2 ~v~a ~! = 16 T 2 +1 2c 2 Spin-Orbit Interaction of Electron with Nucleus in an Atom Nowwe are set to apply this kinematic e ect to spin precession in an atom. In its own rest frame the electron \sees" the nucleus ying by. ~ The electron's magnetic moment, ~, and spin angular momentum, S , are related by e ~ mu~ = S 17 m c c The torque on the magnetic momentis ~ dS 0 ~ 18 = ~ B ~ = dt 0 ~ where B is the magnetic eld in the e frame. ! ~v e 0 ~ ~ ~ B B = E 19 c ~ ~ Where B is the magnetic eld and E is the electric eld in the nucleus rest frame. v=c << 1 so that 1, ! ~ ~v dS e ~ ~ B 20 = ~ E dt c 4 arises from the interaction energy ! ~v e 0 ~ ~ U = ~ B E 21 c ~ If E is due to a spherically symmetrical charge distribution { as for a one- electron atom or one outside a closed shell { then dV ~r ~ ~ eE = rV r : 22 r dr Then ! dV e e ~r 0 ~ ~ ~ U = S B + S ~v 23 2 m c m c r dr e e ~ ~ ~ ~ S ~v r =+Sf~v e e 1 dV 0 ~ ~ ~ U = SB+ S~r~v 2 2 m c m c r dr e e e e 1 dV ~ ~ ~ ~ = S B + S L 24 2 m c m c r dr e e ~ since m~r ~v = L angular momentum. This second term is the spin-orbit interaction. Now, if the electron rest frame is rotating { Thomas angular velo city ~! , 0 ~ ~ dS =dt 6= ~ B . The general kinematic result from classical physics is: @ @ j = j ~! 25 rotation co ordinates inertial co ordinates @t @t as an op erator on any vector. So ~ ~ @ S @ S ~ j = j ~! S 26 rotation co ordinates inertial co ordinates @t @t With this expression the interation energy is changed to: 0 ~ U = U S ~! 27 T where ~! is prop ortional to the centrip etal acceleration due to E . T r 0 1 ~ eE 1 1 @ A ~v ~a = ~v ~! T 2 2 2c 2c m e ! 1 ~r dV = ~v 2 2mc r dr 5 ~ dV 1 dV 1 1 L = ~r ~v = 28 2 2 2 2m c r dr 2m c r dr e e Thus dV 1 1 0 ~ ~ U = U S L 2 2 2m c r dr e e dV 1 1 1 ~ ~ ~ ~ = S B +1 S L 29 2 2 m c 2 m c r dr e e The -1/2 is the famous one half. Including it, the observed ne-structure spacings in atomic sp ectra, due to electron spin, are correctly predicted. This schematic gives a heuristic indication of how the torque arises. ~ E 6 +q s ~ ` -q The force on eachcharge p ositive and negative is F = qE . The magnetic momentis =g`. The net torque is = q E `sin = E sin The energy relativeto = is ` ~ cos = ~ E E = 2qE 2 6 3 A Simple Derivation of the Thomas Precession The follwoing derivation is based up on a suggestion by E.M. Purcell. Imagine an aricraft ying in a large circular orbit. Approximate the orbit bya p olygon of N sides, with N avery large numb er. As the aircraft traverses eachofthe N sides, it changes its angle of ightby the angle =2=N as shown in the gure. BM Z} Z B Z W B Z BM B B B B side of L B polygon B B B B B B 6 After the aircraft has own N segments, it is back at its starting p oint. IN the lab oratory frame, the aircraft has rotated through an angle of 2 radians. However in the aircraft's instantaneous rest frame, the triangles shown have a Lorentz-contraction along the direction it is ying but not transversely.Thus at the end of each segment, in the aircraft frame, the aircraft turns by a larger angle than the lab oratory =2=N , 0 but by an angle = = W=L= =2 =N .