Physics 139 Relativity. Thomas Precession February 1998 G. F. SMOOT. Department ofphysics, University of California, Berkeley, USA 94720

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1 Physics 139 Relatiity Thomas Precession February 1998 G. F. SMOOT Department ofphysics, Uniersity of California, erkeley, USA Thomas Precession Thomas Precession is a kinematic eect discoered by L. T. Thomas in 192 (L. T. Thomas Phil. Mag. 3, 1 (1927)). It is fairly subtle and mathematically sophisticated but it has great importance in atomic physics in connection with spinorbit 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 eect for graitational elds. The eect is connected with the fact that two successie Lorentz transformations in dierent directions are equialent to a Lorentz transformation plus a three dimensional rotation. This rotation of the local frame of rest is the kinematic eect that causes the Thomas precession. For the lecture we will not do the full mathematical treatment, since it is rather inoled. Instead we will show by a simple example how the rotation and thus precession comes about. Make two successie Lorentz transformations in orthogonal directions: from S to S 0 with elocity along the x axis, followed by a transformation from S 0 to S 00 with elocity 0 along the y 0 axis, as shown by the following diagram. y 00 S 00 > O > y 0 y O 0 - O x S S 0 x 00 x 0 The line from the origin O of S to the origin O 00 of S 00 making and angle in S and an angle 00 in S 00.We can calculate the angles in the two frames by applying 1

2 the Lorentz transformations and ealuating them in each frame. x 0 = (x, t) x = (x 0 + t 0 ) t 0 = (t, x=c 2 ) t = (t 0 + x 0 =c 2 ) y 0 = y y = y 0 y 00 = 0 (y 0, 0 t 0 ) y 0 = 0 (y 00 + t 00 ) x 00 = x 0 x 0 = x 00 where q q =1= 1, 2 =c 2 0 =1= 1,( 0 =c) 2 Combing these equations one nds: (1) y 00 = 0 [y, 0 (x, t)] x 00 = (x, t) (2) Now we can calculate the angle made by the line between origins. For a Galilean transform one would hae tan = y x = 0 t t = 0 but Special Relatiity shows us that 3-D elocities do not transform like 3-D ectors. So we must calculate carefully. so that tan = y x = y0 t = 0 (y t 00 ) j y 00 =0 = 0 0 t 00 t t t = (t 0 + x 0 =c 2 )j x 00 =x 0 =0 = 0 (t y 00 =c 2 )j y 00 =0 = 0 t 00 (5) tan = 0 0 t 00 0 = 0 00 t Note that this answer is ery near the Galilean result but with the factor of 1/ which reminds us of aberration. Now we calculate 00 : (3) (4) () tan 00 = y00 x 00 = 0 [y 0, 0 t 0 ] x 0 (7) where x 00 and y 00 are the coordinates of the origin O of system S in the S 00 system. Thus tan 00 = 0 [y, 0 ] j x 0 y=0 =, 0 0 t 0 =, 0 0 t 0 x 0 (x, t) j x=0 =,0 0 t 0 (8),t t 0 = (t, x=c 2 )j x=0 = t; (9) tan 00 = (10)

3 This looks again similar to the Galilean angle except for the extra factor of 0. Now consider a particle on a cured path y C CC C CC C CC XX? C CW x At a certain time it is at the origin O of our system S. Put the x axis parallel to the path, and y axis toward the center of curature. At t = 0, the rest frame S 0 is moing in the x direction with elocity. At a slightly later time its rest frame S 00 is moing perpendicular to x 0 in the y direction with elocity 0 =. Dene = 00, = tan,1, tan,1 (11) For a ery short time interal the motion is circular. That is t the local cure with a tangent circle with appropriate radius of curature. so Choose to be ery small; Then In a circle the acceleration is x = Rcos y = Rsin (12) x = y = = 0 tan = 0 = 00, = tan,1 ( 0 tan), tan,1 = S R = t R t R T = t R 0, 1 0, 1 tan (13) (14) a = 2 R so that R = a 3

4 giing T = a 0, 1 Suppose we are in a non-relatiistic region <<c, like an electron in an atom: 0, 1 = 1, q1, (=c) q , ( 0 =c) 2 2 (0 c )2, ( c )2 1 2 ( c )2 since tan = 0 = << 1. Putting this back into the expression for T 2 T a 2c 2 = a 2c 2 Thus 00 >,thus a counter-clockwise rotation, implying The rigorous result is ~ T = ~~a 2c 2 (15) ~ T = 2 ~~a (1) +1 2c 2 2 Spin-Orbit Interaction of Electron with Nucleus in an Atom Now we are set to apply this kinematic eect 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 ~mu = e ~S (17) m c c The torque on the magnetic moment is ~ = d~ S dt = ~ ~ 0 (18) where ~ 0 is the magnetic eld in the e, frame. ~ 0 = ~, ~ e c ~ E (19) Where ~ is the magnetic eld and E ~ is the electric eld in the nucleus rest frame. =c << 1 so that 1, d S ~ dt = ~ ~, ~ e c E ~ (20) 4

5 arises from the interaction energy U 0 =,~ ~, ~ e c ~ E (21) If ~ E is due to a spherically symmetrical charge distribution { as for a oneelectron atom or one outside a closed shell { then Then e E ~ =, rv ~ (r), ~r dv r dr : (22) U 0 =, e S m e c ~ ~ + e ~ ~r dv S ~, (23) m e c 2 r dr ~S ~ ~ (, r) =+ ~ S ~ f~ U 0 =, e ~S + m e c ~ e ~ 1dV S(~r~) m 2 c2 r dr =, e ~S m e c ~ + e S ~ ~ 1 dv L m e c 2 r dr (24) since m~r~ = L ~ angular momentum. This second term is the spin-orbit interaction. Now, if the electron rest frame is rotating { Thomas angular elocity ~, d S=dt ~ = ~ ~ 0. The general kinematic result from classical physics is: as an operator on any j rotation j inertial coordinates, ~ S j rotation coordinates S j inertial coordinates, ~ S ~ (2) With this expression the interation energy is changed to: U = U 0, ~ S ~ T (27) where ~ T is proportional to the centripetal acceleration due to E r. 0 1 ~ T 1 2c 2~~a = e E ~ A 2c 2~ m e = 1 2mc 2~,~r dv r dr 5

6 = 1 2m e c 2 (~r~) 1 r dv dr = L ~ 1 dv 2m 2 e c2 r dr (28) Thus U = U 0, 1 S ~ ~ 1 dv L 2m 2 e c2 r dr =, e ~S ~ +(1, 1 m e c 2 ) 1 S ~ ~ 1dV L m 2 e c2 r dr (29) The -1/2 is the famous one half. Including it, the obsered ne-structure spacings in atomic spectra, due to electron spin, are correctly predicted. This schematic gies a heuristic indication of how the torque arises. ~E -q +q s ` The force on each charge (positie and negatie) is F = qe. The magnetic moment is=g`. The net torque is The energy relatie to = is ~ = qe`sin = Esin E =,2qE ` 2 cos =,~ ~ E

7 3 A Simple Deriation of the Thomas Precession The follwoing deriation is based upon a suggestion by E.M. Purcell. Imagine an aricraft ying in a large circular orbit. Approximate the orbit by a polygon of N sides, with N a ery large number. As the aircraft traerses each of the N sides, it changes its angle of ight by the angle =2=N as shown in the gure. Z} M Z Z M W side of polygon L After the aircraft has own N segments, it is back at its starting point. IN the laboratory frame, the aircraft has rotated through an angle of 2 radians. Howeer in the aircraft's instantaneous rest frame, the triangles shown hae a Lorentz-contraction along the direction it is ying but not transersely. Thus at the end of each segment, in the aircraft frame, the aircraft turns by a larger angle than the laboratory = 2=N, but by an angle 0 = = W=(L=) =2=N. After all N segements in the aircraft instanteous rest frame the total angle of rotation is 2. The dierence in the reference frame is =2(,1) Since N has dropped out of the formula for the angle and angle dierence, one can let it go to innity and the motion is circular and the formula is for the rate of precession. P = =T 2=T =, 1 This equation, dispite the simplicity of the deriation, is the exact expression for the Thomas precession. The equation does not include the oscillationg term because the deriation neglected the fact that the front and rear of the inertial bars are not accelerated simultaneously. 7

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