Physics 139 Relativity

Thomas February 1998

G. F. SMOOT

DepartmentofPhysics,

University of California, Berkeley, USA 94720

1

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 in connection with -

orbit interaction. Without including Thomas Precession, the rate of spin precession

of an atomic 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 plus

a three dimensional . 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 on a curved path

C

C

C

y

C

C

C

C



C

C





C



CW

X 

X ?

x

6

 v

At a certain 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 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 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<

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

Nowwe are 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 , S , are

related by

e

~

mu~ = S 17

m c

c

The 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

 !

~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 . 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 . 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 . After all N segements in the aircraft

instanteous rest frame the total angle of rotation is 2 .

The di erence in the reference frame is

 =2 1

Since N has dropp ed out of the formula for the angle and angle di erence, one can let

it go to in nity and the motion is circular and the formula is for the rate of precession.

 =T !

P

= = 1

! 2=T

This equation, dispite the simplicity of the derivation, is the exact expression

for the Thomas precession . The equation do es not include the oscillationg term

b ecause the derivation neglected the fact that the front and rear of the inertial bars

are not accelerated simultaneously. 7