Physics 139 Relativity
Thomas Precession February 1998
G. F. SMOOT
DepartmentofPhysics,
University of California, Berkeley, USA 94720
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= 1 v =c =1= 1 v=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< q 0 1 1 1 v 1 v 1 v 0 2 2 2 2 q