Bookmark 5: spin
S
, spin
S
CM
and acceleration
a
We must remark that in the quantum Dirac’s analysis, the only variable that defines the position of the electron
r
, is contained in Dirac’s spinor ψ(t,
r
), it is moving at the speed c and it is also the point where the external electromagnetic field Aμ(t,
r
) is defined. Clearly, the spin with respect to the CC,
S
, represents the classical equivalent of Dirac’s spin operator.

11. The Center of Charge circles around the Center of Mass

Lets study the dynamical equation in the Center of Mass frame to discover how the Center of Charge moves relative to it.The Center of Mass frame is defined as that frame where
P
=0 and
q
=0, and thus the velocity
v
=0. From (24) and (21) we see that both
S
=
S
CM
are constant vectors in this frame,

S
CM
t
=0
. The equation (20) written in the Center of Mass frame looks (
q
=0,
v
=0)
S
=-m
r

u
=
S
CM
(
30
)
It represents the dynamical equation of the point
r
of the free particle in the Center of Mass frame.
If we take the time derivative of this expression we get:

S
CM
t
=0=

S
t
=-m

t
(
r

u
)=-m((
u

u
)+(
r

a
))
r

a
=0
and this implies that the vector
r
is collinear with the acceleration
a
.
The acceleration
a
is a vector that describes the rate of change of the velocity, either in speed or in direction or both. Its component parallel to velocity changes the speed and its component orthogonal to velocity changes the direction. As the speed 
u
=c is of constant absolute value, doesn’t change, the
a
component parallel to
u
is 0, so the acceleration is orthogonal to the velocity
u
.​
u
2

=
u
u
= c =>

t
u
u
= 2
u

t
u
= 2
u
a
= 0 =>
a
is orthogonal to
u
And thus
a
·
u
=0 and
r
⋅
u
=0, then we can calculate the change of the module of the vector
r
as​​
(
r
r
)
t
= 2
r
u
= 0​this means that 
r
 is constant and the motion of point
r
describes a circle of radius
R
0
at the speed c, orthogonal (30) to the constant spin
S
direction. The solution of this equation is depicted in the the following figure.
We have also depicted de angular velocity
𝝎
=
𝝎
0
of the comoving frame (x,y,z) attached to the point
r
. The spin
S
and the angular velocity in this frame
𝝎
0
, have opposite directions. 
u
=c
r
=
R
0
​From (30) the constant parameter 
S
=mc
R
0
When we quantize this model the parameter
S
=
ℏ
2
, so the radius of this motion is
R
0
=
ℏ
2mc
, or m=
ℏ
2
R
0
c
and the angular velocity is
𝝎
0
=
Δϕ
Δt
=
2π
2
πR
0
c
𝝎
0
=
2m
2
c
ℏ
inthisframe.
the separation between the CC and CM for the electron :ℏ = 1.054571817×
-34
10
kg⋅
2
m
⋅
-1
s
(reduced Planck constant)𝑚 = 9.1093837139 ×
-31
10
 kg (electron mass)𝑐 = 2.99792458 × 1
8
0
 m/s (speed of light)​​then
R
0
=
ℏ
2mc
= 1.932 ×
-13
10
m (separation between the CC and CM for the electron observed from its Center of Mass frame) ​and
𝝎
0
= 1.551 ×
21
10
-1
s