Bookmark 6: This model represents the circular motion, at the speed of light, of the Center of Charge
r
of the Dirac particle in the Center of Mass frame (
q
=0,
v
=0), as described by the dynamical equation (30). The Center of Mass is always a different point than the Center of Charge(16). The spin
S
in this frame has the opposite direction to the angular velocity
𝝎
0
due to the minus sign in (30). The trajectory is flat and the angular velocity has no component along the velocity
u
,
𝝎
u
=0. In general, for an arbitrary inertial observer that is not in the Center of Mass frame, the spin
S
and the angular velocity 𝝎 are not collinear vectors. The angular velocity has another component
𝝎
u
along the velocity
u
which produces the torsion of the CC motion, and which vanishes if the CC trajectory is flat.
The curvature of a trajectory tells us how quickly the curve deviates from a straight line, whereas the torsion tells us how quickly the curve deviates from being planar, how quickly the curve is twisting out of the plane that best fits the curve at a given point, and depends on the third time-derivative of the point
r
.

12. The angular velocity
ω

In this section we will obtain the expressions for the components of the angular velocity
ω
=
ω
p
+
ω
u
.The angular velocity corresponds to the rotation of the comoving Cartesian frame linked to the point
r
. But this frame is completely arbitrary. Once the CC is moving, its dynamics defines the acceleration a, orthogonal to the velocity
u
, and therefore these two orthogonal vectors with the vector
u
×a define an orthogonal comoving frame (the Frenet-Serret frame).
The Frenet-Serret frame is a moving coordinate system that describes the geometric properties of a curve in three-dimensional space. It consists of three mutually orthogonal unit vectors: Tangent vector (T=
u
): Points in the direction of the curve’s velocity and indicates the curve’s instantaneous direction. Normal vector (N=
a
): Points toward the center of curvature and describes how the curve bends. Binormal vector (B=
u

a
): Completes the right-handed coordinate system and is perpendicular to both the tangent and normal vectors.
As the acceleration
a
is orthogonal to the velocity
u
, we have a uniform circular motion and it can be written as a function of the constant angular velocity
ω
that is rotating the Frenet-Serret frame comoving with the CC
a
=
ω

u
The cross product of this expression with
u
gives:
a

u
=(
ω
u
)
u
-
2
c
ω
the first term is the component of the angular velocity along the velocity
u
(
ω
u
) times
2
c
, so that
ω
-
ω
u
=
ω
p
=
1
2
c
(
u

a
)
(
31
)
Then the perpendicular component of the angular velocity
ω
p
to the plane defined by the vectors
u
and
a
, can be expressed as
ω
p
=
1
2
c
(
u

a
)=
1
2
c

r
t

2

r

2
t
(
32
)
This perpendicular component has the opposite direction than the CC spin (26)
S
∝
a

u
. In general neither the CM spin
S
CM
, nor the CC spin
S
, is along the angular velocity
ω
. This perpendicular component
ω
p
is responsible for the curvature of the trajectory, while the
ω
u
is related to the torsion of the trajectory. In the Center of Mass frame the trajectory of the point
r
is flat, there is no torsion and thus
ω
u
=0.The component
ω
u
produces the torsion of the trajectory and depends on the third order derivative of the vector
r
. This component of the angular velocity is, as demonstrated in the book [1],
ω
u
=
1
2
c
2
a
(
u

a
)

a
t
u
=
1
2
c
2
a

r
t

2

r

2
t
3

r

3
t

r
t
(
33
)
The evolution of the angular velocity is completely determined by the motion of the CC.
The evolution of the CC determines both the trajectories of the CC and CM and the rotation of the body frame (Frenet-Serret frame) attached to the point
r
.
We can find an alternative expression for the angular velocity
ω
u
if we use the dynamical equation (29):
a
=

u
t
=
2
c
-
v
.
u
2

q
-
r


q
-
r

Taking the next order time derivative we have:

a
t
=
3

r

3
t
=

t
2
c
-
v
.
u
2

q
-
r


q
-
r
=
-
a
CM
.
u
-
v
.
a

2

q
-
r

+(
2
c
-
v
.
u
)2
q
-
r
(
v
-
u
)
4

q
-
r


q
-
r
+
2
c
-
v
.
u
2

q
-
r

(
v
-
u
)
(
34
)
Then the term in brackets in (33)
Using (
a

b
)·

c
=(

c

a
)·
b
= (
b


c
)·
a

r
t

2

r

2
t
3

r

3
t
=
2

r

2
t

3

r

3
t

r
t
=
a

3

r

3
t
u
(
35
)
and since from (29)
a
∝
q
-
r

the first term when substituting

a
t
as (33) in (34) is nulled and then (34) is:

r
t

2

r

2
t
3

r

3
t
=
2
c
-
v
.
u
2

q
-
r

(
a
(
v
-
u
))
u
and here we can simplify:
(
a
(
v
-
u
))
u
=((
v
-
u
)
u
)
a
=(
v

u
)
a
=(
u

a
)
v​​
Collecting terms, instead of the expression (32) we obtain
ω
u
=
1
2
c
2
a
2
c
-
v
.
u
2

q
-
r

(
u

a
)
v​​

u
and using (29) squared:
1
2
c
-
v
.
u
=
2
c
-
v
.
u
2
a
4

q
-
r

2

q
-
r

=
2
c
-
v
.
u
2
a
2

q
-
r

using this and (31) in
ω
u
:
ω
u
=
ω
p
v
2
c
-
v
.
u
u
(
36
)
If the CM velocity
v
=0, or along to the zitterbewegung plane, (
ω
p
⋅
v
)=0 and thus
ω
u
=0, the CC trajectory is a flat curve and there is no torsion. If
ω
p
and
v
form an angle α<π/2,
ω
u
has the direction of
u
(in (35)
ω
p
.
v
∝ cos
α
>0) and the opposite direction if α>π/2.In the Bookmark 7 we represent the motion of the free Dirac particle when integrating the dynamical equations with the Mathematica code in the last section below.