Figure 5. Dirac particle in the center of mass reference frame, with the spin along the OZ axis .
The initial position and velocity of the CC on the XOY plane is determined by the phase ψ . The radius of this motion is
R
0
= 1/2, in natural units .
If we initially have the CC at the position (1/2, 0, 0) on the OX axis, we rotate first an angle ψ, around the OZ axis to modify the CC position.Next, we change the orientation of the spin by modifying the zenithal angle θ, and azimuthal angle ϕ, so that the arbitrary rotation is the composition of the rotations:​​
R(ψ,θ,ϕ)=
R
OZ
(ϕ)
R
OY
(θ)
R
OZ
(ψ)
​​​
R
OZ
(ψ)
=
Cos[ψ]
-Sin[ψ]
0
Sin[ψ]
Cos[ψ]
0
0
0
1
​​​
R
OY
(θ)
=
Cos[θ]
0
Sin[θ]
0
1
0
-Sin[θ]
0
Cos[θ]
​​​
R
OZ
(ϕ)
=
Cos[ϕ]
-Sin[ϕ]
0
Sin[ϕ]
Cos[ϕ]
0
0
0
1
​​If initially at time
*
t
0
, and in natural units the CC is at the point (1/2, 0, 0), then the position, velocity and acceleration of the CC, before the rotation, since c = 1 anda =
2
c
R
0
= 2, are:the CC is at the point
(1/2,0,0)
, then the position and velocity of the CC, before the rotation, are:​
*
r
(
*
t
0
)=
1
2
0
0
*
u
(
*
t
0
)=
0
-1
0
*
a
(
*
t
0
)=
-2
0
0
These variables, after the rotation, become

r
0
=
R
OZ
(ϕ)
R
OY
(θ)
R
OZ
(ψ)

r
(
t
0
)
:
*
r
0
=
1
2
Cos[ϕ]Cos[θ]Cos[ψ]-Sin[ϕ]Sin[ψ]
Sin[ϕ]Cos[θ]Cos[ψ]+Cos[ϕ]Sin[ψ]
-Sin[θ]Cos[ψ]
(
42
)
andthesamefor
u
0
=
R
OZ
(ϕ)
R
OY
(θ)
R
OZ
(ψ)
u
(
t
0
),andthus:
*
u
0
=
Cos[ϕ]Cos[θ]Sin[ψ]+Sin[ϕ]Cos[ψ]
Sin[ϕ]Cos[θ]Sin[ψ]-Cos[ϕ]Cos[ψ]
-Sin[θ]Sin[ψ]
(
43
)
and
*
a
0
=-2
Cos[ϕ]Cos[θ]Cos[ψ]-Sin[ϕ]Sin[ψ]
Sin[ϕ]Cos[θ]Cos[ψ]+Cos[ϕ]Sin[ψ]
-Sin[θ]Cos[ψ]
=-4
*
r
0
(
44
)
when we apply the Lorentz Boost
v
to calculate the coordinates in the laboratory frame, and time translation b and space translations
d
, the corresponding time
t
0
and position
r
0
of the CC of the same event, for any arbitrary inertial observer in natural units (c=1) are, according to (3) & (4) and with
*
r
0
=
R(α)
r
:
t
0
=
*
γt
0
+γ
v
*
r
0
+b
r
0
=
*
r
0
+γ
v
*
t
0
+
2
γ
(1+γ)

v
*
r
0

v
+
d
where b is the time translation,
d
is the space translation in natural units and
v
is the velocity of the Center of Mass observer
*
O
as measured by the observer O. It represents, therefore, the velocity of the Center of Mass of the particle for the arbitrary laboratory observer O.If the initial instant to integrate the equations in the reference frame of the Laboratory O, is the time
t
0
= 0, this corresponds to
*
γt
0
=-γ
v
*
r
0
-b
, for the Center of Mass observer of the particle, and therefore the initial position of the CC for the laboratory observer at the initial instant
t
0
= 0, is:
r
0
=
*
r
0
-
γ
(1+γ)

v
*
r
0

v
-b
v
+
d
(
45
)
To transform the Center of Charge velocity, we use (7) with c=1 and
*
u
0
=
R(α)·
u
:
u
0
=
*
u
0
+γ
v
+
2
γ
(1+γ)

v
·
*
u
0

v
γ1+
v
*
u
0

(
46
)
And to transform the Center of Charge acceleration, we use (9) with c=1 and
*
u
0
=
R(
u
)
and
*
a
0
=
R(
a
)
:
a
0
=
1
2
γ
3
1+
v
*
u
0

1+
v
*
u
0

*
a
0
-
v
*
a
0

*
u
0
-
γ
(1+γ)

v
*
a
0

v
(
47
)
We have to calculate now the position of the Center of Mass in this reference frame. The definition of the center of mass position of the particle in any reference frame at any time is given in (27)
q
=
r
+
(
2
c
-
v
.
u
)
2
a
a
then the initial value of the CM
q
0
, in the laboratory frame is :
q
0
=
r
0
+
1-
v
.
u
0

2
a
0
a
0
in terms of the position, velocity and acceleration of the CC in this reference frame and of the velocity of the CM, all these variables defined at the initial time
t
0
= 0.Since from (44)
a
0
=-4
*
r
0
, and
*
a
0
*
u
0
=0
since they are perpendicular, taking the squared of (47) we get
2
a
0
=
1
4
γ
6
1+
v
*
u
0

16
2
1+
v
*
u
0

2
*
r
0
+16
2

v
*
r
0

1+
16
2
γ
2
(1+γ)
2

v
*
r
0

2
v
-2*16*1+
v
*
u
0

2

v
*
r
0

γ
(1+γ)
+2*16*
γ
(1+γ)
2

v
*
r
0


v
*
u
0
​​since
2
γ
2
v
=
2
γ
-1=(γ-1)(1+γ)wecanexpandthethirdterminto:​​
16
2
γ
2
v
2
(1+γ)
2

v
*
r
0

=
16(γ-1)
(1+γ)
2

v
*
r
0

​​​​
2
a
0
=
16
4
γ
6
1+
v
*
u
0

2
1+
v
*
u
0

2
*
r
0
+
2

v
*
r
0

+
(γ-1)
(1+γ)
2

v
*
r
0

-2
2

v
*
r
0

γ
(1+γ)
​​andaddingthetermsin
2

v
*
r
0

​​
2
a
0
=
16
4
γ
6
1+
v
*
u
0

2
1+
v
*
u
0

2
*
r
0
+
2

v
*
r
0

1+
γ-1
(1+γ)
-
2γ
(1+γ)
=
16
4
γ
6
1+
v
*
u
0

2
1+
v
*
u
0

2
*
r
0
​​andas
*
r
0
=
1
2
2
a
0
=
4
4
γ
4
1+
v
*
u
0

From (50) the term
1-
v
u
0
=1-
*
u
0
v
+γ
2
v
+
2
γ
(1+γ)

v
·
*
u
0

2
v
γ1+
v
*
u
0

=
γ1+
v
*
u
0
-
*
u
0
v
-γ
2
v
-(γ-1)
v
·
*
u
0

γ1+
v
*
u
0

=
γ-γ
2
v
γ1+
v
*
u
0

=
γ(1-
2
v
)
γ1+
v
*
u
0

​​andas(1-
2
v
)=
1
2
γ
​​1-
v
u
0
=
1
2
γ
1+
v
*
u
0

and then
1-
v
u
0

2
a
0
a
0
=
1
4
2
γ
3
1+
v
*
u
0

a
0
and if we substitute this expression in
q
0
derived above from (27) and for
a
0
the expression (47), where
a
0
=-4
*
r
0
, the initial position of the CM, , is
q
0
=
r
0
+
1-
v
.
u
0

2
a
0
a
0
=
r
0
+
1
4
2
γ
3
1+
v
*
u
0

1
2
γ
3
1+
v
*
u
0

1+
v
*
u
0

*
a
0
-
v
*
a
0

*
u
0
-
γ
(1+γ)

v
*
a
0

v
​​
q
0
=
r
0
+
1
4
-41+
v
*
u
0

*
r
0
+4
v
*
r
0

*
u
0
+
4γ
(1+γ)

v
*
r
0

v
​​andsubstituting
r
0
from(45)​​
q
0
=
*
r
0
-
γ
(1+γ)

v
*
r
0

v
-b
v
+
d
-1+
v
*
u
0

*
r
0
+
v
*
r
0

*
u
0
+
γ
(1+γ)

v
*
r
0

v
​​
q
0
=-
v
*
u
0

*
r
0
+
v
*
r
0

*
u
0
-b
v
+
d
​​andthen:
q
0
=
v

*
u
0

*
r
0
-b
v
+
d
​​
q
0
=
v
*
r
0

*
u
0
-
v
*
u
0

*
r
0
-b
v
+
d
(
48
)
As a summary, the boundary conditions at
t
0
= 0, in the laboratory frame, are :
r
0
=
*
r
0
-
γ
(1+γ)

v
*
r
0

v
-b
v
+
d
​​
u
0
=
*
u
0
+γ
v
+
2
γ
(1+γ)

v
·
*
u
0

v
γ1+
v
*
u
0

​​
q
0
=
v
*
r
0

*
u
0
-
v
*
u
0

*
r
0
-b
v
+
d
​​
v
0
=
v
(
49
)

20. Code for the dynamic simulation of the Spinning Particle Model

Output

CITE THIS NOTEBOOK

Spinning Particle Model (I): a classical model of the electron spin​
by Juan Barandiaran, Martin Rivas
Wolfram Community, STAFF PICKS, February 12, 2025
https://community.wolfram.com/groups/-/m/t/3521996