Bookmark 4: The angular momentum of the elementary particle can be measured with respect to different points: spin
S
, angular momentum with respect to the CC (
r
), spin
S
CM
, angular momentum with respect to the CM (
q
) and
J
angular momentum with respect to the origin
​If the particle is free then
v
is constant and

J
t
=0, and

P
t
=0
and from (22) this leads to

S
t
=
P

u
(
23
)

S
CM
t
=0
(
24
)
The CM spin
S
CM
, is conserved, is constant for the free particle, while the spin with respect to the CC,
S
, satisfies the same dynamical equation as Dirac’s spin operator in the quantum case [4]. ​Its time derivative is always orthogonal to the direction of the linear momentum. ​
S
precesses around the Center of Mass spin
S
CM
, that is constant.
Taking the time derivative of the kinematical momentum (12) we get, because in the free case H,
P
,
J
and
K
are constants of the motion:

K
t
=0=
H
2
c
u
-
P
-
1
2
c

S
t

u
-
1
2
c
S


u
t
(
25
)
If in (25) we replace (23)

S
t
=
P

u
and using (19)
0=
H
2
c
u
-
P
-
1
2
c

P

u

u
-
1
2
c
S


u
t
=
H
2
c
u
-
P
-
1
2
c

P
.
u

u
+
1
2
c
2
u
P
-
1
2
c
S


u
t
=H-
P
u

u
-
S


u
t
and making the cross product with
a
=

u
t
, we get
0=H-
P
u

u


u
t
-
S


u
t


u
t
=H-
P
u

u


u
t
-
S

u
t

u
t
+

u
t

u
t
S
a
=

u
t
is perpendicular to
u
because
u
is of constant absolute value c.
a
is also perpendicular to because
S
from (20) is orthogonal to both
u
and to (

r
-
q
), which is along the acceleration
a
, as we shall see later (28), and thus also orthogonal to the acceleration,
S
·

u
t
=0 and
S
=
H-
P
.
u

2

u
t

u
t

u
S
=
H-
P
.
u

2
a
(
a

u
)
(
26
)
where we see that the spin
S
is orthogonal to the velocity and acceleration of
r
.
From (16) and (14)
q
=
r
-
1
H
S

u
, by substituting this expression of
S
we finally obtain the definition of the position of the CM:
q
=

r
-
1
H
H-
P
.
u

2
a
(
a

u
)
u
=

r
-
1
H
H-
P
.
u

2
a
((
a
.
u
)
u
-
a
.
2
u
)​​​​as
a
isperpendicularto
u
a
.
u
=0,substitutingH(17)and
P
(18)​​​​
q
=

r
+
1
H
H-
P
.
u

2
a
a
.
2
c
=

r
+
1-
γm

v
.
u
2
γmc
2
a
a
.
2
c
q
=
r
+
(
2
c
-
v
.
u
)
2
a
a
(
27
)
If we isolate
a
a
=
2
a
(
2
c
-
v
.
u
)

q
-
r

(
28
)
and produce the squared of this expression and isolate
2
a
we get:
2
a
=
2
(
2
c
-
v
.
u
)
2

q
-
r

and using this in (28) to eliminate
2
a
we get
a
=

u
t
=
2

r
2

t
=
2
c
-
v
.
u
2

q
-
r


q
-
r

(
29
)
and represents the acceleration of the CC in terms of the position and velocity of both points CM (
q
,
v
) & CC (
r
,
u
).​The expressions of the Center of Mass
q
(27) and of the spins
S
(20) and SCM (21), have been obtained without the explicit knowledge of the free Lagrangian
L
0
, in terms of the positions and velocities of both points, the CC and the CM.