the magnetic top of universe as a model of quantum...
TRANSCRIPT
The magnetic top of Universe as a model of quantum spinSource file of A.O. Barut, M. Bozic and Z. Maric
Substitution, conversion and transformation by Dusan Stosic
Abstract The magnetic top is defined by the property that the external magnetic field B coupled to the angular velocity as distinct from the top fhose magnetic moment is independent of angular velocity. This allows one to construct a "gauge" theory of the top where the caninical angular momentum of the ooint particle and the B field plays the role of the gauge potential. Magnetic top has four constants of motion so that Lagrange equations for Euler angles ,, (wich define the orientation of the top) are solvable, and are solved here. Although the Euk=ler angles have comlicated motion.,the canonical angular momentum s, interpreted as spin , obeys precisely a simple precession equation. The Poisson brackets of si allow us further to make an unambiguous quantization of spin , leading to the Pauli spin Hamiltonian. The use of canonical angular momentumalleviates the ambiguity in the ordering of the variables P P P in the Hamiltonian. A detailed gauge theory of the asimmetric magnetic top is alsou given.si
Euler angles - The xyz (fixed) system is shown in blue, the XYZ (rotated) system is shown in red. The line of nodes, labelled N, is shown in green.Contents page3 Introduction I. II . Lagrangian and Hamiltonian of the symmetric magnetic top 6III. Lagrange equation for the magnetic topand their solutions for constant magnetic field1 10 IV. The torque equation and its equivalence with the Lagrange equations 17V. Hamilton's equations for the magnetic top 18VI. Quantum magnetic top 21 VII. The states of the quantum magnetic top 26VIII. The Asymmetric Magnetic top 29 Appendix A.Top with magnetic moment fixed in the body frame 36 I. InroductionReferences 41Whereas the coordinates and momenta of quantum particles have a classical origin or a classical counterpart,the spin is generally thought to have no classical origin. It is, in Pauli words,"a calassicay non-explanable two-valuedness"{1} .Thus, the spin and coordinates are not on the same footing as far as thepicture of the particles is concerned. In atomic physics the role of spin is enormous due to the Pauli-principle and spin statistics connection,althougt the numerical values of spin orbit terms are small.In nuclear and particle physics and in very high energy physics, there spin hyperfine
2
terms turned out to play an essential role, whose theoretical understandig is still lacking (2). Even in the interpretation and foundations of quantum theory, the nature of spin seems to be rather crucial, and a need for a classical model of spin has long been felt (3). Our knowledge about the importance of spin in all these areas comes from the widespread and succesfull applicability of Pauli and Dirac matrices and spin representation of Galillei and Poincare groups. Although there is no mystery is actually some mystery in the physical origin and in the visualization of spin. (It cocerns the spin 1/2 as well as the higher spins). Because of all those reasons there has been in the past many attempts to identify internal spin variables and to main clssical models of spin, both of Pauli (4-12) as well as of Dirac spin (13-18) But , none ofthe nonrelativistic spin models has been generally accepted, either because none of the propsed models is without shorthcomings and difficulties or because the prevailling attitude of physicists towards internal spin variables is, in Schulman's words: a general unconfortablenes at the mention of internal spin variables and a reliance on the more formal, but nevertheless completely adequate, spinor wave functions which are labelled basis vectors for a representation of so*3) but are endowed with no further properties"(10) In this paper, we shall consider the nonrelativistic Pauli spin, and a minimal classical model - in the sense of the smallest possible phase space dimension - underlying the Pauli equation. Our classical model of quantum spin is based on magnetic top , wich we define as a top whose mafnetic moment is proportional to the angular velocity(Chapter II) By solving the classical equation of motion of the magnetic top we shall show that it has, by virtue of the special coupling to the magnetic field, a unique property that the motion of its magnetic moment is one dimensional (i.e ptecessio around the magnetic field) whereas the top itself performs a complicated three-dimensional motion (Chapters III and IV). The motion of the magnetic moment of the magnetic top is different in an essential way from the motion of the top which carries magnetic moment fixed in the body frame. Namely, a magnetic moment which is fixed to the top preform a three-dimensional motion (precession with nutation) since it shares the motion of the body to which it is attached (Apendix A). This distinction is the consecuence of the differnce in the form of the two Lagrangian. The potential in the Lagrangian of magnetic top (Chapter II) is angular velocity dependent whereas the potential of the top which carries magnetic moment is velocity indeoendent (Apendix A). Also, Hamiltonian of the latter top is simple sum of kinetic and angular velocity independent potential wheras Hamiltonian of magnetic top is not of this form(Chapter II). It is necessery to relize those differences in order to understand the difference between our work and previous works (8,9,10( on the classical models of spin which were also based on the top. In Rosen work, classical model of spin is in fact the top with angular velocity independent potential (8). In our oppinion this model is unsatisfactory because for quantum spin there exists the linear relation
s
between magnetic
3
moment operator
and spin angular momentum operator s ,whereas, such
a relation does not characterize Rosen's classical model in which it is assumed
that Hamiltonian is a sum of kinetic energy I 22
and potential energy B is
independent of angular velocity. But this is possible only if
is independent of spin angular momentum.
The Lagrangian of the magnetic top is identical with the Lagrangian of the Bopp and Haag (9) model of spin. But the procedure of the construction of the Hamiltonian and subsequent quantization procedures differ in our and in the Bopp and Haag aproach (Chapter VI). Certain authors have arged in the past that the top is not an appropriate model of spin, because its configuration space (which is three dimensional ) is larger than it is necessert. Namely, in Nielsen and Rohrlich words (11) "quantum-mechanical perticle of definite spin is essentially one-dimension (since it is completelt bythe eigenstates of one coordinate) so Schulman's formulation seems over complicated". It follous from our analysis that this remark is not applicable to the magnetic top because although its configuration space is three-dimensional, the magnetic moment of magnetic top precesses around constant magnetic field (Chapter III). Moreover, in the light of this result it becomes understadable why Pauli theory of the spin motion in a magnetic field has been so succseful despite the fact that it avoids to answer the question as to what the internal spin variables are and what the variables conjugate to spin are. The explanation is simple. It is a satisfactory theory for those phenomena for which only thr motion of magnetic moment is relevant. But, are there phenomena determined by the motion of the magnetic top itself. Our answer is positive. One example is the phase change of spinors in magnetic fields (Chapter VII).
II Lagrangian and Hamiltonian of the symetric magnetic top
As stated in the Inroduction we shell use the word"top" to denote the mechanicalobject whose orientation in the reference frame is discribed by Euler angles ,,.Magnetic top by definition has a magnetic moment proportional to its angularmomentum
Mtopsv gsv sv
sv 1.171 1095 erg sec
gsv 5.166 10 4 cm1 gm0 sec-1
(1)
Mtopsv gsv sv
Mtopsv 6.05 1091erg
stattesla
4
gsv sv
6.05 1091erg
stattesla
The angular momentum itself is proportional to the vector of angular velocity (2)
sv
Isv sv
Isvi
sv..i ei Isv x.sv i y.sv j z.sv k
ei are unit vectors of the coordinates system attached to the body and whose orientation iz the Laboratory frame are three Euler anglesx.sv sv
i 1 1ei 1 1
y.sv sv
sv..i sv
z.sv sv
k 1 1j 1 1
Isvi
sv ei
1.171·10 95 gm cm2 sec-1
Isv x.sv i y.sv j z.sv k 3
1.171·10 95 gm cm2 sec-1
sv 1.171 1095 gm cm2 sec-1
Isv sv
1.171 1095 gm cm2 sec-1
i j k are unit vectors along the axis of the Laboratory referenceframe. The components of
in the Laboratory frame are :
x cos ' sin sin '
3 y sin ' cos sin '
z ' cos '
The components of
in the body-fixsed frame, on the other hand are:1 sin sin cos '
4 2 sin cos ' sin '
5
3 cos ' '
The kinetic energy T.sv of the free symmetrical top is a simple function of
(or
)
Ha Hb1
5
TsvIsv sv
2
2
sv
2
2 Isv1
2' 2 ' 2 sin 2 ' ' cos
2
'Ha
dd
'Ha
dd
'Ha
dd
sv
2
2 Isv1.031 1077 gm cm2 sec-2
gm cm22
' 2 ' 2 sin 2 ' ' cos 2 0 gm cm2 sec-2
gm cm22
' 2 ' 2 sin 2 ' ' cos 2
sv
2
2 Isv 1.031 1077 gm cm2 sec-2
1
2 xd
d
2
z
dd
2
sin 2zd
d
y
dd
cos
2
sv
2
2 Isv
TsvIsv sv
2
2
According to classical electrodynamics the potential energy of the magnetic moment Min a magnetic field B is:Bsv 6.816 10 15 stattesla
6 Vsv Mtopsv Bsv
Vsv Mtopsv Bsv
Vsv 4.123 1077 gm cm2 sec-2
Conseqently , the Lagrangian takes the form:G Msv
2
Rgsv2.062 1077 gm cm2 sec-2
6
Isvsv
2
2Mtopsv Bsv 5.154 1077 gm cm2 sec-2
Conseqently , the Lagrangian takes the form:7
Lsv Tsv VsvIsv sv
2
2Mtopsv Bsv
Tsv Vsv 5.154 1077 gm cm2 sec-2
Isv sv
2
2Mtopsv Bsv 5.154 1077 gm cm2 sec-2
But , for our magnetic top we assume that the relation(1) is valid. By incorporatingthis relation into the Lagrangian we get:
LIsv sv
2
2gsv Isv sv
Bsv
LsvIsv sv
2
2gsv Isv sv
Bsv
Lsv 5.154 1077 gm cm2 sec-2
Isv sv
2
2gsv Isv sv
Bsv 5.154 1077 gm cm2 sec-2
It is important to realize that this Lagrangian is different, in an essential way , fromthe Lagrangian studied in classical electromagnetism, where M is a fixed vectorin body frame and 90 45
P Ixd
dI B1 B
130
Hb 1.76 10 18 sec-1
Ha Hb1
'Ha
dd
p 'Ld
d
Isv ' Isv gsv Bsv
7
Isv ' Isv gsv Bsv 2.342 1095 gm cm2 sec-1
P Iy
dd
cos xd
d
I g1Bz
P Izd
dcos
zd
d
I g1Bz
where sin cos B Bx cos By sin
R1cos sin
0
sin cos
0
0
0
0
1
0
0
0
cos sin
0
sin cos
cos sin
0
sin cos
0
0
0
1
cos sin
0
sin cos
0
0
0
0
1
0
0
0
cos sin
0
sin cos
cos sin
0
sin cos
0
0
0
1
0.044
0.554
0
0.61
0.6740
0.7910.489
0
Bz1 Bx sin sin By cos sin Bz cos
Following general procedures we need now to express ,, and in terms of P , P and P
R10.044
0.554
0
0.61
0.6740
0.7910.4890
xd
d
PI
g1 B
yd
d
P P cos I g1 Bz cos Bz I sin 2
zd
d
P P cos I g1 Bz1 cos Bz I sin 2
Bicause the dependence of the Lagrangian on , and istrough angular velocity
it is usefull ro express angular velocity through
the cannonical moment P , P and P
sv.x Isv sv.x cos Psin sin P
sin cos
sin P gsv Isv Bsv.x
sv.y Isv sv.y sin Pcos sin P
cos cos
sin P gsv Isv Bsv.y
sv.x sv
Isv sv.x 1.171 1095 gm cm2 sec-1
sv.y sv
P Isv ' cos ' Isv gsv Bsv
8
2
2
P P
Bsv.y Bsv
P P
Isv sv.y sin Pcos sin P
cos cos
sin P gsv Isv Bsv.y
sin Pcos sin P
cos cos
sin P gsv Isv Bsv.y 0 gm cm2 sec-1
P 2.342 1095 gm cm2 sec-1
x cos Psin sin P
sin cos
sin P gsv Isv Bsv
11
y sin Pcos sin P
cos cos
sin P gsv Isv Bsv
z P gsv Isv Bsv
z 0 gm cm2 sec-1
sx cos Psin sin P
sin cos
sin P
x 0 gm cm2 sec-1
y 0 gm cm2 sec-1
sy sin Pcos sin P
cos cos
sin P
sz P
sx 2.342 1095 gm cm2 sec-1
Isv gsv Bsv 2.342 1095 gm cm2 sec-1
Isv sv.y 1.171 1095 gm cm2 sec-1
sx gsv Isv Bsv 0 gm cm2 sec-1
z I x P g1 IBz
We shallnow define a new vector quantity - cannonical angularmomentum s, by
sx cos Psin sin P
sin cos
sin P
sy sin Pcos sin P
cos cos
sin P
sz PIt is seen ...take the form
9
s g1 I BThe latter relation is analogous to the relation between thekinetic momentum in the electromagnetic field of the vector potential A L 1
2m q 2 e A q
1 m q1
P1 q1
Ldd
m q p e A p m q e A
m q exp 1 ANow we are ready to write the Hamiltonian of the magnetic topaccording to
2
P 2.342 1095 gm cm2 sec-1
H P P I
2
2 g1 I
B
P 3.513 1095 gm cm2 sec-1
Psec
Psec
Isvsv
2
2 gsv Isv sv
Bsv
1.428 1019
Msv c2
2
Psec
Psec
Isv sv
2
2 gsv Isv sv
Bsv
1.428 1019
1.947 1085 gm cm sec-25.293 10 9 cm
10
Isvsv
2
2 gsv Isv sv
Bsv
Msv c2
2
3
Psec
Psec
1.472 1096 gm cm2 sec-2
P 1.104 1096 gm cm2 sec-1
2
After some algebra we obtainy a0 1
x a01
z a01
ssv
2
2
s2
2 Ig1 s B g12 I B2
s2
2 Ig1 s B g12 I B2
H I
2
2
2
2 IM2
2 I g12
s g1 I B( )2
2 Is2
2 Ig1 s B g12I B2
gsv 5.166 10 4 cm1 gm0 sec-1
s2
2 Ig1 s B g12 I B2
I
2
11
Isvsv
2
2
sv
2
2 Isv
Mtopsv2
2 Isv gsv2
s gsv Isv Bsv 2
2 Isv
s2
2 Isvgsv s Bsv gsv
2 Isv Bsv2
5
1.031 1077
1.031 1077
1.031 1077
2.319 1077
1.289 1077
erg
Mtopsv2
2 Isv gsv2
1.031 1077 gm cm2 sec-2
gsv 5.166 10 4 cm1 gm0 sec-1
s2
2 Isvgsv s Bsv gsv
2 Isv Bsv2
6.251.031 1077 gm cm2 sec-2
Isvsv
2
2
sv
2
2 Isv
Mtopsv2
2 Isv gsv2
s gsv Isv Bsv 2
Isv
4.5 s2
2 Isvgsv s Bsv gsv
2 Isv Bsv2
6.25
1.031 1077
1.031 1077
1.031 1077
1.031 1077
1.031 1077
erg
12
.50 Isv sv2
.50sv
2
Isv
.50Mtopsv
2
Isv gsv2
.50s 1. gsv Isv Bsv 2
Isv
.10s2
Isv .20 gsv s Bsv .20 gsv
2 Isv Bsv2
4
Msv c2
2
1
4 gsv2 Bsv
2 2 gsv s Bsv 5 Msv c2 4 gsv
2 Bsv2 s2 20 gsv s Bsv Msv c2 25 Msv
2 c4 1
2
1
4 gsv2 Bsv
2 2 gsv s Bsv 5 Msv c2 4 gsv2 Bsv
2 s2 20 gsv s Bsv Msv c2 25 Msv2 c4
1
2
5.572 10112
2.481 10111
gm cm2
Msv c2
21.031 1077 gm cm2 sec-2
Isv
5.572 10112 gm cm2 1.194
Mtopsvsv
Mtopsv
sv
5.166 10 4
5.166 10 4
cm1 gm0 sec-1
s2
2 Isvgsv s Bsv gsv
2 Isv Bsv2
51.289 1077 gm cm2 sec-2
2 6.283So ,again the form of the Hamiltonian ......
1
2 xd
d
2
z
dd
2
sin 2zd
d
y
dd
cos
2
me c2
2
1
2 mep e A( )2
2
2 I
H 1
2 mp e A( )2
gm
2 sec 2 xd
d
2
z
dd
2
sin 2zd
d
y
dd
cos
2
me c2 12
2 2.18 10 11 erg
13
1
2 mep e A( )2
s g1 I B( )2
2I
L 1
2me q 2 e A q
1
me2
me g h1 1
2
c
1me
2me g h1
1
2
c
1.589 10 9
1.589 10 9
cm0 sec0
Ovde dodje tekstg h1 1.034 10 24 gm cm3 sec-3
2 gh1
c2 12 4.322 10 41 gm cm sec-1
III . Lagrange equations for the magnetic top and their solutions for constant magnetic fieldsWe shell now write and solve Lagrange equations of motion for magnetic top in a constantmagnetic field, assumed to be directed along the z-axis of the space-fixed reference frame. This assumption does not reduce the generality of our solution, since the orientation of the Laboratory frame may be chosen convenniently. With this assumption the Lagrangian (8) takes the form : 18
Lsv1Isv
2'2 '2 '2 2 ' ' cos gsv Bsv Isv ' ' cos
sec gsv Bsv Isv
Lsv1 2.342 1095 gm cm2 sec-1
Because this Lagrangian does not depende on f and c the momenta P and P are integrals of motions :
Ha 'Lsv1
dd
dd
Lsv1dd
Ha
P secdd
19
Ha 'Lsv1
dd
dd
Lsv1dd
Ha
P secdd
Hence the corresponding two Lagrange equations reduce to two first order differential equations :20
' ' cos gsv BsvPIsv
21
' cos ' gsv Bsv cos PIsv
The third Lagrange equation is a second order differential equation 22
14
''2Had
d
2
Ha 'Lsv
dd
dd
Lsvdd
0 gm cm2 sec-2
'' ' ' sin gsv Bsv ' sin gm cm2 0 gm cm2 sec-2
In order to solve the latter equation we shall substitute into it the following expressions 23
1'P P cos
Isv sin 2gsv Bsv
24 P P cos
Isv sin 2gsv Bsv 0 sec-1
1'P P cos
Isv sin 2
P P cos
Isv sin 23.521 10 18 sec-1
obtained from eqs.(20) P P cos
Isv sin 2
P P cos
Isv sin 1.24 10 35 sec-2
25
''P P cos
Isv sin 2
P P cos
Isv sin 1.24 10 35 sec-2
Now we note the remarkable identities P P cos
sin 2.342 1095 gm cm2 sec-1
26
'
P P cos
sin sec 1dd
0 gm cm2 sec-1
P P cos
sin 22.342 1095 gm cm2 sec-1
27
'
P P cos
sin dd
0 gm cm2
P P cos
sin 22.342 1095 gm cm2 sec-1
With the aid of those identities we transforme equation (25) to any one of following two forms :
15
''P P cos
Isv sin '
P P cos
Isv sin dd
P P cos
Isv sin '
P P cos
Isv sin dd
0 sec-1
'' 0 sec-1
''P P cos
Isv sin '
P P cos
Isv sin dd
P P cos
Isv sin '
P P cos
Isv sin dd
0 sec-1
Now multiplying bots equations with ' dHa = d we find
d'2P P cos
Isv sin
2
d'2P P cos
Isv sin
2
P P cos
Isv sin
2
1.24 10 35 sec-2
'2P P cos
Isv sin
2
1.24 10 35 sec-2
'0 '
A '2P P cos
Isv sin
2
'02
P P cos
Isv sin
2
1.24 10 35 sec-2
'02
P P cos
Isv sin
2
1.24 10 35 sec-2
A21.24 10 35 sec-2
BP P cos
Isv sin
2
So, we found two other integrals of motion. In order to find (t). It is sufficient to use of themd
1P P cos
A Isv sin
2
A dt
16
d
1P P cos
A Isv sin
2
A dt
or After some algebraic operations we recognize on the left hand site an integrable function ;x=sin()32
dcos a b cos c cos 2
dt
0
'0
1.571where sin 1
a '02
P2 P
2 cos 0 2 2 P P cos 0 Isv sin 0 2
a '02
gm cm2 sec-2 P
2 P2 cos 0 2 2 P P cos 0
Isv sin 0 2
'0 '
b2 P P
Isv2
'0 '
2 P P
Isv2
2.479 10 35 sec-2
b 2 cos 0 '02 '0 gsv Bsv 2 ' '0 gsv Bsv 1 cos 2
2 cos 0 '02 '0 gsv Bsv 2 ' '0 gsv Bsv 1 cos 2 1.518 10 51 sec-2
33
c1'0
2 gm2 cm4 sec 2 P2
2 P P cos 0 P2
Isv2 sin 0 2
c1 2.479 10 35 sec-233
'02 sec 2 '0 gsv Bsv 2 '0
2 2 cos '0 '0 gsv Bsv 2.467 sec-2
b b gm cmThe solution reads 4 a c1 b2
17
cos 2 c
sin c t asin2 c cos 0 b
b
2 c
34
cos 2 c
sin c t asin2 c cos 0 b
b
2 c
where 35 4 a c1 b2 4 '0
4 '02 sin 0 2 '0
2 '0 gsv Bsv 2 sin 0 4 '02 '0 gsv Bsv 2
4 a c1 b2
Therefore cos( ) oscillates with the period T0 2
c between the two values
cos 1 and cos 2 determined by 36
cos 2 b2 c1
cos 1 b2 c1
Consequently, oscillates with the same period T0 between the corresponding values 1 and 2 : 2 T0 1 depending on the initial condition. Now we are ready to determine t( ) and t( ) . By integrating the equation (23) we find :t Ha
gsv Bsv t
0
t
tP P cos
Isv sin 2
d
gsv Bsv t
0
P P
Isv sin 2
1
td
d
d 2
37
gsv Bsv t
0
P P
A Isv sin cos
1P P cos
A Isv sin
2
P P
Isv sin 2
d
0 1.571
18
P P
A Isv sin ( )
1P P
A Isv sin
2
0
0
38 a
0 gsv Bsv t asinP P
A Isv sin cos
asinP P
A Isv sin 0 cos 0
38 b
0 gsv Bsv t asinP P
A Isv sin cos
asinP P
A Isv sin 0 cos 0
In an analogous way we obtain39 a
0 asinP P
A Isv sin cos
asinP P
A Isv sin 0 cos 0
39 b
0 asinP P
A Isv sin cos
asinP P
A Isv sin 0 cos 0
Not that only (t) depends on the magnetic field BsvImplicit assumption that sin( )=0 , there exist particular solution of Lagrange equation, which are characterised by : sin(t) for any value of t. Such solution exist for initial conditios : 0 0
or 0 , '0 0 ,P0Isv
P0Isv
'0 '0 gsv Bsv. Lagrange equation 20-22 are then equvalent
to :P0 P
P0 P
'0
' '0
P0Isv
P0Isv
'0 '0 gsv Bsv 1
P P0
41
' ' gsv BsvP0Isv
P0Isv
PIsv
PIsv
1
' 0 0
The solution of the latter equations are :
19
PIsv
gsv Bsv
t 0 0 3.142
t( ) 0
2
1.571
t( ) 1.571
PIsv
gsv Bsv
t 0 0
z0 Isv Isv t P t gsv Bsv Isv 0 Isv
Isv1.571
The fact that for those initial conditions the equations give only the dependence of (+) on t and do not give the dependence on t of each angle separately is understandable. When the z-axis of the body frame coincides with the z-axis of laboratory frame the rotation described by (t) and (t) are rotations about the same axis (z-axis) and consequently the angles (t) and (t) do not appear separately but together in a sum. Having determined the solution of Lagrange equations of motion we may now determine the time dependence of the most important quantity for our purpouse, i.e. kinetic angular momentum (2) and cannonical (spin) - (12). By virtue of the equations (23) and (24) we find that z is a constant of motion
z IsvPIsv
gsv Bsv
P gsv Bsv Isv z0
41 P gsv Bsv Isv 0 gm cm2 sec-1
IsvPIsv
gsv Bsv
0 gm cm2 sec-1
z0Further, taking into account the relation (24) and (30) and introducing the angle such that : ' A cos ' sin A sin
asinP P cos
A Isv sin
' 0
asinP P cos
A Isv sin
' 0
we can write x and y in the form x A Isv cos
y A Isv sin
Taking into account the solution t( ) given in (37) we obtain a simple dependence of -on
20
t.
t( ) 0 gsv Bsv t asinP P cos 0
A Isv sin 0
'0 0
t( ) 0 gsv Bsv t asinP P cos 0
A Isv sin 0
'0 0
Cosequently, the dependence of x and y on tis simple too. The vector
precesses around
the z-axis with the frequency L gsv Bsv forming fixed angle s with the x-axis.
s atanx
2 y2
z
A Isv
P gsv Bsv Isv
Taking into account the relation (13) between and s we find that the canonical angular momentum also precesses around time-independent magnetic field B Bksz z P
P 2.342 1095 gm cm2 sec-1
sz 2.342 1095 gm cm2 sec-1
sx x A Isv cos 0 gsv Bsv t asinP P cos 0
A Isv sin
2
1 sign 0
44
sy y A Isv cos 0 gsv Bsv t asinP P cos 0
A Isv sin
2
1 sign 0
In the case of motion described by the solution (40a ) neither nor s precess because for sin 0 and ' t( ) 0
we have:41
x 0 y 0 z P gsv Bsv Isv
sx 0 sy 0 sz P
But , at the same time the body rotates around z with the frequency gsv BsvPIsv
which is different from Larmor frequency L gsv Bsv As we are going to prove in the following section, this result reflects the fact that the potential V gsv Bsv
in the Lagrangian(8) comes from the torqe N Mtopsv Bsv which governs the motion of Isv sv
according to the well known torqe equation. In fact we shall prove that the Lagrange equations are equivalent to differential equations for (t) ,t and (t) resulting from the torqe equation, an give two other proofs of the spin precession equation.IV .The Torque equation and its
21
equivalence with Lagrange equationWe are going to demonstrate the equivalence of the torque equationMtopsv Bsv 4.123 1077 gm cm2 sec-2
gsv sv Bsv 4.123 1077 gm cm2 sec-2
' 0 sec-1
tsv
dd
N Mtopsv Bsv gsv sv
Bsv
sin wits the Lagrange equation(20,21) and (22) by substituing into the torque equation the expressions (3)for the components of the vector .sv assuming B=Bk' 0 sec-1t Ha
46
tcos ' sin sin ' gsv Bsv sin ' cos sin '
dd
Ha 5.68 1017 sec
0 sec-147
tsin sec ' cos sec sin sec ' gsv Bsv cos ' sec sin sin ' sec ' sec
dd48
t' cos ' d
dFrom eq.(48) we obtains immediately one of the integral of motion49
' cos 'zIsv
x0Isv
But , this equation is equvalent to the Lagrange equation (20) , the relation between the constants being :50 z P gsv Bsv Isv
Next , by multiplying the equation (46) and (47) by cos() and sin() respectively, and summing the resultant expression, we find the Lagrange (22).The third equvalence between the Lagrange and torque equations may be established after the following operations. First, we multiply (20) with -cos( ) and sum with (21).This given :51
' sin PIsv
PIsv
cos
'' Differentiation of the latter equation gives : 52
22
'' sin 2 cos ' 'PIsv
'
On obtains the same equation by multiplaying Eqs. (46) and (47) with cos() and -sin(), respectively and then summing the resultant expression. Hence the equvalence is proved.V. Hamilton's equations for the magnetic topFrom (16) we eassily derive Hamilton's equations for the magnetic top
'P
Hdd
PIsv
gsv Bx cos By sin
'P
Hdd
P P cos
Isv sin 2gsv
Bx sin By cos sin cos Bz
53a
'P
Hdd
P P cos
Isv sin 2gsv
Bx sin By cos
sin
P' Hd
dP
2 cos P2 cos P P 1 cos 2 gsv Bx
sin P P cos sin 2
gsv Bycos P cos P
sin 2
P2 cos P
2 cos P P 1 gm cm2 cos 2 gsv Bxsin P P cos
sin 2
gsv Bycos P cos P
sin 2
P' Hd
d
53b
P' Hd
d
By taking B along z axis, we obtain the simpler equations
'PIsv
P'P P cos P cos P
Isv sin 2
'P P cos
Isv sin 2gsv Isv Bsv
P const
P' 0
23
'P P cos
Isv sin 2
P const
P' 0
We see that Hamilton equations for ' and ' are identical with the equations (23) and (24) which were derived form the Lagrange equations (20) and (21). By combining the equations for ' and P' through
''P'Isv
we find the Lagrange equations (25). Now we shall show that Hamilton's formalisme for magnetic top leads also to the torque equation for the motion of spin for this purpose we shall use the Poisson-bracket formalism. By applying the general dynamical for any quantity u{q.a,p.a) in phase space (q,a,p,a) for the equations of motion of the angular momentum we find :
t1
dd
i H q
idd
i
Hdd
p
idd
q
idd
j
Hdd
p
1dd j
H i jdd
For the Poisson brackets of spin components we after some calculationx y z gsv Isv Bsvz
z gsv Isv Bsv 2.342 1095 gm cm2 sec-1
x z y gsv Isv Bsvy
z y x gsv Isv Bsv.x
We have also from(16)56
xHd
d
xIsv
yHd
d
yIsv
zHd
d
zIsv
By supstitution (56) and (57) into (55) we find again the torque equation (45), i.e.57
tx
dd
gsv y Bx z By
ty
dd
gsv z Bx x Bx
58
tz
dd
gsv x By y Bx
24
It is well known that it follows from (58) that 2 is a constant of motion59
t 2d
d0
Before we start to quantiye this system let us note that due to the equalities
q i
dd q
si gsv Isv Bi dd q
sidd
60
p i
dd p
si gsv Isv Bi dd p
sidd
we have the follwing important relations61 i j si sj
Taking this relation into account we find the Poisson brackest of the components of the canonical angular momentum or spin vector s.62
si sj i j k( ) sk
as wellas the dynamical equation for s58 '
tsd
dgsv s Bsv
VI. Quantum magnetic topIn order to quantze the motion, we shall aply two standard quantization procedures.1) Cannonical quantization and 2) Schrodinger quantization. The third form of quantization, the path integral formalism, will be discussed separately. 1) Canonical quantization
It is well known that in the framework of this formalism one passes from the classical to the quantum case by replacing the classical dynamical variables f(p,q) , g(p,q), etc. by operators F,G, etc.in some Hilbert space of states, in such a way that the Lie product in the space of classical functions, defined as a Poisson bracket :
f g( )q
fdd
p
gdd
p
fdd
q
gdd
is replaced by the Dirac commutator (quantum Poisson bracket)F G( )0 i h( ) 1 F G G F( ) i h( ) 1 F G( )
which now plays the role of the Lie product in the space of operators.The Dirac Lie product conserves the structure of Lie algebra of classical functions with Poisson bracket as the Lie product. The equation of motion for a dynamical variable F now reads
tFd
d1
i hF H( ) F H( )Q
where H is Hamilton operator associeted with the classical Hamiltonian H(p,q). The basic quantity of the magnetic top is cannonical angular momentum s. Taking into account the Poisson bracket (62 )of the components of s and the requirement that the quantum Poisson bracket (s.i,s.j)^0 have to
25
conserve the structure of the classical Lie algebra we may immediatly write the Dirac bracket of the components s.i of the operator of cannonical angular momentum s.
si sj i j k( ) sk
It follows strainghtforwardly that the commutators of the components of s have to be :si sj si sj sj si i h i j k( ) sk
One further step leads now to Hamilton operator of the quantum magnetic top. Inthe classical Hamiltonian (16) canonical angular momentum s has to be substituted by the operator s.
Hs2
2 Isvgsv s Bsv
gsv2 Isv Bsv
2
2
s2
2 Isvgsv s Bsv
gsv2 Isv Bsv
2
2 2.319 1077 gm cm2 sec-2
The components of the well known Pauli spin operatporx
0
1
1
0
z1
0
0
1
y0
1
10
sv2
x0
5.855 1094
5.855 1094
0
gm cm2 sec-1
sv2
y0
5.855 1094
5.855 1094
0
gm cm2 sec-1
sv2
z5.855 1094
0
0
5.855 1094
gm cm2 sec-1
satisfy the commutation relations (65) and therefore Pauli operators represent one possible representation of quantum canonical angular momentum operators. But of cource there are many other bigher dimensional representationss. In the two-dimensional spin space spanned by two eigenstates of s.zs
1
0
s0
1
the cotribution of the term s2
2 Isvgsv s Bsv
gsv2 Isv Bsv
2
2 to the eigenstates is constant
(independent of the state) and we argue that those two terms in the quantum Hamiltonian give a constant energy shift. In this way we conclude that Pauli Hamiltonian 69
HP gsv s Bsv gsvsv
2
Bsv
gsvsv
2
x Bsv0
2.062 1077
2.062 1077
0
gm cm2 sec-2
is the dynamical part of the Hamiltonan and one of the quantum representation of the magnetic
26
quantum top One shorthcoming of this representation is that it does not contain quantum analogues of ,, and p., p. and p. . But this shorthcoming may be removed by applyng the Schrodinger quantization (22).
ii^0) Schrodinger quantization
In Schrodinger quantization, with canonical momenta P.,P. and P. one associates the operators of canonical momenta P.,P.,P.
P i sv
dd
70
P i sv
dd
P i sv
dd
By substituing into (12) the canonical momenta P.,P.,P. in terms of the above operators, we find the differential representation of the s.x,s.y,s.z.
sx cos
i hdd
sin cos sin
i hd
d
sin sin
i hdd
71
sy sin
i hdd
cos cos sin
i hd
d
cos sin
i hdd
sz i hd
dIt is eqsy to see that commutators of the above differential operators sartisfy the commutation relations (65) . By squaring the operators (71) and by summing the resultant expression we obtain the differential representation of the operator s^2.
s2 sx2 sy
2 sz2 2
h2d
d
2
cot
h2dd
1
sin 2 2h2d
d
2
2h2d
d
2
2cot sin
2 dh2d
d
2
The differential representation of the Hamilton operator (66) reads :
H 12 Isv 2
h2d
d
2
cot
h2dd
1
sin 2 2h2d
d
2
2h2d
d
2
2cot sin
2 dh2d
d
2
gsv Bsv i
h2dd
gsv2 Isv Bsv
2
2
As in the case of Pauli representation, in the subspace spanned by the eigenstates of s^2 associated with the eigenvalue s*(s+1), the contribution of the first two terms to energy eigenvalues is independent of the states. Thr ramaininig term is another possible representation of the Pauli HamiltonianBsv Bsv k
27
s 1
2
HP gsv i h Bsv
dd
We want to stress here that s is quantum analogue of the canonical angular momentum s and not of the kinetic angular momentum . In the absence of the field the angular momentum coincides with the canonical angular momentum s. In the works of Bopp and Haag (9) and Dahl (13) the operators (71) and (72) have been derived starting form the free top and from the angular momentum =I.sv*.sv expressed trough the momenta P.1 and P.2 of two point particles at point with radius vectors r.1 and r.2 (with constant mutual angle u).73 Isv sv P1 P2 P2 r2
Judd (23) associetedthe same differential operators with =I.sv*.sv of the free top using the corresondence rule (70). Rosen also uses those differential operators (8). The subsequent procedure of Bopp and Haag in the presence of the field consists in the following steps : Yhey substituted the expressions (9) for P.,P.,P. valid in the presence of the field into the following relation between angular momentum components .x,.y,.z (denoted in their paper by M= (m.x,M.y,M.z)) and canonical momenta P.,P.,P.
Mx x cos Psin sin P
sin cos
sin P
My y sin Pcos sin P
cos cos
sin P
74 Mz z PBut , as it is seen from (11) this latter relation is valid in the absence of the field. In this way Bopp and Haag obtained the relation75 M 'M M' Isv sv gsv Isv Bsv
Isv sv gsv Isv Bsv 3.513 1095 gm cm2 sec-1
which they substitued into H expressed through ,,,',',' (H=I.sv*.sv^2/2)In this way they found76
HM2
2 Isv
gsv M Bsvgsv
2 Isv Bsv2
2
In the next step Bopp and Haag claim that the quantum analogue of M is the operator (71) In the above reasoning the justification of the use of the relation (74) in the presence of the field is missing. Consequently, the theoretical meaning of the relation (75) (the relation (36) in Bopp and Haag paper) is missing too. In our reasoning, which strictly follows the standard procedure for the construction of the Hamiltonian (which has to be considered as a function in phase space ,,,P.,P.,P. and not as a function of ,,,',',') we obtain the relation (11) which takes the place of Bopp and Haag relation (360. But , then we define in (12) a new quantity s and we look for the quantum analogue of this quantity. In this way we make a clear distinction between angular momentum .sv=I.sv*.sv and canonical angular momentum s and this distinction is theoretically justified in the framework of Hamiltonian formalism. Moreover,
28
the analogousdistinction betweein the kinetic momentum mv and canonical momentum p is standard in the gauge theory of point particles. On the other hand, theoretical status of Bopp and Haag quantity M,M' and'M has not been established. The quantization based on the form (66) of the Hamiltonian has one more advantage. One discovers this advantage if one tries to quantize on the basis of the Hamiltonian expressed through phase space variables ,,,P.',P.',P.' .
H 1
2 IsvP
2P
2
sin 2 P
2
2 P Pcos sin 2
gsv Bsv Pgsv
2 Bsv2 Isv
2
gsv Bsv Pgsv
2 Bsv2 Isv
2 1.237 1078 gm cm2 sec-2
2 P Pcos sin 2
cm 3 1.476i 10271 gm2 cm2 sec-2
1
2 IsvP
2P
2
sin 2 P
2
2.749 10109 gm cm2 sec-2
The direct substitution of the phase space variables by operators (70) into the above form of H leads to the operator which differs from Hamilton operator (66a) by the absence of the terms -
(h^2 /2*I.sv)*cotg.^
dd
This difference is due to the ambiguity in the ordering of ,,,P.,P.,P. in (16') It seems that the use of canonical angular momentum implicitly alleviates this ambiguity and provides the correct orderingVII The states of the quantum magnetic tops 1
2
With Pauli representation of the spin operators, the associeted quantum states are the spinors
which are linear combinations of two basic states 1
0
and
0
1
, namely the eigenstates
of .z
1
0
0
1
The two eigenstates of the Pauli Hamiltonian are very often written in terms of the polar (.B) and asimuthal (.B) angle of the vector B.
B B e
I B
2 cos2
1
0
eI B
sin2
0
1
78
B B e
i B
2 2
sin1
0
ei B
cos2
0
1
In this way of writting one stresses the fact that the eigenstates of the spin Hamiltonian in a magnetic field are the eigenstates of the component s.B of the spin operator s. As is well known, the differential operators (71) and (72) can act on larger spaces of states than the space of Pauli spinors and these spaces are richer in informations than are Pauli states.
29
The operator s^2 has the eigenvalues s*(s+1) where s takes all integer and half integer values. In the corresponding subspaces D^s thetwo-valued representations of the Rotation group are realized (9). In the case of s=2/2, which is of interest to us here, the basic states of D^1/2 are usually chosen to be the eigenstates of s.z which are the following functions of,,, (9,13)
u12
i ei
2
2
cos
2
2 2
79 a
u 1
2 i e
i2
2
sin
2
2 2
or
u12
i ei
2
2
cos
2
2 2
79 b
u12
i ei
2
2
cos
2
2 2
Therefore ,the use of differential operators (71) instead of Pauli operators (67) implies the description of spin states by probabillity amplitudes u.n and their linear combinations insread by
matrices 1
0
and
0
1
and their linear combinations.
Is there any advantage of using wave functions U.n(,,,) to describe spin states rather than
Pauli spinor
?
The first advantage is that with u.n(,,),the spin is no longer a strange and abstruse quantum-mechanical object fitted into the general quantum-mechanical framework. From this advantage follows the second one. It is telated to the understanding of the law of transformation of spin states under rotation.
The property of spinors
to change sign under 2* rotation which is a cocequence of
the law of transformation of spinor under rotation for an arbitrary angle
Rz ei
2
x
ei
2
x
e1 B
cosB2
ei B sin
B2
cosB
2
30
Rz ei
2
x
ei
2
x
e1 B
cosB2
ei B sin
B2
cosB
2
has been the subject of studies (both theoretically and experimentally), discussions and controversies (25-31). The source of controversies lies in the difficulties to physically
understandthis property. Namely, if one uses for the states 1
0
and
0
1
the usual physical
picture of the spin vector alongz-axis, one can hardly understand what is the physical reason for the phase changes by - and under 2 rotation (32,33). It seems that these difficuilties are removed if one interpretes the spin property as a modification of the interaction between a magnetic field and magnetic top such that u.n(,,) is the probability amplitude of the angles ,, under this motion. The effect ofR.z() on U.n(,,)
Rz u12
i e1
2 ( )
cos
2
2 2
81
Rz u1
2 i e
1
2 ( )
sin
2
2 2
is seen to be due to the change of the angle by - In our study of the classical magnetic top we saw that to the simple precession of spin with frequency -g.sv*B.sv corresponds a more complicated motion of the magnetic top, in which the angles and a very complicated functions of time. In the absence of precession of spin (when spin is along the z-axis) the body rotates with frequency -g.sv*B.sv+P./I.sv wich is different from Larmor frequecy .L=-g.sv*B.sv. Consequently, when the azimuthal angle (t)-(t) of spin vector changes by - (or does not change at all) the orientation angles ,,,change by 2 (or zero) ,, do not necessarlly leads to the initial orientation. We expect that those differeces in the motion of the angular momentum and of the body, in the classical case, have their counerparts in the quantized motions. They might explain the strange transformation properties of spinors under rotation. But, the full understanding requires more detailed study of the quantized motion of the magnetic top.
gsv Bsv 3.521 10 18 sec-1
gsv BsvPIsv
7.042 10 18 sec-1
PIsv
3.521 10 18 sec-1
VIII. The Asymmetric Magnetic TopIt seems worthwhile to generalize the above study to the case of an asymmetric top for which the simple relation (2) betweein the kinetic angular momentum and the angular velocity is no longer valid. Istead the following relation holds
31
82
i
Ii i eiwhere e.i are unit vectors along the body fixed frame for which the moment inertia tensor is diagonal. In addition we shall assume, insread of relation (1), the more general relation between the kinetic angular momentum and magnetic momentumM
i
gi i eii
gi Ii i ei83 Consequently , the Lagrangian of the magnetic top in a magnetic field B=Bk readsi 1 184
L
i
Isv sv2
2 Mtopsv Bsv
i
Isv sv2
2i
gsv Isv sv Bsv
i
Isv sv2
2i
sv Bsv1
where B.i are the components of B in body-fixed frame85
B
i
Bi ei B
sin sin
sin cos
cos
andBi Ii gi Bi
i
Isv sv2
2 Mtopsv Bsv 5.154 1077 gm cm2 sec-2
Bsv1 Isv gsv Bsv
i
Isv sv2
2i
gsv Isv sv Bsv 5.154 1077 gm cm2 sec-2
i
Isv sv2
2i
sv Bsv1 3.217 1089 eV
Bohr radius by coeficient
i
Isv sv2
2i
sv Bsv1
9.74 1085 gm cm sec-25.292 10 9 cm
As in the case of symmetric top, the three coordinates which determine the orientation of the top in the laboratory frame are :q1
86 q2
q3
32
Since the components of angular velocity are linear functions, eq.(4), of the time derivative of the angles, it is appropriate to write the set of relations (4) in matrix form87 C q( ) q'where
C q( )
cos sin
0
sin sin
sin cos
cos
0
0
1
88
q'
'
'
'
Using this notation we write the Lagrangian as84 a
L 1
2i
Iin
Cin q( ) q1n
2
in
Cimq'n
Bi
Consequently ,the canonical momenta are :89
Pk k'kLd
di
I Cin q( ) q'n Cik Ak
90
A B
g1 I1 sin cos g2 I2 sin sin cos
g1 I1 sin 2 sin 2 g2 I2 sin 2 cos 2 g3 I3 cos 2
g3 I3 cos
A
A
A
We shall define the quantity :.k Pk Ak
i
Iin
Cin q( ) q'n Cinin
CinTIi Cik q'n
.kni
CniT Cik q'n
or in matrix form CT C q'n
where Cij Ii Cij
and
CT
is the transposed matrix of C
C
I1 cos
I2 sin
0
I1 sin sin
I2 sin cos
I3 cos
0
0
I3
33
It follows from (91) thatq' C 1 CT 1
where
C 1
cos I1
sin I1 sin
sin cos
I1 sin
sin I2
cos I2 sin 0
cos cos
I2 sin
0
0
1
I3
CT 1
cos
sin
0
sin sin cos sin
0
sin
sin
cos
cos
sin cos
1
In order to express q'.1^ in terms of '.1^ we need the product C^-1*(C^T)^-1 wich reads :
g C 1 CT 1
cos 2
I1
sin 2
I2
cos sin
sin 1
I1
1
I2
cos sin cos
sin 1
I1
1
I2
cos sin sin
1
I1
1
I2
1
sin 2
sin 2
I1
cos 2
I2
cos
sin 2
sin 2
I1
cos I2
2
cos sin cos
sin 1
I1
1
I2
cos
sin 2
sin 2
I1
cos 2
I2
cos 2
sin 2sin 2I1
cos 2
I2
1
I3
94 with the aid of matrix elements g.ik, the relation (92) readq'k
i
gki .ii
gki Pi Ai Having expressed velocities q'.k in terms of momenta P.i we are now ready in construct the Hamiltonian of the asymmetric top starting from the general relation
H p q' L p q' A q'( )
i
I12
C q'( )i[ ]2
Using (92) the first two terms take the form
p q' A q'( ) q' C 1 CT 1 .1
2
i
1
2k
.k Cki1
j
CiT 1
.j
1
2 C 1 CT 1
It folows now that96
H 1
2 C 1 CT 1
i
Ii2
i 2 T 1
2p A( ) C 1 CT 1
p A( ) 1
2p A( ) G p A( )
H 1
2p A( ) G p A( ) 1
2gik q( ) p A( ) .i p A( ) .k
Eqs. (92a) and (96) suggest to interpret g.ik(q) as the metric tensor in the space of the kinetic
34
momenta The more explicit form of H reads ;
H I sin 2 I1cos 2
I2
P A 2 1
sin 2P A 2
cos 2
sin 2P A 2
1
2 I3P A 2
cos sin
sin I2 I1
I1 I2 p A P A
cos sin cos
sin I2 I1
I1 I2p A P A
cos
sin 2
sin 2
I1
cos 2
I2
p A P A
For the symmetric top (I.1=I.2=I.3) and for g.1=g.2=g.3=g, the above Hamiltonian reduces to the form given in (18'). Hamilton and Lagrange equations follow directly from the above expressions for Hamiltonian and Lagrangian.Canonical angular momentumLet us now express the canonical angular momentum s through the canonical momenta P.,P.,P. . For this purpose we shall substitute the relations (87) and (92a) into (82).
i
I1 1 e1i
I1k
Cik q'k e1i
I1.kk
Cik C 1 CT 1
CT 1 CT 1
P CT 1A
97
cos
sin
0
sin sin cos sin
0
sin
sin cos
cos
sin cos
0
P A
P A
P A
If we compare the latter relation with the relation (12) and (13) we conclude that in the phase space expressios of for asymmetric top the first term is the same function of phase space variables as is the function s defined in (12). So,we shall call the quantity
s CT 1p
cos Psin sin P
sin
sin cos P
sin Pcos sin P
cos
sin cos P
P
i
si ei
the canonical angular momentum, or simply spin of the asymmetric top. The components of s in the laboratory frame are identical with the ones given in (12). To the quantity .i((C^T)^-1*A)*e.i we shall give the namea
i
CT 1A .i ei
Its components in the body frame are :a1 g1 I1 sin sin
35
a2 g2 I2 sin cos
a3 g3 I3 cos
The relation (97) turns into : s aNow it is matter of simple algebra to express the Hamiltonian of the asymmetric magnetic top in terms of its spin100
H 1
2CT C 1
jik
1
2Cji j Cik
1 k 1
2jk
j k
i
Cjk Cik1 1
2jk
j kjkIj
j
j2
2 Ijj
sj aj 2
2 IjA remarkable simplification occurs if we choose the constants g.1 that we introduces in (83) to satisfy g.1^2 =g^2/I.1. Then the Lagrangian and Hamiltonian becomeL 1
2 g22 B
101 H 1
2g 2
Components of the spin vector in the body frame satisfy the folowing equations of motion :
s'j H s( )
i i
Hdd
k
pk i
dd
qk
sidd pk
sidd
qk
idd
i i
Hdd
i si
i si si ai si ai si
n
ijn sn ai sj
ai sj k
qksi
dd
pk
sjdd
s'ji
si ai Ii n
ij sn ai sj
Appendix A: Top with magnetic moment fixed in the body frameA top is fully characterized and specified by its coupling. In this paper we have defined and studied magnetic top characterized by a velocity dependent magnetic moment. In order to make more clear our argumentation that the magnetic top is the more appropriate classical model of spin we shall present here a theory of the top which carries the magnetic moment M attached to the body. That implies that the magnetic moment M is independent of the angular velocity (for if M were dependent on angular velocity it could not be constant in the body frame).Consequently, the coupling with magnetic field B is velocity independent.V M BThis potential has the smae form as the potential top (with mass M.sv and center of mass coordinate R) in the gravitational field g.V Msv R gsv
M being analogous to M.0*R playing the role of gravitationald field g. We shall deal here with the axialy symmetric top (I.1=I.2) and shall assume that M is along the body z-axis, i.e. M=M.z. Then, making B along the z-axis of the laboratory frame (B=Bk), wich does not reduce thr generality of our results,we write the ineraction potential V in the formA3
36
V M B cos The Lagrangian is differnce of kinetic and potential energy termsA4
L T VI12
12 2
2 I32
32 M cos
I12
'2 '2 sin 2 I32
' ' cos 2 M B cos
In order to construct the Hamiltonian we folow the usual procedure. Canonocal momenta are the following functions of , and
P 'Ld
dI1 '
P 'Ld
dI1 sin 2 ' I3 ' ' cos cos
P 'Ld
dI3 ' cos '
Now one express velocities ,, and ' in the terms of canonical momenta
'PI.1
'P P cos
I1 sin 2
'PI3
P P cos cos I2 sin 2
By substituting the latter expression into T andp q ' P ' P ' P
one obtain
p q 2 T 2P
2
2 I1
P2
2 I3
P2
2 I1 sin 2
P2 cos 2
2 I3 sin 2 P P
cos I1 sin 2
and consequently
H P P P ' P ' P ' P P P P V T VP
2
2 I1
P2
2 I3
P2
2 I1 sin 2
P2 cos 2
2 I3 sin 2P P
cos I1 sin 2
M B cos
Therefore , in agreement with the general theory, to the Lagrangian with velocity independent potential there corresponds a Hamiltonian which is a simple sum of kinetic and potential energy terms. The Hamiltonian (16) of the magnetic top does not have this property, again in agreement with the general theory, since the interaction term in The Lagrangian (7) is dependent on velocities ',', and 'Hamilton's equations of motionThe first three Hamilton1s equations are the eqs.(A6) .The remaining three read :
P'
Hdd
P2 cos P cos P P 1 cos 2
I1 sin 2M B sin
37
P'
Hdd
A9
P'
Hdd
Comparing the equations (A6) and (A9) with Hamilton's equations (54) of the magnetic top we find that equations for ',P'. and P'. in both sets are the same whereas the equations for ',', and P'. are different. Since the Hamiltonian of the top with velocity independent magnetic moment is identical to the gravitational top, the corresponding Hamilton's equation are to be found in literature (24). Here we shall review the well known qualitative analysis (19) Two immediate first integrals of motion are :A10aP I3 ' ' cos I3 3 P0
A10b
P I1 sin 2 ' I3 ' I3 ' ' cos P0
Since the system is conservative the total energy is the third integral o f motion.A10c
E T VI12
'2 '2 sin 2 I32
P0
I32
M B cos
Only three additional quadratures are needed to solve the problrm. From the above three integrals it is possible to express ',' and ' as functions of ' and constant of motionA11
'P0 P0 cos
I1 sin 2
A12
'P0I3
cos P0 P0 cos
I1 sin 2
A13 I12
'2P0 P0 cos
2I1 sin 2
M B cos EP0
2
2 I3
E'
The equation (A13) differs from the corresponding equation (31) of the magnetic top by presence of the term M*B*cos(). As we are going to see, due to the presence of this term the equation (A13) leads to an eltptic integral (with cubic polynomial under the integral sign) On the other hand the equation (30) leads to the equation (32) with square polynomial and therefore is integrable. From (A13) it follows : A14
' I1 sin sin 22 I1 E' M B cos P0 P0 cos 2
1
2
A15
38
T
COS 0( )( )
COS ( ) 1( )
xdcos .
2 I1 1 cos 2 E' M B cos P0 P0 cos 2
1
2
d
Since the solution of the equation (A15) cannot be writwn in an analytic form, the sme is valid for the solution of equation (A11) and (A12). But, the qualitative featrures of the solution (t) of the equation *A13) are known (19). They are pictured on Fig.3 in which the possible shapes for the locus of the body axis on the unit sphere are indicated. Recalling that M was assumed to be along e.x. this figure presents also the motion of the magnetic moment M fixed with the body. Therefore, M follows the motion of the body, quite different from M and of the magnetic top wich move with respect to the body. So, M performs a complicated motion (precession with nutation) whereas and M simpli precess.AcknowlegmentsOne of the authors (M.B) would like to thank Professor Abdus Salam, the International Centre for Theoretical Physics, Trieste.
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