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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