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arXiv:physics/0004008v1 [physics.class-ph] 3 Apr 2000

Dirac monopole with Feynman brackets

Alain. B´ erard

LPLI-Institut de Physique, 1 blvd D.Arago, F-57070 Metz, France

Y. Grandati

LPLI-Institut de Physique, 1 blvd D.Arago, F-57070 Metz, France

Herv´ e Mohrbach

M.I.T, Center for Theoretical Physics, 77 Massachusetts Avenue,

Cambridge, MA 02139-4307 USA

and

LPLI-Institut de Physique, 1 blvd D.Arago, F-57070 Metz, France

Abstract

We introduce the magnetic angular momentum as a consequence of the struc-

ture of the sO(3) Lie algebra defined by the Feynman brackets. The Poincar´ e

momentum and Dirac magnetic monopole appears as a direct result of this

framework.

I. INTRODUCTION

In 1990, Dyson [1] published a proof due to Feynman of the Maxwell equations, assuming

only commutation relations between position and velocity. In this article we don’t use the

commutation relations explicitly. In fact what we call a commutation law is a structure of

algebra between position and velocity called in this letter Feynman’s brackets. With this

minimal assumption Feynman never supposed the existence of an Hamiltonian or Lagrangian

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formalism and didn’t need the not gauge invariant momentum. Tanimura [2] extended

Feynman’s derivation to the case of the relativistic particle.

In this letter one concentrates only on the following point: the study of a nonrelativistic

particle using Feynman brackets. We show that Poincare’s magnetic angular momentum is

the consequence of the structure of the sO(3) Lie algebra defined by Feynman’s brackets.

II. FEYNMAN BRACKETS

Assume a particle of mass m moving in a three dimensional Euclidean space with position:

xi(t) (i = 1,2,3) depending on time. As Feynman we consider a non associative internal

structure (Feynman brackets) between the position and the velocity. The starting point is

the bracket between the various components of the coordinate:

[xi,xj] = 0(1)

We suppose that the brackets have the same properties than in Tanimura’s article [2], that

is:

[A,B] = −[A,B] (2)

[A,BC] = [A,B]C + [A,C]B

(3)

d

dt[A,B] = [

.

A,B] + [A,

.

B] (4)

where the arguments A, B and C are the positions or the velocities.

The following Jacobi identity between positions is also trivially satisfied:

[xi,[xj,xk]] + [xj,[xk,xi]] + [xk,[xi,xj]] = 0(5)

In addition we will need also a “Jacobi identity” mixing position and velocity such that:

[

.xi,[

.xj,xk]] + [

.xj,[xk,

.xi]] + [xk,[

.xi,

.xj]] = 0 (6)

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Deriving (1) gives:

[

.xi,xj] + [xi,

.xj] = 0 (7)

This implies:

[xi,

.xj] = gij(xk),

(8)

where gij(xk) is a symmetric tensor. We consider here only the case where:

gij=δij

m

(9)

this gives the following relations:

[xi,f(xj)] = 0(10)

[xi,f(xj,

.xj)] =1

m

∂f(

∂

.xj)

.xi

(11)

[

.xi,f(xj)] = −1

m

∂f(xj)

∂xi

(12)

III. ANGULAR MOMENTUM

Suppose first the following relation:

[

.xi,

.xj] = 0(13)

which permits to say that the force law is velocity independent:

..xi=

..xi(xj)(14)

By definition the orbital angular momentum is:

Li= mεijkxj

.xk

(15)

which satisfies the standard sO(3) Lie algebra for Feynman’s brackets:

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[Li,Lj] = εijkLk

(16)

The transformation law of the position and velocity under this symmetry is:

[xi,Lj] = εijkxk

(17)

[

.xi,Lj] = εijk

.xk

(18)

We consider as Feynman [1], the case with a ”gauge curvature”:

[

.xi,

.xj] =

α

m2Fij

(19)

where F must be an antisymmetric tensor (electromagnetic tensor for our example) and α

a constant. The goal of our work is to see what happens if we keep the structure of the Lie

algebra of the angular momentum and the transformation law of the position and velocity.

Using (6) we get the relations:

α∂Fjk

∂

.xi

= −m2[xi,[

.xj

.

,xk]](20)

= −m2[

.xj,[xi,

.xk]] + [

.xk,[

.xj,xi]] = 0

then the electromagnetic tensor is independent of the velocity:

Fjk= Fjk(xi)(21)

By deriving (8) we have:

[xi,

..xj] = −[

.xi,

.xj] = −αFij

m2

(22)

then:

m∂

..xj

.xi

∂

= αFji(xk)(23)

or:

m

..xi= α(Ei(xk) + Fij(xk)

.xj) (24)

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We get the ” Lorentz force’s law”, where the electric field appears as a constant of integration

(this is not the case for the relativistic problem, see [2]). Now the force law is velocity

dependent:

..xi=

..xi(xj,

.xj) (25)

For the case (19), the equations (16), (17)and (18) become :

[xi,Lj] = εijkxk

(26)

[

.xi,Lj] = εijk

.xk+αεjklxkFil

m

(27)

[Li,Lj] = εijkLk+ αεiklεjmsxkxmFls

(28)

Introducing the magnetic field we write F in the following form:

Fij= εijkBk,

(29)

We get then the new relations:

[

.xi,Lj] = εijk

.xk+α

m{xiB − δij(

→r .

→

B)}(30)

[Li,Lj] = εijk{Lk+ αxk(

→r .

→

B)} (31)

To keep the standard relations we introduce a generalized angular momentum:

L?= L?+ M?

(32)

We call Mi the magnetic angular momentum because it depends on the field

→

B. It has

no connection with the spin of the particle, which can be introduced by looking at the

spinorial representations of the sO(3) algebra. Now we impose for the {αj}’s the following

commutation relations:

[

.xi,L|] = ε?|?§?

(33)

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