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Revista mexicana de física

versión impresa ISSN 0035-001X

Rev. mex. fis. vol.66 no.4 México jul./ago. 2020  Epub 31-Ene-2022

https://doi.org/10.31349/revmexfis.66.411 

Research

Gravitation, Mathematical Physics and Field Theory

Deviation of inverse square law based on Dunkl derivative: deformed Coulomb’s law

Y. Kima 

W. Sang Chunga, † 

H. Hassanabadib, ‡ 

aDepartment of Physics and Research Institute of Natural Science, College of Natural Science, Gyeongsang National University, Jinju 660-701, Korea. e-mail: †ymimip44@naver.com

bFaculty of physics, Shahrood University of Technology, Shahrood, Iran ‡h.hasanabadi@shahroodut.ac.ir


Abstract

In this paper, we consider the Coulomb’s law with deviation. We use the Dunkl derivative to derive the deformed Gauss law for the electric field and the electrostatic potential, which gives a new deformed electrostatics called a Dunkl-deformed electrostatics. We modify the Dunkl derivative for the electric field for multi-sources or continuous charge distribution. We discuss some examples of the Dunkl-deformed electrostatics.

Keywords: Coulomb’s law; Dunkl derivative; deformed Gauss law

PACS: 03.65.Ge; 03.65.Fd; 02.30.Gp

1.Introduction

The inverse square law arises in the Coulomb’s law. This states that the electric field r due to the charge qlocated at origin is given by

E(r)=q|r|2r^=q|r|3\r. (1)

Recently, some measurements were accomplished for the accuracy of inverse square law for Coulomb’s law by considering the small deviation of the inverse square law of the form

E(r)=q|r|2+σr^=q|r|3+σr, (2)

where σcan be regarded as a deviation from the inverse square law [1-3]. Plimpton and Lawton [1] charged an outer sphere with a slowly varying alternating current and detected the potential difference between the inner and outer spheres. They found σ = 2x 10-9. In 1970, Bartlett, Goldhagen, and Phillips [2] also achieved an upper limit of σ=1.3×10-13. One year later, Williams and Faller [3] estimated σ=(2.7±3.1)×10-16. From these experimental data for σ, we know that the deviation from the inverse square law is sufficiently small but can be considered to be non-zero.

As a theoretical background of the Eq. (2), we will invoke Wigner deformation. In 1950, Wigner [4] proposed a new deformed Heisenberg algebra of the form

x^,p^=i1+2vR,   vR (3)

where Ris the reflection operator obeying

Rx^=-x^R, (4)

and we set =1. It is well known that algebra (3) gives the same equation of motion for the harmonic potential, which implies that algebra is the general second quantization scheme for the classical harmonic potential problem. The coordinate representation for the Eq. (3) is then given by

p^=1iDx,   x^=x, (5)

where Dunkl derivative (DD) [5] Dxis given by

Dx=x+νx(1-Rx) (6)

and the reflection operator obeys

Rxf(x)=f(-x). (7)

It is well known that Wigner algebra (3) is linked to the two-particle Calogero model [6-8] when the Wigner parameter is related to the Calogero coupling constant. Wigner Hamiltonian, which comes from Wigner algebra, has the potential which consists of a simple harmonic potential and an additional inverse square potential. This potential was shown to possess the conformal symmetry and has been the subject of much interest [9-14]. Especially, Macfarlane [14] found the relation between Wigner algebra and para-statistics.

Here we have a fundamental question. If Coulomb’s law does not obey the inverse square law, what will happen? Is it possible to construct the electrostatics in such case? The answer is YES, which is the main purpose of this work. In this paper, we use DD to derive the deviation of the inverse square law. We apply the deviation of the inverse square law to the electrostatics. This paper is organized as follows. In Sec. 2, we discuss the deformed electrostatics with DD. In Sec. 3, we discuss the modification of DD and deformed electrostatics when the point charge is not located at origin. In Sec. 4, we discuss the deformed electrostatics with a multi-source and continuous source. In Sec. 5, we discuss some examples of electrostatics with discrete charges. In Sec. 6, we discuss some examples of Dunkl-deformed electrostatics with continuous charge distribution.

2.Deformed electrostatics with DD

In this section, we consider the electrostatics with DD. To do so, we introduce the vector DD’s as

D=+νP (8)

where

P=e^11x11-R1+e^21x21-R2+e^31x31-R3 (9)

and

Rixj=-xjRi     (i=j)xjRi     (ij). (10)

We consider the electromagnetic theory in which the vector differential operator is replaced with the vector DD operator D. Outside the point charge qlocated at the origin, the electric field is assumed to obey the deformed Gauss law outside source

DE=0. (11)

Now let us solve the Eq. (11) by setting

E=f(r)r, (12)

where

r=|r|=x12+x22+x32. (13)

Inserting the Eq. (12) into the Eq. (11) we get

DE=rf'(r)+(3+6ν)f(r)=0. (14)

Solving the Eq. (14) we get

f(r)=qr3+6ν. (15)

Thus, the deformed electric field reads

E=qr3+6νr, (16)

which reveals the law of inverse square with deviation. One can easily check that the deformed electric field obeys

D×E=0, (17)

which holds because P×E=0. Thus, the electrostatic potential is defined through

E=-DV(r)=-V(r). (18)

Now let us derive the deformed Gauss law. We assume that the Gauss law for a point charge located at the origin takes the form

(DE)μ(r)d3r=4πq, (19)

where μ(r)is the weight function. The Eq. (19) can be written as

Eμrd3r+vPEμrd3r=4πq (20)

or

Eμrd3r-Eμ´rrrd3r+vPEμrd3r=4πq (21)

Inserting the Eq. (16) into the Eq. (21) we get

4πqr-6vμr+q6vμ(r)r2-μ´(r)r2d3r=4πq, (22)

which gives

μ(r)=r6ν. (23)

Thus, the deformed Gauss law reads

r6νDE=4πρ(r), (24)

where ρdenotes the charge density, and for a point charge it is

ρ(r)=qδ(r). (25)

If we locate a charge Qon r, it will be subject to the force of the form

F=QE=qQ|r|3+6νr. (26)

Now let us find the potential energy for the deformed Coulomb force (26). The Eq. (26) indicates that the deformed Coulomb force is also conservative. Thus, the potential energy is defined through

F=-U(r). (27)

Solving the Eq. (27) we get

U(r)=qQ(1+6ν)|r|1+6ν. (28)

Similarly, the electrostatic potential Vis defined through

E=-V(r), (29)

which gives

V(r)=q(1+6ν)|r|1+6ν. (30)

Thus, the electrostatic potential obeys the Dunkl-Laplace equation

DV=0.   (31)

3.Modification of DD and deformed electrostatics when the point charge is not located at the origin

In ordinary electro-magnetics, consider the Coulomb force. Let F denote the force acting on a electrically charged particle, with charge qlocated at r, due to the presence of a charge q'located at r'. According to Coulomb’s law, this force is, in a vacuum, given by the expression

F=qq´(r-r´)r-r´3=-qq´1r-r´=qq´1r-r´ (32)

The vector DD defined in the Eq. (8) does not obey the above property. Thus, we need a new definition for vector DD when we consider two points, source point, and field point, and this derivative is acted on the function in r - r'. Here we define the DD gradient of the function f(|r-r'|)as

DDgradf(|r-r'|)=Dr,r'f(|r-r'|), (33)

where

Dr,r'=+νPr,r' (34)

and

(Pr,r')i=1xi-xi'(1-Rxi,xi'), (35)

and the exchange operator obeys

Rxix'i(xi-xi')=(xi'-xi)Rxixi'. (36)

Similarly, we define the DD divergence, and DD curl acted on the vector function of the form f(|r-r'|)(r-r')as

DDdivfr-r´r-r´=Dr,r´fr-r´r-r´ (37)

and

DDcurlfr-r´r-r´=Dr,r´×fr-r´r-r´ (38)

The electric field at $\ra$ due to charge q'located at r' is then given by

E(r-r')=q'|r-r'|3+6ν(r-r'), (39)

which satisfies

|r-r'|6νDr,r'E(r-r')=4πq'δ(r-r') (40)

Dr,r'×E(r-r')=0. (41)

The deformed electrostatic potential is given by

E(r-r')=-V(|r-r'|), (42)

which gives

Vr-r´=q´1+6v1r-r´1+6v (43)

4.Deformed electrostatics with multi source and continuous source

Now let us consider the multi source case. Consider the situation that there are Ncharges q1,q2,,qN, each of which is located at r1,r2,,rN. The electric field at r due to Ncharges is defined as the sum of Nsub electric fields as follows:

E(r)=j=1NE(r-rj), (44)

where the sub electric field due to charge qjis

E(r-rj)=qj|r-rj|3+6ν(r-rj). (45)

The sub electric field obeys

|r-rj|6νDr,rjE(r-rj)=4πqjδ(r-rj), (46)

and

Dr,rj×E(r-rj)=0. (47)

The electrostatic potential is defined through

Etot(r)=-V(r), (48)

which gives

V(r)=j=1Nqj(1+6ν)|r-rj|1+6ν. (49)

If the discrete electric charges are small and numerous enough, we introduce the electric charge density ρlocated at r' within a volume V'of a limited extent and replace summation with integration over this volume. This allows us to describe the total electric field as integral of the infinitesimal electric field as follows:

E(r)=V'd3\ra'E(r-r'), (50)

where the infinitesimal electric field due to the infinitesimal charge located at r' is

E(r-r')=ρ(r')r-r'|r-r'|3+6ν. (51)

The infinitesimal electric filed obeys

|r-r'|6νDr,r'E(r-r')=4πρ(r')δ(r-r'), (52)

and

Dr,r'×E(r-r')=0. (53)

The electrostatic potential is defined through

E(r)=-V(r), (54)

which gives

Vr=Vp(r´)1+6v1r-r´1+6v (55)

5.Some examples of Dunkl-deformed electrostatics with discrete charges

Now let us discuss some examples of Dunkl-deformed electrostatics with discrete charges.

5.1.Monopole

Let us consider the case that a charge qis located at ak, where k denotes the unit vector in zdirection. Then the electrostatic potential is

V=q1+6v1r-ak1+6v (56)

and the electric field is

E=-V=q(r-ak)|r-ak|3+6ν. (57)

If we express the Eq. (56) in terms of the spherical coordinates, we get

V=q(r2+a2-2ar cosθ)1/2+3ν. (58)

Let us consider the Gegenbauer polynomial defined through the following generating function

1(1-2xt+t2)α=n=0Cn(α)(x)tn. (59)

The generating function for Gegenbauer polynomial is a generalization of the generating function for Legendre polynomials. Indeed, when a =1/2, the Gegenbauer polynomials become Legendre polynomials. Thus, with the help of generating function for Gegenbauer polynomials, the electrostatic potential reads

V=q1+6vrn=0Cn12+3v(cosθ)arn (60)

for r>a, while we get

V=q1+6van=0Cn12+3v(cosθ)ran (61)

for r<a. We know that the Gegenbauer polynomial satisfies the recurrence relation

C0ax=1 (62)

C1ax=2ax (63)

Cnax=1n2xn+a-1Cn-1ax-(n+2a-2)Cn-2a(x) (64)

5.2.Dipole

Let us consider the case that a charge qis located at ak, and another charge -qis located at ak. Then the electrostatic potential is composed of the electrostatic potential due to q, Vq, and the electrostatic potential due to -q, V-q,

V=q1+6v1r-ak1+6v-q1+6v1r+ak1+6v (65)

Thus, the electric field is

E=-V=q(r-ak)|r-ak|3+6ν-q(r+ak)|r+ak|3+6ν. (66)

If we express the Eq. (65) in terms of the spherical coordinates we get

V=q1+6v1r2+a2-2arcosθ12+3v-1r2+a2-2arcosθ12+3v (67)

The electric field is then given by

E=Err^+Eθθ^, (68)

where

Er=qr-acosθr2+a2+2arcosθ32+3v-r-acosθr2+a2+2arcosθ32+3v (69)

Eθ=qasinθ1r2+a2+2arcosθ32+3v+1r2+a2+2arcosθ32+3v (70)

For ra, we have

Vp cosθr, (71)

where the electric dipole moment is p=2qa. Thus we have the same form as ordinary electro-magnetics.

5.2.Quadrupole

Let us consider the case that a charge qis located at ak, another charge qis located at - ak, and the third charge -2qat origin. Then the electrostatic potential is

V=-2q1+6v1r1+6v+q1+6v1r-ak1+6v+q1+6v1r+ak1+6v (72)

Thus, the electric field is

E=-V=qr-akr-ak3+6v+qr+akr+ak3+6v-2qrr3+6v (73)

If we express the Eq. (72) in terms of the spherical coordinates we get

V=q1+6v1r2+a2-2 arcosθ12+3v+1r2+a2-2 arcosθ12+3v-2r1+6v (74)

The electric field is then given by

E=Err^+Eθθ^, (75)

where

Er=q2r2+6v-r-acosθr2+a2-2arcosθ32+3v-r-acosθr2+a2-2arcosθ32+3v (76)

Eθ=qasinθ1r2+a2-2arcosθ32+3v+1r2+a2+2arcosθ32+3v (77)

For ra, we have

VQr332+3vcos2θ-12 (78)

where the electric quadrupole moment is Q=2qa2. Thus we know that the $\ta$-dependency of the potential is deformed.

6.Some examples of Dunkl-deformed electrostatics with continuous charge distribution

Let us discuss some examples of Dunkl-deformed electrostatics with continuous charge distribution.

6.1.Electric field on the axis of a finite line charge

A charge Qis uniformly distributed along the xaxis from x=0to x=L>0. The linear charge density for this charge is λ=Q/L. We wish to find the electric field produced by this line charge at some field point P(x)with x>Lon the x-axis. The electrostatic potential is then given by

VPx=11+6vy=0Lλdyx-y1+6v=λ6v1+6vx-L-6v-x-6v (79)

For a small ν, we get

VPxλ Inxx-L+3λv Inx-Lx2+Inxx-L (80)

For xLwe have

VP(x)λL(1+6ν)x1+6ν. (81)

The electric field is

EPx=-ddxVx=λ1-6v1x-L1+6v-1x1+6v (82)

For a small ν, we get

EPxλLxx-L-6vx(x-L)×L+L In x+x Inx-Lx (83)

We can see that if xis much larger than L, the electric field at Pis approximately

EPQx2+6ν. (84)

That is, if we are sufficiently far away from the line charge, it approaches that of a point charge Qat the origin.

6.2.Electric field on the perpendicular bisector of a uniformly charged line segment

A charge Qis uniformly distributed on a straight-line segment of length Lon the y -axis. The charge density is given by

λ(y)=λ      (|y|L/2)0       (|y|>L/2). (85)

Now let us find the electrostatic potential and electric field at point Pon the x -axis, P(x,0), where we consider the case of x>0. The electrostatic potential is

VP=11+6νy=-L/2L/2λdy|r-r'|1+6ν, (86)

where r=xi, r'=yj, (|y|L/2). Then we have

VP=2λ1+6vy=0L/2dyx2+y212+3v=λL1+6vx1+6v2F112+3v,12;32;L24x2 (87)

For xLwe have

VP(x)λL(1+6ν)x1+6ν. (88)

The electric field is

EPx=-ddxVx=λLx2+6v2F132+3v,12;32;L24x2 (89)

We can see that if xis much larger than L, the electric field at Pis approximately

EPQx2+6ν. (90)

6.3.Electric field on the axis of a uniformly charged disk

Consider a uniformly charged disk of radius Rcentered at the origin and total charge Q, which lies on xy-plane. We can calculate the electrostatic potential and electric field at the point P(0,0,z)on the z-axis with z>0. In this case, the surface charge density (the charge per unit area) is given by σ=Q/(πR2). Then the electrostatic potential is

VP=2πσ1+6v0Rrdrz2+r212+3v=2πσ1-36v2z2+R212-3v-z1-6v (91)

and the electric field is

Epz=2πσ1-6vz-6v-zz2+R2-12-3v (92)

For a small νwe have

V2πσz2+R2-z+2πvσ6z In z-3z2+R2 Inz2+R2 (93)

and

EPz2πσ1-zz2+R2+2πvσ-6+6zz2+R2-6 In z+3zz2+R2 In R2+z2 (94)

7.Conclusion

According to Cavendish type experiments [15], let us consider that the radii of the two concentric spheres are R1and R2, (R1<R2), with the charges Q1and Q2spread uniformly over them, respectively. Then, the potential at r [16] is given by

Vr=Q12R1rf(r+R1-fr-R1+Q22R2rfr+R2-f(r-R2 (95)

where ris the distance from the origin ( center of concentric spheres). According to Ref. [17], we get the potential on the inner shell and outer shell as

VR1=Q12R12f2R1+Q22R1R2fR1+R2-fR2-R1 (96)

VR2=Q22R22f2R1+Q12R1R2fR1+R2-fR2-R1 (97)

where

f(r)=0rsU(s)ds (98)

and U(r)is the electrostatic potential for a unit charge. For the ordinary electrostatics, we have U=1/r, which gives f=r. In the deformed electrostatics with DD, from the Eq. (30), we get

U=11+6ν1r1+6ν, (99)

For small ν, we have

f(r)r(1-6νlnr). (100)

After the outer shell was charged by a potential V0, a part of the charge would pass through the connecting wire into the inner shell until the equilibrium V(R1)=V(R2)=V0was reached. Then, the charges on the inner shell could be determined as

VcR16vR2R2-R1V0MR1,R2 (101)

where

M(R1,R2)=12R2R1lnR2+R1R2-R1-ln4R22R22-R12. (102)

In the ordinary electrostatics ( ν=0), we have Vc(R1)=0, while Vc(R1)0for ν0. So, by detecting the electrostatic potential on the inner shell directly, one could obtain the deviation from Coulomb’s inverse square law, and then the ordinary derivative in the Maxwell equation could be replaced with DD.

There remains much to be studied in this direction. We should investigate the Maxwell equation with DD and derive the electromagnetic wave equation for the completion of this deformed electro-magnetics. The study of magnetic monopole with deviation seems to be possible with the help of DD. We think that these topics and their related content will be clear soon.

Acknowledgement

The authors thank the referee for a careful reading of the manuscript and constructive remarks.

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Received: March 18, 2020; Accepted: March 19, 2020

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