Research
Gravitation, Mathematical Physics and Field Theory
Deviation of inverse square law based on Dunkl derivative: deformed
Coulomb’s law
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, v∈R
(3)
where Ris the reflection operator obeying
Rx^=-x^R,
(4)
and we set ℏ=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 (i≠j).
(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
D⋅E=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
D⋅E=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
∫(D⋅E)μ(r)d3r=4πq,
(19)
where μ(r)is the weight function. The Eq. (19) can be written as
∫∇∙Eμrd3r+v∫P∙Eμrd3r=4πq
(20)
or
∫∇∙Eμrd3r-∫E∙μ´rrrd3r+v∫P∙Eμrd3r=4πq
(21)
Inserting the Eq. (16) into the Eq. (21) we get
4πqr-6vμr+q∫6vμ(r)r2-μ´(r)r2d3r=4πq,
(22)
which gives
μ(r)=r6ν.
(23)
Thus, the deformed Gauss law reads
r6νD⋅E=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
D⋅∇V=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,rj⋅E(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=0∞Cn(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+6vr∑n=0∞Cn12+3v(cosθ)arn
(60)
for r>a, while we get
V=q1+6va∑n=0∞Cn12+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 r≫a, we have
V≈p 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 r≫a, we have
V≈Qr332+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+6v∫y=0Lλdyx-y1+6v=λ6v1+6vx-L-6v-x-6v
(79)
For a small ν, we get
VPx≈λ Inxx-L+3λv Inx-Lx2+Inxx-L
(80)
For x≫Lwe 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
EPx≈λLxx-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
EP≈Qx2+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+6ν∫y=-L/2L/2λdy|r-r'|1+6ν,
(86)
where r=xi, r'=yj, (|y|≤L/2). Then we have
VP=2λ1+6v∫y=0L/2dyx2+y212+3v=λL1+6vx1+6v2F112+3v,12;32;L24x2
(87)
For x≫Lwe 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
EP≈Qx2+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+6v∫0Rrdrz2+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
V≈2πσz2+R2-z+2πvσ6z In z-3z2+R2 Inz2+R2
(93)
and
EPz≈2πσ1-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
VcR1≈6vR2R2-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 ν≠0. 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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