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INSTITUTE OF PHYSICS PUBLISHING
JOURNAL OF PHYSICS A: MATHEMATICAL AND GENERAL
J. Phys. A: Math. Gen. 34 (2001) 6143–6156
PII: S0305-4470(01)16876-9
The effect of radial acceleration on the electric and
magnetic fields of circular currents and rotating
charges
Oleg D Jefimenko
Physics Department, West Virginia University, PO Box 6315, Morgantown, WV 26506-6315,
USA
E-mail: [email protected]
Received 5 September 2000, in final form 12 April 2001
Published 27 July 2001
Online at stacks.iop.org/JPhysA/34/6143
Abstract
It is shown that time-independent circular currents and uniformly rotating
charge distributions create heretofore unreported constant electric and magnetic
fields associated with radial acceleration of the charges forming the circular
currents and with radial acceleration of the charges comprising the rotating
charge distributions. These fields are computed for several types of rotating
charge distributions and for several types of circular currents. One of the
consequences of the existence of these fields is that the Aharonov–Bohm effect
can now be explained on the basis of classical electrodynamics.
PACS numbers: 41.20.-q, 03.50.De
1. Introduction
It is well known that the electric field of an electric charge in the state of uniform translational
motion is different from the electric field of the same charge at rest. Therefore it is reasonable
to expect that the electric field of a charge in the state of uniform rotational motion should
also be different from the electric field of the same stationary charge. The theoretical analysis
presented in this paper shows that uniformly rotating charge distributions do indeed produce
constant electric fields different from those of the same stationary charge distributions. This
analysis is based on the ‘causal’ solutions of Maxwell’s equations representing the electric
field E and the magnetic flux density field B in a vacuum in terms of their causative sources—
electric charges (charge density ρ) and electric currents (current density J ) [1–6, 7 pp 514–6]:
[ρ]
1 ∂J
1 ∂ρ
1
1
+
r dV −
dV
(1)
E=
r 3 r 2 c ∂t
4π ε0 c2
r ∂t
4πε0
0305-4470/01/316143+14$30.00
© 2001 IOP Publishing Ltd
Printed in the UK
6143
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O D Jefimenko
Figure 1. A ring carrying a uniformly distributed electric charge of density ρ rotates with constant
linear velocity u about its symmetry axis (y axis). The point of observation P (x, y) is in the xy
plane.
and
µ0
B=
4π
[J ]
µ0
× r dV +
r3
4π c
1
r2
∂J
× r dV .
∂t
(2)
In these integrals the brackets are the ‘retardation symbol’, indicating that the quantities
between the brackets are to be evaluated for the ‘retarded’ time t = t − r/c, where t is
the time for which E and B are computed, r is the distance from the source point (volume
element dV ) to the field point (the point for which E and B are computed), c is the velocity of
light, ε0 is the permittivity of space and µ0 is the permeability of space. Equations (1) and (2)
are of a relatively recent origin and provide a new means for analysing and discussing electric
and magnetic phenomena.
Particularly significant for the calculations that follow is the last integral in equation (1).
It represents the so-called ‘electrokinetic field’ Ek [8], so named because it arises from the
motion of electric charges constituting the current density J . Although this integral makes
an important contribution to the electric field of uniformly moving charges [9, 10], its main
function is to represent the effect of charge acceleration on the electric field of the charge
(or current) under consideration. In the calculations that follow, we shall use this integral for
determining the electric field of electric charge distributions in the state of circular and rotational
motion. This type of motion is only possible if the individual charges comprising the charge
distribution under consideration experience a radial acceleration. Therefore, even when the
speed of the charges is constant, the derivative ∂ J /∂t does not vanish, and an electrokinetic
field Ek is inevitably created.
The derivative ∂ J /∂t appearing in equation (2) indicates that the acceleration-related
electrokinetic fields of rotating charges and circular currents are accompanied by accelerationrelated magnetic fields. These fields will also be considered in this paper.
Electric and magnetic fields of circular currents and rotating charges
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2. Theory I. Electric field
Consider a charged ring of radius b and cross-sectional area S carrying a uniformly distributed
charge of density ρ and rotating about its symmetry axis (y axis) with constant linear velocity u
(figure 1). Let us find the electric field produced by this ring at a point P (x, y) at a distance
R from the centre of the ring.
According to equation (1), there are two different contributions to the field at P . The first
contribution comes from the first integral in equation (1). Since the charge density ρ of the
ring does not depend on time, the derivative ∂ρ/∂t vanishes. Furthermore, since the charge
density is the same at all times at all points of the ring, the retardation may be assumed to have
no effect, so that the first integral in equation (1) represents the ordinary Coulomb field of the
ring:
ρ
1
EC =
r dV .
(3)
4πε0
r3
The second contribution to the electric field is from the last integral in equation (1), which
represents the electrokinetic field Ek . Also in this integral the retardation may be assumed to
have no effect because at any given point of the ring the derivative ∂ J /∂t is the same at all
times, so that ∂ J /∂t at t = t − r/c is the same as ∂ J /∂t at t. Thus the electrokinetic field
created by the ring is
1 ∂J
1
Ek = −
dV .
(4)
4πε0 c2
r ∂t
Only Ek depends on the acceleration of the charges, and therefore in the calculations that
follow we shall be primarily interested in Ek .
Writing the three position vectors shown in figure 1 as
R = xi + yj
(5)
b = −b cos ϕ i − b sin ϕ k
(6)
r =b+R
(7)
and
we have
r = −b cos ϕ i − b sin ϕ k + x i + y j
= (x − b cos ϕ)i − b sin ϕ k + y j
(8)
where i, j and k are unit vectors in the direction of the x, y and z axis, respectively. The
magnitude of r is
r = (r · r )1/2 = [(x − b cos ϕ)2 + b2 sin2 ϕ + y 2 ]1/2
= (x 2 − 2bx cos ϕ + b2 + y 2 )1/2 .
The derivative ∂ J /∂t for the shaded element of the ring shown in figure 1 is
∂u
∂ρ u
∂J
=ρ
= ρa
=
∂t
∂t
∂t
where a is the radial (centripetal) acceleration of the element1 . Since
u2
u2
a = 2 b = 2 (−b cos ϕ i − b sin ϕ k)
b
b
(9)
(10)
(11)
The reason that, for the charge element under consideration, ∂ u/∂t = a is as follows. The partial derivative ∂ u/∂t
is the rate of change of u at a fixed point of the trajectory of the charge element. At the moment when the charge
element arrives at a particular point of its trajectory, its velocity vector is in one direction. When, after a time interval
t, the charge element leaves this point, the velocity vector is in a slightly different direction. The change in the
velocity vector is u. The partial derivative is the limit of the ratio u/t for t → 0 (very short charge element).
The acceleration a is the limit of the same ratio.
1
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O D Jefimenko
we have
∂J
u2
= −ρ (cos ϕ i + sin ϕ k).
∂t
b
(12)
Substituting equations (9) and (12) into (4), we find the contribution of the shaded element
shown in figure 1 to the electrokinetic field Ek observed at the point P (x, y):
d Ek =
ρu2 (cos ϕ i + sin ϕ k)
dV
4πε0 c2 b(x 2 − 2bx cos ϕ + b2 + y 2 )1/2
(13)
where dV is the volume of the shaded element. The electrokinetic field produced by the entire
ring is therefore
(cos ϕ i + sin ϕ k)
ρu2
Ek =
dV
(14)
2
2
4πε0 bc
(x − 2bx cos ϕ + b2 + y 2 )1/2
where the integration is over the volume of the ring. Expressing dV as dV = Sb dϕ, we have
2π
(cos ϕ i + sin ϕ k)
ρu2 S
Ek =
dϕ.
(15)
4πε0 c2 0 (x 2 − 2bx cos ϕ + b2 + y 2 )1/2
By the symmetry of the system, the z component of Ek vanishes, so that we are left with
2π
ρu2 S
cos ϕ
Ek = i
dϕ.
(16)
2
2
4πε0 c 0 (x − 2bx cos ϕ + b2 + y 2 )1/2
A closed form solution of the integral in equation (16) does not exist. Therefore we shall
apply equation (16) to several special cases for which an approximate solution of equation (16)
can be obtained.
2.1. A small rotating ring
The first special case that we shall consider is a ring as in figure 1 except that its radius is much
smaller than the distance between the ring and the point of observation, b R. In this case
we can write
1/(x 2 − 2bx cos ϕ + b2 + y 2 )1/2 ≈ 1/{(x 2 + y 2 )1/2 [1 − 2bx cos ϕ/(x 2 + y 2 )]1/2 }
1 + bx cos ϕ/R 2
= 1/[R(1 − 2bx cos ϕ/R 2 )1/2 ] ≈
R
which, by equation (16), gives
2π
ρu2 S
bx cos ϕ
dϕ
Ek ≈ i
cos
ϕ
1
+
4πε0 c2 R 0
R2
(17)
(18)
and, after integration,
Ek ≈ i
ρu2 bxS
ρu2 bS sin θ
=
i
4ε0 R 3 c2
4ε0 R 2 c2
(19)
or, in terms of spherical coordinates R and θ,
Ek ≈
ρu2 bS
(sin θ Ru + sin θ cos θ Θu )
4ε0 R 2 c2
where Ru and Θu are unit vectors in the directions of R and θ , respectively.
(20)
Electric and magnetic fields of circular currents and rotating charges
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Using the same approximation as in equation (17), we have for r /r 3 in equation (3)
r
1 + 3bx cos ϕ/R 2
≈
R
r3
R3
(21)
which gives for the Coulomb field of the ring
EC ≈
ρbS
ρbS
R=
Ru .
2ε0 R 3
2ε0 R 2
(22)
The total electric field of the ring observed in the xy plane is obtained by adding
equations (20) and (22):
u2
ρbS
ρbSu2
2
1
+
E≈
sin
θ
R
+
sin(2θ)Θu .
(23)
u
2ε0 R 2
2c2
8ε0 R 2 c2
Expressing E in terms of the charge of the ring, q = 2π bSρ, we have
u2
q
qu2
2
1
+
E≈
sin
θ
R
+
sin(2θ)Θu .
u
4πε0 R 2
2c2
16π ε0 R 2 c2
(24)
2.2. A small rotating disc
Expressing in equation (20) the linear velocity u in terms of the angular velocity ω, replacing
b by x and S by τ dx , and considering the ring to be a differential element of a disc, we can
write
d Ek ≈
ρω2 x 3
(sin2 θ Ru + sin θ cos θ Θu )τ dx 4ε0 R 2 c2
(25)
where τ is the thickness of the disc (the same as that of the ring). Integrating equation (25)
between 0 and b, we then obtain for the electrokinetic field of a disc of radius b R
Ek ≈
ρω2 b4 τ
(sin2 θ Ru + sin θ cos θ Θu ).
16ε0 R 2 c2
(26)
For the Coulomb field we similarly obtain
EC ≈
ρb2 τ
Ru .
4ε0 R 2
(27)
Adding equations (26) and (27), we obtain for the total electric field of the rotating disc
ω2 b2
ρb2 τ
ρω2 b4 τ
2
1
+
E≈
sin
θ
R
+
sin(2θ)Θu .
(28)
u
4ε0 R 2
4c2
32ε0 R 2 c2
Expressing E in terms of the charge of the disc, q = π b2 τρ, we have
ω2 b2
q
qω2 b2
2
1
+
sin
θ
R
+
sin(2θ)Θu
E≈
u
4πε0 R 2
4c2
32π ε0 R 2 c2
u2
q
qu2
2
1
+
=
sin
θ
Ru +
sin(2θ)Θu
2
2
4πε0 R
4c
32π ε0 R 2 c2
where u is the linear velocity of the rim of the disc.
(29)
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O D Jefimenko
2.3. A small rotating sphere
Replacing in equation (26) b4 by (b2 − y 2 )2 and τ by dy , and assuming that the disc whose
electrokinetic field is represented by equation (26) is a differential element of a sphere, we find
by integrating over y from −b to b that the electrokinetic field of a rotating sphere of radius
b R is
Ek ≈
ρω2 b5
(sin2 θ Ru + sin θ cos θ Θu ).
15ε0 R 2 c2
(30)
For the Coulomb field we have
EC =
ρb3
Ru .
3ε0 R 2
(31)
Adding equations (30) and (31), we obtain for the total electric field of the rotating sphere
ω2 b2
ρb3
ρω2 b5
2
1
+
E≈
sin
θ
Ru +
sin(2θ)Θu .
(32)
2
2
3ε0 R
5c
30ε0 R 2 c2
Expressing E in terms of the charge of the sphere, q = 4π b3 ρ/3, we have
ω2 b2
q
qω2 b2
2
1
+
E≈
sin
θ
Ru +
sin(2θ)Θu
2
2
4πε0 R
5c
40π ε0 R 2 c2
u2
q
qu2
2
1
+
=
sin
θ
R
+
sin(2θ)Θu
u
4πε0 R 2
5c2
40π ε0 R 2 c2
(33)
where u is the equatorial linear velocity of the sphere.
2.4. A long rotating hollow cylinder
Consider a long hollow cylinder of radius b rotating about its symmetry axis. Let the cylinder
carry a uniformly distributed charge of density ρ and let the length of the cylinder be 2L.
The electrokinetic field outside the cylinder can be obtained by assuming that the ring whose
electrokinetic field is represented by equation (16) is a differential element of the cylinder. Let
the thickness of the cylinder wall be w b. The electrokinetic field of the cylinder can then
be found by replacing in equation (16) S by w dy , y by y , and by integrating the resulting
equation over the length of the cylinder.
Let us assume that the point of observation is in the middle plane of the cylinder and that
L x (‘long’ cylinder). We then have
2π L
ρu2 w
cos ϕ
Ek = i
dϕ dy .
(34)
2
2
4πε0 c 0 −L (x − 2bx cos ϕ + b2 + y 2 )1/2
Integrating by parts over ϕ, we have
ρu2 wxb 2π L
sin2 ϕ
Ek = i
dϕ dy .
4πε0 c2 0 −L (x 2 − 2bx cos ϕ + b2 + y 2 )3/2
Integrating over y and taking into account that L x, we obtain
ρu2 wxb 2π
sin2 ϕ
Ek = i
dϕ.
2πε0 c2 0 (x 2 − 2bx cos ϕ + b2 )
(35)
(36)
Electric and magnetic fields of circular currents and rotating charges
6149
The integral in equation (36) is just π/x 2 . The electrokinetic field of the cylinder is therefore
Ek =
ρu2 wb
i.
2ε0 xc2
(37)
For the Coulomb field we have (see, for example, [7, pp 89–90])
EC =
ρwb
i.
ε0 x
Adding equations (37) and (38) and using cylindrical coordinates, we obtain
u2
ρwb
1
+
E=
r0
2c2
ε0 r02
(38)
(39)
where r0 is a radius vector normal to the axis of the cylinder and directed from the axis to
the point of observation; r0 (the magnitude of r0 ) is the distance from the axis to the point of
observation. Expressing E in terms of the charge of the cylinder, q = 4π bwLρ, we have
u2
ω2 b2
q
q
1 + 2 r0 =
1+
E=
r0
(40)
2c
2c2
4πε0 Lr02
4π ε0 Lr02
where ω is the angular velocity of the cylinder.
2.5. A long rotating solid cylinder
Expressing in equation (37) the linear velocity u in terms of the angular velocity ω, replacing
b by x and w by dx , and considering the hollow cylinder to be a differential element of the
solid cylinder, we can write
dE k = i
ρω2 x 3 dx .
2ε0 xc2
(41)
Integrating equation (41) between 0 and b, we obtain for the electrokinetic field of a solid
rotating cylinder of radius b
Ek =
ρω2 b4
i.
8ε0 xc2
(42)
For the Coulomb field we have [7, pp 89–90]
EC =
ρb2
i.
2ε0 x
Adding equations (42) and (43) and using cylindrical coordinates we obtain
ω2 b2
ρb2
1+
E=
r0 .
4c2
2ε0 r02
Expressing E in terms of the charge of the cylinder, q = 2π b2 Lρ, we have
ω2 b2
u2
q
q
1
+
1
+
E=
r
=
r0
0
4c2
4c2
4πε0 Lr02
4π ε0 Lr02
where u is the linear velocity at the surface of the cylinder.
(43)
(44)
(45)
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O D Jefimenko
2.6. A long solenoid
Consider a long solenoid whose symmetry axis is the y axis. Let the length of the solenoid be
2L and let it carry a current I . The electrokinetic field outside the solenoid can be obtained from
equation (37) for the long thin rotating cylinder. Let the thickness of the current-carrying wall
of the solenoid be w, let the current in each turn of the solenoid be I and let the solenoid have n
turns. The charge density ρ associated with the current in the solenoid is then ρ = nI /2Luw.
Expressing Ek in terms of I , we then obtain
Ek =
nI ub
nI ub
i=
r0
4ε0 Lxc2
4ε0 Lr02 c2
(46)
where r0 is a radius vector normal to the axis of the solenoid and directed from the axis to
the point of observation; r0 (the magnitude of r0 ) is the distance from the axis of the solenoid
to the point of observation. Since ordinarily only negative charges produce the current in the
solenoid, whereas an equal number of positive charges are at rest in the solenoid, the positive
charges make no contribution to the electrokinetic field of the solenoid, and the Coulomb fields
of the positive and negative charges of the solenoid cancel each other.
3. Theory II. Magnetic field
Let us designate the magnetic flux density field associated with the second integral in
equation (2) as Bk and, by analogy with the electrokinetic field Ek , let us call it the
‘magnetokinetic field’. We thus have
µ0
1 ∂J
Bk =
× r dV .
(47)
4πc
r 2 ∂t
To determine the magnetokinetic field produced by the rotating ring shown in figure 1, we
proceed as follows. Since at any point of the ring the derivative ∂ J /∂t is the same at all times,
the retardation may be assumed to have no effect and we can write equation (47) as
µ0
1 ∂J
Bk =
(48)
× r dV .
4πc
r 2 ∂t
Just as for the electrokinetic field Ek , r is given by equation (8) and the derivative ∂ J /∂t for
the shaded element in figure 1 is given by equations (10)–(12), so that dBk associated with the
shaded element is
d Bk =
µ0 ρu2 b × r
dV
4πcb2 (x 2 − 2bx cos ϕ + b2 + y 2 )
(49)
and, since by equation (7) r = b + R, we obtain
d Bk =
µ0 ρu2 b × R
dV
4πcb2 (x 2 − 2bx cos ϕ + b2 + y 2 )
(50)
which, with equation (6), yields
d Bk = −
µ0 ρu2 (cos ϕ i + sin ϕ k) × R
dV .
4π cb(x 2 − 2bx cos ϕ + b2 + y 2 )
The magnetokinetic field produced by the ring is therefore
µ0 ρu2
cos ϕ i + sin ϕ k
dV
Bk =
R×
4πcb
(x 2 − 2bx cos ϕ + b2 + y 2 )
(51)
(52)
Electric and magnetic fields of circular currents and rotating charges
where we have factored out the constant vector R. Writing dV as Sb dϕ, we have
2π
cos ϕ i + sin ϕ k
µ0 ρu2 S
dϕ.
Bk =
R×
2 − 2bx cos ϕ + b2 + y 2 )
4πc
(x
0
6151
(53)
By symmetry, the contribution of the z component of the integrand is zero, so that we are left
with
µ0 ρu2 S 2π
cos ϕ
dϕ
(54)
Bk = R × i
2 − 2bx cos ϕ + b2 + y 2 )
4πc
(x
0
and, since R × i = −R cos θ k = −yk, we obtain
µ0 ρu2 yS 2π
cos ϕ
Bk = −k
dϕ.
2
4πc
(x − 2bx cos ϕ + b2 + y 2 )
0
(55)
Let us now apply equation (55) to the special cases considered in section 2.
3.1. A small rotating ring
Assuming that b R (‘small ring’), we have, as in equation (17),
1/(x 2 − 2bx cos ϕ + b2 + y 2 ) ≈
1 + 2bx cos ϕ/R 2
.
R2
Substituting equation (56) into (55), we obtain
µ0 ρu2 yS 2π cos ϕ(1 + 2bx cos ϕ/R 2 )
B k ≈ −k
dϕ
4πc
R2
0
µ0 ρu2 byxS
= −k
2cR 4
or
µ0 ρu2 bS sin(2θ )
Bk ≈ −
k.
4cR 2
In spherical coordinates, Bk is
(56)
(57)
(58)
µ0 ρu2 bS sin(2θ )
Φu
(59)
4cR 2
where Φu is the azimuthal unit vector left-handed relative to the y axis. Written in terms of
the current I = ρuS created by the rotating ring, Bk is
Bk ≈ −
µ0 I ub sin(2θ )
Φu .
(60)
4cR 2
The total magnetic field produced by the ring is the sum of Bk and the field associated
with the first integral in equation (2). However, the latter field (which is the ordinary dipole
field of the ring) is of no interest for the present discussion and we shall not include it in our
derivations.
Bk ≈ −
3.2. A small rotating disc
Expressing in equation (59) the linear velocity u in terms of the angular velocity ω, replacing
b by x and S by τ dx , and considering the ring to be a differential element of a disc, we can
write
µ0 ρω2 x 3 τ sin(2θ) dx .
(61)
dBk ≈ −Φu
4cR 2
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O D Jefimenko
Integrating equation (61) between 0 and b, we then obtain for the magnetokinetic field of a
disc of radius b R
µ0 ρω2 b4 τ
µ0 ρu2 b2 τ
Bk ≈ −
sin(2θ)Φu = −
sin(2θ)Φu
(62)
2
16cR
16cR 2
where u is the linear velocity of the rim of the disc. Expressing Bk in terms of the charge of
the disc, q = πb2 τρ, we obtain
Bk ≈ −
µ0 qω2 b2
µ0 qu2
sin(2θ
)
Φ
=
−
sin(2θ)Φu .
u
16πcR 2
16π cR 2
(63)
3.3. A small rotating sphere
Using equation (62) with angular velocity ω, replacing b4 by (b2 − y 2 )2 and τ by dy , and
assuming that the disc represented by equation (62) is a differential element of a sphere, we
find by integrating from −b to b that the magnetokinetic field of the rotating sphere of radius
b R is
µ0 ρω2 b5 sin(2θ )
µ0 ρu2 b3 sin(2θ)
Bk ≈ −
Φu = −
Φu
(64)
2
15cR
15cR 2
where u is the equatorial linear velocity of the sphere. Expressing Bk in terms of the charge
of the sphere, q = 4πb3 ρ/3, we obtain
Bk ≈ −
µ0 qω2 b2 sin(2θ )
µ0 qu2 sin(2θ)
Φ
=
−
Φu .
u
20πcR 2
20π cR 2
(65)
3.4. Long rotating cylinders and a long solenoid
Consider a long thin rotating cylinder whose symmetry axis is the y axis. Let the length of
the cylinder be 2L, let the thickness of its wall be w, and let it carry a uniformly distributed
charge of density ρ. The magnetokinetic field outside the cylinder can be obtained with the
help of equation (55) by assuming that the ring, whose electrokinetic field is represented by
equation (55), is a differential element of the cylinder. Let the point of observation be in the
middle plane of the cylinder and let L x (‘long’ cylinder). Replacing y in equation (55)
by y , and replacing S by w dy , we then have
L
µ0 ρu2 wy 2π
cos ϕ
B k = −k
dϕ dy 2 − 2bx cos ϕ + b2 + y 2 )
4πc
(x
−L
0
µ0 ρu2 w L 2π
y cos ϕ
= −k
(66)
dϕ dy .
4πc
(x 2 − 2bx cos ϕ + b2 + y 2 )
−L 0
However, the contributions of −y and y cancel when we integrate from −L to L. Thus the
integral in equation (66) is zero and the cylinder does not create a magnetokinetic field in the
midplane of the cylinder. The same considerations apply to a solid rotating cylinder and to a
current-carrying solenoid.
4. Discussion
The calculations of the electrokinetic and magnetokinetic fields Ek and Bk presented above
reveal the existence of several previously unreported and unforeseen electromagnetic effects,
the most important of which are:
(1) A steady-state circular electric current creates a constant radial electric field associated
with radial acceleration of the charges constituting the current.
Electric and magnetic fields of circular currents and rotating charges
6153
(2) A uniformly rotating spherical charge creates not only the usual Coulomb field, but
also a constant radial electric field associated with radial acceleration of the elementary charges
within the spherical charge under consideration.
(3) A steady-state circular electric current creates not only the usual magnetic field
associated with the current, but also a constant circular magnetic field associated with radial
acceleration of the charges constituting the current.
(4) A uniformly rotating spherical charge creates not only the usual magnetic dipole field,
but also a constant circular magnetic field associated with radial acceleration of the elementary
charges within the spherical charge under consideration.
In connection with these effects two questions arise:
(1) Why were these effects not predicted by electromagnetic theory in the past?
(2) Are there any experiments manifesting these effects?
The answer to the first question is quite simple. Crucial for the calculations presented
in this paper are equations (1) and (2). Although these equations were derived almost half a
century ago, they became generally known relatively recently and have not yet been used for
practical applications to any appreciable extent.
The answer to the second question is more complicated. First, it should be noted that,
because of the factor u2 /c2 in the electrokinetic field equations and the factor u2 /c in the
magnetokinetic field equations, these fields are extremely weak, except when u is close to
c, which can hardly happen in macroscopic systems. Furthermore, in order to detect these
very weak fields in macroscopic experiments one needs to use field-detecting devices whose
sensitivity is much greater than that of conventional macroscopic electromagnetic instruments.
The situation is quite different in mesoscopic and microscopic systems. High velocities
are commonplace in such systems, and electric and magnetic fields are detected there by
elementary particles whose motion is strongly affected even by extremely weak fields. In fact,
there is good evidence that the electrokinetic field associated with a circular current has already
been observed in a mesoscopic experiment, in the well known Aharonov–Bohm experiment.
In their now famous 1959 paper, Aharonov and Bohm advanced the theory that the
magnetic vector potential is not just a mathematical device, but an observable physical quantity
having a physical significance of its own [11]. They suggested that the wavefunction of
electrons moving outside a long current-carrying solenoid in a region where (as was generally
assumed) there are no electric or magnetic fields can be altered solely by the magnetic vector
potential. Experiments have apparently supported their theory [12], and the phenomenon
predicted by them became known as the Aharonov–Bohm effect.
Although the Aharonov–Bohm effect is considered to be a quantum mechanical effect
(it manifests itself as a shift of wavefunction interference fringes), it could have profound
and dramatic consequences for classical electrodynamics. In classical electrodynamics, only
electric and magnetic fields can alter the motion of charged particles. They do so by exerting
electric and magnetic forces on the particles. If electromagnetic potentials can exert a
dynamical effect on charged particles in the absence of electric or magnetic fields at the location
of the particles, then the entire Maxwellian electrodynamics is incorrect and must be replaced
by a different theory. However, no physical theory has been validated more convincingly and
more completely than Maxwellian electrodynamics.
It should be noted that the expressions for the fringe shift obtained by Aharonov and
Bohm involve purely classical electrodynamic quantities: the macroscopic magnetic vector
potential of the solenoid and the macroscopic magnetic flux inside the solenoid in particular.
The question arises therefore: is it possible that the fringe shift observed in the Aharonov–
Bohm experiment is actually associated with an electric or magnetic field outside the solenoid
that somehow was overlooked in the formulation of the theory of the effect? Numerous
6154
O D Jefimenko
attempts have been made to interpret this fringe shift in terms of classical electrodynamics [13],
and the role of the magnetic vector potential in the Aharonov–Bohm experiment has been
questioned [14]. Quite clearly, a successful electrodynamic theory of the Aharonov–Bohm
experiment should accomplish two things: it should show that an electric or magnetic field
does indeed exist outside the solenoid in the Aharonov–Bohm experiment and it should explain
the role of the macroscopic magnetic vector potential in that experiment. Until now such a
theory could not be formulated.
The theory of acceleration-related electrokinetic fields presented in this paper finally
makes it possible to reconcile the Aharonov–Bohm experiment with classical electrodynamics.
According to equation (46), there is indeed an electric (electrokinetic) field outside a currentcarrying solenoid:
Ek =
nI ub
r0 .
4ε0 Lr02 c2
(46)
Therefore, according to this equation, the electrons in the Aharonov–Bohm experiment are
moving in the presence of an electric field and, consequently, experience an electric force due
to this field. The question remains, however, how this electric force is related to the magnetic
vector potential of the solenoid. The answer to this question is provided below.
As with any electric field, the electrokinetic field exerts a force
F = q Ek
(67)
on the electric charge q located in this field. If the charge q moves through Ek , it acquires a
momentum p given by
p=
F dt =
q Ek dt.
(68)
Substituting Ek from equation (4) and integrating over time, we have
1 ∂J
J
q
q
p=−
dV
dt
=
−
dV .
2
2
4πε0 c
r ∂t
4π ε0 c
r
However, the last integral in equation (69) is
1
J
dV = A
2
4πε0 c
r
(69)
(70)
where A is the vector potential produced at any point of space by the current J (see, for
example, [7, pp 363–4] noting that 1/ε0 c2 = µ0 ), the current which, according to equation (2),
also produces the magnetic field B in the solenoid. Thus, for an electron in the Aharonov–
Bohm experiment, equation (69) can be written as
p = eA
(71)
where A is the vector potential at the point where the electron (negative charge e) is located.
When the electron moves from its source to the screen on which the interference fringes
are observed, the phase of its wavefunction changes by [15]
p
e
δ=
· dl =
A · d l.
(72)
h̄
h̄
The difference in the phases of the wavefunction of electrons passing the solenoid on two
different sides S1 and S2 is then
e
e
δ1 − δ2 =
A · dl −
A · dl =
A · d l.
(73)
h̄ S1
h̄
S2
Electric and magnetic fields of circular currents and rotating charges
Since
6155
A · dl = $
(74)
where $ is the magnetic flux enclosed by the path of integration (see, for example, [7, p 366])
(the magnetic flux inside the solenoid), we finally obtain
e
δ1 − δ2 = $.
(75)
h̄
This is the same equation as that obtained by Aharonov and Bohm on the basis of their theory.
Thus, equation (75), which is the mathematical expression for phase shift in the Aharonov–
Bohm experiment, is a consequence of an electric force acting on moving electrons. The force
is caused by the electric field produced by the same current which creates the magnetic field
in the solenoid. This force is a strictly classical phenomenon and is a direct consequence
of Maxwell’s electrodynamic equations. Therefore, although Aharonov and Bohm attributed
the phase shift observed in their experiment to the magnetic vector potential, the experiment
does not reveal any special physical significance of the magnetic vector potential2 . In this
connection it may be mentioned that electromagnetic forces can be quite generally described
and computed not only in terms of electromagnetic fields but also in terms of electromagnetic
potentials [17].
In conclusion it should be noted that the study of the electrokinetic and magnetokinetic
fields represented by the second integral of equation (1) and by the second integral of
equation (2) is still in its rudimentary stage. More studies, especially experimental, are needed
for elucidating properties, peculiarities and possible uses of these fields. The calculations
presented in this paper indicate that these fields may give rise to a variety of important
electromagnetic effects not yet described or observed.
References
[1] Griffiths D J and Heald M A 1991 Time-dependent generalization of the Biot–Savart and Coulomb laws Am. J.
Phys. 59 111–7
[2] Tran-Cong Ton 1991 On the time-dependent, generalized Coulomb and Biot–Savart laws Am. J. Phys. 59 520–8
[3] Heald M A and Marion J B 1995 Classical Electromagnetic Radiation (Philadelphia, PA: Saunders) pp 261-2
[4] Rosser W G V 1997 Interpretation of Classical Electromagnetism (Dordrecht: Kluwer) pp 48–9
Rosser W G V 1997 Interpretation of Classical Electromagnetism (Dordrecht: Kluwer) pp 82–4
[5] Jackson J D 1999 Classical Electrodynamics 3rd edn (New York: Wiley) pp 246–7
[6] Griffiths D J 1999 Introduction to Electrodynamics 3rd edn (Englewood Cliffs, NJ: Prentice-Hall) pp 427–8
[7] Jefimenko O D 1992 Electricity and Magnetism 2nd edn (Star City: Electret Scientific) (1st edn 1966 New York:
Appleton-Century-Crofts)
[8] Jefimenko O D 2000 Causality, Electromagnetic Induction, and Gravitation 2nd edn (Star City: Electret
Scientific) pp 28–30
[9] Jefimenko O D 1994 Direct calculation of the electric and magnetic fields of an electric point charge moving
with constant velocity Am. J. Phys. 62 79–85
[10] Jefimenko O D 1997 Electromagnetic Retardation and Theory of Relativity (Star City: Electret Scientific)
pp 62–70
[11] Aharonov Y and Bohm D 1959 Significance of electromagnetic potentials in the quantum theory Phys. Rev. 115
485–91
2 Several variations of the original Aharonov–Bohm experiment have been reported. In all probability the fringe shift
observed in such experiments is caused by a combination of several different factors. Depending on the experimental
arrangement, one or another of these factors may predominate. An experiment that has attracted much attention
involved a ferromagnet enclosed in a superconducting shield [16]. It is quite clear that in this particular experiment
the electrons passing the shielded magnet created electric images in the superconducting shield (see, e.g., [7] pp 161–
8). These images inevitably created an electric image field around the shield, so that, contrary to the supposition, the
electrons were not moving in a field-free region at all.
6156
O D Jefimenko
[12] See, for example, Peshkin M and Tonomura A 1989 The Aharonov–Bohm Effect (New York: Springer)
[13] For a discussion, see Portis A M 1978 Electromagnetic Fields: Sources and Media (New York: Wiley) pp 400–5
and references cited there
[14] See, for example, Boyer T H 2000 Does the Aharonov–Bohm effect exist Found. Phys. 30 893–905
Boyer T H 2000 Classical electromagnetism and the Aharonov–Bohm phase shift Found. Phys. 30 907–32 and
papers cited there
[15] See, for example, Feynman R P, Leighton R B and Sands M 1964 The Feynman Lectures on Physics vol 2
(Reading, MA: Addison-Wesley) section 15 p 9
[16] Osakabe N, Matsuda T, Kawasaki T, Endo J, Tonomura A, Yano S and Yamada 1986 Experimental confirmation
of the Aharonov–Bohm effect using a toroidal magnetic field confined by a superconductor Phys. Rev. A 34
815–22
[17] Jefimenko O D 1990 Direct calculation of electric and magnetic forces from potentials Am. J. Phys. 58 625–31
INSTITUTE OF PHYSICS PUBLISHING
JOURNAL OF PHYSICS A: MATHEMATICAL AND GENERAL
J. Phys. A: Math. Gen. 35 (2002) 2527–2531
PII: S0305-4470(02)29719-X
COMMENT
Comment on ‘The effect of radial acceleration on the
electric and magnetic fields of circular currents and
rotating charges’
W P Healy
Department of Mathematics, RMIT University, GPO Box 2476V, Melbourne, Victoria 3001,
Australia
Received 25 September 2001, in final form 15 January 2002
Published 1 March 2002
Online at stacks.iop.org/JPhysA/35/2527
Abstract
In a recent paper in this journal, Jefimenko (Jefimenko O D 2001 J. Phys.
A: Math. Gen. 34 6143–56) used time-dependent generalizations of the
Coulomb and Biot–Savart laws to infer the existence of ‘electrokinetic’ and
‘magnetokinetic’ fields due to radial acceleration of charges in circular motion.
It is shown here that for the stationary charge and current distributions discussed
in Jefimenko’s paper, the ‘electrokinetic’ and ‘magnetokinetic’ fields are
zero. In particular, there is no ‘electrokinetic’ field outside an infinitely long,
uniformly charged and steadily rotating hollow cylinder, and there is no electric
field whatsoever outside an infinitely long cylindrical solenoid. It follows that
the classical explanation of the Aharonov–Bohm effect proposed by Jefimenko
is invalid.
PACS numbers: 41.20.−q, 03.50.De
1. Introduction
In a recent paper [1], Jefimenko has argued that so-called ‘electrokinetic’ electric fields and
‘magnetokinetic’ magnetic-induction fields can be produced by charges in uniform circular
motion, even though the charge–current distribution is everywhere constant in time. These
fields are supposed to exist in addition to the usual electric and magnetic-induction fields
given by the Coulomb and Biot–Savart laws for stationary sources. It is the purpose of this
comment to point out (a) that Jefimenko’s argument relies on an erroneous identification of
the total and partial time derivatives of the current density and (b) that the ‘electrokinetic’ and
‘magnetokinetic’ fields vanish for the systems considered. It follows that an ‘electrokinetic’
field cannot be invoked as a classical explanation of the Aharonov–Bohm effect due to a long
cylindrical solenoid and it will be shown that the field used for this purpose in [1] does not
satisfy Maxwell’s equations inside the solenoid.
0305-4470/02/102527+05$30.00
© 2002 IOP Publishing Ltd
Printed in the UK
2527
2528
Comment
2. Generalized Coulomb and Biot–Savart laws
The discussion in [1] starts from the following equations, which represent the electric field E
and the magnetic-induction field B as volume integrals involving the charge density ρ, the
current density J and their partial time derivatives ∂ρ/∂t and ∂ J /∂t:
[ρ]
1 ∂J
1
1 ∂ρ
1
dV
(1)
E=
+
r
dV
−
4π0
r 3 r 2 c ∂t
4π 0 c2
r ∂t
µ0
µ0
[J ]
1 ∂J
B=
× r dV +
(2)
× r dV .
4π
r3
4π c
r 2 ∂t
The notation is that used in [1]—0 is the permittivity and µ0 the permeability of the vacuum,
c is the speed of light in vacuo, r is the distance from a source point in the volume element dV to
the field point where E and B are evaluated and the square brackets indicate that the expressions
inside them are to be evaluated at the retarded time t − r/c, where t is the time at which E and
B are evaluated. Equations (1) and (2) constitute time-dependent generalizations [2–4] of the
Coulomb and Biot–Savart laws
ρ
J
1
µ0
E=
r
dV
and
B
=
× r dV
(3)
3
4π0
r
4π
r3
to which they reduce if the charge–current distribution is stationary; that is, if ∂ρ/∂t and ∂ J /∂t
are identically zero and, as a consequence, [ρ] = ρ and [J ] = J . For stationary sources,
∇ · J = 0 and Ampère’s law
∇ × B = µ0 J
(4)
is valid1 .
Jefimenko [1] refers to the contributions to E from the integrals over [∂ρ/∂t] and
[∂ J /∂t] in equation (1) as ‘electrokinetic’ and to the contribution to B from the integral over
[∂ J /∂t] in equation (2) as ‘magnetokinetic’. Although the terminology may suggest that these
contributions to E and B are due to mere motion of the charges2 , the stronger condition that the
charge–current distribution be non-stationary is evidently required. For example, the charge–
current distribution associated with a single moving point charge is non-stationary even when
the velocity is constant [6]. All the distributions considered in [1], however, are stationary,
since uniformly charged rings, cylinders or spheres in steady rotational motion about an axis
of rotational symmetry have charge and current densities that are constant in time at every
space point. There are, therefore, no ‘electrokinetic’ or ‘magnetokinetic’ contributions to the
corresponding fields—E and B are simply given by the Coulomb and Biot–Savart laws (3).
While noting that ∂ρ/∂t in equation (1) is zero for the systems being discussed, Jefimenko
asserts that ∂ J /∂t in both equations (1) and (2), although constant in time, is not identically
zero. In the case of a steadily rotating ring with uniform charge density ρ, the expression ρ a
for ∂ J /∂t is obtained in equation (10) of [1]. Here a is the (centripetal) radial acceleration of
the element of charge on the ring that is instantaneously located at the point where ∂ J /∂t is
being evaluated. In reality, however, ρ a is the total time derivative of J , given by
dJ
∂J
=
+ (u · ∇)J ,
dt
∂t
1
(5)
Griffiths and Heald [4] use the term ‘static’ instead of ‘stationary’ to describe a charge–current distribution for
which both ρ and J are time independent. These authors have also delineated more general distributions, termed
‘semistatic’, for which the Coulomb and Biot–Savart laws (3) still hold, although Ampère’s law (4) fails.
2 It may be of interest to note that in Maxwell’s Treatise on Electricity and Magnetism [5], the electromagnetic energy
of a system of interacting electric currents is called ‘electrokinetic energy’.
Comment
2529
and the partial time derivative ∂ J /∂t is identically zero3 . In equation (5), u is the velocity J /ρ
of the charge element. By using cylindrical polar coordinates and the fact that ρ is constant
in the rotating ring, it may readily be verified that the second term on the right-hand side of
equation (5) gives ρ a. That Jefimenko has calculated the total time derivative of J is clear
from the footnote on page 6145 of [1]. It is also clear from the derivation of equations (1) and
(2) from the retarded electromagnetic potentials in the Lorentz gauge (see [4], for example)
that ∂ρ/∂t and ∂ J /∂t in these equations are partial time derivatives at a fixed point in space.
3. Maxwell’s equations
As the generalized Coulomb and Biot–Savart laws are based on Maxwell’s equations, any fields
E and B obtained by exact evaluation of the integrals in equations (1) and (2) must satisfy
Maxwell’s equations with the prescribed charge and current densities ρ and J as sources. It
will now be shown that in the case of a uniformly charged and steadily rotating hollow cylinder,
Jefimenko’s ‘electrokinetic’ field Ek does not comply with this requirement. In the limit of
an infinitely long cylinder of radius b that rotates about its axis with constant angular velocity
u/b and has constant surface charge density σ ,
 2
u σ r0


r̂0
if 0 r0 < b
 2
Ek = 2c 2 0 b
(6)
u σb



r̂
if
r
>
b
0
0
2c2 0 r0
where r0 is the cylindrical–polar radial coordinate, r̂0 is the outward radial unit vector and the
axis of the cylinder is the z-axis. Again the notation is that of [1] except that the product of
the charge density ρ and the width w of the cylindrical shell used in [1] is written here as σ ;
that is, the thin shell is replaced by a cylindrical surface by letting ρ tend to ∞ and w tend to
zero in such a way that ρw tends to σ . The expression given in the second of equations (6) for
the ‘electrokinetic’ field outside the cylinder r0 = b was obtained in [1] by inserting ρ a for
∂ J /∂t in equation (1) and using the value π/x 2 for the integral4
2π
sin2 φ
dφ
(7)
x 2 − 2bx cos φ + b2
0
in the case where x > b. By interchanging x and b, it can be seen that the integral (7) has the
value π/b2 when 0 < x < b and it is easy to verify that it has this value also when x = 0. The
expression for Ek inside the cylinder r0 = b, given in the first of equations (6), then follows. It
may be shown that ∇ × Ek = 0 when r0 = b. The discontinuity in Ek on the cylinder r0 = b
is removable and hence so is that of its normal component, which is Ek itself.
The Coulomb field EC of the charged cylinder, which corresponds to the first integral in
equation (1), is given by

if 0 r0 < b
0
σb
EC =
(8)

r̂0
if r0 > b.
0 r0
3 In hydrodynamics, the total time derivative is often called the derivative ‘following the motion’ of a fluid element
and the corresponding differential operator is sometimes written as D/Dt instead of as d/dt. The total and partial
derivatives give rates of change with respect to time that correspond respectively to the Lagrangian and Eulerian
descriptions of fluid flow [7].
4 The integrand has period 2π and the integral may be be evaluated over the interval [−π, π ] by changing the
integration variable to tan(φ/2) and using partial fractions or contour integration to integrate the resulting rational
function.
2530
Comment
This field too is radial but, in contrast to Ek , it changes discontinuously by (σ/0 )r̂0 when r0
passes through the point b and hence it satisfies the correct boundary condition for the cylinder
with surface charge density σ . Thus, the given charge on the cylinder acts as a source or a sink
for the Coulomb field EC outside the cylinder.
Now the outward flux of Ek per unit length of cylinder through any cylinder r0 = d with
radius d greater than b is u2 π bσ/(c2 0 ) and is independent of d. By Gauss’s flux theorem,
there must be charge u2 πbσ/c2 per unit length of cylinder inside the cylinder r0 = b. It
follows from equations (6) that
 2
 u σ
if 0 r0 < b
∇ · E k = c 2 0 b
(9)

0
if r0 > b.
The first of equations (9) implies that inside the cylinder there is a constant charge density
u2 σ/(c2 b) and hence a charge u2 πbσ/c2 per unit length of cylinder that acts as a source
or a sink for the field Ek outside. Jefimenko’s field Ek is therefore inconsistent with the
prescribed charge density ρ, which is non-zero only on the cylinder, and Maxwell’s equation
∇ · E = ρ/0 is not satisfied by the total electric field EC + Ek at any point inside the
cylinder. This inconsistency confirms the result of section 2 that there is no field Ek due to
radial acceleration of charges on the cylinder.
4. Aharonov–Bohm effect
In section 4 of [1], an explanation, based on classical electrodynamics, of the Aharonov–
Bohm effect due to an infinitely long cylindrical solenoid was proposed. For the purposes
of discussion, it is convenient here to regard the solenoid as consisting of two contiguous
cylinders, one with positive surface charge density −σ at rest and the other with negative
surface charge density σ rotating about its axis with constant angular velocity u/b, where σ is
a negative constant and b is the radius of both cylinders. This serves as a model for the system
considered in [1], in which a current of conduction electrons with negative charge density ρ
flows through a closely wound solenoid against a fixed background of positive ions with charge
density −ρ. The Coulomb fields of the two cylinders cancel but, according to [1], there is
a radially inward electric field Ek caused by the rotation of the negatively charged cylinder.
Since this field (but no magnetic-induction field) is supposed to exist in the configuration space
r0 > b of a charged particle being scattered by the solenoid, it was argued in [1] that it gives
rise to the Aharonov–Bohm phase shift. There is then apparently no need to attribute a local
but gauge-invariant influence to either the vector potential outside the solenoid or the particle’s
magnetization field inside the solenoid [8]. It has been shown here, however, that there is no
‘electrokinetic’ field Ek accompanying the rotating cylinder and so such a field cannot be used
to explain the Aharonov–Bohm effect.
It is, of course, possible for an electric field of the form given in equations (6) to exist,
but this, as has been seen in section 3, would require a static charge distribution of constant
negative density u2 σ/(c2 b) in the interior region 0 r0 < b. If this charge were present, the
electric field would act directly on a charged particle in the exterior region r0 > b whether
the negatively charged cylinder were at rest or rotating and hence a change in the particle’s
wavefunction due to the electric field would occur whether the magnetic-induction field B
inside the solenoid were zero or not. The Aharonov–Bohm effect, which is the additional
change caused by a non-zero field B in the interior region, would still be manifest as a shift
in the interference pattern of the scattered particles and would still require a non-classical
interpretation.
Comment
2531
References
[1] Jefimenko O D 2001 The effect of radial acceleration on the electric and magnetic fields of circular currents and
rotating charges J. Phys. A: Math. Gen. 34 6143–56
[2] Jefimenko O D 1966 Electricity and Magnetism (New York: Appleton-Century-Crofts)
[3] Tran-Cong Ton 1991 On the time-dependent, generalized Coulomb and Biot–Savart laws Am. J. Phys. 59 520–8
[4] Griffiths D J and Heald M A 1991 Time-dependent generalizations of the Biot–Savart and Coulomb laws Am. J.
Phys. 59 111–7
[5] Maxwell J C 1891 A Treatise on Electricity and Magnetism 3rd edn, vol 2 (reprinted 1954 (New York: Dover))
[6] Jefimenko O D 1994 Direct calculation of the electric and magnetic fields of an electric point charge moving with
constant velocity Am. J. Phys. 62 79–85
[7] Chirgwin B H and Plumpton C 1967 Elementary Classical Hydrodynamics (Oxford: Pergamon)
[8] Healy W P and Samandra R A 1997 Polarization and magnetization fields in the Aharonov–Bohm effect Phys.
Rev. A 55 890–9