Discover Physics, Vol. 2, 5 (2026). https://doi.org/10.1007/s44418-026-00008-w
. https://rdcu.be/fgOlw
Analysis
of solution of equations for magnetic field of rotating ball using polynomials
Sergey G.
Fedosin
Perm State University: Permskij gosudarstvennyj nacional'nyj issledovatel'skij universitet, 614990,
Bukireva 15, Perm, Russian Federation
E-mail: sergey.fedosin@gmail.com
Abstract: The exact form of the solution for
the vector potential and magnetic field of a rotating uniformly charged ball is
explicitly found. Expressions for specific vector spherical polynomials
associated with the corresponding components of the potential are used to
represent the solution. Inside and outside a charged ball uniformly rotating
around its axis, the components of the potential and magnetic field are
determined up to terms containing the sixth power of the speed of light in the
denominator. To do this, it was necessary to use spherical coordinates and
eight polynomials of each type. In addition, the solutions inside and outside
the ball were equated to each other on the surface of the ball, taking into
account the symmetry of the ball. The accuracy of the approach used can be
increased, since it is determined only by the number of polynomials used, whose
contribution to solutions decreases rapidly as the degree of the polynomials
increases. To calculate the vector potential and magnetic field of a rotating
ball within the framework of special relativity, it is sufficient to substitute
the coordinates of the observation point, the invariant volumetric charge
density, the angular velocity of rotation and the radius of the ball into the
formulas.
Keywords: polynomial
expansion; vector potential; magnetic field;
rotating ball.
1. Introduction
There are various methods for determining
the magnetic field of moving charge configurations. Thus, in [1], rotating
charges create currents that are considered as currents in coils with many
turns. In [2-3], the Biot-Savart law is used to
determine the magnetic field. These approaches allow only first-order accuracy
when determining terms containing the square of the speed of light in the
denominator. Calculations using retarded potentials allow improving the accuracy
[4], but the complexity of the calculations quickly increases as the accuracy
increases.
There are known works in which the electromagnetic
field inside and outside a rotating uniformly charged or conducting body is
calculated in various ways [5-8]. In the case when there are arbitrarily
directed currents in a spherical configuration, the solution of the Laplace
equation contains spherical harmonic functions [9]. In order to simplify the
solution for the magnetic field of rotating charge configurations, in [10-11]
the magnetic field was calculated using the electric potential and electric field
strength. In [12], cylindrical and spherical coordinates, as well as spherical
polynomials, were used to determine the magnetic field of a spherical system of
chaotically moving particles.
One of the widely used methods for approximately
solving differential equations in modern engineering is the Finite Element
Method (FEM). Some examples of using this method to determine magnetic fields
can be found in [13-15]. A comparison of the FEM with the harmonic method is
presented in [16], where it is indicated that in some situations, the harmonic
method provides a faster convergence rate for the vector potential and magnetic
field components than the finite element method.
In magnetostatics, a method of determining
the magnetic field in the form of a gradient from a scalar magnetic potential
is often used. Thus, in [17] one can find a solution for the magnetic field of
a uniformly magnetized sphere, which is expressed through Legendre polynomials.
This method is suitable outside the body, as well as in the body's substance in
the case where there are no currents in this substance. Beginning with the
classical work of Gauss [18], the Earth's magnetic field is modeled in a similar
way, and about 150 spherical harmonics were used to study the small-scale
magnetic field of the lithosphere in [19].
A more general approach, suitable for
electric currents, involves the use of a vector magnetic potential. This was
used, for example, in [20], where formulas are presented that relate the scalar
magnetic potential to the vector magnetic potential.
The use of polynomials allows one to find
a solution quite simply also in the case of rotating charged bodies. In this
case, in the solution of the Laplace equation for the vector magnetic potential
outside the rotating body generating a stationary magnetic field, as a rule,
the associated Legendre polynomials appear.
The purpose of this work is detailed derivation of specific formulas
describing vector magnetic potential and magnetic field inside and outside a
uniformly charged rotating ball. In the spherical coordinates
, the radial component
of the vector
potential
turns out to
depend on polynomials that are the product of
by the associated Legendre polynomials
of the first order in the form
. As for the component
, it turns out to depend on the products of
by
. However, solving the inhomogeneous Laplace equation
inside the body leads to an unexpected peculiarity, which lies in the fact that
the component
contains the
polynomial
that does not
coincide with the polynomial
. The difference arises as a consequence of the fact
that the polynomial
is not only one
of the solutions of the Laplace equation for the component
, but also exactly agrees with the Lorentz factor of
rotating charges, which must be taken into account in the special theory of
relativity.
Another purpose is not just to present the solution in
a general form with undetermined coefficients, as is the case in most papers,
but also to describe a specific procedure for determining such coefficients. For
rotating ball, using a sufficient number of polynomials, it is possible to
obtain the values of vector potential and magnetic field at any given point
with the required accuracy. In this case, to determine the components of the
field at an arbitrary point with specified coordinates, it is sufficient to
know only the invariant volume charge density, the angular velocity of
rotation, and the radius of the rotating ball. An essential part of the method
is the use of specific spherical polynomials containing associated Legendre
polynomials and harmonic functions as multipliers.
The results obtained may be
relevant in those models of generation of the Earth's and planets' magnetic
fields, which assume spatial radial separation of charges in matter at high
temperatures and high pressure. The rotation of the positive and negative charges
separated from each other, together with the matter of the rotating planet, can
serve as a source of the observed magnetic field [21-26]. An additional effect
occurs if tidal forces from the Sun or the moons of the planets are taken into
account [27].
Of particular
interest is calculation of the magnetic field of those neutron stars that have
an electric charge like a proton [28-31], and at the same time rotate so fast
that the velocity of the surface of the stars becomes comparable to the speed
of light. In this case, even those terms in the formulas that contain
and
in the denominator cease to be small and
require their own accounting. From the relationship between the charge density,
angular velocity, and radius of the star with the observed magnetic field, one
can obtain an estimate of the charge that such stars could have. In addition,
it becomes possible to calculate the energy and angular momentum of the
electromagnetic field of a neutron star with increased accuracy.
The
formulas obtained for the magnetic field of the ball, given their simple
appearance and sufficiently high accuracy, can be used for calibration of
magnetic field standards and sensors in engineering and applied research. If
necessary, these formulas can easily take into account the magnetic
permeability of the substance of the ball itself and the
magnetic permeability of the medium surrounding the ball.
2. Method
Let us assume
that there is charged matter continuously distributed in a certain volume in
the Cartesian reference frame. Also we will assume
that this matter is in stationary motion in the planes parallel to the plane
,
but it does not move in the direction of the axis
.
This is possible, for example, when the charged matter rotates about the axis
.
Another example is a body that is motionless as a whole, in which electric
currents distributed over the volume carry the charge in the planes parallel to
the plane
.
For the sake of certainty, let us consider the first
case, which corresponds to the rotation of a spherical body uniformly charged
throughout its entire volume. With a uniform rotation of such a ball around its
axis, the vector potential
is independent of time and satisfies the
following equation [32-33], which follows from Maxwell's equations:
where
is the vector Laplace operator acting on the
vector potential
;
is the vacuum magnetic permeability;
is the magnetic permeability;
is the vacuum permittivity;
is the speed of light;
is the current density;
is the Lorentz factor of motion
of a charged element of matter at the observation point;
is the invariant charge density
of the element of matter;
is the velocity of motion of the element of
matter. We will further assume that
and both in the substance of the ball and in
the space outside the ball there are no magnetic moments of particles that
could lead to the magnetization of matter and to a change in the vector
potential and magnetic field.
If equation
(1) is considered in Cartesian coordinates, then the component
of the vector potential directed toward the
axis
, according
to [34], is equal to zero,
. This is
due to the condition that the velocity component
in (1) is equal to zero, since the elements of
charged matter move in planes perpendicular to the axis
.
Based on the
symmetry of the ball, the components of the vector potential should be
expressed in spherical three-dimensional coordinates
as
. Among all
possible solutions, we will limit ourselves to those in which the components
of the vector potential are independent of the
angle
of the observation point. In this case, all
solutions will be symmetrical with respect to the axis of the ball, which is associated with the axis
of the Cartesian coordinate system and has its
origin at the center of the ball.
Taking this
into account, we can write the following for the vector potential:
, (2)
where
,
,
are unit vectors of the spherical
coordinate system. In this case, the unit vector
is directed along the radial
coordinate
; the radius-vector
is directed from the center of
the spherical reference frame to the point of observation; the unit vector
is directed parallel to the line
of meridians (longitude line) in the direction opposite to the axis
, that is, towards the South pole; the unit vector
is directed parallel to the line
of parallels (latitude line) and counterclockwise, when viewed from the side of
the axis
. In (2) ,
the components of the vector potential
,
and
depend only on the radial
coordinate
and on the angle
.
In
general, the vector Laplace operator for a vector
with spherical components
in spherical coordinates has the
following form:
(3)
Replacing the vector
in (3) with the vector
from (2), we rewrite equation (1) in spherical
coordinates as follows:
(4)
In (4), it was taken into account that the Lorentz
factor on the right-hand side of (1) is expressed by the formula
,
and derivatives of the vector potential components
with respect to the angle
are zero. The velocity components of
a charged element of the ball's matter in spherical coordinates have the form
, where
is the angular velocity of the ball's
rotation;
is the radial coordinate, specifying
the distance from the origin at the center of the ball to the element of
matter;
is the angle between the axis
of the Cartesian coordinate system
and the radius vector , specifying the position of the element of matter.
In spherical coordinates, for an arbitrary scalar
function
, the Laplacian of this function for the case
has the form:
. (5)
We will use (5) to represent the Laplacians
,
and
in (4). The
components of the magnetic induction vector can be calculated through the curl
of the vector potential in spherical coordinates using the formula
(6)
3. Results
3.1 External vector potential
Similarly to (2), for the components of the external
vector potential we have:
. (7)
There is no charged matter outside the ball,
therefore, on the right-hand side of (1) we should assume
, so that the equation for the vector potential takes
the form
. In spherical coordinates, taking into account the
equality of the right-hand side of (4) to zero for
, we obtain the following:

(8)
Let us now take into account the fact that according
to (1) the component
of the vector
potential, directed towards the axis
and written in
Cartesian coordinates, is equal to zero,
. We will use the fact that the components of an
arbitrary vector in Cartesian coordinates can be expressed in terms of
components of the same vector in spherical coordinates. We proceed from the
relations between the unit vectors in Cartesian and spherical coordinates:
.
.
. (9)
.
.
. (10)
In Cartesian coordinates, the external vector
potential is written as follows:
. (11)
Let us equate
in (7) to
in (11) and use the expressions for the
unit vectors
,
,
from (10). As a result, for the Cartesian
components of the vector potential outside the rotating charged ball, the
following relations are obtained:
.
.
. (12)
Hence, from (12) with the condition
we obtain the relation
.
(13)
In order for vector equation (8) to be satisfied, it is necessary
that each parenthesis in (8) vanishes. In view of (13), this leads to three equations
. (14)
. (15)
. (16)
Relation (13) has led to the fact that in each equation in (14-16) there is only one of the
components of the external vector potential. Since the components
,
and
do not mix with
each other in the equations, this makes the solution of equations (14-16)
noticeably easier.
Let us substitute into (5) the components
.
,
instead of
and find the
Laplacians
,
,
, respectively, given that the components
.
,
do not depend
on the angle
. Then substituting
,
,
into (14-16), we
arrive at three partial differential equations for the vector potential
components:
. (17)
. (18)
. (19)
3.2 Calculation of component ![]()
In order to solve equation
(17), we assume that a specific solution for the vector potential component
can be represented as follows:
.
(20)
After substituting (20) into
(17) and separating variables, we have:
![]()
The left-hand side (21)
depends only on
, and
the middle part of the equality depends only on
. In this
connection, both parts of the equality, according to the standard procedure for
separating variables, are equated to some constant value
.
This yields two equations for
the functions
and
, expressed
through the parameter
:
. (22)
. (23)
Substituting into (22) a trial
solution of the form
,
where
is some constant, allows us to find two
possible values of the exponent
by solving a quadratic equation. This yields a
solution to equation (22), expressed in terms of
and two undetermined coefficients
and
:
. (24)
In order to find regular
solutions for
in (20), it is necessary to obtain integer
powers in (24). To do this, we should set
,
. (25)
, (26)
where
is an integer, and
;
and
are constant coefficients depending on
.
Equation (23), taking into
account (25), becomes dependent on
as a discrete parameter:
. (27)
Individual solutions of
equation (27) can be represented as
,
where
depend on
and are vector spherical polynomials for
,
are constant coefficients associated with the
corresponding polynomials
.
The first eight such vector
spherical polynomials have the following form:
,
,
.
,
.
.
.
. (28)
To obtain a polynomial
, it
is necessary to find a solution to equation (27) for a given
. We can
assume that each such solution depends on
. The
polynomials in (28) are not normalized and are given up to an arbitrary
constant coefficient. This is due to the fact that equation (27) is linear with
respect to any constant coefficient multiplied by the polynomials in (28), so
that each such coefficient is cancelled out as a result.
Assuming that the function
in (27) depends only on
, we find
.
. (29)
Substituting (29) into (27)
yields the equation
. (30)
Making the substitutions
,
in (30), we obtain:
. (31)
In (31) we replace the
function in the form
,
passing from
to the new function
. For the
derivatives of the function
and equation (31) as a whole, we find:
.

.
(32)
By
definition, associated Legendre polynomials
of degree
and order
are expressed in terms of Legendre polynomials
using the formula [17]:
. (33)
The first
eight Legendre polynomials are:
,
,
,
,
.
,
.
. (34)
Legendre
polynomials satisfy the Legendre equation [35]
, (35)
and are found
according to Rodrigues' formula [36]
. (36)
For the
associated Legendre polynomials
, the
differential equation and Rodrigues' formula for normalized solutions of this
equation of the following form are valid [17]:
. (37)
. (38)
A comparison of equations (32)
and (37) shows that for
and for
the equations have the same form, and the
solutions
in (32), up to the sign and constant
coefficients, are equal to the associated Legendre polynomials
. The first
eight such associated Legendre polynomials have the following form:
,
,
.
,
.
,
.
. (39)
To calculate
the polynomials
in (39), one can use formulas (33) taking into
account (34), or use (38) when
.
In (31-32) a substitution in
the form
was used. Replacing
with polynomials
at
, we obtain
up to constant coefficients
, where
are represented in (39). Making the
substitution
, we arrive
at
. Since
, it turns
out that the polynomials
in (28) up to the sign and constant
coefficients are expressed through the product of
and the associated Legendre polynomials
at order
and at
. The only
exception is the polynomial
in (28), for which the relation
is not satisfied.
The
general solution for the component
can be represented as a sum of partial
solutions (20):
. Substituting
(26) and
into this expression and taking into account
(28), we find
:
, (40)
where the constant coefficients
,
and
in order to distinguish them from each other
are given by two indices.
3.3 Calculation of
component ![]()
A specific
solution for the component
in (18) can be expressed through the product
of two functions:
.
(41)
We substitute (41) into (18)
and separate the variables:
.
(42)
As a consequence of (42), two
equations are obtained for the functions
,
, and the
parameter
:
. (43)
(44)
Regular solutions of equation
(43) similarly to (26) are obtained at
in the form:
, (45)
where
is an integer, and
;
and
are constant coefficients depending on
.
Substitution
into (44) leads to an equation in which the
function
will depend on both the angle
and the parameter
:
(46)
Solutions (46) depending on
can be represented as
, where
are vector spherical polynomials for
,
are constant coefficients associated with the
corresponding polynomials.
. The first
eight such polynomials have the following form:
,
,
.
,
.
.
.
. (47)
Assuming that the function
is a function of
, the
derivatives in equation (46) and the equation itself can be expressed as
follows:
.
.
(48)
Let's make a change of
variables in (48) in the form
and move from the function
to the function
:
.
(49)
Let us represent the function
as
. This gives
the following in (49):
.
.
(50)
Comparison of the last
equation (50) and equation (37) for
and for
yields the coincidence of the equations and
the equality of the functions
.
Consequently,
, where
are presented in (39). After substitution
, we have
. On the
other hand, in (46) we had
. Therefore,
up to the sign and constant coefficients, we obtain an expression for the
polynomials (47) in the form
, that is,
through the associated Legendre polynomials
of order
, in which
. However,
the first polynomial
presented in (47) is not determined in this
way.
The general solution for the
component
can be represented as the sum of specific
solutions (41):
.
Substituting
(45) and
into this expression, taking
into account
(47), we find
:
, (51)
where
,
and
are constant coefficients.
3.4 Calculation of
component ![]()
To solve equation (19) for the
vector potential component
, we
assume that a specific solution for the component
can be represented as
. This leads
to a separation of variables in (19) and to two equations for
and
:
.
. (52)
. (53)
Regular
solutions of equations (52-53) depend on
. The
solution of (52) is the expression
, (54)
where
is an integer, and
;
and
are constant coefficients depending on
.
Denoting a
specific solution of equation (53) as
, where
are constant coefficients associated with the
corresponding polynomials
, for vector
spherical polynomials
we obtain the following:
,
,
.
,
.
,
.
. (55)
In (55) the
first eight vector polynomials
for computing
are presented.
Assuming that
the function
is a function of
in the form
, we
calculate the derivatives in equation (53) and transform the equation itself:
.
.
. (56)
Denoting
, we
make a change of variables in (56), passing from
to
:
. (57)
A comparison
of equation (57) and equation (37) shows that
,
where the polynomials
are given in (39). Therefore, for
we have
. Since
, the
polynomials
in (55), up to a sign and constant
coefficients, are the associated Legendre polynomials
, in which
.
In (55) at
it is clear that
. In this
case, substituting the polynomial
into equation (37) instead of
at
and at
satisfies this equation. It turns out that
is not equal to
in (39), although for the remaining
polynomials at
up to the sign and constant coefficient the
equality holds
.
Summing up
all possible specific solutions
,
taking into account
(54) and the relation
, where
are presented in (55), we
find
:
, (58)
where
,
,
are constant coefficients.
3.5 Expressions for external vector potential and external
magnetic field
Considering
that in (6) the components of the vector potential do not depend on the angle
, the
following is obtained for the components of the external magnetic field:

(59)
Let us
substitute the polynomials
(28) into the expression
(40), using the first eight polynomials:

(60)
Similarly, we
substitute polynomials
(47) into the expression
(51), and also polynomials
(55) into the expression
(58):

(61)

(62)
From the
relationship between Cartesian and spherical coordinates we obtain the relation
.
Since the charged ball rotates about the axis
, it follows
from symmetry that when
replaced by
the components of the vector potential in
(60-62) must remain unchanged. Only
and other even powers of the form
satisfy this condition. Therefore, in (60-62)
we should choose the corresponding coefficients equal to zero in those terms
that contain
.
All this
gives the following:
(63)
(64)
(65)
We substitute
(63-65) into (59) and after differentiation with respect to spherical
coordinates we determine components of external magnetic field
,
and
:
(66)
(67)

(68)
The
expressions for the components of the vector potential in (63-65) and for the
components of the magnetic field in (66-68) can be simplified further. Indeed,
the vector potential and the magnetic field outside the ball cannot be directly
proportional to the radial coordinate
to avoid infinite values at large values of
. The
corresponding coefficients in the terms proportional to
must be zero. In addition, all terms in the
components of the vector potential and magnetic field must depend on
, otherwise
even at infinity there will be nonzero terms depending only on the angle
. Consequently, the coefficients
,
,
and
for the external field must
be zero. Taking this into account, we have:
(69)

(70)
(71)
(72)
(73)

(74)
3.6 Components of
vector potential inside ball
Let's take a
closer look at how the expression
for the velocity of a charged element of
matter is derived in spherical coordinates. The velocity
arises due to the rotation of the ball around
the axis
. The
Cartesian coordinates of the rotating element of matter are determined in terms
of spherical coordinates by the expressions:
,
,
. (75)
When the ball rotates, the radial coordinate
and angle
of an arbitrary particle of the ball remain
constant. If we denote the angular velocity of rotation as
, then the
components of the velocity, taking into account (75), will be equal to:
,
,
.
(76)
Using (76),
we express the velocity through Cartesian unit vectors:
. (77)
Similarly, in
spherical coordinates, velocity can be represented in terms of spherical unit
vectors:
. (78)
Replacing the
Cartesian unit vectors in (77) with unit vectors (9) and comparing with (78),
we find the velocity components in spherical coordinates:
.
(79)
According to
(79) the components of the velocity
and
are equal to zero, and the non-zero component
of the velocity at each point is directed
along the unit vector
, that is,
along the parallels of the corresponding spherical surface.
The vector
potential inside the ball is written similarly to (2):
. (80)
Due to
symmetry, in the case under consideration, the components of the vector
potential do not depend on the angle
.
From (4)
follows the equation for the components of the vector potential inside the
ball:
(81)
According to
(1), the component of the vector potential in Cartesian coordinates directed
along the axis
must be equal to zero. This leads to relation
(13), which will also be valid for the components of the vector potential
inside the ball in the following form:
.
(82)
Taking into
account (82), three equations for the components of the vector potential follow
from (81):
. (83)
. (84)
. (85)
Equations
(83-84), as well as equation (85) with the zero right-hand side, have the same
form as equations (14-16). Therefore, for solutions of equations (83-84), as
well as equation (85) with the zero right-hand side, we can use previously
found solutions of equations (14-16).
The general
solutions of equations (14-16) are presented in (63-65). We replace the
coefficients
in (63-65) with coefficients
and write down the corresponding solutions of
equations (83-84), as well as equation (85) with the
zero right-hand side:
(86)
(87)
(88)
In (88) the
quantity
is the general solution of equation (85) with the
zero right-hand side. Now we need to find a specific solution
of the equation with the non-zero right-hand
side in (85).
To determine
in (85) in spherical coordinates, we use
expression (5), in which
instead of
should be substituted taking into account that
and
do not depend on the angle
and therefore
. For the
quantity
according to (5) and (85), the following
equation follows:
. (89)
To find a specific
solution to the inhomogeneous equation (89), we will assume that
is a function of the new variable
, that is
. In this
case, it will be
.

.

(90)
Taking into
account (90), relation (89) is transformed into an equation for the function
:
. (91)
To find the
solution to equation (91), it is convenient to proceed as follows. We replace
with the function
and first consider the homogeneous equation
that is part of (91). This equation must correspond to some function
:
. (92)
Let us
substitute a trial solution in the form
into (92), assuming that
depends only on
. We obtain
two admissible values for the exponent
, so
and
turn out to be two
independent specific solutions of (92). Moreover,
, where
and
are arbitrary constant coefficients, will be
the general solution of (92).
Next, we
substitute
in (91) instead of
, and since
depends only on
, we replace
the partial derivatives with ordinary derivatives:
To find a specific solution to equation (93), we
use the variation of parameters method. To do this, we replace the constant
coefficients
and
with the functions
and
, to be determined, in the form
. (94)
The derivatives of the function
are equal to
.
. (95)
In (95), notations of the form
,
,
are used.
Substituting (94-95) into (93) yields:
(96)
In (96) we assume that
,
. (97)
.
(98)
. (99)
Relations (97) are satisfied because
and
are two independent specific solutions of the
homogeneous equation (92). Substituting
,
,
,
into (98-99) yields:
.
(100)
. (101)
Expressing
from (100) and substituting into (101), we
arrive at a first-order differential equation for determining
and solution of this equation:
. (102)
. (103)
From (100) and (102) it follows:
.
. (104)
From (94), (103-104) we find a specific solution
to equation (93):
. (105)
Let's transform the bracket in the third term on
the right-hand side into (105):
.
This yields in (105) the following:
. (106)
Taking into account the correspondence
and
in (91) and (93), and making the substitution
in (106), for a specific solution of the
inhomogeneous equation (89) we find:
. (107)
The complete
solution of equation (85) will then be equal to the sum
. Taking
into account (88) and (107), we find:

(108)
In the
expressions for the components of the vector potential (86-87) and (108), the
coefficients of those terms that contain
,
,
etc. in the denominator should be equated to
zero. This is necessary so that the components of the vector potential do not
turn to infinity inside the body at
. Taking
this into account, we have:
(109)
(110)
(111)
In (110-111)
the terms containing
and
were retained, which depend on the angle
but do not depend on
. Similar
terms
and
in (64-65) in the components of the vector
potential outside the ball were equated to zero based on the fact that at
infinity the vector potential in any case vanishes. But inside the ball, due to
the limited size of its dimensions, the radial coordinate cannot be infinite.
In this case, the terms containing
and
are admissible.
3.7 Components of internal
magnetic field induction vector
Based on
formula (6) for the magnetic field induction in spherical coordinates, for the
components of the magnetic field inside a rotating ball, similar to (59), we
have:
(112)
Substituting
the components of the vector potential inside the ball (109-111) into (112) we
find the components of the internal magnetic field:

(113)

(114)

(115)
3.8 Clarification of components
of vector potential and magnetic field induction vector
We transform
in (111), using the expansion of the following
function in the form
,
(116)
and then
expressing
only through
:

(117)
Similarly, we
express
in (71) only through
:

(118)
On the
surface of a rotating charged ball of radius
, the
external and internal vector potentials must coincide and transform into each
other. In this connection, at an arbitrarily chosen angle
and at
, we equate
the components of the external vector potential (69-70) to the components of
the internal vector potential (109-110), and also equate (117) and (118). Considering terms with identical
dependencies on
and
, we obtain
the following equalities:
,
,
,
.
,
,
,
.
,
,
.
(119)

(120)
(121)
. (122)
. (123)
From (120-123) we find the following:
,
.
.
. (124)
Thus, the
constant coefficients in the components of the external vector potential are
expressed through the coefficients in the components of the internal vector
potential.
Let us now
compare the components of magnetic field on the surface of the ball at
. Analogously
to (116), we expand
in the form

(125)
and then we
substitute (125) into (113) and into (114). In this case, we
express
in (113) only through
, and
in (114) only through
:

(126)

(127)
Similarly, we
express
in (72) through powers of
, and also
in (73) through powers of
:

(128)

(129)
Comparison of
(126) and (128), (127) and (129), (74) and (115) at
gives the relationships between the
coefficients:


.
.


.
. (130)
,
,
,
.
,
,
.
,
. (131)
In (130) the coefficients can be separated from each other:
.
,
.
. (132)
,
.
,
. (133)
By
substituting the coefficients (132-133) into the relations (124), we can verify that these
relations are satisfied.
In order for
the values of the coefficients in (119) to coincide with the values of the coefficients found in (131), it is necessary to assume that
,
,
,
,
,
. (134)
Taking into
account (119) and (132-134), we can clarify the vector potential
components (109-111) and the magnetic field components (113-115) inside the
rotating charged ball:
(135)
.
(136)
(137)
(138)
(139)
(140)
In the same
way, taking into account (119) and (132-134), we can clarify the vector potential
components (69-71) and the magnetic field components (72-74) outside the ball:
(141)
.
(142)
(143)
(144)
(145)
(146)
Note that in
(143-145) there are terms with round
brackets containing quantities of the type
,
etc. These quantities appeared as a result of
taking into account the Lorentz factor
in the inhomogeneous Laplace equation (81) and
applying the procedure of comparing the external and internal field components
on the surface of the ball. As a result, the components of the external
magnetic field in (144-145) acquire an additional and
nonlinear dependence on the radius
of the ball and on the angular velocity
of rotation. Quantities of the type
also appeared in [4] when calculating the
field components using retarded potentials, but for a different reason. related
to integration over the volume of the ball.
According to
(136) and (142), the expressions for the component of the internal vector potential
and for the component of the external vector
potential
are identical in form. However, the component
in (136) at
and at an angle
tends to infinity, which should not be the
case for a ball. Therefore, in (119), it is necessary to choose
,
. In this
case, instead of (136) and (142), the following will be obtained:
,
. (147)
If we take
into account (116) in (137), we get the following:

(148)
In (148) the component
at
and at the angle
turns to infinity, which is not permissible
for a ball. Consequently, the coefficients must satisfy the condition
. (149)
Substituting
(149) into (137) and into (143) gives the following:
(150)
(151)
Now note that
all terms
in (140) contain
as a factor. It is known that
. It turns
out that if on the globe we lay off an angle
from the north pole towards the equator and it
turns out
to be positive, then if we lay off the same
angle
from the south pole towards the equator,
getting an angle
, the value
will be negative.
A
contradiction arises: when a charged ball rotates, currents arise from the
moving charges of the ball, which are all directed in one direction, determined
by the angular velocity of rotation. In this case, it should be expected that
the component
directed along the parallels of the ball
should also be symmetrical relative to the plane of the equator and have the
same sign everywhere. But according to (140), the signs of the component
in the northern and southern hemispheres are
opposite. From the point of view of magnetic lines, the following picture is
obtained: in one of the hemispheres, the magnetic lines pass inside the ball,
go outside and then go to the equator, twisting relative to the axis
of rotation of the ball. After the magnetic
field lines pass the equator, they begin to twist in the opposite direction,
returning to their original state inside the ball.
To avoid a
contradiction, we must assume that
.
This requires that the coefficients
,
,
and
. be equal to zero. This leads to changes in (135), (140-141), (146) and gives the following:
,
,
,
. (152)
From (138-152) the following is obtained:
,
.

(153)


.
(154)
,
.
(155)


.
(156)
The solutions
(153-154) for the vector potential and the magnetic field inside a rotating
charged ball are smooth and do not contain discontinuities and infinities.
These solutions contain functions such as
,
, and
associated Legendre polynomials
, which have
continuous derivatives of any order. The same can be said about the solutions
(155-156) for the vector potential and the magnetic field outside the rotating
charged ball.
4. Discussion
Our results can be compared with the results in [37],
where the magnetic field of axisymmetric current configurations was studied and
cases where circular currents flowed in a certain layer inside a stationary
body were considered. In this case, it was found, similarly to (153) and (155),
that the main component of the vector potential is expressed in terms of the
sum of terms in the form
where the functions
depend on the radial coordinate
and on the degree
of the associated Legendre
polynomials
.
In [2], the following was found inside a rotating
uniformly charged ball:
,
. (157)
. (158)
If we
decompose the square root in (154) by the formula (125), then it becomes clear
that the expressions for
and
coincide with expressions (157) in the first
approximation, given that in (157)
is the vacuum
permeability.
Similarly, if we decompose the expression
in (153) by the formula
(116), then in (153)
will coincide with (158) in the first
approximation. However, expressions (154) and (153) are much more accurate,
since they contain additional terms containing
in the denominator. In
addition, in [2], the Lorentz factor
of moving charges was not
taken into account when calculating the field.
In order to verify the obtained solutions for the
magnetic field induction
of a rotating ball, we substitute
these solutions into Maxwell's equation with a source in the form of a current
density
:
.
(159)
In spherical coordinates, equation (159),
taking into account the relations
,
,
,
, can be written as follows:
(160)
Considering that the components of the magnetic field
induction vector do not depend on the angle
, three equations follow from (160):
,
.
. (161)
There is no charged matter outside the ball,
, and for the external magnetic field (161) is
simplified:
.
(162)
The corresponding components of the magnetic field
induction inside the ball (154) and outside the ball
(156) exactly satisfy equations (161-162).
One of Maxwell's equations for the magnetic field
is that the divergence of this field
at any point in space is equal to zero. In spherical coordinates, it looks like
this:
. (163)
Equation
(163) is satisfied both for the magnetic field induction components (154)
inside the ball and for the magnetic field induction components (156) outside
the ball.
In curved
space-time, the equation for the electromagnetic field tensor
has the following form [32]:
, (164)
where
is
the charge four-current with covariant index;
is the invariant charge density;
is the four-velocity with covariant index;
is the Ricci tensor with mixed indices;
is the Riemann curvature
tensor.
When deriving
equation (164), it was assumed that, according to [38], for the signature
of the metric for the curvature tensor and the
Ricci tensor, the following relations are satisfied:
,
. (165)
If we use the
relations according to [39] for the curvature tensor and the Ricci tensor in
the form
,
, (166)
then the
equation for the tensor
is written as:
. (167)
The
components of the tensor
in the Cartesian frame of reference are
expressed in terms of the components of the electric
and magnetic
fields as follows:
. (168)
In special
relativity, space-time is not curved, so the tensors
and
are zeroed. In this case, the covariant
derivatives
are transformed into partial derivatives
with respect to four-dimensional coordinates
. The
four-velocity with covariant index becomes equal to
, where
is the Lorentz factor,
is the velocity of a small element of charge.
Accordingly, the charge four-current with covariant index will be equal to
, (169)
where
is the current density.
All this,
taking into account (168-169), leads to a simplification of equations (164) and
(167):
. (170)
In (170), the
operator
is the d'Alembert operator, and the operator
is the Laplacian. With this in mind, it
follows from (170):
. (171)
Substituting
(168) into (171), taking into account the relation
,
gives two inhomogeneous wave equations, for electric field strength
and for magnetic field induction
:
.
. (172)
In the case
under consideration, the rotation of the charged body is constant, so that the
magnetic field
and the Lorentz factor
of the moving charges of the body are
independent of time. However, inside the body,
depends on the coordinates of the rotating
charges, which coincide with the coordinates of the observation point at which
the magnetic field is being searched. Therefore, the rotor operator also acts
on the Lorentz factor
. Then the
inhomogeneous Laplace equation for the magnetic field follows from (172).:
.
(173)
Equation
(173) is written in vector form and is therefore valid in any coordinate
system.
In spherical
coordinates, the Laplace vector operator
is expressed by formula (3), where instead of
the spherical components of the vector
, the
spherical components of the magnetic field vector
should be substituted. To
calculate the curl
, we use
(6), where instead of
we substitute the product
:

(174)
Inside a
rotating charged body, the velocity
has spherical components according to (79) in
the form
, so that
the components of velocity
and
in (174) are zero. All this, taking into
account (174), gives in (173) the following:
(175)
Outside the
body, where there is no current density
and the charge density
is zero, according to (173), the following
equation is obtained:
(176)
In (175-176)
it can be taken into account that for a uniformly rotating axisymmetric charged
body none of the components of the magnetic field depends on the angle
.
Substituting into (175) the expression for the velocity component
and the expression for the Lorentz factor
, we arrive
at three equations for the components of internal magnetic field:
.
.
.
(177)
The
right-hand side of the equations (177) contains the magnetic vacuum permeability
.
It is easy to
verify that, taking into account the expression for the Laplacian (5) in
spherical coordinates, the components of the external magnetic field (156)
satisfy the vector equation (176), and the components of the internal magnetic
field (154) satisfy the equations (177).
5. Conclusions
Using the harmonic method to find solutions to
equations for magnetic field of a rotating uniformly charged ball rotating
around its axis leads to solutions (153-156). These solutions yield formulas
for the components of the vector potential and the magnetic induction vector,
expressed in spherical coordinates.
In the course of the solution, the fact was used that
the vector potential of an axisymmetric rotating body does not depend on the
angle
of rotation. The use of symmetry
considerations when replacing
with
in solutions
makes it possible to exclude some of the undefined constant coefficients and
consider them equal to zero, which simplifies solutions. This is also due to
the fact that at infinity the potentials and fields should be zeroed, and in
the center of the rotating body they should not give infinity. Some of the
coefficients can also be excluded based on the condition that the external and
internal components of the vector potential and the magnetic field on the
surface of the body must be equal to each other.
With this in mind, we
calculated the components of the vector potential and the magnetic field inside
and outside a uniformly charged ball, rotating with a constant angular
velocity, with an accuracy of terms containing
in the denominator. The calculation method
allows us to explicitly find the components of fields with any given accuracy,
it is only necessary to use a sufficient number of vector spherical polynomials
(28),
(47) and
(55). In this
case, the polynomials
are expressed through the product
by the associated Legendre
polynomials
, the polynomials
are expressed through the product
by
, and the polynomials
are equal to
up to a constant coefficient.
The feature
we found is that the polynomials
,
and
of degree 0 have values that are not expressed
through the polynomial
. However,
the polynomial
for uniformly rotating axisymmetric bodies
turns out to be necessary, since it is present in the component
of the vector potential inside the body, as
indicated by the presence of
in the denominator of the first terms in
(153). This suggests that when calculating the vector potential, it is
insufficient to use the associated Legendre polynomials
and their products with
and
. Instead,
it is more appropriate to take into account the spherical polynomials
,
and
as specific, independent vector polynomials when
solving problems with vector functions in curved coordinates.
It should be noted the limitations that were used
in the presented method. The main assumptions include the constancy of angular
rotation of the ball around its axis and uniform distribution of electric
charge over the entire volume of the ball. In order to simplify calculations,
all possible solutions were limited to solutions in which the components
of the vector potential do not depend on the
angle
of the observation point and are symmetrical
with respect to the axis of rotation and the center of the ball.
The
calculations also assumed that the magnetic permeability of the substance of
the ball itself and the magnetic permeability of the medium surrounding the
ball are the same as in vacuum, that is
. If necessary, the formulas obtained for the vector potential and the
magnetic field can be easily adjusted to take into account the values of the
magnetic permeability and magnetic susceptibility of the substance of the ball
and the medium around it.
Declarations
Availability of data
and materials
All data generated or analysed during this study are included in this published
article.
Competing interests
The authors have no relevant
financial or non-financial interests to disclose.
Funding
No funds, grants, or other
support was received.
Authors'
contributions
Not applicable.
Acknowledgements
Not applicable.
Clinical trial
number
Not applicable.’
Ethical approval
Not applicable.
Consent to
participate
Not applicable.
Not applicable.
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