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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:

 

.                                                (1)

 

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:

 

(21)

 

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:

 

.                                       (93)

 

 

 

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.

 

Consent to publish

Not applicable.

 

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