# Relativistic uniform system

Relativistic uniform system is an ideal physical system, in which the mass density (or any other physical quantity) depends on the Lorentz factor of the system’s particles, but is constant in the reference frames associated with the moving particles.

## Difference from the classical uniform system

In classical physics, the ideal uniform body model is widely used, in which the mass density is constant throughout the volume of the body or is given as the volume-averaged quantity. This model simplifies solution of physical problems and allows us to quickly estimate different physical quantities. For example, the body mass is calculated by simply multiplying the mass density by the body volume, which is easier than integrating the density over the volume in case of dependence of the density on the coordinates. The disadvantage of the classical model is that the majority of real physical systems are far from this ideal uniformity.

The use of the concept of the relativistic uniform system is based on the special theory of relativity (STR) and is the next step towards a more precise description of physical systems. In STR particular importance is given to invariant physical quantities, which can be calculated in each inertial reference frame and are equal to the values that these quantities have in the proper reference frame of the body. For example, multiplication of the invariant mass by the four-velocity gives the four-momentum of the body containing the invariant energy, and the multiplication of the corresponding invariant quantities by the four-velocity allows us in the case of motion of solid point particles to find the four-potentials of any vector fields and to develop their complete theory. [1] Another example is that for determination of the four-velocity or four-acceleration as a rule the operator of proper-time-derivative is used instead of the time derivative. Therefore, the use of the invariant mass density and charge density of the moving particles that make up the system does not only conform to the principles of STR but also significantly simplifies solution of the relativistic equations of motion.

## Field functions for the bodies of spherical shape

Field equations are most easily solved in case of spherical symmetry in the absence of general rotation of particles. In this case all the physical quantities depend only on the current radius, which starts at the center of the sphere. Below the solutions are provided of the equations for different fields in the framework of STR, including the scalar potentials, field strengths and solenoidal vectors. Due to the random motion of the particles in the system, the vector field potentials become equal to zero. This leads to zeroing of the solenoidal vectors of fields, including the magnetic field and the gravitational torsion field.

### Acceleration field

The four-potential ${\displaystyle ~U_{\mu }=\left({\frac {\vartheta }{c}},-\mathbf {U} \right)}$  of the acceleration field includes the scalar potential ${\displaystyle ~\vartheta }$  and the vector potential ${\displaystyle ~\mathbf {U} }$ . Applying the four-curl to the four-potential gives the acceleration tensor ${\displaystyle ~u_{\mu \nu }=\nabla _{\mu }U_{\nu }-\nabla _{\nu }U_{\mu }}$ . In the curved spacetime the acceleration field equation with the field sources is derived from the principle of least action: [1]

${\displaystyle ~\nabla ^{\nu }u_{\mu \nu }=-{\frac {4\pi \eta }{c^{2}}}J_{\mu }.}$

This equation after expressing the acceleration tensor ${\displaystyle ~u_{\mu \nu }}$  in terms of the four-potential turns into the wave equation for finding the four-potential of the acceleration field:

${\displaystyle ~\nabla ^{\nu }\nabla _{\mu }U_{\nu }-\nabla ^{\nu }\nabla _{\nu }U_{\mu }=-{\frac {4\pi \eta }{c^{2}}}J_{\mu },}$

which, taking into account the calibration condition of the four-potential ${\displaystyle ~\nabla ^{\mu }U_{\mu }=0}$ , can be transformed as follows:

${\displaystyle ~\nabla ^{\nu }\nabla _{\nu }U_{\mu }+R_{\mu \nu }U^{\nu }={\frac {4\pi \eta }{c^{2}}}J_{\mu },}$

where ${\displaystyle ~c}$  is the speed of light, ${\displaystyle ~\eta }$  is the acceleration field coefficient, ${\displaystyle ~J_{\mu }=g_{\mu \nu }J^{\nu }=g_{\mu \nu }\rho _{0}u^{\nu }}$  is the mass four-current with the covariant index, ${\displaystyle ~g_{\mu \nu }}$  is the metric tensor, ${\displaystyle ~R_{\mu \nu }}$  is the Ricci tensor, ${\displaystyle ~u^{\nu }}$  is the four-velocity, ${\displaystyle ~\rho _{0}}$  is the invariant mass density of the particles in the comoving reference frames, which is the same for all the particles.

In Minkowski spacetime within the framework of STR, the covariant derivatives of the form ${\displaystyle ~\nabla _{\mu }}$  turn into the partial derivatives of the form ${\displaystyle ~\partial _{\mu }}$ , while the result of the action of the partial derivatives does not depend on the order of their action. As a consequence of the calibration of the 4-potential, the equality holds: ${\displaystyle ~\partial ^{\nu }\partial _{\mu }U_{\nu }=\partial _{\mu }\partial ^{\nu }U_{\nu }=0}$ . As a result, the four-potential of the acceleration field can be found from the wave equation:

${\displaystyle ~\partial ^{\nu }\partial _{\nu }U_{\mu }={\frac {4\pi \eta }{c^{2}}}J_{\mu }.}$

This equation can be divided into two equations – one for the scalar potential and the other for the vector potential of the acceleration field. In this system under consideration the vector potential is equal to zero, and the scalar potential of the acceleration field is given by:

${\displaystyle ~\vartheta =cg_{0\mu }u^{\mu }=\gamma 'c^{2},}$

where ${\displaystyle ~g_{0\mu }}$  are the time components of the metric tensor, ${\displaystyle ~\gamma '}$  is the Lorentz factor of the particles in the reference frame K' associated with the center of the sphere.

Since the scalar potential of the stationary system does not depend on time, the wave equation for the scalar potential turns into the Poisson equation: [2]

${\displaystyle ~\triangle \vartheta =-4\pi \eta \rho _{0}\gamma '}$

and the following formula is obtained for the Lorentz factor of the particles: [3]

${\displaystyle ~\gamma '={\frac {c\gamma _{c}}{r{\sqrt {4\pi \eta \rho _{0}}}}}\sin \left({\frac {r}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\approx \gamma _{c}-{\frac {2\pi \eta \rho _{0}r^{2}\gamma _{c}}{3c^{2}}},\qquad \qquad (1)}$

where ${\displaystyle ~\gamma _{c}}$  is the Lorentz factor of the particles at the center of the sphere, ${\displaystyle ~r}$  is the current radius.

The acceleration field strength and the corresponding solenoidal vector are expressed by the formulas:

${\displaystyle ~\mathbf {S} =-\nabla \vartheta -{\frac {\partial \mathbf {U} }{\partial t}}={\frac {c^{2}\gamma _{c}\mathbf {r} }{r^{3}}}\left[{\frac {c}{\sqrt {4\pi \eta \rho _{0}}}}\sin \left({\frac {r}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)-r\cos \left({\frac {r}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\right]\approx }$
${\displaystyle ~\approx {\frac {4\pi \eta \rho _{0}\gamma _{c}\mathbf {r} }{3}}\left(1-{\frac {4\pi \eta \rho _{0}r^{2}}{10c^{2}}}\right).}$
${\displaystyle ~\mathbf {N} =\nabla \times \mathbf {U} =0.}$

### Pressure field

The four-potential ${\displaystyle ~\pi _{\mu }=\left({\frac {\wp }{c}},-\mathbf {\Pi } \right)}$  of the pressure field includes the scalar potential ${\displaystyle ~\wp }$  and the vector potential ${\displaystyle ~\mathbf {\Pi } }$ , and obeys the calibration condition: ${\displaystyle ~\nabla ^{\mu }\pi _{\mu }=0}$ .

The pressure field equation with the field sources, the pressure field tensor ${\displaystyle ~f_{\mu \nu }}$  and the equation for finding the four-potential of the pressure field have the form: [1]

${\displaystyle ~\nabla ^{\nu }f_{\mu \nu }=-{\frac {4\pi \sigma }{c^{2}}}J_{\mu },\quad f_{\mu \nu }=\nabla _{\mu }\pi _{\nu }-\nabla _{\nu }\pi _{\mu },\quad \nabla ^{\nu }\nabla _{\nu }\pi _{\mu }+R_{\mu \nu }\pi ^{\nu }={\frac {4\pi \sigma }{c^{2}}}J_{\mu },}$

where ${\displaystyle ~\sigma }$  is the pressure field coefficient.

In STR the latter equation turns into the wave equation:

${\displaystyle ~\partial ^{\nu }\partial _{\nu }\pi _{\mu }={\frac {4\pi \sigma }{c^{2}}}J_{\mu }.}$

In the stationary case the potentials do not depend on time and the time component of the wave equation turns into the Poisson equation for the scalar potential of the pressure field:

${\displaystyle ~\triangle \wp =-4\pi \sigma \rho _{0}\gamma '.}$

The solution of this equation inside the sphere with particles is as follows: [3]

${\displaystyle ~\wp =\wp _{c}-{\frac {\sigma c^{2}\gamma _{c}}{\eta }}+{\frac {\sigma c^{3}\gamma _{c}}{\eta r{\sqrt {4\pi \eta \rho _{0}}}}}\sin \left({\frac {r}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\approx \wp _{c}-{\frac {2\pi \sigma \rho _{0}r^{2}\gamma _{c}}{3}}.}$

where ${\displaystyle ~\wp _{c}}$  is the scalar potential at the center of the sphere. This potential is approximately equal to: [4]

${\displaystyle ~\wp _{c}\approx {\frac {3\sigma m}{10a}}\left(1+{\frac {9}{2{\sqrt {14}}}}\right),}$

where the acceleration field constant ${\displaystyle ~\eta }$  and the pressure field constant ${\displaystyle ~\sigma }$  are expressed by the formulas:

${\displaystyle ~\eta ={\frac {3}{5}}\left(G-{\frac {\rho _{0q}^{2}}{4\pi \varepsilon _{0}\rho _{0}^{2}}}\right),\qquad \qquad \sigma ={\frac {2}{5}}\left(G-{\frac {\rho _{0q}^{2}}{4\pi \varepsilon _{0}\rho _{0}^{2}}}\right).}$

The strength of the pressure field and the corresponding solenoidal vector are found as follows:

${\displaystyle ~\mathbf {C} =-\nabla \wp -{\frac {\partial \mathbf {\Pi } }{\partial t}}={\frac {\sigma c^{2}\gamma _{c}\mathbf {r} }{\eta r^{3}}}\left[{\frac {c}{\sqrt {4\pi \eta \rho _{0}}}}\sin \left({\frac {r}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)-r\cos \left({\frac {r}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\right]\approx }$
${\displaystyle ~\approx {\frac {4\pi \sigma \rho _{0}\gamma _{c}\mathbf {r} }{3}}\left(1-{\frac {4\pi \eta \rho _{0}r^{2}}{10c^{2}}}\right).}$
${\displaystyle ~\mathbf {I} =\nabla \times \mathbf {\Pi } =0.}$

### Gravitational field

The gravitational four-potential ${\displaystyle ~D_{\mu }=\left({\frac {\psi }{c}},-\mathbf {D} \right)}$  of the gravitational field is made up with the use of the scalar ${\displaystyle ~\psi }$  and vector ${\displaystyle ~\mathbf {D} }$  potentials. The calibration condition of the four-potential is: ${\displaystyle ~\nabla ^{\mu }D_{\mu }=0}$ .

The gravitational field equation with the field sources, the gravitational tensor ${\displaystyle ~\Phi _{\mu \nu }}$  and the equation for finding the four-potential of the gravitational field in the covariant theory of gravitation have the form: [5] [6]

${\displaystyle ~\nabla ^{\nu }\Phi _{\mu \nu }={\frac {4\pi G}{c^{2}}}J_{\mu },\quad \Phi _{\mu \nu }=\nabla _{\mu }D_{\nu }-\nabla _{\nu }D_{\mu },\quad \nabla ^{\nu }\nabla _{\nu }D_{\mu }+R_{\mu \nu }D^{\nu }=-{\frac {4\pi G}{c^{2}}}J_{\mu },}$

where ${\displaystyle ~G}$  is the gravitational constant.

In STR the latter equation is simplified and becomes the wave equation:

${\displaystyle ~\partial ^{\nu }\partial _{\nu }D_{\mu }=-{\frac {4\pi G}{c^{2}}}J_{\mu }.}$

From the wave equation in the stationary case, the Poisson equation follows for the scalar potential inside the sphere with randomly moving particles in the framework of the Lorentz-invariant theory of gravitation (LITG):

${\displaystyle ~\triangle \psi _{i}=4\pi G\rho _{0}\gamma '.}$

The right-hand side of this equation contains the Lorentz factor ${\displaystyle ~\gamma '}$ , which depends on the radius according to (1). In addition, the internal scalar potential near the surface of the sphere must coincide with the scalar potential of the external field of the system, in view of the standard potential gauge, that is with equality of the potential to zero at infinity.

As a result, the dependence of the scalar potential on the current radius differs from the dependence in the classical case of the uniform sphere with the radius ${\displaystyle ~a}$  and is equal to it only approximately: [3]

${\displaystyle ~\psi _{i}=-{\frac {Gc^{2}\gamma _{c}}{\eta r}}\left[{\frac {c}{\sqrt {4\pi \eta \rho _{0}}}}\sin \left({\frac {r}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)-r\cos \left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\right]\approx -{\frac {2\pi G\rho _{0}\gamma _{c}(3a^{2}-r^{2})}{3}}.}$

For the gravitational field strength and the gravitational torsion field inside the sphere we obtain the following: [7]

${\displaystyle ~\mathbf {\Gamma } _{i}=-\nabla \psi _{i}-{\frac {\partial \mathbf {D} _{i}}{\partial t}}=-{\frac {Gc^{2}\gamma _{c}\mathbf {r} }{\eta r^{3}}}\left[{\frac {c}{\sqrt {4\pi \eta \rho _{0}}}}\sin \left({\frac {r}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)-r\cos \left({\frac {r}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\right]\approx }$
${\displaystyle ~\approx -{\frac {4\pi G\rho _{0}\gamma _{c}\mathbf {r} }{3}}\left(1-{\frac {4\pi \eta \rho _{0}r^{2}}{10c^{2}}}\right).}$
${\displaystyle ~\mathbf {\Omega } _{i}=\nabla \times \mathbf {D} _{i}=0.}$

The solutions for the external gravitational field potential and for the field strength ${\displaystyle ~\Gamma _{o}}$  according to LITG are as follows:

${\displaystyle ~\psi _{o}=-{\frac {Gc^{2}\gamma _{c}}{\eta r}}\left[{\frac {c}{\sqrt {4\pi \eta \rho _{0}}}}\sin \left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)-a\cos \left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\right]\approx -{\frac {Gm\gamma _{c}}{r}}\left(1-{\frac {3\eta m}{10ac^{2}}}\right).}$
${\displaystyle ~\mathbf {\Gamma } _{o}=-\nabla \psi _{o}-{\frac {\partial \mathbf {D} _{o}}{\partial t}}=-{\frac {Gc^{2}\gamma _{c}\mathbf {r} }{\eta r^{3}}}\left[{\frac {c}{\sqrt {4\pi \eta \rho _{0}}}}\sin \left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)-a\cos \left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\right]\approx }$
${\displaystyle ~\approx -{\frac {Gm\gamma _{c}\mathbf {r} }{r^{3}}}\left(1-{\frac {3\eta m}{10ac^{2}}}\right).}$

Here, the auxiliary mass ${\displaystyle ~m}$  is equal to the product of the mass density ${\displaystyle ~\rho _{0}}$  by the volume of the sphere: ${\displaystyle ~m={\frac {4\pi \rho _{0}a^{3}}{3}}}$ . From the expressions for the potential and strength of the external gravitational field we can see that the role of the gravitational mass is played by the mass ${\displaystyle ~m_{g}\approx m\gamma _{c}\left(1-{\frac {3\eta m}{10ac^{2}}}\right).}$  Since ${\displaystyle ~\gamma _{c}>1}$  then the relation ${\displaystyle ~m_{g}>m}$  is satisfied.

To understand the difference between these masses we should calculate the total relativistic mass ${\displaystyle ~m_{b}}$  of the particles moving inside the sphere. For the motion of particles there should be some voids between them. Both the average accelerations and the average velocities of the particles inside the sphere are functions of the current radius. Dividing the particles’ velocities by their acceleration, we can find the dependence of the average period of the oscillatory motion of particles on the radius. Finally, multiplying the velocity by the average period of motion, we can obtain an estimate of the size of the voids between the particles.

In order to calculate the volume of the sphere, it is necessary to sum up the volumes of all the typical particles moving inside the sphere, as well as the volumes of the voids between them. Suppose now that the sizes of the typical particles are much larger than the voids between the particles, and the volume of the voids is substantially less than the total volume of the particles. In this case, we can use the approximation of continuous medium, so that the unit of the mass of matter inside the sphere will be given by the approximate expression ${\displaystyle ~dm\approx \rho _{0}\gamma 'dV}$ , where ${\displaystyle ~\rho _{0}}$  is the mass density in the reference frames associated with the particles, ${\displaystyle ~\gamma '}$  is the Lorentz factor of the moving particles, the product ${\displaystyle ~\rho _{0}\gamma '}$  gives the mass density of the particles from the viewpoint of an observer, who is stationary with respect to the sphere, and the volume element ${\displaystyle ~dV}$  inside the sphere corresponds to the volume of the particle from the viewpoint of this observer. This leads to the fact that the total volume of the particles moving inside the sphere becomes approximately equal to the volume of the sphere. For the mass, in view of the Lorentz factor (1), the following relation is obtained:

${\displaystyle ~m_{b}=\int dm=\int \rho _{0}\gamma 'dV={\frac {c^{2}\gamma _{c}}{\eta }}\left[{\frac {c}{\sqrt {4\pi \eta \rho _{0}}}}\sin \left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)-a\cos \left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\right]\approx }$
${\displaystyle ~\approx m\gamma _{c}\left(1-{\frac {3\eta m}{10ac^{2}}}\right).}$

This implies the equality of the gravitational mass ${\displaystyle ~m_{g}}$  and the total relativistic mass ${\displaystyle ~m_{b}}$  of the particles moving inside the sphere. The both masses are greater than the mass ${\displaystyle ~m}$ . By the method of its calculation, the mass ${\displaystyle ~m_{b}}$  is equal to the sum of the invariant masses of the particles that make up the system.

The external torsion field is equal to zero:

${\displaystyle ~\mathbf {\Omega } _{o}=\nabla \times \mathbf {D} _{o}=0.}$

### Electromagnetic field

The electromagnetic four-potential ${\displaystyle ~A_{\mu }=\left({\frac {\varphi }{c}},-\mathbf {A} \right)}$  of the electromagnetic field includes the scalar potential ${\displaystyle ~\varphi }$  and the vector potential ${\displaystyle ~\mathbf {A} }$ . The covariant Lorentz calibration for four-potential is: ${\displaystyle ~\nabla ^{\mu }A_{\mu }=0}$ . For a fixed uniformly charged spherical body with random motion of charges the total electromagnetic field on the average is purely electric and the vector potential is equal to zero.

The electromagnetic field equation with the field sources, the electromagnetic tensor ${\displaystyle ~F_{\mu \nu }}$  and the equation for finding the four-potential are expressed as follows:

${\displaystyle ~\nabla ^{\nu }F_{\mu \nu }=-{\frac {1}{\varepsilon _{0}c^{2}}}j_{\mu },\quad F_{\mu \nu }=\nabla _{\mu }A_{\nu }-\nabla _{\nu }A_{\mu },\quad \nabla ^{\nu }\nabla _{\nu }A_{\mu }+R_{\mu \nu }A^{\nu }={\frac {1}{\varepsilon _{0}c^{2}}}j_{\mu },}$

where ${\displaystyle ~\varepsilon _{0}}$  is the electric constant, ${\displaystyle ~j_{\mu }}$  is the electromagnetic four-current.

The latter equation in STR turns into the wave equation:

${\displaystyle ~\partial ^{\nu }\partial _{\nu }A_{\mu }={\frac {1}{\varepsilon _{0}c^{2}}}j_{\mu }.}$

Due to the absence of time-dependence in the case under consideration, the wave equation becomes the Poisson equation for the scalar potential ${\displaystyle ~\varphi _{i}}$  inside the sphere:

${\displaystyle ~\triangle \varphi _{i}=-{\frac {\rho _{0q}\gamma '}{\varepsilon _{0}}},}$

where ${\displaystyle ~\rho _{0q}}$  is the charge density in the reference frames associated with the charges.

The dependence of the scalar potential on the current radius in the general case differs from the dependence in the classical case of the potential of a uniformly charged sphere with the radius ${\displaystyle ~a}$ , coinciding with it only in the first approximation: [8]

${\displaystyle ~\varphi _{i}={\frac {\rho _{0q}c^{2}\gamma _{c}}{4\pi \varepsilon _{0}\eta \rho _{0}r}}\left[{\frac {c}{\sqrt {4\pi \eta \rho _{0}}}}\sin \left({\frac {r}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)-r\cos \left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\right]\approx {\frac {\rho _{0q}\gamma _{c}(3a^{2}-r^{2})}{6\varepsilon _{0}}}.}$

The field strength of the electric field and the magnetic field inside the sphere have the form:

${\displaystyle ~\mathbf {E} _{i}=-\nabla \varphi _{i}-{\frac {\partial \mathbf {A} _{i}}{\partial t}}={\frac {\rho _{0q}c^{2}\gamma _{c}\mathbf {r} }{4\pi \varepsilon _{0}\eta \rho _{0}r^{3}}}\left[{\frac {c}{\sqrt {4\pi \eta \rho _{0}}}}\sin \left({\frac {r}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)-r\cos \left({\frac {r}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\right]\approx }$
${\displaystyle ~\approx {\frac {\rho _{0q}\gamma _{c}\mathbf {r} }{3\varepsilon _{0}}}\left(1-{\frac {4\pi \eta \rho _{0}r^{2}}{10c^{2}}}\right).}$
${\displaystyle ~\mathbf {B} _{i}=\nabla \times \mathbf {A} _{i}=0.}$

Outside the system under consideration the charge density is equal to zero and the Poisson equation for the scalar potential turns into the Laplace equation:

${\displaystyle ~\triangle \varphi _{o}=0.}$

The solution for the external electric field potential, corresponding to the potential gauge and the Maxwell's equations for the electric field strength ${\displaystyle ~E_{o}}$  is given by:

${\displaystyle ~\varphi _{o}={\frac {\rho _{0q}c^{2}\gamma _{c}}{4\pi \varepsilon _{0}\eta \rho _{0}r}}\left[{\frac {c}{\sqrt {4\pi \eta \rho _{0}}}}\sin \left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)-a\cos \left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\right]\approx {\frac {q\gamma _{c}}{4\pi \varepsilon _{0}r}}\left(1-{\frac {3\eta m}{10ac^{2}}}\right).}$
${\displaystyle ~\mathbf {E} _{o}=-\nabla \varphi _{o}-{\frac {\partial \mathbf {A} _{o}}{\partial t}}={\frac {\rho _{0q}c^{2}\gamma _{c}\mathbf {r} }{4\pi \varepsilon _{0}\eta \rho _{0}r^{3}}}\left[{\frac {c}{\sqrt {4\pi \eta \rho _{0}}}}\sin \left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)-a\cos \left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\right]\approx }$
${\displaystyle ~\approx {\frac {q\gamma _{c}\mathbf {r} }{4\pi \varepsilon _{0}r^{3}}}\left(1-{\frac {3\eta m}{10ac^{2}}}\right).}$

The external magnetic field is equal to zero:

${\displaystyle ~\mathbf {B} _{o}=\nabla \times \mathbf {A} _{o}=0.}$

In these expressions, the charge ${\displaystyle ~q}$  is an auxiliary quantity equal to the product of the charge density ${\displaystyle ~\rho _{0q}}$  by the volume of the sphere: ${\displaystyle ~q={\frac {4\pi \rho _{0q}a^{3}}{3}}}$ .

In this case, the following quantity serves as the total charge of the system:

${\displaystyle ~q_{b}=\int \rho _{0q}\gamma 'dV={\frac {\rho _{0q}c^{2}\gamma _{c}}{\eta \rho _{0}}}\left[{\frac {c}{\sqrt {4\pi \eta \rho _{0}}}}\sin \left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)-a\cos \left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\right]\approx }$
${\displaystyle ~\approx q\gamma _{c}\left(1-{\frac {3\eta m}{10ac^{2}}}\right),}$

while ${\displaystyle ~q_{b}>q.}$  The charge ${\displaystyle ~q_{b}}$  is calculated in the same way as the mass ${\displaystyle ~m_{b}}$  and has the meaning of the sum of the charges of all the system’s particles.

## Tensor field invariants

The knowledge of the field strengths and the solenoidal components of fields allows us to find the tensor components of the corresponding fields with the covariant indices. To pass on to the field tensors with the contravariant indices we need to know the metric tensor. In STR the metric tensor does not depend on the coordinates and time, is uniquely defined, and in Cartesian coordinates consists of zeros and unities. As a result, it is easy to find the tensor field invariants ${\displaystyle ~u_{\mu \nu }u^{\mu \nu }}$ , ${\displaystyle ~f_{\mu \nu }f^{\mu \nu }}$ , ${\displaystyle ~\Phi _{\mu \nu }\Phi ^{\mu \nu }}$  and ${\displaystyle ~F_{\mu \nu }F^{\mu \nu }}$ , where ${\displaystyle ~u_{\mu \nu }}$ , ${\displaystyle ~f_{\mu \nu }}$ , ${\displaystyle ~\Phi _{\mu \nu }}$  and ${\displaystyle ~F_{\mu \nu }}$  are the acceleration tensor, the pressure field tensor, the gravitational tensor and the electromagnetic tensor, respectively.

The tensor field invariants are included in the Lagrangian, the Hamiltonian. the action function and the relativistic energy of the system, and they are located there inside the integrals over the space volume. In addition, they are included in the corresponding stress-energy tensors of the fields. [2] Since in the system under consideration the solenoidal vectors are zero, the tensor invariants depend only on the field strengths:

${\displaystyle ~u_{\mu \nu }u^{\mu \nu }=-{\frac {2}{c^{2}}}(S^{2}-c^{2}N^{2})=-{\frac {2}{c^{2}}}S^{2}.}$
${\displaystyle ~f_{\mu \nu }f^{\mu \nu }=-{\frac {2}{c^{2}}}(C^{2}-c^{2}I^{2})=-{\frac {2}{c^{2}}}C^{2}.}$
${\displaystyle ~\Phi _{\mu \nu }\Phi ^{\mu \nu }=-{\frac {2}{c^{2}}}(\Gamma ^{2}-c^{2}\Omega ^{2})=-{\frac {2}{c^{2}}}\Gamma ^{2}.}$
${\displaystyle ~F_{\mu \nu }F^{\mu \nu }=-{\frac {2}{c^{2}}}(E^{2}-c^{2}B^{2})=-{\frac {2}{c^{2}}}E^{2}.}$

The volume integrals of the tensor invariants multiplied by the corresponding factors were calculated in the article. [7] For the acceleration field and the pressure field the integrals are taken only over the volume of the sphere:

${\displaystyle ~\int \limits _{r=0}^{a}{\frac {c^{2}}{16\pi \eta }}u_{\mu \nu }u^{\mu \nu }dV=-{\frac {c^{4}\gamma _{c}^{2}}{2\eta }}\left[{\frac {a}{2}}+{\frac {c}{4{\sqrt {4\pi \eta \rho _{0}}}}}\sin \left({\frac {2a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)-{\frac {c^{2}}{4\pi \eta \rho _{0}a}}\sin ^{2}\left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\right]\approx }$
${\displaystyle ~\approx -{\frac {\eta m^{2}\gamma _{c}^{2}}{10a}}\left(1-{\frac {3\eta m}{7ac^{2}}}\right).}$
${\displaystyle ~\int \limits _{r=0}^{a}{\frac {c^{2}}{16\pi \sigma }}f_{\mu \nu }f^{\mu \nu }dV=-{\frac {\sigma c^{4}\gamma _{c}^{2}}{2\eta ^{2}}}\left[{\frac {a}{2}}+{\frac {c}{4{\sqrt {4\pi \eta \rho _{0}}}}}\sin \left({\frac {2a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)-{\frac {c^{2}}{4\pi \eta \rho _{0}a}}\sin ^{2}\left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\right]\approx }$
${\displaystyle ~\approx -{\frac {\sigma m^{2}\gamma _{c}^{2}}{10a}}\left(1-{\frac {3\eta m}{7ac^{2}}}\right).}$

The gravitational and electromagnetic fields of the system are present not only inside but also outside the sphere, where they extend to infinity, while the field strengths of the internal and external fields behave differently. The field strengths ${\displaystyle ~\mathbf {\Gamma } _{i}}$  and ${\displaystyle ~\mathbf {E} _{i}}$  are substituted respectively into the integrals of the tensor invariants of these fields taken over the volume of the sphere, which gives the following:

${\displaystyle ~-\int \limits _{r=0}^{a}{\frac {c^{2}}{16\pi G}}\Phi _{\mu \nu }\Phi ^{\mu \nu }dV=}$
${\displaystyle ~={\frac {Gc^{4}\gamma _{c}^{2}}{2\eta ^{2}}}\left[{\frac {a}{2}}+{\frac {c}{4{\sqrt {4\pi \eta \rho _{0}}}}}\sin \left({\frac {2a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)-{\frac {c^{2}}{4\pi \eta \rho _{0}a}}\sin ^{2}\left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\right]\approx {\frac {Gm^{2}\gamma _{c}^{2}}{10a}}\left(1-{\frac {3\eta m}{7ac^{2}}}\right).}$

${\displaystyle ~\int \limits _{r=0}^{a}{\frac {c^{2}\varepsilon _{0}}{4}}F_{\mu \nu }F^{\mu \nu }dV=}$
${\displaystyle ~=-{\frac {\rho _{0q}^{2}c^{4}\gamma _{c}^{2}}{8\pi \varepsilon _{0}\eta ^{2}\rho _{0}^{2}}}\left[{\frac {a}{2}}+{\frac {c}{4{\sqrt {4\pi \eta \rho _{0}}}}}\sin \left({\frac {2a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)-{\frac {c^{2}}{4\pi \eta \rho _{0}a}}\sin ^{2}\left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\right]\approx }$
${\displaystyle ~\approx -{\frac {q^{2}\gamma _{c}^{2}}{40\pi \varepsilon _{0}a}}\left(1-{\frac {3\eta m}{7ac^{2}}}\right).}$

Into the volume integrals of the tensor invariants of the gravitational and electromagnetic fields of the system outside the sphere the field strengths ${\displaystyle ~\mathbf {\Gamma } _{o}}$  and ${\displaystyle ~\mathbf {E} _{o}}$  are substituted, respectively:

${\displaystyle ~-\int \limits _{r=a}^{\infty }{\frac {c^{2}}{16\pi G}}\Phi _{\mu \nu }\Phi ^{\mu \nu }dV=}$
${\displaystyle ~={\frac {Gc^{4}\gamma _{c}^{2}}{2\eta ^{2}a}}\left[{\frac {c}{\sqrt {4\pi \eta \rho _{0}}}}\sin \left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)-a\cos \left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\right]^{2}\approx {\frac {Gm^{2}\gamma _{c}^{2}}{2a}}\left(1-{\frac {3\eta m}{5ac^{2}}}\right).}$

${\displaystyle ~\int \limits _{r=a}^{\infty }{\frac {c^{2}\varepsilon _{0}}{4}}F_{\mu \nu }F^{\mu \nu }dV=}$
${\displaystyle ~=-{\frac {\rho _{0q}^{2}c^{4}\gamma _{c}^{2}}{8\pi \varepsilon _{0}\eta ^{2}\rho _{0}^{2}a}}\left[{\frac {c}{\sqrt {4\pi \eta \rho _{0}}}}\sin \left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)-a\cos \left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\right]^{2}\approx -{\frac {q^{2}\gamma _{c}^{2}}{8\pi \varepsilon _{0}a}}\left(1-{\frac {3\eta m}{5ac^{2}}}\right).}$

## Energies of particles in the field potentials

All the four fields act on the particles inside the sphere, and therefore each particle of the system acquires the corresponding energy in a particular field. The energy of the particle in the field is calculated as the volume integral of the product of the effective mass density ${\displaystyle ~\rho =\rho _{0}\gamma '}$  by the corresponding scalar potential, and for the electric field the energy is determined as the volume integral of the product of the effective charge density ${\displaystyle ~\rho _{q}=\rho _{0q}\gamma '}$  by the scalar potential ${\displaystyle ~\varphi }$ , where the Lorentz factor ${\displaystyle ~\gamma '}$  from (1) is used. In STR the energies of particles in the acceleration field, pressure field, gravitational and electric fields in the uniform relativistic spherical system, in view of the expressions for the field potentials [7] and the corrections to calculations, [8] [9] [10] [11] are, respectively:

${\displaystyle ~\int \rho \vartheta dV=\rho _{0}c^{2}\int \gamma '^{2}dV={\frac {c^{4}\gamma _{c}^{2}}{\eta }}\left[{\frac {a}{2}}-{\frac {c}{4{\sqrt {4\pi \eta \rho _{0}}}}}\sin \left({\frac {2a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\right]\approx }$
${\displaystyle ~\approx mc^{2}\gamma _{c}^{2}-{\frac {3\eta m^{2}\gamma _{c}^{2}}{5a}}\left(1-{\frac {2\eta m}{7ac^{2}}}\right).}$

${\displaystyle ~\int \rho \wp dV=\rho _{0}\int \gamma '\wp dV={\frac {c^{2}\gamma _{c}}{\eta }}\left(\wp _{c}-{\frac {\sigma c^{2}\gamma _{c}}{\eta }}\right)\left[{\frac {c}{\sqrt {4\pi \eta \rho _{0}}}}\sin \left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)-a\cos \left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\right]+}$
${\displaystyle ~+{\frac {\sigma c^{4}\gamma _{c}^{2}}{\eta ^{2}}}\left[{\frac {a}{2}}-{\frac {c}{4{\sqrt {4\pi \eta \rho _{0}}}}}\sin \left({\frac {2a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\right]\approx m\wp _{c}\gamma _{c}\left(1-{\frac {3\eta m}{10ac^{2}}}\right)-{\frac {3\sigma m^{2}\gamma _{c}^{2}}{10a}}\left(1-{\frac {13\eta m}{28ac^{2}}}\right).}$
${\displaystyle ~\int \rho \psi _{i}dV=\rho _{0}\int \gamma '\psi _{i}dV=}$
${\displaystyle ~={\frac {Gc^{4}\gamma _{c}^{2}}{\eta ^{2}}}\cos \left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\left[{\frac {c}{\sqrt {4\pi \eta \rho _{0}}}}\sin \left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)-a\cos \left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\right]-}$
${\displaystyle ~-{\frac {Gc^{4}\gamma _{c}^{2}}{\eta ^{2}}}\left[{\frac {a}{2}}-{\frac {c}{4{\sqrt {4\pi \eta \rho _{0}}}}}\sin \left({\frac {2a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\right]\approx -{\frac {6Gm^{2}\gamma _{c}^{2}}{5a}}\left(1-{\frac {4\eta m}{7ac^{2}}}\right).}$
${\displaystyle ~\int \rho _{q}\varphi _{i}dV=\rho _{0q}\int \gamma '\varphi _{i}dV=}$
${\displaystyle ~=-{\frac {\rho _{0q}^{2}c^{4}\gamma _{c}^{2}}{4\pi \varepsilon _{0}\eta ^{2}\rho _{0}^{2}}}\cos \left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\left[{\frac {c}{\sqrt {4\pi \eta \rho _{0}}}}\sin \left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)-a\cos \left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\right]+}$
${\displaystyle ~+{\frac {\rho _{0q}^{2}c^{4}\gamma _{c}^{2}}{4\pi \varepsilon _{0}\eta ^{2}\rho _{0}^{2}}}\left[{\frac {a}{2}}-{\frac {c}{4{\sqrt {4\pi \eta \rho _{0}}}}}\sin \left({\frac {2a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)\right]\approx {\frac {3q^{2}\gamma _{c}^{2}}{10\pi \varepsilon _{0}a}}\left(1-{\frac {4\eta m}{7ac^{2}}}\right).}$

It should be noted that all the fields, which contain particles, are not the fields arising from the external sources, but are generated by the particles themselves. As a result, the particles’ energies calculated above in the scalar potentials of the fields are twice as large as the potential energy of one or another interaction. For example, in order to calculate the electrostatic energy of the system of two charges, it is sufficient to take the potential of the first charge at the location of the second charge and to multiply it by the value of the second charge. But if we use the formula for the energy in the form of an integral, then the electrostatic energy will be taken into account twice, because the term is added, which contains the potential of the second charge at the location of the first charge multiplied by the value of the first charge. On the other hand, the electrostatic energy must consist of two components that take into account both the energy of particles in each other’s fields and the energy of the electric field itself.

Instead, in electrostatics, the electrostatic energy is calculated either through the scalar potential or through the field strength by integrating the time component of the stress-energy tensor over the volume. Both methods provide the same result, but the connection between the field energy and the energy of particles in the potential is lost in this case, and it is not clear why these energies should coincide.

## Relation between the field coefficients

For the four fields under consideration the equation of motion of matter in the concept of the general field is as follows: [12] [13]

${\displaystyle ~u_{\mu \nu }J^{\nu }+f_{\mu \nu }J^{\nu }+\Phi _{\mu \nu }J^{\nu }+F_{\mu \nu }j^{\nu }=0,}$

where ${\displaystyle ~J_{\mu }}$  is the mass four-current, ${\displaystyle ~j^{\nu }}$  is the electromagnetic four-current.

The components of the field tensors are the field strengths and the corresponding solenoidal vectors, but in the physical system under consideration the latter are equal to zero. As a result, the space component of the equation of motion is reduced to the relation:

${\displaystyle ~\mathbf {S} +\mathbf {C} +\mathbf {\Gamma } _{i}+{\frac {\rho _{0q}}{\rho _{0}}}\mathbf {E} _{i}=0.}$

If we substitute here the expression for the field strengths inside the sphere, we obtain the relation between the field coefficients: [14]

${\displaystyle ~\eta +\sigma =G-{\frac {\rho _{0q}^{2}}{4\pi \varepsilon _{0}\rho _{0}^{2}}}=G-{\frac {q^{2}}{4\pi \varepsilon _{0}m^{2}}}.\qquad \qquad (2)}$

The same is obtained for the time component of the equation of motion, which leads to the generalized Poynting theorem. [9]

## Relation between the energies of the internal and external fields

In article [15] it was found that the energy of particles in the gravitational field inside the stationary sphere is up to a sign two times greater than the total energy associated with the tensor invariants of the gravitational field inside and outside the body. A similar situation takes place in the system under consideration with the random motion of particles and zero solenoidal vectors both for the gravitational [10] and electromagnetic fields. [8] In particular, we can write the following:

${\displaystyle ~\int \limits _{r=0}^{a}\rho \psi _{i}dV=2\int \limits _{r=0}^{a}{\frac {c^{2}}{16\pi G}}\Phi _{\mu \nu }\Phi ^{\mu \nu }dV+2\int \limits _{r=a}^{\infty }{\frac {c^{2}}{16\pi G}}\Phi _{\mu \nu }\Phi ^{\mu \nu }dV=2\int \limits _{r=0}^{\infty }{\frac {c^{2}}{16\pi G}}\Phi _{\mu \nu }\Phi ^{\mu \nu }dV.}$
${\displaystyle ~\int \limits _{r=0}^{a}\rho _{q}\varphi _{i}dV=-2\int \limits _{r=0}^{a}{\frac {c^{2}\varepsilon _{0}}{4}}F_{\mu \nu }F^{\mu \nu }dV-2\int \limits _{r=a}^{\infty }{\frac {c^{2}\varepsilon _{0}}{4}}F_{\mu \nu }F^{\mu \nu }dV=-2\int \limits _{r=0}^{\infty }{\frac {c^{2}\varepsilon _{0}}{4}}F_{\mu \nu }F^{\mu \nu }dV.}$

These expressions relate the energy of particles in the scalar field potentials with the energy found with the help of the field strengths.

## Relativistic energy

In the curved spacetime the system’s energy for the continuously distributed matter is given by the formula: [2] [4]

${\displaystyle ~E_{r}={\frac {1}{c}}\int {(\rho _{0}\vartheta +\rho _{0}\wp +\rho _{0}\psi +\rho _{0q}\varphi )u^{0}{\sqrt {-g}}dx^{1}dx^{2}dx^{3}+}}$
${\displaystyle ~+\int {\left({\frac {c^{2}}{16\pi \eta }}u_{\mu \nu }u^{\mu \nu }+{\frac {c^{2}}{16\pi \sigma }}f_{\mu \nu }f^{\mu \nu }-{\frac {c^{2}}{16\pi G}}\Phi _{\mu \nu }\Phi ^{\mu \nu }+{\frac {c^{2}\varepsilon _{0}}{4}}F_{\mu \nu }F^{\mu \nu }\right){\sqrt {-g}}dx^{1}dx^{2}dx^{3}}.\qquad \qquad (5)}$

This formula is valid in the case when we can assume that the potentials and strengths of the fields at each point of space do not depend on the velocities of the motion of the individual particles of the system.

In the STR the metric tensor determinant is ${\displaystyle ~g=-1}$ , the time component of the four-velocity is ${\displaystyle ~u^{0}=c\gamma '}$ , and in order to calculate the energy of the spherical system with particles, taking into account the fields’ energies, we can use the above-mentioned energies of particles in the field potentials and the energies in the form of the tensor invariants of the fields:

${\displaystyle ~E_{r}\approx mc^{2}\gamma _{c}^{2}-{\frac {3\eta m^{2}\gamma _{c}^{2}}{5a}}\left(1-{\frac {2\eta m}{7ac^{2}}}\right)+m\wp _{c}\gamma _{c}\left(1-{\frac {3\eta m}{10ac^{2}}}\right)-{\frac {3\sigma m^{2}\gamma _{c}^{2}}{10a}}\left(1-{\frac {13\eta m}{28ac^{2}}}\right)-}$
${\displaystyle ~-{\frac {6Gm^{2}\gamma _{c}^{2}}{5a}}\left(1-{\frac {4\eta m}{7ac^{2}}}\right)+{\frac {3q^{2}\gamma _{c}^{2}}{10\pi \varepsilon _{0}a}}\left(1-{\frac {4\eta m}{7ac^{2}}}\right)-{\frac {\eta m^{2}\gamma _{c}^{2}}{10a}}\left(1-{\frac {3\eta m}{7ac^{2}}}\right)-{\frac {\sigma m^{2}\gamma _{c}^{2}}{10a}}\left(1-{\frac {3\eta m}{7ac^{2}}}\right)+}$
${\displaystyle ~+{\frac {Gm^{2}\gamma _{c}^{2}}{10a}}\left(1-{\frac {3\eta m}{7ac^{2}}}\right)-{\frac {q^{2}\gamma _{c}^{2}}{40\pi \varepsilon _{0}a}}\left(1-{\frac {3\eta m}{7ac^{2}}}\right)+{\frac {Gm^{2}\gamma _{c}^{2}}{2a}}\left(1-{\frac {3\eta m}{5ac^{2}}}\right)-{\frac {q^{2}\gamma _{c}^{2}}{8\pi \varepsilon _{0}a}}\left(1-{\frac {3\eta m}{5ac^{2}}}\right).}$

The expression for the energy is simplified if we use the relation between the field coefficients (2):

${\displaystyle ~E_{r}\approx mc^{2}\gamma _{c}^{2}-{\frac {3\eta m^{2}\gamma _{c}^{2}}{5a}}\left(1-{\frac {2\eta m}{7ac^{2}}}\right)+m\wp _{c}\gamma _{c}\left(1-{\frac {3\eta m}{10ac^{2}}}\right)-{\frac {3\sigma m^{2}\gamma _{c}^{2}}{10a}}\left(1-{\frac {13\eta m}{28ac^{2}}}\right)-}$
${\displaystyle ~-{\frac {6Gm^{2}\gamma _{c}^{2}}{5a}}\left(1-{\frac {4\eta m}{7ac^{2}}}\right)+{\frac {3q^{2}\gamma _{c}^{2}}{10\pi \varepsilon _{0}a}}\left(1-{\frac {4\eta m}{7ac^{2}}}\right)+{\frac {Gm^{2}\gamma _{c}^{2}}{2a}}\left(1-{\frac {3\eta m}{5ac^{2}}}\right)-{\frac {q^{2}\gamma _{c}^{2}}{8\pi \varepsilon _{0}a}}\left(1-{\frac {3\eta m}{5ac^{2}}}\right).}$

Taking into account the relations between the energies of the internal and external fields also simplifies the expression for the system’s energy:

${\displaystyle ~E_{r}\approx mc^{2}\gamma _{c}^{2}-{\frac {3\eta m^{2}\gamma _{c}^{2}}{5a}}\left(1-{\frac {2\eta m}{7ac^{2}}}\right)+m\wp _{c}\gamma _{c}\left(1-{\frac {3\eta m}{10ac^{2}}}\right)-{\frac {3\sigma m^{2}\gamma _{c}^{2}}{10a}}\left(1-{\frac {13\eta m}{28ac^{2}}}\right)-}$
${\displaystyle ~-{\frac {6Gm^{2}\gamma _{c}^{2}}{10a}}\left(1-{\frac {4\eta m}{7ac^{2}}}\right)+{\frac {3q^{2}\gamma _{c}^{2}}{20\pi \varepsilon _{0}a}}\left(1-{\frac {4\eta m}{7ac^{2}}}\right)-{\frac {\eta m^{2}\gamma _{c}^{2}}{10a}}\left(1-{\frac {3\eta m}{7ac^{2}}}\right)-{\frac {\sigma m^{2}\gamma _{c}^{2}}{10a}}\left(1-{\frac {3\eta m}{7ac^{2}}}\right).}$

## Relation between the energy and the cosmological constant

In the approach under consideration, the relativistic energy of the system is not an absolute value and requires gauging. For this purpose the cosmological constant ${\displaystyle ~\Lambda }$  is used. The gauge condition for the four main fields is related to the sum of the products of the fields’ four-potentials by the corresponding four-currents and has the following form: [2] [4]

${\displaystyle ~-ck\Lambda =A_{\mu }j^{\mu }+(D_{\mu }+U_{\mu }+\pi _{\mu })J^{\mu },\qquad \qquad (6)}$

where for large cosmic systems ${\displaystyle ~-ck={\frac {c^{4}}{16\pi G\beta }}}$ , and ${\displaystyle ~\beta }$  is the constant of the order of unity.

Within the framework of the STR the gauge condition has the following form:

${\displaystyle ~-ck\Lambda =\gamma \rho _{0q}(\varphi -\mathbf {A} \cdot \mathbf {v} )+\gamma \rho _{0}(\psi -\mathbf {D} \cdot \mathbf {v} +\vartheta -\mathbf {U} \cdot \mathbf {v} +\wp -\mathbf {\Pi } \cdot \mathbf {v} ).}$

If we divide the system’s particles and remove them to infinity and leave there at rest, the terms with the products of the vector field potentials by the velocity of particles ${\displaystyle ~\mathbf {v} }$  would vanish, and the Lorentz factor of an arbitrary particle would be ${\displaystyle ~\gamma =1}$ . On the right-hand side we will have only the sum of the terms specifying the energy density of the particles located in the potentials of their proper fields. Since ${\displaystyle ~\vartheta \approx \gamma _{c}c^{2}}$ , we see that the cosmological constant for each system’s particle is up to the multiplier ${\displaystyle ~-ck}$  equal to the rest energy density of this particle with a certain addition from its proper fields. Then the integral over the volume of all the particles gives a certain energy:

${\displaystyle ~-ck\int \Lambda dV=m'c^{2},}$

where the gauge mass ${\displaystyle ~m'}$  is related to the gauge condition of the energy.

In the process of gravitational clustering the particles that were initially far from each other are united into closely bound systems, in which the field potentials increase manyfold. In the system under consideration ${\displaystyle ~\gamma =\gamma '}$ , the solenoidal vectors of the fields are considered equal to zero due to the random motion of the particles, which gives the following:

${\displaystyle ~m'c^{2}=\int [\gamma '\rho _{0q}\varphi _{i}+\gamma '\rho _{0}(\psi _{i}+\vartheta +\wp )]dV.}$

The expression on the right-hand side is part of the relativistic energy of the system, so that the energy can be written as follows:

${\displaystyle ~E_{r}=Mc^{2}\approx m'c^{2}-{\frac {\eta m^{2}\gamma _{c}^{2}}{10a}}\left(1-{\frac {3\eta m}{7ac^{2}}}\right)-{\frac {\sigma m^{2}\gamma _{c}^{2}}{10a}}\left(1-{\frac {3\eta m}{7ac^{2}}}\right)+}$
${\displaystyle ~+{\frac {Gm^{2}\gamma _{c}^{2}}{10a}}\left(1-{\frac {3\eta m}{7ac^{2}}}\right)-{\frac {q^{2}\gamma _{c}^{2}}{40\pi \varepsilon _{0}a}}\left(1-{\frac {3\eta m}{7ac^{2}}}\right)+{\frac {Gm^{2}\gamma _{c}^{2}}{2a}}\left(1-{\frac {3\eta m}{5ac^{2}}}\right)-{\frac {q^{2}\gamma _{c}^{2}}{8\pi \varepsilon _{0}a}}\left(1-{\frac {3\eta m}{5ac^{2}}}\right).}$

The mass ${\displaystyle ~M}$  is related to the relativistic energy of the generally stationary system and is the inertial mass of the system. In view of (2), the energy will be equal to:

${\displaystyle ~E_{r}=Mc^{2}\approx m'c^{2}+{\frac {Gm^{2}\gamma _{c}^{2}}{2a}}\left(1-{\frac {3\eta m}{5ac^{2}}}\right)-{\frac {q^{2}\gamma _{c}^{2}}{8\pi \varepsilon _{0}a}}\left(1-{\frac {3\eta m}{5ac^{2}}}\right).}$

This shows that the relativistic energy of this system is equal to the gauge mass-energy ${\displaystyle ~m'c^{2}}$ , from which the gravitational and electromagnetic energy of the fields outside the system should be subtracted.

## Lagrange function and motion integrals

The Lagrange function for a system of particles and four main vector fields has the following form:[1] [2]

${\displaystyle ~L=-\int {(U_{\mu }J^{\mu }+\pi _{\mu }J^{\mu }+D_{\mu }J^{\mu }+A_{\mu }j^{\mu }){\sqrt {-g}}dx^{1}dx^{2}dx^{3}+}}$
${\displaystyle ~+\int {\left(ckR-2ck\Lambda -{\frac {c^{2}}{16\pi \eta }}u_{\mu \nu }u^{\mu \nu }-{\frac {c^{2}}{16\pi \sigma }}f_{\mu \nu }f^{\mu \nu }+{\frac {c^{2}}{16\pi G}}\Phi _{\mu \nu }\Phi ^{\mu \nu }-{\frac {c^{2}\varepsilon _{0}}{4}}F_{\mu \nu }F^{\mu \nu }\right){\sqrt {-g}}dx^{1}dx^{2}dx^{3}}.}$

Here ${\displaystyle ~R}$  is the scalar curvature. With the help of such Lagrange function, one can calculate the generalized momentum of the system:[16]

${\displaystyle ~\mathbf {p} ={\frac {1}{c}}\int {(\rho _{0}\mathbf {U} +\rho _{0}\mathbf {\Pi } +\rho _{0}\mathbf {D} +\rho _{0q}\mathbf {A} )u^{0}{\sqrt {-g}}dx^{1}dx^{2}dx^{3}}.}$

This vector depends on the vector potentials of all four fields and is preserved in a closed physical system, that is, it is an integral of motion. Another integral of motion is the relativistic energy of the system ${\displaystyle ~E_{r}}$ , which is found by formula (5). Further, it is assumed that one can neglect the contributions from the gravitational and electromagnetic fields outside the matter and take into account only the generalized momentum. Then we can assume that these values form a four-momentum of the system, written with a covariant index:

${\displaystyle ~p_{\mu }=\left({\frac {E_{r}}{c}},-\mathbf {p} \right).}$

The angular momentum of the system is also an integral of motion:

${\displaystyle ~\mathbf {M} ={\frac {1}{c}}\int {(\rho _{0}[\mathbf {r} \times \mathbf {U} ]+\rho _{0}[\mathbf {r} \times \mathbf {\Pi } ]+\rho _{0}[\mathbf {r} \times \mathbf {D} ]+\rho _{0q}[\mathbf {r} \times \mathbf {A} ])u^{0}{\sqrt {-g}}dx^{1}dx^{2}dx^{3}}.}$

The antisymmetric angular momentum pseudotensor is determined through the four-radius ${\displaystyle ~x_{\mu }}$ , taken with a covariant index, and through the four-momentum ${\displaystyle ~p_{\mu }}$ :

${\displaystyle ~M_{\mu \nu }=\int {(x_{\mu }dp_{\nu }-x_{\nu }dp_{\mu })}.}$

The spatial components of the angular momentum pseudotensor ${\displaystyle ~M_{\mu \nu }}$  are the components of the angular momentum ${\displaystyle ~\mathbf {M} }$  of the system:

${\displaystyle ~M_{12}=-M_{21}=-M_{z},\qquad M_{13}=-M_{31}=M_{y},\qquad M_{23}=-M_{32}=-M_{x}.}$

The radius-vector of the center of momentum of a physical system is determined by the formula:

${\displaystyle ~\mathbf {R} _{m}={\frac {1}{cE_{r}}}\int {(\rho _{0}\vartheta +\rho _{0}\wp +\rho _{0}\psi +\rho _{0q}\varphi )\mathbf {r} u^{0}{\sqrt {-g}}dx^{1}dx^{2}dx^{3}+}}$
${\displaystyle ~+{\frac {1}{E_{r}}}\int {\left({\frac {c^{2}}{16\pi \eta }}u_{\mu \nu }u^{\mu \nu }+{\frac {c^{2}}{16\pi \sigma }}f_{\mu \nu }f^{\mu \nu }-{\frac {c^{2}}{16\pi G}}\Phi _{\mu \nu }\Phi ^{\mu \nu }+{\frac {c^{2}\varepsilon _{0}}{4}}F_{\mu \nu }F^{\mu \nu }\right)\mathbf {r} {\sqrt {-g}}dx^{1}dx^{2}dx^{3}}.}$

The time components of the pseudotensor ${\displaystyle ~M_{\mu \nu }}$  are the components of the three-dimensional vector ${\displaystyle ~\mathbf {\mathbb {C} } }$ , which is often called the time-varying dynamic mass moment:

${\displaystyle ~M_{01}=-M_{10}=-\mathbb {C} _{x},\qquad M_{02}=-M_{20}=-\mathbb {C} _{y},\qquad M_{03}=-M_{30}=-\mathbb {C} _{z}.}$

If we take into account definition of the radius-vector of the center of momentum and the relationship between the momentum and the velocity of the center of the momentum in the form ${\displaystyle ~\mathbf {p} ={\frac {E_{r}}{c^{2}}}\mathbf {V} }$ , we get the relationship:

${\displaystyle ~\mathbf {\mathbb {C} } ={\frac {E_{r}}{c}}(\mathbf {V} t-\mathbf {R} _{m}).}$

In a closed system the pseudotensor ${\displaystyle ~M_{\mu \nu }}$  must be conserved, and its components must be some constants. For the space components of the pseudotensor this results in conservation of the angular momentum: ${\displaystyle ~\mathbf {M} =const}$ . From the equality of the pseudotensor’s time components and the components of the vector ${\displaystyle ~\mathbf {\mathbb {C} } }$  it follows that it should be ${\displaystyle ~\mathbf {\mathbb {C} } =const}$ . Given the expression for ${\displaystyle ~\mathbf {\mathbb {C} } }$ , it can be written as ${\displaystyle ~\mathbf {R} _{m}=\mathbf {R} _{m0}+\mathbf {V} t}$ , where the constant vector ${\displaystyle ~\mathbf {R} _{m0}}$  specifies the position of the system’s center of momentum at ${\displaystyle ~t=0}$ . Thus, in this reference frame we obtain the equation of motion of the center of momentum at the constant velocity ${\displaystyle ~\mathbf {V} }$ , as a property of the motion of a closed system.

## Integral vector

The equation used to find the metric tensor components in the covariant theory of gravitation for the tensors with mixed indices has the following form:[2]

${\displaystyle ~R_{\alpha }^{\ \beta }-{\frac {1}{4}}R\delta _{\alpha }^{\ \beta }=-{\frac {1}{2ck}}\left(B_{\alpha }^{\ \beta }+P_{\alpha }^{\ \beta }+U_{\alpha }^{\ \beta }+W_{\alpha }^{\ \beta }\right).}$

here ${\displaystyle ~R_{\alpha }^{\ \beta }}$  is the Ricci tensor with mixed indices; ${\displaystyle ~\delta _{\alpha }^{\ \beta }}$  is the unit tensor or the Kronecker delta; ${\displaystyle ~B_{\alpha }^{\ \beta }}$ , ${\displaystyle ~P_{\alpha }^{\ \beta }}$ , ${\displaystyle ~U_{\alpha }^{\ \beta }}$  and ${\displaystyle ~W_{\alpha }^{\ \beta }}$  are the stress-energy tensors of the acceleration field and pressure field, gravitational and electromagnetic fields, respectively.

With the help of the covariant derivative ${\displaystyle ~\nabla _{\beta }}$  we can find the four-divergence of both sides of the above equation for the metric. The divergence of the left-hand side is zero due to equality to zero of the divergence of the Einstein tensor, ${\displaystyle ~\nabla _{\beta }\left(R_{\alpha }^{\ \beta }-{\frac {1}{2}}R\delta _{\alpha }^{\ \beta }\right)=0}$ , and also as a consequence of the fact that outside the body the scalar curvature vanishes, ${\displaystyle ~R=0}$ , and inside the body it is constant. The latter follows from the gauge condition of the energy of the closed system. The divergence of the right-hand side of equation for the metric is also zero:

${\displaystyle ~\nabla _{\beta }\left(B_{\alpha }^{\ \beta }+P_{\alpha }^{\ \beta }+U_{\alpha }^{\ \beta }+W_{\alpha }^{\ \beta }\right)=\nabla _{\beta }T_{\alpha }^{\ \beta }=0,}$

where the tensor ${\displaystyle ~T_{\alpha }^{\ \beta }}$  with mixed indices represents the sum of the stress-energy tensors of all the fields acting in the system.

The resulting expression for the tensors’ space components is nothing but the differential equation of the matter’s motion under the action of forces generated by the fields, which is written in a covariant form. [13] As for the tensors’ time components, for them the expression is expression of the generalized Poynting theorem for all the fields. [9]

In a weak field and at low velocities of motion of the particles, the equation ${\displaystyle ~\nabla _{\beta }T_{\alpha }^{\ \beta }\approx \partial _{\beta }T_{\alpha }^{\ \beta }=0}$  can be integrated over the four-volume, taking into account the divergence theorem. As a result, at the initial moment of time for the system under consideration, the following relation will be valid:

${\displaystyle ~J_{\alpha }=\int {T_{\alpha }^{\ 0}dx^{1}dx^{2}dx^{3}}=const.}$

In a closed system, the four-dimensional integral vector ${\displaystyle ~J_{\alpha }}$  must be constant. [16] For a stationary sphere with randomly moving particles in the continuous medium approximation, the energy fluxes of the fields defining the components ${\displaystyle ~T_{j}^{\ 0}}$ , where ${\displaystyle ~j=1,2,3}$ , are missing , so that the spatial components are zero, ${\displaystyle ~J_{j}=0}$ . As for the time component ${\displaystyle ~J_{0}}$  of the integral vector, then for the volume occupied by the matter inside the sphere, it also vanishes due to relation (4) for the field coefficients. However, outside the sphere, where there are only gravitational and electromagnetic fields, the time component of the integral vector is not equal to zero. As a result, the contribution to this component is made by the energies of the external fields:

${\displaystyle ~J_{0}=-{\frac {Gm_{g}^{2}}{2a}}+{\frac {q_{b}^{2}}{8\pi \varepsilon _{0}a}}.}$

It follows from the above that the integral vector shows the distribution of energy and energy fluxes in the system under consideration. For the nonzero space components ${\displaystyle ~J_{j}}$  of the integral vector to appear some stationary motion of the matter and fields is required, for example, general rotation, volume pulsations or mixing of matter. In this case, solenoidal vectors and the fields’ energy fluxes appear in the system.

Since the integral vector ${\displaystyle ~J_{\alpha }}$  is associated with the energies and energy fluxes of the fields in the energy-momentum tensors, it differs from the four-momentum ${\displaystyle ~p_{\mu }}$ , which includes the invariant mass and proportional to its rest energy. It turns out that the difference between ${\displaystyle ~J_{\alpha }}$  and ${\displaystyle ~p_{\mu }}$  is due to the fundamental difference between particles and fields, they cannot be reduced to each other, although they are interrelated with each other.

## Virial theorem and the kinetic energy of particles

In article [17] the kinetic energy of the particles of the system under consideration is estimated by three methods: from the virial theorem, from the relativistic definition of energy and using the generalized momenta and the proper fields of the particles. In the limit of low velocities, all these methods give for the kinetic energy the following:

${\displaystyle ~E_{k}\approx {\frac {0.3608\eta m^{2}\gamma _{c}}{a}}.}$

The possibility to use the generalized momenta to calculate the energy of the particles’ motion is associated with the fact that despite zeroing of the vector potentials and the solenoidal vectors on the large scale, in the volume of each randomly moving particle these potentials and vectors are not equal to zero. As a result, the energy of motion of the system’s particles can be found as the half-sum of the scalar products of the vector field potentials by the particles’ momentum, while for the electromagnetic field we should take not the momentum, but the product of the charge by the velocity and the Lorentz factor.

If we square the equation for ${\displaystyle ~\gamma '}$  in (1), we can obtain the dependence of the squared velocity of the particles’ random motion on the current radius:

${\displaystyle ~{v'}^{2}\approx v_{c}^{2}-{\frac {4\pi \eta \rho _{0}r^{2}}{3}}.}$

On the other hand, we can assume that ${\displaystyle ~\mathbf {v} '=\mathbf {v} _{r}+\mathbf {v} _{\perp },}$  where ${\displaystyle ~\mathbf {v} _{r}}$  denotes the averaged velocity component directed along the radius, and ${\displaystyle ~\mathbf {v} _{\perp }}$  is the averaged velocity component perpendicular to the current radius. In addition, from statistical considerations, it follows that

${\displaystyle ~{v'}^{2}=v_{r}^{2}+v_{\perp }^{2}=3v_{r}^{2}.}$

${\displaystyle ~v_{r}\approx {\frac {v_{c}}{\sqrt {3}}}\left(1-{\frac {2\pi \eta \rho _{0}r^{2}}{3v_{c}^{2}}}\right).}$

Next, from the virial theorem we find the squared velocity of the particles at the center of the sphere:

${\displaystyle ~v_{c}^{2}\approx {\frac {3\eta m}{5a}}\left(1+{\frac {9}{\sqrt {56}}}\right)\approx {\frac {1.3216\eta m}{a}}.}$

This makes it possible to estimate the Lorentz factor in the center:

${\displaystyle ~\gamma _{c}={\frac {1}{\sqrt {1-{\frac {v_{c}^{2}}{c^{2}}}}}}\approx 1+{\frac {v_{c}^{2}}{2c^{2}}}+{\frac {3v_{c}^{4}}{8c^{4}}}\approx 1+{\frac {3\eta m}{10ac^{2}}}\left(1+{\frac {9}{2{\sqrt {14}}}}\right)+{\frac {27\eta ^{2}m^{2}}{200a^{2}c^{4}}}\left(1+{\frac {9}{2{\sqrt {14}}}}\right)^{2}.}$

In the ordinary interpretation of the virial theorem the time-averaged kinetic energy of the system of particles must be two times less than the averaged energy associated with the forces ${\displaystyle ~\mathbf {F} _{i}}$  holding the particles at the radius-vectors ${\displaystyle ~\mathbf {r} _{i}}$  :

${\displaystyle ~\langle W_{k}\rangle _{m}=-0.5\langle \sum _{i=1}^{N}\mathbf {F} _{i}\cdot \mathbf {r} _{i}\rangle .}$

However, in the relativistic uniform system this equation is changed:

${\displaystyle ~\langle W_{k}\rangle \approx -0.6\langle \sum _{i=1}^{N}\mathbf {F} _{i}\cdot \mathbf {r} _{i}\rangle ,}$

while the quantity ${\displaystyle ~W_{k}}$  exceeds the kinetic energy of particles, ${\displaystyle ~W_{k}\approx \gamma _{c}E_{k}}$ , and it becomes equal to it only in the limit of low velocities.

In contrast to the classical case, the total time derivative of the virial in the stationary system is other than zero due to the virial’s dependence on the radius:

${\displaystyle ~{\frac {dG_{V}}{dt}}\approx \mathbf {v} \cdot \nabla {G_{V}}\approx {\frac {0.1216\eta m^{2}\gamma _{c}^{2}}{a}}.}$

An analysis of the integral theorem of generalized virial makes it possible to find, on the basis of field theory, a formula for the root-mean-square speed of typical particles of a system without using the notion of temperature: [18]

${\displaystyle v_{\mathrm {rms} }=c{\sqrt {1-{\frac {4\pi \eta \rho _{0}r^{2}}{c^{2}\gamma _{c}^{2}\sin ^{2}{\left({\frac {r}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)}}}}}.}$

## Extreme objects

In formula (2) for the gravitational field strength ${\displaystyle ~\mathbf {\Gamma } _{o}}$  outside the body there is a quantity ${\displaystyle ~A=\sin \delta -\delta \cos \delta }$ , where ${\displaystyle ~\delta ={\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}}$ . As was shown in article, [11] at the value ${\displaystyle ~\delta =\delta _{0}=4.494}$  radians the gravitational field strength ${\displaystyle ~\mathbf {\Gamma } _{o}}$  vanishes and the gravitational acceleration disappears. Therefore, in real physical objects the following condition must hold: ${\displaystyle ~\delta ={\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}<\delta _{0}}$ . If the angle ${\displaystyle ~\delta }$  is increased, then the quantity ${\displaystyle ~A}$  would first increase, and then would begin to decrease and even change its sign. So, at ${\displaystyle ~\delta ={\frac {\pi }{2}}}$  we will have ${\displaystyle ~A=1}$ , at ${\displaystyle ~\delta =\pi }$  we will have ${\displaystyle ~A=\pi }$ , at ${\displaystyle ~\delta ={\frac {3\pi }{2}}}$  we will have ${\displaystyle ~A=-1}$ .

Let us now consider the observable Universe, which on a scale 100 Mpc or more can be considered as a relativistic uniform system. The total mass-energy density of the Universe is close to the critical value ${\displaystyle ~\rho _{c}\approx 10^{-26}}$  kg/m3 and the size of the Universe can be estimated as the Hubble length ${\displaystyle ~R_{H}=c/H_{0}\approx 10^{26}}$  m, where ${\displaystyle ~H_{0}}$  is the Hubble parameter.

Using the approximate equality ${\displaystyle ~\eta \approx {\frac {3}{5}}G}$  according to, [14] we find the value ${\displaystyle ~\delta _{U}={\frac {R_{H}}{c}}{\sqrt {4\pi \eta \rho _{c}}}\approx 1.7>{\frac {\pi }{2}}}$  radians. Since the angle ${\displaystyle ~\delta _{U}}$  is sufficiently large, then for modeling of the gravitational field of the Universe it is necessary to use refined formulas with sines and cosines. For example, if we take the size of the observable Universe equal to ${\displaystyle ~2.64R_{H}}$ , then we will have ${\displaystyle ~\delta _{U}=\delta _{0}}$ , and the gravitational field at the boundaries of the Universe will tend to zero. This is what we observe in the form of a large-scale cellular structure consisting of clusters of galaxies. The reason for the gravitation action weakening is assumed to be graviton scattering by the particles of the space medium. [19]

Another extreme object is a proton, in which the mass density in the entire volume changes approximately by 1.5 times. As a result, in the first approximation a proton is a relativistic uniform system. The proton radius ${\displaystyle ~r_{p}}$  is of the order of 0.873 fm, [20] and the average density is of the order of ${\displaystyle ~\rho _{p}=6\cdot 10^{17}}$  kg/m3. As a gravitational constant at the atomic level the strong gravitational constant ${\displaystyle ~G_{s}}$  should be used. An estimate of the quantity ${\displaystyle ~\delta }$  for a proton at ${\displaystyle ~\eta \approx {\frac {3}{5}}G_{s}}$  gives: ${\displaystyle ~\delta _{p}={\frac {r_{p}}{c}}{\sqrt {4\pi \eta \rho _{p}}}\approx 2.4<\delta _{0}}$  radians. This shows that a proton is an extreme object from the point of view of weakening of its gravitational field.

In article, [11] a method is provided for estimating the Lorentz factor of the matter’s motion at the center of a proton, which gives ${\displaystyle ~\gamma _{c}=1.9}$ . In addition, the radius of action of the strong gravitation in the matter with the critical mass density ${\displaystyle ~\rho _{c}\approx 10^{-26}}$  kg/m3 in the observable Universe is estimated: ${\displaystyle ~r_{G}<1.3\cdot 10^{7}}$  m. On a large scale in the Universe not the strong gravitation, but the ordinary gravitation is acting with the radius of action of the order of the Hubble length.

Let us suppose that ${\displaystyle ~r_{G}}$  corresponds to the radius of a certain black hole for the strong gravitation, calculated by the Schwarzschild formula: ${\displaystyle ~r_{G}={\frac {2G_{s}m}{c^{2}}}}$ . If the mass is ${\displaystyle ~m={\frac {4\pi \rho _{c}r_{G}^{3}}{3}}}$ , then for the radius of a black hole with such mass we obtain${\displaystyle ~r_{G}=c{\sqrt {\frac {3}{8\pi G_{s}\rho _{c}}}}=2.7\cdot 10^{6}}$  m, and the mass is ${\displaystyle ~m=8\cdot 10^{-7}}$  kg. The Schwarzschild formula admits a black hole for the strong gravitation at small mass of the order of the proton mass, large mass density and a radius smaller than the proton radius. In addition, substitution of the mass ${\displaystyle ~m}$  and the radius ${\displaystyle ~r_{G}}$  into the Schwarzschild formula formally corresponds to a black hole with a large radius and low density ${\displaystyle ~\rho _{c}}$ . However, for an external observer, such a black hole would rather correspond not to a black hole, but to an object, containing strongly rarefied hydrogen gas of the cosmic space. Similarly, the Metagalaxy with the radius of order of ${\displaystyle ~r_{H}}$  and the mass density ${\displaystyle ~\rho _{c}}$  is not a black hole, although it corresponds to the Schwarzschild formula for the ordinary gravitation. Hence, in accordance with the theory of infinite nesting of matter, the conclusion follows – at each level of matter the corresponding gravitation forms only one type of the most compact and stable object. So, at the level of nucleons a proton appears under the action of the strong gravitation, and at the level of stars the ordinary gravitation generates a neutron star. If we multiply the radius of a neutron star by the coefficient of similarity in size ${\displaystyle ~P=1.4\cdot 10^{19}}$ , which is equal to the ratio of the stellar radius to the proton radius, we will obtain the radius of the order of ${\displaystyle 1.7\cdot 10^{23}}$  m. This radius must correspond to a compact object of a neutron star-type at the level of metagalaxies, which can emerge under the action of gravitation at this matter level. In the first approximation, the gravitational constant for metagalaxies is determined with the help of the similarity theory: ${\displaystyle ~G_{M}={\frac {GPS^{2}}{\Phi }}=3\cdot 10^{-50}}$  m3•s–2•kg–1, where ${\displaystyle ~S=0.23}$  is the coefficient of similarity in velocities, ${\displaystyle ~\Phi =1.62\cdot 10^{57}}$  is the coefficient of similarity in mass.

By analogy with the case of a proton, a neutron star is also considered as a relativistic uniform system. For a star with the mass of 1.35 Solar masses, the radius ${\displaystyle ~R_{s}=12}$  km and the average density ${\displaystyle ~\rho _{s}\approx 3.7\cdot 10^{17}}$  kg/m3, at ${\displaystyle ~\eta \approx {\frac {3}{5}}G}$  we obtain the angle ${\displaystyle ~\delta _{s}={\frac {R_{s}}{c}}{\sqrt {4\pi \eta \rho _{s}}}\approx 0.546}$  radians. With this in mind, if we substitute into (3) the stellar mass instead of ${\displaystyle ~m_{b}}$  and the stellar radius instead of ${\displaystyle ~a}$ , we can estimate the Lorentz factor at the center of the star: ${\displaystyle ~\gamma _{cs}=1.04}$ . This allows us to estimate the temperature at the center of the star: ${\displaystyle ~T_{s}\approx 2.8\cdot 10^{11}}$  K, which is close enough to calculation of the temperature at the center of a newly formed star. [14]

Thus, the dependences of the gravitational field inside and outside the bodies in article [11] are in good agreement with the conclusions of the Le Sage’s theory of gravitation and the theory of Infinite Hierarchical Nesting of Matter, with the strong gravitation at the level of nucleons and with the concept of a dynamic force vacuum field in electrogravitational vacuum.

## Cosmological constant and scalar curvature

According to (6), outside the body, where the four-currents are equal to zero, the cosmological constant ${\displaystyle ~\Lambda }$  becomes equal to zero. In addition, the scalar curvature ${\displaystyle ~R}$  also becomes equal to zero. [4] Inside the body the relation ${\displaystyle ~R=2\Lambda }$  holds true, so that in the matter with higher density both the scalar curvature and the cosmological constant increase. These quantities can be calculated using (6) as the averaged values for typical particles of the physical system. For the cosmic space we obtain approximately the following: ${\displaystyle ~\Lambda _{0}\approx {\frac {16\pi G\rho _{0}}{c^{2}}}\approx 10^{-52}}$  m-2, where the average mass density is ${\displaystyle ~\rho _{0}\approx 2.7\cdot 10^{-27}}$  kg/m3.

A similar formula for a proton gives the following: ${\displaystyle ~\Lambda \approx {\frac {16\pi G\rho _{p}}{c^{2}}}\approx 2.2\cdot 10^{-8}}$  m-2. However, for a proton in the calculations we should use the strong gravitational constant ${\displaystyle ~G_{s}}$ . In this case, we find: ${\displaystyle ~\Lambda _{p}\approx {\frac {16\pi G_{s}\rho _{p}}{c^{2}}}\approx 5.1\cdot 10^{31}}$  m-2. The obtained value is almost 84 orders of magnitude greater than the value of the cosmological constant for the cosmic space. The difference between the cosmological constants for the cosmic space and for a proton is associated with the averaging procedure: the cosmological constant inside a proton is large, but in the cosmic space the matter containing protons, neutrons and electrons is very rarefied, the main place is occupied by the void, so that the cosmological constant averaged over the entire space becomes a small value. Thus one of the paradoxes of the general theory of relativity is solved, in which the cosmological constant is associated with zero vacuum energy and therefore it must be very large, but in fact the cosmological constant turns out to be a small value.

For the relativistic uniform system with four fields acting in it, the average value ${\displaystyle ~{\stackrel {-}{\Lambda }}}$  of the cosmological constant in the matter is constant and can be written as follows:

${\displaystyle ~-ck{\stackrel {-}{\Lambda }}={\frac {G\rho _{0}c^{2}\gamma _{c}}{\eta }}\cos \left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)-{\frac {\rho _{0q}^{2}c^{2}\gamma _{c}}{4\pi \varepsilon _{0}\eta \rho _{0}}}\cos \left({\frac {a}{c}}{\sqrt {4\pi \eta \rho _{0}}}\right)+\rho _{0}\wp _{c}-{\frac {\sigma \rho _{0}c^{2}\gamma _{c}}{\eta }}.}$

This expression can be simplified by using the scalar potential of the gravitational field ${\displaystyle ~\psi _{a}=-{\frac {Gm_{g}}{a}}}$  and the scalar potential of the electric field ${\displaystyle ~\varphi _{a}={\frac {q_{b}}{4\pi \varepsilon _{0}a}}}$  on the surface of the body at ${\displaystyle ~r=a}$  :

${\displaystyle ~-ck{\stackrel {-}{\Lambda }}\approx \rho _{0}\psi _{a}-{\frac {Gm\rho _{0}\gamma _{c}}{2a}}+\rho _{0}c^{2}\gamma _{c}+\rho _{0q}\varphi _{a}+{\frac {q\rho _{0q}\gamma _{c}}{8\pi \varepsilon _{0}a}}+\rho _{0}\wp _{c}.}$

## Field energy theorem

In a relativistic uniform system, the exact values of the strengths and potentials of all active fields are known. This allows us to check the field energy theorem for such a system and verify the theorem.[21] This theorem explains, in particular, why electrostatic energy can be calculated either through the field strength, included in the electromagnetic field tensor, or in another way, through the field potential.

The kinetic energy and potential energy of the field are defined as follows:

${\displaystyle ~E_{kf}=\int {A_{\alpha }j^{\alpha }{\sqrt {-g}}dx^{1}dx^{2}dx^{3}}.}$
${\displaystyle ~W_{f}={\frac {1}{4\mu _{0}}}\int {F_{\mu \nu }F^{\mu \nu }{\sqrt {-g}}dx^{1}dx^{2}dx^{3}}.}$

If we take the entire infinite volume both inside and outside the matter of the system, then in the framework of the special theory of relativity and in the absence of magnetic fields, these expressions are simplified:

${\displaystyle ~E_{kf}=\int \rho _{q}\varphi _{i}dV\approx {\frac {3q^{2}\gamma _{c}^{2}}{10\pi \varepsilon _{0}a}}\left(1-{\frac {4\eta m}{7ac^{2}}}\right).}$
${\displaystyle ~W_{fi}=\int \limits _{r=0}^{a}{\frac {c^{2}\varepsilon _{0}}{4}}F_{\mu \nu }F^{\mu \nu }dV\approx -{\frac {q^{2}\gamma _{c}^{2}}{40\pi \varepsilon _{0}a}}\left(1-{\frac {3\eta m}{7ac^{2}}}\right).}$
${\displaystyle ~W_{fo}=\int \limits _{r=a}^{\infty }{\frac {c^{2}\varepsilon _{0}}{4}}F_{\mu \nu }F^{\mu \nu }dV\approx -{\frac {q^{2}\gamma _{c}^{2}}{8\pi \varepsilon _{0}a}}\left(1-{\frac {3\eta m}{5ac^{2}}}\right).}$
${\displaystyle ~W_{f}=W_{fi}+W_{fo}\approx -{\frac {3q^{2}\gamma _{c}^{2}}{20\pi \varepsilon _{0}a}}\left(1-{\frac {4\eta m}{7ac^{2}}}\right).}$

By virtue of the field energy theorem, the following relation will be satisfied:

${\displaystyle ~E_{kf}+2W_{f}=0.}$

In the general case, the tensor invariant is expressed in terms of the square of the electric field strength and the square of the magnetic field induction: ${\displaystyle ~F_{\mu \nu }F^{\mu \nu }=-{\frac {2}{c^{2}}}(E^{2}-c^{2}B^{2})}$ . The field energy density is found through the time component of the energy-momentum tensor: ${\displaystyle ~W^{00}={\frac {1}{2}}(\varepsilon _{0}E^{2}+{\frac {1}{\mu _{0}}}B^{2})}$ . In electrostatics, when there are no magnetic fields and ${\displaystyle ~B=0}$ , the integral over the volume of the tensor invariant becomes proportional to the integral over the volume of the component ${\displaystyle ~W^{00}}$ . As a result, electrostatic energy can be calculated in different ways:

${\displaystyle ~U_{e}=E_{kf}+W_{f}\approx {\frac {3q^{2}\gamma _{c}^{2}}{20\pi \varepsilon _{0}a}}\left(1-{\frac {4\eta m}{7ac^{2}}}\right).}$

Besides:

${\displaystyle ~U_{e}=-W_{f}={\frac {1}{2}}E_{kf}=\int W^{00}dV.}$

## Binding energy

With the help of the covariant theory of gravitation the total energy, binding energy, energy of fields, pressure energy and the potential energy of the system consisting of particles and four fields is precisely calculated in the relativistic uniform model. [22] A noticeable difference is shown between the obtained results and the relations for simple systems in classical mechanics, in which the acceleration field and pressure field are not taken into account or the pressure is considered to be a simple scalar quantity. In this case the inertial mass of the massive system is less than the total inertial mass of the system’s parts.

## System mass

The article [23] shows that the relativistic uniform system with continuous matter distribution is characterized by five types of mass: the gauge mass ${\displaystyle ~m'}$  is related to the cosmological constant and represents the mass-energy of the matter’s particles in the four-potentials of the system’s fields; the inertial mass ${\displaystyle ~M}$ ; the auxiliary mass ${\displaystyle ~m}$  is equal to the product of the particles’ mass density by the volume of the system; the mass ${\displaystyle ~m_{b}}$  is the sum of the invariant masses (rest masses) of the system’s particles, which is equal in value to the gravitational mass ${\displaystyle ~m_{g}}$ . The relation for these masses is as follows:

${\displaystyle ~m'

## Solution of 4/3 problem

For the electromagnetic and gravitational fields, the 4/3 problem consists in inequality of mass-energy extracted from the energy of the field of a body at rest, and mass-energy resulting from the field momentum of the moving body. If such a body is a relativistic uniform system of spherical shape, then the mass-energy associated with electrostatic energy of the system is:

${\displaystyle ~m_{f}={\frac {E_{e}}{c^{2}}}\approx {\frac {3q^{2}\gamma _{c}^{2}}{20\pi \varepsilon _{0}ac^{2}}}.}$

The momentum of electromagnetic field of a moving sphere is calculated through the Pointing vector. If ${\displaystyle ~\gamma }$  is the Lorentz factor, and ${\displaystyle ~v}$  is the velocity of the sphere, then the field momentum inside and outside the sphere, as well as the total momentum are equal: [9]

${\displaystyle ~g_{pi}\approx {\frac {\gamma q^{2}\gamma _{c}^{2}v}{30\pi \varepsilon _{0}ac^{2}}}.}$
${\displaystyle ~g_{po}\approx {\frac {\gamma q^{2}\gamma _{c}^{2}v}{6\pi \varepsilon _{0}ac^{2}}}.}$
${\displaystyle ~g_{p}=g_{pi}+g_{po}\approx {\frac {\gamma q^{2}\gamma _{c}^{2}v}{5\pi \varepsilon _{0}ac^{2}}}.}$

From here is the mass-energy associated with the field momentum:

${\displaystyle ~m_{p}={\frac {g_{p}}{\gamma v}}\approx {\frac {q^{2}\gamma _{c}^{2}}{5\pi \varepsilon _{0}ac^{2}}}.}$

For mass-energies, a ratio describing the 4/3 problem is obtained:

${\displaystyle ~m_{p}={\frac {4}{3}}m_{f}.}$

If we consider the energy and momentum of the electromagnetic field only inside the sphere, or only outside the sphere, similar correlations are obtained for the corresponding mass-energies.

As indicated in the article, [9] the mass-energy mismatch is a consequence of the fact that the time components of the electromagnetic stress-energy tensor and their integrals over volume do not together form any four-vector. In contrast, the four-momentum of the system is a four-vector, so that the same inertial mass enters both the energy and the momentum of the system. On the other hand, the energy and momentum of the electromagnetic field are included only as components in the energy and momentum of the entire system under consideration, and therefore they themselves do not have to form a four-vector.

To calculate a four-momentum of the system, it is necessary to add energy and momentum of other fields operating in the system to the energy and momentum of the electromagnetic field. In addition to the electromagnetic field, the minimum set of fields of the system includes the acceleration field, the pressure field and the gravitational field, and therefore it is necessary to take into account their energy and momentum. In this case, inside the sphere, the sum of the energies of all fields found through tensor invariants and through the stress-energy tensors is zeroed out. The total energy flow and the total momentum of the fields inside the sphere are also zero, so that within the sphere, the 4/3 problem as applied to general field disappears. The equality to zero of the sum of the energies and the sum of the momenta of the fields inside the sphere with randomly moving particles is a consequence of the fact that the particles and fields have the opportunity to exchange energy and momentum with each other. As a result, contribution to the relativistic energy of the system is made only by the particle energies in the scalar potentials of the fields, and the energies of the electromagnetic and gravitational fields outside the sphere.

The 4/3 problem shows in particular why the energy and momentum of an electron and any other body cannot be reduced only to the action of its own electromagnetic field. Despite the fact that an electron has a maximum charge per unit mass and is extremely charged, there are other fields in the electron's matter, for example strong gravitation. These fields have their own energy and momentum, which contribute to the four-momentum of the electron.

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