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 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 result, the equality holds: ${\displaystyle ~\partial ^{\nu }\partial _{\mu }u_{\nu }=\partial _{\mu }\partial ^{\nu }u_{\nu }=0}$, if we also take into account the continuity equation of the mass four-current in the form of ${\displaystyle ~\partial ^{\nu }u_{\nu }=0}$, this equality is valid in STR when ${\displaystyle ~\rho _{0}=const}$ . 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.

The field strength of the pressure field and the corresponding solenoid 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: [4] [5]

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

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: [6]

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

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: [7]

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

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. [6] 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 [6] and the corrections to calculations, [8] [9] [7] 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: [10]

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

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: [11]

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

The same is obtained for the time component of the equation of motion, which leads to the generalized Poynting theorem. [8]

Relation between the energies of the internal and external fields

In article [12] 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 [9] and electromagnetic fields. [7] 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]

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

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]

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

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:

${\displaystyle ~-ck\int \Lambda dV=M_{0}c^{2},}$

where the gauge mass ${\displaystyle ~M_{0}}$ 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. Despite this, the mass ${\displaystyle ~M_{0}}$ as a consequence of the energy gauging remains unchanged. 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:

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

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:

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

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:

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

This shows that the relativistic energy of this system is equal to the gauge mass-energy ${\displaystyle ~M_{0}c^{2}}$, from which the gravitational and electromagnetic energy of the fields outside the system should be subtracted.

Virial theorem and the kinetic energy of particles

In article [13] 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:

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

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:

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

References

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