The behavior of continuum within the framework of the Newton gravitation theory is described by motion and compatibility equations which have the following form in spherical coordinates [12] :

$\frac{\partial p}{\partial r}+\rho \frac{\text{d}\phi }{\text{d}r}+\rho \left(\frac{\partial v}{\partial t}+v\frac{\partial v}{\partial r}\right)=0,\frac{\partial }{\partial r}\left(\rho v\right)+\frac{2}{r}\rho v+\frac{\partial \rho }{\partial t}=0$ (3)

in which p, $\rho$ and v are pressure, density and velocity, whereas $\phi$ is the Newton gravitational potential that satisfies the Poisson equation

$\frac{1}{r^2}\frac{\text{d}}{\text{d}r}\left(r^2\frac{\text{d}\phi }{\text{d}r}\right)=4\pi G\rho$ (4)

In the general case, $\rho$ depends on r and t. However, we further use the special model of the continuum which consists of a system of isolated and not non-interacting particles such that the pressure is zero. In this case, it is natural to assume that the density which is initially uniform, i.e. $\rho \left(t=0\right)={\rho }_0$, does not depend on r in the process of gravitational collapse. Thus, we take $\rho =\rho \left(t\right)$. Using equations for the sphere mass for the current time and $t=0$

$m=\frac{4}{3}\pi R^3\left(t\right)\rho \left(t\right),m_0=\frac{4}{3}\pi R_0^3{\rho }_0$ (5)

and taking into account that $m=m_0$, we arrive at the following conditions:

$R\left(t\right)=R_0\sqrt[3]{{\rho }_0/\rho \left(t\right)}$ (6)

Now, we can integrate Equation (4). Applying Equations (1) and (5) to eliminate G, we can present the first integral of this equation as

$\frac{\text{d}\phi }{\text{d}r}=a^{2}r,a^2=\frac{r_g{c}^2}{2R^3\left(t\right)}$ (7)

and write the first equation of Equations (3) in the following form:

$\frac{\partial v}{\partial t}+v\frac{\partial v}{\partial r}=-a^{2}r$ (8)

As follows from Equations (7), a = a(t), i.e. in the process of collapse, the sphere radius decreases, the density increases and the gravitation force depends on time.

To obtain the solution in the first approximation that can be useful for the further more rigorous analysis, assume that $R=R_0$ and $\rho ={\rho }_0$ which means that the gravitation force does not depend on time and corresponds to the initial parameters of the sphere. Then, Equation (8) takes the form

$\frac{\partial v_0}{\partial t}+v_0\frac{\partial v_0}{\partial r}=-a_0^{2}r,a_0^2=\frac{r_g{c}^2}{2R_0^3}$ (9)

Assume that the solution of this equation is

$v\left(r,t\right)=rf\left(t\right)$ (10)

Then, Equation (9) yields

$f^{″}+f^2=-a_0^2,{(\cdot )}^{\prime }=\text{d}(\cdot )/\text{d}t$

This is the Riccati equation [13] whose solution allows us to find the velocity

$v\left(r,t\right)=-a_{0}r\tan \left(C_0{a}_0+a_{0}t\right)$ (11)

Using the initial condition $v\left(r,t=0\right)=0$, we finally get

$v\left(r,t\right)=-a_{0}r\tan \left(a_{0}t\right)$ (12)

Now, the second equation in Equations (3) takes the form

${\rho }^{\prime }-3\rho a_0\tan \left(a_{0}t\right)=0$

and has the following solution:

$\rho =\frac{B_0}{{\cos }^3\left(a_{0}t\right)}$ (13)

Using the initial condition $\rho \left(t=0\right)={\rho }_0$, we arrive at

$\rho =\frac{{\rho }_0}{{\cos }^3\left(a_{0}t\right)}$ (14)

Finally, Equation (6) allows us to determine the sphere radius

$R\left(t\right)=R_0\cos \left(a_{0}t\right)$ (15)

Collapse results in concentration of the sphere mass at a point and an infinitely high density. As follows from the foregoing solution, this happens if $t=t_c^{\left(0\right)}=\pi /2a_0$. Substituting $a_0$ from Equations (9), we get

$t_c^{\left(0\right)}=\frac{\pi R_0}{c}\sqrt{\frac{R_0}{2r_g}}$ (16)

Equations (12) and (15) allow us to obtain the following useful result:

$v\left(r,t\right)=r\frac{R^{\prime }\left(t\right)}{R_0}$ (17)

As an example, consider the sphere with the parameters analogous to Sun, i.e., $R_0=6.96\times {10}^8\text{m}$, $r_g=2960\text{m}$. For this sphere, Equation (16) yields $t_c^{\left(0\right)}=2500\sec$. Dependence $R\left(t\right)/R_0$ is shown in Figure 1 (line 1).

Now recall that the foregoing solution does not take into account that parameter a in Equation (8) depends on t which means that the gravitational force does not change in the process of the sphere collapse. To allow for this change, we use the following step-by-step procedure. Divide the collapse time $t_c^{\left(0\right)}$ into a system of increments $t_1,t_2,\cdots ,t_n$ and assume that the foregoing solution corresponding to $a=a_0$ is valid for $0\le t\le t_1$. Then, for $t=t_1$, we have the expressions following from Equations (12), (14) and (15), i.e.,

$v_0\left(r,t_1\right)=-a_{0}r\tan \left(a_0{t}_1\right),{\rho }^{\left(0\right)}\left(t_1\right)=\frac{{\rho }_0}{{\cos }^3\left(a_0{t}_1\right)},R\left(t_1\right)=R_1=R_0\cos \left(a_0{t}_1\right)$ (18)

For $t_1\le t\le t_2$, we change $R_0$ to $R_1$ and Equation (8) becomes

$\frac{\partial v_1}{\partial t}+v_1\frac{\partial v_1}{\partial r}=-a_1^{2}r,a_1^2=\frac{r_g{c}^2}{2R_1^3}$

The solution is specified by Equation (11), i.e.,

$v_1\left(r,t\right)=-a_{1}r\tan \left(C_1{a}_1+a_{1}t\right)$

Constant $C_1$ is determined from the following condition: $v_0\left(r,t_1\right)=v_1\left(r,t_1\right)$ in which $v_0\left(r,t_1\right)$ is specified by Equations (18). The result is

$C_1=\frac{1}{a_1}{\tan }^{-1}\frac{a_0\tan \left(a_0{t}_1\right)-a_1\tan \left(a_1{t}_1\right)}{a_1-a_0\tan \left(a_0{t}_1\right)\tan \left(a_1{t}_1\right)}$

The second equation in Equations (3) yields

$\frac{\text{d}{\rho }_1}{\text{d}t}-3{\rho }_1{a}_1\tan \left(C_1{a}_1+a_{1}t\right)=0$

The solution is similar to Equation (13), i.e.,

${\rho }_1=\frac{B_1}{{\cos }^3\left(C_1{a}_1+a_{1}t\right)}$

Constant $B_1$ can be found from the condition ${\rho }^{\left(0\right)}\left(t_1\right)={\rho }_1\left(t_1\right)$ in which ${\rho }^{\left(0\right)}\left(t_1\right)$ is specified by Equations (17). The result is

$B_1={\rho }_0\frac{{\cos }^3\left(C_1{a}_1+a_1{t}_1\right)}{{\cos }^3\left(a_0{t}_1\right)}$

The sphere radius for $t=t_2$ can be found from Equation (6) which takes the form $R_0^3{\rho }_0=R_2^3{\rho }_1\left(t_2\right)$. Finally, we get

$R_2=R_0\frac{\cos \left(C_1{a}_1+a_1{t}_2\right)}{\cos \left(C_1{a}_1+a_1{t}_1\right)}\cos \left(a_0{t}_1\right)$

The described step-by-step process proceeds until $R_n$ becomes zero which corresponds to the collapse time $t_c^{\left(n\right)}$. The results of calculation for the sphere with parameters of Sun are presented in Figure 1 for $n=20$ (line 2) and $n=80$ (line 3). The collapse times are $t_c^{\left(20\right)}=1880\sec$ and $t_c^{\left(80\right)}=1825\sec$.

The foregoing results allow us to conclude that the sequence of collapse times $t_c^n$ obtained by step-by-step calculation converges in the Newton gravitation theory to the collapse time $t_c^N$. To determine this time, return to Equation (8) and transform parameter $a^2$ using Equations (1) and (5) for $r_g$ and $m$. The result is $a^2=\left(4/3\right)\pi G\rho \left(t\right)$ and Equation (8) reduces to

$\frac{\partial v}{\partial t}+v\frac{\partial v}{\partial r}=-k_n^{2}r\rho \left(t\right),k_n^2=\frac{4}{3}\pi G$ (19)

Applying Equation (6), we get

$\rho \left(t\right)=\frac{{\rho }_0}{{\bar{R}}^3\left(t\right)},\bar{R}\left(t\right)=\frac{R\left(t\right)}{R_0}.$ (20)

Substituting this result in the second equation of Equations (3), we have

$\frac{\partial v}{\partial r}+\frac{2}{r}v-3\frac{{\bar{R}}^{\prime }}{\bar{R}}=0$ (21)

Now, assume that $v\left(r,t\right)$ is specified by equation similar to Equation (17), i.e.,

$v\left(r,t\right)=r\frac{{\bar{R}}^{\prime }\left(t\right)}{\bar{R}\left(t\right)}$ (22)

Substituting Equation (22) in Equation (21), we can conclude that this equation is satisfied identically. Finally, substituting Equations (20) and (22) in Equation (19), we arrive at

${\bar{R}}^{″}+\frac{k_n^2{\rho }_0}{{\bar{R}}^2}=0$

This is the well-known equation which describes the free-fall motion and has the following first integral [11] :

${\bar{R}}^{\prime }=\pm \sqrt{\left(C_1+\frac{2k_n^2{\rho }_0}{\bar{R}}\right)}$ (23)

For the collapse problem, we should take sign “−” in Equation (23). As follows from the foregoing analysis, in this case we have $v\left(t=0\right)=0$ which means that ${\bar{R}}^{\prime }\left(t=0\right)=0$ and $\bar{R}\left(t=0\right)=1$. Then, $C_1=-2k^2{\rho }_0$ and

${\bar{R}}^{\prime }=-k_n\sqrt{2{\rho }_0\left(\frac{1}{\bar{R}}-1\right)}$ (25)

Integration yields the following implicit expression for $\bar{R}\left(t\right)$:

$\frac{1}{2}{\sin }^{-1}\left(1-2\bar{R}\right)+\sqrt{\bar{R}\left(1-\bar{R}\right)}+\frac{\pi }{4}=k_{n}t\sqrt{2{\rho }_0}$ (26)

Determine the collapse time $t_c^{\left(N\right)}$ corresponding to the Newton theory. Taking into account that $\bar{R}\left(t=t_c^{\left(R\right)}\right)=0$, using Equations (10) and (19) and applying Equation (26), we get

$t_c^{\left(N\right)}=\frac{\pi }{2k_n\sqrt{2{\rho }_0}}=\frac{\pi R_0}{2c}\sqrt{\frac{R_0}{r_g}}$ (27)

Equations (22) and (25) allow us to determine the velocity as

$v=\frac{k_{n}r}{\bar{R}}\sqrt{2{\rho }_0\left(\frac{1}{\bar{R}}-1\right)}$

For $t=0$, we have $\bar{R}=1$ and $v=0$ which means that the static initial state exists for the collapse problem in the Newton theory.

As an example, consider as earlier the sphere with parameter of Sun. For this sphere, calculation yields $t_c^{\left(N\right)}=1766$ sec. Numerical solution of Equation (26) for $\bar{R}\left(t\right)$ is shown in Figure 1 (line 4)

It should be noted that the forgoing formulation of the collapse problem in the absence of pressure is not realistic since pressure can prevent the collapse. This scenario is supported by the existence of the known static solution of Equations (3) for the constant pressure which is

$p_0=\frac{2\pi }{3}G{R}^2{\rho }^2\left(1-\frac{r^2}{R^2}\right)$

Now return to Equation (23) and take sign “+”, i.e.,

${\bar{R}}^{\prime }=\sqrt{C_1+\frac{2k_n^2{\rho }_0}{\bar{R}}}$ (28)

This solution corresponds to the sphere expansion. The applied pressure-free model of continuum looks more realistic for the expansion problem than for the collapse problem considered above. For $t\to \infty$ we have $\bar{R}\to \infty$ and ${\bar{R}}^{\prime }\to 0$. Then, $C_1=0$ in Equation (28) and

${\bar{R}}^{\prime }\sqrt{\bar{R}}=k_n\sqrt{2{\rho }_0}$

Integration yields

$\frac{2}{3}{\bar{R}}^{3/2}=k_{n}t\sqrt{2{\rho }_0}+C$

Taking into account that $\bar{R}\left(t=0\right)=1$, we get $C=2/3$ and

$\bar{R}=\sqrt[3]{{\left(1+\frac{3k_{n}t}{2}\sqrt{2{\rho }_0}\right)}^2}=\sqrt[3]{{\left(1+t\sqrt{6\pi G{\rho }_0}\right)}^2}$ (29)

In the second part of this equation coefficient $k_n$ is substituted from Equation (19). Using Equations (20) and (26), we can find the density, i.e.,

$\rho =\frac{{\rho }_0}{{\left(1+t\sqrt{6\pi G{\rho }_0}\right)}^2}$ (30)

Finally, Equations (22) and (29) allow us to determine the velocity as

$v=\frac{2r\sqrt{6\pi G{\rho }_0}}{1+t\sqrt{6\pi G{\rho }_0}}$ (31)

Note that $v\left(t=0\right)\ne 0$, which means that the initial static state in the expansion problem does not exist. For $t\to \infty$ we have $\rho \to 0$, $v\to 0$ and $R\to \infty$ which corresponds to an infinite homogeneous static space.

The study is undertaken within the mechanical interpretation of general relativity as a phenomenological theory based on the traditional model of space as a homogeneous isotropic continuum whose physical microstructure is ignored. The collapse and the expansion problems are considered further for the special four-dimensional space which is Euclidean with respect to space spherical coordinates $r,\theta ,\phi$ and Riemannian with respect to the time coordinate only [11] . To introduce the special geometry, consider first the metric coefficients corresponding to the Newton gravitation theory [14] , i.e.

$g_{11}=1,g_{22}=r^2,g_{44}=1-\frac{r_g}{r}$

As can be seen, the space coefficients $g_{11}$ and $g_{22}$ belong to the Euclidean space, whereas the Riemannian space appears only in the time coefficient $g_{44}$ which is associated with gravitation. Indeed, in the absence of gravitation $r_g=0$ and the space becomes Euclidean. Thus, the space is Euclidean with respect to the space coordinates and is Riemannian with respect to the time coordinate only. The idea is to extend this result to the general case. Undertake the following virtual experiment. Assume that we observe in a traditional three-dimensional Euclidean space a solid body loaded with self-balanced forces inducing a certain stressed state inside the body in the absence of gravitation. As known, the general relativity equations contain differential operators acting on a metric tensor of a Riemannian space in the left-hand sides and the so-called energy-momentum tensor in the right-hand sides. The energy-momentum tensor, in general, consists of the stress tensor and kinematic terms [15] . Thus, the general relativity equations describe the Riemannian space which is induced inside the body by the stresses only. However, such situation is not possible in Riemannian geometry, the dimension of the Euclidean space ( $n_E$) in which the Riemannian space with $n_R$ dimensions can exist is $n_E=n_R\left(n_R+1\right)/2$ [16] . Taking $n_R=3$, we get $n_E=6$. However, the stresses solid is in a three-dimensional Euclidean space which means that in the absence of gravitation, the space is Euclidean, whereas gravitation induces the Riemannian space associated with time only. So, we arrive at the special model of the four-dimensional space in which the space is Euclidean with respect to space coordinates and Riemannian with respect to time. This model allows us to study the problems which cannot be solved within the traditional Riemannian space model. One of such problems is associated with gravitation in solids. For static problems in elastic solids, the system of general relativity equations includes three conservation equations which are actually the equilibrium equations. These equations contain, in general, six components of the stress tensor which cannot be found from three equations. In the theory of elasticity [17] , the equilibrium equations are supplemented with compatibility equations which require the stressed space to be Euclidean and do not exist in the Riemannian space. Thus, a paradoxical situation occurs, the gravitational stresses in solids which are traditionally found in the classical solid mechanics cannot be determined in the theory of relativity. In the proposed space model, the space is Euclidean with respect to space coordinates, compatibility equations exist and the problem can be solved [18] . Moreover, this model allows us to overcome some principal problems of general relativity in the traditional Riemannian space. As known [2] [3] [14] , the system of the field equations is not complete in this space and should be supplemented with additional coordinate conditions. In the proposed model the number of the unknown metric coefficients is less than in the general Riemannian space and no additional conditions are required. This property of equations is demonstrated further for a spherically symmetric problem.

The line element for the proposed four-dimensional space in spherical coordinates has the following form [11] :

$\text{d}{s}^2=\text{d}{r}^2+r^2\text{d}{\Omega }^2+2g_{14}c\text{d}r\text{d}t-g_{44}{c}^2\text{d}{t}^2,\text{d}{\Omega }^2=\text{d}{\theta }^2+{\sin }^2\theta \text{d}{\phi }^2$ (32)

Here, $g_{14}$ $g_{44}$ depends on r and t. In contrast to traditional coordinate frames, space is not “orthogonal” to time in the four-dimensional space corresponding to Equation (32). The metric form in Equation (32) was derived in 1921-22 by Gullstrand and Painleve [19] [20] from the Schwarzschild form

$\text{d}{s}^2=g_{11}\text{d}{r}^2+r^2\text{d}{\Omega }^2-g_{44}{c}^2\text{d}{t}^2$ (33)

by coordinate transformation. Indeed, identical transformations in Equation (32) allow us to reduce it to

$\text{d}{s}^2=\left(1+\frac{g_{14}^2}{g_{44}}\right)\text{d}{r}^2+r^2\text{d}{\Omega }^2-g_{44}{\left(c-\frac{g_{14}}{g_{44}}\frac{\text{d}r}{\text{d}t}\right)}^2\text{d}{t}^2$ (34)

In static problems, $\text{d}r/\text{d}t=0$ and Equation (34) reduces to Equation (33) in which

$g_{11}=1+\frac{g_{14}^2}{g_{44}}$

In dynamic problems, the derivative $\text{d}r/\text{d}t$ is not known and direct reduction of Equation (32) to Equation (33) is not possible. Here, Equation (32) is not the results of coordinate transformation and follows from the proposed model of a four-dimensional space.

For the line element in Equation (32), the Einstein field equations are

$E_1^1=-\frac{1}{r^2{g}^2}\left[g\left(r{g^{\prime }}_{44}-g_{14}^2\right)+2r{g}_{44}{\stackrel{\dot{}}{g}}_{14}-r{g}_{14}{\stackrel{\dot{}}{g}}_{44}\right]$ (35)

$\begin{array}{c}{E}_2^2=\frac{1}{4r{g}^2}(4g_{14}{g}_{44}{g^{\prime }}_{14}-4g_{14}^2{g^{\prime }}_{44}-2g_{44}{g^{\prime }}_{44}-2r{g}_{44}{g^{″}}_{44}+r{g^{\prime }}_{44}{}^2-2r{g}_{14}^2{g^{″}}_{44}\\ +2r{g}_{14}{g^{\prime }}_{14}{g^{\prime }}_{44}-4g_{44}{\stackrel{\dot{}}{g}}_{14}+2g_{14}{\stackrel{\dot{}}{g}}_{44}+4r{g}_{14}{g^{\prime }}_{14}{\stackrel{\dot{}}{g}}_{14}+2r{g^{\prime }}_{14}{\stackrel{\dot{}}{g}}_{44}-4rg{{\stackrel{\dot{}}{g}}^{\prime }}_{14})\end{array}$ (36)

$E_4^4=\frac{1}{r^2{g}^2}\left(r{g}_{14}{g^{\prime }}_{44}-g{g}_{14}-2r{g}_{44}{g^{\prime }}_{14}\right)=\frac{1}{r^2}\frac{\partial }{\partial r}\left(\frac{r{g}_{14}^2}{g}\right)$ (37)

$E_1^4=-\frac{g_{14}}{r{g}^2}\left(2g_{14}{g^{\prime }}_{14}+{g^{\prime }}_{44}\right)=\frac{g_{14}}{r}\frac{\partial }{\partial r}\left(\frac{1}{g}\right)$ (38)

$E_4^1=-\frac{g_{14}}{r{g}^2}\left(2g_{44}{\stackrel{\dot{}}{g}}_{14}-g_{14}{\stackrel{\dot{}}{g}}_{44}\right)=\frac{g_{14}^2{g}_{44}}{r{g}^2}\frac{\partial }{c\partial t}\ln \frac{g_{14}}{g_{44}^2}$, $g=g_{44}+g_{14}^2$ (39)

Here, $(\cdot {)}^{\prime }=\partial (\cdot )/\partial r$ and $(\cdot \stackrel{\dot{}}{)}=\partial (\cdot )/c\partial t$. We use mixed tensor components because in spherical coordinates they coincide with physical components. Einstein tensor is proportional to the energy-momentum tensor, i.e.,

$E_i^j=\chi T_i^j,\chi =\frac{8\pi G}{c^4}$ (40)

In the absence of pressure, the energy momentum tensor has the following form [16] :

$T_1^1=\rho v_1{v}^1,T_2^2=T_3^3=0,T_4^4=\rho c^2,T_1^4=-\rho v_{1}c,T_4^1=\rho v^{1}c$ (41)

Consider briefly the empty space surrounding the sphere [11] for which $T_i^j=0$ and Equations (35)-(39) are homogeneous. Equation (39) in which $E_4^1=0$ allows us to conclude that $g_{14}$ and $g_{44}$ do not depend on time. This result is analogous to the Birkhoff theorem in the Schwarzschild solution [15] . Then, Equations (15) and (17) yield

$r{g^{\prime }}_{44}-g_{14}^2=0,g=g_{44}+g_{14}^2=C_1$

The solution of these equations is

${\left(g_{14}^e\right)}^2=\frac{C_1{C}_2}{r},g_{44}^e=C_2-\frac{C_1{C}_2}{r}$

Here, index “e” corresponds to the external space. Integration constants $C_1$ and $C_2$ can be found from the asymptotic condition according to which the obtained solution must reduce for $r\to \infty$ to the Newton theory, i.e.,

$g_{14}^e=0,g_{44}^e=1-\frac{r_g}{r}$

where $r_g$ is given by Equation (2). The final solution becomes

$g_{14}^e=\pm \sqrt{\frac{r_g}{r}},g_{44}^e=1-\frac{r_g}{r}$ (42)

As can be seen, the obtained solution specifies two spaces corresponding to signs “+” and “−” in the first of these equations. It is important for the further analysis that

$g_e=g_{44}^e+{\left(g_{14}^e\right)}^2=1$ (43)

Parameter g has a simple geometrical meaning, the determinant of the metric tensor in the four-dimensional space is $D=-r^{4}g{\sin }^2\theta$.

As can be proved [11] , the escape velocity for the sphere with radius R is specified by the following equation:

$v_e=c\sqrt{\frac{r_g}{R}}$ (44)

which means that for the sphere with radius $R=r_g$ the escape velocity $v_e=c$ and the sphere becomes invisible. Such object is analogous to a black hole, but the Schwarzschild singularity does not appear and the space inside the sphere is Euclidian with respect to space spherical coordinates.

Consider the internal space for the sphere with radius R. Equations (37), (40) and (41) yield

$\frac{\partial }{\partial r}\left(\frac{r{g}_{14}^2}{g}\right)=\chi \rho c^2$ (45)

As in the Newton theory, we assume that in the absence of pressure $\rho =\rho \left(t\right)$. The solution of Equation (45) is

$g_{14}^2=g\left(k_r\rho c^2{r}^2+\frac{C}{r}\right),k_r=\frac{1}{3}\chi c^2$ (46)

The integration constant $C=0$. Otherwise, $g_{14}$ becomes singular at the center of a sphere with any dimensions. Taking into account that $g=g_{44}+g_{14}^2$, we finally get

$g_{14}^2=kg\rho r^2,g_{44}=g\left(1-k\rho r^2\right)$ (47)

Consider Equations (39), (40) and (41) which yield

$\frac{g_{14}^2{g}_{44}}{r{g}^2}\frac{\partial }{c\partial t}\ln \frac{g_{44}}{g_{14}^2}=\chi \rho c{v}^1$

Substituting Equations (47) and taking into account Equation (29) for k, we arrive at

$v^1=-\frac{r}{3\rho }\frac{\text{d}\rho }{\text{d}t}$ (48)

Applying Equations (38), (40) and (21), we have the following equation

$\frac{g_{14}}{r}\frac{\partial }{\partial r}\left(\frac{1}{g}\right)=-\rho v_{1}c$

which yields

$v_1=\pm \frac{g^{\prime }}{cg}\sqrt{\frac{k_r}{\rho g}}$ (49)

Thus, Equations (47), (48) and (49) allow us to express $g_{14},g_{44},v^1$ and $v_1$ in terms of two functions $\rho \left(t\right)$ and $g\left(r,t\right)$.

Now, introduce the basic assumption of this analysis. For the sphere with radius R, the metric coefficients must be continuous and satisfy the following boundary conditions on the sphere surface:

$g_{14}\left(R\right)=g_{14}^e\left(R\right),g_{44}\left(R\right)=g_{44}^e\left(R\right)$ (50)

However, in the collapse problem, R depends on t. So, the right-hand parts of Equations (32) should be written as [3]

$g_{14}^e\left(R\right)=\pm \sqrt{\frac{r_g}{R\left(t\right)}},g_{44}^e\left(R\right)=1-\frac{r_g}{R\left(t\right)}$ (51)

As follows from Equation (43) and (51), $g=1$ for the external space and for the boundary of the internal space. Thus, it is natural to assume that $g=1$ for the whole internal space. Taking $g=1$ in Equations (47), (48) and (49), we get

$g_{14}^2=k_r\rho r^2,g_{44}=1-k_r\rho r^2,v^1=-\frac{r}{3\rho }\frac{\text{d}\rho }{\text{d}t},v_1=0$ (52)

As can be seen, at the sphere center $g_{14}=0$, $g_{44}=1$, the space is Euclidean and $v^1=v_1=0$.

Now, return to the field equations, Equations (35)-(39). To obtain the solution in Equations (52), we used Equations (37), (38) and (39). For one unknown function $\rho \left(t\right)$, we have two remaining equations, Equations (35) and (36). Substituting Equations (52) in Equations (35) and (36), we arrive at one and the same equation for density

$\frac{\text{d}\rho }{\text{d}t}\mp 3\rho c\sqrt{k\rho }=0$ (53)

Thus, the system of the field equation turns out to be complete and reduces to Equation (53). The solution of this equation which satisfies the initial condition $\rho \left(t=0\right)={\rho }_0$ is

$\rho \left(t\right)=\frac{{\rho }_0}{{\left(1\mp \frac{3}{2}ct\sqrt{k_r{\rho }_0}\right)}^2}$

Taking into account Equation (46) for $k_r$ and Equations (1), (5), (40) for $r_g,m_0,\chi$, we find

$k_r=\left(1/3\right)\chi c^2=8\pi G/3c^2=r_g/{\rho }_0{R}_0^3$

and finally arrive at

$\rho \left(t\right)=\frac{{\rho }_0}{{\left(1\mp t\sqrt{6\pi G{\rho }_0}\right)}^2}=\frac{{\rho }_0}{{\left(1\mp \frac{3ct}{2R_0}\sqrt{\frac{r_g}{R_0}}\right)}^2}$ (54)

The sphere radius follows from Equations (6) and (56) which yield

$R\left(t\right)=R_0\sqrt[3]{{\left(1\mp t\sqrt{6\pi G{\rho }_0}\right)}^2}=R_0\sqrt[3]{{\left(1\mp \frac{3ct}{2R_0}\sqrt{\frac{r_g}{R_0}}\right)}^2}$ (55)

Now check the boundary conditions. Taking $r=R$ in Equations (54) and using Equations (56), (57), we arrive at Equations (53) for the external space which means that the boundary conditions on the sphere surface are satisfied. The velocity follows from Equations (34) and (56), i.e.,

$v^1\left(r,t\right)=-\frac{2r\sqrt{6\pi G{\rho }_0}}{3\left(1\mp t\sqrt{6\pi G{\rho }_0}\right)}=-\frac{cr\sqrt{r_g/R_0}}{1\mp \frac{3ct}{2R_0}\sqrt{\frac{r_g}{R_0}}}$ (56)

As follows from Equations (55) and (56), sign “−” corresponds to the sphere collapse under gravitation forces, whereas the solution with sign “+” describes the sphere expansion which can start from the collapsed sphere under the action of internal forces the nature of which is not discussed here.

To study the collapse problem, we should take sign “−” in the forgoing solution. Then

$\rho \left(t\right)=\frac{{\rho }_0}{{\left(1-t/t_c^{\left(R\right)}\right)}^2},R\left(t\right)=R_0\sqrt[3]{{\left(1-t/t_c^{\left(R\right)}\right)}^2},v^1\left(r,t\right)=-\frac{2r}{t_c^{(R)}\left(1-t/t_c^{\left(R\right)}\right)}$ (59)

in which

$t_c^{\left(R\right)}=\frac{1}{\sqrt{6\pi G{\rho }_0}}=\frac{2R_0}{3c}\sqrt{\frac{R_0}{r_g}}$ (60)

is the collapse time. It is interesting to note that $v^1\left(t=0\right)=2r/t_c^{\left(R\right)}$, i.e., that the initial velocity is not zero which means that, as in the Newton theory, a static initial solution does not exist for collapse problem in General Relativity.

As earlier, consider the sphere with the parameters of Sun for which Equation (60) yields $t_c^{\left(R\right)}=750\sec$. This time is about one half of the result corresponding to the Newton theory ( $t_c^{\left(N\right)}=1766\sec$). The obtained result does not coincide with Equation (2) presented in the literature [2] [3] which gives the same collapse time that in the Newton theory, i.e. $t_c^{\left(W\right)}=t_c^{\left(N\right)}$. The difference is most probably associated with different metric forms used for analysis, the time $t_c^{\left(W\right)}$ is found from equations following from the Schwarzschild metric form presented in the co-moving coordinates in which $g_{44}=1$.

The dependence $R\left(t\right)/R_0$ corresponding to Equations (59) is shown in Figure 1 (line 5). It should be emphasized that the scenario of the collapse with zero pressure, as it has been already noted, is not realistic. Pressure resists gravitational contraction and can prevent it. This scenario is supported by the existence of the static solution for the general relativity equations incorporating pressure [21] [22] . However, even in the absence of pressure, there is the effect that stops collapse. Indeed, as follows from Equation (46), the sphere radius cannot be smaller that $r_g$. Taking $R\left(t\right)=r_g$ in the second equation of Equations (59), we can find the lower limit of time for which the foregoing solution exists

$t^{\ast }=\left(1-\sqrt{\frac{r_g^3}{R_0^3}}\right)t_c^{\left(R\right)}$

For the sphere with the parameters of Sun, $t^{\ast }$ is rather close to $t_c^{\left(R\right)}$, i.e.,

$t^{\ast }=\left(1-8.69\times {10}^{-9}\right)t_c^{\left(R\right)}$

Now consider the second case corresponding to sign “+” in the forgoing equations which yield

$\rho \left(t\right)=\frac{{\rho }_0}{{\left(1+t\sqrt{6\pi G{\rho }_0}\right)}^2},R\left(t\right)=R_0\sqrt[3]{{\left(1+t\sqrt{6\pi G{\rho }_0}\right)}^2},v^1=\frac{2r\sqrt{6\pi G{\rho }_0}}{3\left(1+t\sqrt{6\pi G{\rho }_0}\right)}$ (61)

This solution describes the sphere expansion. It looks more realistic than the collapse solution considered above because the absence of pressure is a reasonable assumption for the expansion problem. It is interesting to note Equations (61) coincide with Equations (29), (30) and (31) obtained in the previous Section for the Newton theory.

Suppose that the initial density is infinitely high and $R_0=0$. Taking ${\rho }_0\to \infty$ in the first and the last equations of Equations (61), we arrive at

$\rho \left(t\right)=\frac{1}{6\pi G{t}^2},v^1=\frac{2r}{3t}$ (62)

The limit version of the second equation in Equations (61) can be obtained if we use Equation (5) for the sphere mass which does not change in the process of expansion. The result is

$R^3\left(t\right)=\frac{9}{2}mG{t}^2$

As an example, consider the universe with the age $t_u=13.787\times {10}^9$ years [23] . Taking into account that $G=6.67\times {10}^{-11}$ m³/kg sec² and using the first equation in Equations (62), we get $\rho \left(t_u\right)=0.421\times {10}^{-26}$ kg/m³. This result is in fair agreement with existing evaluations which have rather high scatter. The second equation in Equations (62) shows that the velocity is proportional to the radial coordinate which also corresponds to the existing observations. This equation describes a specific motion with zero coordinate acceleration, i.e.,

$\frac{\text{d}{v}^1}{\text{d}t}=\frac{\partial v^1}{\partial t}+v^1\frac{\partial v^1}{\partial r}=0$

For $t\to \infty$, we have $\rho \to 0$, $v^1\to 0$, $R\to \infty$ and $g_{14}\to 0$, $g_{44}\to 1$ which corresponds to an infinite static Euclidean space.