New Relativistic Equations of Motion and Their Comparison with the Equations from the General Relativity
Kostadin Tren\v{c}evski$^*$
Faculty of Natural Sciences and Mathematics
P.O.Box 162
1000 Skopje
Macedonia
In this paper is presented the infinitesimally locally Lorentzian theory of gravitation (ILLTG), which predicts the same experimental results which confirm the GR. Motion in a neighborhood of the center of gravitation is considered. The geodesics between the ILLTG and GR are compared. ILLTG differs from GR mainly in the equations of motion and for $v<<c$ the trajectories are almost the same in both theories. Two tests in the solar system are proposed, such that ILLTG and GR predict different results. The mass and energy with respect to ILLTG is considered. The periastron shift of the binary stars is calculated and a cosmological problem is considered.
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\def\npag{\thanks{%\hspace* {-18pt} 
This paper was done under the auspices of a CNCSIS Grant.}}
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\title{New Relativistic Equations of Motion and Their Comparison with the 
Equations from the General Relativity}

\author{Kostadin Tren\v{c}evski$^*$\\ 
Faculty of Natural Sciences and Mathematics, \\
P.O.Box 162, 1000 Skopje, Macedonia} 

\begin{document}
\maketitle

{\renewcommand{\thefootnote}{}%
\footnote{$^*$Electronic address: kostatre@iunona.pmf.ukim.edu.mk}%
\setcounter{footnote}{0}}%

%Keywords: ILLTG, General relativity, Hubble constant, 
%gravitational potential, fifth dimension. 
PACS numbers: 04.50.+h, 04.25.-g, 04.80.Cc, 98.80.Jk
\bigskip 

\begin{abstract}
In this paper is presented the 
infinitesimally locally Lorentzian theory of gravitation (ILLTG), 
which predicts the same experimental results which confirm the GR. 
Motion in a neighborhood 
of the center of gravitation is considered. 
The geodesics between the ILLTG and GR are compared. 
ILLTG differs from GR mainly in the equations of motion and for $v<<c$ 
the trajectories are almost the same in both theories. 
Two tests in the solar system are proposed, such that 
ILLTG and GR predict different results. 
The mass and energy with respect to ILLTG is considered. 
The periastron shift of the binary stars is calculated and 
a cosmological problem is considered. 

\end{abstract} 

\vskip 1 cm
\section{Introduction}

In the paper [8] was presented a new model of gravitational theory, 
which yields to the same experimental results for motion of 
a particle in a gravitational field predicted by the 
general relativity (GR) and which are experimentally 
verified. In this paper that theory is 
studied now in many details and a lot new results are presented. 
In this section will be repeated the basic concepts of the 
theory given in [8] with some improvements. We convenient 
to use that $x^{4}=ict$ for the time coordinate and orthogonal matrix 
$A$ with complex elements will mean that $AA^{T}=I$. 

First we underline the main difference between the 
GR and the proposed theory. 
Let us start from the metric in GR 
$$ds^2=g_{11}dX^2+g_{22}dY^2+g_{33}dZ^2+g_{44}d(icT)^2,$$
where $g_{ij}$ can approximately be written in the form 
$$g_{11}=g_{22}=g_{33}=1+2{V\over c^2},\quad g_{44}=1-2{V\over c^2}, 
\quad g_{ij}=0\;\;\; (i\neq j).$$
Here $V$ is the gravitational potential, for example 
$V={GM\over rc^2}$ in case of gravitation of a spherical body. Note that 
the coefficients $g_{ij}$ may not be determined very precisely like 
in the GR, because we shall use them for metric proportions, but not 
for equations of motion. Thus, the expressions of type $c^{-4}$ and less 
are neglected. Hence, the experiments for the red shift of the spectral 
lines and the Shapiro time delay, where the Christoffel symbols are not used,
are explained in the same way like in the GR. 

The above coordinates $X,Y,Z,T$ will be called {\em ordinary coordinates}. 
In order to consider the equations of motion, we introduce the 
so called {\em orthonormal coordinates} $x,y,z,t$ such that 
\vspace{0.5cm}
$$dx=\sqrt{{1\over g_{11}}}dX\approx \Bigl (1-{V\over c^2}\Bigr )dX, \quad 
dy=\sqrt{{1\over g_{22}}}dY\approx \Bigl (1-{V\over c^2}\Bigr )dY, \quad 
dz=\sqrt{{1\over g_{33}}}dZ\approx \Bigl (1-{V\over c^2}\Bigr )dZ, $$
$$dt=\sqrt{{1\over g_{44}}}dT\approx \Bigl (1+{V\over c^2}\Bigr )dT.
\eqno{(1.1)}$$
Further we shall deal mainly with 
orthonormal coordinates, otherwise it will be emphasized that the 
coordinates are ordinary. 
According to the ordinary coordinates, the velocity 
of the light is equal to $c\Bigl (1-2{V\over c^2}\Bigr )$, like in GR, 
while according to the orthonormal coordinates the velocity of the light 
is universal constant. Indeed, according to the orthonormal coordinates 
the space-time in an infinitesimally small neighborhood is flat like in 
the special theory of relativity (STR), and hence the proposed theory 
will be called infinitesimally locally Lorentzian 
theory of gravitation (ILLTG). Note that ILLTG, using orthonormal 
coordinates, does not describes only gravitation, but also inertial forces 
(see Ref. [9]). 
The ordinary coordinates $X,Y,Z,T$ from ILLTG are analogous to the 
usual coordinates $x,y,z,t$ from the GR, while 
the orthonormal coordinates from ILLTG mean that the observer 
is far from gravitation, where the metric is like in the STR. 

Instead of the metric $g_{ij}$ we can use only a potential $\mu$. 
For example, this potential caused by a spherical body with mass M, 
whose center is at the coordinate origin and 
which rests with respect to the chosen coordinate system is 
$\mu =1+{GM\over rc^2}$. Thus, the equations (1.1) will be replaced by 
$$dx=\mu^{-1} dX, \quad 
dy=\mu^{-1} dY, \quad 
dz=\mu^{-1} dZ, \quad 
dt=\mu dT.\eqno{(1.1')}$$
Further we shall present the basic results 
about the equations of motion in orthonormal coordinates, and naturally 
then we should return back in ordinary coordinates as natural. This 
transformation in an infinitesimally small neighborhood is always possible,
but in large we can deal with the nonholonomic coordinates. 

If two gravitational fields are given by the 
gravitational potentials $\mu _1 $ and $\mu _2 $, then the superposition
of those two fields is a gravitational field with potential $\mu _1\mu _2 $.
In special case if $\mu_2 =C$ is a constant in the space-time, then the 
motions of particles should be the same 
like for the field given by $\mu_1 $. 

ILLTG is built using a 4-vector potential $U_i $. 
If the magnitude of this vector changes but its direction is a constant, 
then there exists a gravitational field and the source of gravitational 
field is moving with 4-vector of velocity $U_i/\Vert U\Vert $; if 
the direction of this vector changes but its magnitude is a constant, 
then there exist some inertial forces, for example centrifugal. 
This 4-vector field can be written in the form 
$$ (U_{1},U_{2},U_{3},U_{4}) = 
{\mu \over \sqrt {1-u^{2}/c^{2}}}\Bigl ( 
{u_{1}\over ic}, 
{u_{2}\over ic}, 
{u_{3}\over ic}, 
1\Bigr ) ,\eqno{(1.2)}$$
where 
${\bf u}=(u_{1},u_{2},u_{3})$ is a 3-vector of velocity and $\mu $ is the 
gravitational potential. 

{\em Example 1.} The STR is characterized by $\mu =const.$ and 
${\bf u}$ is a constant vector field. 

{\em Example 2.} The gravitational field of a spherical body 
with mass $M$ which is at the origin of the coordinate system 
is given by the vector field 
$(0,0,0,\mu )$, where $\mu =1+{GM\over rc^2}$. 

{\em Example 3.} If we measure with respect to a noninertial system 
which rotates
around the $x^{3}$ axis with angular velocity $w$, then the 4-vector 
potential obtains by putting $\mu =1$ and 
${\bf u}=(x^{2}\omega ,-x^{1}\omega ,0)$ [9]. 

Further we introduce a non-linear connection induced by the 
4-vector potential $U_i $. Let us denote by $S^{i}_{j}ds$ the form of 
connection in the direction of the unit vector $V^{i}=dx^{i}/ds$. The 
necessary and sufficient condition for the metric 
tensor $g_{ij}=\delta _{ij}$ to be parallel, is 
$$S^{i}_{j}=-S^{j}_{i}.\eqno{(1.3)}$$ 

Now let us introduce the following tensor 
$$F_{ij}={1\over \mu }(\partial U_{i}/\partial x^{j}- 
\partial U_{j}/\partial x^{i}).\eqno{(1.4)}$$
If we denote by $V_{i}$ the 4-vector of velocity of the 
moving particle, i.e. 
$$(V_i)=\Bigl ( 
{v_{1}\over ic\sqrt {1-v^{2}/c^{2}}}, 
{v_{2}\over ic\sqrt {1-v^{2}/c^{2}}}, 
{v_{3}\over ic\sqrt {1-v^{2}/c^{2}}}, 
{1\over \sqrt {1-v^{2}/c^{2}}}\Bigr ), \eqno{(1.5)}$$
one can easily verify that the following simple equations 
$$ dV_{i}/ds = F_{ij}V_{j},\eqno{(1.6)}$$
which are analogous to the Lorentz force in the theory of electromagnetism, 
contain inertial accelerations (Coriolis acceleration, 
centrifugal and transversely acceleration) and gravitation acceleration 
for Newton approach. In order to obtain a precise theory, we consider 
the following two tensors 
$$\phi _{ij}={1\over 2}F_{ij} - {1\over 2}\mu ^{-2}
(U_{i}U_{k}F_{jk}- U_{j}U_{k}F_{ik}),\eqno{(1.7)}$$
$$P_{ij}=\delta _{ij} - {1\over \mu + U_{s}V_{s}}
(\mu V_{i}V_{j}+V_{i}U_{j}+U_{i}V_{j}+{1\over \mu }U_{i}U_{j}) + 
{2\over \mu }U_{j}V_{i}.\eqno{(1.8)}$$
These two tensors together with $F_{ij}$ are invariant under the gauge 
transformation $\mu \rightarrow C\mu$, where $C$ is a constant. The tensors 
$F_{ij}$ and $\phi _{ij}$ are antisymmetric and $P_{ij}$ is an orthogonal 
matrix. It can be verified by using the identities 
$U_{i}U_{i}=\mu ^{2}$ and $V_{i}V_{i}=1$. Now the required connection 
$S_{ij}$ is introduced by 
$$S_{ij} = - P_{ri}\phi _{rk}P_{kj},\eqno{(1.9)}$$
or in matrix form $S=-P^{t}\phi P$. Although this connection seems 
to be very complicated, now we will show the opposite. Indeed, assume that 
$(U_{i})=(0,0,0,1)$ at the considered point. Then 
$$ F = \left [\matrix{
0 & -2i\omega _{3}/c & 2i\omega _{2}/c & -a_{1}/c^{2}\cr 
2i\omega _{3}/c & 0 & -2i\omega _{1}/c & -a_{2}/c^{2}\cr 
-2i\omega _{2}/c & 2i\omega _{1}/c & 0 & -a_{3}/c^{2}\cr 
a_{1}/c^{2} & a_{2}/c^{2} & a_{3}/c^{2} & 0\cr }\right ], \eqno{(1.10)}$$
$$ \phi = \left [\matrix{
0 & -i\omega _{3}/c & i\omega _{2}/c & -a_{1}/c^{2}\cr 
i\omega _{3}/c & 0 & -i\omega _{1}/c & -a_{2}/c^{2}\cr 
-i\omega _{2}/c & i\omega _{1}/c & 0 & -a_{3}/c^{2}\cr 
a_{1}/c^{2} & a_{2}/c^{2} & a_{3}/c^{2} & 0\cr }\right ], \eqno{(1.11)}$$
where 
$$ {\bf a}=(a_{1},a_{2},a_{3})=c^{2}\hbox {grad}\; U_{4} + 
\partial {\bf u}/\partial t \quad \hbox { and }\quad 
{\bf w}= -{1\over 2}\hbox {rot}\; {\bf u}\eqno{(1.12)}$$
represent the 3-vector of acceleration and the 3-vector of angular velocity. 
In that special case $(U_{i})=(0,0,0,1)$, the tensor $P$ is given by 
$$ P = \left [\matrix{
1-{1\over \nu }V_{1}^{2} & -{1\over \nu }V_{1}V_{2} &
-{1\over \nu }V_{1}V_{3} & V_{1}\cr 
& & & \cr
-{1\over \nu }V_{2}V_{1} & 1-{1\over \nu }V_{2}^{2} &
-{1\over \nu }V_{2}V_{3} & V_{2}\cr 
& & & \cr
-{1\over \nu }V_{3}V_{1} & -{1\over \nu }V_{3}V_{2} & 
1-{1\over \nu }V_{3}^{2} & V_{3}\cr 
& & & \cr
-V_{1} & -V_{2} & -V_{3} & V_{4}\cr }\right ], \eqno{(1.13)}$$
where $V_{1}, V_{2}, V_{3}, V_{4}$ are given by (1.5) and 
$\nu =1+V_{4}$. Note that the matrix $P$ represents just a Lorentz 
transformation for the 3-vector $-{\bf v}=(-v_{1},-v_{2},-v_{3})$. Now the 
connection $S_{ij}$ becomes much more clear and acceptable. 

{\em Remark.} Although $\phi$ is a tensor, its 
components have physical meaning given by (1.11) only for 
$U_1=U_2=U_3=0$. On the other hand, the tensor $F$ defined by (1.4)
always has components given approximately by (1.10). In case of 
gravitational field, $U_1=U_2=U_3=0$ means that the source of 
gravitation is in rest with respect to the chosen coordinate system. 
In this system it is $F_{ij}=\phi_{ij}$ assuming that ${\bf w}=0$, 
and hence it is true in each coordinate system. 

Note that the introduced connection $S_{ij}$ is non-linear. The first step 
of approximation yields to the following linear connection 
$\Gamma _{k}=(\Gamma ^{i}_{jk})$ such that 
$S^i_j\approx \Gamma ^i_{js}V_{s}$, where 
$$ \Gamma _{1} = \left [ \matrix{
0 & a_{2}/c^{2} & a_{3}/c^{2} & 0\cr
-a_{2}/c^{2} & 0 & 0 & -iw_{3}/c \cr 
-a_{3}/c^{2} & 0 & 0 & iw_{2}/c \cr 
0 & iw_{3}/c & -iw_{2}/c &0\cr }\right ],\eqno{(1.14a)} $$
$$ \Gamma _{2} = \left [ \matrix{
0 & -a_{1}/c^{2} & 0 &  iw_{3}/c\cr
a_{1}/c^{2} & 0 & a_{3}/c^{2} & 0\cr 
0 & -a_{3}/c^{2} & 0 & -iw_{1}/c \cr 
-iw_{3}/c & 0 & iw_{1}/c & 0\cr }\right ],\eqno{(1.14b)} $$
$$ \Gamma _{3} = \left [ \matrix{
0 & 0 & -a_{1}/c^{2} & -iw_{2}/c\cr
0 & 0 & -a_{2}/c^{2} & iw_{1}/c\cr
a_{1}/c^{2} & a_{2}/c^{2} & 0 & 0 \cr 
iw_{2}/c & -iw_{1}/c &0&0\cr }\right ], \eqno{(1.14c)}$$
$$ \Gamma _{4} = \left [ \matrix{
0 & iw_{3}/c & -iw_{2}/c & a_{1}/c^{2}\cr 
-iw_{3}/c & 0 & iw_{1}/c & a_{2}/c^{2}\cr 
iw_{2}/c & -iw_{1}/c & 0 & a_{3}/c^{2}\cr 
-a_{1}/c^{2} & -a_{2}/c^{2} & -a_{3}/c^{2} & 0 \cr }\right ],
\eqno{(1.14d)}$$
and where we assumed that $(U_{i})=(0,0,0,1)$ at the considered point. 
The Newton's approximation obtains by replacing the components 
$\Gamma ^{i}_{jk}$ for $i,j,k\in \{1,2,3\}$ with zero. 
Using this linear connection 
the curvature tensor is easy to be calculated and in [8] it is shown that 
the Einstein equations of gravitational field 
$$ R_{ij} - {1\over 2} g_{ij}R = 8G \pi c^{-2} T_{ij}\eqno{(1.15)}$$
are satisfied, where the energy-momentum tensor $T_{ij}$ is given by 
$T_{ij}=\rho U_{i}U_{j}$. It is also shown in [8] that the following 
equations 
$$(\mu F_{ij})_{;k}+(\mu F_{jk})_{;i}+(\mu F_{ki})_{;j}=0 ,\eqno{(1.16)}$$
$$ (\mu F_{ij})_{;j} = 4G \pi c^{-2} J_{i} ,\eqno{(1.17)}$$
analogous to the Maxwell equations, also hold at that order of 
approximation. Here $J_{i}=\rho U_{i}$
is the 4-current density for gravitation, analogous to the 4-current 
density for the electromagnetism. 

\section{Some results for the geodesics according to the ILLTG}

In this section the planetary orbits in the solar system 
as well as the trajectories of the light rays will be considered. 
Our coordinate origin will be chosen to be at the center of the Sun. 
The source of gravitation is in rest to our coordinates system, and we can 
put $U_1=U_2=U_3=0$. We shall assume that the mass of the moving body 
is negligible with respect to the Sun. The general equations of motion of a 
body with nonzero mass are given in section 6. We shall deal with the 
orthonormal coordinates. 

Let $\mu =1+{G M\over rc^2}$, where $M$ is the mass of the Sun. Then 
$$(F_{ij})=\left [\matrix{ 0 & 0 & 0 &{G M\over r^3c^2}x\cr
& & & \cr
0 & 0 & 0 &{G M\over r^3c^2}y\cr
& & & \cr
0 & 0 & 0 &{G M\over r^3c^2}z\cr
& & & \cr
-{G M\over r^3c^2}x&-{G M\over r^3c^2}y&
-{G M\over r^3c^2}&0\cr}\right ] \eqno{(2.1)}$$
and the equations of motion ${dV_i\over ds} = -P_{ri}\phi_{rk}P_{kj}$ for 
$i=1,2,3,4$ without approximation reduce to [8] 
$${d^2x\over dt^2}={G M\over \mu r^3}
\Bigl [(\beta ^{-2}-2)x + {v_x\over c^2}(xv_x+yv_y)
(2+{1\over \beta (\beta +1)})\Bigr ],\eqno{(2.2a)}$$
$${d^2y\over dt^2}={G M\over \mu r^3}
\Bigl [(\beta ^{-2}-2)y + {v_y\over c^2}(xv_x+yv_y)
(2+{1\over \beta (\beta +1)})\Bigr ],\eqno{(2.2b)}$$
$$\beta - \ln \mu = C_1 (=const.),\eqno{(2.2c)}$$
where $\beta = \Bigl (1-{v^2\over c^2}\Bigr )^{-1/2}$. 
Without loss of generality we assumed that $v_z=0$ and that the 
particle moves in the $xy$-plane. Thus we omit the equation for $i=3$. 

{\em Remark.} Using the formulas (1.1'), and the fact that $\mu$ is a 
scalar which does not depend from the coordinate system (orthonormal or 
ordinary), it is easy to see that in ordinary coordinates the equations 
of motion (2.2) become:
$${d^2X\over dT^2}={2\over\mu}{d\mu\over dT}{dX\over dT}
+\mu^3{d^2x\over dt^2},
\eqno{(2.2'a)}$$
$${d^2Y\over dT^2}={2\over\mu}{d\mu\over dT}{dY\over dT}+
\mu^3{d^2y\over dt^2}
\eqno{(2.2'b)}$$
$$\beta - \ln \mu = C_1 (=const.),\eqno{(2.2'c)}$$
where ${d^2x\over dt^2}$ and ${d^2y\over dt^2}$ are given by (2.2a,b). 
Note that 
the equation (2.2c) remains unchanged. Indeed, $\mu$ appears to be the same 
in both coordinate systems and also the quotient $v^2/c^2$ takes the same
value in both coordinate systems. 

{\em Remark.} Note that if we have not only a gravitational field with 
a potential $\mu$, but we also have a 3-vector of acceleration 
$(a_x,a_y,a_z)$ and a 3-vector of angular velocity $(w_x,w_y,w_z)$, 
then in orthonormal coordinates it is 
$${dv_x\over dt}={1\over \mu }
\Bigl [(2-\beta ^{-2})a_x - {v_x\over c^2}(a_xv_x+a_yv_y+a_zv_z)
(2+{1\over \beta (\beta +1)})+2(v_yw_z-v_zw_y)\Bigr ],\eqno{(2.3a)}$$
$${dv_y\over dt}={1\over \mu }
\Bigl [(2-\beta ^{-2})a_y - {v_y\over c^2}(a_xv_x+a_yv_y+a_zv_z)
(2+{1\over \beta (\beta +1)})+2(v_zw_x-v_xw_z)\Bigr ],\eqno{(2.3b)}$$
$${dv_z\over dt}={1\over \mu }
\Bigl [(2-\beta ^{-2})a_z - {v_z\over c^2}(a_xv_x+a_yv_y+a_zv_z)
(2+{1\over \beta (\beta +1)})+2(v_xw_y-v_yw_x)\Bigr ],\eqno{(2.3c)}$$
$${d\over dt}{1\over \sqrt{1-{v^2\over c^2}}} = {1\over \mu c^2}
(a_xv_x+a_yv_y+a_zv_z).\eqno{(2.3d)} $$
The acceleration which is caused by the gravitational 
potential $\mu$ is included in the 3-vector of acceleration 
$(a_x,a_y,a_z)$. 

Now, using that 
$${d\varphi \over dt}={d\over dt}\arctan {y\over x} = {v_yx-v_xy\over r^2},$$
from (2.2a) and (2.2b), we obtain 
$${d\over dt}\Bigl (r^2{d\varphi \over dt}\Bigr ) = 
{d^2y\over dt^2}x-{d^2x\over dt^2}y = {G M\over \mu r^3c^2}(v_yx-v_xy)
(xv_x+yv_y)(2+{1\over \beta (\beta +1)}) =$$
$$=-r^2{d\varphi \over dt}{G M\over \mu c^2}\Bigl (
2+{1\over \beta (\beta +1)}\Bigr ){d\over dt}\Bigl ({1\over r}\Bigr ).$$
Two cases will be considered. 

(i) Assume that $v<<c$, i.e. consider a planetary orbit. Then 
$2+{1\over \beta (\beta +1)} \approx 2,5$ and hence 
$${d\over dt}\Bigl (r^2{d\varphi \over dt}\Bigr ) = -{5\over 2}
{G M\over c^2}\Bigl (r^2{d\varphi \over dt}\Bigr ) 
{d\over dt}\Bigl ({1\over r}\Bigr ),$$
by neglecting the expressions of order $c^{-4}$. The solution 
of the previous equation is 
$$r^2{d\varphi \over dt}=
C_2 \exp \Bigl ({-5\over 2}{G M\over rc^2}\Bigr ), 
\quad C_2=const.\eqno{(2.4)}$$

Further, using the metric $(dr)^2+r^2(d\varphi )^2-c^2t^2=ds^2$ 
in orthonormal coordinates, we obtain 
$$\Bigl ({dr\over dt}\Bigr )^2 + \Bigl ({rd\varphi \over dt}\Bigr )^2 = v^2, 
\quad 
\Bigl (r^{-2} {dr\over d\varphi }\Bigr )^2 + r^{-2} = v^2
\Bigl (r^2{d\varphi \over dt}\Bigr )^{-2},$$ 
and get 
$$\Bigl ({d\rho \over d\varphi }\Bigr )^2+\rho^2 = v^2C_2^{-2} 
\exp \Bigl ({5G M\rho\over c^2}\Bigr ),\eqno{(2.5)}$$
where $\rho =r^{-1}$. {From} (2.2c) we obtain approximately 
$$\Bigl ( 1-{v^2\over c^2}\Bigr )^{-1/2} - 
{G M\over rc^2}\Bigl (1-{1\over 2}{G M\over rc^2}\Bigr )=
1+c^{-2}C_3,$$
where $C_3=c^2(C_1-1)=const$. Hence we obtain 
$$v^2=c^2\Bigl [ 1-\Bigl (1+c^{-2}\Bigl (C_3+G M\rho \Bigl (1-
{1\over 2}{G M\rho\over c^2}\Bigr )\Bigr )\Bigr )^{-2}\Bigr ]=$$
$$=2C_3+2G M\rho\Bigl (1-{1\over 2}{G M\rho\over c^2}\Bigr )
-3c^{-2}C_3^2-6C_3{G M\rho\over c^2}-3{G^2 M^2\rho^2\over c^2}.$$
Replacing this value of $v^2$ into (2.5), after neglecting the summands of 
order $c^{-4}$, we obtain the following differential equation
$$\Bigl ({d\rho\over d\varphi}\Bigr )^2+\rho^2=A+B\rho +
{6G ^2M^2\over c^2C_2^2} \rho^2\eqno{(2.6)}$$
and hence 
$$d\varphi = -\Bigl ( 1+{3G ^2M^2\over c^2C_2^2}\Bigr )
{d\rho\over \sqrt{-\rho^2+p\rho-q}},$$
where $p$ and $q$ are constants. Indeed, $p=\rho_1+\rho_2$, 
$q=\rho_1\rho_2$, where $\rho_1={1\over r_1}$, $\rho_2={1\over r_2}$ 
and $r_1$ and $r_2$ are extreme distances $(r_1<r_2)$ from the planet to 
the center of the Sun. It is well known that 
$C_2=\sqrt{G Ma(1-\epsilon^2)}$, where $a=(r_1+r_2)/2$ and $\epsilon$ 
is the eccentricity of the approximate ellipse. Thus, 
$$d\varphi = -\Bigl ( 1+{3G M\over ac^2(1-\epsilon^2)}\Bigr )
{d\rho\over \sqrt{-\rho^2+p\rho-q}}.\eqno{(2.7)}$$
Hence we obtain 
$$\varphi = -\Bigl ( 1+{3G M\over ac^2(1-\epsilon^2)}\Bigr )
\Bigl (\arcsin {\rho-{p\over 2}\over \sqrt{{p^2\over 4}-q}}-C_4\Bigr ) =
-\Bigl ( 1+{3G M\over ac^2(1-\epsilon^2)}\Bigr )
\Bigl (\arcsin {2\rho-\rho_1-\rho_2\over \rho_1-\rho_2}-C_4\Bigr ) .$$
Assume that $\varphi =0$ if $\rho=\rho_1$, and hence $C_4=\pi /2$. This 
equation can be solved by $\rho $ and its solution is 
$$\rho = {\rho_1+\rho_2\over 2}+{\rho_1-\rho_2\over 2}
\cos \Bigl [\varphi \Bigl (1-{3G M\over ac^2(1-\epsilon^2)}\Bigr )
\Bigr ]. \eqno{(2.8)}$$

Now it follows that the angle between the points for which 
$\rho=\rho_1$ and $\rho=\rho_2$ is equal to 
$\pi +{3G M\pi\over ac^2(1-\epsilon^2)}$ and hence the angle 
for the perihelion shift is equal to $\Delta \varphi = 
{6G M\pi\over ac^2(1-\epsilon^2)}$. 

(ii) Let us consider the trajectory of a light ray near the Sun. We denote 
by $R$ the radius of the Sun. In this case $\beta =\infty$, 
$${d\over dt}\Bigl (r^2{d\varphi\over dt}\Bigr ) = -2{G M\over c^2}
\Bigl (r^2{d\varphi\over dt}\Bigr ){d\over dt}\Bigl ({1\over r}\Bigr )$$
and its solution is 
$$r^2{d\varphi\over dt}=C_2\exp \Bigl (-2{G M\over rc^2}\Bigr ),
\quad C_2=const.\eqno{(2.9)}$$
Analogously to (2.5) we obtain 
$$\Bigl ({d\rho \over d\varphi}\Bigr )^2 + \rho^2=v^2C_2^{-2}
\exp \Bigl ({4G M\rho\over c^2}\Bigr )$$
and replacing $v=c$, we get 
$$\Bigl ({d\rho \over d\varphi}\Bigr )^2 + \rho^2=c^2C_2^{-2}
\exp \Bigl ({4G M\rho\over c^2}\Bigr ).\eqno{(2.10)}$$

We shall prove now that the light has a constant velocity $c$ in a 
gravitational field, using orthonormal coordinates. 
Summing up the equation (2.2a) multiplied by 
$2v_x$ and (2.2b) multiplied by $2v_y$ we obtain 
$${dv^2\over dt}={2G M\over \mu r^3}\Bigl [ -\Bigl (1+{v^2\over c^2}
\Bigr )(xv_x+yv_y)+(xv_x+yv_y)
\Bigl (2+{1\over \beta (\beta +1)}\Bigr ){v^2\over c^2}
\Bigr ] =$$
$$=-{2G M\over \mu r^3}(xv_x+yv_y){\Bigl (1-{v^2\over c^2}\Bigr )^{3/2}
\over 1+\sqrt{1-{v^2\over c^2}}}.$$
Hence, if $v^2=c^2$ at one point, then $v^2=c^2$ holds on the whole 
trajectory, because $v^2=c^2$ satisfies the previous differential equation 
and the initial condition. 

If $r=R$, then 
$R{d\varphi \over dt}=c$ and from (2.9) we get $C_2=Rc\exp ({2G M\over 
Rc^2})$. By replacing this value of $C_2$ into (2.10) we get 
$$\Bigl ({d\rho \over d\varphi}\Bigr )^2 + \rho^2={1\over R^2}
\exp \Bigl ({4G M\over c^2}\Bigl (\rho -{1\over R}\Bigr )\Bigr ).
\eqno{(2.11)}$$
By neglecting the expressions of order $c^{-4}$ we obtain 
$$\Bigl ({d\rho \over d\varphi}\Bigr )^2 + \rho^2={1\over R^2}
-{4G M\over R^3c^2}+{4G M\over R^2c^2}\rho ,\eqno{(2.12)}$$
which is equation of a hyperbola. Note that according to the Newton's theory 
the trajectory is a different hyperbola: 
$\Bigl ({d\rho \over d\varphi}\Bigr )^2 + \rho^2={1\over R^2}
-{2G M\over R^3c^2}+{2G M\over R^2c^2}\rho $. 

{From} (2.12) $\varphi $ can be determined as a function of $\rho$:
$$\varphi = \arccos {\rho R^2c^2-2G M\over Rc^2-2G M},\eqno{(2.13)}
$$
such that $\varphi =0$ if $\rho ={1\over R}$. Hence from (2.13) we obtain 
$$\rho = {2G M\over R^2c^2}+\Bigl ({1\over R}-{2G M\over R^2c^2}
\Bigr )\cos \varphi. \eqno{(2.14)}$$

It is easy to conclude from (2.13) that the angle of deflection of a 
light ray near the Sun is equal to ${4G M\over Rc^2}$. This  conclusion 
can be obtained also by the following way. Assume that the light ray is 
parallel to the $x$-axis at $(0,R,0,0)$. Let us put 
${dx\over dt}=v\cos \alpha$, ${dy\over dt}=-v\sin \alpha$, $x=vt$, 
$y=R$ and $r=\sqrt{R^2+v^2t^2}$ in (2.2a) and (2.2b). Then 
$${dv\over dt}\cos \alpha -v\sin \alpha {d\alpha \over dt} = 
{G M\over \mu r^3}\Bigl [-\Bigl (1+{v^2\over c^2}\Bigr )vt + 
{v\cos \alpha \over c^2}(xv_x+yv_y)(2+{1\over \beta (\beta +1)})\Bigr ],
\eqno{(2.15a)}$$
$$-{dv\over dt}\sin \alpha -v\cos \alpha {d\alpha \over dt} = 
{G M\over \mu r^3}\Bigl [-\Bigl (1+{v^2\over c^2}\Bigr )R - 
{v\sin \alpha \over c^2}(xv_x+yv_y)(2+{1\over \beta (\beta +1)})\Bigr ].
\eqno{(2.15b)}$$
Summing up the first equation multiplied by $\sin \alpha$ and the second 
equation multiplied by $\cos \alpha$, we obtain 
$$v{d\alpha \over dt}={G M\over \mu r^3}\Bigl (1+{v^2\over c^2}\Bigr )
(R\cos \alpha +vt \sin \alpha ).$$
By putting $v=c$, $\mu \approx 1$ and $R\cos \alpha +vt\sin \alpha\approx R$,
we get 
$${d\alpha \over dt}={2G MR\over c(R^2+c^2t^2)^{3/2}},$$
$$\Delta \alpha =\int_{-\infty}^{\infty}{2G MRdt\over c(R^2+c^2t^2)^{3/2}}
={4G M\over Rc^2}.\eqno{(2.16)}$$

Although we deduced the main formulas 
$\Delta \varphi = {6GM\pi \over ac^2(1-\epsilon ^2)}$ and 
$\Delta \alpha = {4G M\over Rc^2}$ by using orthonormal coordinates, 
they remain unchanged if we use ordinary coordinates. 

To the end of this section we shall consider the motion of a particle 
(with zero mass) under the gravitation of $k$ gravitational objects. 
Indeed, we have in special case a motion in the solar system with 
perturbations of the planets and other gravitational bodies. 
Let $m_i$ be the mass of the $i$-th gravitational object and $r_i$ 
be the distance from its center to the particle. Further, let 
the 3-vector of velocity of the particle is $v=(v_x,v_y,v_z)$ and the 
3-vector of velocity of the $i$-th gravitational object is 
$u_i=(u_{ix},u_{iy},u_{iz})$ and let us put 
$$U_i=-c^2\ln \Bigl (1+{G m_i\over r_ic^2}\Bigr ).\eqno{(2.17)}$$
Let us denote the 4-vector of velocity of the $i$-th object by
$$(U_{i1},U_{i2},U_{i3},U_{i4})=
\Bigl ({{u_{ix}\over ic}\over \sqrt{1-{u_{i}^2\over c^2}}},
{{u_{iy}\over ic}\over \sqrt{1-{u_{i}^2\over c^2}}},
{{u_{iz}\over ic}\over \sqrt{1-{u_{i}^2\over c^2}}},
{1\over \sqrt{1-{u_{i}^2\over c^2}}}\Bigr ),$$
such that the 4-vector potential is 
$\Bigl (1+{G m_i\over r_ic^2}\Bigr )(U_{i1},U_{i2},U_{i3},U_{i4})$. 

The tensors $F_{pq}$ and $\phi_{pq}$ considered for the $i$-th object are 
$$F^i_{pq}=\phi^i_{pq}={1\over c^2}\Bigl [
{\partial U_i\over \partial x_p}U_{iq}-{\partial U_i\over \partial x_q}U_{ip}
\Bigr ]$$
(do not sum over $i$). 
Now, analogously to (2.3) the equations of motion under the gravitation of 
the $i$-th object is given by 
$${dv_x\over dt}=
\Bigl [(2-\beta ^{-2})a_x - {v_x\over c^2}(a_xv_x+a_yv_y+a_zv_z)
(2+{1\over \beta (\beta +1)})+2(v_yw_z-v_zw_y)\Bigr ],\eqno{(2.18a)}$$
$${dv_y\over dt}=
\Bigl [(2-\beta ^{-2})a_y - {v_y\over c^2}(a_xv_x+a_yv_y+a_zv_z)
(2+{1\over \beta (\beta +1)})+2(v_zw_x-v_xw_z)\Bigr ],\eqno{(2.18b)}$$
$${dv_z\over dt}=
\Bigl [(2-\beta ^{-2})a_z - {v_z\over c^2}(a_xv_x+a_yv_y+a_zv_z)
(2+{1\over \beta (\beta +1)})+2(v_xw_y-v_yw_x)\Bigr ],\eqno{(2.18c)}$$
$${d\over dt}{1\over \sqrt{1-{v^2\over c^2}}} = {1\over c^2}
(a_xv_x+a_yv_y+a_zv_z),\eqno{(2.18d)} $$
where 
$$a_x=\Bigl [ U_{i1}{\partial U_i\over \partial x_4}-
U_{i4}{\partial U_i\over \partial x_1}\Bigr ],\quad 
a_y=\Bigl [ U_{i2}{\partial U_i\over \partial x_4}-
U_{i4}{\partial U_i\over \partial x_2}\Bigr ],\quad 
a_z=\Bigl [ U_{i3}{\partial U_i\over \partial x_4}-
U_{i4}{\partial U_i\over \partial x_3}\Bigr ],$$
$$w_x={i\over c}\Bigl [ U_{i3}{\partial U_i\over \partial x_2}-
U_{i2}{\partial U_i\over \partial x_3}\Bigr ],\quad 
w_y={i\over c}\Bigl [ U_{i1}{\partial U_i\over \partial x_3}-
U_{i3}{\partial U_i\over \partial x_1}\Bigr ],\quad 
w_z={i\over c}\Bigl [ U_{i2}{\partial U_i\over \partial x_1}-
U_{i1}{\partial U_i\over \partial x_2}\Bigr ].$$
If the potential $U_i$ is not time dependent, then 
$$a_x=-{1\over \sqrt{1-{u_i^2\over c^2}}}
{\partial U_i\over \partial x},\quad 
a_y=-{1\over \sqrt{1-{u_i^2\over c^2}}}
{\partial U_i\over \partial y},\quad 
a_z=-{1\over \sqrt{1-{u_i^2\over c^2}}}
{\partial U_i\over \partial z}.$$
Further, after some algebraic transformations, the final equations 
for the motion of the particle under the gravitation of all objects 
become 
$${dv_x\over dt}=\sum_{i=1}^k\Bigl [{\partial U_i\over \partial x}
(\beta ^{-2}-2+2\frac{u_{ix}v_x+u_{iy}v_y+u_{iz}v_z}{c^2}) + 
{\partial U_i\over \partial x}{Mv_x^2-2u_{ix}v_x\over c^2}+$$
$$\qquad +
{\partial U_i\over \partial y}{Mv_xv_y-2u_{ix}v_y\over c^2}+
{\partial U_i\over \partial z}{Mv_xv_z-2u_{ix}v_z\over c^2}\Bigr ]
{1\over \sqrt{1-{u_i^2\over c^2}}},\eqno{(2.19a)}$$
$${dv_y\over dt}=\sum_{i=1}^k\Bigl [{\partial U_i\over \partial y}
(\beta ^{-2}-2+2\frac{u_{ix}v_x+u_{iy}v_y+u_{iz}v_z}{c^2}) + 
{\partial U_i\over \partial x}{Mv_xv_y-2u_{iy}v_x\over c^2}+$$
$$\qquad +
{\partial U_i\over \partial y}{Mv_y^2-2u_{iy}v_y\over c^2}+
{\partial U_i\over \partial z}{Mv_yv_z-2u_{iy}v_z\over c^2}\Bigr ]
{1\over \sqrt{1-{u_i^2\over c^2}}},\eqno{(2.19b)}$$
$${dv_z\over dt}=\sum_{i=1}^k\Bigl [{\partial U_i\over \partial z}
(\beta ^{-2}-2+2\frac{u_{ix}v_x+u_{iy}v_y+u_{iz}v_z}{c^2}) + 
{\partial U_i\over \partial x}{Mv_xv_z-2u_{iz}v_x\over c^2}+$$
$$\qquad +
{\partial U_i\over \partial y}{Mv_yv_z-2u_{iz}v_y\over c^2}+
{\partial U_i\over \partial z}{Mv_z^2-2u_{iz}v_z\over c^2}\Bigr ]
{1\over \sqrt{1-{u_i^2\over c^2}}},\eqno{(2.19c)}$$
$${d\over dt}{1\over \sqrt{1-{v^2\over c^2}}}+{1\over c^2}\sum_{i=1}^k
{1\over \sqrt{1-{u_i^2\over c^2}}}\Bigl (
{\partial U_i\over \partial x}v_x+{\partial U_i\over \partial y}v_y+
{\partial U_i\over \partial z}v_z\Bigr ) = 0, \qquad \eqno{(2.19d)}$$
where $M=2+{1\over \beta (\beta +1)}$. 
It is important to emphasize that $\frac{\partial U_i}{\partial x}v_x+
\frac{\partial U_i}{\partial y}v_y+\frac{\partial U_i}{\partial z}v_z$ 
is not a total differential of $U_i$, 
because $U_i$ depends also on the coordinates of the $i$-th body. 

\section{Representations of the geodesics near the Schwarzschild 
radii according to the ILLTG}

In section 2 were considered the geodesics in polar coordinates 
with approximation, neglecting 
the expressions of order $c^{-4}$ and smaller. In this section we shall 
consider the geodesics in polar coordinates without approximation, 
using orthonormal coordinates. 
Starting from the equation (section 2) 
$${d\over dt}\Bigl (r^2{d\varphi \over dt}\Bigr )=-
\Bigl (r^2{d\varphi \over dt}\Bigr ){G M\over \mu c^2}\Bigl (
2+{1\over \beta (\beta +1)}\Bigr ){d\over dt}\Bigl ({1\over r}\Bigr ) ,$$
we obtain 
$${d\over dt}\Bigl (\ln \Bigl (r^2{d\varphi \over dt}\Bigr )\Bigr )
=-{G M\over \mu c^2}\Bigl (2+{1\over \beta (\beta +1)}\Bigr )
{d\over dt}\Bigl ({1\over r}\Bigr ).$$ 
Replacing in this equation that 
$\beta =C_1+\ln \Bigl (1+{G M\over rc^2}\Bigr )=
C_1+\ln \Bigl (1+{G M\rho \over c^2}\Bigr )$ as a function of 
$\rho$, we obtain 
$${d\over dt}\Bigl (\ln \Bigl (r^2{d\varphi \over dt}\Bigr )\Bigr )=
{d\over dt}\int \Bigl (-{G M\over \mu c^2}
\Bigl (2+{1\over \beta (\beta +1)}\Bigr )\Bigr )d\Bigr ({1\over r}\Bigr ),$$
$$r^2{d\varphi \over dt}=C_2
\exp \Bigl (-\int {G M\over \mu c^2}
\Bigl (2+{1\over \beta (\beta +1)}\Bigr )d\rho \Bigr )$$
and further like in section 2 we obtain 
$$\Bigl ({d\rho \over d\varphi}\Bigr )^2+\rho^2=v^2C_2^{-2} 
\exp \Bigl (\int {2G M\over \mu c^2}
\Bigl (2+{1\over \beta (\beta +1)}\Bigr )d\rho \Bigr ),$$
where 
$$v^2=c^2\Bigl [1-\Bigl (C_1+\ln \mu 
\Bigr )^{-2}\Bigr ], \qquad \mu =1+{G M\rho\over c^2}.$$
The constant $C_2^{-2}$ will be determined, assuming that 
${d\rho \over d\varphi} =0$ at the considered point. Indeed, 
$$\rho_0^2=v_0^2C_2^{-2}
\exp \Bigl (\int {2G M\over \mu c^2}
\Bigl (2+{1\over \beta (\beta +1)}\Bigr )
d\rho \Bigl \vert_{\rho =\rho_0}\Bigr ),$$
and hence 
$$\Bigl ({d\rho \over d\varphi}\Bigr )^2+\rho^2=v^2{\rho_0^2\over v_0^2}
\exp \Bigl (\int 2{G M\over \mu c^2}
\Bigl (2+{1\over \beta (\beta +1)}\Bigr )d\rho -\int 2{G M\over\mu c^2}
\Bigl (2+{1\over \beta (\beta +1)}\Bigr )
d\rho\Bigl \vert_{\rho =\rho_0}\Bigr ).\eqno{(3.1)}$$
We convenient to use the short notations
$$\int 2{G M\over \mu c^2}\Bigl (2+{1\over \beta (\beta +1)}\Bigr )d\rho
=I\quad \hbox {and} \quad \int 2{G M\over\mu c^2}
\Bigl (2+{1\over \beta (\beta +1)}\Bigr )
d\rho\Bigl \vert_{\rho =\rho_0}=I_0.$$
Now, (3.1) can be written in the following forms 
$${{1\over v^2}\Bigl ({d\rho \over d\varphi}\Bigr )^2+{\rho^2\over v^2}
\over e^I}= {\rho_0^2\over v_0^2}{1\over e^{I_0}},\eqno{(3.2)}$$
$$\Bigl ({d\rho \over d\varphi}\Bigr )^2=
\Bigl [{\rho_0^2\over v_0^2}{1\over e^{I_0}}-
{\rho^2\over v^2}{1\over e^I}\Bigr ]v^2e^I.\eqno{(3.3)}$$

Let us denote by $f$ the following {\em characteristic function}
$$f(\rho )={\rho^2\over v^2e^I}.\eqno{(3.4)}$$
Note that $v^2=c^2[1-(C_1+\ln \mu )^{-2}]$ and we can consider the 
function (3.4) only for those $\rho$, for which $v^2\ge 0$. Two cases are 
possible and we shall consider them separately. 

(i) Assume that $C_1\ge 1$. The case $C_1=\infty$, i.e. $v=c$ is included. 
Physically $C_1>1$ means that if the particle left the gravitational filed 
and goes into infinity ($r=\infty$), then it will move with velocity 
$v_0$ such that $C_1=\Bigl (1-{v_0^2\over c^2}\Bigr )^{-1/2}$. 
Specially if $C_1=1$, then $v_0=0$. 
The characteristic function 
$f(\rho )$ is defined for each $\rho >0$ and obviously 
$\lim_{\rho\rightarrow 0}f(\rho )= 0$. 
Since 
$2\le 2+{1\over \beta (\beta +1)}<2{1\over 2}$, we get $\mu^4\le e^I<\mu^5$. 
Thus, if $\rho \rightarrow \infty$, then $v\rightarrow c$ and 
using the previous inequalities we obtain 
$\lim_{\rho\rightarrow \infty }f(\rho )= 0$.
We shall find the maximal value of $f(\rho )$. Indeed, $f'(\rho )=0$ 
can easily be written in an equivalent form 
$$\Bigl [ {dv\over d\rho }+{vG M\over \mu c^2}
\Bigl (2+{1\over \beta (\beta +1)}\Bigr )\Bigr ]\rho ^2=v\rho.$$
By differentiating the equality $\beta -\ln \mu = C_1$, it is easy to get 
${dv\over d\rho}={G M\over \mu}{1\over v}\Bigl (1-{v^2\over c^2}
\Bigr )^{3/2}$. Thus, $f'(\rho )=0$ if and only if 
$${G M\rho \over \mu }{1\over v}\Bigl (1-{v^2\over c^2}\Bigr )^{3/2}
+v{G M\rho\over \mu c^2}\Bigl (2+{1\over \beta (\beta +1)}\Bigr )=v$$
and after some algebraic transformations it it is equivalent to 
$$v^2=G M\rho.\eqno{(3.5)}$$

The physical meaning of (3.5) is very simple. Indeed, it is easy to 
verify that the equations (2.2) are verified if a particle moves with 
velocity $v=\sqrt{G M\rho }$ on a distance $r={1\over \rho}$ on a 
circle. It is just like in the Newton's theory, but remember that 
$C_1>1$ which is analogous with ${1\over 2}v^2-G M\rho>0$ in the 
Newton's theory and according to it the trajectory is hyperbola, but not a 
circle. We shall prove that this effect in ILLTG is a "relativistic effect". 
The equation (3.5) is equivalent to 
$$C_1={1\over \sqrt{1-{G M\rho\over c^2}}}-
\ln \Bigl (1+{G M\rho\over c^2}\Bigr ). \eqno{(3.6)}$$
The graphic of the right side of (3.6) is given on Fig. 1. 

\begin{figure}
\begin{center}
\includegraphics{fig1.eps}
\end{center}
\end{figure}

Each of these circles has radius in the interval 
$\Bigl [{c^2\over G M}, {1\over 0,5030}{c^2\over G M}\Bigr ]$
and will be called {\em Schwarzschild radius} which corresponds to the 
chosen value $C_1>1$. We shall denote these radii by index $s$. The 
graphic of the function $f(\rho )$ is given on figure 2. 
On fig. 3 is given the graphic of $f(\rho )$ when $C_1=1$. 
The graphics are compared with the corresponding Newton's case 
$f_N(\rho )={\rho^2\over v^2}={\rho^2\over const. +2G M\rho}$. 

\begin{figure}
\begin{center}
\includegraphics{fig2.eps}
\end{center}
\end{figure}

\begin{figure}
\begin{center}
\includegraphics{fig3.eps}
\end{center}
\end{figure}

Now according to (3.2) we are able to draw some conclusions for the 
trajectory. Let be given $\rho $, $v$ and ${d\rho \over d\varphi}$ 
at a chosen point. 
Since $\rho $ and $v$ are known, the constant 
$C_1\; (\ge 1)$ is known and also the function $f(\rho )$ is known. 
We have the following 
options and the conclusions follow by Propositions 1 and 2 at the end 
of this section. 

- If ${({d\rho\over d\varphi})^2+\rho^2\over v^2e^I}>f(\rho_s)$, then the 
equation (3.2) has no solutions for $\rho_0$ and hence 
${d\rho\over d\varphi}=0$ will never be achieved and the particle will tend 
to the center $r=0$. This easily can happen for large initial values of 
${d\rho \over d\varphi}$. 

- If ${({d\rho\over d\varphi})^2+\rho^2\over v^2e^I}=f(\rho_s)$, then the 
particle will move on a circle. 

- If ${({d\rho\over d\varphi})^2+\rho^2\over v^2e^I}<f(\rho_s)$, and 
$\rho<\rho_s$, i.e. $r>r_s$, then the trajectory of the particle will 
be out of the Schwarzschild sphere (S.s.), it will look like a hyperbola 
and ${d\rho \over d\varphi} =0$ will be achieved once. 

- If ${({d\rho\over d\varphi})^2+\rho^2\over v^2e^I}<f(\rho_s)$, but 
$\rho>\rho_s$, i.e. $r<r_s$, then the trajectory of the particle will 
be only inside of the S.s., and it will tend to the 
center $r=0$. The condition ${d\rho \over d\varphi}=0$ will be achieved once. 

According to the previous discussion we see that the S.s. with radius 
$r=r_s$ is a sphere of separation. We shall consider further in more 
details the motion of a light ray. In this case $C_1=\infty$, $v=c$ and 
$\rho_s={c^2\over G M}$. On fig. 4 in the $xy$-plane are given different
trajectories of a light ray assuming that ${d\rho \over d\varphi}=0$ at the 
initial point. 

\begin{figure}
\begin{center}
\includegraphics{fig4.eps}
\end{center}
\end{figure}

If the light ray cut the S.s., 
then it will tend to the center $r=0$, because 
the equation (3.2) has no solution. The equations of motion of a light ray 
are simpler, because $\beta =\infty$. 
The equation (3.3) for light rays becomes 
$$\Bigl ({d\rho \over d\varphi}\Bigr )^2 = 
\Bigl [{\rho_0^2\over (1+{G M\rho_0\over c^2})^4} -
{\rho^2\over (1+{G M\rho\over c^2})^4} \Bigr ] 
\Bigl (1+{G M\rho\over c^2}\Bigr )^4,\eqno{(3.7)}$$
while the equation 
$$r^2{d\varphi \over dt}=C_2
\exp \Bigl (-\int {G M\over \mu c^2}
\Bigl (2+{1\over \beta (\beta +1)}\Bigr )d\rho \Bigr )$$
becomes 
$$r^2{d\varphi \over dt}=C_2\Bigl (1+{G M\rho\over c^2}\Bigr )^{-2}.$$
Assuming that ${d\rho \over d\varphi }=0$ if $r=r_0$, we find 
$$C_2=r_0^2{d\varphi \over dt}\Bigr (1+{G M\rho_0\over c^2}\Bigr )^2=
r_0c\Bigr (1+{G M\rho_0\over c^2}\Bigr )^2,$$
and hence 
$$r^2{d\varphi \over dt}=r_0c\Bigl ({1+{G M\over r_0c^2}\over 
1+{G M\over rc^2}}\Bigr )^2.\eqno{(3.8)}$$
The total angle of deflection of a light ray outside the S.s. is equal to 
$$\Delta \alpha =2\int_0^{\rho_0}
{d\rho\over \sqrt{{\rho_0^2\over (1+G M\rho_0/c^2)^4}
(1+G M\rho/c^2)^4-\rho^2}}-\pi . \eqno{(3.9)}$$

(ii) Assume that $C_1<1$. Physically it means that the particle can not 
leave the gravitational field because of small energy. 
In this case $f(\rho )$ is 
defined only for $\rho >\rho _*$, where $\rho_*$ is the solution of 
the equation 

$$1-\ln \Bigl (1+{G M\rho \over c^2}\Bigr )=C_1,\qquad 
\rho_*={c^2\over G M}\Bigl (e^{1-C_1}-1\Bigr ).$$

Note that $\lim_{\rho\rightarrow \rho*+}f(\rho )=\infty$ and 
$\lim_{\rho\rightarrow \infty}f(\rho )= 0$. Similarly to the case (i), the 
condition ${df\over d\rho }=0$ is equivalent to (3.5) and further it reduces 
to (3.6), but now $C_1<1$. Now three subcases are possible. 

(ii$_a$) Assume that $0,9313 < C_1<1$. Then the equation (3.6) has two 
solutions (fig. 1), $\rho_c$ and $\rho_s$. The smaller solution $\rho_c$ has
just the same meaning as in the Newton's theory: Each particle on a distance
$r_c={1\over \rho_c}$ will move on a circle with velocity 
$\sqrt{G M\rho_c}$. The larger solution $\rho_s$ has the same property 
but it is a relativistic effect, because $\rho_s>0,2637 {c^2\over G M}$,
$(r_s<3,7921{G M\over c^2})$, and $r_s={1\over \rho_s}$ will be called 
again Schwarzschild radius because it separates the curves just like on 
fig. 4. The graphic of the function $f(\rho )$ is presented on fig. 5. 

\begin{figure}
\begin{center}
\includegraphics{fig5.eps}
\end{center}
\end{figure}

Analogously as in the previous cases we have the following conclusions. 

- If ${({d\rho\over d\varphi})^2+\rho^2\over v^2e^I}>f(\rho_s )$, 
then the 
equation (3.2) has no solutions, such that on the trajectory will never be 
${d\rho\over d\varphi}=0$. Thus, the particle will tend to the 
center $r=0$. This easily can be achieved if we throw a particle 
with small velocity which is almost radial motion, and then the particle 
will come to the center of the Earth, assuming that mass of the Earth is 
concentrated at a point. According to the Newton's theory the  particle  will 
move on an ellipse with large eccentricity. 

- If ${({d\rho\over d\varphi})^2+\rho^2\over v^2e^I}=f(\rho_s )$, then two 
cases are possible. If $\rho =\rho_s$, then the particle will move on 
a circle on the 
S.s., while if $\rho <\rho_s$, then the particle will tend to the center 
$r=0$. 

- If $f(\rho_c)<{({d\rho\over d\varphi})^2+\rho^2\over v^2e^I}<f(\rho_s )$, 
then two cases are possible. If $\rho <\rho_s$, then the equation (3.3) has 
two solutions smaller than $\rho_s$ and then the trajectory will be like 
an ellipse, analogously to the planetary orbits. The distance $r$ to the 
center of the gravitation satisfies $r_1\le r\le r_2$, where 
$\rho_1={1\over r_1}$ and $\rho_2={1\over r_2}$ can be found as solutions  of 
(3.3). On the other hand, if $\rho >\rho_s$, then the particle will remain
in the S.s. and will tend to the center $r=0$. 

- If ${({d\rho\over d\varphi})^2+\rho^2\over v^2e^I}=f(\rho_c )$, then two 
cases are possible. If $\rho <\rho_s$, then the particle will move on a 
circle, while if $\rho >\rho_s$, the particle will remain in the S.s. and 
will tend to the center $r=0$. 

- If ${({d\rho\over d\varphi})^2+\rho^2\over v^2e^I}<f(\rho_c  )$,  then 
the particle must remain in the S.s. and will tend to the center $r=0$. 

(ii$_b$) Assume that $C_1=0,9313 $. Then the equation (3.6) has a double 
solution $\rho_s=\rho_c=0,2637 {c^2\over G M}$. In this case $\rho_s$ 
and $\rho_c$ coincide, the graphic of $f(\rho )$ is given on 
fig. 6. The discussion is analogous as previous. 

\begin{figure}
\begin{center}
\includegraphics{fig6.eps}
\end{center}
\end{figure}

\begin{figure}
\begin{center}
\includegraphics{fig7.eps}
\end{center}
\end{figure}

(ii$_c$) Assume that $C_1<0,9313 $. Then the equation (3.2) has no solution
and the particle tends to the center $r=0$. The graphic of $f(\rho )$ 
is given on fig. 7. 

At the end of this section we give some global conclusions in 
condensed form. First note that the Schwarzschild radius is defined 
for each constant $C_1\ge 0,9313 $, $r_s$ belongs to the following segment
$${G M\over c^2}\le r_s \le {1\over 0,2637 }{G M\over c^2}
=3,7921{G M\over c^2}$$
and $r_s$ for the light rays is equal to ${G M\over c^2}$. 

- If at the initial moment the particle is inside the S.s. and the 
equation (3.2) has a solution $\rho >\rho_s$, then the particle has 
never been outside the S.s., it will remain in the S.s. and will tend 
to the center $r=0$.

- Suppose that at the initial moment the particle is outside the S.s. 
If there exists an unique solution $\rho$ of (3.2), such that 
$\rho <\rho_s$, then the trajectory will look like a parabola or hyperbola 
and will never pass inside the S.s. If there exist two solutions 
$\rho_1$, $\rho_2$, ($\rho_1, \rho_2 >\rho_s$), then the motion will look 
like an ellipse. 

- If the equation (3.2) has no solution, then it will never be 
${d\rho\over d\varphi}=0$ on the trajectory. The particle will tend to the 
center $r=0$. In this case the particle can cut the S.s. 

We shall explain why the other cases are not possible. If the equation 
(3.2) has at least one solution, the trajectory will be axially symmetric
with respect to the line which connects the center $r=0$ and the point 
where ${d\rho\over d\varphi}=0$. Moreover, the trajectories are 
"time reversible", 
which means that if a particle moves in one direction on the trajectory, then
it can move also in the opposite direction of that trajectory. 

We shall prove the following statements. 

{\bf Proposition 1.} {\em If a particle cut the S.s., then inside the S.s. 
can never be ${d\rho\over d\varphi}=0$, and the particle tends to the center 
of gravitation. }

{\em Proof.} Assume that $\rho$, $v$ and ${d\rho\over d\varphi}$ are given.
Then $\rho$ and $v$ determine the value of $C_1$. Now the equation (3.2) 
can not have a solution, because if such a solution $\rho '$ exists, then
$\rho '<\rho$, because $f(\rho )$ is a decreasing function for 
$\rho >\rho _s$. This finishes the proof. 

{\bf Corollary 1.} 
{\em The trajectories of "elliptic type" can not cut the S.s. } 

{\bf Proposition 2.} {\em Inside the S.s. there does not exist a 
trajectory of "elliptic type". } 

{\em Proof.} Assume that there is a trajectory of elliptic type which 
is inside the S.s. and let the extreme distances of the particle to the 
center of gravitation be $r_1={1\over \rho_1}$ and $r_2={1\over \rho_2}$. 
Then there will exist a radius $r\in (r_1,r_2)$, such that for the same 
value of $C_1$, a new particle can move on a circle of radius $r$. This is 
a contradiction with the fact that each S.s. is not contained inside another 
S.s. because the radius of the S.s. is minimal such that a particle can 
move on a circle for the same value of $C_1$. 

\section{Some relationships between the geodesics in ILLTG and GR} 

In this section we shall find some connections between the planetary orbits 
according to ILLTG using orthonormal coordinates and GR using the 
Schwarzschild solution. 
Namely, we shall show that there are simple 
transformations which the geodesics from ILLTG map into the geodesics from 
GR and conversely. 

Let us denote by $r$, $\rho={1\over r}$, $\varphi$, $t$ the polar 
coordinates in ILLTG, and by $r_*$, $\rho_*={1\over r_*}$, $\varphi_*$, $t_*$
we shall denote the polar coordinates in GR. Note that in both tests 
in the solar system (deflection of the light ray near the Sun and the 
perihelion shift) were measured angles and these angles 
were the same for ILLTG and GR. This associates us to assume that 
$\varphi_*=\varphi$, i.e. we can choose the angle $\varphi$ as a common
parameter, and then we can find the relationships between $\rho$ and 
$\rho_*$, as well as between $t$ and $t_*$. These relationships will be 
found by neglecting the expressions of order $c^{-4}$. We convenient to 
write $\varphi$ instead of $\varphi_*$. Although we shall consider 
ellipses as planetary trajectories, the obtained results are also true if 
the trajectory of a planet is a hyperbola or a parabola. 

Let us start with the equations 
$$d\varphi = -{(1+3{G M\over ac^2(1-\epsilon^2)})d\rho\over 
\sqrt{(\rho_1-\rho)(\rho-\rho_2)}},\eqno{(4.1)}$$
$$d\varphi = -{d\rho_*\over \sqrt{(\rho_1-\rho_*)(\rho_*-\rho_2)}}
\Bigl [1+{G M\over c^2}\rho_*+{2G M\over ac^2(1-\epsilon^2)}
\Bigr ]\eqno{(4.2)}$$
from the ILLTG and GR respectively. The solution of (4.1) we found to be 
$$\rho ={\rho_1+\rho_2\over 2}+{\rho_1-\rho_2\over 2}\cos \Bigl [\varphi 
\Bigl (1-{3G M\over ac^2(1-\epsilon^2)}\Bigr )\Bigr ]\eqno{(4.3)}$$
and the solution of (4.2) is given by 
$$\varphi = -\Bigl (1+{2G M\over ac^2(1-\epsilon^2)}\Bigr )
\Bigl (\arcsin {2\rho_*-\rho_1-\rho_2\over \rho_1-\rho_2}-{\pi\over 2}\Bigr )
+$$
$$+{G M\over c^2}\Bigl [-\sqrt{(\rho_1-\rho_*)(\rho_*-\rho_2)}+
{\rho_1+\rho_2\over 2}\Bigl (-\arcsin 
{2\rho_*-\rho_1-\rho_2\over \rho_1-\rho_2}+{\pi\over 2}\Bigl )\Bigr ]=$$
$$=-\Bigl (1+{3G M\over ac^2(1-\epsilon^2)}\Bigr )\Bigl (\arcsin 
{2\rho_*-\rho_1-\rho_2\over \rho_1-\rho_2}-{\pi\over 2}\Bigr )-
{G M\over c^2}\sqrt{(\rho_1-\rho_*)(\rho_*-\rho_2)}$$
and hence 
$$\cos \Bigl [\varphi \Bigl (1-{3G M\over ac^2(1-\epsilon^2)}\Bigl )
\Bigr ] ={2\rho_*-\rho_1-\rho_2\over \rho_1-\rho_2}+{G M\over c^2}
\sqrt{(\rho_1-\rho_*)(\rho_*-\rho_2)}\sqrt{1-
\Bigl ({2\rho_*-\rho_1-\rho_2\over \rho_1-\rho_2}\Bigr )^2},$$
i.e. 
$$\cos \Bigl [\varphi \Bigl (1-{3G M\over ac^2(1-\epsilon^2)}\Bigl )
\Bigr ] ={2\rho_*-\rho_1-\rho_2\over \rho_1-\rho_2}+{G M\over c^2}
{2(\rho_1-\rho_*)(\rho_*-\rho_2)\over \rho_1-\rho_2}.\eqno{(4.4)}$$
By eliminating the angle $\varphi$ from (4.3) and (4.4), we get 
$$\rho =\rho_*+{G M\over c^2}(\rho_1-\rho_*)(\rho_*-\rho_2)
\eqno{(4.5)}$$
and also the converse relation 
$$\rho _*=\rho -{G M\over c^2}(\rho_1-\rho )(\rho -\rho_2).
\eqno{(4.5')}$$

Now we shall find the analogous equation of (4.1) written in 
ordinary coordinates. We shall use the letter $R$ for radial distance 
in ordinary coordinates. Fist let us write the solution (4.3) in 
the following form 
$$r=\Bigl [{\rho_1+\rho_2\over 2}+{\rho_1-\rho_2\over 2}\cos \Bigl [\varphi 
\Bigl (1-{3G M\over ac^2(1-\epsilon^2)}\Bigr )\Bigr ]\Bigr ]^{-1}$$
and after differentiating, using that 
$dr=\Bigl (1-{GM\over rc^2}\Bigr )dR\approx 
\Bigl (1-{GM\over Rc^2}\Bigr )dR$, 
we obtain 
$$\Bigl (1-{GM\over Rc^2}\Bigr )dR=d\Bigl \{ 
\Bigl [{\rho_1+\rho_2\over 2}+{\rho_1-\rho_2\over 2}\cos \Bigl [\varphi 
\Bigl (1-{3G M\over ac^2(1-\epsilon^2)}\Bigr )\Bigr ]\Bigr ]^{-1}\Bigr \}.$$
After integration of this equation we get 
$$R-{GM\over c^2}\ln {R\over R_1}=
\Bigl [{\rho_1+\rho_2\over 2}+{\rho_1-\rho_2\over 2}\cos \Bigl [\varphi 
\Bigl (1-{3G M\over ac^2(1-\epsilon^2)}\Bigr )\Bigr ]\Bigr ]^{-1},$$
where the constant of integration is chosen such that 
$R=R_1=(\rho_1)^{-1}$ for $\varphi =0$. 
Further, for elliptic trajectories with small eccentricity we can put 
$\ln {R\over R_1}\approx {R\over R_1}-1,$
and hence 
$$R=-{GM\over c^2}+\Bigl (1+{GM\rho_1\over c^2}\Bigr )
\Bigl [{\rho_1+\rho_2\over 2}+{\rho_1-\rho_2\over 2}\cos \Bigl [\varphi 
\Bigl (1-{3G M\over ac^2(1-\epsilon^2)}\Bigr )\Bigr ]\Bigr ]^{-1}.
\eqno{(4.6)}$$
This function $R$ for $\varphi=0$ and 
$\varphi={\pi\over 2}\Bigl (1+{3GM\over ac^2(1-\epsilon )^2)}\Bigr )$ 
takes values $(\rho_1)^{-1}$ and 
$(\rho_2)^{-1}\Bigl (1+{GM\rho_1\over c^2}\Bigr )-{GM\over c^2}\approx 
(\rho_2)^{-1}\Bigl (1+{GM(\rho_2-\rho_1)\over c^2}\Bigr )^{-1}$ respectively.
Thus, in (4.6) $\rho_2$ should be replaced by 
$\rho_2\Bigl (1-{GM(\rho_2-\rho_1)\over c^2}\Bigr )$. Hence we obtain 
$$R=-{GM\over c^2}+$$
$$+\Bigl (1+{GM\rho_1\over c^2}\Bigr )
\Bigl \{ {1\over 2}\Bigl [\rho_1+\rho_2-\rho_2{GM(\rho_2-\rho_1)\over c^2}
\Bigr ]
+{1\over 2}\Bigl [\rho_1-\rho_2+\rho_2{GM(\rho_2-\rho_1)\over c^2}\Bigr ]
\cos \varphi \Bigl ( 1-{3GM\over ac^2(1-\epsilon^2)}\Bigr )\Bigr \} ^{-1}.$$
After some calculations, one can verify that the function 
$R^{-1}$ satisfies the equality 
$$R^{-1}=\rho -{GM\over c^2}(\rho_1-\rho )(\rho -\rho_2),
\eqno{(4.7)}$$
i.e. 
$$\rho_*=R^{-1} .\eqno{(4.8)}$$
The differential equation for $R^{-1}$, analogous to (4.1) is given by 
$$d\varphi = -{d{1\over R}\over 
\sqrt{(\rho_1-R^{-1})(R^{-1}-\rho_2)}}
\Bigl [1+{G M\over c^2}{1\over R}+{2G M\over ac^2(1-\epsilon^2)}
\Bigr ] .\eqno{(4.9)}$$

Note that if we put 
$r=r_*\Bigl (1-{GM\over r_*c^2}\Bigr )=r_*-{GM\over c^2}$, 
then the equation (4.1) becomes 
$$d\varphi = -{d\rho_*\over \sqrt{(\rho_{1*}-\rho_*)(\rho_*-\rho_{2*})}}
\Bigl [1+{G M\over c^2}\rho_*+{2G M\over ac^2(1-\epsilon^2)}
\Bigr ],\eqno{(4.9')}$$
where $\rho_{1*}=(r_1+{GM\over c^2})^{-1}$ and 
$\rho_{2*}=(r_2+{GM\over c^2})^{-1}$. 

The equation (4.8) shows that 
{\bf for $v<<c$, the trajectory of motion in the 
$(r,\varphi )$ plane is the same with the trajectory according ILLTG 
using ordinary coordinates}. 

Further we shall find the relationships between $t$, $t_*$ and $T$. 
According to GR it is 
$$r_*^2{d\varphi\over dt_*} = C_1\Bigl (1-{2G M\over c^2}\rho_*\Bigr )$$
and according to ILLTG it is 
$$r^2{d\varphi\over dt} = C_2\Bigl (1-{2G M\over c^2}\rho 
-{1\over \beta (\beta +1)}{G M\rho\over c^2}\Bigr ).$$
The quotient of these two equations yields to 
$$\Bigl ({r_*\over r}\Bigr )^2{dt\over dt_*}=
C_3\Bigl (1+{1\over \beta (\beta +1)}{G M\rho\over c^2}\Bigr ).$$
Since ${r_*\over r}=1+{GM\over \rho c^2}(\rho_1-\rho )(\rho -\rho _2)$ 
according to (4.5'), and 
${1\over \beta (\beta +1)}\approx {1\over 2}$ 
we obtain 
$${dt\over dt_*}
=C_3\Bigl (1+{1\over 2}{GM\rho\over c^2}
-2{GM\over \rho c^2}(\rho_1-\rho )(\rho -\rho _2)
\Bigr ).$$
Moreover, $C_3=1$, because in case of parabola or hyperbola, 
${dt\over dt_*}\rightarrow 1$ if $\rho \rightarrow 0$. Hence we get 
$${dt\over dt_*}=1+{1\over 2}{GM\rho\over c^2}
-2{GM\over \rho c^2}(\rho_1-\rho )(\rho -\rho _2).$$
Assuming that the trajectory is elliptic with eccentricity close to 0, then 
$${dt\over dt_*}=1+{1\over 2}{GM\rho\over c^2}.$$
Moreover, we can find the connection between $t_*$ and the ordinary 
time coordinate $T$ from ILLTG using that 
$dt=\Bigl (1+{GM\rho\over c^2}\Bigr )dT$, i.e. 
$${dT\over dt_*}=1-{1\over 2}{GM\rho\over c^2}.\eqno{(4.10)}$$

The formula (4.10) does not show how much the time according to GR
is "faster" (or "slower") than the time according to ILLTG, 
but only a factor 
for transition from the geodesic of GR into geodesic of ILLTG 
in ordinary coordinates. 

\medskip
Further we shall compare also the null geodesics, i.e. 
the trajectories of the light rays. According to (2.14) in ILLTG it is 
$$\rho ={1\over R}\cos \varphi +{G M\over R^2c^2}(2-2\cos \varphi ),
\eqno{(4.11)}$$
while according to GR it is [5] 
$$\rho_* ={1\over R}\cos \varphi +{G M\over R^2c^2}(2-\cos^2\varphi ).
\eqno{(4.12)}$$
Hence we obtain 
$$\rho_* = \rho +{G M\over R^2c^2}\cos \varphi (2-\cos \varphi ),
\eqno{(4.13)}$$
$${\rho_*\over \rho}=1+{G M\over \rho R^2c^2}
\cos \varphi (2-\cos \varphi ),$$
i.e. 
$${r_*\over r}=1-{G Mr\over R^2c^2}
\cos \varphi (2-\cos \varphi ). \eqno{(4.14)}$$

Further, using (4.14) as well as the following two equalities 
$$r^2{d\varphi \over dt}=C\Bigl (1-2{G M\over rc^2}\Bigr ),$$
$$r_*^2{d\varphi \over dt_*}=C\Bigl (1-2{G M\over r_*c^2}\Bigr ),$$
which are almost the same in both theories, we get 
$$\Bigl ({r_*\over r}\Bigr )^2 {dt\over dt_*}=1,$$
$${dt\over dt_*}=1+2{G Mr\over R^2c^2}\cos \varphi (2-\cos \varphi ).
\eqno{(4.15)}$$

The angle of deflection 
from a point $A$ to a point $B$ when the ray passes near the Sun
according to (2.16) in ILLTG is equal to 
$$\Delta \alpha _1={G M\over Rc^2}\Bigl (2\sqrt{1-{R^2\over r_1^2}}+
2\sqrt{1-{R^2\over r_2^2}}\Bigr ),\eqno{(4.16)}$$
and also using PPN coordinates GR yields to the same formula (4.16). 
This value (4.16) is confirmed 
with a very precise measurements in the recent test with Cassini spacecraft
[4]. According to ILLTG, the angle of deflection is always twice 
larger than the classical Newton's result, and can be explained with the 
coefficient $2-\beta^{-2}=1+v^2/c^2=2$ in (2.2) when $v=c$. 
Note that 
this deflection of the light is not a "gravitational effect", because it will
appear if we have only an acceleration without gravitation, and ILLTG 
differs slightly in these two cases when there is a gravitation and there
is only simulation of gravitation using inertial forces. 
About this will be said more at the end of this section. 

We shall compare now the radial motions of a particle 
with large velocity $v$ in a gravitational field. The Newton's relation 
${1\over 2}v^2-G M\rho=const.$ according to ILLTG should be replaced by
(2.2c), i.e. 
$${1\over \sqrt{1-{v^2\over c^2}}}-
\ln \Bigl (1+{G M\rho\over c^2}\Bigr )= const. \eqno{(4.17)}$$
According to GR, for a radial motion in a gravitational field we get 
$${\sqrt{1-{2G M\rho\over c^2}}\over \sqrt{1-{v^2\over c_*^2}}}=const.
\eqno{(4.18)}$$
where $c_*=c(1-{2G M\rho\over c^2})$ is the velocity of the light 
measured in the gravitational field. The physical interpretation for both 
formulas (4.17) and (4.18) obtains if we multiply them by $m_0c^2$. {From} 
(4.17) we obtain 
$$E_k + E_p=const.\eqno{(4.19)}$$
because in section 6 will be proved that the potential energy for 
a particle with small mass is equal to 
$E_p=-m_0c^2\ln \Bigl (1+{G M\rho\over c^2}\Bigr )$. The equation 
(4.18) can be written as 
$$E_{total} \sqrt{1-{2G M\rho\over c^2}}=const.\eqno{(4.20)}$$

If ${G M\rho\over c^2}$ is very small, i.e. we consider a weak 
gravitational field, (4.18) can be written in the form 
$${1\over \sqrt{1-{v^2\over c_*^2}}}=C{G M\rho\over c^2}+C,  
\qquad C=const.$$
and hence 
$${d\over dt}{1\over \sqrt{1-{v^2\over c_*^2}}}=
C{G M\over c^2}{d\over dt}{1\over r},$$
i.e. 
$${d\over dt}{1\over \sqrt{1-{v^2\over c_*^2}}}=C{a\over c^2}v,
\eqno{(4.21)}$$
where $a=-{G M\over r^2}$ is the acceleration and $v=dr/dt$. The 
analogous equation from (4.17) in weak gravitational field is the following 
$${d\over dt}{1\over \sqrt{1-{v^2\over c^2}}}={a\over c^2}v.\eqno{(4.22)}$$ 

In a short time interval the acceleration $a$ is almost constant and the 
equations (4.21) and (4.22) should be close to the equation of motion of a 
particle in STR under a constant acceleration $a$. The STR motion 
$${d\over \sqrt{1-{v^2\over c^2}}dt}{m_0c^2\over 
\sqrt{1-{v^2\over c^2}}} = a {m_0v\over \sqrt{1-{v^2\over c^2}}}
\eqno{(4.23)}$$
is just the equation (4.22). Indeed, in (4.23) $a$ is a constant, because 
the acceleration $a$ is unchanged under the Lorentz transformations 
of motion in the $x$ direction, because the tensor with components 
$$\left [\matrix{0&0&0&a\cr
0&0&0&0\cr 0&0&0&0\cr -a&0&0&0\cr }\right ]$$
is unchanged under such Lorentz transformations, analogously as the 
component $E_x$ of the electric field is constant under such Lorentz 
transformations. 

Both of equations (4.21) and (4.22)
in the $(xt)$-plane are hyperbolas, but with different 
parameters. The parameter $C$ has no physical meaning according 
to STR for a motion under a constant acceleration, although this constant 
in (4.18) has a good physical meaning. Thus, the equation of 
radial motion is not close to the STR. Moreover, from (4.18) one can 
conclude that ${dv\over dt}\approx a(1-3{v^2\over c^2})$, such that the 
acceleration changes its sign if $v>c/\sqrt{3}$. 

We note that ILLTG is very close to the STR. Namely, if we consider a 
noninertial system without gravitation but there is only a 3-vector of 
acceleration $(a_x,a_y,a_z)$ and a 3-vector of angular velocity 
$(w_x,w_y,w_z)$, then in the equations of motion (2.3) we should only put 
$\mu =1$. This is obtained in the frame of STR, under the assumption 
that the Jacobi 
matrices with respect to the inertial systems are orthogonal 
and hence the coordinates in such a system are unholonomic [9]. The 
influence of the coefficient $\mu$ in (2.2) and (2.3) is very small. 
For example, if we have a noninertial system without gravitation with the 
same acceleration $(a_x,a_y,a_z)=\Bigl 
(-{G M\over r^3}x,-{G M\over r^3}y,-{G M\over r^3}z\Bigr )$, 
then the angle of deflection of a light ray near the Sun will be again 
$\Delta \alpha ={4G M\over Rc^2}$, while the angle between two 
perihelions will be 
$\Delta \alpha = {7G M\pi\over ac^2(1-\epsilon^2)}$, instead of 
$\Delta \alpha = {6G M\pi\over ac^2(1-\epsilon^2)}$. The 
equation (2.4) will not be changed, and the equation (4.17) will be 
replaced by 
$${1\over \sqrt{1-{v^2\over c^2}}}-
{G M\rho\over c^2}= const. $$

\section{Two proposal tests} 

After the discussion in section 4 we propose the following 
tests for which we have different results from ILLTG and GR. 

{\bf Test 1.} Let $A_1$, $A_2$ and $A_3$ are three not close points 
of a planet orbit or a spacecraft trajectory, 
and let $\angle A_0OA_1=\varphi_0$, 
$\angle A_0OA_2=\varphi_1$, $\angle A_0OA_3=\varphi_2$, where $A_0$ is the 
perihelion of the planet orbit and O is the center of the Sun. We denote by 
$T_1$ and $T_2$ the times for which the planet comes from $A_1$ to $A_2$ 
and from $A_2$ to $A_3$ respectively. We will calculate $T_1:T_2$ according 
to ILLTG and GR, and later can be compared with the experimental results. 
We shall find $T_1:T_2$ where $T_1$ and $T_2$ are the proper times 
for the Earth, where the trajectory of the Earth is observed. 
Let us denote the proper time for the Earth by $\tau$. 

First let us calculate $T_1:T_2$ according to ILLTG. {From} (2.4) we obtain 
$$r^2{d\varphi \over d\tau}=r^2{d\varphi \over dt}{dt\over dT}
{dT\over d\tau} = C_2\Bigl (1-{5\over 2}{G M\over rc^2}\Bigr )
\Bigl (1+ {G M\over rc^2}\Bigr )\Bigl (1+ {G M\over rc^2}\Bigr ) =
C_2\Bigl (1-{1\over 2}{G M\over rc^2}\Bigr ). 
$$
Thus 
$$d\tau ={d\varphi\over \rho^2C_2(1-{1\over 2}{G M\rho\over c^2})} = 
{1\over C_2}\rho^{-2}\Bigl (1+{1\over 2}{G M\rho\over c^2}
\Bigr )d\varphi,$$
and hence 
$$T_1:T_2=\int _{\varphi_0}^{\varphi_1}
\Bigl (\rho^{-2}+{1\over 2}{G M\rho^{-1}\over c^2}\Bigr )d\varphi :
\int_{\varphi_1}^{\varphi_2}
\Bigl (\rho^{-2}+{1\over 2}{G M\rho^{-1}\over c^2}\Bigr )d\varphi ,$$
where $\rho$ is given by (4.3). 
Using that ${G M\over c^2}$ 
is close to 0, and replacing 
$\varphi\Bigl (1-{3G M\over ac^2(1-\epsilon^2)}\Bigr )$ with a new 
variable, one obtains 
$$T_1:T_2={A\over C}-{3G M\over ac^2(1-\epsilon^2)} 
{C(\rho^{-2}(\varphi_1)\varphi_1-\rho^{-2}(\varphi_0)\varphi_0)-
A(\rho^{-2}(\varphi_2)\varphi_2-\rho^{-2}(\varphi_1)\varphi_1)\over C^2}+$$
$$+{1\over 2}{G M\over c^2}{BC-AD\over C^2},\eqno{(5.1)}$$
where 
$$A=\int_{\varphi_0}^{\varphi_1}\rho^{-2}d\varphi, \quad 
B=\int_{\varphi_0}^{\varphi_1}\rho^{-1}d\varphi, \quad 
C=\int_{\varphi_1}^{\varphi_2}\rho^{-2}d\varphi, \quad 
D=\int_{\varphi_1}^{\varphi_2}\rho^{-1}d\varphi. $$

After solving these integrals, where 
$\rho ={1\over 2}[(\rho_1+\rho_2)+(\rho_1-\rho_2)\cos \varphi ]$, 
we obtain 
$$A=-{\rho_1+\rho_2\over (\rho_1\rho_2)^{3/2}}
\arctan {\sqrt{\rho_1\rho_2}(\cot {\varphi_1\over 2}-\cot {\varphi_0\over 2})
\over \rho_2+\rho_1\cot {\varphi_1\over 2}\cot {\varphi_0\over 2}}+
{\rho_2-\rho_1\over \rho_1\rho_2}
\Bigl ({\cot {\varphi _1\over 2}\over \rho_2+\rho_1\cot^2{\varphi_1\over 2}}-
{\cot {\varphi _0\over 2}\over \rho_2+\rho_1\cot^2{\varphi_0\over 2}}\Bigr ),
\eqno{(5.2)}$$
$$C=-{\rho_1+\rho_2\over (\rho_1\rho_2)^{3/2}}
\arctan {\sqrt{\rho_1\rho_2}(\cot {\varphi_2\over 2}-
\cot {\varphi_1\over 2})
\over \rho_2+\rho_1\cot {\varphi_2\over 2}\cot {\varphi_1\over 2}}+
{\rho_2-\rho_1\over \rho_1\rho_2}
\Bigl ({\cot {\varphi _2\over 2}\over \rho_2+\rho_1\cot^2{\varphi_2\over 2}}
-{\cot {\varphi _1\over 2}\over \rho_2+\rho_1\cot^2{\varphi_1\over 2}}
\Bigr ),\eqno{(5.3)}$$
$$B={2\over \sqrt{\rho_1\rho_2}}
\arctan {\sqrt{\rho_1\rho_2}(\tan {\varphi_1\over 2}-\tan {\varphi_0\over 2})
\over \rho_1+\rho_2\tan {\varphi_1\over 2}\tan {\varphi_0\over 2}},
\eqno{(5.4)}$$
$$D={2\over \sqrt{\rho_1\rho_2}}
\arctan {\sqrt{\rho_1\rho_2}(\tan {\varphi_2\over 2}-\tan {\varphi_1\over 2})
\over \rho_1+\rho_2\tan {\varphi_2\over 2}\tan {\varphi_1\over 2}}.
\eqno{(5.5)}$$

Now let us calculate $T_1:T_2$ according to GR. Using that 
$r_*^2{d\varphi \over d\tau_*} = C_1$, 
and (4.5') we obtain 
$$d\tau_*={d\varphi\over \rho_*^2C_1}=
{d\varphi\over C_1\rho^2
\Bigl [1-{2G M\rho\over c^2}({\rho_1\over \rho}-1)
(1-{\rho_2\over \rho})\Bigr ]},$$
$$\tau_*={1\over C_1}\int \rho^{-2}
\Bigl [1+{2G M\rho\over c^2}({\rho_1\over \rho}-1)
(1-{\rho_2\over \rho})\Bigr ]d\varphi ,$$
where $\rho$ is given by (4.3). 
If the eccentricity of the ellipse is small, then the expression 
${2G M\rho\over c^2}({\rho_1\over \rho}-1)(1-{\rho_2\over \rho})$
can be neglected. 
Then, analogously to (5.1) we obtain 
$$T_1:T_2={A\over C}-{3G M\over ac^2(1-\epsilon^2)} 
{C(\rho^{-2}(\varphi_1)\varphi_1-\rho^{-2}(\varphi_0)\varphi_0)-
A(\rho^{-2}(\varphi_2)\varphi_2-\rho^{-2}(\varphi_1)\varphi_1)\over C^2}
\eqno{(5.6)}$$
where $A$, $C$, $B$ and $D$ are given by (5.2-5). 

It is convenient to assume that $\varphi_0=0$, $\varphi_1=\pi /2$ and 
$\varphi_2=\pi$. Then 
$$A={\rho_1+\rho_2\over (\rho_1\rho_2)^{3/2}}\arctan \sqrt
{{\rho_2\over \rho_1}}+{\rho_2-\rho_1\over (\rho_1+\rho_2)\rho_1\rho_2},
\quad C={\rho_1+\rho_2\over (\rho_1\rho_2)^{3/2}}\arctan \sqrt
{{\rho_1\over \rho_2}}-{\rho_2-\rho_1\over (\rho_1+\rho_2)\rho_1\rho_2},
\eqno{(5.7)}$$
$$B={2\over \sqrt{\rho_1\rho_2}}\arctan \sqrt{{\rho_2\over \rho_1}},
\quad D={2\over \sqrt{\rho_1\rho_2}}\arctan \sqrt{{\rho_1\over \rho_2}}.
\eqno{(5.8)}$$

Remember that both formulas (5.1) and (5.6) are obtained 
for the observers on the planet whose trajectory is observed. Indeed, 
it means that these formulas are valid only for the trajectory of the Earth,
because we observe from the Earth. Further let us modify these two 
formulas assuming that the observer is far from gravitation  
in the solar system. Then, according to ILLTG we obtain 
$$dt={d\varphi (1+2{G M\rho\over c^2})\over \rho^2C_2(1-{1\over 2}
{G M\rho\over c^2})} = 
{1\over C_2}\rho^{-2}\Bigl (1+{5\over 2}{G M\rho\over c^2}
\Bigr )d\varphi,$$
$$T_1:T_2=\int _{\varphi_0}^{\varphi_1}
\Bigl (\rho^{-2}+{5\over 2}{G M\rho^{-1}\over c^2}\Bigr )d\varphi :
\int _{\varphi_1}^{\varphi_2}
\Bigl (\rho^{-2}+{5\over 2}{G M\rho^{-1}\over c^2}\Bigr )d\varphi ,$$
and hence 
$$T_1:T_2={A\over C}-{3G M\over ac^2(1-\epsilon^2)} 
{C(\rho^{-2}(\varphi_1)\varphi_1-\rho^{-2}(\varphi_0)\varphi_0)-
A(\rho^{-2}(\varphi_2)\varphi_2-\rho^{-2}(\varphi_1)\varphi_1)\over C^2}+$$
$$+{5\over 2}{G M\over c^2}{BC-AD\over C^2},\eqno{(5.1')}$$
where $A$, $B$, $C$ and $D$ were previously determined. Analogously, 
according to GR instead of (5.6) we have the formula 
$$T_1:T_2={A\over C}-{3G M\over ac^2(1-\epsilon^2)} 
{C(\rho^{-2}(\varphi_1)\varphi_1-\rho^{-2}(\varphi_0)\varphi_0)-
A(\rho^{-2}(\varphi_2)\varphi_2-\rho^{-2}(\varphi_1)\varphi_1)\over C^2}+$$
$$+2{G M\over c^2}{BC-AD\over C^2},\eqno{(5.6')}$$
where $A$, $C$, $B$ and $D$ were previously determined. 

Finally, if we measure from the Earth the trajectory of another planet,
then we should modify the formulas (5.1') and (5.6') because the 
trajectory of the Earth is not a circle, the "speed" of the time is 
proportional to 
$(1-{G M\over rc^2})^2$, where $r$ is the distance from the 
Earth to the center of the Sun. Thus, let us denote by $\nu_1$ and ${\nu_2}$
the average values of the "speed" of the time on the Earth during the first 
and the second time interval. Thus, on the Earth we shall measure the 
quotient 
$${\nu_1T_1\over \nu_2T_2}={\nu_1\over \nu_2} {T_1\over T_2}
\eqno{(5.9)}$$
where $T_1:T_2$ is given by (5.1') and (5.6') respectively 
according to ILLTG and GR. 

Note that in this test we should measure only the quotient of long 
time intervals, large angles and we need also the quotient 
$\rho_1:\rho_2=(1+\epsilon ):(1-\epsilon )$. The perturbations from the 
other planets, Moon, stars, galaxy and universe 
(see sec. 8 and Ref. [10,11]) are not included in the final results. 

\smallskip 
{\bf Test 2.} 
Let us consider the second Kepler's law, i.e. consider the expression 
$r^2{d\varphi \over d\tau}$ on the trajectory, 
where $\tau$ is the proper time. 
We shall present the formulas assuming that we observe the 
trajectory of the Earth from the Earth. 

According to ILLTG we have 
$$r^2{d\varphi \over d\tau}=r^2{d\varphi \over dt}{dt\over dT}
{dT\over d\tau} =
C'\Bigl (1-{5\over 2}{G M\over rc^2}\Bigr )
\Bigl (1+ {G M\over rc^2}\Bigr )\Bigl (1+ {G M\over rc^2}\Bigr )
=C'\Bigl (1-{1\over 2}{G M\over rc^2}\Bigr )
$$
and according to GR (omitting the star indices), it is 
$$r^2{d\varphi \over d\tau} = C'',$$
where $C'$ and $C''$ are constants close to 
$C=\sqrt{G Ma(1-\epsilon^2)}$. Let $P$ and $Q$ be two points on the 
orbit and let us denote the corresponding distances to the center of the Sun 
by $r_1$ and $r_2$ respectively. For example, $P$ can be the perihelion of 
the Earth and $Q$ aphelion. Using the previous formulas 
according to ILLTG we obtain 
$$r_2^2\Bigl ({d\varphi \over d\tau}\Bigr )_Q-
r_1^2\Bigl ({d\varphi \over d\tau}\Bigr )_P = 
C\Bigl (-{1\over 2}{G M\over r_2c^2}+
{1\over 2}{G M\over r_1c^2}\Bigr )= 
{1\over 2}{G M(r_2-r_1)\over r_1r_2c^2} 
\sqrt{G Ma(1-\epsilon ^2)}\eqno{(5.10)}$$
and according to GR we obtain 
$$r_2^2\Bigl ({d\varphi \over d\tau}\Bigr )_Q-
r_1^2\Bigl ({d\varphi \over d\tau}\Bigr )_P =0.
\eqno{(5.11)}$$

The perturbations from the other gravitational objects 
are not included in (5.10) and (5.11). 

Note that in the tests 1 and 2 the departure between the GR and ILLTG 
is proportional to ${1\over 2}{GM\over rc^2}$. The coefficient ${1\over 2}$
here is just ${1\over \beta (\beta +1)}$. 

\section{Mass and energy in gravitational field according to ILLTG}

If a particle with mass $m$, measured when it is 
far from gravitational field, has come in a gravitational field with 
potential $\mu =1+{G M\over rc^2}$, then its mass measured from an 
observer far from gravitation, approximately is 
equal to ${m\over 1+{G M\over rc^2}}$. 
We convenient in this section 
that $M,m_1, m_2,...$ will mean proper mass, measured 
when both the observer and the mass are far from gravitation. 

Let us consider two particles with masses $m_1$ and $m_2$. Assume that the 
distance between their centers is $R$. Then the gravitational force between 
them is given approximately by 
$$f=G {m_1\over 1+{G m_2\over Rc^2}}
{m_2\over 1+{G m_1\over Rc^2}}{1\over R^2},\eqno{(6.1)}$$
because each of the masses changes under the gravitational field of the 
other particle. This formula becomes more precise if one of the masses 
is very close to zero, but (6.1) is of sufficient accuracy. For example, 
if $m_1\approx 0$ is mass of a small particle and $m_2=M$ is the mass of the 
Sun, then 
$f={{G mM\over R^2}\over 1+{G M\over Rc^2}}$
and the acceleration for the particle is 
$a={{G M\over R^2}\over 1+{G M\over Rc^2}}$, which is in 
accordance with the formulas in section 2. 

The energy which needs to distinguish both objects on a 
large distance is equal to 
$$E=\int _R^{\infty}fdr = \int_R^{\infty}
G {m_1\over 1+{G m_2\over rc^2}}
{m_2\over 1+{G m_1\over rc^2}}{1\over r^2}dr,$$
$$E={m_1m_2c^2\over m_1-m_2}\Bigl [\ln \Bigl (1+{G m_1\over Rc^2}\Bigr )
- \ln \Bigl (1+{G m_2\over Rc^2}\Bigr )\Bigr ],\eqno{(6.2)}$$
which corresponds to the defect of mass 
$$\Delta m=
{m_1m_2\over m_1-m_2}\Bigl [\ln \Bigl (1+{G m_1\over Rc^2}\Bigr )
- \ln \Bigl (1+{G m_2\over Rc^2}\Bigr )\Bigr ].\eqno{(6.3)}$$
If we assume in (6.2) that $m_1=M$ 
is a mass of a large body for example a star, 
and $m_2=m$ has negligible mass compared with $M$, then according to 
(6.3) for the potential energy of the particle we get 
$$E_p=-mc^2\ln \Bigl (1+{G M\over Rc^2}\Bigr ).\eqno{(6.4)}$$
Later we will return to this formula. In opposite case, if $m_1=m_2=m$, 
analogously to (6.2) the potential energy is equal to 
$$E_p=-{G m^2\over R+{G m\over c^2}}\eqno{(6.5)}$$
and the corresponding defect of mass is equal to 
$$\Delta m={G m^2\over Rc^2+G m}.\eqno{(6.6)}$$
This energy can be written in the form 
$$E_p={mc^2\over 1+{G m\over Rc^2}}-mc^2,\eqno{(6.7)}$$
which is analogous to the formula for the 
kinetic energy according to the STR. 

Further, let us consider the case when $R\rightarrow 0$. Namely, we shall 
consider an "ideal fusion", assuming that both particles are concentrated 
into points and assuming that $R$ tends to 0. {From} (6.1) it just follows 
that the force tends to 
$$f={c^4\over G },\eqno{(6.8)}$$
while the acceleration for each particle is of type 
$$a\sim {c^2\over R}. \eqno{(6.9)}$$
Note that this acceleration does not depend on each of the masses. 
Probably it does not depend on the nature of these particles,
for example their electricity if they are charged. In this case, 
when $R\rightarrow 0$, according to (6.2) for the defect of the mass we 
obtain 
$$\Delta m={m_1m_2\over m_1-m_2}\ln {m_1\over m_2},\eqno{(6.10)}$$
such that the mass after an ideal fusion is equal to 
$$m=m_1+m_2-{m_1m_2\over m_1-m_2}\ln {m_1\over m_2}.\eqno{(6.11)}$$
Using the differential calculus, it is easy to obtain the following 
inequalities:
$$m_2\le {m_1m_2\over m_1-m_2}\ln {m_1\over m_2}\le {m_1+m_2\over 2}\le m_1,
\eqno{(6.12)}$$
assuming that $m_2\le m_1$. Specially, if $m_1=m_2=m$, then 
according to (6.6), the defect of mass is equal to $\Delta m=m$. Hence the 
conclusion that in "ideal fusion" we get the largest quantity of energy 
if $m_1=m_2=m$. Note also that if 
$m_2\rightarrow 0$ in (6.12), then 
${m_1m_2\over m_1-m_2}\ln {m_1\over m_2}\rightarrow 0$. 

Analogously to the inequalities (6.12), in general case one can prove the 
following inequalities 
$$m_2-{m_2\over 1+{G m_1\over rc^2}}\le 
{m_1m_2\over m_1-m_2}\ln {1+{G m_1\over rc^2}\over 
1+{G m_2\over rc^2}}\le $$
$$\le {1\over 2}\Bigl [\Bigl (m_1-{m_1\over 1+{G m_2\over rc^2}}\Bigr )+
\Bigl (m_2-{m_2\over 1+{G m_1\over rc^2}}\Bigr )\Bigr ] \le 
m_1-{m_1\over 1+{G m_2\over rc^2}},\eqno{(6.13)}$$
if $m_2\le m_1$. Indeed, in special case if we put $r=0$, we obtain the 
inequalities (6.12). 

According to (6.4) the equation (4.17) can be written as 
$$E_k+E_p=const.\eqno{(6.14)}$$
where $E_k$ is the kinetic energy. The equation (6.4) was obtained under the 
assumption that the mass of the particle is almost 0. Now, if we want to 
obtain the equations of motion using that the mass of the particle 
is not almost 0, for example motion of binary stars, 
we should use the potential 
$$U=-\Bigl (1+{3Gm\over rc^2}\Bigr )
{c^2\over 1-{m\over M}}\ln {1+{G M\over rc^2}\over 1+{G m\over rc^2}},
\eqno{(6.15)}$$
which is obtained from (6.2), 
where $m$ and $M$ are the masses of the two bodies. 
The "correcting multiplier" 
$1+{3Gm\over rc^2}$ appears if we want to use this 
potential to derive the equations of motion for the body with mass $m$. 
Indeed, the body with nonzero mass $m$ influences on the gravitational 
potential of the other body. This potential (6.15) is sufficient 
to derive the approximation of the first post-Newtonian order (sec. 7). 
At the end of section 7 will be proved that the "correcting multiplier" 
must be $1+{3Gm\over rc^2}$. 

If we consider the motion of a body with nonzero mass, 
under the gravitation of $k$ bodies, then we can use the equations 
(2.19a,b,c,d), but the potential $U_i$ instead of (2.17) is given by 
$$U_i=-\Bigl (1+{3Gm\over r_ic^2}\Bigr )
{c^2\over 1-{m\over m_i}}\ln {1+{G m_i\over r_ic^2}\over 
1+{G m\over r_ic^2}},
\eqno{(6.16)}$$
where $m$ is the mass of the moving test body. 

Note that (2.19) does not consider the interaction among the given 
$k$ bodies, but only assumes that their positions and velocities are given. 
But it does not matter, because 
if it is given a system of $n$ bodies with masses 
$m_1$, $m_2$, $\ldots $, $m_n$, with initial positions and initial 
velocities, then we can find the coordinates of all bodies as 
functions of $t$ by considering a system of $3n+n$ equations, 
i.e. 4 equations 
(2.19a,b,c,d) for each body, where the potential for each pair of bodies 
is given analogously to (6.16). Indeed, then we have the following system 
$${d^2x_j\over dt^2}=\sum_{i=1,i\neq j}^n
\Bigl [{\partial U_{ij}\over \partial x_j}
\Bigl (\beta_j^{-2}-2+\frac{2}{c^2}
\Bigl ({dx_i\over dt}{dx_j\over dt}+{dy_i\over dt}{dy_j\over dt}+
{dz_i\over dt}{dz_j\over dt}\Bigr )\Bigr )+
{1\over c^2}{\partial U_{ij}\over \partial x_j}
\Bigl (M({dx_j\over dt})^2-2{dx_i\over dt}{dx_j\over dt}\Bigr )$$
$$\quad +
{1\over c^2}{\partial U_{ij}\over \partial y_j}
\Bigl (M{dx_j\over dt}{dy_j\over dt}-2{dx_i\over dt}{dy_j\over dt}\Bigr ) +
{1\over c^2}{\partial U_{ij}\over \partial z_j}
\Bigl (M{dx_j\over dt}{dz_j\over dt}-2{dx_i\over dt}{dz_j\over dt}\Bigr )
\Bigr ]\beta_i, \quad 1\le j\le n,\eqno{(6.17a)}$$
$${d^2y_j\over dt^2}=\sum_{i=1,i\neq j}^n
\Bigl [{\partial U_{ij}\over \partial y_j}
\Bigl (\beta_j^{-2}-2+\frac{2}{c^2}
\Bigl ({dx_i\over dt}{dx_j\over dt}+{dy_i\over dt}{dy_j\over dt}+
{dz_i\over dt}{dz_j\over dt}\Bigr )\Bigr )+
{1\over c^2}{\partial U_{ij}\over \partial x_j}
\Bigl (M{dx_j\over dt}{dy_j\over dt}-2{dy_i\over dt}{dx_j\over dt}\Bigr )$$
$$\quad +
{1\over c^2}{\partial U_{ij}\over \partial y_j}
\Bigl (M({dy_j\over dt})^2-2{dy_i\over dt}{dy_j\over dt}\Bigr ) +
{1\over c^2}{\partial U_{ij}\over \partial z_j}
\Bigl (M{dy_j\over dt}{dz_j\over dt}-2{dy_i\over dt}{dz_j\over dt}\Bigr )
\Bigr ]\beta_i, \quad 1\le j\le n,\eqno{(6.17b)}$$
$${d^2z_j\over dt^2}=\sum_{i=1,i\neq j}^n
\Bigl [{\partial U_{ij}\over \partial z_j}
\Bigl (\beta_j^{-2}-2+\frac{2}{c^2}
\Bigl ({dx_i\over dt}{dx_j\over dt}+{dy_i\over dt}{dy_j\over dt}+
{dz_i\over dt}{dz_j\over dt}\Bigr )\Bigr )+
{1\over c^2}{\partial U_{ij}\over \partial x_j}
\Bigl (M{dx_j\over dt}{dz_j\over dt}-2{dz_i\over dt}{dx_j\over dt}\Bigr )$$
$$\quad +
{1\over c^2}{\partial U_{ij}\over \partial y_j}
\Bigl (M{dy_j\over dt}{dz_j\over dt}-2{dz_i\over dt}{dy_j\over dt}\Bigr ) +
{1\over c^2}{\partial U_{ij}\over \partial z_j}
\Bigl (M({dz_j\over dt})^2-2{dz_i\over dt}{dz_j\over dt}\Bigr )
\Bigr ]\beta_i, \quad 1\le j\le n,\eqno{(6.17c)}$$
$${d\over dt}\beta_j +{1\over c^2}\sum_{i=1,i\neq j}^n \beta_i
\Bigl (
{\partial U_{ij}\over \partial x_j}{dx_j\over dt}+
{\partial U_{ij}\over \partial y_j}{dy_j\over dt}+
{\partial U_{ij}\over \partial z_j}{dz_j\over dt}\Bigr ) = 0, 
\quad 1\le j\le n, \eqno{(6.17d)}$$
where $(x_j,y_j,z_j)$ are the coordinates 
of the $j$-th body, $\beta_j =
\Bigl (1-{1\over c^2}\Bigl ( ({dx_j\over dt})^2+({dy_j\over dt})^2+
({dz_j\over dt})^2\Bigr ) \Bigr )^{-1/2}$, and 
$$U_{ij}=
-{c^2m_i\over m_i-m_j}\ln {1+{G m_i\over r_{ij}c^2}\over 
1+{G m_j\over r_{ij}c^2}}\prod_{s\neq i}
\Bigl (1+{3Gm_j\over r_{sj}c^2}\Bigr ),
\quad r_{ij}=\sqrt{(x_i-x_j)^2+(y_i-y_j)^2+(z_i-z_j)^2}.\eqno{(6.18)}$$
Note that 
${\partial U_{ij}\over \partial x_j}{dx_j\over dt}+
{\partial U_{ij}\over \partial y_j}{dy_j\over dt}+
{\partial U_{ij}\over \partial z_j}{dz_j\over dt}$ 
is not a total differential like in (2.19d). 

In the general equations of motion (6.17) are not 
included the multiple moments, the gravitational radiation and the 
decrease of the orbit periods of the binary systems. Indeed, 
these equations lead only to the first post-Newtonian order, because of the 
weak choice of the potentials (6.18). But, using the ordinary coordinates 
we have the same calculations like in GR 
about the gravitational radiation and the 
decrease of the energy and orbit period of the binary systems. 

At the end of this section we shall make some considerations about the atoms 
of hydrogen and deuterium. First we accept the following convention. 
If we have a particle with mass $m$ and electricity $e$, then we convert
this electricity into mass $\pm {e\over \sqrt{4\pi\epsilon_0G}}$ 
(the sign depends on the charge positive or negative), such that 
the Coulon's force $f={1\over 4\pi \epsilon_0}{e_1e_2\over r^2}$ can be 
written as a gravitational force $f=G {m_1m_2\over r^2}$. 
This mass will be called "charged mass". In order to distinguish the 
charged mass from the ordinary mass we can write 
$m\pm i{e\over \sqrt{4\pi \epsilon_0}}$, where $i$ is the imaginary unit. 
According to this identification, the mass of the proton is equal to 
$$m_{p}=(1,672415\times 10^{-24}+i\times 1,8594077\times 10^{-6})\hbox{g}.$$

Let us consider the fusion of two atoms of hydrogen into atom of deuterium. 
According to the previous discussion, we will obtain the energy
$$E=2{e^2G\over Rc^2+eG }c^2
\approx 2{e^2G \over R},$$
where $e=1,8594077\times 10^{-6}$g. If we assume that 
$R=2\times 10^{-13}$cm is the distance between the centers of the hydrogen 
cores just before the fusion, we get energy 
$E=2,5670184\times 10^{-27}$ g$\times c^2$. 
On the other hand, it is known from the nuclear physics that this energy
is about $E=2,569\times 10^{-27}$ g$\times c^2$. 

Further let us consider the question for calculation of the coefficient 
$\mu $ inside the atom. First we should remember that in case of gravitation 
it is $\mu =1+{G M\over rc^2}$, and remember that 
${G M\over r}=v^2$ if the particle moves on a circle with velocity $v$.
Hence it is better to write $\mu =1+{v^2\over c^2}$. Now, let 
us consider the potential $\mu $, when an electron is moving around the core 
of the atom. If its velocity is approximately $v=c/100$, then 
$\mu \approx 1+100^{-2}=1,0001$. If we consider another particle 
instead of electron, for example a non charged particle, since the 
gravitational force is much smaller than the Coulon's force, its velocity 
around the core will be much smaller and hence $\mu$ will be almost 1. 
Hence the 
conclusion that {\em the potential $\mu$ depends on the nature of the 
particle}, i.e. it is not the same for all particles as it was in the 
previous sections. 

Let us calculate the potential $\mu$ for the proton near the core of the 
hydrogen on a distance $2\times 10^{-13}$cm. We can not use the formula 
$\mu =1+{G e\over Rc^2}$ where $e$ is charged mass, but should use the 
following revised formula 
$$\mu = 1+{G e\over Rc^2}{\lambda},\eqno{(6.19)}$$
where the coefficient $\lambda$ should be such that ${G m\over R}=v^2$. 
It is easy to see that the coefficient 
$\lambda $ should be the quotient between the charged mass and the 
gravitational mass, i.e. 
$$\lambda ={1,8594077\times 10^{-6}\over 1,672415\times 10^{-24}}
\approx 1,11181\times 10^{18}.$$
Hence, if we replace in (6.19) that $e=1,8594077\times 10^{-6}$g, because 
the ordinary mass is negligible, and $R=2\times 10^{-13}$cm, we 
get $\mu =1,0007674584$. 

If we calculate the coefficient $\mu $ for the electron close to 
the core, then $\lambda$ will be 1836,12 times larger than the previous, 
because the mass of the electron is 1836,12 times smaller than the mass of 
the proton, while they have the same charge. In this case, for an 
electron on distance $2,81794\times 10^{-13}$cm, i.e. radius of the 
electron, we obtain $\mu =2$ and $\mu =2$ achieves 
just for the Schwarzschild radius for the light ray (sec. 3). 

Now, let us return to the potential $\mu$ for a proton near the core of the 
hydrogen. Since the both masses are equal, the energy obtained by fusion 
is equal to 
$$2\Bigl (m-{m\over \mu}\Bigr )c^2=2\times 1,672415\times 
10^{-24}\hbox{g}\times 
{\mu -1\over \mu}c^2 = 2,56505 \times 10^{-27}\hbox{g}\times c^2.$$

\section{Periastron shift of the binary systems}

In this section will be calculated the periastron shift of arbitrary 
binary system. For example, we shall assume that it is a binary system of 
a pulsar and its companion. 
Applying the equations of motion (6.17), or simply the equations 
(2.18) or (2.19) with the potential (6.15) or (6.18), 
we shall obtain the required periastron shift. 

Let $m$ be the mass of the pulsar and $M$ be the mass of its companion. 
Let us choose a coordinate system 
having origin at their barycenter, 
in the moment when they are at the periastron. 
Let $(x,y,0)$ be the coordinates of the pulsar, 
and let $(x^*,y^*,0)$ be the coordinates of its companion. 
The gravitational potential 
at $(x,y,0)$ caused by the companion is given by (sec. 6) 
$$U=-\Bigl (1+\frac{3Gm}{Rc^2}\Bigr )\frac{c^2M}{M-m}
\ln\frac{1+\frac{GM}{Rc^2}}{1+\frac{Gm}{Rc^2}}, 
\eqno{(7.1)}$$
where $R=\sqrt{(x-x^*)^2+(y-y^*)^2}$. 
Since the following potentials 
$$U'=-\Bigl (1+\frac{3Gm}{Rc^2}\Bigr )
\frac{c^2M}{M+m}\ln\Bigl (1+\frac{G(M+m)}
{Rc^2}\Bigr ) \eqno{(7.1')}$$
and 
$$U''=-\frac{GM}{R}+\frac{1}{2}\frac{G^2M(M-5m)}{R^2c^2},\eqno{(7.1'')}$$
do not differ from $U$ up to $c^{-2}$, 
the same result obtains up to $c^{-2}$ 
if we use alternatively the potential $U'$ or $U''$. 

Let us denote the 4-vector of velocity of the pulsar by 
$$(V_i)=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} 
\Bigl (\frac{v_x}{ic},\frac{v_y}{ic},\frac{v_z}{ic},1\Bigr ),$$
and the 4-vector of velocity of its companion by 
$$(U_i)=\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}
\Bigl (\frac{u_x}{ic},\frac{u_y}{ic},\frac{u_z}{ic},1\Bigr ) .$$

Using the potential $(7.1')$ 
we can calculate the components $a_x,a_y,a_z,w_x,w_y,w_z$, which 
appear in (2.18). It is convenient to use the 
notation $r=\sqrt{x^2+y^2}$ and $r\approx \frac{M}{M+m}R$. 
Let us put $\cos \alpha =\frac{x-x^*}{R}$ and 
$\sin \alpha =\frac{y-y^*}{R}$. Then, up to $c^{-2}$ we have 
$$a_x=\frac{-1}{\sqrt{1-\frac{u^2}{c^2}}} 
\frac{\partial U}{\partial x} = 
\frac{-1}{\sqrt{1-\frac{u^2}{c^2}}}\frac{GM}{R^2+\frac{G(M+m)R}{c^2}}
\frac{1}{1-6\frac{Gm}{Rc^2}}
\frac{x-x^*}{\sqrt{(x-x^*)^2+(y-y^*)^2}} =$$
$$= 
\frac{-1}{\sqrt{1-\frac{u^2}{c^2}}}\frac{GM}{R^2+\frac{G(M-5m)R}{c^2}}
\cos \alpha ,$$
i.e. 
$$a_x=\frac{-1}{\sqrt{1-\frac{u^2}{c^2}}}\frac{GM}{R^2+\frac{G(M-5m)R}{c^2}}
\cos \alpha ,$$
and analogously 
$$a_y=\frac{-1}{\sqrt{1-\frac{u^2}{c^2}}}\frac{GM}{R^2+\frac{G(M-5m)R}{c^2}}
\sin \alpha ,$$
$$a_z=w_x=w_y=0,$$
$$w_z=\frac{i}{c} 
\frac{\frac{u_y}{ic}}{\sqrt{1-\frac{u^2}{c^2}}}
\frac{GM}{{R^2}+\frac{G(M-5m)R}{c^2}}\cos \alpha -\frac{i}{c}
\frac{\frac{u_x}{ic}}{\sqrt{1-\frac{u^2}{c^2}}}
\frac{GM}{{R^2}+\frac{G(M-5m)R}{c^2}}\cos \alpha =$$
$$=\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}
\frac{Gm}{{R^2}+\frac{G(M-5m)R}{c^2}}\frac{v_xy-v_yx}{rc^2}.$$
In the last equality it is used that $\cos \alpha \approx \frac{x}{r}$, 
$\sin \alpha \approx \frac{y}{r}$, $u_x\approx v_x\frac{m}{M}$ and 
$u_y\approx v_y\frac{m}{M}$. We may assume 
approximately that $x^*/y^*=x/y$, because the barycenter is very close to 
coordinate origin. Hence 
$\cos \alpha =x/r$ and $\sin \alpha =y/r$. 

By replacing these values in (2.18a,b), and neglecting the terms 
of order $c^{-4}$ and smaller, we obtain 
$$\frac{d^2x}{dt^2}=\Bigl (1+\frac{v^2}{c^2}\Bigr ) 
\frac{-1}{\sqrt{1-\frac{u^2}{c^2}}} 
\frac{GM}{{R^2}+\frac{G(M-5m)R}{c^2}}\frac{x-x*}{R}+ 
\frac{5}{2}\frac{v_x}{c^2}\frac{GM}{R^2}\frac{xv_x+yv_y}{r} 
-\frac{2v_y}{c^2}\frac{Gm}{R^2}\frac{xv_y-yv_x}{r},\eqno{(7.2a)}$$
$$\frac{d^2y}{dt^2}=\Bigl (1+\frac{v^2}{c^2}\Bigr )
\frac{-1}{\sqrt{1-\frac{u^2}{c^2}}}
\frac{GM}{{R^2}+\frac{G(M-5m)R}{c^2}}\frac{y-y^*}{R}+
\frac{5}{2}\frac{v_y}{c^2}\frac{GM}{R^2}\frac{xv_x+yv_y}{r}
+\frac{2v_x}{c^2}\frac{Gm}{R^2}\frac{xv_y-yv_x}{r},\eqno{(7.2b)}$$

These equations are analogous to (2.2), and continuing to deal like for the 
perihelion shift in section 2, we obtain 
$$\frac{d}{dt}\Bigl (r^2\frac{d\varphi}{dt}\Bigr ) = 
\frac{d^2y}{dt^2}x-\frac{d^2x}{dt^2}y = $$
$$=\frac{5}{2}\frac{GM}{R^2} 
\frac{xv_y-yv_x}{c^2}\frac{xv_x+yv_y}{r} + 2\frac{Gm}{R^2}
\frac{xv_x+yv_y}{c^2}\frac{xv_y-yv_x}{r}=$$
$$=\Bigl (\frac{5}{2}M+2m\Bigr )
\frac{G}{R^2}\frac{xv_x+yv_y}{c^2}\frac{xv_y-yv_x}{r} = 
-\Bigl (\frac{5}{2}M+2m\Bigr )\frac{Gr^2}{c^2R^2}
\Bigl (r^2\frac{d\varphi}{dt}\Bigr )
\frac{d}{dt}\Bigl (\frac{1}{r}\Bigr )=$$
$$=-\Bigl (\frac{5}{2}M+2m\Bigr )\frac{GM^2}{c^2(M+m)^2}
\Bigl (r^2\frac{d\varphi}{dt}\Bigr )\frac{d\rho}{dt},$$ 
where $\rho =1/r$. 
The solution of this differential equation is given by 
$$r^2\frac{d\varphi}{dt} = 
C_2\Bigl [1-\Bigl (\frac{5}{2}M+2m\Bigr )
\frac{G}{c^2}\frac{M^2}{(M+m)^2}\rho \Bigr ] ,\eqno{(7.3)}$$
and hence 
$$\Bigl (r^2\frac{d\varphi}{dt}\Bigr )^{-2} \approx C_2^{-2}
\Bigl [1+(5M+4m)\frac{G}{c^2}\frac{M^2}{(M+m)^2}\rho\Bigr ] .$$
Analogously to (2.5), now we have 
$$\Bigl (\frac{d\rho}{d\varphi}\Bigr )^2+\rho^2 =v^2 C_2^{-2}
\Bigl [1+(5M+4m)\frac{G}{c^2}\frac{M^2}{(M+m)^2}\rho\Bigr ] .\eqno{(7.4)}$$

Further we should find $v$ as a function of $R$ or $r$. We can replace 
$$\sqrt{1-\frac{u^2}{c^2}}\approx \Bigl (1-\frac{v^2}{c^2}\Bigr ) 
^{\frac{u^2}{2v^2}}\approx \Bigl (1-\frac{v^2}{c^2}\Bigr )
^{\frac{m^2}{2M^2}}$$
into the terms of $a_x$ and $a_y$. 
According to this approximation, the solution of (2.18d) is given by 
$$\Bigl (1-\frac{v^2}{c^2}\Bigr )^{-\frac{1}{2}(1-\frac{m^2}{M^2})}
\Bigl (1-\frac{m^2}{M^2}\Bigr )^{-1}-\frac{M^2}{(M+m)(M-5m)}
\ln \Bigl (1+\frac{M-5m}{M+m}\frac{GM\rho}{c^2}\Bigr )=$$
$$=\Bigl (1-\frac{m^2}{M^2}\Bigr )^{-1}+\frac{Const.}{c^2}.$$
Hence after some transformations we can express $v^2$ as a function 
of $\rho$ up to $c^{-2}$, and obtain 
$$v^2=2\frac{GM^3\rho}{(M+m)^2}-
\frac{G^2M^4\rho^2(4M^2-4mM-6m^2)}{c^2(M+m)^4}.$$
Note that the constant of integration $Const.$ is neglected in the last 
formula, because one can verify that it has no role in the final formula 
for the periastron shift, like for the perihelion shift in section 2. 
Substituting this value of $v^2$ into (7.4) we obtain 
$$\Bigl (\frac{d\rho}{d\varphi}\Bigr )^2+\rho^2= 
A+B\rho +C\rho^2 ,$$
where 
$$C=(10M+8m)\frac{G^2M^5}{C_2^2(M+m)^4c^2}-
\frac{G^2M^4(4M^2-4mM-6m^2)}{C_2^2(M+m)^4c^2}=
\frac{6G^2M^4}{C_2^2(M+m)^2c^2}.$$
Since $C_2=\frac{2}{P}\pi a^2\sqrt{1-\epsilon^2}=\frac{2\pi}{P}
a_r^2\frac{M^2}{(M+m)^2}\sqrt{1-\epsilon^2}$, 
where $a_r$ is the semi-major axis of the relative orbit 
and $P$ is the orbit period, we obtain 
$$C=\frac{3P^2G^2}{2\pi^2c^2}\frac{(M+m)^2}{a_r^4(1-\epsilon^2)},$$
and analogously to the calculations for the perihelion shift in section 2, 
for the periastron shift in this case we obtain 
$$\Delta \varphi = \frac{3G^2(M+m)^2P^2}{2\pi c^2a_r^4(1-\epsilon^2)}.
\eqno{(7.5)}$$
Using that $a_r^3=P^2G(M+m)/(4\pi ^2)$, we obtain finally 
$$\Delta \varphi = \frac{6\pi (4\pi^2)^{1/3}G^{2/3}}
{P^{2/3}c^2(1-\epsilon^2)}(M+m)^{2/3}.\eqno{(7.6)}$$
This value (7.6) of the periastron shift is the same which predicts the GR. 

Note that if we did not take into account that the pulsar with mass $m$ 
has an influence on the gravitation from the companion with mass $M$, i.e. 
if we omit the expression $1+\frac{3Gm}{Rc^2}$ in (7.1), 
then instead of (7.6) the periastron shift would be 
$\frac{M}{M+m}\Delta \varphi ,$
where $\Delta \varphi$ is given by (7.6). It means that the companion 
star participates with $\frac{M}{M+m}\Delta \varphi$ in the total periastron 
shift given by (7.6). Symmetrically, the pulsar with mass $m$, 
participates with 
$\frac{m}{M+m}\Delta \varphi$ in the total periastron shift. 

Now let us assume that 
$$a_x=\frac{-1}{\sqrt{1-\frac{u^2}{c^2}}}\frac{GM}{R^2+
\frac{G(\alpha M+\beta m)R}{c^2}}\frac{x}{r},
\quad \hbox {and} \quad 
a_y=\frac{-1}{\sqrt{1-\frac{u^2}{c^2}}}\frac{GM}{R^2+
\frac{G(\alpha M+\beta m)R}{c^2}}\frac{y}{r}.$$
Then the previous calculation leads that the periastron shift 
instead of (7.6) is equal to 
$$\Delta \varphi = \frac{6\pi (4\pi^2)^{1/3}G^{2/3}}
{P^{2/3}c^2(1-\epsilon^2)}(M+m)^{2/3}\times 
\frac{(7-\alpha )M^2+(8-\alpha -\beta )mM+(1-\beta )m^2}{6(M+m)^2}.$$
Since the angle $\Delta \varphi$ 
should be symmetric with respect to $M$ and $m$, it must be 
$7-\alpha =1-\beta$, i.e. $\alpha -\beta =6$. But this is satisfied if 
and only if the "correcting multiplier" is $1+{3Gm\over rc^2}$. Further, 
in that case 
$$\Delta \varphi = \frac{6\pi (4\pi^2)^{1/3}G^{2/3}}
{P^{2/3}c^2(1-\epsilon^2)}(M+m)^{2/3}\times \frac{7-\alpha}{6},$$
and it depends on the total mass $M+m$, but not on $M$ and $m$ separately. 
Finally, $\alpha$ must be 1 (and hence $\beta =-5$) in order to be 
$$\Delta \varphi = \frac{6\pi (4\pi^2)^{1/3}G^{2/3}}
{P^{2/3}c^2(1-\epsilon^2)}M^{2/3}$$
for $m=0$. 

Note that the formula $(7.3)$ does not depend on $\alpha$ and $\beta$. 
Thus, (7.3) for the pulsar's orbit can be important test for ILLTG, 
like the tests 1 and 2 in section 5. For example, from (7.3) it just 
follows that 
$$\Bigl (r^2\frac{d\varphi}{dt}\Bigr )_{periastron}:
\Bigl (r^2\frac{d\varphi}{dt}\Bigr )_{apastron} = 1-
\frac{(5M+4m)GM^2\epsilon}{a_pc^2(M+m)^2(1-\epsilon^2)},\eqno{(7.7)}$$
where $a_p$ is the semi-major axis of the pulsar's orbit. 
Using the recent data about the Hulse-Taylor pulsar PSR  B1913+16 
[12], we get 
$$\Bigl (r^2\frac{d\varphi}{dt}\Bigr )_{periastron}:
\Bigl (r^2\frac{d\varphi}{dt}\Bigr )_{apastron}=1-0,47\times 10^{-5}.$$

Probably using the fact that $\Delta \varphi$ 
should be a symmetric function with respect to 
$M$ and $m$, or some similar conditions, it can be used to obtain the 
gravitational potential in the form 
$$-\frac{GM}{R}+\frac{G^2M(M-5m)}{2R^2c^2}-
\frac{G^3M(\alpha M^2+\beta mM+\gamma m^2)}{3R^3c^4}+\cdots $$
and hence to obtain the post-Newtonian approximations of 2,3,... order. For 
post-Newtonian approximations of order 2,5; 3,5;... should be considered 
the multiple moments. 

\section{ILLTG and the Hubble constant } 

The representations in sections 1-7 would not be complete if we skip 
some cosmological consequences considered in [10,11]. 
In this section the basic arguments of [10] 
from viewpoint of ILLTG will be considered. 

In [10] is given a new view of the Hubble law, showing that 
the Hubble red shift and the anomalous 
acceleration of Pioneer spacecraft (see Ref. [1,2,3]) 
are consequences of the change of the 
potential in the universe. 
It is detected red shift such that 
$\nu =\nu_0 \Bigl (1-{v\over c}\Bigr )$, and replacing $v=RH$, 
indeed it is observed that 
$$\nu =\nu_0 \Bigl ( 1-{RH\over c}\Bigr ).\eqno{(8.1)}$$
On the other hand, if the potential changes linearly with the time, after 
time $t=R/c$ we have additional potential 
${R\over c}{\partial V\over \partial t}$, 
and hence we have a frequency shift such that 
$$\nu =\nu_0 \Bigl ( 1+{1\over c^2}{R\over c}{\partial V\over \partial t}
\Bigr ).\eqno{(8.2)}$$
We should remark here that the sign of the potential $V$ is  such 
that near to the spherical gravitational body the potential 
$(V=\frac{GM}{R})$ 
is larger than the potential far from gravitation $(V=0)$. 

Comparing the formulas (8.1) and (8.2) we obtain 
$$H=-{1\over c^2}{\partial V\over \partial t}.\eqno{(8.3)}$$
The change of the 
potential energy in the universe is probably related with the 
dark energy, which is about 67\% of the whole energy in the universe [6]. 
Some recent measurements show that $H$ is about 72 km/s/Mpc, but we shall 
accept value of about 82 km/s/Mpc. 

Since ${\partial V\over \partial t}<0$, in the past the time in the universe 
was slower than now, and in future it will be faster than now. 
Indeed, each second the time is faster approximately 
$\lambda = (1+H\times 1$s$)=1+{1\over 3,76\times 10^{17}}$ times. 

On the other hand, in the last decades are considered the motions 
of the spacecraft Pioneer 10 and 11, by comparing the initial 
frequency of a signal which is sent from the Earth to the spacecraft 
and the frequency of the signal which comes back. The frequency of the 
received signal on the Earth does not fit with the predicted frequency 
modeled by using the Doppler effect, the position and motion of the 
spacecraft using the GR and a lot of perturbations [3]. 
The observations show that appears almost a constant apparent unmodeled 
acceleration $a_P\approx (8,74\pm 1,33)\times 10^{-8}$cm/s$^2$ 
which seems to act to the spacecraft toward the Sun. 

The acceleration $a_P$ is introduced by formula (15) in [3], i.e. 
$$[\nu_{obs}(t)-\nu_{model}(t)]_{DSN} =-\nu_0{2a_Pt\over c},\;
\; \hbox {where}\; \; \nu_{model}=
\nu_0\Bigl [1-2{v_{model}(t)\over c}\Bigr ],\eqno{(8.4)}$$
where $\nu_0$ is the reference frequency, the factor 2 is because of two way 
data, $v_{model}$ is the modeled velocity of the spacecraft due to the 
gravitational and other forces, and $\nu_{obs}$ is frequency of the 
re-transmitted  signal  observed  by  DSN  antennae.  
After time $2t$ has been detected a small blue shift on the top of the 
red shift caused by the motion of the spacecraft 
outwards the Sun. Form (8.4), an unexplained blue shift 
$$\nu = \nu_0\Bigl (1+{2a_Pt\over c}\Bigr ),\eqno{(8.5)}$$
is detected, where $2t$ is the time of the light signal in two directions. 
Without using 
the acceleration $a_P$, the anomalous frequency shift can be 
interpreted by 
"clock acceleration" $-a_t=-2,8\times 10^{-18}$s/s$^2$ [3]. 
This causes just the blue shift given by (8.5). This model assumes that the 
time is uniform, but there is only technical problem with the 
clocks. Although this model gives good explanations for both frequency shift 
and the trajectories, it is rejected [3]. 

Many people have noticed that the acceleration $a_P$ and the Hubble 
constant $H$ are related by [3] 
$$a_P=cH.\eqno{(8.6)}$$
Indeed, if we assume that $H$=82km/s/Mpc, then from (8.6) it follows $a_P=
8\times 10^{-8}$cm/s$^2$. Thus we assumed that $H$=82km/s/Mpc. 

Now we shall explain the apparent acceleration $a_P$ 
connected with the frequency shift, according to ILLTG. 
We denote by $x,y,z,t$ the orthonormal coordinates 
and denote by $X,Y,Z,T$ ordinary coordinates 
where there is time dependent gravitational potential 
$V=-c^2Ht$. According to (1.1), where $V=-Hc^2t$,
we have the following equalities 
$$dx=\Bigl (1+{V\over c^2}\Bigr )^{-1}dX 
= \Bigl (1+tH\Bigr )dX,\qquad 
dy=\Bigl (1+{V\over c^2}\Bigr )^{-1}dY = \Bigl (1+tH\Bigr )dY,$$
$$dz=\Bigl (1+{V\over c^2}\Bigr )^{-1}dZ 
= \Bigl (1+tH\Bigr )dZ,\qquad 
dt=\Bigl (1-{V\over c^2}\Bigr )^{-1}dT = \Bigl (1-tH\Bigr )dT.
\eqno{(8.7)}$$

{\em Remark.} Note that it is sufficient in this section to assume that 
(8.7) are satisfied. Then it is not necessary to speak about 
the time dependent gravitational potential. 
The coefficients $1+Ht$ and $1-Ht$ in (8.7) should be exponential 
functions, but neglecting $H^2$ and smaller quantities we accept 
these linear functions. 

If we use orthonormal coordinates $x,y,z,t$, then 
the motion of the spacecraft 
does not depend on the time dependent potential $-Hc^2t$. 
Indeed, it is easy to see that the tensor $F_{ij}$ does not depend on this 
potential, and hence the motion of the particles (2.2) does not depend 
on the time dependent potential, which means that 
$({d^2x\over dt^2},{d^2y\over dt^2},{d^2z\over dt^2})=(0,0,0)$,
if we consider only the time dependent potential. 
Using the equalities (8.7), (2.2') and (8.6) we have an additional 
3-acceleration 
$$
-3HT\Bigl ({d^2X\over dT^2},{d^2Y\over dT^2},{d^2Z\over dT^2}\Bigr ) 
-2\Bigl (H{dX\over dT},H{dY\over dT},H{dZ\over dT}\Bigr ).\eqno{(8.8)}$$
In the explanation of the shift frequency of Pioneers, 
the first component 
$-3HT\Bigl ({d^2X\over dT^2},{d^2Y\over dT^2},{d^2Z\over dT^2}\Bigr ) $
is smaller, compared with the second and main component 
$$
-2\Bigl (H{dX\over dT},H{dY\over dT},H{dZ\over dT}\Bigr )=
-2\Bigl ({a_P\over c}{dX\over dT},
{a_P\over c}{dY\over dT},{a_P\over c}{dZ\over dT}\Bigr ).$$

We shall explain why it is measured the blue shift given by (8.5), 
instead of the red shift like in (8.1). It will be explained 
without using the acceleration (8.8), but only the definition (8.4) 
where $a_P$ is introduced, and our model of time dependent 
gravitational potential. 

Let $R=R(T)$ be the distance from the spacecraft to the DSN antenna. Let us 
calculate ${\Delta \nu\over \nu_0}=(\nu_{obs}-\nu_0)/\nu_0$. 
According to our model, neglecting the terms containing $H^2$, 
this expression is equal to 
$${\Delta \nu\over \nu_0} = -2H{R\over c}-2{{dR\over dT}\over c},$$
where the expression $-2H{R\over c}$ corresponds to the Hubble red shift 
from (8.2). 
On the other side, according to (8.4), where $v_{model}$ it determined 
without using the Hubble constant $H$, we have 
$${\Delta \nu\over \nu_0} = 
2a_P{R\over c^2}-2{({dR\over dT})_{H=0}\over c}.$$ 
{From} the last two equalities we are able to find 
the expression for $a_P$ 
$$a_P=-c\Bigl ({1\over R}{dR\over dT}-{1\over R}
\Bigl ({dR\over dT}\Bigr )_{H=0}+H\Bigr ).$$
Using that 
${dR\over dT}={dr\over dt}(1-2HT)=
\Bigl ({dR\over dT}\Bigr )_{H=0}(1-2HT)$ according to (8.7), 
we obtain finally 
$$a_P=cH\Bigl (2{dR\over dT}{T\over R}-1\Bigr ).\eqno{(8.9)}$$
The equality (8.9) gives the required expression of the 
acceleration $a_P$. Now we are not able to determine the initial 
value of $T$, i.e. when we should start to measure the time $T$ in (8.9). 
It depends on 
the estimation of the initial values for the motion of the spacecraft, i.e. 
it depends on the departure of the real initial values. Thus, for different 
spacecraft are obtained close but different values of $a_P$. Note also that 
for each initial value of $T$, when $T$ tends to infinity (or $R$ tends to 
infinity), then $a_P$ tends to $cH$. 
In an ideal case, if the experiment is done in an 
inertial system, then $dR/dT=R/T$ is a constant velocity and then 
$a_P\equiv cH$. 

The change of the gravitational potential 
causes slight changes in the results of the known 
experiments about GR, but they are negligible. 
The change of the results in the test of 
Shapiro time delay is negligible if we consider radio 
signals on short distances like in the solar system. 
Indeed, the change of the potential in the universe is about 
$2,4\times 10^3$cm$^2$/s$^2$ each second, while the gravitational 
potential on 1AU from the Sun is equal to $30^2$km$^2$/s$^2=
9\times 10^{12}$cm$^2$/s$^2$. This potential difference will be achieved 
from the universe after 118 years, which is too long compared with the 
time needed for an experiment about Shapiro time delay. The change of the 
deflection of the light rays near the Sun is completely negligible up to 
$c^{-2}$. The angle between two perihelions is changed additionally of 
order ${\Theta^2H^2\over \epsilon}$, 
where $\Theta$ is the orbit period of the planet 
and $\epsilon$ is the eccentricity of the elliptic orbit. 
This change of the angle is negligible with respect to the measured angle 
${6GM\pi \over ac^2(1-\epsilon^2)}$. 
%Although there is no perihelion shift 
%caused by the time dependent gravitational potential, 
%it appears the following new phenomena. 
%In the paper [11] (without using relativistic corrections) 
%it is shown that the planetary 
%orbits are not axially symmetric and the angle from the perihelion to 
%the aphelion is $\pi-\frac{H\Theta}{3\epsilon\pi}$, 
%while the angle from the 
%aphelion to the perihelion is $\pi+\frac{H\Theta}{3\epsilon\pi}$. 

Note that if we replace $V$ by $V-c^2Ht$ $(tH<<1)$ in the metric of GR 
$$(g_{ij})=\hbox{diag} \Bigl (-1+{2V\over c^2}-{2V^2\over c^4}, 
1+{2V\over c^2}, 1+{2V\over c^2}, 1+{2V\over c^2}\Bigr )$$
the additional acceleration is not convenient. 
The situation becomes much worse if $t$ is very large. 
While the Newton's theory is 
invariant under the gauge transformation $V\mapsto V+C$, and ILLTG is 
invariant under the gauge transformation $\mu \mapsto C\mu$, 
GR does not posses similar property. The equations of motion 
are unchanged if we replace the metric $g_{ij}$ by $Cg_{ij}$, but 
it can not be correlated with the change of the gravitational potential. 

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