Energy of the Moving Particles in 3-Body Problem of Classical Electrodynamics

1. Introduction

The main purpose of the present paper is to continue the investigation of the 3-body problem of classical electrodynamics in the 3D Kepler formulation from [1,2]. The number of equations of motion for 3 bodies in the Minkowski space is 12, while the unknown functions are 9. It is proved in [2] that the last three equations are consequences from the first 9 once. These three equations (two in the Kepler formulation) give estimations of the energy of the moving particles. Our main goal here is to calculate this energy and compare it with the values from quantum mechanics [3,4]. The results obtained refer to the helium atom where both electrons circle on the first (ground) stationary state.

The paper consists of six sections and References. Section 1 is an Introduction. In Section 2, a derivation of the explicit form of the equations of motion for the three-body problem is made. Every vector equation contains in the right-hand side the Lorentz force and a radiation term. The Lorentz force is a sum of two terms each of which shows the influence of the other two particles on the first one. In Section 3, the formalism of the transition to the Euclidean coordinates is described, introducing delays depending on the unknown trajectories. In Section 4, we obtain the radiation part of the energy terms in the energy equations. In Section 5 we obtain inequalities which allow us to estimate the energy of the moving particles. Section 6 is a Conclusion. We calculate the energy of the moving electrons rounding the nuclei. We establish that with the natural parameters of electron motion, the energies coincide with the values obtained from quantum mechanics [3,4].

The equations of motion for the 3-body problem introduced in [2] are:

$$m_1{\displaystyle \frac{d{\lambda }_r^{(1)}}{d{s}_1}}={\displaystyle \frac{e_1}{c^2}}\left(F_{rl}^{(12)}{\lambda }_l^{(1)}+F_{rl}^{(13)}{\lambda }_l^{(1)}+F_{rl}^{(1)rad}{\lambda }_l^{(1)}\right)$$

$$m_2{\displaystyle \frac{d{\lambda }_r^{(2)}}{d{s}_2}}={\displaystyle \frac{e_2}{c^2}}\left(F_{rl}^{(21)}{\lambda }_l^{(2)}+F_{rl}^{(23)}{\lambda }_l^{(2)}+F_{rl}^{(2)rad}{\lambda }_l^{(2)}\right)$$

(1)

$$\begin{array}{l}{m}_3{\displaystyle \frac{d{\lambda }_r^{(3)}}{d{s}_3}}={\displaystyle \frac{e_3}{c^2}}\left(F_{rl}^{(31)}{\lambda }_l^{(3)}+F_{rl}^{(32)}{\lambda }_l^{(3)}+F_{rl}^{(3)rad}{\lambda }_l^{(3)}\right)\\ (r=1,2,3,4).\end{array}$$

As usually Latin indices run from 1 to 4, while Greek – from 1 to 3.

Let us note the number of equations in (1) is 12, while the unknown functions (trajectories) are 9. It is proved in [2] that every 4-th equation is a consequence of the previous three ones. In this way one obtains that the independent equations in (1) are 9 in number.

2. Derivation the Explicit Form of the Energy Equation

We follow the procedure introduced in [2]. By $\left(x_1^{(k)}(t),x_2^{(k)}(t),x_3^{(k)}(t),x_4^{(k)}=ict\right),(k=1,2,3)$ we denote the space-time coordinates of the charged particles. The dot product in the Minkowski space is ${〈a,b〉}_4=a_r{b}_r={\displaystyle \sum _{r=1}^4{a}_r{b}_r}$, while in the 3-dimensional subspace ― $〈a,b〉=a_{\alpha }{b}_{\alpha }={\displaystyle \sum _{\alpha =1}^3{a}_{\alpha }{b}_{\alpha }};$ c is the vacuum speed of light; $m_k(k=1,2,3)$ are the proper masses of the particles; $e_k(k=1,2,3)$ ― their charges. The elements of proper times are $d{s}_k(k=1,2,3)$ and the unit tangent vectors to the world lines are ${\lambda }^{(k)}=\left({\lambda }_1^{(k)},{\lambda }_2^{(k)},{\lambda }_3^{(k)},{\lambda }_4^{(k)}\right)=\left({\displaystyle \frac{u_1^{(1)}(t)}{{\Delta }_k}},{\displaystyle \frac{u_2^{(1)}(t)}{{\Delta }_k}},{\displaystyle \frac{u_3^{(1)}(t)}{{\Delta }_k}},{\displaystyle \frac{ic}{{\Delta }_k}}\right)$, where ${\overrightarrow{u}}^{(k)}=\left(u_1^{(k)}(t),u_2^{(k)}(t),u_3^{(k)}(t)\right)=\left({\dot{x}}_1^{(k)}(t),{\dot{x}}_2^{(k)}(t),{\dot{x}}_3^{(k)}(t)\right)$, ${\Delta }_k=\sqrt{c^2-〈{\overrightarrow{u}}^{(k)}(t),{\overrightarrow{u}}^{(k)}(t)〉}$.

To compare denotations with originally accepted ones we recall that (cf. [5]):

$${\gamma }_k={\displaystyle \frac{1}{\sqrt{1-\left(〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}〉/c^2\right)}}}; {\displaystyle \frac{d}{d{s}_k}}={\displaystyle \frac{{\gamma }_k}{c}}{\displaystyle \frac{d}{dt}}; {\lambda }_{\alpha }^{(k)}={\displaystyle \frac{{\gamma }_k{u}_{\alpha }^{(k)}(t)}{c}}={\displaystyle \frac{u_{\alpha }^{(k)}(t)}{{\Delta }_k}}(\alpha =1,2,3); {\lambda }_4^{(k)}=i{\gamma }_k={\displaystyle \frac{ic}{{\Delta }_k}}.$$

The components of accelerations are ${\displaystyle \frac{d{\lambda }^{(k)}}{d{s}_k}}=\left({\displaystyle \frac{{\gamma }_k}{c}}{\displaystyle \frac{d}{dt}}{\displaystyle \frac{{\gamma }_k{\lambda }_{\alpha }^{(k)}}{c}},i{\displaystyle \frac{{\gamma }_k}{c}}{\displaystyle \frac{d{\gamma }_k}{dt}}\right)=\left({\displaystyle \frac{1}{{\Delta }_k}}{\displaystyle \frac{d}{dt}}{\displaystyle \frac{u_{\alpha }^{(k)}(t)}{{\Delta }_k}},{\displaystyle \frac{ic}{{\Delta }_k}}{\displaystyle \frac{d}{dt}}{\displaystyle \frac{1}{{\Delta }_k}}\right)$.

The isotropic vectors ${\xi }^{(kn)}(k=1,2,3;n\ne k)$ are obtained as in [2] (cf. [5,6]). Namely, fix any event on the world line of the k-th particle and draw the light cone into the past. This cone intersects the world line of the n-th particle in any other (past) event. Then

$${\xi }^{(kn)}=\left({{\xi }_1}^{(kn)},{{\xi }_2}^{(kn)},{{\xi }_3}^{(kn)},{{\xi }_4}^{(kn)}\right)=\left(x_1^{(k)}(t)-x_1^{(n)}(t-{\tau }_{kn}),x_2^{(k)}(t)-x_2^{(n)}(t-{\tau }_{kn}),x_3^{(k)}(t)-x_{13}^{(n)}(t-{\tau }_{kn}),ic{\tau }_{kn}\right).$$

Since ${〈{\xi }^{(kn)},{\xi }^{(kn)}〉}_4=0$, the retarded functions ${\tau }_{kn}(t)$ should satisfy the following 6 functional equations

$${\tau }_{kn}(t)={\displaystyle \frac{1}{c}}\sqrt{{〈{\xi }^{(kn)},{\xi }^{(kn)}〉}_4}\equiv {\displaystyle \frac{1}{c}}\sqrt{{\displaystyle \sum _{\alpha =1}^3{\left[x_{\alpha }^{(k)}(t)-x_{\alpha }^{(n)}(t-{\tau }_{kn}(t)\right]}^2}}.$$

For the velocities vectors we have

$${\lambda }^{(n)}=\left({\displaystyle \frac{u_1^{(n)}(t-{\tau }_{kn})}{c}},{\displaystyle \frac{u_2^{(n)}(t-{\tau }_{kn})}{c}},{\displaystyle \frac{u_3^{(n)}(t-{\tau }_{kn})}{c}},{\displaystyle \frac{ic}{{\Delta }_{kn}}}\right)=\left({\displaystyle \frac{{\overrightarrow{u}}^{(n)}(t-{\tau }_{kn})}{c}},{\displaystyle \frac{ic}{{\Delta }_{kn}}}\right),$$

where ${\Delta }_{kn}=\sqrt{c^2-〈{\overrightarrow{u}}^{(n)}(t-{\tau }_{kn}),{\overrightarrow{u}}^{(n)}(t-{\tau }_{kn})〉}$; ${\displaystyle \frac{d}{d{s}_k}}={\displaystyle \frac{1}{{\Delta }_k}}{\displaystyle \frac{d}{dt}};{\displaystyle \frac{d}{d{s}_n}}={\displaystyle \frac{1}{{\Delta }_{kn}}}{\displaystyle \frac{d}{d{t}_{kn}}}={\displaystyle \frac{1}{{\Delta }_{kn}}}{\displaystyle \frac{dt}{d{t}_{kn}}}{\displaystyle \frac{d}{dt}},$${\displaystyle \frac{dt}{d{t}_{kn}}}={\displaystyle \frac{c^2{\tau }_{kn}-〈{\overrightarrow{\xi }}^{(kn)},{\overrightarrow{u}}^{(n)}〉}{c^2{\tau }_{kn}-〈{\overrightarrow{\xi }}^{(kn)},{\overrightarrow{u}}^{(k)}〉}}=D_{kn}$.

To derive the explicit form of the radiation terms (cf. [7]), we consider a charge $e_k(k=1,2,3)$ describing any curve $L_k(k=1,2,3)$ in the space-time. Let $A_k\left(x_1^{(k)}(t),x_2^{(k)}(t),x_3^{(k)}(t),ict\right)$ be any event and let

$A_k^{ret}\left(x_1^{(k)}({\stackrel{⌣}{t}}_k),x_2^{(k)}({\stackrel{⌣}{t}}_k),x_3^{(k)}({\stackrel{⌣}{t}}_k),ic{\stackrel{⌣}{t}}_k\right),{\stackrel{⌣}{t}}_k<t$ be the intersection of $L_k(k=1,2,3)$ with the null-cone drawn into the past from $A_k(k=1,2,3)$, and $A_k^{adv}\left(x_1^{(k)}({\stackrel{⌢}{t}}_k),x_2^{(k)}({\stackrel{⌢}{t}}_k),x_3^{(k)}({\stackrel{⌢}{t}}_k),ic{\stackrel{⌢}{t}}_k\right),{\stackrel{⌢}{t}}_k>t$ be the intersection of $L_k(k=1,2,3)$ with the null-cone drawn into the future from $A_k(k=1,2,3)$. Let $A_k^{ret}{A}_k$ and $A_k^{adv}{A}_k$ be the isotropic vectors

${\xi }^{(k)ret}=\left({\xi }_1^{(k)ret},{\xi }_2^{(k)ret},{\xi }_3^{(k)ret},{\xi }_4^{(k)ret}\right)$, ${\xi }^{(k)adv}=\left({\xi }_1^{(k)adv},{\xi }_2^{(k)adv},{\xi }_3^{(k)adv},{\xi }_4^{(k)adv}\right)$. Then we set

${\tau }_k^{ret}(t)=t-{\stackrel{⌣}{t}}_k\Rightarrow {\stackrel{⌣}{t}}_k=t-{\tau }_k^{ret}(t)$ and ${\tau }_k^{adv}(t)={\stackrel{⌢}{t}}_k-t\Rightarrow {\stackrel{⌢}{t}}_k=t+{\tau }_k^{adv}(t)$ and obtain

$$\begin{gathered} {\xi }^{(k)ret}=\left({\xi }_1^{(k)ret},{\xi }_2^{(k)ret},{\xi }_3^{(k)ret},ic{\tau }_k^{ret}\right)=\left({\overrightarrow{\xi }}^{(k)ret},ic{\tau }_k^{ret}\right)=\left(x_1^{(k)}(t)-x_1^{(k)}(t-{\tau }_k^{ret}),x_2^{(k)}(t)-x_2^{(k)}(t-{\tau }_k^{ret}),x_3^{(k)}(t)-x_3^{(k)}(t-{\tau }_k^{ret}),ic{\tau }_k^{ret}\right); \\ \begin{array}{l}{\xi }^{(k)adv}=\left({{\xi }_1}^{(k)adv},{{\xi }_2}^{(k)adv},{{\xi }_3}^{(k)adv},ic{\tau }_k^{adv}\right)=\left({\overrightarrow{\xi }}^{(k)adv},ic{\tau }_k^{adv}\right)=\\ =\left(x_1^{(k)}(t)-x_1^{(k)}(t+{\tau }_k^{adv}),x_2^{(k)}(t)-x_2^{(k)}(t+{\tau }_k^{adv}),x_3^{(k)}(t)-x_3^{(k)}(t+{\tau }_k^{adv}),ic{\tau }_k^{adv}\right).\end{array} \end{gathered}$$

The velocity vectors to the world-line $L_k(k=1,2,3)$ at $A_k^{ret}$ and $A_k^{adv}$ are:

$${\lambda }^{(k)ret}=\left({\lambda }_1^{(k)ret},{\lambda }_2^{(k)ret},{\lambda }_3^{(k)ret},{\lambda }_4^{(k)ret}\right)=\left({\displaystyle \frac{u_1^{(k)}({\stackrel{⌣}{t}}_k)}{{\Delta }_k^{ret}}},{\displaystyle \frac{u_2^{(k)}({\stackrel{⌣}{t}}_k)}{{\Delta }_k^{ret}}},{\displaystyle \frac{u_3^{(k)}({\stackrel{⌣}{t}}_k)}{{\Delta }_k^{ret}}},{\displaystyle \frac{ic}{{\Delta }_k^{ret}}}\right)=\left({\displaystyle \frac{{\overrightarrow{u}}^{(k)}({\stackrel{⌣}{t}}_k)}{{\Delta }_k^{ret}}},{\displaystyle \frac{ic}{{\Delta }_k^{ret}}}\right);$$

$${\Delta }_k^{ret}=\sqrt{c^2-〈{\overrightarrow{u}}^{(k)}({\stackrel{⌣}{t}}_k),{\overrightarrow{u}}^{(k)}({\stackrel{⌣}{t}}_k)〉};$$

$${\lambda }^{(k)adv}=\left({\lambda }_1^{(k)adv},{\lambda }_2^{(k)adv},{\lambda }_3^{(k)adv},{\lambda }_4^{(k)adv}\right)=\left({\displaystyle \frac{u_1^{(k)}({\stackrel{⌢}{t}}_k)}{{\Delta }_k^{adv}}},{\displaystyle \frac{u_2^{(k)}({\stackrel{⌢}{t}}_k)}{{\Delta }_k^{adv}}},{\displaystyle \frac{u_3^{(k)}({\stackrel{⌢}{t}}_k)}{{\Delta }_k^{adv}}},{\displaystyle \frac{ic}{{\Delta }_k^{adv}}}\right)=\left({\displaystyle \frac{{\overrightarrow{u}}^{(k)}({\stackrel{⌢}{t}}_k)}{{\Delta }_k^{adv}}},{\displaystyle \frac{ic}{{\Delta }_k^{adv}}}\right);$$

$${\Delta }_k^{adv}=\sqrt{c^2-〈{\overrightarrow{u}}^{(k)}({\stackrel{⌢}{t}}_k),{\overrightarrow{u}}^{(k)}({\stackrel{⌢}{t}}_k)〉}.$$

Following the Dirac physical assumptions we define the radiation term as a half of the difference between both retarded and advanced potentials (cf. [7]) $F_{mn}^{(k)rad}={\displaystyle \frac{1}{2}}\left[\left({\displaystyle \frac{\partial A_n^{(k)ret}}{\partial x_m^{(k)ret}}}-{\displaystyle \frac{\partial A_m^{(n)ret}}{\partial x_n^{(k)ret}}}\right)-\left({\displaystyle \frac{\partial A_n^{(k)adv}}{\partial x_m^{(k)adv}}}-{\displaystyle \frac{\partial A_m^{(n)adv}}{\partial x_n^{(k)adv}}}\right)\right]$, where $A_n^{(k)ret}=-{\displaystyle \frac{e_k{\lambda }_n^{(k)ret}}{{〈{\lambda }^{(k)ret},{\xi }^{(k)ret}〉}_4}}$ and $A_n^{(k)adv}=-{\displaystyle \frac{e_k{\lambda }_n^{(k)adv}}{{〈{\lambda }^{(k)adv},{\xi }^{(k)adv}〉}_4}}$.

Then the equations of motion become $(r=1,2,3,4)$:

$$\begin{gathered} m_1{\displaystyle \frac{d{\lambda }_r^{(1)}}{d{s}_1}}={\displaystyle \frac{e_1}{c^2}}\left(F_{rl}^{(12)}{\lambda }_l^{(1)}+F_{rl}^{(13)}{\lambda }_l^{(1)}+{\displaystyle \frac{1}{2}}\left[{\displaystyle \frac{\partial A_l^{(1)ret}}{\partial x_r^{(1)ret}}}-{\displaystyle \frac{\partial A_r^{(1)ret}}{\partial x_l^{(1)ret}}}-\left({\displaystyle \frac{\partial A_l^{(1)adv}}{\partial x_r^{(1)adv}}}-{\displaystyle \frac{\partial A_r^{(1)adv}}{\partial x_l^{(1)adv}}}\right)\right]{\lambda }_l^{(1)}\right), \\ {m}_2{\displaystyle \frac{d{\lambda }_r^{(2)}}{d{s}_2}}={\displaystyle \frac{e_2}{c^2}}\left(F_{rl}^{(21)}{\lambda }_l^{(2)}+F_{rl}^{(23)}{\lambda }_l^{(2)}+{\displaystyle \frac{1}{2}}\left[{\displaystyle \frac{\partial A_l^{(2)ret}}{\partial x_r^{(2)ret}}}-{\displaystyle \frac{\partial A_r^{(2)ret}}{\partial x_l^{(2)ret}}}-\left({\displaystyle \frac{\partial A_l^{(2)adv}}{\partial x_r^{(2)adv}}}-{\displaystyle \frac{\partial A_r^{(2)adv}}{\partial x_l^{(2)adv}}}\right)\right]{\lambda }_l^{(2)}\right), \\ {m}_3{\displaystyle \frac{d{\lambda }_r^{(3)}}{d{s}_3}}={\displaystyle \frac{e_3}{c^2}}\left(F_{rl}^{(31)}{\lambda }_l^{(3)}+F_{rl}^{(32)}{\lambda }_l^{(3)}+{\displaystyle \frac{1}{2}}\left[{\displaystyle \frac{\partial A_l^{(3)ret}}{\partial x_r^{(3)ret}}}-{\displaystyle \frac{\partial A_r^{(3)ret}}{\partial x_l^{(3)ret}}}-\left({\displaystyle \frac{\partial A_l^{(3)adv}}{\partial x_r^{(3)adv}}}-{\displaystyle \frac{\partial A_r^{(3)adv}}{\partial x_l^{(3)adv}}}\right)\right]{\lambda }_l^{(3)}\right).. \end{gathered}$$

In the present paper we investigate only the fourth, eight and twelfth equations $(k=1,2,3)$:

$$\begin{gathered} {\displaystyle \frac{d{\lambda }_4^{(k)}}{d{s}_k}}={\displaystyle \sum _{n=1,n\ne k}^3}{\displaystyle \frac{e_k{e}_n}{m_k{c}^2}}\left\{{\displaystyle \frac{{\xi }_4^{(kn)}{〈{\lambda }^{(k)},{\lambda }^{(n)}〉}_4-{\lambda }_4^{(n)}{〈{\lambda }^{(k)},{\xi }^{(kn)}〉}_4}{{〈{\lambda }^{(n)},{\xi }^{(kn)}〉}_4^3}}\right.\left(1+{〈{\displaystyle \frac{d{\lambda }^{(n)}}{d{s}_n}},{\xi }^{(kn)}〉}_4\right)+ \\ +\left.{\displaystyle \frac{1}{{〈{\lambda }^{(n)},{\xi }^{(kn)}〉}_4^2}}\left[{\displaystyle \frac{d{\lambda }_4^{(n)}}{d{s}_n}}{〈{\lambda }^{(k)},{\xi }^{(kn)}〉}_4-{\xi }_4^{(kn)}{〈{\lambda }^{(k)},{\displaystyle \frac{d{\lambda }^{(n)}}{d{s}_n}}〉}_4\right]\right\}+ \\ +{\displaystyle \frac{{е}_k^2}{2m_k{c}^2}}\left[{\displaystyle \frac{{\xi }_4^{(k)ret}{\langle {\lambda }^{(k)},{\lambda }^{(k)ret}\rangle }_4-{\lambda }_4^{(k)ret}{\langle {\xi }^{(k)ret},{\lambda }^{(k)}\rangle }_4}{{\langle {\lambda }^{(k)ret},{\xi }^{(k)ret}\rangle }_4^3}}\left(1+{〈{\xi }^{(k)ret},{\displaystyle \frac{d{\lambda }^{(k)ret}}{d{s}_{ret}}}〉}_4\right)\right.+ \\ \left.+{\displaystyle \frac{1}{{\langle {\lambda }^{(k)ret},{\xi }^{(k)ret}\rangle }_4^2}}\left({\langle {\xi }^{(k)ret},{\lambda }^{(k)}\rangle }_4{\displaystyle \frac{d{\lambda }_4^{(k)ret}}{d{s}_{ret}}}-{〈{\lambda }^{(k)},{\displaystyle \frac{d{\lambda }^{(k)ret}}{d{s}_{ret}}}〉}_4{\xi }_4^{(k)ret}\right)\right]- \\ -{\displaystyle \frac{{е}_k^2}{2m_k{c}^2}}\left[{\displaystyle \frac{{\xi }_{\alpha }^{(k)adv}{\langle {\lambda }^{(k)},{\lambda }^{(k)adv}\rangle }_4-{\lambda }_{\alpha }^{(k)adv}{\langle {\xi }^{(k)adv},{\lambda }^{(k)}\rangle }_4}{{\langle {\lambda }^{(k)adv},{\xi }^{(k)adv}\rangle }_4^3}}\left(1+{〈{\xi }^{(k)adv},{\displaystyle \frac{d{\lambda }^{(k)adv}}{d{s}_{adv}}}〉}_4\right)\right.+ \\ +\left.{\displaystyle \frac{1}{{\langle {\lambda }^{(k)adv},{\xi }^{(k)adv}\rangle }_4^2}}\left({\langle {\xi }^{(k)adv},{\lambda }^{(k)}\rangle }_4{\displaystyle \frac{d{\lambda }_4^{(k)adv}}{d{s}_{adv}}}-{〈{\lambda }^{(k)},{\displaystyle \frac{d{\lambda }^{(k)adv}}{d{s}_{adv}}}〉}_4{\xi }_4^{(k)adv}\right)\right]. \end{gathered}$$

3. Transition to the Euclidean Coordinates

Following reasoning from [7] the relations ${〈{\xi }^{(k)ret},{\xi }^{(k)ret}〉}_4=0,{〈{\xi }^{(k)adv},{\xi }^{(k)adv}〉}_4=0$ yield

$${\tau }_k^{ret}={\displaystyle \frac{1}{c}}\sqrt{〈{\overrightarrow{\xi }}^{(k)ret},{\overrightarrow{\xi }}^{(k)ret}〉},{\tau }_k^{adv}={\displaystyle \frac{1}{c}}\sqrt{〈{\overrightarrow{\xi }}^{(k)adv},{\overrightarrow{\xi }}^{(k)adv}〉}.$$

For the retarded part of radiation term we differentiate $t-{\stackrel{⌣}{t}}_k={\displaystyle \frac{1}{c}}\sqrt{{\displaystyle \sum _{\gamma =1}^3{\left[x_{\gamma }^{(k)}(t)-x_{\gamma }^{(k)}({\stackrel{⌣}{t}}_k)\right]}^2}}$ with respect to ${\stackrel{⌣}{t}}_k$ and obtain $D_k^{ret}={\displaystyle \frac{dt}{d{\stackrel{⌣}{t}}_k}}=={\displaystyle \frac{c^2{\tau }_k^{ret}-〈{\overrightarrow{\xi }}^{(k)ret},{\overrightarrow{u}}^{(k)}(t-{\tau }_k^{ret})〉}{c^2{\tau }_k^{ret}-〈{\overrightarrow{\xi }}^{(k)ret},{\overrightarrow{u}}^{(k)}(t)〉}}$ and in a similar way for advanced term we obtain $D_k^{adv}={\displaystyle \frac{dt}{d{\stackrel{⌢}{t}}_k}}={\displaystyle \frac{c^2{\tau }_k^{adv}-〈{\overrightarrow{\xi }}^{(k)adv},{\overrightarrow{u}}^{(k)}(t+{\tau }_k^{adv})〉}{c^2{\tau }_k^{adv}-〈{\overrightarrow{\xi }}^{(k)adv},{\overrightarrow{u}}^{(k)}(t)〉}}$.

Further on,

$$\begin{gathered} {\displaystyle \frac{d{\lambda }_{\alpha }^{(k)}}{d{s}_k}}={\displaystyle \frac{1}{{\Delta }_k}}{\displaystyle \frac{d}{dt}}\left({\displaystyle \frac{u_{\alpha }^{(k)}(t)}{{\Delta }_k}}\right)={\displaystyle \frac{{\dot{u}}_{\alpha }^{(k)}(t)}{{\Delta }_k^2}}+{\displaystyle \frac{u_{\alpha }^{(k)}(t)〈{\overrightarrow{u}}^{(k)}(t),{\dot{\overrightarrow{u}}}^{(k)}(t)〉}{{\Delta }_k^4}}; {\displaystyle \frac{d{\lambda }_4^{(k)}}{d{s}_k}}={\displaystyle \frac{ic}{{\Delta }_k}}{\displaystyle \frac{d}{dt}}\left({\displaystyle \frac{1}{{\Delta }_k}}\right)={\displaystyle \frac{ic〈{\overrightarrow{u}}^{(k)}(t),{\dot{\overrightarrow{u}}}^{(k)}(t)〉}{{\Delta }_k^4}}; \\ {\displaystyle \frac{d{\lambda }_{\alpha }^{(n)}}{d{s}_n}}=D_{kn}\left({\displaystyle \frac{{\dot{u}}_{\alpha }^{(n)}(t-{\tau }_{kn})}{{\Delta }_{kn}^2}}+{\displaystyle \frac{u_{\alpha }^{(n)}(t-{\tau }_{kn})〈{\overrightarrow{u}}^{(n)}(t-{\tau }_{kn}),{\dot{\overrightarrow{u}}}^{(n)}(t-{\tau }_{kn})〉}{{\Delta }_{kn}^4}}\right); \\ {\displaystyle \frac{d{\lambda }_4^{(n)}}{d{s}_n}}={\displaystyle \frac{ic{D}_{kn}〈{\overrightarrow{u}}^{(n)}(t-{\tau }_{kn}),{\dot{\overrightarrow{u}}}^{(n)}(t-{\tau }_{kn})〉}{{\Delta }_{kn}^4}}; \\ {〈{\lambda }^{(k)},{\lambda }^{(n)}〉}_4={\displaystyle \frac{〈{\overrightarrow{u}}^{(k)}(t),{\overrightarrow{u}}^{(n)}(t-{\tau }_{kn})〉-c^2}{{\Delta }_k{\Delta }_{kn}}}; {〈{\lambda }^{(k)},{\xi }^{(kn)}〉}_4={\displaystyle \frac{〈{\overrightarrow{u}}^{(k)}(t),{\overrightarrow{\xi }}^{(kn)}〉-c^2{\tau }_{kn}}{{\Delta }_k}}; \\ {〈{\lambda }^{(n)},{\xi }^{(kn)}〉}_4={\displaystyle \frac{〈{\overrightarrow{u}}^{(n)}(t-{\tau }_{kn}),{\overrightarrow{\xi }}^{(kn)}〉-c^2{\tau }_{kn}}{{\Delta }_{kn}}}; \\ {〈{\xi }^{(kn)},{\displaystyle \frac{d{\lambda }^{(n)}}{d{s}_n}}〉}_4=D_{kn}\left({\displaystyle \frac{〈{\overrightarrow{\xi }}^{(kn)},{\dot{\overrightarrow{u}}}^{(n)}(t-{\tau }_{kn})〉}{{\Delta }_{kn}^2}}+{\displaystyle \frac{〈{\overrightarrow{\xi }}^{(kn)},{\overrightarrow{u}}^{(n)}(t-{\tau }_{kn})〉-c^2{\tau }_{kn}}{{\Delta }_{kn}^4}}〈{\overrightarrow{u}}^{(n)}(t-{\tau }_{kn}),{\dot{\overrightarrow{u}}}^{(n)}(t-{\tau }_{kn})〉\right); \\ {〈{\lambda }^{(k)},{\displaystyle \frac{d{\lambda }^{(n)}}{d{s}_n}}〉}_4={\displaystyle \frac{D_{kn}}{{\Delta }_k}}\left({\displaystyle \frac{〈{\overrightarrow{u}}^{(k)}(t),{\dot{\overrightarrow{u}}}^{(n)}(t-{\tau }_{kn})〉}{{\Delta }_{kn}^2}}+{\displaystyle \frac{〈{\overrightarrow{u}}^{(k)}(t),{\overrightarrow{u}}^{(n)}(t-{\tau }_{kn})〉-c^2{\tau }_{kn}}{{\Delta }_{kn}^4}}〈{\overrightarrow{u}}^{(n)}(t-{\tau }_{kn}),{\dot{\overrightarrow{u}}}^{(n)}(t-{\tau }_{kn})〉\right); \\ {〈{\lambda }^{(k)},{\lambda }^{(k)ret}〉}_4={\displaystyle \frac{〈{\overrightarrow{u}}^{(k)}(t),{\overrightarrow{u}}^{(k)}(t-{\tau }_k^{ret}〉-c^2}{{\Delta }_k{\Delta }_k^{ret}}}; {〈{\lambda }^{(k)},{\lambda }^{(k)adv}〉}_4={\displaystyle \frac{〈{\overrightarrow{u}}^{(k)}(t),{\overrightarrow{u}}^{(k)}(t+{\tau }_k^{adv}〉-c^2}{{\Delta }_k{\Delta }_k^{adv}}}; \\ {〈{\xi }^{(k)ret},{\lambda }^{(k)}〉}_4={\displaystyle \frac{〈{\overrightarrow{u}}^{(k)}(t),{\overrightarrow{\xi }}^{(k)ret}〉-c^2{\tau }_k^{ret}}{{\Delta }_k}};{〈{\xi }^{(k)adv},{\lambda }^{(k)}〉}_4={\displaystyle \frac{〈{\overrightarrow{u}}^{(k)}(t),{\overrightarrow{\xi }}^{(k)adv}〉-c^2{\tau }_k^{adv}}{{\Delta }_k}}; \\ {〈{\xi }^{(k)ret},{\lambda }^{(k)ret}〉}_4={\displaystyle \frac{〈{\overrightarrow{u}}^{(k)ret}(t-{\tau }_k^{ret}),{\overrightarrow{\xi }}^{(k)ret}〉-c^2{\tau }_k^{ret}}{{\Delta }_k^{ret}}};{〈{\xi }^{(k)adv},{\lambda }^{(k)adv}〉}_4={\displaystyle \frac{〈{\overrightarrow{u}}^{(k)adv}(t+{\tau }_k^{adv}),{\overrightarrow{\xi }}^{(k)adv}〉-c^2{\tau }_k^{adv}}{{\Delta }_k^{adv}}}; \\ {\displaystyle \frac{d}{d{s}_{ret}}}={\displaystyle \frac{1}{{\Delta }_k^{ret}}}{\displaystyle \frac{d}{d\stackrel{⌣}{t}}}={\displaystyle \frac{1}{{\Delta }_k^{ret}}}{\displaystyle \frac{dt}{d{\stackrel{⌣}{t}}_k}}{\displaystyle \frac{d}{dt}}={\displaystyle \frac{D_k^{ret}}{{\Delta }_k^{ret}}}{\displaystyle \frac{d}{dt}};{\displaystyle \frac{d}{d{s}_{adv}}}={\displaystyle \frac{1}{{\Delta }_k^{adv}}}{\displaystyle \frac{d}{d\stackrel{⌣}{t}}}={\displaystyle \frac{1}{{\Delta }_k^{adv}}}{\displaystyle \frac{dt}{d{\stackrel{⌣}{t}}_k}}{\displaystyle \frac{d}{dt}}={\displaystyle \frac{D_k^{adv}}{{\Delta }_k^{adv}}}{\displaystyle \frac{d}{dt}}; \\ {\displaystyle \frac{d{\lambda }_{\alpha }^{(k)ret}}{d{s}_{ret}}}=D_k^{ret}\left({\displaystyle \frac{{\dot{u}}_{\alpha }^{(k)}(t-{\tau }_k^{ret})}{{({\Delta }_k^{ret})}^2}}+{\displaystyle \frac{u_{\alpha }^{(k)}(t-{\tau }_k^{ret})〈{\overrightarrow{u}}^{(k)}(t-{\tau }_k^{ret}),{\dot{\overrightarrow{u}}}^{(k)}(t-{\tau }_k^{ret})〉}{{({\Delta }_k^{ret})}^4}}\right); \\ {\displaystyle \frac{d{\lambda }_4^{(k)ret}}{d{s}_{ret}}}={\displaystyle \frac{ic{D}_k^{ret}〈{\overrightarrow{u}}^{(k)}(t-{\tau }_k^{ret}),{\dot{\overrightarrow{u}}}^{(k)}(t-{\tau }_k^{ret})〉}{{({\Delta }_k^{ret})}^4}}; \\ {\displaystyle \frac{d{\lambda }_{\alpha }^{(k)adv}}{d{s}_{adv}}}=D_k^{adv}\left({\displaystyle \frac{{\dot{u}}_{\alpha }^{(k)}(t+{\tau }_k^{adv})}{{({\Delta }_k^{adv})}^2}}+{\displaystyle \frac{u_{\alpha }^{(k)}(t+{\tau }_k^{adv})〈{\overrightarrow{u}}^{(k)}(t+{\tau }_k^{adv}),{\dot{\overrightarrow{u}}}^{(k)}(t+{\tau }_k^{adv})〉}{{({\Delta }_k^{adv})}^4}}\right); \\ {\displaystyle \frac{d{\lambda }_4^{(k)adv}}{d{s}_{adv}}}={\displaystyle \frac{ic{D}_k^{adv}〈{\overrightarrow{u}}^{(k)}(t+{\tau }_k^{adv}),{\dot{\overrightarrow{u}}}^{(k)}(t+{\tau }_k^{adv})〉}{{({\Delta }_k^{adv})}^4}}; \\ {〈{\xi }^{(k)ret},{\displaystyle \frac{d{\lambda }^{(k)ret}}{d{s}_{ret}}}〉}_4=D_k^{ret}\left({\displaystyle \frac{〈{\overrightarrow{\xi }}^{(k)ret},{\dot{\overrightarrow{u}}}^{(k)}(t-{\tau }_k^{ret})〉}{{({\Delta }_k^{ret})}^2}}+{\displaystyle \frac{〈{\overrightarrow{\xi }}^{(k)ret},{\overrightarrow{u}}^{(k)}(t-{\tau }_k^{ret})〉-c^2{\tau }_k^{ret}}{{({\Delta }_k^{ret})}^4}}〈{\overrightarrow{u}}^{(k)}(t-{\tau }_k^{ret}),{\dot{\overrightarrow{u}}}^{(k)}(t-{\tau }_k^{ret})〉\right); \\ {〈{\lambda }^{(k)},{\displaystyle \frac{d{\lambda }^{(k)ret}}{d{s}_{ret}}}〉}_4={\displaystyle \frac{D_k^{ret}}{{\Delta }_k}}\left({\displaystyle \frac{〈{\overrightarrow{u}}^{(k)}(t),{\dot{\overrightarrow{u}}}^{(k)}(t-{\tau }_k^{ret})〉}{{({\Delta }_k^{ret})}^2}}+{\displaystyle \frac{〈{\overrightarrow{u}}^{(k)}(t),{\overrightarrow{u}}^{(k)}(t-{\tau }_k^{ret})〉-c^2{\tau }_k^{ret}}{{({\Delta }_k^{ret})}^4}}〈{\overrightarrow{u}}^{(k)}(t),{\dot{\overrightarrow{u}}}^{(k)}(t-{\tau }_k^{ret})〉\right); \\ {〈{\xi }^{(k)adv},{\displaystyle \frac{d{\lambda }^{(k)adv}}{d{s}_{adv}}}〉}_4=D_k^{adv}\left({\displaystyle \frac{〈{\overrightarrow{\xi }}^{(k)adv},{\dot{\overrightarrow{u}}}^{(k)}(t+{\tau }_k^{adv})〉}{{({\Delta }_k^{adv})}^2}}+{\displaystyle \frac{〈{\overrightarrow{\xi }}^{(k)adv},{\overrightarrow{u}}^{(k)}(t+{\tau }_k^{adv})〉-c^2{\tau }_k^{adv}}{{({\Delta }_k^{adv})}^4}}〈{\overrightarrow{u}}^{(k)}(t+{\tau }_k^{adv}),{\dot{\overrightarrow{u}}}^{(k)}(t+{\tau }_k^{adv})〉\right); \\ {〈{\lambda }^{(k)},{\displaystyle \frac{d{\lambda }^{(k)adv}}{d{s}_{adv}}}〉}_4={\displaystyle \frac{D_k^{adv}}{{\Delta }_k}}\left({\displaystyle \frac{〈{\overrightarrow{u}}^{(k)}(t),{\dot{\overrightarrow{u}}}^{(k)}(t+{\tau }_k^{adv})〉}{{({\Delta }_k^{adv})}^2}}+{\displaystyle \frac{〈{\overrightarrow{u}}^{(k)}(t),{\overrightarrow{u}}^{(k)}(t+{\tau }_k^{adv})〉-c^2{\tau }^{(k)adv}}{{({\Delta }_k^{adv})}^4}}〈{\overrightarrow{u}}^{(k)}(t),{\dot{\overrightarrow{u}}}^{(k)}(t+{\tau }_k^{adv})〉\right); \\ {H}_{kn}=1+D_{kn}\left({\displaystyle \frac{〈{\overrightarrow{\xi }}^{(kn)},{\dot{\overrightarrow{u}}}^{(n)}〉}{{\Delta }_{kn}^2}}+{\displaystyle \frac{\left(〈{\overrightarrow{\xi }}^{(kn)},{\overrightarrow{u}}^{(n)}〉-c^2{\tau }_{kn}\right)〈{\overrightarrow{u}}^{(n)},{\dot{\overrightarrow{u}}}^{(n)}〉}{{\Delta }_{kn}^4}}\right),(k=1,2,3),n\ne k; \\ {H}_k^{ret}=1+D_k^{ret}\left({\displaystyle \frac{〈{\overrightarrow{\xi }}^{(k)ret},{\dot{\overrightarrow{u}}}^{(k)ret}〉}{{({\Delta }_k^{ret})}^2}}+{\displaystyle \frac{\left(〈{\overrightarrow{\xi }}^{(k)ret},{\overrightarrow{u}}^{(k)ret}〉-c^2{\tau }_k^{ret}\right)〈{\overrightarrow{u}}^{(k)ret},{\dot{\overrightarrow{u}}}^{(k)ret}〉}{{({\Delta }_k^{ret})}^4}}\right) \\ {H}_k^{adv}=1+D_k^{adv}\left({\displaystyle \frac{〈{\overrightarrow{\xi }}^{(k)adv},{\dot{\overrightarrow{u}}}^{(k)adv}〉}{{({\Delta }_k^{adv})}^2}}+{\displaystyle \frac{\left(〈{\overrightarrow{\xi }}^{(k)adv},{\overrightarrow{u}}^{(k)adv}〉-c^2{\tau }_k^{adv}\right)〈{\overrightarrow{u}}^{(k)adv},{\dot{\overrightarrow{u}}}^{(k)adv}〉}{{({\Delta }_k^{adv})}^4}}\right). \end{gathered}$$

Then the energy equations become $(k=1,2,3)$:

$${\displaystyle \frac{〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(k)}〉}{{\Delta }_k^4}}={\displaystyle \sum _{n=1,n\ne k}^3}{\displaystyle \frac{e_k{e}_n}{m_k{c}^2}}\left\{{\displaystyle \frac{\frac{\left(c^2-〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(n)}〉\right)}{{\Delta }_k{\Delta }_{kn}}{\tau }_{kn}-\frac{\left(c^2{\tau }_{kn}-〈{\overrightarrow{\xi }}^{(kn)},{\overrightarrow{u}}^{(k)}〉\right)}{{\Delta }_{kn}{\Delta }_k}}{{\left(\frac{〈{\overrightarrow{\xi }}^{(kn)},{\overrightarrow{u}}^{(n)}〉-c^2{\tau }_{kn}}{{\Delta }_{kn}}\right)}^3}}{H}_{kn}+\right.$$

(2)

$$\begin{gathered} +\left.{\Delta }_{kn}^2\left[{\displaystyle \frac{\frac{〈{\xi }^{(kn)},{\overrightarrow{u}}^{(k)}〉-c^2{\tau }_{kn}}{{\Delta }_k}\frac{D_{kn}}{{\Delta }_{kn}^4}〈{\overrightarrow{u}}^{(n)},{\dot{\overrightarrow{u}}}^{(n)}〉}{{\left(〈{\overrightarrow{\xi }}^{(kn)},{\overrightarrow{u}}^{(n)}〉-c^2{\tau }_{kn}\right)}^2}}-{\displaystyle \frac{\frac{{\tau }_{kn}{D}_{kn}}{{\Delta }_k}\left(\frac{〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(n)}〉}{{\Delta }_{kn}^2}+\frac{\left(〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(n)}〉-c^2\right)〈{\overrightarrow{u}}^{(n)},{\dot{\overrightarrow{u}}}^{(n)}〉}{{\Delta }_{kn}^4}\right)}{{\left(〈{\overrightarrow{\xi }}^{(kn)},{\overrightarrow{u}}^{(n)}〉-c^2{\tau }_{kn}\right)}^2}}\right]\right\}+ \\ +{\displaystyle \frac{{е}_k^2}{2m_k{c}^2}}\left\{{\displaystyle \frac{{\tau }_k^{ret}\frac{〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)ret}〉-c^2}{{\Delta }_k{\Delta }_{(k)ret}}-\frac{〈{\overrightarrow{u}}^{(k)},{\overrightarrow{\xi }}^{(k)r}et〉-c^2{\tau }_k^{ret}}{{\Delta }_{(k)ret}{\Delta }_k}}{{\left(\frac{〈{\overrightarrow{u}}^{(k)ret},{\overrightarrow{\xi }}^{(k)ret}〉-c^2{\tau }_k^{ret}}{{\Delta }_{(k)ret}}\right)}^3}}{H}_k^{ret}\right.+ \\ \begin{array}{l}+{\Delta }_{(k)ret}^2\left[{\displaystyle \frac{\left[\frac{〈{\overrightarrow{\xi }}^{(k)ret},{\overrightarrow{u}}^{(k)}〉-c^2{\tau }_k^{ret}}{{\Delta }_k}\right.\frac{D_k^{ret}〈{\stackrel{\rightharpoonup }{u}}^{(k)ret},{\dot{\overrightarrow{u}}}^{(k)ret}〉}{{\Delta }_{(k)ret}^4}}{{\left(c^2{\tau }_k^{ret}-〈{\overrightarrow{\xi }}^{(k)ret},{\overrightarrow{u}}^{(k)ret}〉\right)}^2}}\right.-{\displaystyle \frac{{\tau }_k^{ret}{D}_k^{ret}}{{\Delta }_k}}\left.\left.{\displaystyle \frac{\left(\frac{〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(k)ret}〉}{{\Delta }_{(k)ret}^2}+\frac{\left(\langle {\overrightarrow{u}}^{(k)},{\stackrel{\rightharpoonup }{u}}^{(k)ret}\rangle -c^2\right)〈{\overrightarrow{u}}^{(k)ret},{\dot{\overrightarrow{u}}}^{(k)ret}〉}{{\Delta }_{(k)ret}^4}\right)}{{\left(c^2{\tau }_k^{ret}-〈{\overrightarrow{\xi }}^{(k)ret},{\overrightarrow{u}}^{(k)ret}〉\right)}^2}}\right]\right\}-\\ \end{array} \\ -{\displaystyle \frac{{е}_k^2}{2m_k{c}^2}}\left\{{\displaystyle \frac{{\tau }_k^{adv}\frac{〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)adv}〉-c^2}{{\Delta }_k{\Delta }_{(k)adv}}-\frac{〈{\overrightarrow{u}}^{(k)},{\overrightarrow{\xi }}^{(k)adv}〉-c^2{\tau }_k^{adv}}{{\Delta }_{(k)adv}{\Delta }_k}}{{\left(\frac{〈{\overrightarrow{u}}^{(k)adv},{\overrightarrow{\xi }}^{(k)adv}〉-c^2{\tau }_k^{adv}}{{\Delta }_{(k)adv}}\right)}^3}}{H}_k^{adv}+\right. \\ +\left.{\Delta }_{(k)adv}^2\left[{\displaystyle \frac{\frac{〈{\overrightarrow{\xi }}^{(k)adv},{\overrightarrow{u}}^{(k)}〉-c^2{\tau }_k^{adv}}{{\Delta }_k}\frac{D_k^{adv}〈{\overrightarrow{u}}^{(k)adv},{\dot{\overrightarrow{u}}}^{(k)adv}〉}{{\Delta }_{(k)adv}^4}}{{\left(c^2{\tau }_k^{adv}-〈{\overrightarrow{\xi }}^{(k)adv},{\overrightarrow{u}}^{(k)adv}〉\right)}^2}}-{\displaystyle \frac{{\tau }_k^{adv}\frac{D_k^{adv}}{{\Delta }_k}\left(\frac{〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(k)adv}〉}{{\Delta }_{(k)adv}^2}+\frac{\left(〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)adv}〉-c^2\right)〈{\overrightarrow{u}}^{(k)adv},{\dot{\overrightarrow{u}}}^{(k)adv}〉}{{\Delta }_{(k)adv}^4}\right)}{{\left(c^2{\tau }_k^{adv}-〈{\overrightarrow{\xi }}^{(k)adv},{\overrightarrow{u}}^{(k)adv}〉\right)}^2}}\right]\right\} \end{gathered}$$

or

$${\displaystyle \frac{m_k〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(k)}〉}{{\Delta }_k^3}}=$$

(3)

$$\begin{gathered} ={\displaystyle \sum _{n=1,n\ne k}^3}{\displaystyle \frac{e_k{e}_n}{c^2}}\left\{{\displaystyle \frac{〈{\overrightarrow{\xi }}^{(kn)},{\overrightarrow{u}}^{(k)}〉-{\tau }_{kn}〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(n)}〉}{{\left(〈{\overrightarrow{\xi }}^{(kn)},{\overrightarrow{u}}^{(n)}〉-c^2{\tau }_{kn}\right)}^3}}{{\Delta }_{kn}}^2{H}_{kn}+\right.\left.{\displaystyle \frac{\left(〈{\xi }^{(kn)},{\overrightarrow{u}}^{(k)}〉-c^2{\tau }_{kn}\right)D_{kn}〈{\overrightarrow{u}}^{(n)},{\dot{\overrightarrow{u}}}^{(n)}〉}{{\Delta }_{kn}^2{\left(c^2{\tau }_{kn}-〈{\overrightarrow{\xi }}^{(kn)},{\overrightarrow{u}}^{(n)}〉\right)}^2}}-{\tau }_{kn}{D}_{kn}{\displaystyle \frac{{\Delta }_{{\tau }_{kn}}^2〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(n)}〉+\left(〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(n)}〉-c^2\right)〈{\overrightarrow{u}}^{(n)},{\dot{\overrightarrow{u}}}^{(n)}〉}{{\Delta }_{{\tau }_{kn}}^2{\left(c^2{\tau }_{kn}-〈{\overrightarrow{\xi }}^{(kn)},{\overrightarrow{u}}^{(n)}〉\right)}^2}}\right\}+ \\ +{\displaystyle \frac{{е}_k^2}{2c^2}}\left\{{\displaystyle \frac{〈{\overrightarrow{u}}^{(k)},{\overrightarrow{\xi }}^{(k)ret}〉-{\tau }_k^{ret}〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)ret}〉}{{\left(〈{\overrightarrow{u}}^{(k)ret},{\overrightarrow{\xi }}^{(k)ret}〉-c^2{\tau }_k^{ret}\right)}^3}}{({\Delta }_k^{ret})}^2{H}_k^{ret}+{\displaystyle \frac{\left(〈{\overrightarrow{u}}^{(k)},{\overrightarrow{\xi }}^{(k)ret}〉-c^2{\tau }_k^{ret}\right)D_k^{ret}〈{\overrightarrow{u}}^{(k)ret},{\dot{\overrightarrow{u}}}^{(k)ret}〉}{{({\Delta }_k^{ret})}^2{\left(c^2{\tau }_k^{ret}-〈{\overrightarrow{u}}^{(k)ret},{\overrightarrow{\xi }}^{(k)ret}〉\right)}^2}}-{\tau }_k^{ret}{D}_k^{ret}{\displaystyle \frac{{({\Delta }_k^{ret})}^2〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(k)ret}〉+\left(〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)ret}〉-c^2\right)〈{\overrightarrow{u}}^{(k)ret},{\dot{\overrightarrow{u}}}^{(k)ret}〉}{{({\Delta }_k^{ret})}^2{\left(c^2{\tau }_k^{ret}-〈{\overrightarrow{u}}^{(k)ret},{\overrightarrow{\xi }}^{(k)ret}〉\right)}^2}}\right\}- \\ -{\displaystyle \frac{{е}_k^2}{2c^2}}\left\{{\displaystyle \frac{〈{\overrightarrow{u}}^{(k)},{\overrightarrow{\xi }}^{(k)adv}〉-{\tau }_k^{adv}〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)adv}〉}{{\left(〈{\overrightarrow{u}}^{(k)adv},{\overrightarrow{\xi }}^{(k)adv}〉-c^2{\tau }_k^{adv}\right)}^3}}{({\Delta }_k^{adv})}^2{H}_k^{adv}+{\displaystyle \frac{\left(〈{\overrightarrow{u}}^{(k)},{\overrightarrow{\xi }}^{(k)advt}〉-c^2{\tau }_k^{adv}\right)D_k^{adv}〈{\overrightarrow{u}}^{(k)adv},{\dot{\overrightarrow{u}}}^{(k)adv}〉}{{({\Delta }_k^{adv})}^2{\left(c^2{\tau }_k^{adv}-〈{\overrightarrow{u}}^{(k)adv},{\overrightarrow{\xi }}^{(k)adv}〉\right)}^2}}-{\tau }_k^{adv}{D}_k^{adv}{\displaystyle \frac{{({\Delta }_k^{adv})}^2〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(k)adv}〉+\left(〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)adv}〉-c^2\right)〈{\overrightarrow{u}}^{(k)adv},{\dot{\overrightarrow{u}}}^{(k)adv}〉}{{({\Delta }_k^{adv})}^2{\left(c^2{\tau }_k^{ret}-〈{\overrightarrow{u}}^{(k)adv},{\overrightarrow{\xi }}^{(k)adv}〉\right)}^2}}\right\}. \end{gathered}$$

4. Transformation of the Radiation Part of the Energy Terms

Here we follow the Dirac assumption ${\tau }_k^{ret}={\tau }_k^{adv}=\tau$, $\tau$ is a small parameter. This assumption is motivated by the fact $\tau ={\tau }_0\sqrt{1-{\beta }^2},\left({\tau }_0=r_e/c\approx {10}^{-24}\sec \right)$. Since $u_{\alpha }^{(k)}(t)$ are infinitely smooth functions, using the Taylor expansions we have:

$$\begin{gathered} {\xi }_{\alpha }^{(k)adv}=x_{\alpha }^{(k)}(t+\tau )-x_{\alpha }^{(k)}(t)=\tau u_{\alpha }^{(k)}(t)+O({\tau }^2)\Rightarrow {\xi }_{\alpha }^{(k)adv}\approx \tau u_{\alpha }^{(k)}(t);{\xi }_{\alpha }^{(k)ret}=x_{\alpha }^{(k)}(t)-x_{\alpha }^{(k)}(t-\tau )\approx u_{\alpha }^{(k)}\tau ; \\ {u}_{\alpha }^{(k)}(t+\tau )=u_{\alpha }^{(k)}(t)+O(\tau );u_{\alpha }^{(k)}(t-\tau )=u_{\alpha }^{(k)}(t)-O(\tau );u_{\alpha }^{(k)}(t)u_{\alpha }^{(k)}(t+\tau )={\left(u_{\alpha }^{(k)}(t)\right)}^2+O(\tau ); \\ {u}_{\alpha }^{(k)}(t)u_{\alpha }^{(k)}(t-\tau )={\left(u_{\alpha }^{(k)}(t)\right)}^2-O(\tau ); \\ 〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)adv}〉={\displaystyle \sum _{\gamma =1}^3{u}_{\gamma }^{(k)}(t)u_{\gamma }^{(k)}(t+\tau )}\approx {\displaystyle \sum _{\gamma =1}^3{u}_{\gamma }^{(k)}(t)u_{\gamma }^{(k)}(t)}=〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}〉; \\ 〈{\overrightarrow{u}}^{(k)adv},{\overrightarrow{u}}^{(k)adv}〉\approx 〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}〉;〈{\overrightarrow{u}}^{(k)ret},{\overrightarrow{u}}^{(k)ret}〉\approx 〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}〉; \\ {c}^2{\tau }_k^{ret}-〈{\overrightarrow{\xi }}^{(k)ret},{\overrightarrow{u}}^{(k)ret}〉=c^2\tau -\tau 〈{\overrightarrow{u}}^{(k)}(t),{\overrightarrow{u}}^{(k)}(t-\tau )〉\approx \tau \left(c^2-〈{\overrightarrow{u}}^{(k)}(t),{\overrightarrow{u}}^{(k)}(t)〉\right); \\ {c}^2{\tau }_k^{adv}-〈{\overrightarrow{\xi }}^{(k)adv},{\overrightarrow{u}}^{(k)adv}〉=c^2\tau -\tau 〈{\overrightarrow{u}}^{(k)}(t),{\overrightarrow{u}}^{(k)}(t+\tau )〉\approx \tau \left(c^2-〈{\overrightarrow{u}}^{(k)}(t),{\overrightarrow{u}}^{(k)}(t)〉\right); \\ {D}_{kn}={\displaystyle \frac{c^2{\tau }_{kn}-〈{\overrightarrow{\xi }}^{(kn)},{\overrightarrow{u}}^{(n)}〉}{c^2{\tau }_{kn}-〈{\overrightarrow{\xi }}^{(kn)},{\overrightarrow{u}}^{(k)}〉}}\approx 1;D_k^{ret}={\displaystyle \frac{c{\tau }_k^{ret}-〈{\overrightarrow{\xi }}^{(k)ret},{\overrightarrow{u}}^{(k)}(t-{\tau }_k^{ret})〉}{c{\tau }_k^{ret}-〈{\overrightarrow{\xi }}^{(k)ret},{\overrightarrow{u}}^{(k)}(t)〉}}\approx {\displaystyle \frac{\tau \left(c^2-〈{\overrightarrow{u}}^{(k)}(t),{\overrightarrow{u}}^{(k)}(t)〉\right)}{\tau \left(c^2-〈{\overrightarrow{u}}^{(k)}(t),{\overrightarrow{u}}^{(k)}(t)〉\right)}}=1; \\ {D}_k^{adv}={\displaystyle \frac{c{\tau }_k^{adv}-〈{\overrightarrow{\xi }}^{(k)adv},{\overrightarrow{u}}^{(k)}(t+{\tau }_k^{adv})〉}{c{\tau }_k^{adv}-〈{\overrightarrow{\xi }}^{(k)adv},{\overrightarrow{u}}^{(k)}(t)〉}}\approx {\displaystyle \frac{\tau \left(c^2-〈{\overrightarrow{u}}^{(k)}(t),{\overrightarrow{u}}^{(k)}(t)〉\right)}{\tau \left(c^2-〈{\overrightarrow{u}}^{(k)}(t),{\overrightarrow{u}}^{(k)}(t)〉\right)}}=1; \\ {H}_k^{ret}\approx 1+\left({\displaystyle \frac{〈\tau {\overrightarrow{u}}^{(k)}(t),{\dot{\overrightarrow{u}}}^{(k)}(t-\tau )〉}{c^2-〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}〉}}-{\displaystyle \frac{\tau \left(c^2-〈{\overrightarrow{u}}^{(k)}(t),{\overrightarrow{u}}^{(k)}(t)〉\right)〈{\overrightarrow{u}}^{(k)}(t),{\dot{\overrightarrow{u}}}^{(k)}(t-\tau )〉}{{\left(c^2-〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}〉\right)}^2}}\right)=1; \\ {H}_k^{adv}=1+\left({\displaystyle \frac{〈\tau {\overrightarrow{u}}^{(k)}(t),{\dot{\overrightarrow{u}}}^{(k)}(t+\tau )〉}{c^2-〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}〉}}-{\displaystyle \frac{\tau \left(c^2-〈{\overrightarrow{u}}^{(k)}(t),{\overrightarrow{u}}^{(k)}(t)〉\right)〈{\overrightarrow{u}}^{(k)}(t),{\dot{\overrightarrow{u}}}^{(k)}(t+\tau )〉}{{\left(c^2-〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}〉\right)}^2}}\right)=1. \end{gathered}$$

Then the radiation part becomes

$$\begin{gathered} \begin{array}{l}{R}_{(k)}^{rad}={\displaystyle \frac{{е}_k^2}{2c^2}}\left\{{\displaystyle \frac{〈{\overrightarrow{u}}^{(k)},{\overrightarrow{\xi }}^{(k)ret}〉-{\tau }_k^{ret}〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)ret}〉}{{\left(〈{\overrightarrow{u}}^{(k)ret},{\overrightarrow{\xi }}^{(k)ret}〉-c^2{\tau }_k^{ret}\right)}^3}}{({\Delta }_k^{ret})}^2{H}_k^{ret}+{\displaystyle \frac{\left(〈{\overrightarrow{u}}^{(k)},{\overrightarrow{\xi }}^{(k)ret}〉-c^2{\tau }_k^{ret}\right)D_k^{ret}〈{\overrightarrow{u}}^{(k)ret},{\dot{\overrightarrow{u}}}^{(k)ret}〉}{{({\Delta }_k^{ret})}^2{\left(c^2{\tau }_k^{ret}-〈{\overrightarrow{u}}^{(k)ret},{\overrightarrow{\xi }}^{(k)ret}〉\right)}^2}}-{\tau }_k^{ret}{D}_k^{ret}{\displaystyle \frac{{({\Delta }_k^{ret})}^2〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(k)ret}〉+\left(〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)ret}〉-c^2\right)〈{\overrightarrow{u}}^{(k)ret},{\dot{\overrightarrow{u}}}^{(k)ret}〉}{{\Delta }_k^{ret}{)}^2{\left(c^2{\tau }_k^{ret}-〈{\overrightarrow{u}}^{(k)ret},{\overrightarrow{\xi }}^{(k)ret}〉\right)}^2}}-\right.\\ -{\displaystyle \frac{〈{\overrightarrow{u}}^{(k)},{\overrightarrow{\xi }}^{(k)adv}〉-{\tau }_k^{adv}〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)adv}〉}{{\left(〈{\overrightarrow{u}}^{(k)adv},{\overrightarrow{\xi }}^{(k)adv}〉-c^2{\tau }_k^{adv}\right)}^3}}\left.{({\Delta }_k^{adv})}^2{H}_k^{adv}-{\displaystyle \frac{\left(〈{\overrightarrow{u}}^{(k)},{\overrightarrow{\xi }}^{(k)advt}〉-c^2{\tau }_k^{adv}\right)D_k^{adv}〈{\overrightarrow{u}}^{(k)adv},{\dot{\overrightarrow{u}}}^{(k)adv}〉}{{({\Delta }_k^{adv})}^2{\left(c^2{\tau }_k^{adv}-〈{\overrightarrow{u}}^{(k)adv},{\overrightarrow{\xi }}^{(k)adv}〉\right)}^2}}+{\tau }_k^{adv}{D}_k^{adv}{\displaystyle \frac{{({\Delta }_k^{adv})}^2〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(k)adv}〉+\left(〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)adv}〉-c^2\right)〈{\overrightarrow{u}}^{(k)adv},{\dot{\overrightarrow{u}}}^{(k)adv}〉}{{({\Delta }_k^{adv})}^2{\left(c^2{\tau }_k^{ret}-〈{\overrightarrow{u}}^{(k)adv},{\overrightarrow{\xi }}^{(k)adv}〉\right)}^2}}\right\}=\end{array} \\ \begin{array}{l}={\displaystyle \frac{{е}_k^2}{2c^2}}\left\{{\displaystyle \frac{\tau 〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}〉-〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}\tau 〉}{{\left(c^2\tau -〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}\tau 〉\right)}^3}}{({\Delta }_k^{ret})}^2+{\displaystyle \frac{\left(〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}\tau 〉-c^2\tau \right)〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(k)ret}〉}{{({\Delta }_k^{ret})}^2{\left(c^2\tau -〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}\tau 〉\right)}^2}}-\tau {\displaystyle \frac{{({\Delta }_k^{ret})}^2〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(k)ret}〉+\left(〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}〉-c^2\right)〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(k)ret}〉}{{({\Delta }_k^{ret})}^2{\left(c^2\tau -〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}\tau 〉\right)}^2}}-\right.\\ -{\displaystyle \frac{\tau 〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}〉-〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}\tau 〉}{{\left(c^2\tau -〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}\tau 〉\right)}^3}}\left.{({\Delta }_k^{adv})}^2-{\displaystyle \frac{\left(〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}\tau 〉-c^2\tau \right)〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(k)adv}〉}{{({\Delta }_k^{adv})}^2{\left(c^2\tau -〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}\tau 〉\right)}^2}}+\tau {\displaystyle \frac{{({\Delta }_k^{adv})}^2〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(k)adv}〉+\left(〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}〉-c^2\right)〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(k)adv}〉}{{({\Delta }_k^{adv})}^2{\left(c^2\tau -〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}\tau 〉\right)}^2}}\right\}=\end{array} \\ \begin{array}{l}={\displaystyle \frac{{е}_k^2}{2c^2}}\left\{-{\displaystyle \frac{\left(c^2-〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}〉\right)〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(k)ret}〉}{{{\Delta }_k}^2\tau {\left(c^2-〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}〉\right)}^2}}-\tau {\displaystyle \frac{{{\Delta }_k}^2〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(k)ret}〉-\left(c^2-〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}〉\right)〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(k)ret}〉}{{{\Delta }_k}^2{\tau }^2{\left(c^2-〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}〉\right)}^2}}-\right.\\ \left.-{\displaystyle \frac{\left(c^2-〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}〉\right)〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(k)adv}〉}{{{\Delta }_k}^2\tau {\left(c^2-〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}〉\right)}^2}}+\tau {\displaystyle \frac{{{\Delta }_k}^2〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(k)adv}〉-\left(c^2-〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}〉\right)〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(k)adv}〉}{{{\Delta }_k}^2{\tau }^2{\left(c^2-〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}〉\right)}^2}}\right\}=\end{array} \\ ={\displaystyle \frac{{е}_k^2}{2c^2{{\Delta }_k}^4}}\left\{{\displaystyle \frac{〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(k)adv}〉}{\tau }}-{\displaystyle \frac{〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(k)ret}〉}{\tau }}+{\displaystyle \frac{〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(k)adv}〉-〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(k)adv}〉}{\tau }}-{\displaystyle \frac{〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(k)ret}〉-〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(k)ret}〉}{\tau }}\right\}={\displaystyle \frac{{е}_k^2}{c^2{{\Delta }_k}^4}}〈{\overrightarrow{u}}^{(k)},{\displaystyle \frac{{\dot{\overrightarrow{u}}}^{(k)adv}-{\dot{\overrightarrow{u}}}^{(k)ret}}{2\tau }}〉\approx {\displaystyle \frac{{е}_k^2}{c^2{{\Delta }_k}^4}}〈{\overrightarrow{u}}^{(k)},{\ddot{\overrightarrow{u}}}^{(k)}〉. \end{gathered}$$

Besides

$$\begin{array}{l}{H}_{21}=1+{\displaystyle \frac{〈{\overrightarrow{\xi }}^{(21)},{\dot{\overrightarrow{u}}}^{(1)}〉}{{\Delta }_{21}^2}}+{\displaystyle \frac{\left(〈{\overrightarrow{\xi }}^{(21)},{\overrightarrow{u}}^{(1)}〉-c^2{\tau }_{21}\right)〈{\overrightarrow{u}}^{(1)},{\dot{\overrightarrow{u}}}^{(1)}〉}{{\Delta }_{21}^4}}=1,\\ H_{23}=1+{\displaystyle \frac{〈{\overrightarrow{\xi }}^{(23)},{\dot{\overrightarrow{u}}}^{(3)}〉}{{\Delta }_{23}^2}}+{\displaystyle \frac{\left(〈{\overrightarrow{\xi }}^{(23)},{\overrightarrow{u}}^{(3)}〉-c^2{\tau }_{23}\right)〈{\overrightarrow{u}}^{(3)},{\dot{\overrightarrow{u}}}^{(3)}〉}{{\Delta }_{23}^4}}\end{array}.$$

It is known that (cf. [5]):

$$E_k=m_k{c}^2/\sqrt{1-{\displaystyle \frac{〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}〉}{c^2}}}={\displaystyle \frac{m_k{c}^3}{{\Delta }_k}},{\displaystyle \frac{d{E}_k}{dt}}=-{\displaystyle \frac{1}{2}}{\displaystyle \frac{m_k{c}^3(-2〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(k)}〉)}{{{\Delta }_k}^3}}={\displaystyle \frac{m_k{c}^3〈{\overrightarrow{u}}^{(k)},{\dot{\overrightarrow{u}}}^{(k)}〉}{{{\Delta }_k}^3}}.$$

Then from (3) we obtain

$$\begin{gathered} \begin{array}{l}{\displaystyle \frac{d{E}_2}{dt}}=c{e}_2{e}_1\left[{\displaystyle \frac{{\tau }_{21}〈{\overrightarrow{u}}^{(2)},{\overrightarrow{u}}^{(1)}〉-〈{\overrightarrow{\xi }}^{(21)},{\overrightarrow{u}}^{(2)}〉}{{\left(c^2{\tau }_{21}-〈{\overrightarrow{\xi }}^{(21)},{\overrightarrow{u}}^{(1)}〉\right)}^3}}{{\Delta }_{21}}^2+{\displaystyle \frac{\left(〈{\xi }^{(21)},{\overrightarrow{u}}^{(2)}〉-{\tau }_{21}〈{\overrightarrow{u}}^{(2)},{\overrightarrow{u}}^{(1)}〉\right)〈{\overrightarrow{u}}^{(1)},{\dot{\overrightarrow{u}}}^{(1)}〉-{\tau }_{21}{\Delta }_{21}^2〈{\overrightarrow{u}}^{(2)},{\dot{\overrightarrow{u}}}^{(1)}〉}{{\Delta }_{21}^2{\left(c^2{\tau }_{21}-〈{\overrightarrow{\xi }}^{(21)},{\overrightarrow{u}}^{(1)}〉\right)}^2}}\right]+\\ +c{e}_2{e}_3\left[{\displaystyle \frac{{\tau }_{23}〈{\overrightarrow{u}}^{(2)},{\overrightarrow{u}}^{(3)}〉-〈{\overrightarrow{\xi }}^{(23)},{\overrightarrow{u}}^{(2)}〉}{{\left(c^2{\tau }_{23}-〈{\overrightarrow{\xi }}^{(23)},{\overrightarrow{u}}^{(3)}〉\right)}^3}}{{\Delta }_{23}}^2{H}_{23}+{\displaystyle \frac{\left(〈{\xi }^{(23)},{\overrightarrow{u}}^{(2)}〉-{\tau }_{23}〈{\overrightarrow{u}}^{(2)},{\overrightarrow{u}}^{(3)}〉\right)〈{\overrightarrow{u}}^{(3)},{\dot{\overrightarrow{u}}}^{(3)}〉-{\tau }_{23}{\Delta }_{23}^2〈{\overrightarrow{u}}^{(2)},{\dot{\overrightarrow{u}}}^{(31}〉}{{\Delta }_{23}^2{\left(c^2{\tau }_{23}-〈{\overrightarrow{\xi }}^{(23)},{\overrightarrow{u}}^{(3)}〉\right)}^2}}\right]+\\ +{\displaystyle \frac{c{е}_2^2〈{\overrightarrow{u}}^{(2)},{\ddot{\overrightarrow{u}}}^{(2)}〉}{{{\Delta }_2}^4}};\end{array} \\ \begin{array}{l}{\displaystyle \frac{d{E}_3}{dt}}=c{e}_3{e}_1\left[{\displaystyle \frac{{\tau }_{21}〈{\overrightarrow{u}}^{(3)},{\overrightarrow{u}}^{(1)}〉-〈{\overrightarrow{\xi }}^{(31)},{\overrightarrow{u}}^{(3)}〉}{{\left(c^2{\tau }_{31}-〈{\overrightarrow{\xi }}^{(31)},{\overrightarrow{u}}^{(1)}〉\right)}^3}}{{\Delta }_{31}}^2+{\displaystyle \frac{\left(〈{\xi }^{(31)},{\overrightarrow{u}}^{(3)}〉-{\tau }_{31}〈{\overrightarrow{u}}^{(3)},{\overrightarrow{u}}^{(1)}〉\right)〈{\overrightarrow{u}}^{(1)},{\dot{\overrightarrow{u}}}^{(1)}〉-{\tau }_{31}{\Delta }_{31}^2〈{\overrightarrow{u}}^{(3)},{\dot{\overrightarrow{u}}}^{(1)}〉}{{\Delta }_{31}^2{\left(c^2{\tau }_{31}-〈{\overrightarrow{\xi }}^{(31)},{\overrightarrow{u}}^{(1)}〉\right)}^2}}\right]+\\ +c{e}_3{e}_2\left[{\displaystyle \frac{{\tau }_{32}〈{\overrightarrow{u}}^{(3)},{\overrightarrow{u}}^{(2)}〉-〈{\overrightarrow{\xi }}^{(32)},{\overrightarrow{u}}^{(3)}〉}{{\left(c^2{\tau }_{32}-〈{\overrightarrow{\xi }}^{(32)},{\overrightarrow{u}}^{(3)}〉\right)}^3}}{{\Delta }_{32}}^2{H}_{32}+{\displaystyle \frac{\left(〈{\overrightarrow{\xi }}^{(32)},{\overrightarrow{u}}^{(3)}〉-{\tau }_{32}〈{\overrightarrow{u}}^{(3)},{\overrightarrow{u}}^{(2)}〉\right)〈{\overrightarrow{u}}^{(3)},{\dot{\overrightarrow{u}}}^{(3)}〉-{\tau }_{32}{\Delta }_{32}^2〈{\overrightarrow{u}}^{(3)},{\dot{\overrightarrow{u}}}^{(2)}〉}{{\Delta }_{32}^2{\left(c^2{\tau }_{32}-〈{\overrightarrow{\xi }}^{(32)},{\overrightarrow{u}}^{(3)}〉\right)}^2}}\right]+\\ +{\displaystyle \frac{c{е}_3^2〈{\overrightarrow{u}}^{(3)},{\ddot{\overrightarrow{u}}}^{(3)}〉}{{{\Delta }_3}^4}}.\end{array} \end{gathered}$$

In the Keper formulation one obtains ${\overrightarrow{u}}^{(1)}=\overrightarrow{0}$ and ${\Delta }_{31}=\sqrt{c^2-〈\overrightarrow{u}(t-{\tau }_{31}),\overrightarrow{u}(t-{\tau }_{31})〉}\approx c\sqrt{1-{\displaystyle \frac{〈\overrightarrow{u}(t-{\tau }_{31}),\overrightarrow{u}(t-{\tau }_{31})〉}{c^2}}}\approx c$, because we consider the case $\left|〈\overrightarrow{u}(t-{\tau }_{31}),\overrightarrow{u}(t-{\tau }_{31})〉\right|\le {\overline{c}}^2<c^2$and then $\beta =\overline{c}/c=1/137$- Sommerfeld fine structure constant (cf. [3]).

Further on,

$$\begin{gathered} {\xi }^{(21)}=\left({{\xi }_1}^{(21)},{{\xi }_2}^{(21)},{{\xi }_3}^{(21)},{{\xi }_4}^{(21)}\right)=\left(x_1^{(2)}(t),x_2^{(2)}(t),x_3^{(2)}(t),ic{\tau }_{21}\right), \\ {\xi }^{(23)}=\left({{\xi }_1}^{(23)},{{\xi }_2}^{(23)},{{\xi }_3}^{(23)},{{\xi }_4}^{(23)}\right)=\left(x_1^{(2)}(t)-x_1^{(3)}(t-{\tau }_{23}),x_2^{(2)}(t)-x_2^{(3)}(t-{\tau }_{23}),x_3^{(2)}(t)-x_3^{(3)}(t-{\tau }_{23}),ic{\tau }_{23}\right), \\ {\xi }^{(31)}=\left({{\xi }_1}^{(31)},{{\xi }_2}^{(31)},{{\xi }_3}^{(31)},{{\xi }_4}^{(31)}\right)=\left(x_1^{(3)}(t),x_2^{(3)}(t),x_3^{(3)}(t),ic{\tau }_{31}\right), \\ {\xi }^{(32)}=\left({{\xi }_1}^{(32)},{{\xi }_2}^{(32)},{{\xi }_3}^{(32)},{{\xi }_4}^{(32)}\right)=\left(x_1^{(3)}(t)-x_1^{(2)}(t-{\tau }_{32}),x_2^{(3)}(t)-x_2^{(2)}(t-{\tau }_{32}),x_3^{(3)}(t)-x_3^{(2)}(t-{\tau }_{32}),ic{\tau }_{32}\right), \\ \left|〈{\overrightarrow{\xi }}^{(23)},{\overrightarrow{u}}^{(3)}〉\right|\le \sqrt{〈{\overrightarrow{\xi }}^{(23)},{\overrightarrow{\xi }}^{(23)}〉}\sqrt{〈{\overrightarrow{u}}^{(3)},{\overrightarrow{u}}^{(3)}〉}=c{\tau }_{23}\overline{c}, \\ {H}_{23}\approx 1+{\displaystyle \frac{〈{\overrightarrow{\xi }}^{(23)},{\dot{\overrightarrow{u}}}^{(3)}〉}{c^2}}+{\displaystyle \frac{\left(〈{\overrightarrow{\xi }}^{(23)},{\overrightarrow{u}}^{(3)}〉-c^2{\tau }_{23}\right)〈{\overrightarrow{u}}^{(3)},{\dot{\overrightarrow{u}}}^{(3)}〉}{c^4}}, \\ {H}_{32}\approx 1+{\displaystyle \frac{〈{\overrightarrow{\xi }}^{(32)},{\dot{\overrightarrow{u}}}^{(2)}〉}{c^2}}+{\displaystyle \frac{\left(〈{\overrightarrow{\xi }}^{(32)},{\overrightarrow{u}}^{(2)}〉-c^2{\tau }_{32}\right)〈{\overrightarrow{u}}^{(2)},{\dot{\overrightarrow{u}}}^{(2)}〉}{c^4}}, \\ \begin{array}{l}{\displaystyle \frac{d{E}_2}{dt}}=e_2{e}_1{\displaystyle \frac{-〈{\overrightarrow{x}}^{(2)},{\overrightarrow{u}}^{(2)}〉}{c^3{{\tau }_{21}}^3}}+\\ +e_2{e}_3\left[{\displaystyle \frac{{\tau }_{23}〈{\overrightarrow{u}}^{(2)},{\overrightarrow{u}}^{(3)}〉-〈{\overrightarrow{\xi }}^{(23)},{\overrightarrow{u}}^{(2)}〉}{{\left(c^2{\tau }_{23}-〈{\overrightarrow{\xi }}^{(23)},{\overrightarrow{u}}^{(3)}〉\right)}^3}}{c}^3\left(1+{\displaystyle \frac{〈{\overrightarrow{\xi }}^{(23)},{\dot{\overrightarrow{u}}}^{(3)}〉}{c^2}}+{\displaystyle \frac{\left(〈{\overrightarrow{\xi }}^{(23)},{\overrightarrow{u}}^{(3)}〉-c^2{\tau }_{23}\right)〈{\overrightarrow{u}}^{(3)},{\dot{\overrightarrow{u}}}^{(3)}〉}{c^4}}\right)+\right.\\ \left.+{\displaystyle \frac{\left(〈{\xi }^{(23)},{\overrightarrow{u}}^{(2)}〉-{\tau }_{23}〈{\overrightarrow{u}}^{(2)},{\overrightarrow{u}}^{(3)}〉\right)〈{\overrightarrow{u}}^{(3)},{\dot{\overrightarrow{u}}}^{(3)}〉-{\tau }_{23}{c}^2〈{\overrightarrow{u}}^{(2)},{\dot{\overrightarrow{u}}}^{(3)}〉}{c{\left(c^2{\tau }_{23}-〈{\overrightarrow{\xi }}^{(23)},{\overrightarrow{u}}^{(3)}〉\right)}^2}}\right]+{\displaystyle \frac{{е}_2^2〈{\overrightarrow{u}}^{(2)},{\ddot{\overrightarrow{u}}}^{(2)}〉}{c^3}},\end{array} \\ \begin{array}{l}{\displaystyle \frac{d{E}_3}{dt}}=e_3{e}_1{\displaystyle \frac{-〈{\overrightarrow{x}}^{(3)},{\overrightarrow{u}}^{(3)}〉}{c^3{{\tau }_{31}}^3}}+e_3{e}_2\left[{\displaystyle \frac{{\tau }_{32}〈{\overrightarrow{u}}^{(3)},{\overrightarrow{u}}^{(2)}〉-〈{\overrightarrow{\xi }}^{(32)},{\overrightarrow{u}}^{(3)}〉}{{\left(c^2{\tau }_{32}-〈{\overrightarrow{\xi }}^{(32)},{\overrightarrow{u}}^{(3)}〉\right)}^3}}{c}^3\left(1+{\displaystyle \frac{〈{\overrightarrow{\xi }}^{(32)},{\dot{\overrightarrow{u}}}^{(2)}〉}{c^2}}+{\displaystyle \frac{\left(〈{\overrightarrow{\xi }}^{(32)},{\overrightarrow{u}}^{(2)}〉-c^2{\tau }_{32}\right)〈{\overrightarrow{u}}^{(2)},{\dot{\overrightarrow{u}}}^{(2)}〉}{c^4}}\right)+\right.\\ \left.+{\displaystyle \frac{\left(〈{\overrightarrow{\xi }}^{(32)},{\overrightarrow{u}}^{(3)}〉-{\tau }_{32}〈{\overrightarrow{u}}^{(3)},{\overrightarrow{u}}^{(2)}〉\right)〈{\overrightarrow{u}}^{(3)},{\dot{\overrightarrow{u}}}^{(3)}〉-{\tau }_{32}{c}^2〈{\overrightarrow{u}}^{(3)},{\dot{\overrightarrow{u}}}^{(2)}〉}{c{\left(c^2{\tau }_{32}-〈{\overrightarrow{\xi }}^{(32)},{\overrightarrow{u}}^{(3)}〉\right)}^2}}\right]+{\displaystyle \frac{{е}_3^2〈{\overrightarrow{u}}^{(3)},{\ddot{\overrightarrow{u}}}^{(3)}〉}{c^3}}.\end{array}. \end{gathered}$$

5. Energy Inequalities

We recall that $‖{\overrightarrow{u}}^{(k)}‖=\sqrt{〈{\overrightarrow{u}}^{(k)},{\overrightarrow{u}}^{(k)}〉},(k=2,3)$,${\rho }_k(t)=\sqrt{〈{\overrightarrow{x}}^{(k)},{\overrightarrow{x}}^{(k)}〉},(k=2,3)$,$c{\tau }_{kn}=\sqrt{〈{\overrightarrow{\xi }}^{(kn)},{\overrightarrow{\xi }}^{(kn)}〉}$.

Then

$$\begin{gathered} \begin{array}{l}\left|E_2(t)\right|\le \left|e_2{e}_1\right|{\displaystyle \underset{0}{\overset{t}{\int }}\frac{{\rho }_2‖{\overrightarrow{u}}^{(2)}‖}{c^3{{\tau }_{21}}^3}}ds+\\ +e_2{e}_3{\displaystyle \underset{0}{\overset{t}{\int }}}\left[{\displaystyle \frac{{\tau }_{23}‖{\overrightarrow{u}}^{(2)}‖‖{\overrightarrow{u}}^{(3)}‖+\sqrt{〈{\overrightarrow{\xi }}^{(23)},{\overrightarrow{\xi }}^{(23)}〉}‖{\overrightarrow{u}}^{(2)}‖}{{\left(c^2{\tau }_{23}-〈{\overrightarrow{\xi }}^{(23)},{\overrightarrow{u}}^{(3)}〉\right)}^3}}{c}^3\left(1+{\displaystyle \frac{\sqrt{〈{\overrightarrow{\xi }}^{(23)},{\overrightarrow{\xi }}^{(23)}〉}‖{\dot{\overrightarrow{u}}}^{(3)}‖}{c^2}}+{\displaystyle \frac{\left(\sqrt{〈{\overrightarrow{\xi }}^{(23)},{\overrightarrow{\xi }}^{(23)}〉}‖{\overrightarrow{u}}^{(3)}‖+c^2{\tau }_{23}\right)‖{\overrightarrow{u}}^{(3)}‖‖{\dot{\overrightarrow{u}}}^{(3)}‖}{c^4}}\right)+\right.\\ \left.+{\displaystyle \frac{\left(\sqrt{〈{\overrightarrow{\xi }}^{(23)},{\overrightarrow{\xi }}^{(23)}〉}‖{\overrightarrow{u}}^{(2)}‖+{\tau }_{23}‖{\overrightarrow{u}}^{(2)}‖‖{\overrightarrow{u}}^{(3)}‖\right)‖{\overrightarrow{u}}^{(3)}‖‖{\dot{\overrightarrow{u}}}^{(3)}‖+{\tau }_{23}{c}^2‖{\overrightarrow{u}}^{(2)}‖‖{\dot{\overrightarrow{u}}}^{(3)}‖}{c{\left(c^2{\tau }_{23}-〈{\overrightarrow{\xi }}^{(23)},{\overrightarrow{u}}^{(3)}〉\right)}^2}}\right]ds+{\displaystyle \underset{0}{\overset{t}{\int }}}{\displaystyle \frac{{е}_2^2‖{\overrightarrow{u}}^{(2)}‖\sqrt{〈{\ddot{\overrightarrow{u}}}^{(2)},{\ddot{\overrightarrow{u}}}^{(2)}〉}}{c^3}}ds\le \\ \le \left|e_2{e}_1\right|{\displaystyle \underset{0}{\overset{t}{\int }}}{\displaystyle \frac{{\rho }_2\overline{c}}{c^3{{\tau }_{21}}^3}}ds+\left|e_2{e}_3\right|{\displaystyle \underset{0}{\overset{t}{\int }}}\left[{\displaystyle \frac{{\tau }_{23}({\overline{c}}^2+c\overline{c})}{{\left(c^2{\tau }_{23}-c{\tau }_{23}\overline{c}\right)}^3}}{c}^3\left(1+{\displaystyle \frac{c{\tau }_{23}‖{\dot{\overrightarrow{u}}}^{(3)}‖}{c^2}}+{\displaystyle \frac{{\tau }_{23}\left(c\overline{c}+c^2\right)\overline{c}‖{\dot{\overrightarrow{u}}}^{(3)}‖}{c^4}}\right)+\right.\\ \left.+{\displaystyle \frac{\left(c{\tau }_{23}\overline{c}+{\tau }_{23}{\overline{c}}^2\right)\overline{c}‖{\dot{\overrightarrow{u}}}^{(3)}‖+{\tau }_{23}{c}^2\overline{c}‖{\dot{\overrightarrow{u}}}^{(3)}‖}{c{\left(c^2{\tau }_{23}-c{\tau }_{23}\overline{c}\right)}^2}}\right]ds+{\displaystyle \underset{0}{\overset{t}{\int }}\frac{{е}_2^2\overline{c}‖{\ddot{\overrightarrow{u}}}^{(2)}‖}{c^3}ds}\cong \\ \cong \beta {\displaystyle \underset{0}{\overset{t}{\int }}}\left[\left|e_2{e}_1\right|{\displaystyle \frac{{\rho }_2}{c^2{{\tau }_{21}}^3}}+\left|e_2{e}_3\right|{\displaystyle \frac{(1+\beta )}{{\left(1-\beta \right)}^3}}\left({\displaystyle \frac{1}{c{{\tau }_{23}}^2}}+{\displaystyle \frac{2‖{\dot{\overrightarrow{u}}}^{(3)}‖}{c^2{\tau }_{23}}}\right)+{\displaystyle \frac{{е}_2^2‖{\ddot{\overrightarrow{u}}}^{(2)}‖}{c^2}}\right]ds,\end{array} \\ \begin{array}{l}\left|E_3(t)\right|\le \left|e_3{e}_1\right|{\displaystyle \underset{0}{\overset{t}{\int }}}{\displaystyle \frac{{\rho }_3\overline{c}}{c^3{{\tau }_{31}}^3}}ds+\left|e_3{e}_2\right|{\displaystyle \underset{0}{\overset{t}{\int }}}\left[{\displaystyle \frac{{\tau }_{32}{\overline{c}}^2+c{\tau }_{32}\overline{c}}{{\left(c^2{\tau }_{32}-c{\tau }_{32}\overline{c}\right)}^3}}{c}^3\left(1+{\displaystyle \frac{c^3+c{\overline{c}}^2+\overline{c}{c}^2}{c^4}}{\tau }_{32}‖{\dot{\overrightarrow{u}}}^{(2)}‖\right)+\right.\\ \left.+{\displaystyle \frac{\left(c{\tau }_{32}\overline{c}+{\tau }_{32}{\overline{c}}^2\right)\overline{c}‖{\dot{\overrightarrow{u}}}^{(3}‖+{\tau }_{32}{c}^2\overline{c}‖{\dot{\overrightarrow{u}}}^{(2)}‖}{c{\left(c^2{\tau }_{32}-c{\tau }_{32}\overline{c}\right)}^2}}\right]ds+{\displaystyle \underset{0}{\overset{t}{\int }}}{\displaystyle \frac{{е}_3^2\overline{c}‖{\ddot{\overrightarrow{u}}}^{(3}‖}{c^3}}ds\cong \\ \cong \beta {\displaystyle \underset{0}{\overset{t}{\int }}}\left[\left|e_3{e}_1\right|{\displaystyle \frac{{\rho }_3}{c^2{{\tau }_{31}}^3}}+\left|e_2{e}_3\right|{\displaystyle \frac{(1+\beta )}{{\left(1-\beta \right)}^3}}\left({\displaystyle \frac{1}{c{{\tau }_{32}}^2}}+{\displaystyle \frac{2‖{\dot{\overrightarrow{u}}}^{(2)}‖}{c^2{\tau }_{32}}}\right)+{\displaystyle \frac{{е}_3^2‖{\ddot{\overrightarrow{u}}}^{(3)}‖}{c^2}}\right]ds.\end{array} \end{gathered}$$

In the 3D-Kepler formulation the first particle P₁ is put at the origin O(0,0,0), that is, P₁$\left(x_1^{\left(1\right)}(t)=0,x_2^{\left(1\right)}(t)=0,x_3^{\left(1\right)}(t)=0\right),t\in [0,\infty )$. Then $E_{kin}^{(1)}(t)=0$.

We pass to the spherical coordinates, that is, the particle $P_n(n=2,3)$ is located at the point

$${x_1}^{(n)}={\rho }_n\cos {\phi }_n\cos {\lambda }_n;{x_2}^{(n)}={\rho }_n\sin {\phi }_n\cos {\lambda }_n;{x_3}^{(n)}={\rho }_n\sin {\lambda }_n$$

where ${\rho }_n(\theta )\ge 0;{\phi }_n(\theta )\ge 0;{\lambda }_n(\theta )\in \left[-{\displaystyle \frac{\pi }{2}}+\delta ,{\displaystyle \frac{\pi }{2}}-\delta \right],0<\delta <{\displaystyle \frac{\pi }{2}}$ and then the velocities are:

$$\begin{array}{l}{u_1}^{(n)}={\dot{\rho }}_n\cos {\phi }_n\cos {\lambda }_n-{\rho }_n{\dot{\phi }}_n\sin {\phi }_n\cos {\lambda }_n-{\rho }_n{\dot{\lambda }}_n\cos {\phi }_n\sin {\lambda }_n\\ {u_2}^{(n)}={\dot{\rho }}_n\sin {\phi }_n\cos {\lambda }_n+{\rho }_n{\dot{\phi }}_n\cos {\phi }_n\cos {\lambda }_n-{\rho }_n{\dot{\lambda }}_n\sin {\phi }_n\sin {\lambda }_n\\ {u_3}^{(n)}={\dot{\rho }}_n\sin {\lambda }_n+{\rho }_n{\dot{\lambda }}_n\cos {\lambda }_n\end{array}$$

where ${\dot{\rho }}^{(2)}=d{\rho }^{(2)}/dt$. For the accelerations we have (cf. [1,2]):

$$\begin{gathered} {{\dot{u}}_2}^{(n)}\approx {\ddot{\rho }}_n\sin {\phi }_n\cos {\lambda }_n+{\rho }_n{\ddot{\phi }}_n\cos {\phi }_n\cos {\lambda }_n-{\rho }_n{\ddot{\lambda }}_n\sin {\phi }_n\sin {\lambda }_n \\ {{\dot{u}}_3}^{(n)}\approx {\ddot{\rho }}_n\sin {\lambda }_n+{\rho }_n{\ddot{\lambda }}_n\cos {\lambda }_n. \end{gathered}$$

In the same manner we obtain the expressions for the second derivatives $(n=2,3)$:

$$\begin{array}{l}{\ddot{u}}_1^{(n)}\approx {\stackrel{⃛}{\rho }}_n\cos {\phi }_n\cos {\lambda }_n-{\stackrel{⃛}{\phi }}_n{\rho }_n\sin {\phi }_n\cos {\lambda }_n-{\stackrel{⃛}{\lambda }}_n{\rho }_n\cos {\phi }_n\sin {\lambda }_n\\ {\ddot{u}}_2^{(n)}\approx {\stackrel{⃛}{\rho }}_n\sin {\phi }_n\cos {\lambda }_n+{\stackrel{⃛}{\phi }}_n{\rho }_n\cos {\phi }_n\cos {\lambda }_n-{\stackrel{⃛}{\lambda }}_n{\rho }_n\sin {\phi }_n\sin {\lambda }_n\\ {\ddot{u}}_3^{(n)}\approx {\stackrel{⃛}{\rho }}_n\sin {\lambda }_n+{\stackrel{⃛}{\lambda }}_n{\rho }_n\cos {\lambda }_n.\end{array}$$

Then ${\rho }_2={\rho }_2(t)$; ${\varphi }_2={\varphi }_2(t)$;${\rho }_3={\rho }_3(t-{\tau }_{23})$; ${\varphi }_3={\varphi }_3(t-{\tau }_{23})$;

$$\begin{gathered} \begin{array}{l}{\tau }_{21}(t)={\displaystyle \frac{1}{c}}\sqrt{{\displaystyle \sum _{\alpha =1}^3{\left[x_{\alpha }^{(2)}(t)-x_{\alpha }^{(1)}(t-{\tau }_{21}(t)\right]}^2}}={\displaystyle \frac{{\rho }_2(t)}{c}};{\tau }_{31}(t)={\displaystyle \frac{1}{c}}\sqrt{{\displaystyle \sum _{\alpha =1}^3{\left[x_{\alpha }^{(3)}(t)-x_{\alpha }^{(1)}(t-{\tau }_{21}(t)\right]}^2}}={\displaystyle \frac{{\rho }_3(t)}{c}};\\ {\rho }_2(t)={\rho }_{20}+{\displaystyle \underset{0}{\overset{t}{\int }}{r}_2(s)ds\ge }{\rho }_{20}-R_2{\displaystyle \frac{e^{\mu T}-1}{\mu }}\approx {\rho }_{20};{\rho }_3(t)={\rho }_{30}+{\displaystyle \underset{0}{\overset{t}{\int }}{r}_3(s)ds\ge }{\rho }_{30}-R_3{\displaystyle \frac{e^{\mu T}-1}{\mu }}\approx {\rho }_{30};\end{array} \\ {\overrightarrow{\xi }}^{(23)}=\left(0,{\rho }_2\cos {\varphi }_2-{\rho }_3\cos {\varphi }_3,{\rho }_2\sin {\varphi }_2-{\rho }_3\sin {\varphi }_3\right); \\ {\tau }_{23}(t)={\displaystyle \frac{1}{c}}\sqrt{〈{\overrightarrow{\xi }}^{(23)},{\overrightarrow{\xi }}^{(23)}〉}\equiv {\displaystyle \frac{1}{c}}\sqrt{{\displaystyle \sum _{\alpha =1}^3{\left[x_{\alpha }^{(2)}(t)-x_{\alpha }^{(3)}(t-{\tau }_{23}(t))\right]}^2}}={\displaystyle \frac{1}{c}}\sqrt{{\rho }_2^2+{\rho }_3^2-2{\rho }_2{\rho }_3\cos ({\phi }_2-{\phi }_3)}\ge {\displaystyle \frac{\left|{\rho }_2(t)-{\rho }_3(t-{\tau }_{23})\right|}{c}}; \\ \begin{array}{l}\left|{\rho }_2(t)-{\rho }_3(t-{\tau }_{23})\right|=\left|{\rho }_{20}+{\displaystyle \underset{0}{\overset{t}{\int }}{r}_2(s)ds-{\rho }_{30}}-{\displaystyle \underset{0}{\overset{t-{\tau }_{23}}{\int }}{r}_3(s)ds}\right|=\left|{\rho }_{20}+{\displaystyle \underset{0}{\overset{t}{\int }}{r}_2(s)ds-{\rho }_{30}}-{\displaystyle \underset{0}{\overset{t}{\int }}{r}_3(s)ds-{\displaystyle \underset{t-{\tau }_{23}}{\overset{t}{\int }}{r}_3(s)ds}}\right|\ge \\ \ge \left|{\rho }_2(t)-{\rho }_3(t)\right|-R_3{\displaystyle \frac{e^{\mu t}-e^{\mu (t-{\tau }_{23})}}{\mu }}=\left|{\rho }_2(t)-{\rho }_3(t)\right|-R_3{e}^{\mu t}{\displaystyle \frac{1-e^{-\mu {\tau }_{23}}}{\mu }}\ge \left|{\rho }_2(t)-{\rho }_3(t)\right|-(R_3{e}^{\mu T})/\mu \approx \left|{\rho }_3(t)-{\rho }_2(t)\right|;\end{array} \\ \left|{\rho }_2(t)-{\rho }_3(t)\right|\ge \left|{\rho }_{20}-{\rho }_{30}\right|-(R_2+R_3)e^{\mu T}/\mu >0\approx \left|{\rho }_{20}-{\rho }_{30}\right|;{\displaystyle \frac{1}{{\tau }_{23}(t)}}\le {\displaystyle \frac{c}{\left|{\rho }_{20}-{\rho }_{30}\right|}}; \\ {\tau }_{32}(t)={\displaystyle \frac{1}{c}}\sqrt{〈{\overrightarrow{\xi }}^{(32)},{\overrightarrow{\xi }}^{(32)}〉}={\displaystyle \frac{1}{c}}\sqrt{{\rho }_3^2+{\rho }_2^2-2{\rho }_2{\rho }_3\cos ({\phi }_3-{\phi }_2)}\ge {\displaystyle \frac{\left|{\rho }_3(t)-{\rho }_2(t-{\tau }_{32})\right|}{c}}; \\ \begin{array}{l}\left|{\rho }_3(t)-{\rho }_2(t-{\tau }_{32})\right|=\left|{\rho }_{30}+{\displaystyle \underset{0}{\overset{t}{\int }}{r}_3(s)ds-{\rho }_{20}}-{\displaystyle \underset{0}{\overset{t-{\tau }_{32}}{\int }}{r}_2(s)ds}\right|=\left|{\rho }_{30}+{\displaystyle \underset{0}{\overset{t}{\int }}{r}_3(s)ds-{\rho }_{20}}-{\displaystyle \underset{0}{\overset{t}{\int }}{r}_2(s)ds-{\displaystyle \underset{t-{\tau }_{32}}{\overset{t}{\int }}{r}_2(s)ds}}\right|\ge \\ \ge \left|{\rho }_3(t)-{\rho }_2(t)\right|-R_2{\displaystyle \frac{e^{\mu t}-e^{\mu (t-{\tau }_{32})}}{\mu }}=\left|{\rho }_3(t)-{\rho }_2(t)\right|-R_2{e}^{\mu t}{\displaystyle \frac{1-e^{-\mu {\tau }_{32}}}{\mu }}\ge \left|{\rho }_3(t)-{\rho }_2(t)\right|-(R_2{e}^{\mu T})/\mu \approx \left|{\rho }_3(t)-{\rho }_2(t)\right|;\end{array} \\ \left|{\rho }_3(t)-{\rho }_2(t)\right|\ge \left|{\rho }_{30}-{\rho }_{20}\right|-(R_3+R_2)e^{\mu T}/\mu >0\approx \left|{\rho }_{30}-{\rho }_{20}\right|;{\displaystyle \frac{1}{{\tau }_{32}(t)}}\le {\displaystyle \frac{c}{\left|{\rho }_{30}-{\rho }_{20}\right|}}. \end{gathered}$$

Recall that $\sqrt{\left|〈{\overrightarrow{u}}^{(n)},{\overrightarrow{u}}^{(n)}〉\right|}\le \overline{c}$ implies $\sqrt{\left|〈{\overrightarrow{u}}^{(n)},{\overrightarrow{u}}^{(n)}〉\right|}=\sqrt{{r_n}^2+{{\rho }_n}^2{{\varphi }_n}^2+{{\rho }_n}^2{{\eta }_n}^2}\le e^{\mu T}\sqrt{{R_n}^2+{{\rho }_n}^2{{\Phi }_n}^2+{{\rho }_n}^2{Y_n}^2}\le \overline{c}$ and

$$\begin{gathered} \sqrt{\left|〈{\dot{\overrightarrow{u}}}^{(n)},{\dot{\overrightarrow{u}}}^{(n)}〉\right|}=\sqrt{\left|{{\ddot{\rho }}_n}^2+{{\rho }_n}^2{{\ddot{\phi }}_n}^2{\cos }^2{\lambda }_n+{{\rho }_n}^2{{\ddot{\lambda }}_n}^2\right|}\le e^{\mu T}{\omega }_n\sqrt{{R_n}^2+{{\rho }_n}^2{{\Phi }_n}^2+{{\rho }_n}^2{Y_n}^2}\le {\omega }_n\overline{c}; \\ \sqrt{\left|〈{\ddot{\overrightarrow{u}}}^{(n)},{\ddot{\overrightarrow{u}}}^{(n)}〉\right|}=\sqrt{\left|{{\stackrel{⃛}{\rho }}_n}^2+{{\rho }_n}^2{{\stackrel{⃛}{\phi }}_n}^2{\cos }^2{\lambda }_n+{{\rho }_n}^2{{\stackrel{⃛}{\lambda }}_n}^2\right|}\le e^{\mu T}{{\omega }_n}^2\sqrt{{R_n}^2+{{\rho }_n}^2{{\Phi }_n}^2+{{\rho }_n}^2{Y_n}^2}\le {{\omega }_n}^2\overline{c}. \end{gathered}$$

Since ${\displaystyle \frac{(1+\beta )}{{\left(1-\beta \right)}^3}}=1.03$, then

$$\begin{gathered} \left|E_2(t)\right|\le \beta {\displaystyle \underset{0}{\overset{t}{\int }}}\left[\left|e_2{e}_1\right|{\displaystyle \frac{{\rho }_2}{c^2{{\tau }_{21}}^3}}+\left|e_2{e}_3\right|{\displaystyle \frac{(1+\beta )}{{\left(1-\beta \right)}^3}}\left({\displaystyle \frac{1}{{{\tau }_{23}}^{2}c}}+{\displaystyle \frac{2‖{\dot{\overrightarrow{u}}}^{(3)}‖}{c^2{\tau }_{23}}}\right)+{\displaystyle \frac{{е}_2^2‖{\ddot{\overrightarrow{u}}}^{(2)}‖}{c^2}}\right]ds\le \\ \le {\displaystyle \frac{\left|e_2{e}_1\right|}{137}}{\displaystyle \underset{0}{\overset{t}{\int }}}\left[{\displaystyle \frac{c}{{{\rho }_{20}}^2}}+1.03\left({\displaystyle \frac{c^2}{{\left|{\rho }_2(t)-{\rho }_3(t)\right|}^2}}{\displaystyle \frac{1}{c}}+{\displaystyle \frac{2c‖{\dot{\overrightarrow{u}}}^{(3)}‖}{c^2\left|{\rho }_2(t)-{\rho }_3(t)\right|}}\right)+{\displaystyle \frac{‖{\ddot{\overrightarrow{u}}}^{(2)}‖}{c^2}}\right]ds\le \\ \le {\displaystyle \frac{\left|e_2{e}_1\right|}{137}}{\displaystyle \underset{0}{\overset{t}{\int }}}\left[{\displaystyle \frac{c}{{{\rho }_{20}}^2}}+1.03\left({\displaystyle \frac{c}{{\left|{\rho }_{20}-{\rho }_{30}\right|}^2}}+{\displaystyle \frac{2{\omega }_3\overline{c}}{c\left|{\rho }_{20}-{\rho }_{30}\right|}}\right)+{\displaystyle \frac{{{\omega }_2}^2\overline{c}}{c^2}}\right]ds\le \\ \begin{array}{l}\le {\displaystyle \frac{\left|e_2{e}_1\right|}{137}}\left[{\displaystyle \frac{c}{{{\rho }_{20}}^2}}+1.03\left({\displaystyle \frac{c}{{\left|{\rho }_{20}-{\rho }_{30}\right|}^2}}+{\displaystyle \frac{2{\omega }_3\overline{c}}{c\left|{\rho }_{20}-{\rho }_{30}\right|}}\right)+{\displaystyle \frac{{{\omega }_2}^2\overline{c}}{c^2}}\right]{\displaystyle \underset{0}{\overset{t}{\int }}{e}^{\mu s}}ds\le \\ \le {\displaystyle \frac{\left|e_2{e}_1\right|}{137}}\left[{\displaystyle \frac{c}{{{\rho }_{20}}^2}}+1.03\left({\displaystyle \frac{c}{{\left|{\rho }_{20}-{\rho }_{30}\right|}^2}}+{\displaystyle \frac{2{\omega }_3\overline{c}}{c\left|{\rho }_{20}-{\rho }_{30}\right|}}\right)+{\displaystyle \frac{{{\omega }_2}^2\overline{c}}{c^2}}\right]{\displaystyle \frac{e^{\mu T}-1}{\mu }}\end{array} \end{gathered}$$

and

$$\begin{array}{l}\left|E_3(t)\right|\le \beta {\displaystyle \underset{0}{\overset{t}{\int }}}\left[\left|e_3{e}_1\right|{\displaystyle \frac{{\rho }_3}{c^2{{\tau }_{31}}^3}}+\left|e_2{e}_3\right|{\displaystyle \frac{(1+\beta )}{{\left(1-\beta \right)}^3}}\left({\displaystyle \frac{1}{{{\tau }_{32}}^{2}c}}+{\displaystyle \frac{2‖{\dot{\overrightarrow{u}}}^{(2)}‖}{c^2{\tau }_{32}}}\right)+{\displaystyle \frac{{е}_3^2‖{\ddot{\overrightarrow{u}}}^{(3)}‖}{c^2}}\right]ds\le \\ \le {\displaystyle \frac{\left|e_2{e}_1\right|}{137}}\left[{\displaystyle \frac{c}{{{\rho }_{30}}^2}}+1.03\left({\displaystyle \frac{c}{{\left|{\rho }_{30}-{\rho }_{20}\right|}^2}}+{\displaystyle \frac{2{\omega }_2\overline{c}}{c\left|{\rho }_{30}-{\rho }_{20}\right|}}\right)+{\displaystyle \frac{{{\omega }_3}^2\overline{c}}{c^2}}\right]{\displaystyle \underset{0}{\overset{t}{\int }}{e}^{\mu s}}ds\le \\ \le {\displaystyle \frac{\left|e_2{e}_1\right|}{137}}\left[{\displaystyle \frac{c}{{{\rho }_{30}}^2}}+1.03\left({\displaystyle \frac{c}{{\left|{\rho }_{30}-{\rho }_{20}\right|}^2}}+{\displaystyle \frac{2{\omega }_2\overline{c}}{c\left|{\rho }_{30}-{\rho }_{20}\right|}}\right)+{\displaystyle \frac{{{\omega }_3}^2\overline{c}}{c^2}}\right]{\displaystyle \frac{e^{\mu T}-1}{\mu }}.\end{array}$$

6. Conclusion - Numerical Verification of the Above Formulas

For the He-atom both electrons are on the first stationary state, that is, one can assume

${\omega }_2={\omega }_3=\omega =4.1\times {10}^{16},T=2\pi /\omega =1.53\times {10}^{-16}$. Let us choose $\mu =4.12\times {10}^{16}>\omega =4.1\times {10}^{16}$ such that

$\mu T=4.12\times {10}^{16}\times 1.53\times {10}^{-16}\approx 6,3\Rightarrow e^{\mu T}=e^{6,3}\approx 545$ and ${\Phi }_2={\Phi }_3=\Phi \le \omega .e^{-\mu T}=4.1\times {10}^{16}\times 0.0018=7.38\times {10}^{13}$ and $R_2=R_3\le \overline{c}.e^{-\mu T}={\displaystyle \frac{1}{137}}\times 3\times {10}^8\times 1.8\times {10}^{-3}\approx 0.039\times {10}^5=3.9\times {10}^3$. It is known that $e_2=e_3\approx 1.6\times {10}^{-19}C$, $\beta =1/137$, $c\approx 3\times {10}^{8}m/\sec$, ${\displaystyle \frac{(1+\beta )}{{\left(1-\beta \right)}^3}}={\displaystyle \frac{138\times {137}^2}{{136}^3}}=1.03$. Consequently for $n=2$:

$$\begin{array}{l}\left|E_2(t)\right|\le {\displaystyle \frac{\left|e_2{e}_1\right|}{137}}\left[{\displaystyle \frac{c}{{{\rho }_{20}}^2}}+1.03\left({\displaystyle \frac{c}{{\left|{\rho }_{20}-{\rho }_{30}\right|}^2}}+{\displaystyle \frac{2\omega \overline{c}}{c\left|{\rho }_{20}-{\rho }_{30}\right|}}\right)+{\displaystyle \frac{{\omega }^2\overline{c}}{c^2}}\right]{\displaystyle \frac{e^{\mu T}-1}{\mu }}=\\ ={\displaystyle \frac{\left|e_2{e}_1\right|}{137}}\left({\displaystyle \frac{c}{{{\rho }_{20}}^2}}+{\displaystyle \frac{1.03\times c}{{\left|{\rho }_{20}-{\rho }_{30}\right|}^2}}+\left({\displaystyle \frac{2.06}{\left|{\rho }_{20}-{\rho }_{30}\right|}}+{\displaystyle \frac{\omega }{c}}\right)\omega \beta \right){\displaystyle \frac{e^{\mu T}-1}{\mu }}=\\ ={10}^{-46}\left({\displaystyle \frac{7.41}{{{\rho }_{20}}^2}}+{\displaystyle \frac{7.657}{{\left|{\rho }_{20}-{\rho }_{30}\right|}^2}}+{\displaystyle \frac{1.54\times {10}^7}{\left|{\rho }_{20}-{\rho }_{30}\right|}}\right)+{10}^{-46}\times {10}^{15}\approx {10}^{-46}\left({\displaystyle \frac{7.41}{{{\rho }_{20}}^2}}+{\displaystyle \frac{7.66}{{\left|{\rho }_{20}-{\rho }_{30}\right|}^2}}+{\displaystyle \frac{1.54\times {10}^7}{\left|{\rho }_{20}-{\rho }_{30}\right|}}\right).\end{array}$$

Since ${\rho }_{20}=5.3\times {10}^{-11}$ then

$$\begin{array}{l}{10}^{-46}\left({\displaystyle \frac{7.41}{{{\rho }_{20}}^2}}+{\displaystyle \frac{7.66}{{\left|{\rho }_{20}-{\rho }_{30}\right|}^2}}+{\displaystyle \frac{1.54\times {10}^7}{\left|{\rho }_{20}-{\rho }_{30}\right|}}\right)={10}^{-46}\left({\displaystyle \frac{7.41\times {10}^{22}}{{5.3}^2}}+{\displaystyle \frac{7.66}{{\left|{\rho }_{20}-{\rho }_{30}\right|}^2}}+{\displaystyle \frac{1.54\times {10}^7}{\left|{\rho }_{20}-{\rho }_{30}\right|}}\right)\approx \\ \approx {10}^{-46}\left({\displaystyle \frac{7.66}{{\left|{\rho }_{20}-{\rho }_{30}\right|}^2}}+{\displaystyle \frac{1.54\times {10}^7}{\left|{\rho }_{20}-{\rho }_{30}\right|}}\right)={\overline{E}}_2.\end{array}$$

Let us compare the obtained energy with the energy of the first stationary state from quantum mechanics (cf. [4]). Since $W_p=-{\displaystyle \frac{1}{{{\epsilon }_0}^2}}.{\displaystyle \frac{e_2^4{m}_2}{8{\hslash }^2}}{\displaystyle \frac{1}{p^2}}(p=1,2,\dots )$ for $p=1$ one obtains

$$W_1=-{\displaystyle \frac{1}{{{\epsilon }_0}^2}}.{\displaystyle \frac{e_2^4{m}_2}{8{\hslash }^2}}=-{\displaystyle \frac{1}{{8.86}^2\times {10}^{-24}}}.{\displaystyle \frac{{(1.6\times {10}^{-19})}^4\times 9.1\times {10}^{-31}}{{\mathrm{8.6.62}}^2\times {10}^{-68}}}=-2.2\times {10}^{-18}.$$

It is easy to see that one can choose $\left|{\rho }_{02}-{\rho }_{03}\right|$ in such a way that ${\overline{E}}_2=2.2\times {10}^{-22}$. Indeed, for $\left|{\rho }_{02}-{\rho }_{03}\right|=\alpha \times {10}^{-12}$ we obtain

$$\begin{array}{l}{\overline{E}}_2={10}^{-46}\left({\displaystyle \frac{7.66}{{\left|{\rho }_{20}-{\rho }_{30}\right|}^2}}+{\displaystyle \frac{1.54\times {10}^7}{\left|{\rho }_{20}-{\rho }_{30}\right|}}\right)={10}^{-46}\left({\displaystyle \frac{7.66}{{\alpha }^2}}\times {10}^{24}+{\displaystyle \frac{1.54\times {10}^{19}}{\alpha }}\right)=\\ ={10}^{-46}{\displaystyle \frac{7.66}{{\alpha }^2}}\times {10}^{24}+{10}^{-46}\left({\displaystyle \frac{1.54\times {10}^{19}}{\alpha }}\right)\approx {10}^{-22}{\displaystyle \frac{7.66}{{\alpha }^2}}=2.2\times {10}^{-22}\end{array}$$

for $\alpha =1.8659$.

We note that similar results can be obtained if ${\omega }_2\ne {\omega }_3$ but ${\omega }_2\approx {\omega }_3$.