## 6 Propagation Time for a Signal, Emitted and Intercepted by Satellites on a Space-Oriented Orbit
### Second Formulation of the Theorem for Four-Dimensional Uniformization
From the comparison of the corresponding expressions under the cubic and square roots of (235) and (244), it followsthat the equality is fulfilled when the following relations between the Weierstrass invariants \(g_{2},g_{3}\) and \(\widehat{g}_{2},\widehat{g}_{3}\) are fulfilled:  
\[\overline{s}_{3}=\frac{1}{6}a_{0}a_{2}a_{4}+\frac{1}{48}a_{1}a_{2}a_{3}-\frac {a_{2}^{3}}{6^{3}}-\frac{1}{16}(a_{0}a_{3}+a_{1}^{2}a_{4}). \tag{224}\]1  
The last relation can be rewritten, expressing the dependance of \(\widehat{g}_{2},\widehat{g}_{3}\) on \(g_{2},g_{3}\):  
\[\overline{s}_{3}=\frac{1}{6}a_{0}a_{2}a_{4}+\frac{1}{48}a_{1}a_{2}a_{3}-\frac {a_{2}^{3}}{6^{3}}-\frac{1}{16}(a_{0}a_{3}+a_{1}^{2}a_{4}). \tag{224}\]2  
It can be noted that these relations are different from the ones obtained in a previous paper [33]:  
\[\overline{s}_{3}=\frac{1}{6}a_{0}a_{2}a_{4}+\frac{1}{48}a_{1}a_{2}a_{3}-\frac {a_{2}^{3}}{6^{3}}-\frac{1}{16}(a_{0}a_{3}+a_{1}^{2}a_{4}). \tag{224}\]3  
In fact, the correct relations are (245) and (246) (and not those in (247), which are wrong), because they are obtained from the relation for the roots (242). This relation takes into account the multiplication both of the roots and the variables in (242) with the conformal factor \(a^{\frac{3}{9}}\) and not with \((\sigma_{0},\sigma_{1},\sigma_{2})=\sqrt[3]{a}(\sigma_{0},\sigma_{1},\sigma_{ 2})\), as this is applied in the paper [33]. Finding the Relation Between the Other Two Roots \((\beta_{2},\gamma_{2})\) and \((\beta_{1},\gamma_{1})\) for the Case When \(\alpha_{2}=\alpha_{1}\)--The Case of Positive Discriminant  
The other two roots \((\beta_{2},\gamma_{2})\) of the second Equation (241) \(4\sigma^{3}-\widehat{g}_{2}\sigma-\widehat{g}_{3}=0\) can be written similarly to the expressions (239) and (240) for \(\beta_{1}\) and \(\gamma_{1}\), but with indices \"2\":  
\[\overline{s}_{3}=\frac{1}{6}a_{0}a_{2}a_{4}+\frac{1}{48}a_{1}a_{2}a_{3}-\frac {a_{2}^{3}}{6^{3}}-\frac{1}{16}(a_{0}a_{3}+a_{1}^{2}a_{4}). \tag{224}\]4  
\[\overline{s}_{3}=\frac{1}{6}a_{0}a_{2}a_{4}+\frac{1}{48}a_{1}a_{2}a_{3}-\frac {a_{2}^{3}}{6^{3}}-\frac{1}{16}(a_{0}a_{3}+a_{1}^{2}a_{4}). \tag{224}\]5  
Taking into account that \(\alpha_{1}=\alpha_{2}\), and also the equality (244), it can be obtained that  
\[\overline{s}_{3}=\frac{1}{6}a_{0}a_{2}a_{4}+\frac{1}{48}a_{1}a_{2}a_{3}-\frac {a_{2}^{3}}{6^{3}}-\frac{1}{16}(a_{0}a_{3}+a_{1}^{2}a_{4}). \tag{224}\]6  
If, substituted in the formulaes (248) and (249) for \(\beta_{2}\) and \(\gamma_{2}\), the following relations are found between the roots \((\beta_{2},\gamma_{2})\) and \((\beta_{1},\gamma_{1})\):  
\[\overline{s}_{3}=\frac{1}{6}a_{0}a_{2}a_{4}+\frac{1}{48}a_{1}a_{2}a_{3}-\frac {a_{2}^{3}}{6^{3}}-\frac{1}{16}(a_{0}a_{3}+a_{1}^{2}a_{4}). \tag{224}\]7  
where \((\beta_{1},\gamma_{1})\) are the complex roots, given by the formulaes (239) and (240). Expressing the Ratio of the Weierstrass Invariants \(g_{2}\) and \(g_{3}\) as a Rational Function of Higher-Order Polynomials, Depending on the Modulus Parameter \(q\) of the Elliptic Integral in the Legendre Form--The Case of Positive Discriminant  
Finding the relation between the Weierstrass invariants \(g_{2},g_{3}\) and \(\widehat{g}_{2},\widehat{g}_{3}\) in (245) and (246) is rather insufficient and does not give any valuable information because of the following two reasons:  
A. The variable \(a\) is not just a parameter, but represents a complicated function, depending on the ratio \(\frac{g_{3}}{g_{2}}\) and on the rational function \(K(q)\), and is determined in (146) as \(a=\frac{9\sigma_{1}}{\widehat{g}_{2}}K(q)\), where (148) \(K(q)\equiv\frac{q^{4}-q^{2}+1}{2q^{4}-5q^{2}+2}\). B.