First, let us recall some of the conclusions that we have already received in the first Chapter. For example, as we Vassili from the previous work, the length of the vector \(\mathbf{R}\), obtained by the layout of the Lorentz factor will always be equal to unity: \[\mathbf{R} \cdot \mathbf{R} = 1, \quad |\mathbf{R}| = 1 \qquad (1.1)\] And the vector itself, in General terms, as follows: \[\mathbf{R} = \frac{1}{\gamma} \left\{\pm 1,\, \pm \beta,\, \pm \beta^2,\, \ldots,\, \pm \beta^n \right\} \qquad (1.2)\] where: \(\beta = v/c\), and \(\gamma = 1/\sqrt{1 - \beta^2}\). If you multiply this vector by the speed of light and to choose the positive direction (plus the coefficients), we get a reflection of our world in the first two coordinates: \[\mathbf{V} = c\mathbf{R} = \frac{1}{\gamma} (\mathbf{i} c + \mathbf{j} v + \ldots) \qquad (1.3)\] where: \(\mathbf{i}, \mathbf{j}\) — the unit vectors of the orthonormal basis, \(\mathbf{V}\) — vector of total or global speed, \(v\) — familiar to us speed in our world. Hence we automatically get the global energy conservation law: \[\mathbf{V} \cdot \mathbf{V} = c^2 \qquad (1.4)\]

If we assume that our speed changes over time according to a certain law: \(v = v(t)\), then the derivative of the vector \(\mathbf{R}\) for \(t\) will consist of a total acceleration vector.

Fig.1. The total velocity vector is always perpendicular to the total vector acceleration.

We need to prove that the total acceleration vector \(\mathbf{R}'_t\) is always perpendicular to the total velocity vector, i.e.: \[\mathbf{R} \cdot \mathbf{R}'_t = 0 \qquad (1.5)\] you Can lead a very complex proof, but we found a more simple solution. We prove this property: \[\mathbf{R} \cdot \mathbf{R} = 1\, \Rightarrow\, [\mathbf{R} \cdot \mathbf{R}]'_t = 0\] \[\Downarrow\] \[[\mathbf{R} \cdot \mathbf{R}]'_t = \mathbf{R} \cdot 2 \mathbf{R}'_t = 0\] Let's check this property on a simple example. Assume that the speed changes according to a linear law: \(v(t) = at\) where \(a\) is the acceleration. Then: \(\beta(t) = at/c\). Direction to all coefficients, for simplicity, we choose everywhere positive. A unit vector will be \[\mathbf{R}(t) = \sqrt{1 - \beta(t)^2} \left\{1,\, \beta(t)\, \beta(t)^2,\, \ldots,\, \beta(t)^n \right\} \qquad (1.6)\] and the vector of its time derivative is: \[\mathbf{R}'_t = {- k^2 t \over \sqrt{1 - \beta(t)^2}} \left\{1,\, \beta(t)\, \beta(t)^2,\, \ldots,\, \beta(t)^n \right\} + \sqrt{1 - \beta(t)^2} \left\{0,\, k\, 2k^2 t,\, \ldots,\, n k^n t^{n-1} \right\} \qquad (1.7)\] where we have introduced the following coefficient: \(k = a/c\). For further calculations we take into account that the following series is convergent: \[1 + 2x + 3x^2 + \ldots + nx^{n-1} = {1 \over (1-x)^2} \qquad (1.8)\] Multiply the two resulting vectors according to the rules of scalar multiplication, and simplifying the expression by using (1.8), we get: \[\mathbf{R}(t) \cdot \mathbf{R}'_t = {- k^2 t \over 1 - \beta(t)^2} + {k^2 t \over 1 - \beta(t)^2} = 0 \qquad (1.9)\]

Please note that receipt of the above property works not only in Lorentzian space (obtained from the Lorentz factor), but in any other.