If the First property works in any space, the action of the second and third — will pertains only to Lorenzo a single space. Recall that we call such a space described by a General vector \(\mathbf{R}\), which is obtained by transformations of the Lorentz factor: \[\mathbf{R} = \frac{1}{\gamma} \left\{\pm 1,\, \pm \beta,\, \pm \beta^2,\, \ldots,\, \pm \beta^n \right\}, \quad \gamma = 1/\sqrt{1 - \beta^2} \qquad (2.1)\]

The first property implies that the total velocity vector \(\mathbf{R}(t)\) is always perpendicular to the total acceleration vector \(\mathbf{R}'(t)\) where: \(\mathbf{R}(t)\cdot \mathbf{R}'(t) = 0\). Hereinafter, the derivative is taken at \(t\). Let us now prove that the length of this new vector of acceleration will be equal to: \[|\mathbf{R}'(t)| = {\beta'(t) \over 1 - \beta(t)^2}, \quad \beta(t) = {v(t) \over c} \qquad (2.2)\] To do this, first we introduce some notation: \[\mathbf{r_1} = \left\{1,\, \beta(t),\, \beta(t)^2,\, \ldots,\, \beta(t)^n \right\} \] \[\mathbf{r_2} = \left\{0,\, 1,\, 2\beta(t)\, \ldots,\, n \beta(t)^{n-1} \right\} \] Then the derivative of the total velocity vector will be like this: \[\mathbf{R}'(t) = - \gamma(t) \beta(t) \beta'(t) \mathbf{r_1} + {\beta'(t) \over \gamma(t)} \mathbf{r_2} \qquad (2.3)\] Now multiply the resulting vector on itself and get the square of its module: \[\mathbf{R}'(t)\cdot \mathbf{R}'(t) = |\mathbf{R}'(t)|^2 \qquad (2.4)\] Multiply the vector (2.3) according to the rules for scalar multiplication: \[|\mathbf{R}'(t)|^2 = \beta'(t)^2 \left( \gamma(t)^2 \beta(t)^2 \mathbf{r_1}^2 - 2 \beta(t) \mathbf{r_1} \mathbf{r_2} + {1 \over \gamma(t)^2} \mathbf{r_2}^2 \right) \qquad (2.5)\] For the final result we need to obtain the scalar sum of products of vectors: \[\mathbf{r_1}^2 = \gamma(t)^2 \] \[\mathbf{r_1} \mathbf{r_2} = \beta(t) \left(1 + 2\beta(t)^2 + \ldots + n\beta(t)^{2(n-1)} \right) = \gamma(t)^4 \beta(t) \] \[\mathbf{r_2}^2 = \left(1 + 4\beta(t)^2 + \ldots + n^2\beta(t)^{2(n-1)} \right) = \gamma(t)^6 (1 + \beta(t)^2) \] the Last product of vectors forms the sum of the following power series: \[1 + 4x^2 + 9x^4 + \ldots + n^2 x^{2(n-1)} = {1 + x^2 \over (1-x^2)^3} \qquad (2.6)\] Substituting these in (2.5), we obtain: \[|\mathbf{R}'(t)|^2 = \beta'(t)^2 \gamma(t)^4 \left[ \beta(t)^2 - 2\beta(t)^2 + (1 + \beta(t)^2) \right] \qquad (2.7)\] Abbreviated terms in brackets and get the final result: \[|\mathbf{R}'(t)|^2 = \beta'(t)^2 \gamma(t)^4 = {\beta'(t)^2 \over \left(1 - \beta(t)^2 \right)^2} \qquad (2.8)\] \[|\mathbf{R}'(t)| = {\beta'(t) \over 1 - \beta(t)^2} \qquad (2.9)\] To a full global vector of acceleration, as we already know, sufficiently normalized is to multiply by \(c\): \[|\mathbf{V}'(t)| = {c\beta'(t) \over 1 - \beta(t)^2} \qquad (2.10)\] This property is proved. By the way, the last formula can be written like this: \[|\mathbf{V}'(t)| = a\,\gamma^2 \qquad (2.11)\] where: \(a\) is the acceleration. Equations (2.9-2.11) are of great importance both in themselves, and to output the following properties.