In this paper we will introduce our readers to unusual properties of the global velocity vector (GVV). But first, let us recall some conclusions obtained previously. For example, the length of the vector \(\mathbf{R}\), obtained by decomposing the scalar Lorentz factor into the global vector, will always be equal to one: \[\mathbf{R} \cdot \mathbf{R} = 1, \quad |\mathbf{R}| = 1 \tag{1.1}\] And the normalized GVV itself, in general, looks like this: \[\mathbf{R} = \frac{1}{\gamma} \left\{\pm 1,\, \pm \beta,\, \pm \beta^2,\, \ldots,\, \pm \beta^n \right\} \tag{1.2}\] where: \(\beta = v/c\), and \(\gamma = 1/\sqrt{1 - \beta^2}\) -- Lorentz factor, responsible for the law of conservation of energy of a mathematical point (without mass).

If we multiply this vector by the speed of light and choose a positive direction (plus before the coefficients), we will get some reflection of our world in the first two coordinates: \[\mathbf{V} = c\mathbf{R} = \frac{1}{\gamma} (\mathbf{i} c + \mathbf{j} v + \ldots) \tag{1.3}\] where: \(\mathbf{i}, \mathbf{j}\) -- unit vectors of the orthonormal basis, \(\mathbf{V}\) -- vector of the total or global velocity, \(v\) -- the velocity we are accustomed to in our world. From here we automatically obtain the global law of conservation of energy: \[\mathbf{V} \cdot \mathbf{V} = c^2 \tag{1.4}\]

This property is well known in mechanics. It states that when a point moves, the velocity vector is perpendicular to the acceleration vector [1], albeit under some limiting conditions. Here we will consider the general case with a global vector and show that the global velocity vector is always perpendicular to the global acceleration vector.

If we assume that our velocity changes over time according to a certain law: \(v = v(t)\), then the derivative of the vector \(\mathbf{R}\) with respect to \(t\) will be the global acceleration vector \(\mathbf{R}'_t\), which we will abbreviate as follows: GVA. We need to prove that GVA will always be perpendicular to GVV, that is: \[\mathbf{R} \cdot \mathbf{R}'_t = 0 \tag{1.5}\] A very complex proof can be given, but we have found a simpler solution. Let us prove this property as follows: \[\mathbf{R} \cdot \mathbf{R} = 1\, \Rightarrow\, [\mathbf{R} \cdot \mathbf{R}]'_t = 0\] \[\Downarrow\] \[[\mathbf{R} \cdot \mathbf{R}]'_t = \mathbf{R} \cdot 2 \mathbf{R}'_t = 0\] Let us check this property using a simple example. Let us assume that the velocity changes linearly: \(v(t) = at\), where \(a\) is the acceleration. Then: \(\beta(t) = at/c\). For simplicity, we will choose the direction in front of all coefficients to be positive everywhere. The unit vector will be: \[\mathbf{R}(t) = \sqrt{1 - \beta(t)^2} \left\{1,\, \beta(t),\, \beta(t)^2,\, \ldots,\, \beta(t)^n \right\} \tag{1.6}\] and its time derivative vector will be: \[\mathbf{R}'_t = {- k^2 t \over \sqrt{1 - \beta(t)^2}} \left\{1,\, \beta(t),\, \beta(t)^2,\, \ldots,\, \beta(t)^n \right\} + \sqrt{1 - \beta(t)^2} \left\{0,\, k,\, 2k^2 t,\, \ldots,\, n k^n t^{n-1} \right\} \tag{1.7}\] where we introduced the following coefficient: \(k = a/c\). For further calculations, we take into account that the following series is convergent: \[1 + 2x + 3x^2 + \ldots + nx^{n-1} = {1 \over (1-x)^2} \tag{1.8}\] We multiply the two resulting vectors according to the rules of scalar multiplication, and simplifying the expression using (1.8), we obtain: \[\mathbf{R}(t) \cdot \mathbf{R}'_t = {- k^2 t \over 1 - \beta(t)^2} + {k^2 t \over 1 - \beta(t)^2} = 0 \tag{1.9}\]

We draw your attention to the fact that the property obtained above works not only in the Lorentz space (obtained from Lorentz factor), but also in any other Cartesian.

The above property works not only for the general form of GVV (1.2), but also for its projections onto the world with the number of dimensions \(2^n,\, n=0,1,2,3,\ldots\) Let's see how it works for 2D, that is, when the worldconsists of one spatial coordinate and one time coordinate. In this case, we can obtain the projection of GVV onto such a world: \[ \mathbf{R_{\perp}} = \mathbf{j}_0 \frac{1}{\gamma(t)} + \mathbf{j}_1\, \beta(t) \tag{1.10}\] Taking the derivative with respect to time, we obtain the projection of GVA: \[ \mathbf{R_{\perp}'} = - \mathbf{j}_0\, \beta(t) \beta_t' \gamma(t) + \mathbf{j}_1\, \beta_t' \tag{1.11}\] Scalarly multiplying the projections of GVV and GVA, we obtain: \[ \mathbf{R_{\perp}} \cdot \mathbf{R_{\perp}'} = \beta(t) \beta_t' + \beta(t) \beta_t' = 0 \tag{1.12}\] This is a proof that property No. 1 also works in the 2D world, and here the velocity vector is also perpendicular to the acceleration vector. Note the simplicity of the proof of this property.

Now let's see how it works for 4D, that is, when the world consists of three spatial coordinates and one time coordinate, that is, this is our world. Using the above algorithm, we first obtain the GVV projection: \[ \mathbf{R_{\perp}} = \mathbf{j}_0\, \frac{1}{\gamma(t)} + \mathbf{j}_1\, \frac{1}{\gamma(t)} \beta(t) + \mathbf{j}_2\, \beta(t)^2 \tag{1.13}\] Here it should be explained that in fact the unit vector \(\mathbf{j}_2\) splits into two equally probable unit vectors, but this property will work in any case. Therefore, we can safely work with the vector from (1.13). Taking its derivative, we obtain the projection of GVA: \[ \mathbf{R_{\perp}'} = - \mathbf{j}_0\, \beta(t) \beta_t' \gamma(t) + \mathbf{j}_1\, \beta_t' \left[\frac{1}{\gamma(t)} - \beta(t)^2 \gamma(t) \right] + \mathbf{j}_2\, 2 \beta(t) \beta_t' \tag{1.14}\] Scalarly multiplying the projections of GVV and GVA, reducing similar terms, we also obtain zero here: \[ \mathbf{R_{\perp}} \cdot \mathbf{R_{\perp}'} = 0 \tag{1.15}\] In our four-dimensional world, property No. 1 applies, and the velocity vector is still perpendicular to the acceleration vector.