This Appendix provides an example of a single space and the global velocity vector with respect to the ground. In further works we will try to give a definition of mass based on this space, but for now, in this article, we "potreniruemsya" on known data about it and show you the two easiest withdrawal option most famous formula of the 20th century through a global vector.

As we already know, a vector is obtained by converting the scalar Lorentz factor: \[\mathbf{V} = \frac{c}{\gamma} \left\{\pm 1,\, \pm \beta,\, \pm \beta^2,\, \ldots,\, \pm \beta^n \right\} \qquad (1.1)\] where: \(\gamma = 1 / \sqrt{1 - \beta^2}\) and \(\beta = v/c\). In this case, \(c\) is the speed of light, and \(v\) is the velocity in the real space. For more visual evidence will present the global velocity vector as \[\mathbf{V} = \frac{1}{\gamma} \mathbf{v}, \quad \mathbf{v} = c \left\{\pm 1,\, \pm \beta,\, \pm \beta^2,\, \ldots,\, \pm \beta^n \right\} \qquad (1.2)\] Now remember the formula for finding momentum from school physics course, but instead of normal speed it will take the global vector. Then we automatically obtain a global vector of momentum: \[\mathbf{P} = m \mathbf{V} \qquad (1.3)\] As we look forward at any speed, you should be aware that weight can vary depending on its size [1]: \[m = m_0 \gamma \qquad (1.4)\] Then the global vector momentum will be such: \[\mathbf{P} = m_0 c \left\{\pm 1,\, \pm \beta,\, \pm \beta^2,\, \ldots,\, \pm \beta^n \right\} = m_0 \mathbf{v} \qquad (1.5)\] Let us write the value of a global vector momentum in two different ways. Such formulas will be useful in future calculations: \[\mathbf{P} = m_0 \mathbf{v} = m_0 c \sum \limits_{n=0}^{\infty} \pm \mathbf{j_n} \beta^n \qquad (1.6)\] Left to remember, how are the velocity, momentum and energy, and take the integral, given the formula (1.6): \[E = \int \mathbf{V} \Bbb{d} \mathbf{P} = \int \frac{1}{\gamma} \mathbf{v}\, \Bbb{d} (m_0 \mathbf{v}) \qquad (1.7)\] Next, we continue to transform the integrand \[E = \frac{m_0}{2} \int \frac{1}{\gamma} \Bbb{d} (\mathbf{v}\cdot \mathbf{v}) = \frac{m_0}{2} \int \frac{1}{\gamma} \Bbb{d} (c \gamma)^2 \qquad (1.8)\] and finally we get the final result: \[E = m_0 c^2 \int \Bbb{d} \gamma = m_0 c^2 \gamma = m c^2 \qquad (1.9)\] In fact, the first formula was the preparation, and the proof occupies just three expressions (1.7-1.9).

The same result can be obtained quite simply, if a scalar multiply of the global vector of momentum (1.6) the global velocity vector (1.1): \[E = \mathbf{P}\cdot \mathbf{V} = m_0\, \mathbf{v}\cdot \mathbf{V} = m_0 c^2 \gamma = m c^2 \qquad (1.10)\] As we can see, when working with global vectors and isolated space, integration and differentiation are often not required quite simple algebraic operations (example).