Electron

An electron (from Ancient Greek ήλεκτρον, ēlektron, "amber") is a subatomic particle with a charge of -e (negative elementary charge), and a spin of 1/2. It has about 1/1836 of the mass of a proton.

An electron appears in different orbits around atomic nuclei depending on energy levels. It is further away from the nucleus when it has more energy, but that means it must be orbiting faster. That's why pilitron physics predicts that its rotational speed must be constant. It is noted by X (capital X). Its value is:


 * $$X \approx 6.568 \times 10^{15} \mbox{rev}/\mbox{s}$$

Based on X, we can also define the alpha constant.

Dynamics
In pilitron dynamics, the electron is contained in the electron field, whose momentum can be summarised as:


 * $$M\left(\tilde{x}\right) = \frac{\mu_e}{F\left(\tilde{x}\right) \sqrt{1-\frac{F\left(\tilde{x}\right)^2}{c^2}}} - \frac{\mu_e}{F(\tilde{x})}$$

Where $$\mu_e = m_ec^2$$.

Energy
Remember that the momentum function is defined as:


 * $$M(\tilde{x}) = \frac{E(\tilde{x})-\mu_e}{F(\tilde{x})}$$

From this we can derive that the energy function is:


 * $$E(\tilde{x}) = M(\tilde{x})F(\tilde{x}) + \mu_e$$

Substituting in the value of the momentum function gives us:


 * $$E(\tilde{x}) = \frac{\mu_e}{\sqrt{1-\frac{F\left(\tilde{x}\right)^2}{c^2}}}$$

Lagrangian
The Lagrangian of an electron, in a field containing one other electrically charged particle, is given by:


 * $$\mathcal{L} = pv\tau^2 - \frac{q_1 q_2 \tau^2 \cos \theta}{4\pi\epsilon_0\mu_0 d}$$

Where $$q_1$$ and $$q_2$$ are the charges of the 2 particles, $$d$$ is the distance between them, $$\tau$$ is the tension factor, and $$\theta$$ is the angle between the direction towards the other particle and the direction of the electron's motion. Therefore, if the direction of the other particle from the electron is $$\mathbf{R}$$ and the direction of the electron's motion is $$\mathbf{V}$$, then we have:


 * $$\cos \theta = \mathbf{R} \cdot \mathbf{V}$$

If we substitute in the definitions of the variables, we get:


 * $$\mathcal{L} = \frac{m_e v^2 \tau^2 q_1 q_2 \cos \theta}{4\pi \epsilon_0 \mu_0 d \sqrt{1-\frac{v^2}{c^2}}}$$