Science Forums 2

KickAssClown · 06-14-2006

Properties of the Fundemental particles, Charge Quanta:
$E = 1, spin = 1, parity = +$
$B = 1, spin = 1, parity = -$
$\gamma = (1, 1), spin (1, 1), parity (+,-)$

How does one describe a relative co-ordinace system? Such that body is always (x, y, z) = (1, 1, 1) to itself.

$\begin{bmatrix}1 & x, y & x, z & x, t \\ y, x & 1 & y, z & y, t \\ z, x & z, y & 1 & z, t \\ t, x & t, y & t, z & 1\end{bmatrix}$
$\begin{bmatrix}+ & - & + & - \\ - & + & - & + \\ + & - & + & - \\ - & + & - & +\end{bmatrix}$

Lorentzian Transformation Matrix:
$\begin{bmatrix} c t_1 \\x_1 \\y_1 \\z_1 \end{bmatrix} = \begin{bmatrix} \gamma & - \beta \gamma =& 0 =& 0\\ - \beta \gamma =& \gamma =& 0 =& 0 \\ 0 =& 0 =& 1 =& 0 \\ 0 =& 0 =& 0 =& 1 \\ \end{bmatrix} \begin{bmatrix} c t_2 \\x_2 \\ y_2 \\ z_2 \end{bmatrix}=$

Such that:
$\frac{\begin{bmatrix} c t_1 \\x_1 \\y_1 \\z_1 \end{bmatrix}}{\begin{bmatrix} c t_2 \\x_2 \\ y_2 \\ z_2 \end{bmatrix}} = \begin{bmatrix} \gamma & - \beta \gamma =& 0 =& 0\\ - \beta \gamma =& \gamma =& 0 =& 0 \\ 0 =& 0 =& 1 =& 0 \\ 0 =& 0 =& 0 =& 1 \\ \end{bmatrix} =$

Maxwell's Equations with Magnetic Monopoles:
$\nabla \cdot \mathbf{E} = \frac{\rho_e}{\epsilon_0}$
$\nabla \cdot \mathbf{B} = \mu_0 c \rho_m$
$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} - \mu_0 c \mathbf{J}_m$
$\nabla \times \mathbf{B} = \frac{1}{c^2} \frac{\partial \mathbf{E}} {\partial t} + \mu_0 \mathbf{J}_e$

Quote:

Originally Posted by Inaccurate, Needs further examination

It has come to my attention through careful scrutiny and the help of a friend of mine by the name of Mack, that it can be expressed that:
$q_1, q_2, r_1, t_1$
where
$q_1$ is the net charge of a given body, which is the rest frame for this interaction.
$q_2$ is the net charge of the body in motion about $q_1$ .
$r_1$ is the distance to $q_2$ from $q_1$ .
$t_1 = 2\pi r_1$
These can be given the relationship of, if electromagnetic in nature:
$|F| = k_C \frac{|q_1| |q_2|}{r_1^2}$

In our example we will assume for now that $q_1$ is 1 electric quanta and $q_2$ is 1 magnetic Quanta at a quantum distance, $r_1$ , from one another of 1. Now if we introduce a second state of particles into this, thus that we get:
$q_3, q_4, r_2, t_2, r_3, t_3$
Where:
$q_3 = 1$
$q_4 = 1$
$r_2 = 1$
$t_2 = 2\pi r_2$
$r_3 = n$ which is the distance of this system, $E_2$ , to the other, $E_1$ . As usual it must be a whole, non-zero, non-negative number.
$t_3 = 2\pi r_3$
So that:
$E_1 = q_1q_2r_1^2/t_1^2$
$E_2 = q_3q_4r_2^2/t_2^2$

then the sum of the states should be such that:
$\sum_{i=E_n}^{n}E_n = E_1+E_2+E_3+E_{...}+E_n$

Science Forums 2

Hypography?

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