**The geometric analysis of the medium**

Terms are used here that are explained in the page Matrix

A medium that applies to all aspects of our physical world is a centuries-old goal of physics. As long as this is not recognized, our physical world consists of separate particles whose distances are perceived as empty space. The question of one medium for the particles and another for space remains open. Time and power remain unexplained if their effect continues to be viewed as tunneling under the Absolute Void. In my paper "The Matrix" a theory is presented that describes a medium that generates distance-time-force and thus becomes the basis of all physical events.

This matrix is a relational network of distances as a result of oscillations in the elasticity of scaled distances down to zero magnitude. This is not a circular argument here, but a generalization of the medium in that it is based on smaller scaled media, which theoretically lead to zero distances. The process of oscillation generated by this in turn at zero time and taking into account the Planck constant, to infinity force. The ranges of distances, times and forces are thus defined by this matrix from zero to infinity.

The basic idea was to find a spatial structure in which all values in terms of time and energy are scalable.

Space: Take a plastic bag with glass balls, connect their centers and you will see a hexagonal structure made up of tetrahedrons.

Zeit: Die sich ergebenen immer gleichen Distanzen werden durch die Medium-Geschwindigkeit dividiert und erhalten das Zeitmaß per Einheit.

Force: The elasticity with which the original state of harmony is restored creates the impulse. With the unit time measure we get the frequency. Impulse x frequency creates the energy.

Until then, it sounds embarrassingly easy. However, it is only the tip of
the iceberg. This is about a geometry that also includes time and power.
Therefore, my paper “The Matrix” must always be consulted. It is all about a
double oscillation in 4 dimensions, where push/pull can be found in the
known 3 dimensions and time in the 4th dimension. The double oscillation
produces 4 parities:

++ +-
-+ - - .
It is the 4 parities that add up to zero in every tetrahedron. So the
tetrahedron is the unit of this matrix. The intervening octahedra (as a
result of the arrangement of oscillations of these tetrahedra) consist
entirely of parities of surrounding tetrahedra. The medium of the
tetrahedron and octahedron are the smaller scales of the same matrix.
Consult here
space is oscillations

How are the base types tetrahedron / octahedron distributed in space? This is a purely geometric comparison.

Visible here is a cube, 2 nested tetrahedrons with an intersection in
form of an octahedron and the residual forms a; b; c to fill up the cube.
For the sake of math, a cube with side lengths √2 has been assumed, thereby
producing interior distances of hexagonal directions of 1. We see that a
central octahedron has a tetrahedron (green and magenta) on all 8 sides. We
recognize that the parts (a) above (c) below and the parts (b) on the 4
edges are required to fill this structure into a cube. With a little
imagination we see that the parts a; b; c are always ¼ octahedron in size.
If the space were filled with cubes, then the ¼ octahedron with the
adjoining neighboring cubes would again produce whole octahedrons on each
side length and edge. Now the only question that remains is What is the
volume of a tetrahedron.

This we finde out :

[Cube (√2)^3] - [4·a] - [4·b] - [4·c] - [1·octahedron] = 8·tetrahedra.
Because of [4·a];[4·b];[4·c] will become [1·octahedron], the cube will have
4 octahedra and 8
tetrahedra. Becaus all internal parts of a cube are =1 , it derives:

Volume (octahedron) = 1^2·(√2)/3 = √2/3 = 0.4714; 4·octahedra = 1.8856

Vol.(cube) = (√2)^3 = 2.8284

Vol.(tetrahedron) = (cube) 2.8284 – (octahedron) 1.8856 = (tetrahedron) 0.9428

1·tetrahedron=0.9428/8=0.1178 = 1·part a; b; c; = (octahedron) 0.4714/4 = 0.1178

Here we recognize the mystical volume is equality of tetrahedra and the
parts a; b; c; .

Although mathematics based on partial voluma 0.1178 and an orthogonal space structure would be possible, this does not make sense.

Due
to the fact that the orthogonal space cannot add up to zero per tetrahedron
units, its cube units never become transparent. It's an opaque space.
Nevertheless, it is latently present as an orthogonal view in a hexagonal
space, but energetically shows no effect. It consists of octahedrons, but
these are energetically = zero, since all corners of the octahedron do not
belong to it, but to the surrounding tetrahedrons. However, we recognize an
orthogonal space structure based on (√2) in the central octahedron with its
diagonals. This will be explained as QUARKS in my paper
Das LHC-Feuerwerk
(**L**arge **H**adron **C**ollider)

Gunter Michaelis, 22.7.2022