Within the time region the force which the atoms of matter exert upon each other because of their rotational velocities acts in the same *natural* direction as the gravitational force in the time-space region; that is, toward unity. But this direction, toward unity, which is inward in the time-space region and therefore makes the inter-atomic force which we term gravitation a force of attraction, is outward in the time region, and the corresponding inter-atomic force in this region is a force of repulsion, even though it is merely gravitation in a different environment.

This reversal of direction at the unit level makes possible the establishment of an equilibrium in which the atoms of matter can maintain the same relative positions in space indefinitely. Such an equilibrium cannot be established in the time-space region because in this region the effect of a change in the distance between the atoms is to accentuate any unbalance of forces. Here the rotational force (gravitation) is directed inward and the space-time force outward. If the rotational force exceeds the force of the space-time progression an inward motion takes place, making the effective rotational force still greater. Conversely, if the rotational force is the smaller the resulting motion is outward, which further weakens the already inadequate inward force. In either case there can be no establishment of equilibrium.

In the time region, however, the effect of a change in relative position opposes the unbalanced force which caused the change. If the rotational force is the greater an outward motion takes place, weakening this rotational force and ultimately reducing it to an equality with the space-time force. Similarly if the rotational force is the smaller the oppositely directed space-time force causes an inward motion. This strengthens the rotational force and again produces an equilibrium. The separation between any two atoms under these equilibrium conditions is the * inter-atomic distance*.

In order to calculate these inter-atomic distances it will first be necessary to determine the magnitudes of the corresponding inter-atomic forces. Since the inter-atomic rotational force in the time region is merely a different aspect of gravitation we may utilize the gravitational equation for its evaluation, providing that we replace the space-time region terms with the appropriate time region terms. We have already noted that velocity in the time region is 1/t^{2}. Energy, the one-dimensional equivalent of mass, which will take the place of mass in the time region expression of the gravitational equation because of the directional characteristics of the rotations in this region, is the reciprocal of this expression, or t^{2}. Acceleration is velocity divided by time, *1/t ^{3}*. The time region equivalent of the equation F = ma is therefore F = Ea = t

As previously explained, the value *1/t* applies only to the last of the *t* units of time, whereas in calculating the effective rotational force we will want the total. To obtain the latter we integrate *1/t* from unity to *t*, the initial point of the integration being taken at unity rather than at zero because unit velocity is the natural datum, the true physical zero.

The force computed in this manner is the inherent rotational force of the individual atom; that is, the force which it exerts against a single unit of force. The force between two interacting atoms is then:

F = ln _{r}t ln _{A}t_{B} | (4) |

The equivalent distance *s’* between the two atoms will be measured in the time-space region as *s ^{2}*, because of the inter-regional relationship previously discussed. The force at distance

F = ln _{r}t ln _{A}t) / _{B}s^{4} | (5) |

To evaluate the inter-atomic distance from this force equation we take advantage of the fact that at the equilibrium point the force of the space-time progression and the component of the rotational force in the direction opposite to that of the progression are necessarily equal. Since time is three-dimensional the rotational force in the time region is distributed three-dimensionally. The space-time progression, however, is one-dimensional and only that portion of the rotational force in the dimension of the progression is effective in the force equilibrium. It is therefore necessary to introduce a factor into the equilibrium equation representing the ratio of effective to total rotation. In determining this ratio we note that the first effective unit of rotation (the first displacement unit) is equal to the space-time progression, since space-time progresses at a unit rate. This one displacement unit (two total units of rotation) therefore constitutes the time region maximum if the units are disposed linearly. If these units are distributed three-dimensionally there can be two units of rotation in each dimension, raising the allowable total to 2^{3} or 8. Only one of these 8 units, the one displacement unit in the direction of the progression, is effective in opposition to the space-time force.

The same situation prevails in each dimension of the two-dimensional magnetic rotation except that in this case there are two effective units per dimension, one for each of the two rotational systems of the atom, and the ratio of effective to total rotational units in each dimension is 1 to 4. It may be somewhat confusing to speak of distributing the displacements in each space-time dimension in a three-dimensional manner, but it should be remembered that the three time region dimensions are dimensions of time, not of space-time, and the total time displacement of a rotation in any one space-time dimension may be disposed three-dimensionally in the time region without in any way affecting same situation the other space-time dimensions. We will encounter this again later in connection with other physical properties. On this basis one unit of rotation out of every 4 x 4 x 8 = 128 is effective against the space-time force. This ratio is further modified by the initial one unit negative level of the rotation due to the oppositely directed motion of the basic oscillation, as the portion of the rotational force required to overcome the negative initial level is not available to oppose the force of the progression. This initial unit is distributed over three dimensions and the one-third unit in the dimension parallel to the space-time progression is again distributed over the three dimensions of the time region. The resultant is 1/9 unit in each magnetic dimension, a total of 2/9 units. The electric rotation does not affect the initial level since it is merely a secondary rotation of the existing magnetic rotational structure. Other phenomena resulting from the rotational forces are similarly affected by the presence of the oppositely directed basic oscillation and we will encounter initial levels of one kind or another in a great many of the physical properties which we will examine.

Because of this negative initial level another 2/9 unit of displacement must be added to each of the 128 units in order to obtain one full unit in opposition to the space-time progression. This increases the ratio of total to effective units to 156.44 to 1. The one-dimensional rotational force applicable to each atom is therefore divided by 156.44 in setting up the equilibrium equation. For the two-dimensional magnetic rotation this factor becomes (156.44)^{2} and for two interacting magnetic rotations it increases to (156.44)^{4}. Applying this factor to the square of the one-dimensional rotational force, equation 5, we obtain the effective magnetic rotational force.

F = (1/(156.44)_{m}^{4}s) ln^{4}^{2}t ln_{A}^{2}t_{B} | (6) |

The distance factor does not apply to the space-time force as this force is omnipresent and unlike the rotational force is not altered as the objects to which it is applied change their relative positions. At the point of equilibrium, therefore, the rotational force is equal to the unit space-time force. Substituting unity for *F _{m}* in equation 6 and solving for the equilibrium distance, we obtain:

s = (1/156.44) ln_{o}^{½}t ln_{A}^{½}t_{B} | (7) |

The inter-atomic distances for those elements which have no electric rotation, the inert gas series, may be calculated directly from this equation. In the elements, however, *t _{A}* =

s = 1/156.44 ln _{o}t | (8) |

In cgs units this is:

s = 2.914 X 10_{o}^{-8} ln t cm | (9) |

As brought out in the discussion of the general characteristics of the atomic rotation, the two magnetic displacements may be unequal and in this case the velocity distribution takes the form of a spheroid with the principal rotation effective in two dimensions and the subordinate rotation in one. The average effective rotation under these conditions is (*t _{1}^{2} t_{2}*)

Atomic No. | Element | Magnetic Rotation | Inter-atomic Distance | |
---|---|---|---|---|

Calculated | Observed | |||

10 | Neon | 3-3 | 3.20 | 3.20 |

18 | Argon | 4-3 | 3.74 | 3.94 |

36 | Krypton | 4-4 | 4.04 | 4.02 |

54 | Xenon | 5-4, 4½-4 | 4.36 | 4.41 |

Helium, which also belongs to the inert gas series, has some special characteristics due to its low rotational displacement and will be discussed in connection with other elements affected by the same factors. The reason for the appearance of the 4½ value in the xenon rotation will also be explained later.

Turning now to the elements which have electric as well as magnetic displacement, we note again that the electric rotation is one-dimensional and opposes the magnetic rotation. We may therefore obtain an expression for the effect of the electric rotational force on the magnetically rotating photon by inverting the one-dimensional force term of equation 4.

F = 1 / (ln _{e}t’ ln _{A}t’)_{B} | (10) |

Because of the fact that the electric rotation is not an independent motion of the basic photon but a rotation of the magnetic velocities in the reverse direction, combining the electric rotational force from equation 10 with the magnetic rotational force of equation 6 modifies the rotational terms (the functions of *t*) only and leaves the remainder of equation 6 unchanged.

F = (1/(156.44)^{4}) (ln^{2}t ln_{A}^{2}t) / (_{B}s ln ^{4}t’ ln _{A}t_{B} | (11) |

Here again the effective rotational (outward) and space-time (inward) forces are necessarily equal at the equilibrium point. Since the space-time force is unity we substitute this unit value for *F* in equation 11 and solve for so, the equilibrium distance.

s = (1/156.44) (ln_{o}^{½}t ln_{A}^{½}t) / (ln_{B}^{1/4}t’ ln_{A}^{1/4}t’)_{B} | (12) |

Again simplifying for application to the elements, where A is generally equal to B,

s = (1/156.44) (ln _{o}t) / (ln^{½}t’) | (13) |

In cgs units this becomes

s = 2.914×10_{o}^{-8} ln t / ln^{½ }t’ cm | (14) |

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