Quantum Nanotechnology/Back To The Beginning

Back to the BeginningEdit

We are stuck again. No thanks to positional prohibitions, the absence of simultaneity in straightway travel degrees of freedom, relativity, Quantum unknowns time and of course failure after failure.

As we know a mass existing at a single position cannot be actual

Where:

\frac{1\Rho}{M}=i

If we attempt to delude ourselves and assume that we can cause FTL using anything within the physical there is a 100% sure chance that energy will ultimately cause that failure because it just won’t succeed.

For a moment let us try to incorporate the principles of simultaneity from a light-ring to line segment traversion. When we do this, \Delta\nu is either increasing or decreasing. Positive, gaining additional velocity, which is non-uniform. Now the light ring constantly changes direction in 1 direction and is repetitive but no \Delta\nu is apparent unless of course some other force may cause it to become compressed. So to cause repetition to \Delta\nu=a, would mean that \Delta\nu must become uniform.


If “a” were uniform, then \Delta\nu\neq\Delta\nu because no increase would be present by a positive change-in (\Delta) or decrease by negative change-in (-\Delta). Similarly, renewal in RPS changes, (the outer circumference distance) the value for RPS changes to compensate respectively. However in the sinusoidal propagation of a photon, deceleration leads to momentum where a very rapid interchange is; -\nu=+Ρe-Ρe+Κe.


If we assume this to be true then excess momentum is the residual amount of G retained in every increase in \nu when \Delta\nu is induced. Because of the difference in \nu2-v_1 every “time” an increase in \nu by \Delta is intended. Thus to cause a repeat-like the light ring-and intentional disengagement of F, in \nu is initiated temporarily (not in the opposing direction) or a slight drop-back to manifest a rapid interchange between Pe and Ke and hence uniform forward motion for every \Delta\nu.

Now we have a repeat going in straight-line motion but only initiative after \nu_1-\nu_0. We could go back to \nu_0 but what would this purpose serve lest to accomplish inversion.

\frac{\Delta\nu_2-\nu_1}{-dv_2+\Delta\nu} = (-a) \nu_2-\nu_1

Now we have a repetition in straight-line propagation and if \Delta\nu is unequal to the temporary disengagement of F then it might be assumed that the changes in \nu increase \nu in percentages past \nu_2.The uniform motion-being repetitive now may be assumed to be equal to \nu_2 where no increase in \nu_2 maybe assumed to be entirely absent. Thus acceleration is eliminated by the repetitive act.

Henceforth this is an equal and opposite interaction. So we want this repetition to be as close as it can be made to be to the renewal of E in the light ring. Not necessarily that one must be unequal to the other but that one occurs faster than the other-timitically. However what has been intentionally done, is reversed the process of acceleration to restart it anew.

We have therefore:

\nu_1-\nu_0\nu_2-\nu_1-\nu_2=\nu_1\Delta\nu_1=\nu_2

Initially then what we have done is caused a repeat in \nu_2 -– an oscillatory one. And unlike the light ring is constantly changing direction back upon itself. An oscillation is a change in direction (\Delta d) back upon itself. An oscillation is a change in direction but not constantly. In -\nu-\nu_2 and \Delta\nu-\nu_1=\nu_2.

Now then we have to make these changes in direction very, very small so that in continuing the process the repeats will be fractionated taking up as it were the very positions themselves.

Quantum.

Do you recall what we were saying about a mass at 1 position?

\frac{1\Rho}{M}=i

What this shows is that it possesses no energy as well. Because energy in debt to potential energy can only be possible with the manifestation of 2 or more positions but is manifest by Ε.


Then it is equally true that:

\int v\Delta v=\frac{1P}{E}=i

Thus:

\int v\Delta v = \frac{1P}{E} = -E = i

where:

-E \frac{PN}{i} = \infty

One position only because energy when established, is established via 2 positions that constitute 1 possibility. There can be no other way. And even though the positions themselves are not present without energy this is the way it works. If in assuming that such an interaction could be constrained to 1 position at a time and much more mass itself comprising a ship, how then would one move from position to position?

The answer lies in the fact that E automatically disintegrates to a state of in-actuality for the mass if it were to occupy a solitary position. Suggesting that Ε becomes actual only with 2 or more positions while inversely 2 or more df become actual in the presence of energy actualizing. (2- dimensional line segment) In other words:

                              Ε =Δd ÷2Ρ
                                 OR
                              Ε=ΔΡ / d
Last modified on 24 September 2011, at 03:41