Quantum Quadrant Mechanics ?

[arkain101]

Preface

 

I've been working on a principle to do with energy and a quantifiable geometry, at least, that is what it appears to be, for what I can describe at this point.

 

What I plan to do here is explain 'how it works', 'what it is', and request you experts out there to analyze and check the work.

 

Principle

 

An angular momentum closed system of rotating, or orbiting mass is equivalent to 4 times(per cycle) the kinetic energy of that same mass moving in a vector.

 

Another way to put this is: The energy and/or momentum required to stop a moving object moving is exactly equivalent to a 90 degree non-accelerated-vector change of trajectory.

 

 

Illustration of the above principle.

 

 

 

 

As you can visualize, the red arrows set at 90 degree direction represent the original course of a moving body before and after the turn. Based on the laws of conservation of momentum and energy, during the 'turn', equal opposite forces exert within the system.

 

Distance at 90 degrees(circumference = distance):

 

[math]d = \frac {1}{2}\pi {r} [/math]

 

[math]d [/math] is distance

[math]r [/math] is radius

 

Show of work:

 

Circumference of circle is:

 

[math]C = 2 \pi {r} [/math]

 

Where a quadrant is equal to 1/4 the circumference

 

[math]C = \frac {2 \pi r}{4} = \frac {1}{2}\pi {r} [/math]

 

 

 

This principle assumes you forces can not be created or destroyed, but must be part of an existing geometric dimension or some reason to this effect (Ie, its not acceptable to just leave it at "the force is there).

 

 

I begin with describing the equation to calculate the work value along a curved path for one quadrant, ie: 90 degrees.

f is force

d is distance

E = energy

W = work

 

 

 

[math]E = W = f \cdot d [/math]

 

where distance is:

 

[math]d = \frac {1}{2}\pi \cdot r [/math]

 

 

The equation becomes:

 

 

[math]E = W = f \cdot \frac {1}{2}\pi \cdot {r} [/math]

 

However, as shown in the illustration the total force exertion within the system is equal and opposite for a total of two times force, 2f.

 

 

 

[math]E = W = 2f \cdot \frac {1}{2}\pi \cdot {r} [/math]

 

We can then reduce the fraction which leaves us with:

 

[math]E = W = f \cdot \pi {r} [/math]

 

expanding on force we can can write:

 

[math]E = W = m\cdot a \cdot \pi \cdot {r} [/math]

 

Keeping in mind this expresses only a 90 degree quadrant, or 1/4 of a rotation of an angular momentum system.

 

 

 

Full 1 cycle equation for circumference

 

based on this proof

 

[math]Ke = \frac {1}{2}mv^2 \rightarrow 4 \left(\frac {1}{2}mv^2\right) = 2mv^2 = E_{total} = E [/math]

 

We can then express the work/energy equation for one full cycle as:

 

[math]E = 2mv^2 = 2f \cdot d [/math]

 

Where distance of one cycle is:

[math]E = 2mv^2 = 2f \cdot \frac {4}{2} \pi r [/math]

 

[math]2mv^2 = 2f \cdot 2 \pi r [/math]

 

now, solving for v

 

[math]mv^2 = f \cdot 2 \pi r [/math]

 

[math]mv^2 = m a \cdot 2 \pi r [/math]

 

[math]v^2 = a \cdot 2 \pi r [/math]

 

[math]v = \sqrt {a \cdot 2 \pi r }[/math]

 

 

What we see is that the Energy contained in the system must apply in two forms.

1)As kinetic energy of a moving mass

and

2)Energy manifest in some type of form.

 

a)radiation: In the radiation of the 2nd half of the energy total in 90 degree quadrants, energy is dispersed over a distance of the curve at a velocity "V", direction perpendicular to the source at each specific location. Throughout the 90degree / 1/4 circle / full energy conservation transformation, the dispersion of that energy travels out like a wave.

 

 

I think we can say, though I can not specifically calculate that, the momentum of that energy in its total will be of its greatest vector direction in relation to at the 45dgree position. That is, because the momentum is dispersed in opposing directions at the 0 degree and 90 degree position, momentum of those two values is reduced(if not nullified) when combined to a neutral vector. Therefore, the less the angle of separation of momentum, the greater the momentum is provided at the 45 degree vector. As such, what we ended up with when treating this radiation wave like a particle is a frequency orientated particle.

 

 

 

 

 

 

b) As a force.

 

deriving 1 cycle equation:

 

[math]2mv^2 = 2f 2\pi r[/math]

 

[math]f = \frac{2mv^2}{ 4\pi r}[/math]

 

 

 

 

The work can then go on into 1/4cycle to 1 cycle for area of circle, surface area of sphere, volume of sphere. Although, as in all of this math I am unsure of its usefulness or accuracy.

 

We may have:

 

 

Equation of energy per unit Surface Area sphere:

 

?

 

[math]2mv^4 = 4f^2 \pi r^2[/math]

 

 

 

Equation of energy per unit Volume Sphere:

 

?

 

[math]2mv^5 = 4f^3 \pi r^3[/math]