## Derivation of s / t = 1

Discussion concerning the first major re-evaluation of Dewey B. Larson's Reciprocal System of theory, updated to include counterspace (Etheric spaces), projective geometry, and the non-local aspects of time/space.
user737
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### Derivation of s / t = 1

This is not meant to be a formal definition.
That is beyond me at this point in my development.
There exists the distinct possibility this derivation is wrong and/or false (hope not!).
Please check my work.

Simple integration over time yields this Newtonian classic:
v (s/t) = a (s/t2) × t

Given:

acceleration (a) = 10 m/s2
distance (x) = 250 m
time (t) = 5 s

Solve:

velocity (normalized to time): vt = a × t = 10 m/s2 × 5 s = 50 m/s
velocity (normalized to distance): vx = a ÷ x = 10 m/s2 ÷ 250 m = 0.04 s-2

These, of course, are averages as we are dealing with acceleration, or change in velocity over time (or conjugate velocity as change in energy, our "forcing function" [Ed: lulz], over distance).

If we multiply an acceleration by a time, we get a velocity (s/t2 × t = s/t → v)... magnitude, no direction.

If we divide an acceleration by a distance, we also get a velocity... but how can this be? The units don't work? Or do they...

Yes, inverse seconds squared is an equivalent velocity. We define it directly as 1/t2 being the equivalent motion in unit time and so being an angular velocity is a velocity nonetheless.
• In Case 1 (velocity normalized to time)... we do this all the time and is our standard practice. 50 m/s is 50 meters per 1 second. We have normalized time (to unity) by setting 1 = 5 seconds.
• Conversely, in Case 2 (velocity normalized to distance) we invert our operation (as we also invert our operand, i.e. space for time) where we then normalize distance (to unity) by setting 1 = 250 meters. This is only valid if what is stated above is true. i.e. we take inverse time to the second power to indeed be an equivalent velocity
In either case we're simply creating a ratio (defining a motion).
In the case we are most familiar, we create that ratio by normalizing to time and not to space (gee, I wonder why).

Therefore:

acceleration (spacial): ax = vx × x = 0.04 s-2 × 250 m = 10 m/s2
acceleration (temporal): at = vt ÷ t = 50 m/s ÷ 5 s = 10 m/s2

Sidebar: temporal acceleration would be the conjugate of spacial acceleration and therefore would be our "causal" "force" (in actuality just another aspect of motion and neither causal nor a-causal in its own right)

Assert:

a = at = ax = 10 m/s2

Ergo:

vx × x = vt ÷ t

but vx = vt as the kinetics described remain the same, we only change our perspective of motion (and have shown in both cases to have created an equivalent motion, either temporal or spacial)

so 1 × x = 1 ÷ t

or

x ÷ t = 1

put otherwise, F = ma Why mass though? Why is mass the fulcrum?
Is this an inherent bias from our spacial viewpoint? It must be as mass ceases to have traditional meaning outside of a larger gravitational field (i.e. 3D space with clock time). Would its conjugate be ether?
More questions than answers.

"Thanks to Einstein and others, we know that time and distance are equivalent and interchangeable, in many ways." - Miles Mathis
Light and love.