In a recent post, I defined levamentum as the lenticity divided by the mass (or times the vass). Here I show that levamentum is conserved, as momentum is. I will do this in 1D with a result that may be generalized to 3D time.
Consider the equation of motion for a particle:
(1/m) dℓ/dr = Y,
where Y is the release. Multiply both sides by dr:
(1/m) dℓ = dq = Y dr,
where dq is a differential of the particle’s levamentum, q = ℓ/m.
In integral form:
q2 – q1 = Δq = ∫ Y dr.
Consider two interacting particles. For particle 1 we have
dq1/dr = Y12,
where Y12 is the allowance exerted on particle 1 by particle 2. For particle 2 we have
dq2/dr = Y21,
where Y21 is the release exerted on particle 2 by particle 1. By addition,
d(q1 + q2)/dr = Y12 + Y21 = 0,
using Newton’s third law applied to releases. After integration, we find
q = q1 + q2 = constant,
so that the total levamentum of the system is a constant of the motion. That is, levamentum is conserved.
Note: fulmentum was changed to levamentum; legerity/celerity/progressity/allegrity/tempocity was changed to lenticity, and rush was changed to release.