In a recent post, I defined *fulmentum* as the *legerity* divided by the mass. Here I show that fulmentum 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/d*r* = M,

where M is the *surge*. Multiply both sides by d*r*:

(1/*m*) dℓ = d*q* = M d*r*,

where d*q* is a differential of the particle’s fulmentum, *q* = ℓ/m.

In integral form:

*q*_{2} – *q*_{1} = Δ*q* = ∫ M d*r*.

Consider two interacting particles. For particle 1 we have

d*q*_{1}/d*r* = M_{12},

where M_{12} is the allowance exerted on particle 1 by particle 2. For particle 2 we have

d*q*_{2}/d*r* = M_{2}_{1},

where M_{2}_{1} is the *surge* exerted on particle 2 by particle 1. By addition,

d(*q*_{1} + *q*_{2})/d*r* = M_{12} + M_{2}_{1} = 0,

using Newton’s third law applied to surge. After integration, we find

q = *q*_{1} + *q*_{2} = constant,

so that the total fulmentum of the system is a constant of the motion. That is, fulmentum is conserved.

Note: *celerity*/*progressity*/*allegrity*/*tempo* was changed to *legerity*, and *chrondyne* was changed to *surge*.