# Dual dynamics equations

(1) Newton’s Second Law Dual Equations

Momentum is defined as the product of mass m and velocity v. The mass of a body is a scalar, though not necessarily a constant. Velocity is a vector equal to the time rate of change of location, v = ds/dt.

The time rate of change in momentum is dp/dt = m dv/dt + v dm/dt = ma + v dm/dt by the rules of differential calculus and the definition of acceleration, a.

If mass is constant, then v dm/dt equals zero and the equation reduces to dp/dt = ma. If we define F = dp/dt, then we get Newton’s famous F = ma.

The dual equation is derived similarly:

Levamentum is defined as the product of vass n and lenticity u. The vass of a body is a scalar, though not necessarily a constant. Lenticity is a vector equal to the space rate of change of chronation, u = dt/ds.

The space rate of change in levamentum is dq/ds = n du/ds + u dn/ds = nb + u dn/ds by the rules of differential calculus and the definition of relentation, b.

If vass is constant, then u dn/ds equals zero and the equation reduces to dq/dt = nb. If we define R = dq/ds, then we get the dual of Newton’s second law, R = nb.

(2) Work and kinetic energy

Work is defined as a force F applied over a displacement s: WF · s, and for a constant force and displacement in the same direction: W = Fs. From Newton’s second law, F = ma, so we have W = mas.

The third law of motion states that v² – v0² = 2as. Substitute this in the above equation to get

W = m (v² – v0²)/2 = mv²/2 – mv0²/2 = ΔEk.

This is a form of the work-energy theorem, which states that the work done on a body is equal to the change in that body’s kinetic energy.

If the initial kinetic energy is zero, then the kinetic energy Ek = mv²/2.

Dual to this is repose and kinetic lethargy:

Repose is defined as a release R applied over a dischronment t: YR · t, and for a constant release and dischronment in the same direction: Y = R t. From the dual to Newton’s second law, R = nb, so we have R = nbt.

The third law of motion states that u² – u0² = 2bt. Substitute this in the above equation to get

Y = n (u² – u0²)/2 = nu²/2 – nu0²/2 = ΔQk.

This is a form of a repose-lethargy theorem, which states that the repose done applied to a body is equal to the change in that body’s kinetic lethargy.

If the initial kinetic lethargy is zero, then the kinetic lethargy Qk = nu²/2.