One question is how to translate from time rates to space rates and vice versa. Consider the scalars base and time, and designate the stantial position, s, initial stantial position, s0, temporal position, t, initial temporal position, t0, velocity, v, initial velocity, v0, acceleration a (assumed constant over time), lenticity, w, initial lenticity, w0, and relentation, b (assumed constant over space). Then the linear equations of motion are as follows (see also the equations of motion at the top menu or here):
Start with the definition of acceleration and simple differential equation a = d²s / dt². By integration we get
v = v0 + at; integrate again to get s = s0 + v0t + ½at².
From these we get another equation of motion: v² = v0² + 2a(s – s0).
Take the definition of relentation and differential equation b = d²t / ds². By integration we get
w = w0 + bs; integrate again to get t = t0 + w0s + ½bs².
From these we get another equation of motion: w² = w0² + 2b(t – t0). So we have the following derivatives:
ds/dt = v = 1/w; dv/dt = a; dt/ds = w = 1/v; dw/ds = b;
dw/dv = –1/v² = –w² and dv/dw = –1/w² = –v².
If we are given the acceleration, a, what is the relentation, b? This may be determined as follows:
b = dw/ds = dv/dt * dt/ds * dw/dv = a * (1/v) * (–1/v²) = –a/v³ = –a/(v0 + at)³.
If v0 = 0, then b(t) = –1/(a²t³).
So the relentation is related to the square of the acceleration. Similarly, if we are given the relentation, b, what is the acceleration, a? This may be determined as follows:
a = dv/dt = dw/ds * ds/dt * dv/dw = b * (1/w) * (–1/w²) = –b/w³ = –b/(w0 + bs)³.
If w0 = 0, then a(s) = –1/(b²s³).
So the acceleration is related to the square of the relentation.