iSoul In the beginning is reality.

Space, time, arc length, and arc time

Let the vector r be a displacement vector of motion K that is a parametric function of its arc time t so that r = r(t). Then define s as the arc length of K such that

s = s(t) = ∫ || r′(τ) || ,

where the integral is from 0 to t. Let us further assume that s is bijective so that its inverse function is t(s).

Let the vector w be a dischronment vector of K that is a parametric function of arc length s so that w = w(s). Then t is the arc time of w if

t = t(s) = ∫ || w′(σ) || ,

where the integral is from 0 to s.

Now the arc time derivative of r, that is, the derivative of s with respect to t is

s′(t) = ds/dt = || r(t) || = || v(t) ||,

where v(t) is the speed function of K. And the arc length derivative of w, that is, the derivative of t with respect to s is

t′(s) = dt/ds = || w′(s) || = || u(s) ||,

where u(s) is the pace function of K.

From the inverse function theorem we have that

s′(t) = 1/|| w′(t) || = 1/|| u(t) || and

t′(s) = 1/|| r′(s) || = 1/|| v(s) ||.

Putting these together we find

s′(t) = || r′(t) || = || v(t) || = 1/|| u(t) || = 1/|| w′(t) || and

t′(s) = || w′(s) || = || u(s) || = 1/|| v(s) || = 1/|| r′(s) ||.

We then have

s = s(t) = ∫ || r′(τ) || = ∫ || v(τ) || = ∫ 1/|| u(σ) || = ∫ 1/|| w′(σ) || dσ,

where the integrals are from 0 to t. And also that

t = t(s) = ∫ || w′(σ) || = ∫ || u(σ) || = ∫ 1/|| v(τ) || = ∫ 1/|| r′(τ) || ,

where the integrals are from 0 to s.

Because of the difficulty of inverting s(t) and s(t) this shows a bypass is available. That is,

t(s) = ∫ 1/|| r′(τ) || = ∫ 1/|| v(τ) || .

and

s(t) = ∫ 1/|| w′(σ) || = ∫ 1/|| u(σ) || .

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