iSoul In the beginning is reality

Space, time, arc length, and arc time

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

s = s(t) = ∫ || r′(τ) || ,

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

Let there also be a distimement vector w 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′ = ds/dt = || r′(t) ||.

And the arc length derivative of w, that is, the derivative of t with respect to s is

t′ = dt/ds = || w′(s) ||.

From the inverse function theorem we have that

t′ = dt/ds = 1/|| r′(t) || = 1/(ds/dt).

And also that

s′ = ds/dt = 1/|| w′(s) || = 1/(dt/ds).

Putting these together we find

s′ = ds/dt = || r′(t) || = 1/|| w′(s) || = 1/(dt/ds).

And also that

t′ = dt/ds = || w′(s) || = 1/|| r′(t) || = 1/(ds/dt).

We then have

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

where the first integral is from 0 to t and the second integral is from 0 to s. And also that

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

where the first integral is from 0 to s and the second integral is from 0 to t.

Because of the difficulty of inverting s(t), this shows a bypass is available. That is, from r(t) we find

t(s) = ∫ 1/|| r′(τ) || ,

and from w(s) we find

s(t) = ∫ 1/|| w′(σ) || .

The interpretation is that the space vector, r, is a position vector function of the arc time, t, and the time vector, w, is a time vector function of the arc length, s.

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