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(s).

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†(t) = ds/dt = || r(t) || = || v(t) ||,

where v(t) is the curve’s speed function. 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 curve’s pace function.

Next put these equations in terms of a generic variable, x, to get

s′(x) = ds/dx = || r′(x) || = || v(x) || and

t′(x) = dt/dx = || w′(x) || = || u(x) ||.

From the inverse function theorem we have that

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

t′(x) = 1/|| r′(x) || = 1/|| v(x) || = 1/s†(x).

Putting these together we find

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

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

We then have

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

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

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

where the integrals are from 0 to x.

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

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

and from w(x) we find

s(x) = ∫ 1/|| w′(σ) || = ∫ 1/|| u(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.