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*) || *dτ*,

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*) || *dσ*,

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*) || *dτ* = ∫ || **v**(*τ*) || *dτ* = ∫ 1/|| **u**(*σ*) || *dσ* = ∫ 1/|| **w**′(*σ*) || d*σ*,

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

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

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*) || *dτ* = ∫ 1/|| **v**(*τ*) || *dτ*.

and

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