# [Differential Geometry] On the Smooth Equivalence of Parametrised Differentiable Curves.

We say that two regular parametric curves $$\phi$$ and $$\psi$$ are smoothly equivalent, if there is a strictly monotonically increasing differentiable function $$s$$ such that $$\psi(s(t)) = \phi(t)$$. Reparameterisation means that you have parametrised $$\psi$$ by some new parameter $$s(t)$$ that is a function of $$t$$, instead of the original parameter, $$t$$. For example, you can reparameterise a curve with its arc length, given as a function of $$t$$.

Why does $$s$$ have to be monotonically increasing rather than decreasing? $$s$$ must be monotonically increasing, because we want to preserve the orientation (i.e. direction through which it travels) of the curve.

Are the follwing two curves $\phi(\theta) = (cos(\theta),sin(\theta)),$

$\psi(\theta) = (cos(2\theta),sin(2\theta)),$

smoothly equivalent if the domain of $$\phi$$ is $$[0,2\pi]$$, the domain of $$\psi$$ is $$[0,\pi]$$?

These two curves have the same trace, but are not considered to be equivalent curves. The curves have distinct velocities, which you can easily see when you calculate tangent vectors at each of the points on the curves, $\psi'(\theta)=2\cdot\phi'(\theta),$

so $$\psi(\theta)$$ travels twice as fast as $$\phi(\theta)$$. Note that this means that you cannot distinguish parametrised differentiable curve simply by looking at their image sets. Smoothly equivalent parametrised curves can be reparametrised from one to the other, and thus we obtain an equivalence class of regular curves.

Below is a list of parametrised differentiable curves that all trace out a unit circle in $$\mathbb{R}^2$$.

$\phi_1(t) = \cos t + i\sin t, ~ t \in \left[0 \leq t \leq 2\pi \right],$

$\phi_2(t) = \cos(2\pi t^{2}) + i\sin(2\pi t^{2}),~ t \in \left[ 0 \leq t \leq 1 \right],$

$\phi_3(t) = \cos t + i\sin t, ~t \in \left[ 0 \leq t \leq 4\pi \right],$

$\phi_4(t) = \cos (-t) + i\sin(-t), ~t \in \left[ 0 \leq t \leq 2\pi \right].$

Note that $$\phi_1(t)$$ and $$\phi_2(t)$$ are mutually smoothly equivalent, because they are reparametrisations of each other traced in the same direction. However, $$\phi_3(t)$$ traces the unit circle twice while $$\phi_1(t)$$ traces it once, so they are not smoothly equivalent. Also, $$\phi_4(t)$$ traces the unit circle clockwise, while $$\phi_1(t)$$ traces it anticlockwise, so they are not smoothly equivalent.

Note by Tasha Kim
2 weeks ago

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