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I'm currently studying differential geometry, and I'm trying to solve this proof-based exercise I've got assigned.

  • Show that if α is a parametrised regular curve in R, then there exists a reparametrization β of α such that $| β'(t) |=1$.

Well, I know that all reparametrizations of regular curves are regular themlselves. So, β is regular.

I'm having more trouble with $| β'(t) |=1$. I know that if a derivative is 1, then the variation is constant through all values of t.

Does $| β'(t) |=1$ hold any special property? Can I relate it to β being unitary? Or is it as simple as saying that since α is regular, then there are infinite possibilities of reparametrization and one must be $| β'(t) |=1$ ?

Thanks!

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You can find out the arclength parametrization for a certain regular curve in the following way:

Define $$s(t)=\int_a^t ||\beta ' (u)||du$$

First we want to show that $s(t)$ is bijective. Note that, by the fundamental theorem of calculus,

$$s'(t)=||\beta '(t)||>0, \forall t\in [a,b]$$

So $s(t)$ is monotonically increasing, thus it is bijective, which means there exists an inverse function $t: s\mapsto [a,b]$

Now I claim $\gamma(s)=\beta (t(s))$ is an arclength parametrization for the curve.

Proof:

$$\gamma'(s)=\beta'(t) t'(s)$$

$$||\gamma'(s)||= ||\beta'(t)|| \cdot | t'(s)|$$

Where $t'(s)=\frac 1{s'(t)}$ by inverse function theorem

$$||\gamma'(s)||= \frac{ ||\beta'(t)||}{ ||\beta'(t)||}=1$$

While we showed that there exists such a function $s(t)$, it may not be expressible in elementary function. For example if $s(t)=e^t+t$, the function is certainly bijective, but the arclength parametrization could be messy as there is no simple way to find its inverse.

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Your curve $\alpha$ is a function $[a,b]\to X$. A reparametrization of $\alpha$ is accomplished by a function $\gamma:[c,d]\to[a,b]$ which is differentiable, monotonic, and satisfies $\gamma(c)=a$ and $\gamma(d)=b$. The reparametrization itself is the curve $\beta=\alpha\circ \gamma:[c,d]\to X$. We can apply the chain rule: $$\frac{d}{dt}\alpha(\gamma(t))=\alpha'(\gamma(t))\gamma'(t).$$ You want this derivative to equal $1$ in magnitude, so $$|\alpha'(\gamma(t))|\gamma'(t)=1.$$ Thus, $\gamma$ is a solution to the differential equation: $$\gamma'(t)=\frac{1}{|\alpha'(\gamma(t))|}.$$ Now use the fact that $\alpha$ is regular together with the existence of solutions to differential equations to finish the problem.

The curve $\beta $ is special because it is arc - length parametrized. This means that $\beta $ traces out its image at constant speed. This is useful because many applications of differential geometry involve functions of the form $|\beta'(t)|$. These expressions are considerably simpler if $\beta $ is arc - length parametrized.

The argument that there are an infinite number of reparametrizations doesn't work. Consider the statement "there are an infinite number of integers, so one of them must be equal to $\pi $."

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  • $\begingroup$ Whoa. This suddenly got very, very complicated. Why did we bring up a third parametrization $γ$? $\endgroup$ – AJ44 Mar 30 '16 at 15:06
  • $\begingroup$ There might be an easier way to show that such a reparametrization exists, but this was my first thought. $\endgroup$ – Alex S Mar 30 '16 at 15:10
  • $\begingroup$ $\gamma $ is not a third parametrization, it is the reparametrization. A reparametrization means that we are changing the speed that we trace out the curve. $\gamma $ is how we change that speed. $\endgroup$ – Alex S Mar 30 '16 at 15:46
  • $\begingroup$ I see. I'm having a bit of a hard time relating all the different concepts into a one big whole. So, the next step, since α is regular, is to say there must be a solution? $\endgroup$ – AJ44 Mar 30 '16 at 16:57
  • $\begingroup$ Since $\alpha$ is regular, $|\alpha'(\gamma(t))|\neq 0$, so the right hand side of the differential equation is continuous. The Peano existence theorem says there exists a solution. $\endgroup$ – Alex S Mar 30 '16 at 17:17

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