What's the relationship between $\frac{\text{d}^n\gamma}{\text{d}s^n}$ and $\left(\gamma\circ s^{-1}\right)^{(n)}|_{s(t)}$?

Question

Let $\gamma : I\subset\mathbb{R}\to\mathbb{R}^n$ be a $C^\infty$ curve such that $\|\gamma'\|>0$ everywhere. We let $s(t)$ be the arclength of $\gamma$ at time $t$, which has a smooth inverse everywhere (where defined), meaning $\gamma\circ s^{-1}$ is a unit-speed reparametrization of $\gamma$ (that is, $\|(\gamma\circ s^{-1})'\| = 1$).

Disclaimer: Everywhere I use a $'$ symbol, that indicates differentiation with respect to $t$.

Now, it's easy to see that $\left( \gamma\circ s^{-1} \right)'|_{s(t)}$ (that is, the derivative of $(\gamma\circ s^{-1})$ with respect to $t$, evaluated at $t=s(t)$) is equal to $\frac{\text{d}\gamma}{\text{d}s}$. In fact both are simply $1$, but for good measure $$ \left.(\gamma\circ s^{-1})'\right|_{s(t)} = \left.(\gamma'\circ s^{-1})(s^{-1})'\right|_{s(t)} = \left.\frac{\gamma'\circ s^{-1}}{s'\circ s^{-1}}\right|_{s(t)} = \frac{\gamma'}{s'} = \frac{\text{d}\gamma/\text{d}t}{\text{d}s/\text{d}t} = \frac{\text{d}\gamma}{\text{d}s} $$

Above we used the inverse function theorem to write $(s^{-1})' = \frac1{s'\circ s^{-1}}$. Similarly, $(s^{-1})'' = -\frac{s''\circ s^{-1}}{(s'\circ s^{-1})^3}$, and $s'' = \frac{\text{d}}{\text{d}t}\|\gamma'\| = \frac{\left\langle\gamma',\gamma''\right\rangle}{\|\gamma'\|}$.

We can try to extend this for higher order derivatives. First we get $$ \left.(\gamma\circ s^{-1})''\right|_{s(t)} = \kappa(t) = \frac{\gamma''\left\langle\gamma',\gamma'\right\rangle - \gamma'\left\langle\gamma',\gamma''\right\rangle}{\|\gamma'\|^4} $$ (in $\mathbb{R}^3$ you can simplify this significantly using the triple vector product formula, but we will leave it as this).

For $\frac{\text{d}^2\gamma}{\text{d}s^2}$ we have $$ \frac{\text{d}}{\text{d}s}\left(\frac{\text{d}\gamma}{\text{d}s}\right) = \frac{\text{d}}{\text{d}s}\left(\frac{\text{d}\gamma/\text{d}t}{\text{d}s/\text{d}t}\right) = \frac{\text{d}\left(\frac{\text{d}\gamma/\text{d}t}{\text{d}s/\text{d}t}\right)/\text{d}t}{\text{d}s/\text{d}t} $$ The numerator is given by $$ \begin{aligned} \frac{\text{d}}{\text{d}t}\left(\frac{\text{d}\gamma/\text{d}t}{\text{d}s/\text{d}t}\right) & = \frac{(\text{d}\gamma/\text{d}t)'(\text{d}s/\text{d}t)-(\text{d}\gamma/\text{d}t)(\text{d}s/\text{d}t)'}{(\text{d}s/\text{d}t)^2} \\ & = \frac{\gamma''\|\gamma'\|-\gamma'\left\langle\gamma',\gamma''\right\rangle/\|\gamma'\|}{\|\gamma'\|^2} \\ & = \frac{\gamma''\left\langle\gamma',\gamma'\right\rangle - \gamma'\left\langle\gamma',\gamma''\right\rangle}{\|\gamma'\|^3} \end{aligned} $$ and so we see that $$ \frac{\text{d}^2\gamma}{\text{d}s^2} = \frac{\gamma''\left\langle\gamma',\gamma'\right\rangle - \gamma'\left\langle\gamma',\gamma''\right\rangle}{\|\gamma'\|^4} = \left.(\gamma\circ s^{-1})''\right|_{s(t)} $$

I imagine this continues for higher order derivatives, and I'm almost certain I'm going about this in a really, really, really roundabout way, but I don't have an intuition for what the relationship between the objects $\frac{\text{d}^n\gamma}{\text{d}s^n}$ and $(\gamma\circ s^{-1})^{(n)}(s(t))$ actually are. Can someone give me a thorough explanation as to why these two objects should be the same (or if they aren't, why they aren't, and how they're related), so that I can skip the heavy computation each time I need to use either of the two?

Answer 1

But what does $\frac{d^n\gamma}{ds^n}$ really mean? First $\gamma$ has to be a differentiable function of $s$. It indeed is. How is the function defined? $\gamma$ is originally defined as a differentiable function of $t$, which in turn is defined as a differentiable function of $s$ via $t(s)=\varphi^{-1}(s)$ where $\varphi(t)$ is the length function you originally denoted as $s(t)$ which confuses the variable with the function. So $\gamma(t(s))$ is that function that you can perform the differentiation on, specifically $\frac{d^n\gamma}{ds^n}=\frac{d^n\gamma(t(s))}{ds^n}$. But that $\gamma(t(s))$ is exactly $f\circ\varphi^{-1}(s)$ written in a different way. So your desired equation is nothing but the representation of the same function in two different notations.