At the moment, I am self-studying out of the textbook "Elementary Differential Geometry" by Barret O'Neill.
I am currently working through O'Neill's derivation of the Frenet frame for a 1-manifold curve $\alpha$($t$) provided that |$\alpha$'($t$)| $\neq$$1$
He starts off by defining a new curve $\alpha*$($s(t)$)=$\alpha$($t$), where $s(t)$ is the arclength function.
From the chain rule, it follows that $\alpha'*$($s(t))$=$\alpha'*$($s$)$v(t)$, where $v(t)$=$s'(t)$.
Since $\alpha*$($s$) is a function of $s$, it follows that |$\alpha'*$($s$)|=1. So, he defines the unit tangent vector field to $\alpha*$($s$) as $T*(s)$=$\alpha'*$($s$).
He also defines another unit tangent vector field $T(t)$=$T*(s)$.
$T(t)$ is the unit tangent vector vector field to the curve $\alpha$($t$).
From the chain rule expansion above, we then have that $T(t)$ =$\frac{\alpha'*(s(t))}{v(t)}$. Everything up to this point makes sense to me. However, I have some qualms with the way he calculates $T'(t)$.
He states that $T'(t)$= $T'*(s)$$v(t)$. This would seem to say that $T(t)$=$T*(s(t))$ because $T'*(s(t))$= $T'*(s)$$v(t)$. But from our definition, we have that $T*(s)$=$T(t)$.
This is the issue I am stuck on. How can $T*(s(t))$=$T*(s)$, since $s(t)$ is some function of time while $s$ is just a variable. I know that through his so-called "functional notation" he says that $s$=$s(t)$, but I find it hard to believe that we can just substitute between them at will.
I figure I also ought to mention that I tried to define a new tangent vector $T!(t(s))$=$T*(s)$ and tried to calculate $T'*(s)$ using this substitution, but I have had little success with this method.
Thanks in advance.
In the definition of $T(t)$, $s$ is not “just a variable”. The variable $s$ depends on the variable $t$ by the function $s(t)$. So when he defines $T(t)=T^*(s)$ it means given $t$, in order to find $T(t)$ we need to find $s=s(t)$ and $T(t)=T^*(s)$.