Let $\alpha(s)$ and $\beta(s)$ be two unit speed curves and assume that $\kappa_{\alpha}(s)=\kappa_{\beta}(s)$ and $\tau_{\alpha}(s)=\tau_{\beta}(s)$, where $\kappa$ and $\tau$ are respectively the curvature and torsion. Let $$J(s) = T_{\alpha}(s)\dot\ T_{\beta}(s)+N_{\alpha}(s) \dot\ N_{\beta}(s) +B_{\alpha}(s) \dot\ B_{\beta}(s).$$

Show that:

$J(0)=3$ and $J(s)=3$ implies that the Frenet frames of $\alpha$ and $\beta$ agree at $s$

$J'(s) = 0$ and $\alpha(s) = \beta(s)$ for all $s$.

How can I show this?



For the first question, what is the maximum value that $J(s)$ can possibly take? When can it achieve such a value? (Note that the Frenet frame vectors $T,B,N$ are unit vectors!)

The second statement is false. The correct statement should be that $\alpha(s) = \beta(s) + v$ where $v$ is a fixed vector. (Basically the information contained in $J(s)$ tells you the "derivative" of the curves are the same. When you integrate, you should pick up a $+C$.) By the first part, $J'(s) = 0$ everywhere implies that the Frenet frames agree. Now since $\alpha$ and $\beta$ are unit speed, they are equal to the integral of their tangent vector, up to a choice of initial value.

  • $\begingroup$ Thanks! $T,B,N$ are unit vector. But how can I formally show these statements without using the fundamental theory of curves? $\endgroup$
    – Lays
    Oct 2 '13 at 1:49
  • $\begingroup$ What is "the fundamental theory of curves", and why do you not want to use it? $\endgroup$ Oct 2 '13 at 7:20
  • $\begingroup$ I am sorry, I didn't mean to mention that. But for your hint. So since given any two vectors the dot product $v.w=|v||w|cos(\theta).$ If v and w are unit vectors then that's less than one unless the vectors are parallel. So since $J(0) = 3$ we have that it is parallel and hence are unit vectors that up to $3$? Is it that simple, or am I missing something here? $\endgroup$
    – Lays
    Oct 2 '13 at 17:10
  • $\begingroup$ And similarly for $J(s) = 3$? $\endgroup$
    – Lays
    Oct 2 '13 at 17:11
  • $\begingroup$ @Lays: Yes it is exactly that; you are not missing anything. (If you take parallel to mean $\theta = 0$, and ruling out the case of anti-parallel $\theta = \pi$. Some people don't distinguish when using the terminology 'parallel'.) (Also, in this sense, parallel unit vectors are actually identical.) $\endgroup$ Oct 3 '13 at 8:15

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