in our class we defined the torsion $τ(s)$ of a curve $γ$ parameterized by arc length this way $τ(s) = B'(s) \cdot N(s) $ where $B(s)$ is the binormal vector and $N(s)$ the normal vector in many other pdf's and books it's defined this way ($τ(s) = -B'(s) \cdot N(s)$) but let's stick to the first definiton.
we were given in our class other formulas to compute the torsion :
- $$τ(s) = -\frac{\det(γ'(s),γ''(s),γ'''(s))}{||γ''(s)||^2} $$
- $$τ(t) = -\frac{\det(γ'(t),γ''(t),γ'''(t))}{||γ'(t)\timesγ''(s)||^2}$$
ok the first one is used when the curve is parameterized by arc length and the second one can be used to compute the torsion of any regular curve $γ$ whether $||γ'|| = 1$ or not
I tried proving them both and i think I've been able to prove the first one :
$$\begin{align} τ(s) = B'(s) \cdot N(s) = (T(s)\times N(s))' \times N(s) =(T'(s) \times N(s) + T(s) \times N'(s)) \cdot N(s)\end{align}$$
since the curve is parameterized by arc length $T'(s) = N(s)$ so $T'(s) \times N(s) =0$
$$\begin{align} τ(s) =( T(s) \times N'(s)) \cdot N(s)=\det( T(s) , N'(s),N(s)) \end{align}$$ $$\begin{align} =-\det( T(s) , N(s),N'(s)) =-\det(γ'(s),\frac{γ''(s)}{||γ''(s)||},\frac{γ'''(s)}{||γ''(s)||}) = -\frac{\det(γ'(s),γ''(s),γ'''(s))}{||γ''(s)||^2}\end{align}$$
check this proof and tell me If I proved it right
for the second one I tried replacing $γ'(s)$ by $γ(s^{-1}(t))'$ where $s(t) = \int_0^t ||γ'(u)||du$ did the same thing for $γ'(s)$ and $γ''(s)$ applied the chain rule but got stuck
any help or hints concerning the second formula would be appreciated. Thank you !
Your proof of the first looks fine. The problem is that the more general formula requires a more general definition, which you're missing by trying to generalise from the specialised case where $\lVert \gamma'(s) \rVert = 1$. The easiest way to see this is to differentiate the curve directly, and find the tangent and normal afterwards. Let's look at $\gamma(s(t))$. Then $$ \gamma'(s(t)) = s'\dot{\gamma} = s' T, $$ where $\dot{\gamma}=d\gamma/ds$. Then $$ \gamma'' = s''T + s'^2 \dot{T} = s''T + s'^2 \kappa N, $$ by the definition of $N$ as $\kappa N = \dot{T}$. Differentiating one last time, $$ \gamma''' = s'''T + 3s''s'\kappa N + s'^3 \dot{\kappa} N + s'^3 \kappa \dot{N}, $$ and by the Frenet–Serret equation, $\dot{N} = -\kappa T + \tau B$, and we find $$ \gamma''' = (s'''-\kappa^2s'^3)T + (\dot{\kappa}s'^3+3\kappa s's'')N + \kappa\tau s'^3 B. $$ Now we can find expressions for $\kappa$ and $\tau$ independent of $s$ by dotting and crossing this lot together. In particular, $$ \gamma' \times \gamma'' = s'^3 \kappa T \times N = s'^3 \kappa B, $$ so $$ \kappa = \frac{\lVert \gamma' \times \gamma'' \rVert}{s'^3} = \frac{\lVert \gamma' \times \gamma'' \rVert}{\lVert \gamma'\rVert^3}, $$ and more pertinently, $$ \tau = \frac{\det{(\gamma',\gamma'',\gamma''')}}{(\kappa s'^3)^2} = \frac{\det{(\gamma',\gamma'',\gamma''')}}{\lVert \gamma' \times \gamma'' \rVert^2}, $$ the difference in sign coming from differing conventions for the torsion: this has $\tau = -\dot{B} \cdot N$, but it's easy to fix this by changing the sign of $\tau$ whenever it appears.