deriving the formula of the torsion of a curve

Question

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 !

Answer 1

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.

Answer 2

I post this answer in response the comment of Saru on the question itself. Hopefully, it addresses his/her concerns . . .

If $X(s)$ is a vector and $\sigma(s) \ne 0$ is a scalar function, each differentiable in the parameter $s$, then

$\left ( \dfrac{X(s)}{\sigma(s)} \right )' = \dfrac{\sigma(s)X'(s) - \sigma'(s)X(s)}{\sigma^2(s)}; \tag 1$

for setting

$Y(s) = \dfrac{X(s)}{\sigma(s)}, \tag 2$

we have

$\sigma(s) Y(s) = X(s); \tag 3$

thus

$\sigma'(s)Y(s) + \sigma(s) Y'(s) = X'(s), \tag 4$

whence

$\sigma(s)Y'(s) = X'(s) - \sigma'(s)Y(s)= X'(s) - \sigma'(s) \dfrac{X(s)}{\sigma(s)} = \dfrac{\sigma(s)X'(s) - \sigma'(s)X(s)}{\sigma(s)}; \tag 5$

that is,

$\left ( \dfrac{X(s)}{\sigma(s)} \right )' = Y'(s) = \dfrac{\sigma(s)X'(s) - \sigma'(s)X(s)}{\sigma^2(s)}. \tag 6$

We now apply this formula to

$N(s) = \dfrac{\alpha''(s)}{\vert \alpha''(s) \vert}, \tag 7$

where $\alpha(s)$ is a unit speed curve with nowhere vanishing curvature. Here $\alpha'(s) = T(s)$, the unit tangent field to $\alpha(s)$, and $\alpha''(s) = T'(s) = \kappa(s)N(s)$, where $\kappa(s) > 0$ and $N(s)$ are the curvature and unit normal vector field to $\alpha(s)$, respectively; then $\vert \alpha''(s) \vert = \vert \kappa(s) N(s) \vert = \kappa(s) \vert N(s) \vert = \kappa(s)$, so $N(s) = \alpha''(s) / \kappa(s) = \alpha''(s) / \vert \alpha''(s) \vert$, hence (7); we reach

$N'(s) = \dfrac{\vert \alpha''(s) \vert \alpha'''(s) - \vert \alpha''(s) \vert' \alpha''(s)}{\vert \alpha''(s) \vert^2}, \tag 8$

or

$N'(s) = \dfrac{\alpha'''(s)}{\vert \alpha''(s) \vert} - \dfrac{\vert \alpha''(s) \vert' \alpha''(s)}{\vert \alpha''(s) \vert^2}. \tag 9$

Now consider

$\det(T(s), N(s), N'(s)) = \det \left (\alpha'(s), \dfrac{\alpha''(s)}{\vert \alpha''(s) \vert}, \dfrac{\alpha'''(s)}{\vert \alpha''(s) \vert} - \dfrac{\vert \alpha''(s) \vert' \alpha''(s)}{\vert \alpha''(s) \vert^2} \right )$ $= \det \left (\alpha'(s), \dfrac{\alpha''(s)}{\vert \alpha''(s) \vert}, \dfrac{\alpha'''(s)}{\vert \alpha''(s) \vert} \right ) - \det \left (\alpha'(s), \dfrac{\alpha''(s)}{\vert \alpha''(s) \vert}, \dfrac{\vert \alpha''(s) \vert' \alpha''(s)}{\vert \alpha''(s) \vert^2} \right ); \tag{10}$

also,

$\det \left (\alpha'(s), \dfrac{\alpha''(s)}{\vert \alpha''(s) \vert}, \dfrac{\vert \alpha''(s) \vert' \alpha''(s)}{\vert \alpha''(s) \vert^2} \right ) = 0, \tag{11}$

since its second and third columns are linearly dependent, being scalar multiples of the vector $\alpha''(s)$; therefore (10) becomes

$\det(T(s), N(s), N'(s)) = \det \left (\alpha'(s), \dfrac{\alpha''(s)}{\vert \alpha''(s) \vert}, \dfrac{\alpha'''(s)}{\vert \alpha''(s) \vert} \right ). \tag{12}$

Nota Bene: We observe that (12) applies, not because

$N'(s) = \dfrac{\alpha'''(s)}{\vert \alpha''(s) \vert}, \tag {13}$

which is generally false, but rather by virtue of the fact that determinants vanish when at least two rows are linearly dependent. End of Note.

Answer 3

First you need to convert B and N from the arc length domain to the time domain.

For B = T x N

$$T = \frac{\dot{\sigma}}{\parallel\dot{ \sigma }\parallel} $$

$$ N=\frac{\frac{\text{d}T}{\text{d}s}}{\parallel\frac{\text{d}T}{\text{d}s}\parallel} = \frac{\frac{1}{\parallel \dot{\sigma} \parallel}\frac{\text{d}T}{\text{d}t}}{\parallel\frac{\text{d}T}{\text{d}s}\parallel} = \frac{\frac{\text{d}T}{\text{d}t}}{\parallel \dot{\sigma} \parallel\parallel\frac{\text{d}T}{\text{d}s}\parallel} = \frac{\frac{\text{d}T}{\text{d}t}}{\parallel \dot{\sigma} \parallel k} $$

$$ \frac{\text{d}T}{\text{d}t} =\frac{\text{d}}{\text{d}t}\frac{\dot{\sigma}}{\parallel\dot{ \sigma }\parallel}= \frac{1}{\parallel \dot{\sigma} \parallel^{2}}(\ddot{\sigma}\parallel \dot{\sigma} \parallel - \frac{\text{d}\parallel \dot{\sigma} \parallel}{\text{d}t}\hspace{1mm}\dot{\sigma})= \frac{1}{\parallel \dot{\sigma} \parallel}(\ddot{\sigma} - \frac{\text{d}\parallel \dot{\sigma} \parallel}{\text{d}t}\hspace{1mm}T) $$

$$\therefore N = \frac{1}{k\parallel \dot{\sigma} \parallel^{2}}(\ddot{\sigma} - \frac{\text{d}\parallel \dot{\sigma} \parallel}{\text{d}t}\hspace{1mm}T) $$

$$ T\times N = T \times \frac{1}{k\parallel \dot{\sigma} \parallel^{2}}(\ddot{\sigma} - \frac{\text{d}\parallel \dot{\sigma} \parallel}{\text{d}t}\hspace{1mm}T) =\frac{1}{k\parallel \dot{\sigma} \parallel^{2}}(T \times \ddot{\sigma}) =\frac{1}{k\parallel \dot{\sigma} \parallel^{2}}(\frac{\dot{\sigma}}{\parallel \dot{\sigma} \parallel} \times \ddot{\sigma})= \frac{1}{k\parallel \dot{\sigma} \parallel^{3}}(\dot{\sigma} \times \ddot{\sigma}) $$

Using $ k = \frac{\parallel\dot{\sigma} \times \ddot{\sigma} \parallel}{\parallel \dot{\sigma} \parallel^{3}}$

$$ B = \frac{\dot{\sigma} \times \ddot{\sigma}}{\parallel\dot{\sigma} \times \ddot{\sigma} \parallel} $$

Then using the time domain equation for N $$ N= \frac{(\dot{\sigma}\cdot \dot{\sigma})\ddot{\sigma} - (\ddot{\sigma} \cdot \dot{\sigma}) \dot{\sigma} }{\parallel (\dot{\sigma} \cdot \dot{\sigma})\ddot{\sigma} - (\ddot{\sigma} \cdot \dot{\sigma})\dot{\sigma} \parallel} $$

$$ \frac{\text{d}B}{\text{d}t} = \frac{\text{d}}{\text{d}t}(\frac{1}{\parallel \dot{\sigma} \times \ddot{\sigma} \parallel})(\dot{\sigma} \times \ddot{\sigma}) + \frac{1}{\parallel \dot{\sigma} \times \ddot{\sigma} \parallel}\frac{\text{d}}{\text{d}t}(\dot{\sigma} \times \ddot{\sigma}) = $$ $$ \frac{\text{d}}{\text{d}t}(\frac{1}{\parallel \dot{\sigma} \times \ddot{\sigma} \parallel})(\dot{\sigma} \times \ddot{\sigma}) + \frac{1}{\parallel \dot{\sigma} \times \ddot{\sigma} \parallel}[(\ddot{\sigma} \times \ddot{\sigma}) + (\dot{\sigma} \times \dddot{\sigma})] = $$

$$ \frac{\text{d}}{\text{d}t}(\frac{1}{\parallel \dot{\sigma} \times \ddot{\sigma} \parallel})(\dot{\sigma} \times \ddot{\sigma}) + \frac{1}{\parallel \dot{\sigma} \times \ddot{\sigma} \parallel} (\dot{\sigma} \times \dddot{\sigma}) $$

$$ N \cdot \frac{\text{d}B}{\text{d}t} =[\frac{(\dot{\sigma}\cdot \dot{\sigma})\ddot{\sigma} - (\ddot{\sigma} \cdot \dot{\sigma}) \dot{\sigma} }{\parallel (\dot{\sigma} \cdot \dot{\sigma})\ddot{\sigma} - (\ddot{\sigma} \cdot \dot{\sigma})\dot{\sigma} \parallel}] \cdot [\frac{\text{d}}{\text{d}t}(\frac{1}{\parallel \dot{\sigma} \times \ddot{\sigma} \parallel})(\dot{\sigma} \times \ddot{\sigma}) + \frac{1}{\parallel \dot{\sigma} \times \ddot{\sigma} \parallel} (\dot{\sigma} \times \dddot{\sigma}) ] = $$

using $ (u \times v) \cdot (a \times b) = (u \cdot a)(v \cdot b)- (v \cdot a)(u \cdot b) $

$$\frac{(\dot{\sigma} \cdot \dot{\sigma}) \ddot{\sigma}}{(\dot{\sigma} \cdot \dot{\sigma})^{\frac{1}{2}}\parallel\dot{\sigma} \times \ddot{\sigma} \parallel} \cdot \frac{(\dot{\sigma} \times \dddot{\sigma})}{\parallel\dot{\sigma} \times \ddot{\sigma} \parallel} = $$

$$ \frac{\parallel v \parallel \ddot{\sigma} \cdot(\dot{\sigma} \times \dddot{\sigma})}{\parallel\dot{\sigma} \times \ddot{\sigma} \parallel^{2}} = \frac{\parallel v \parallel (\ddot{\sigma} \times\dot{\sigma}) \cdot \dddot{\sigma}}{\parallel\dot{\sigma} \times \ddot{\sigma} \parallel^{2}} = - \frac{\parallel v \parallel (\dot{\sigma} \times\ddot{\sigma}) \cdot \dddot{\sigma}}{\parallel\dot{\sigma} \times \ddot{\sigma} \parallel^{2}} = -\parallel v \parallel\tau $$

Q.E.D

All the dot products are zero except for $(\dot{\sigma} \cdot \dot{\sigma}) \ddot{\sigma}$ and $ (\dot{\sigma} \times \dddot{\sigma})$

$\parallel (\dot{\sigma} \cdot \dot{\sigma})\ddot{\sigma} - (\ddot{\sigma} \cdot \dot{\sigma})\dot{\sigma} \parallel = [((\dot{\sigma} \cdot \dot{\sigma})\ddot{\sigma} - (\ddot{\sigma} \cdot \dot{\sigma})\dot{\sigma}) \cdot ((\dot{\sigma} \cdot \dot{\sigma})\ddot{\sigma} - (\ddot{\sigma} \cdot \dot{\sigma})\dot{\sigma})]^{\frac{1}{2}}$

=$[ (\dot{\sigma} \cdot \dot{\sigma})^2(\ddot{\sigma} \cdot \ddot{\sigma}) - (\dot{\sigma} \cdot \dot{\sigma})(\ddot{\sigma}\cdot \dot{\sigma})^2]^{\frac{1}{2}}$

$=[ (\dot{\sigma} \cdot \dot{\sigma})^2(\ddot{\sigma} \cdot \ddot{\sigma}) - (\dot{\sigma} \cdot \dot{\sigma})(\ddot{\sigma}\cdot \dot{\sigma})^2]^{\frac{1}{2}} = (\dot{\sigma} \cdot \dot{\sigma})^{\frac{1}{2}}[(\dot{\sigma} \cdot \dot{\sigma}) (\ddot{\sigma} \cdot \ddot{\sigma}) - (\ddot{\sigma} \cdot \dot{\sigma})^2]^{\frac{1}{2}}$

then make the substitutions:

$u = \dot{\sigma}$

$v = \ddot{\sigma}$

$a = \dot{\sigma}$

$b = \ddot{\sigma}$

and use the identity