Suppose that $\gamma : I \to \mathbb{R}^3$ is a regular curve (not necessarily parameterised by arc length). How can we show that the curvature and torsion are given respectively by $\kappa = \frac{|\gamma^{\prime} ∧\gamma^{\prime\prime}|}{|\gamma^{\prime}|^3}$ and $\tau = \frac{(\gamma^{\prime} ∧\gamma^{\prime\prime})\cdot \gamma^{\prime\prime\prime}}{|\gamma^{\prime} ∧\gamma^{\prime\prime}|^2}$?


I'll do the curvature, you try the torsion.

Let $\beta:[s_0,s_1] \to \mathbb R^3$ be the reparametrization of $\gamma:[t_0,t_1] \to \mathbb R^3$ by arc length i.e: $\gamma=\beta \circ s$ where $s:[t_0,t_1]\rightarrow[s_0,s_1]$ ; $s(t)=\int_{t_o}^t||\gamma'(u)||du$.

We have:

$$\gamma' =\frac{\mathrm d\gamma}{\mathrm dt} =\frac{\mathrm d\beta}{\mathrm ds} \cdot \frac{\mathrm ds}{\mathrm dt} =T(s) \cdot||\gamma'(t)||\tag1$$ where $T(s)$ is the tangent vector of $\gamma$. From the equation we see that $T(s)=\gamma'(t)/||\gamma'(t)||$. Then differentiating $\gamma'$ and applying the chain rule, we get:

$$\gamma'' = \frac{\mathrm d^2\gamma}{\mathrm dt^2} =\left( \frac{\mathrm dT}{\mathrm ds} \cdot \frac{\mathrm ds}{\mathrm dt}\right) \cdot ||\gamma'(t)||+ T(s) \cdot \left( \frac{\langle\gamma'(t),\gamma''(t)\rangle}{||\gamma'(t)||}\right)\tag2$$

And since $T'=k \cdot N$, where $N$ is the normal vector of $\gamma$, we get:

$$ \gamma'' =k \cdot N \cdot ||\gamma'||^2+\left( \frac{\langle\gamma',\gamma''\rangle}{||\gamma'||}\right)\cdot T(s)\tag3$$

Computing the binormal $B$:

$$B = \frac{\gamma'(t)\times \gamma''(t)}{||\gamma'(t)\times \gamma''(t)||} =T\times N\tag4$$

We get the following relation:

$$ \gamma'(t)\times \gamma''(t)=k \cdot ||\gamma'(t)||^3 \cdot T\times N\tag5$$


$$N = B\times T = \left( \frac{\gamma'(t)\times \gamma''(t)}{||\gamma'(t)\times \gamma''(t)||}\right)\times \frac{\gamma'(t)}{||\gamma'(t)||}=\frac{k \cdot ||\gamma'(t)||^3 \cdot T\times N}{||\gamma'(t)\times \gamma''(t)||}\times \frac{\gamma'(t)}{||\gamma'(t)||}\tag6$$

Since $||N||=||T\times N||=1$ and $k\geq 0$, we get: $$k=\frac {||\gamma'(t)\times \gamma''(t)||}{||\gamma'(t)||^3}.$$

  • $\begingroup$ @kahen so I know that $\gamma : I \to \mathbb{R}^3$ is a regular curve this means $\gamma'(t)\not=0$ for all t in I. So your function s takes I to a new domain $s_{0},s_{1}$ where t is mapped to its corresponding arclength from $t_0$ along the curve $\gamma$. $\beta$ is as you have defined. $$\gamma' =\frac{\mathrm d\gamma}{\mathrm dt} =\frac{\mathrm d\beta}{\mathrm ds} \cdot \frac{\mathrm ds}{\mathrm dt} =T(s) \cdot||\gamma'(t)||$$ are you sure you the derivative product is correct here? It looks wrong? How did you apply chain rule in second eqn? thanks v much $\endgroup$ – Edwardjones Jones Apr 7 '13 at 10:39
  • $\begingroup$ @EdwardjonesJones I didn't write it. I just cleaned up jandrew's formatting. $\endgroup$ – kahen Apr 7 '13 at 12:21
  • $\begingroup$ @EdwardjonesJones $\gamma = \beta \circ s$ i.e $\beta$ is the parametrization by arc length.To find the curvature you need to work with the curve parametrized by arc length.Since $\beta$ depends on $s$ and $s$ depends on $t$, the chain rule is correctly applied.What do you think is wrong? What would you write? $\endgroup$ – jandrew Apr 7 '13 at 16:09
  • $\begingroup$ @jandrew got it my mistake, could you explain how you got that expression for the binormal - $\frac{\gamma'(t)\times \gamma''(t)}{||\gamma'(t)\times \gamma''(t)||}$ and also how you got the 'following relation' from it. kind thanks $\endgroup$ – Edwardjones Jones Apr 7 '13 at 19:46
  • $\begingroup$ @EdwardjonesJones I've tagged the equations to explain it better.You know that the $B=T\times N$ , since $||B||=1$ we can compute the product of parallel vectors to $T$ and $N$ and normalize it to get $B$.By definition $N$ is parallel to $\mathrm{dT}/\mathrm{ds}$,then from $(2)$ we clear $\mathrm{dT}/\mathrm{ds}$ and compute $V=T\times \mathrm{dT}/\mathrm{ds}$.You'll get that $V$ is parallel to $\gamma' \times \gamma''$ then $N=\gamma' \times \gamma'' /||\gamma' \times \gamma''||$.You get $(5)$ by computing $||\gamma' \times \gamma''||$ using $(1)$,$(3)$ and $(4)$. $\endgroup$ – jandrew Apr 7 '13 at 20:31

Let $I=[a,b]$ and $t(s):[0,L]\to I$ be a differentiable function whit $s: [0,L]\to [a,b]$ inverse differentiable and $$ L=\int_{a}^{b} \|\gamma^\prime(t)\| \; d\, t \qquad\mbox{ and }\qquad s(t)=\int_{0}^{t} \|\gamma^\prime(t)\| \; d\, t $$ Fix the notation $\kappa(t)=\kappa(t(s))=\kappa(s)$ and $\gamma(t)=\gamma(t(s))=\gamma(s)$. We have by definition of curvature $\kappa(s)=\|\gamma^{\prime\prime}(s)\|$. Note that \begin{align} \frac{d^2}{d t^2}\gamma(s)= & \frac{d}{d t}\left[\frac{d}{dt}\gamma(s) \right] \\ = & \frac{d}{d t}\left[\gamma^{\prime}(s)\cdot \left(\frac{d s}{dt}\right)\right] \\ = & \left[ \gamma^{\prime\prime}(s)\cdot \left(\frac{d s}{dt}\right) + \gamma^{\prime}(s)\cdot \frac{d s}{dt}\left(\frac{d s}{dt}\right) \right]. \end{align} Using Frenet-Serret equations $$ \left\{ \begin{array}{rl} \gamma^\prime(s)=&T(s)\\ T^\prime(s)=& \kappa(s)\cdot N(s)\\ N^{\prime}(s)= &-\kappa(s)\cdot T(s) + \tau(s)\cdot B(s)\\ B^\prime(s)=& \tau(s)\cdot B(s) \end{array} \right. $$ we have $$ \left\{ \begin{array}{rl} \frac{d^2}{d\, t^2}\gamma(s)=& \frac{d\,s}{d\,t} T(s) \\ \frac{d^2}{d\,t^2}\gamma(s)=& \kappa(s)N(s)\left(\frac{ds}{dt}\right)+T(s)\left(\frac{d^2 s}{d\,t^2} \right) \end{array} \right. $$ Calculate the cross product below \begin{align} \gamma^{\prime}(t)\times \gamma^{\prime\prime}(t)= & \frac{d }{dt}\gamma(s)\times \frac{d^2 }{dt^2}\gamma(s) \end{align} knowing that $\{T(s), N(s), B(s)\}$ is a basis of orthonormal vectors. After replacing the equations $\frac{d\,s}{d\,t}= \|\gamma^\prime(t)\|$ and $\frac{d^2\,s}{d\,t^2}=-\frac{\|\gamma^\prime(t)\|}{|\langle \gamma^\prime,\gamma^{\prime\prime}(t)\rangle|}$. Use too the equation $$ \langle u\times v, w \rangle = det(u,v,w) \quad \forall w\in\mathbb{R}^3 $$ for simplifications. After doing some algebraic manipulations to obtain the curvature.The formula for calculating the torsion is analogous. Just obtain the torsion use the expression below. $$ \left\langle \gamma^\prime(t)\times \gamma^{\prime\prime}(t),\gamma^{\prime\prime\prime}(t) \right\rangle = \left\langle \frac{d}{d\,t}\gamma(s)\times\frac{d^2}{d\,t^2}\gamma(s) \, ,\, \frac{d^3}{d\,t^3}\gamma(s) \right\rangle $$

  • $\begingroup$ can you explain how you got this line from the line before: $ \left[ \gamma^{\prime\prime}(s)\cdot \left(\frac{d s}{dt}\right) + \gamma^{\prime}(s)\cdot \frac{d s}{dt}\left(\frac{d s}{dt}\right) \right]$ that's great thanks. $\endgroup$ – Edwardjones Jones Apr 7 '13 at 20:09
  • $\begingroup$ the product of ds/dt terms are confusing? i know you used produce rule however. $\endgroup$ – Edwardjones Jones Apr 7 '13 at 20:10
  • $\begingroup$ I think it should be $B'(s)=-\tau(s)\cdot N(s)$. $\endgroup$ – delt3 Apr 15 '17 at 16:47

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