Curvature of curve not parametrized by arclength

Question

If I have a curve that is not parametrized by arclength, is the curvature still $||\gamma''(t)||$? I am not so sure about this, cause then we don't know that $\gamma'' \perp \gamma'$ holds, so the concept of curvature might not be transferable to this situation. So is this only defined for curves with constant speed?

Answer 1

The curvature may of course still be defined if $t$ is not the arc-length along $\gamma(t)$, but the formulas will be different; we will not in general have $\kappa(t) = \Vert \gamma''(t) \Vert$. To find the correct expression for $\kappa(t)$, we recall that the curvature is defined in terms of the arc-length $s$ via the Frenet-Serret equation for $T'(s)$, where $T(s)$ is the unit tangent vector field to $\gamma$ in terms of $s$. We have

$T'(s) = \kappa(s) N(s), \tag{1}$

where $N(s)$ is the unit normal field to $\gamma(s)$, leading to

$\kappa(s) = \Vert T'(s) \Vert. \tag{2}$

To find $\kappa(t)$ from (2), we need to express the quantities occurring in (2) in terms if the parameter $t$; we begin with $T(s)$ itself; since it is the unit tangent vector to $\gamma$, in terms of $t$ we have

$T(t) = \dfrac{1}{\Vert \gamma'(t) \Vert} \gamma'(t). \tag{3}$

We can compute $dT(s)/ds$ from (3) using the chain rule, since

$\dfrac{dT(s)}{ds} = \dfrac{dt}{ds}\dfrac{dT(t)}{dt}, \tag{4}$

wherein we see that

$\dfrac{dt}{ds} = \dfrac{1}{\dfrac{ds}{dt}} = \dfrac{1}{\Vert \gamma'(t) \Vert} = \Vert \gamma'(t) \Vert^{-1}; \tag{5}$

bearing (5) in mind, we proceed to compute

$\dfrac{dT(t)}{dt} = \dfrac{d}{dt}(\Vert \gamma'(t) \Vert^{-1} \gamma'(t))$ $= -\Vert \gamma'(t) \Vert^{-2} \dfrac{d\Vert \gamma'(t) \Vert}{dt} \gamma'(t) + \Vert \gamma'(t) \Vert^{-1}\gamma''(t); \tag{6}$

furthermore,

$\dfrac{d\Vert \gamma'(t) \Vert}{dt} = \dfrac{d \langle \gamma'(t), \gamma'(t) \rangle^{1/2}}{dt} = \dfrac{1}{2}\langle \gamma'(t), \gamma'(t) \rangle^{-1/2}\dfrac{d\langle \gamma'(t), \gamma'(t) \rangle}{dt}$ $= \dfrac{1}{2}\langle \gamma'(t), \gamma'(t) \rangle^{-1/2}(2\langle \gamma''(t), \gamma'(t) \rangle) = \langle \gamma'(t), \gamma'(t) \rangle^{-1/2}\langle \gamma''(t), \gamma'(t) \rangle$ $= \Vert \gamma'(t) \Vert^{-1}\langle \gamma''(t), \gamma'(t) \rangle = \langle \gamma''(t), T(t) \rangle; \tag{7}$

inserting the result of (7) into (6) then yields

$\dfrac{dT(t)}{dt} = -\Vert \gamma'(t) \Vert^{-3}\langle \gamma''(t), \gamma'(t) \rangle \gamma'(t) + \Vert \gamma'(t) \Vert^{-1} \gamma''(t), \tag{8}$

whence, from (4),

$\dfrac{dT(s)}{ds} = \Vert \gamma'(t) \Vert^{-1}(-\Vert \gamma'(t) \Vert^{-3}\langle \gamma''(t), \gamma'(t) \rangle \gamma'(t) + \Vert \gamma'(t) \Vert^{-1} \gamma''(t))$ $= -\Vert \gamma'(t) \Vert^{-4}\langle \gamma''(t), \gamma'(t) \rangle \gamma'(t) + \Vert \gamma'(t) \Vert^{-2} \gamma''(t), \tag{9}$

which after a bit of algebraic re-arranging becomes

$T'(s) = \dfrac{dT(s)}{ds} = \Vert \gamma'(t) \Vert^{-2}(\gamma''(t) - \Vert \gamma'(t) \Vert^{-2}\langle \gamma''(t), \gamma'(t) \rangle \gamma'(t)). \tag{10}$

Taking the norm of (10), and using (2), we obtain

$\kappa(t) = \left \Vert \dfrac{dT(s)}{ds} \right \Vert = \Vert \gamma'(t) \Vert^{-2} \Vert \gamma''(t) - \Vert \gamma'(t) \Vert^{-2}\langle \gamma''(t), \gamma'(t) \rangle \gamma'(t) \Vert. \tag{11}$

Formulas (10) and (11) in fact answer the questions posed, since they show that (i.) the curvature is not in general $\Vert \gamma''(t) \Vert$ when $t$ is not arc-length; (ii.) $\gamma''(t) \not \perp \gamma'(t)$ in general either, for (10) shows $\gamma''(t)$ has a component along $\gamma'(t)$; in fact, we may re-arrange (10), using (1) and (3), to read

$\gamma''(t) = \Vert \gamma'(t) \Vert^2 \kappa(t)N(t) + \Vert \gamma'(t) \Vert^{-2}\langle \gamma''(t), \gamma'(t) \rangle \gamma'(t)$ $= \Vert \gamma'(t) \Vert^2 \kappa(t)N(t) + \langle \gamma''(t), T(t) \rangle T(t), \tag{12}$

which gives an explicit decomposition of $\gamma''(t)$ in terms of tangential and normal components to $\gamma(t)$. And of course, we see from (7) that $\langle \gamma''(t), T(t) \rangle$, the coefficient of $T(t)$ in (12), is non-zero precisely when $\Vert \gamma'(t) \Vert$ is changing as a function of $t$ (i.e. $d/dt(\Vert \gamma'(t) \Vert) \ne 0$); (iii.) it is clear from the above, viz. (11), that the curvature can in fact be defined for non-unit speed curves $\gamma(t)$, and it is still given by (2); but in this case the derivatives must all be expressed in terms of $t$, not $s$; it's just that (iv.) the formulas take a much simpler form when $\gamma(t)$ is expressed in terms of arc-length $s$, $\gamma(s(t))$.

Finally, it must be admitted that our formula (11) does not, on the face of it, look much like the elegant expressions advanced by Xipan Xiao and Tony Piccolo in their comments. Nevertheless, it is possible derive the expression

$\kappa(t) = \dfrac{\Vert \gamma'(t) \times \gamma''(t) \Vert}{\Vert \gamma'(t) \Vert^3} \tag{13}$

from (12) as follows: since $\langle T(t), N(t) \rangle = 0$, (12) implies

$\Vert \gamma''(t) \Vert^2 = \Vert \gamma'(t) \Vert^4 \kappa^2(t) + \langle \gamma''(t), T(t) \rangle^2; \tag{14}$

furthermore,

$\langle \gamma''(t), T(t) \rangle^2 = \Vert \gamma''(t) \Vert^2 \Vert T(t) \Vert^2 \cos^2 \theta = \Vert \gamma''(t) \Vert^2 \cos^2 \theta, \tag{15}$

$\theta$ being the angle 'twixt $T(t)$ or $\gamma'(t)$ and $\gamma''(t)$. We re-arrange (14) with the aid of (15):

$\Vert \gamma'(t) \Vert^4 \kappa^2(t) = \Vert \gamma''(t) \Vert^2(1 - \cos^2 \theta) = \Vert \gamma''(t) \Vert^2 \sin^2 \theta, \tag{16}$

and recalling that

$\Vert \gamma'(t) \times \gamma''(t) \Vert = \Vert \Vert \gamma'(t) \Vert T(t) \times \gamma''(t) \Vert = \Vert \gamma'(t) \Vert \Vert T(t) \times \gamma''(t) \Vert$ $= \Vert \gamma'(t) \Vert \Vert T(t) \Vert \Vert \gamma''(t) \Vert \vert \sin \theta \vert = \Vert \gamma'(t) \Vert \Vert \gamma''(t) \Vert \vert \sin \theta \vert, \tag{17}$

so that

$\Vert \gamma'(t) \times \gamma''(t) \Vert^2 = \Vert \gamma'(t) \Vert^2 \Vert \gamma''(t) \Vert^2 \sin^2 \theta, \tag{18}$

we see that (16) combined with (18) yields

$\Vert \gamma'(t) \Vert^6 \kappa^2(t) = \Vert \gamma'(t) \Vert^2 \Vert \gamma''(t) \Vert^2 \sin^2 \theta = \Vert \gamma'(t) \times \gamma''(t) \Vert^2; \tag{19}$

now taking square roots, assuming $\gamma'(t) \ne 0$ (i.e. $\gamma(t)$ is a regular curve), and noting that $\kappa(t) > 0$ by definiton, we see that (13) is the result.

Whew! Quite a ride, that!

Hope this helps. Cheerio,

and as ever,

Fiat Lux!!!

Answer 2

Here is sort of a simple example. Consider $\lambda_1,\lambda_2>1$ with $\lambda_1\not=\lambda_2$. Consider the unit disk. Both of the curves $(\cos(\lambda_1 t),\sin(\lambda_1t))$ and $(\cos(\lambda_2 t),\sin(\lambda_2t))$ trace it out. It makes intuitive sense to define the curvature as the rate of change of the velocity vector. But doing so without first parametrizing with respect to arc length gives two different curvatures for the same shape.

The answer to your first question is no. The curvature of a curve that isn't unit speed is defined to be the curvature of that curve parametrized by arc length. This is justified as the original curve and its parametrization trace out the same shape.

Answer 3

Let a curve $Γ$ parametrized by an arbitrary velocity $γ:Ι\rightarrowΓ$. Define $Τ(t):=(\frac{γ'}{||γ΄||})(t), \forall t\in I (1)$, be the unitary tangent vector on $γ(t)$. If $s\in J$ denotes the unitary velocity parameter and $t=t(s):J\rightarrow I$ the reparametrization map, we get $\frac{dT}{ds}(t(s))=\frac {dT}{dt}(t(s))\frac{dt}{ds}(s)=\frac{\frac {dT}{dt}(t(s))}{\frac{ds}{dt}(t)}=\frac{\frac {dT}{dt}(t(s))}{||γ΄(t)||} (2)$. Using $(1)$ we get $\frac {dT}{dt}=\frac{(γ'\cdot γ')γ''-(γ'\cdot γ'')γ'}{||γ||^3}$, where we ommitt the parameter $t$. From Grassmann Identity we get $\frac {dT}{dt}=\frac{(γ'\cdot γ')γ''-(γ'\cdot γ'')γ'}{||γ||^3}=\frac {γ'\times(γ''\times γ')}{||γ'||^3}$, thus $||\frac {dT}{dt}||=\frac {||γ'\times(γ''\times γ')||}{||γ'||^3}$. We easily observe that $γ'$ and $γ''\times γ'$ are orthogonal, thus $||γ'\times(γ''\times γ')||=||γ'||$ $||γ''\timesγ'||=||γ'||$ $||γ'\timesγ''||$, so $||\frac {dT}{dt}||=\frac {||γ'\times γ''||}{||γ'||^2} (3)$. Finally from $(2)$ and $(3)$ we have $||\frac {dT}{ds}||=(\frac {||γ'\times γ''||}{||γ'||^3})(t)$, which is by definition the curvature of $Γ$. q.e.d.

Answer 4

This is an old question, but I want to add my perspective to this. Overall, I dislike formulas, where neither the formula nor its proof are geometrically enlightening.

The first geometric thought is that curvature is the rate of change of the direction of the curve, but this is not a good definition because it depends on the speed you travel along the curve. So instead you define the curvature to be the rate of change of direction with respect to distance traveled along the curve. This clearly does not depend on any parameterization.

$\newcommand\R{\mathbb{R}}$ Suppose $\gamma: [0,T] \rightarrow \R^n$ is a curve and $\gamma'$ its velocity. Its speed is therefore $\sigma = |\gamma'|$ and the distance $s(t)$ from $\gamma(0)$ to $\gamma(t)$ is the integral of the speed, $$ s(t) = \int_{\tau=0}^{\tau=t} \sigma(\tau)\,d\tau, $$ i.e, $s' = \sigma$.

The direction of the curve at $\gamma(t)$ is $$ u(t) = \frac{\gamma'(t)}{|\gamma'(t)|} = \frac{\gamma'(t)}{\sigma}. $$ In other words, $$ \gamma' = \sigma u. $$ The acceleration is therefore $$ \gamma'' = \sigma' u + \sigma u' $$ The first term is obviously the acceleration in the direction of the curve, and the second, which is orthogonal to the first, is the rate of change of direction. Each term depends on the parameterization.

We now compare this to the unit speed parameterization. If $\hat\gamma: [0,s(T)] \rightarrow \R^n$ is the same curve parameterized by arclength, then $$ \gamma(t) = \hat\gamma(s(t)). $$

If $\hat{u}(s)$ is the direction of the curve at the point $\hat\gamma(s)$, then $$ u(t) = \hat{u}(s(t))\text{ and }\hat{u}'(s) = \hat\kappa(s)\hat u_2(s), $$ where $\hat{u}_2(s)$ is the unit vector in the direction of $\hat{u}'(s)$ and $\hat\kappa(s)$ is the curvature at $\hat\gamma(s)$. and therefore, $$ u'(t) = \hat{u}'(s(t))s'(t) = \sigma(t)\kappa(t)u_2(t), $$ where $\kappa(t) = \hat\kappa(s(t))$ is the curvature at $\gamma(t)$ and $u_2(t) = \hat{u}_2(s(t))$ is the unit vector in the direction of $u'(t)$. Solving for the curvature, we get, for any parameterization of the curve, $$ \kappa = \sigma^{-1}u_2\cdot u', $$ where $u(t)$ is the direction of the curve at $\gamma(t)$ and $\sigma(t)$ is the speed of the parameterization at $\gamma(t)$. Here, $u_2\cdot u'$ is equal to component normal to the direction of the curve of the rate of change of the direction with respect to the parameterization, and the $\sigma^{-1}$ factor is a normalization to make the curvature independent of the parameterization.