# Is there a reason why curvature is defined as the change in $\mathbf{T}$ with respect to arc length $s$

And not with respect to time $t$? (or whatever parameter one is using)

$\displaystyle |\frac{d\mathbf{T}(t)}{\mathit{dt}}|$ seems more intuitive to me.

I can also see that $\displaystyle |\frac{d\mathbf{T}(t)}{\mathit{ds}}| = |\frac{d\mathbf{r}'(t)}{dt}|$ (because $\displaystyle |\mathbf{r}'(t)| = \frac{ds}{dt}$, which does make sense, but I don't quite understand the implications of $\displaystyle |\frac{d\mathbf{T}(t)}{\mathit{dt}}|$ vs. $\displaystyle |\frac{d\mathbf{T}(t)}{\mathit{ds}}|$ and why the one was chosen over the other.

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You can define curvature without reference to any parametrization if you want. The idea being that the tangent line is the line that intersects the curve and is a good approximation in a 1st order sense. You can then talk about circles that intersect the curve and agree with it in a 2nd order sense. I believe this is called the osculating circle. The radius is $1/\kappa$ where $\kappa$ is the curvature. This is a perfectly good definition. – Ryan Budney Oct 20 '11 at 21:16
What does it mean to divide $T(t)$ by $ds$? Intuitively you are dividing a normal-sized vector by an infinitesimal scalar, which doesn't make sense to me... – user7530 Oct 20 '11 at 21:18
typo, meant dT. Curvature as a function of the parameter t. – iDontKnowBetter Oct 20 '11 at 21:27
You want the notion of curvature to be independent of the parametrization. That is, it should only depend on the image of the curve and not the function defining it. Your definition of curvature depends on the parametrization. – Eric O. Korman Oct 20 '11 at 21:33
@Eric hm.. let me see if understand: if it's independent, then, say (in physical terms) for a path travelled by an object in 2 minutes, and then by an object that travels the same path in 4 minutes, we'd get the same curvature. — but with my equation, the "curvature" would appear different because it would depend on the time t that it took to travel? – iDontKnowBetter Oct 20 '11 at 21:39

The motivation is that we want curvature to be a purely geometric quantity, depending on the set of points making up the line alone and not the parametric formula that happened to generate those points.

$\left|\frac{dT}{dt}\right|$ does not satisfy this property: if I reparameterize by $t\to 2t$ for instance I get a curve that looks exactly the same as my original curve, but has twice the curvature. This isn't desirable.

$\left|\frac{dT}{ds}\right|$ on the other hand has the advantage of being completely invariant, by definition, to parameterization (assuming some regularity conditions on the curve).

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Thanks. Since it measures something other than curvature, is there a name/use for what $T$ with respect to $t$ yields? – iDontKnowBetter Oct 20 '11 at 21:45

I am trying to determine what your question is, and I hope this answers what you are trying to ask.

The tangent and the curvature are quantities that describe something about the curve and not about the parameterization of that curve. To avoid dependency on a particular parameterization, we want notions that look only at the curve as a figure in space. The direction of a curve is one such notion. We can compute the direction by computing the change in position vs a change in distance along the curve. This gives $$T(t)=\frac{\dot{\gamma}(t)}{|\dot{\gamma}(t)|}\tag{1}$$ This leads to the notion of a natural parameterization for a curve: parameterization by arc-length. Arc-length, being the distance measured along the curve, is a function of the figure in space, so using it as a parameter, removes the problems of an arbitrary parameterization. Note that $$\frac{\mathrm{d}s}{\mathrm{d}t}=|\dot{\gamma}(t)|\tag{2}$$ Thus, we can use $(2)$ to take the derivative with respect to arc-length: $$\frac{\mathrm{d}\hphantom{t}}{\mathrm{d}s}=\frac{1}{|\dot{\gamma}(t)|}\frac{\mathrm{d}\hphantom{s}}{\mathrm{d}t}\tag{3}$$ If we apply $(3)$ to the curve itself, we get that \begin{align} \frac{\mathrm{d}\gamma}{\mathrm{d}s} &=\frac{1}{|\dot{\gamma}(t)|}\frac{\mathrm{d}\gamma}{\mathrm{d}t}\\ &=\frac{\dot{\gamma}(t)}{|\dot{\gamma}(t)|}\tag{4} \end{align} Notice that $(1)$ and $(4)$ show that $$\frac{\mathrm{d}\gamma}{\mathrm{d}s}=T(t)\tag{5}$$ In $\mathbb{R}^2$, it can be shown that the radius of a curve is the ratio of change in distance along the curve over the change in direction. In $\mathbb{R}^3$, direction is a vector, as seen in $(1)$. Therefore, we look at the reciprocal of the radius, the curvature, which is the change in direction over the change in distance along the curve: \begin{align} \kappa &=\frac{\mathrm{d}T}{\mathrm{d}s}\tag{6a}\\ &=\frac{\mathrm{d}^2\gamma}{\mathrm{d}s^2}\tag{6b} \end{align} Because $(5)$ and $(6)$ are defined in terms of the shape in space, they are not dependent on a particular parameterization.

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The problem is that if you define it in the "more intuitive way" the curvature depends on the parametrisation. We calculate the curvature of a curve, not of a parametrisation.

Simple question: how do you calculate the curvature of the hyperbola $y^2-x^2=1$ at let's say $(0,1)$?

You could

a) solve for $y$ and use $x=t, y= \sqrt{t^2+1}$.

b) use $x= \tan(t), y= \sec(t)$,

c) use $y= \cosh(t) , y= \sinh(t)$.

Each of these lead to a different $\displaystyle \left\vert\frac{d\mathbf{T}(t)}{\mathit{dt}}\right\vert$. So which one would you pick as the curvature of $y^2-x^2=1$?

Intuitively, when we parametrise a curve, we are basically describing the curve as being the trajectory of a particle, by describing its coordinates at time $t$. I always think about different parametrisations as being different particles moving on the same curve/trajectory.

If a particle moves faster, the calculation of $\displaystyle \left\vert\frac{d\mathbf{T}(t)}{\mathit{dt}}\right\vert$ also takes into acount its velocity. If I am not mistaken, in the arc lenght parametrisation, we simply pick the particle which covers 1 unit of the curve per unit of time. Basically we decide that the speed is constantly 1. This is the most "natural" choice you can make.

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Thanks, makes sense. – iDontKnowBetter Oct 20 '11 at 21:43