Understanding the formula for curvature

Question

In this khan academy article, they discuss how can define curvature as,

$$ \bigg\| \frac{dT}{dS} \bigg\| = \kappa$$

In the post they write,

"However, we don't want differences in the rate at which we move along the curve to influence the value of curvature since it is a statement about the geometry of the curve itself and not the time-dependent trajectory of whatever particle happens to be traversing it. For this reason, curvature requires differentiating T(t) with respect to arc length, S(t), instead of the parameter t"

I feel this is not a sufficient explanation and more explanation is needed to clarify the formula. As it just states a reason that 'the curvature should related to the arclength (geometrical quantity) rather than velocity or time'.

This doesn't really help because that is ruling out other quantities which we could have taken derivative with respect to. How would we motivate that when speaking of curvature of the intuitive idea of curvature (how much you need to turn) as the above equatoion?

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And, even after all this one issue remains for me still, we define unit tangent vector using parameterizations, so the tangent vector in itself is reliant on a property outside the curve. So technically speaking curvature is not fully made of properties intrinsic to curve.

Refrence: $$ T = \frac{ v(t)}{|v(t)|} $$

Answer 1

You are free to derive a formula for curvature in any coordinate system that you want, and with respect to any parameter along the curve that you want. For example, you have probably also seen a formula, expressed in terms the $x$-coordinate parameterization $(x,f(x))$, for the curvature of the graph of a function $y=f(x)$: $$\kappa = \frac{|f''(x)|}{(1+f'(x))^{3/2}} $$ So the question is not why "all other alternatives are shut down", because they aren't (and by the way "shutting things down" is almost never how mathematics works).

Perhaps instead a better question might be

Why is the arc length parameterization the primary one used to express the formula for curvature?

I think the answer to this is simply that the arc length parameterization is so natural from a geometric viewpoint: it can be derived using nothing but Euclidean geometry and a limiting argument, as you learn in a real analysis course. So it might be the first thing a geometer would want to know about curvature: How do you write down a formula for curvature expressed in terms of the arc length parameter?

But let me suggest two still better questions:

Is there a definition of curvature independent of parameterization? And can one use that definition to derive to derive a formula in terms of the arc length parameteriation (or in terms of any other parameterization)?

There is indeed a nice definition which is independent of parameter, and it has three steps:

1. The unit circle $S^1 = \{(x,y) \mid x^2+y^2=1\}$ has curvature $1$ at each point:
2. Curvature varies inversely under similarity: Suppose $C$ and $C'$ are two curves such that $C$ is similar to $C'$. Let $f : \mathbb R^2 \to \mathbb R^2$ be a similarity map such that $f(C)=C'$. Let $r>0$ be the similarity factor, meaning that $d(f(p),f(q)) = r d(p,q)$ for all $p,q \in \mathbb R^2$. Then for all $x \in C$ with corresponding point $x' = f(x) \in C'$, the curvature of $C'$ at $x'$ is equal to $\frac{1}{r}$ times the the curvature of $C$ at $x$. (For example, by combining 1 and 2 one can prove easily that all radius $1$ circles have curvature $1$ at each point, and all radius $r$ circles have curature $1/r$ at each point.
3. Curvature is a second order invariant: For any curve $C$ and $p \in C$, and for any circle $C' \subset \mathbb R$ which matches $C$ to second order at the point $p$, the curvatures of $C$ and $C'$ at $p$ are equal (this is the "osculating circle" condition referred to in the comment of @Kajelad).

Knowing this, one can prove the arc length parameterization formula for curvature, and any other formula you want such as the formula for the $x$-coordinate parameterizaiton given earlier.

Answer 2

We would want any definition of curvature to make intuitive sense when applied to straight lines and circles, so the curvature of a straight line should be 0, and the curvature of a circle of radius $r$ should be $1/r$. A way to define curvature then would be to find the "tangent circle" (if it exists) at each point, then the curvature would be the reciprocal of the radius of this "tangent circle". It turns out that the equations needed to derive the tangent circle are simplified if the tangent vector at each point of the curve has length $1$, which is the case only if the curve is parameterised by arc length.

Answer 3

Let's address and clarify a few aspects of the post. First, it's written:

"I feel this is not a sufficient explanation and more explanation is needed to clarify the formula."

Hopefully, the geometric meaning of the formula $\kappa = \left\Vert \frac{dT}{ds} \right\Vert$ is clear: $\kappa$ is the infinitesimal rate of change of the tangent vector per unit arc length. I think that captures the intuitive notion of "curvature" as "change in direction" pretty well.

But the OP's question seems to be more along the lines of: "Why is $\kappa = \left\Vert \frac{dT}{ds} \right\Vert$ so fundamental, when we could have defined it some other way?" The OP writes:

"This doesn't really help because that is ruling out other quantities which we could have taken derivative with respect to."

The response to this is that there essentially aren't other quantities with respect to which we could have differentiated. In other words, curves can always be reparametrized in terms of arc length $s$, and hence any function (or variable) defined on the curve can be expressed in terms of arc length.

(As an aside, and as others have mentioned, there are other (equivalent) definitions of the curvature $\kappa$ out there, such as the definition via osculating circles.)

Beyond this, the OP has another concern, writing:

"[O]ne issue remains for me still, we define unit tangent vector using parameterizations, so the tangent vector in itself is reliant on a property outside the curve ... So technically speaking curvature is not fully made of properties intrinsic to curve."

Here, it's worth mentioning that a geometric quantity being "intrinsic" or "extrinsic" is a completely different question from whether that quantity is parametrization-dependent. In other words, there are two different things that could be meant by "outside the curve." One could mean:

• (a) The unit tangent vector $T$ is dependent on the parametrization of the domain of the curve --- which is not true (if one ignores orientation) --- or
• (b) The unit tangent vector $T$ is an extrinsic geometric quantity (rather than an intrinic one), in that its definition depends on having a parametrization. (This is true. Indeed, all geometric properties of curves are extrinsic. Surfaces, on the other hand, have both intrinsic and extrinsic geometric properties.)

Answer 4

When we go from $\vec{T}(t)$ to $\frac{d\vec{T}(t)}{dt}$, instead of finding how fast the"direction vector" $($or $\vec{T}(t))$ will change compared to it previously went, which is what we want, it finds how fast the direction vector will change compared to the constantly increasing $t$ value, which is not what we want for a local change in direction. For example, in a quickly decreasing slope in a function $(t,f(t))$, when we make the change from $\vec{T}(t)$ to $\frac{d\vec{T}(t)}{dt}$, the slope goes by quicker because the $t$ is basically $x$. However, since we take $\frac{d\vec{T}(t)}{dt}$, we find "how fast the direction will change as time goes by" instead of "how fast the direction will change as we extend the arc". In our quickly decreasing function, the time spans too much of the arc. When we have $||\frac{d\vec{T}(t)}{dS}||$, you have to remember the $dS$ part itself is the arc length amount greater than $dt$.

Answer 5

I found the most understandable explanation in page-98 of Tristan Needham's Visual Differential Geometry book:

If a bead of unit mass is launched at unit speed along a wire in the form of a plane curve, the curvature κ of that curve is the force F, directed perpendicularly to the curve, exerted by the wire on the bead.

And wow! Everything makes sense now! As a practical example, look back at the centripetal equation formula for circle $$ F= \frac{mv^2}{R}$$ If $m=v=1$ and noting that $\kappa = \frac{1}{R}$, we find:

$$ F= \kappa$$