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Given a space curve $\alpha(s)$, parametrised by arc length (this is $|\alpha'(s)|\equiv 1$), we define its curvature to be: $$\kappa (s)=|\alpha''(s)|$$

Now, I've read that given a curve $\gamma(t)$ not necessarily parametrised by arc length, its curvature is given by the formula:

$$\kappa (t)=\frac{|\gamma'(t) \times \gamma''(t)|}{|\gamma(t)|^3}$$

When trying to prove this, I can't get the variables $s$ and $t$ to match up on both sides of the equation, so if someone could pin point and explain my mistake, that would be great. My reasoning is the following:

If we think of $\gamma(s)$ as $\gamma(t(s))$, then deriving with respect to $s$ gives, by the chain rule:

$$\frac{d}{ds}\gamma(s)=\gamma'(t(s))t'(s)$$

$$\frac{d^2}{ds^2}\gamma(s)=\gamma''(t(s))(t'(s))^2 + \gamma'(t(s))t''(s)$$

It follows that:

$$\left (\frac{d}{ds}\gamma(s) \right )\times \left (\frac{d^2}{ds^2}\gamma(s) \right )=[\gamma'(t(s))\times \gamma''(t(s))](t'(s))^3$$

But we also have:

$$\left \| \left (\frac{d}{ds}\gamma(s) \right )\times \left (\frac{d^2}{ds^2}\gamma(s) \right ) \right \|=\left \| \left (\frac{d}{ds}\gamma(s) \right ) \right \| \left \|\left (\frac{d^2}{ds^2}\gamma(s) \right ) \right \|\sin \left ( \left (\frac{d}{ds}\gamma(s) \right ),\left (\frac{d^2}{ds^2}\gamma(s) \right ) \right )$$

$$=\left \|\frac{d^2}{ds^2}\gamma(s) \right \|=\kappa (s)$$

Putting this all together, and using that $t'(s)=\frac{1}{|\gamma'(t(s))|}$ we get:

$$\kappa(s)=\frac{\left \| \gamma'(t(s)) \times \gamma''(t(s))\right \|}{\left \| \gamma '(t(s))\right \|^3}$$

Not $t=t(s) \iff s=s(t)$, hence:

$$\kappa(s(t))=\frac{\left \| \gamma'(t) \times \gamma''(t)\right \|}{\left \| \gamma '(t)\right \|^3}$$

The right side of this last equation matches what I wanted to prove, but it should be just $t$ instead of $s(t)$ on the left side. Where did I go wrong?

Thanks in advance!

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  • $\begingroup$ Can you explain how the sine part cancels with $\|\frac d{ds}\gamma(s)\|$? Anyway, I think the curvature is defined at a point of the curve and the corrsesponding point is $\gamma(t)=\alpha(s)$, if I understand correctly, and thus your result seems OK. $\endgroup$
    – Berci
    Commented Sep 6, 2015 at 18:46
  • $\begingroup$ @Berci Since $\left \|\frac{d}{ds}\gamma(s)\right \| \equiv 1$, in particular it has constant norm, hence it is orthogonal to its tangent vector, this is $\frac{d}{ds}\gamma(s)$ and $\frac{d^2}{ds^2}\gamma(s)$ form an angle of $\pi/2$ radians. $\endgroup$
    – Reveillark
    Commented Sep 6, 2015 at 18:53
  • $\begingroup$ Ah yes. The notation a bit confused me, so you say $\gamma(s)=\gamma(s(t))=\alpha(s)$? $\endgroup$
    – Berci
    Commented Sep 6, 2015 at 19:13
  • $\begingroup$ @Berci The first curve $\alpha$ has nothing to do with $\gamma$, I was just stating the definition of curvature. What I was saying later is that $\gamma (s)$ can be thought of as $\gamma (t(s))$. We define the arc-length as a function $s(t)$ which has positive derivative, and therefore admits an inverse $t(s)$. To paramatrise $\gamma(t)$ by its arc length $s$ by substitute $t$ by $t(s)$, so we get $\gamma(s)=\gamma (t(s))$. The notation is rather confusing to me too. $\endgroup$
    – Reveillark
    Commented Sep 6, 2015 at 19:23

1 Answer 1

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Your calculation is fine.

Observe that $\kappa $ is defined in terms of the arc length parameterization and so if we are given any other parameterization $\vec r :[a,b]\to \mathbb R^3$ then

$s_0=s(t_0)=\int_{a}^{t_0}\left \| \vec r'(u) \right \|du$ so we can write

$\kappa (s_0)=|\alpha''(s_0)|\Rightarrow \kappa (s_0)=|\frac{\mathrm{d^{2}} }{\mathrm{d} s^{2}}(\alpha(s_0))|=|\frac{\mathrm{d^{2}} }{\mathrm{d} s^{2}}(\alpha(s(t_0)))|$ which says that

$\kappa (s(t_0))=\left | \frac{\mathrm{d^{2}} }{\mathrm{d} s^{2}}(\alpha(s(t_0))) \right |$.

This is true before you do any calculation at all.

The glitch is that the RHS is a second derivative with respect to $s$. Your calculation simply expresses it as a function of $t$, using the chain rule.

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