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!
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.