Curvature of curve not parametrized by arclength 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?

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.
