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?

Here is sort of a simple example. Consider $\lambda_1,\lambda_2>1$ with $\lambda_1\not=\lambda_2$. Consider the unit disk. Both of the curves $(\cos(\lambda_1 t),\sin(\lambda_1t))$ and $(\cos(\lambda_2 t),\sin(\lambda_2t))$ trace it out. It makes intuitive sense to define the curvature as the rate of change of the velocity vector. But doing so without first parametrizing with respect to arc length gives two different curvatures for the same shape.
The answer to your first question is no. The curvature of a curve that isn't unit speed is defined to be the curvature of that curve parametrized by arc length. This is justified as the original curve and its parametrization trace out the same shape.
