In $\mathbb R^n$ the defintion of curvature of a smooth regular curve $\gamma : \mathbb R \to \mathbb R^n$ is $$ \kappa (t) = \|\gamma''(t)\| / \|\gamma '(t)\|$$

In $\mathbb R^2$ the definition for the curvature of an arbitrary smooth regular curve $\gamma : \mathbb R \to \mathbb R^2$ given as $\gamma (t) = (x(t),y(t))$ is

$$ \kappa (t) = {x'(t) y''(t) - x'' (t) y' (t) \over (x^{'2} + y^{'2})^{3/2}}$$

I assumed these should be equal because I thought the second formula was simply for convenience and not in fact different. But calculating a simple example says otherwise:

If $\gamma (t) = (2 \cos t, \sin t)$ is an ellipse then

$$ \|\gamma ''\|/\|\gamma'\| = {\sqrt{4 \cos^2 t + \sin^2 t} \over \sqrt{4 \sin^2 t + \cos^2 t}}$$


$$ \kappa(t) = {2\over (4 \sin^2 t + \cos^2 t)^{3/2}}$$

Similarly, a different curvature for $\gamma (t) = (t,t^2)$.

Why is the curvature in $\mathbb R^2$ defined differently than for $\mathbb R^n$? It seems bizarre to me that they are not equal.


2 Answers 2


Your first formula is wrong. Or, rather, it is only correct when $||\gamma'(t)||$ is constant. Otherwise, its value depends of parametrization: for example, that formula gives non-zero curvature even for a straight line parametrized as $(t^3+t,0)$.


The (not oriented) curvature of $\gamma$ is the length of the derivative of its unit speed vector in respect of arclength, i.e., the length of $$\frac{1}{\|\gamma'\|}\left(\frac{\gamma'}{\|\gamma'\|}\right)'.$$ It’s easy to calculate this length as $$\frac{1}{\|\gamma'\|^3}\sqrt{\|\gamma'\|^2\|\gamma''\|^2-\langle \gamma',\gamma''\rangle^2},$$ which equals $$\frac{1}{\|\gamma'\|^2}\sqrt{\|\gamma''\|^2-\langle\frac{\gamma'}{\|\gamma'\|},\gamma''\rangle}.$$ As the radicand is the Gram-matrix of $\gamma''$ and $\gamma'/\|\gamma'\|$ the curvature admits the following interpretation: It is the area of the parallelogram spanned by the unit speed vector and the acceleration, divided by the squared velocity.

Explanation: the result of the calculation shows that the notion of curvature being the change of the unit speed vector in respect to arclength is in fact the same as Euler's idea of curvature: the normal component of the acceleration divided by the square of the velocity.

The main difference regarding $R^2$ vs. $R^n$ for $n>2$ is that we can't naturally establish an orientated curvature.

Edit: In case of $R^2$ we easily achieve that $$\sqrt{\|\gamma'\|^2\|\gamma''\|^2-\langle \gamma',\gamma''\rangle^2}=\sqrt{(x'y''-x''y')^2}=|x'y''-x''y'|.$$

  • $\begingroup$ Thank you for your answer. I believe that orientation would be represented by a sign. As a consequence taking the absolute value of both definitions of curvature we should get the same result. But we do not. Or am I missing something? $\endgroup$
    – student
    May 9, 2015 at 4:25
  • $\begingroup$ See my edit, please. $\endgroup$ May 9, 2015 at 10:25

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