Consider a particle moving in some curve $\gamma(t)$ , p.t at points of zero curvature $\dot{ \gamma} \cdot \ddot{\gamma (t) }=0$, assume $\frac{d}{dt} | \dot{\gamma} | \neq 0$ for all $t$

I know curvature is defined as:

$$ \kappa = \frac{|a \times v|}{|v|^3}$$

setting $\kappa=0$, I get $|a \times v | =0$, expanding the vector into curvilinear basis:

$$| \left[ a_{\tau} \tau + a_n n \right] \times \left[ v_{\tau} \tau \right]| = 0$$

This means:

$$ (a_n) ( v_{\tau}) n \times \tau = 0$$

Since $v_{\tau}>0$, this means $a_n=0$ suggesting there is no normal acceleration.. this seems to contradict my intuition.

What I really wanted to prove is the following: enter image description here

Here note that near the 'flat point' of the curve, we see that the acceleration and velocity vector must be perpendicular, so I wanted to show it generally.. but the above result contradicts it. I say flat point, because I think the describing feature of the 'valley' point here is that the curvature vanishes (?).

If I have made some conceptual mistake, please comment so I can edit the question.



1 Answer 1


Unfortunately, this is a case where you need to "bite the bullet": your intuition is wrong.

If $\kappa=0$, then $\dot{\gamma}\times\ddot{\gamma}=0$. In 2D, if \begin{gather*} \dot{\gamma}\times\ddot{\gamma}=0 \\ \dot{\gamma}\cdot\ddot{\gamma}=0 \end{gather*} then one of $\dot{\gamma}$ and $\ddot{\gamma}$ must be zero; it is certainly not the former.

In the picture you draw, the labeled point is not "flat"; a tangent circle through that point of sufficiently large radius intersects the curve elsewhere. In fact, your drawn curve looks a lot like a plot of $t\mapsto(t,\frac{1}{2}t^2)$ around $t=0$; at that point, the curvature is precisely $1$.

For an example of a flat curve, consider $t\mapsto(e^t-1,(e^t-1)^4)$ around $t=0$. (A quick plot) There the velocity is $(1,0)$, the acceleration is also $(1,0)$, so that $\dot{\gamma}\cdot\ddot{\gamma}=1$.

On the other hand, if a curve always has constant velocity in a particular direction (say, along the $x$-axis…), then any acceleration must point in a perpendicular direction. So flat points on the curve are places where the acceleration vanishes, and then one can achieve $\dot{\gamma}\cdot\ddot{\gamma}=0$, since the latter factor is just $0$.

  • $\begingroup$ Could you explain what's going on in here then? How else can you explain the arguement given mathematically $\endgroup$ Jun 19, 2021 at 11:16
  • $\begingroup$ @Buraian: That's another not-flat point. There's two different definitions of flat being used here. A function is defined to be flat around an input if its first derivative is zero there. This occurs at extrema, like the lowest point on a graph (in this case, the skier's trajectory). You'll notice that the graph is locally curved: the skier is changing direction. A curve is flat at a point when the curvature is zero there — the curve looks locally like a straight line. $\endgroup$ Jun 19, 2021 at 11:22
  • $\begingroup$ Thanks Jacob, that made sense. I'm not sure which flat I am supposed to use then, but what could be an arguement for the mentioned point in the book using mathematics? @Jacob Manaker I'll post another question if you can guide me on the correct terminology $\endgroup$ Jun 19, 2021 at 11:39
  • $\begingroup$ Oh, I see you already posted while I was asleep. $\endgroup$ Jun 19, 2021 at 20:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.