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 $\endgroup$ Commented 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. $\endgroup$ Commented 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 $\endgroup$ Commented Jun 19, 2021 at 11:39
  • $\begingroup$ Oh, I see you already posted while I was asleep. $\endgroup$ Commented Jun 19, 2021 at 20:30

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