# Are there closed curves for which acceleration is orthogonal to position?

Can we find $\vec{f} : \mathbb{R}\rightarrow \mathbb{R}^3$ such that $\vec{f}(t) \cdot \frac{d^2 \vec{f}(t)}{dt^2} =0$ and $\vec{f}(0) = \vec{f}(T)$ for some $T >0$ ? Exclude the trivial cases. I want $\frac{d \vec{f}}{dt} \neq 0$ for some part of the trajectory.

Define $$x(t) = C_1e^t\cos t + C_2e^t\sin t + C_3e^{-t}\cos t + C_4e^{-t}\sin t$$ and $$y(t) = C_2e^t\cos t - C_1e^t\sin t - C_4e^{-t}\cos t + C_3e^{-t}\sin t$$ for some constants $C_1,C_2,C_3,C_4 \in \mathbb{R}$.
Differentiating twice yields $$x''(t) = -2C_1e^t\sin t + 2C_2e^t\cos t + 2C_3e^{-t}\sin t - 2C_4e^{-t}\cos t = 2y(t)$$ and $$y''(t) = -2C_2e^t\sin t - 2C_1e^t\cos t - 2C_4e^{-t}\sin t - 2C_3e^{-t}\cos t = -2x(t)$$
So if we set $f(t) = \begin{bmatrix}x(t)\\y(t)\\0\end{bmatrix}$, then $f''(t) = \begin{bmatrix}2y(t)\\-2x(t)\\0\end{bmatrix}$, and thus, $f(t) \cdot f''(t) = 0$, as desired.
If we pick $C_1 = 1$, $C_2 = 1$, $C_3 = -e^{\pi}$, and $C_4 = e^{\pi}$, we have $x(0) = x(\pi)$ and $y(0) = y(\pi)$. Then, $f(0) = f(\pi)$, as needed.
Finally, it is easy to see that $f'(t) \neq 0$ for most values of $t$.
• Thanks, but now I'm wondering if there is a smooth periodic motion that satisfies all the criteria as well. That is, $f^{(n)}(0) = f^{(n)}(T)$ for all positive integer $n$. Jul 10, 2015 at 1:01
• ^No there is not. Note that $\dfrac{d^2}{dt^2}[\|f\|^2] = \dfrac{d^2}{dt^2}[f \cdot f] = 2f \cdot f'' + 2 f' \cdot f' = 2\|f'\|^2 \ge 0$. Hence, $\|f(t)\|^2$ is concave up with respect to $t$, and thus, $f(t)$ can only attain any particular value for two values of $t$. (Unless of course $f' \equiv 0$, which is a trivial case). Jul 10, 2015 at 2:39