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I was reading this and this posts. From what I gather

  • In 2D: Constant speed $||\dot x||=const$ and constant positive magnitude of the acceleration $||\ddot x|| = const$ imply circular motion.
  • In 3D: The same assumptions can give rise to helical motion.

Hence my questions:

  1. Q1: Suppose $x\in\mathbb{R}^n$ depends on time $t\in\mathbb{R}_+$. Given the constraints $\|\dot x\| = const$ and $\|\ddot x\|=const\neq 0$ does this give rise to a unique motion/dynamics? (e.g. circular in 2D, helical in 3D, etc)
  2. Q2: If this motion is unique in $\mathbb{R}^n$, is there a name for it? (a general name, not circular or helical, which are specific to 2D or 3D).
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    $\begingroup$ Why would constant acceleration imply circular motion? What about the case $a=0$? $\endgroup$ Commented Jul 14, 2021 at 21:29
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    $\begingroup$ @Jakob You are right, I mean 1) constant speed 2) acceleration with positive and constant magnitude. All I know that constant speed $\|\dot x\|=const$ implies acceleration perpendicular to velocity $\dot x \perp \ddot x$. $\endgroup$ Commented Jul 14, 2021 at 21:34
  • $\begingroup$ @Jakob edited my answer so that now I explicitly consider the case where $\|\ddot x\| =const \neq 0$ $\endgroup$ Commented Jul 14, 2021 at 21:39
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    $\begingroup$ just wanted to mention that I added some edits in my post, I hope it can help you clarify things a bit. $\endgroup$ Commented Jul 16, 2021 at 23:37

2 Answers 2

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For a motion with constant $||\dot x||$ in $n$ dimensions the velocity vector travels on an $(n-1)$-sphere and its "velocity", $\ddot x\,,$ (what we call acceleration) has a constant magnitude (not a constant direction). In 3D I can imagine many trajectories on the $2$-sphere that have constant velocity and acceleration and are not helical. Helical means that one component of $\dot x$ is constant. This corresponds to a circle on the $2$-sphere which is a very special case.

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    $\begingroup$ You probably mean that in the strictly helical case the unit tangent traverses a non-great circle of the unit 2D sphere. When the helix degenerates to a circle, then the unit tangent traverses a great circle. $\endgroup$ Commented Jul 17, 2021 at 0:48
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    $\begingroup$ Right ! Thanks for careful reading. The great circles are those where one component of $\dot x$ is zero (one component of $x$ is constant). Such motions live in $\mathbb R^{n-1}$ (one dimension lower). Will edit the answer. $\endgroup$
    – Kurt G.
    Commented Jul 17, 2021 at 4:04
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  1. Q1. The statement is not true. In 3D, if you impose the restrictions $$\|\dot{x}\| = 1 \,\,\, \text{ and } \,\,\, \|\ddot{x}\| = \kappa_0$$ then there are infinitely many curves with this property that are very far from being helixes or circles and having differing geometric properties.

This follows for example from the canonical Frenet-Serret frame of a smooth curve and the corresponding Frenet-Serret differential equations. Here is the link to an old post of mine about the Frenet-Serret frame and the equations. You can also check out Wikipedia on the topic. The Frenet-Serret frame and the corresponding equations are the motor that drives the so called Fundamental Theorem of the Differential Geometry of 3D Smooth Curves. According to this theorem, a 3D smooth curve is determined uniquely, up to translation and rotation in 3D space, by two functions: the curvature, $\kappa(t)$ and the torsion $\tau(t)$. What you are asking in your question is about the possible geometries of 3D smooth curves under two constraints.

(1) The condition $\|\dot{x}\| = 1$ is equivalent to the fact that the curve $x = x(t)$ is arc-length parametrized, which can always be arranged, so this is not a constraint on the curve's geometry. It is just a very useful coordinate choice of parametrization of the curve. In fact, when studying the geometry of curves, arc-length parametrization is most of the time assumed, in order to simplify calculations and expressions.

(2) The second condition $\|\ddot{x}\| = \kappa_0$ is a true geometric constraint and is equivalent to the fact that the curvature is set to a constant $\kappa(t) = \kappa_0$ (here we are assuming that the curve is arc-length parametrized, so condition (1) is assumed).

However, even after imposing these two restrictions, you are left with the freedom to select any (let's say smooth) function $\tau(t)$ as torsion. If you select the torsion to be zero $\tau(t) = 0$, then you obtain a circle. If you set the torsion to a non-zero constant, you get a helix. But if you choose something much more involved, then... well I think you get my point.

Let me put it this way, under the two constraints of arch-length parametrization and constant curvature, the space of all curves with these two properties is parametrized by the space of, say smooth functions $\tau(t)$ which is an infinite dimensional space. For each choice of torsion function, you get a curve with a unique geometry, up to rotation and translation.

So If things do not work in 3D, then imagine the complexity in higher dimensions.

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