It is very well known that conics, spirals, etc. can represent a realistic trajectories of point particles. However, a physical trajectory can also intersect itself, have a cusp, and other kinds of singularities. Therefore, it is natural to assume that physical trajectories are a subset of algebraic curves. But an algebraic curve can consist of several disconnected curves (e.g. hyperelliptic curve) which cannot possibly describe a real trajectory of a particle (we, of course, forbid teleportation).

So, what would be the most general mathematical curve describing a physical trajectory?

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    $\begingroup$ why not a continuous function $x:[a,b]\subseteq\mathbb{R}\rightarrow\mathbb{R}^3$? $\endgroup$ – marco trevi Oct 10 '14 at 10:30
  • $\begingroup$ Surely some continuous paths are forbidden by Newton's Laws, and whatever conditions are applied to the environment, such as smooth gravitational vector fields, or wind, or whatever. $\endgroup$ – Harry Wilson Oct 10 '14 at 10:43
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    $\begingroup$ Depending on how you interpret the question, this question may or may not be answerable. Consider the motion of particles in a fluid. To find the most general trajectories, we need solve the most general equations of motion. To find the most general EoMs, we need the most general fields describing the force terms. To find the most general fields, we need to find the most general solution to Navier-Stokes. To solve the Navier-Stokes problem, you need to be a special kind of smart. Maybe someone smart can come along and answer this! $\endgroup$ – David H Oct 10 '14 at 10:47
  • $\begingroup$ @marcotrevi Sure, that sounds legitimate. But if we were to eliminate the parameter $x$ to get an implicit equation for a curve, what kind of a curve would it be? $\endgroup$ – user54031 Oct 10 '14 at 11:34
  • $\begingroup$ I don't get your comment... $\endgroup$ – marco trevi Oct 10 '14 at 11:48

If you give me machines that can generate forces, then the answer is "lots". Take your curve, and consider its projection $t \mapsto u(t)$ to the $x$ axis. Let's assume the curve describes the motion of a point-particle of mass $m$. By computing $u''(t)/m$, you can find the force, $F_x(t)$ that must be applied to the particle in the $x$-direction. If you assume, for example, that the particle is charged and floats between two nearly infinite planes whose voltage you can adjust arbitrarily, you just aadjust the voltage so that it varies as $F_x$. You do the same in the $y$-direction, perhaps by dragging large masses to be near/far from your particle, so that only gravitational forces act in the $y$ direction.

The only constraint here is that $u$ and $v$ are twice differentiable, which is essential if the first law is to hold: $F = ma$ makes no sense if $a$ is undefined.

Note that even simple force profiles like $F_x(t) = \sin t$, $F_y(t) = 1$ lead to non-algebraic curves, so the answer is certainly richer than just "algebraic curves".

By the way, the image of $(u, v)$ may contain cusps, but the plot of $(t, u(t), v(t))$ will generally not, except at places where $u'$ and $v'$ are both zero.

All this describes idealized physics of a kind of Physics 1 world (in which point particles exists), and ignores relativistic effects completely. Otherwise I'd need to say that $u'(t)^2 + v'(t)^2 < c^2$ for all $t$, where $c$ is the speed of light, and (possible) some other things as well.

  • $\begingroup$ Is there a geometric condition on a curve that excludes the possibility of disconnected loops? I think the most general curve without this kind of loops would then be an adequate trajectory. $\endgroup$ – user54031 Oct 10 '14 at 11:29
  • $\begingroup$ A curve with gaps would necessarily be discontinuous at the gap-points, hence would not be twice-differentiable. (Differentiable -> continuous !) $\endgroup$ – John Hughes Oct 10 '14 at 11:30
  • $\begingroup$ What do you mean by gaps? Does hyperelliptic curve have gaps? $\endgroup$ – user54031 Oct 10 '14 at 11:32
  • $\begingroup$ When you say "disconnected loops", I assumed you meant something like "a curve that traverses the unit circle and then jumps to the right by 10 units and traverses a unit circle there." At the "jump", there's a clear discontinuity in $u$. If you mean something like the appearance of "ee" when written in cursive, that, or something very like it, is certainly a possible path for a particle in a time-varying force field. $\endgroup$ – John Hughes Oct 10 '14 at 11:59
  • $\begingroup$ Yes, I meant it exactly the way you said it. Is it possible that such kind of loops can't be parametrized by e.g. $(x(t),y(t))$ and are therefore not physical? $\endgroup$ – user54031 Oct 10 '14 at 12:02

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