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Consider a unit-mass particle that is always experiencing a single unit-magnitude force towards the origin. This is a central force, but it is not one of the familiar ones, e.g. gravity whose magnitude is proportional to inverse distance squared, or a spring force whose magnitude is proportional to distance.

So the particle is always accelerating towards the origin with constant acceleration magnitude $1$. Stated as a differential equation, working in the $x$-$y$ plane, the particle's position as a function of time $\mathbf{p}(t){=}(x(t),y(t))$ satisfies: $$\ddot{\mathbf{p}}(t) = -\mathbf{p}(t)/\Vert\mathbf{p}(t)\Vert.$$ If we are additionally given initial position $\mathbf{p}(0)$ and velocity $\mathbf{v}(0)=\dot{\mathbf{p}}(0)$, then the function $\mathbf{p}$ is completely determined, and it can be easily computed numerically to any desired accuracy by simply iterating the following with small enough timestep $dt$: \begin{align} \mathbf{a} &\leftarrow -\mathbf{p}/\Vert\mathbf{p}\Vert \\ \mathbf{v} &\leftarrow \mathbf{v} + \mathbf{a}\,\,dt \\ \mathbf{p} &\leftarrow \mathbf{p} + \mathbf{v}\,\,dt \end{align}

My question: is $\mathbf{p}(t)$ a well-known function, and does it have a closed form?

Of course one case of this is a simple circular orbit of unit radius and speed: $$\mathbf{p}(0){=}(1,0), \,\, \mathbf{v}(0){=}(0,1) \,\,\Rightarrow\,\, \mathbf{p}(t)=\left(\cos t,\sin t\right).$$ More generally, a uniform circular orbit of any radius $r$ and speed $\sqrt{r}$ can be obtained: $$\mathbf{p}(0){=}(r,0), \,\, \mathbf{v}(0){=}(0,\sqrt{r}) \,\,\Rightarrow\,\, \mathbf{p}(t)=\left(r \cos\frac{t}{\sqrt{r}},r \sin\frac{t}{\sqrt{r}}\right).$$ and we check that the desired equation holds: \begin{align} \dot{\mathbf{p}}(t) &= \left(-\sqrt{r} \sin \frac{t}{\sqrt{r}}, \sqrt{r} \cos \frac{t}{\sqrt{r}}\right) \\ \ddot{\mathbf{p}}(t) &= \left(-\cos \frac{t}{\sqrt{r}}, -\sin\frac{t}{\sqrt{r}}\right) \\ &= -\mathbf{p}(t)/\Vert\mathbf{p}(t)\Vert. \end{align}

Another simple case is when the initial velocity is zero or collinear with the position and the origin; in this case it's a 1-dimensional problem and the position can easily be seen to be a simple piecewise quadratic function of time.

But what if the initial conditions are not so nicely aligned?

To get an idea of the shapes that are possible, I've made some plots, using gnuplot, of simulations using the simple evolution algorithm I described earlier, with $dt = 1/10000$ degree $\approx .00000175$.

Figure 1 shows five different initial states, each evolved from $t{=}0$ to $t{=}2 \pi$: $\,\,\mathbf{p}(0){=}(1,0)$, $\mathbf{v}(0){=}(0,v_{0 y})$ for $v_{0 y}{=}0.5,1,1.5,2,2.5$.

Figure 1: p(t) for t=0 to 2*pi

Figure 2 shows the one with $\mathbf{v}(0){=}(0,2)$ evolved farther, to $t{=}20\pi$.

Figure 2: p(t) for t=0 to 20*pi

Figure 3 shows it evolved even farther, to $t{=}60\pi$.

Figure 3: p(t) for t=0 to 60*pi

CLARIFICATION: I am ultimately interested in finding the simplest way of expressing $\mathbf{p}$ as a function of $t$. I.e. I really want to know "what this function looks like" rather than "what the curve looks like". Other parametrizations of the curve, and intuition about shape of the curve, are of interest only if they help lead to this answer.

UPDATE 2015/07/02:
It sure looks like a spirograph hypotrochoid, doesn't it? http://mathworld.wolfram.com/Spirograph.html .

Exploring this possibility, I found by binary search an initial velocity (0,1.662656) (probably accurate to only 4 decimal places or so) yielding a closed orbit in the shape of a 7-petalled flower, and then compared that sim result with the 7-petalled hypotrochoid having the same min and max radii; see Figure 4.

Figure 4: Is it a hypotrochoid?

Conclusion: It's really close, but it's not a hypotrochoid. It moves too fast at the fast parts and too slow at the slow parts, and stays a bit too close to the origin during the in-between parts.

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    $\begingroup$ The shape of an orbit governed by a central force is governed by the Binet equation: en.wikipedia.org/wiki/Binet_equation. (If no one else has written up a solution by tomorrow, I'll try my hand at it.) $\endgroup$ – Travis Willse Jul 1 '15 at 17:49
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This is a nice question, and I don't have a comprehensive answer for you, but perhaps this will help a bit. For conservative forces, it's generally convenient to define the scalar potential $V$. The force will satisfy $\vec{F} = \vec{\nabla}V$, and your potential $V = r$ (yay, system is conservative). Since the kinetic + potential energy is constant, and $V$ is unbounded above, this means that the particle will always be in a bounded orbit around the body.

The orbits you've found are bounded, but not closed (on themselves), resulting in the flower-pattern trajectory. I do know that you only get closed orbits for potentials that are $\tfrac 1r$ (like ours), and $\tfrac 1{r^2}$ (where the force goes as $\tfrac{1}{r^3} \large)$, but I never really understood why (phase-space apparently).

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  • $\begingroup$ I think you said 1/r in one place where you meant r? $\endgroup$ – Don Hatch Jul 1 '15 at 9:52
  • $\begingroup$ Sorry, might have been ambiguous. The world we live in has potential as 1/r and force 1/r^2. This case in question would have a potential or r, and constant (central) force. $\endgroup$ – Maciek Jul 1 '15 at 9:55
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    $\begingroup$ Okay, in that case did you mean to say "you only get closed orbits for potentials that are 1/r (for force 1/$r^2$ such as gravity) or $r^2$ (for force r such as a linear spring)"? I'm reading from the wikipedia article on Bertrand's Theorem... $\endgroup$ – Don Hatch Jul 1 '15 at 10:04
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    $\begingroup$ Also I think you want to say "you are only guaranteed to get closed orbits" rather than "you only get closed orbits", since there are particular initial conditions for which this force gives closed orbits-- for example the circle and straight line examples, but more interestingly, p(0)=(1,0),v(0)=(0,1.6626) or so gives an exact 7-petalled flower. $\endgroup$ – Don Hatch Jul 1 '15 at 11:34
  • $\begingroup$ @DonHatch Do you know what the significance of $\sim 1.6626$ vis-a-vis $7$ is here? $\endgroup$ – Travis Willse Jul 1 '15 at 17:26
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Consider an orbit $$\gamma: \quad t\mapsto z(t)=r(t)e^{i\phi(t)}$$ of such a particle. We are interested in the polar representation $$\phi\mapsto r(\phi)\tag{1}$$ of the resulting curve $\hat\gamma\subset {\mathbb C}$. Denoting the differentiation with respect to $t$ by a $\cdot$ and the differentiation with respect to $\phi$ by a $'$ we have $\dot r=r'\>\dot\phi$.

The kinetic energy of the particle (of mass $1$) is given by $$T={1\over2}(\dot r^2+ r^2\dot\phi^2)={1\over2}(r'^2+r^2)\dot\phi^2\ ,$$ and its potential energy simply by $V(z)=r$. It follows that $$E={1\over2}(r'^2+r^2)\dot\phi^2 + r\tag{2}$$ is constant along $\gamma$. A second invariant is the angular momentum $$L=r^2\>\dot\phi\ ,$$ so that we can eliminate $\dot\phi^2$ in $(2)$ and obtain the following differential equation for the functions $(1)$: $${1\over2}(r'^2+r^2){L^2\over r^4} + r=E\ .$$ This can be rewritten as a standard first order $ODE$ in the following form: $$r'=\pm{r\over L}\sqrt{2Er^2-L^2-2r^3}\qquad(r>0)\ .\tag{3}$$ The ODE $(3)$ can be separated. In fact, one obtains the inverse functions $r\mapsto \phi(r)$ by a mere quadrature: $${d\phi\over dr}=\pm{L\over r}\bigl(2Er^2-L^2-2r^3\bigr)^{-1/2}\ .$$ Unfortunately the resulting integral is nonelementary for general values of $E$ and $L$.

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  • $\begingroup$ This looks like an interesting fact about the curve's shape... however I'm ultimately interested in what you call $\gamma: t\mapsto z(t)$, and don't care as much about $\phi\mapsto r(\phi)$ ... unless of course it's a good stepping stone to get to $\gamma$. Do you think it is? Seems to me it might be the hard way around... $\endgroup$ – Don Hatch Jul 2 '15 at 13:46
  • $\begingroup$ From the title of your question and your pictures I got the impression that you are primarily interested in the shape of these curves. $\endgroup$ – Christian Blatter Jul 2 '15 at 14:20
  • $\begingroup$ Sorry about that, I can see how it looked that way, in spite of me saying in the text that I would like a closed form for $\mathbf{p}(t)$ if possible. I added a CLARIFICATION blurb attempting to make it clearer. I'm not sure how to make the title more precise without drying it out, so I'll leave that as is, unless you or anyone has a better suggestion. $\endgroup$ – Don Hatch Jul 2 '15 at 14:47
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    $\begingroup$ Although the integral is not elementary, it can be solved with known, well-analyzed nonelementary functions according to Wolfram Alpha, in particular the elliptic $\Pi$ integral and Cardano's formula (the cubic formula). It's a hideous mess, but he asked for "well-known" functions and there are more well-known functions than just the elementary functions; the elliptic $\Pi$ integral may be such for suitable values of "well-known" and certainly it has been analyzed quite a bit. I think it deserves a mention. $\endgroup$ – The_Sympathizer Mar 26 '16 at 4:50
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  • A shallow cone opening upward, made of aluminum, makes a real potential well for a ball bearing to roll in, closely simulating a particle moving subject to a central force of constant magnitude, except for friction: the motion is not conservative. The orbit is a flower with 7 petals of diminishing size as the motion decays, with the seventh petal pointing the same way as the first. After about 10 petals the ball drops through a central hole.

  • A TI 83 calculator easily computes the trajectory (path) and the schedule (time table), requiring about a minute for the calculation of the improper integral of the unrefined quadrature. Integrals must be truncated at their upper and lower integrals to make them proper, with corrections added at the limits of integration, for efficient calculation.

  • The unrefined quadrature is an elliptic integral which can be reduced to a canonical form in which the integrals are proper. The trajectory theta(rho) is an elliptic integral of the third kind and the schedule t(rho) is a combination of elliptic integrals of the first and second kind. These formulas are not very complicated but took some doing to find the right transformations to accomplish the reduction. The reduced form runs on a TI 83 calculator in about 3 seconds, much faster than the unrefined quadrature.

Best, Chuck

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