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For which $n$ would an object with a non-zero rotation fall to the center of this field?

$$\alpha >0\\ V(r) = \frac{\alpha}{r^n}$$

(Apparently it should never touch the center if it has non-zero rotation and $n=1$). I am totally stumped, as I cant see why the order should change whether or not it falls into the center. Any ideas/explanations would be greatly appreciated.

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  • $\begingroup$ I assume that $x$ is a position vector in 3D and $r=\|x\|$, the magnitude of the position? $\endgroup$ – fibonatic Oct 4 '16 at 12:00
  • $\begingroup$ Sorry, I fixed it. I dazed out as I wrote it. Thanks for the reminder. $\endgroup$ – Blitz Oct 6 '16 at 10:10
  • $\begingroup$ So the problem is one-dimensional? $\endgroup$ – fibonatic Oct 6 '16 at 11:30
  • $\begingroup$ Only in polar coordinates. In cartesian coordinates it's not. $\endgroup$ – Blitz Oct 6 '16 at 12:35
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Consider an object with non-zero rotation, and velocity vector $\bf{v}$. That it has a non-zero rotation means that, at least as long as $r \neq 0$, $\textbf{v}\times\textbf{r} = rv_\perp \neq 0$, where $\bf{r}$ is the radial position vector and $v_\perp$ is the component of $\bf v$ perpendicular to $\bf r$. As a simple consequence of conservation of angular momentum we must have $$ v_\perp \propto \frac{1}{r}. $$ This imposes $$ \dot{v}_\perp = \frac{dv_\perp}{dt} = \frac{dv_\perp}{dr}\frac{dr}{dt} \propto \frac{1}{r^2}v_r. $$ On the other hand the potential \begin{align} V(\textbf{r}) &\propto \frac{1}{r^n} \end{align} imposes an an acceleration $$ \textbf{a} \propto -\nabla V \propto \frac{1}{r^{n+1}}\textbf{e}_r, $$ whence $\dot{v}_r = |\textbf{a}|$. We thus have $$ \frac{\dot{v}_\perp}{\dot{v}_r} \propto r^{n-1}v_r. $$ For $n > 1$ we have \begin{align} \lim_{r\to 0}\frac{\dot{v}_\perp}{\dot{v}_r} = 0. \end{align} We can interpret this to mean that as the object is brought closer to the center the acceleration gradually becomes purely radial. If the potential acts attractively this means that the object may eventually collide with the center. However, for $n = 1$ we have \begin{align} \lim_{r\to 0} \frac{\dot{v}_\perp}{\dot{v}_r} \propto \lim_{r\to 0} v_r \neq 0, \end{align} which means that even when the object is brought closer to the center the acceleration will never become radial: the object will never accelerate directly towards (nor directly away from) the center. And since it was originally not moving towards the center, it will never collide with the center.

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Another way to put the same answer is that if you write the system in polar coordinates (the motion will be in a plane, it's a central force), and just focus on the radial coordinate, conservation of angular momentum allows you to write the situation as if there was an effective repulsive potential that is proportional to the square of the angular momentum, and falls off with distance like the inverse square of r. (For example, see https://en.wikipedia.org/wiki/Effective_potential) Hence, for any real attractive potential that rises less rapidly than inverse square as the particle approaches the center (so n<2), the repulsive potential will eventually win out, and prevent arrival at the center. But if the attractive potential rises more rapidly, as for n>2, then there is no centrifugal barrier, and the potential will cause a monotonic falling in r. The question is then if it reaches the center in finite time, and it will because eventually v(r) will look like a r to a negative power, which does arrive in finite time.

There are several twists that remain, however. If the particle starts out with enough speed to escape the system, then if its angular momentum is large enough, it can avoid the center even if n>2. Also, if n=2, the combined effective potential is flat, so particles initially going inward will hit the center, but particles initially going outward with go to infinity.

Finally, it should be noted that the above expression for the effective potential assumes the speeds are never anywhere close to the speed of light. That will certainly not be true for a particle that falls to the center if n>2, but when relativity enters, it only weakens the centrifugal barrier, so if the particle is going to hit the center in the nonrelativistic analysis, it will in the relativistic one too. However, the situation gets tricky for 1 < n < 2, because then you might think you will not hit the center nonrelativistically, but relativistically, you do hit the center. I presume the question you are asking is not interested in how relativity can complicate the situation, and I think that's only an issue for 1 < n < 2.

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