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This problem is from Introduction to Classical Mechanics With Problems and Solutions by David Morin. The solution is also given in the book for this particular problem.

Problem #6.6.1

A particle moves in a potential $V(r) = -V_0 \exp(-\lambda^2 x^2)$.

(a) Given angular momentum $L$ find the radius of the stable circular orbit.

(b) It turns out that if $L$ is too large, then a circular orbit doesn't exist. What is the largest value of $L$ for which a circular orbit does indeed exists? What is the value of $V_{eff}(r)$ in this case?

The first part to this problem is straight foward, minimize $V_{eff}(r)$ and solve for $r_0$, which gives the radii for the circular orbit If one solves for $r_0$ then one gets, $L^2 = (2m V_0 \lambda^2)r_0^4 \exp(-\lambda^2 r_0^2)$. Now the subtle point here is that there isn't always going to be a solution to this; for large value of $L$ there is no solution. But in general there is going to be two solution for $r_0$. This can be seen from the plot of $V_{eff}(r)$.

So my question is, how to actually find $r_0$? I want to get a value, and then say that for this value the orbit is stable or unstable. Can someone guide me? Secondly, the relation has $r_0^4$, so how can it have just 2 solutions? What about the other 2?

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  • $\begingroup$ Given an $r_{o}$ of an orbit, do you know the condition that stability imposes on $V_{\rm eff}$ in the neighborhood of $r_{o}$ $\endgroup$ Nov 15, 2017 at 20:11
  • $\begingroup$ $r_0 \ge 0$, so some algebraic solutions could be invalid. $\endgroup$
    – JEB
    Nov 15, 2017 at 20:15
  • $\begingroup$ @JerrySchirmer Yes. I'm trying to find a minima. So $V''_{eff}(r_0) > 0$. I should sit in a well to have stable oscillations. $\endgroup$
    – sbp
    Nov 15, 2017 at 21:05

2 Answers 2

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The effective potential in your case is

$$V_{\rm eff}(r)=\frac{L^2}{2mr^2}-V_{0}\exp(-\lambda^2r^2)$$

and if we differentiate

$$V^{\prime}_{\rm eff}(r)=-\frac{L^2}{mr^3}+2\lambda^2V_{0}r\exp(-\lambda^2r^2)$$

This vanishes if

$$L^2=2m\lambda^2V_{0}r^4\exp(-\lambda^2r^2)$$

as you've already observed. Now lets define $C\equiv2m\lambda^2V_{0}$ and $f(r)\equiv Cr^4\exp(-\lambda^2r^2)$, so our equation is

$$L^2=f(r)$$

This equation does not have a solution that can be expressed explicitly using elementary functions. For $L\leq L_{\rm max}$ (which you're asked to find in (b)), it has four solutions. Two positive solutions: one corresponds to a stable equilibrium and the other to an unstable one, and two mirrored negative solutions that are discarded.

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  • $\begingroup$ What do you mean by two mirrored solutions? $\endgroup$
    – sbp
    Nov 15, 2017 at 20:49
  • $\begingroup$ If $r_{0}$ is a solution, then also $-r_{0}$. This is because $f(r)$ is even function. $\endgroup$
    – eranreches
    Nov 15, 2017 at 20:52
  • $\begingroup$ Yup. Just confirming! $\endgroup$
    – sbp
    Nov 15, 2017 at 20:53
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In the generic case that does have solutions, your pseudopotential looks like this:

Mathematica graphics

The two solutions you've got correspond to the points of equilibrium, i.e. to the minimum and the maximum. One of the two is a stable equilibrium (i.e. if you put a marble there, it'll stay there) and the other one is unstable (i.e. if you put a marble there, it'll roll down the hill). Discriminating between stable and unstable equilibria (particularly when all you have is formulas) is standard fare and it's explored in depth in any mechanics textbook, as well as in Wikipedia.

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  • $\begingroup$ Anyway can I get the values of the constants used to plot this? $\endgroup$
    – sbp
    Nov 16, 2017 at 21:39
  • $\begingroup$ They're quite generic - but it's better if you actively have to search for them, it'll expose you to a better fraction of the parameter space, which you probably need quite a bit. $\endgroup$ Nov 16, 2017 at 21:52

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