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In a mechanical system with some equilibrium point $r_0$, one can do an analysis in order to check if such point is stable or unstable, i.e. the motion of a particle around of it is bounded or unbounded. There are two ways of testing this:

  1. Taking a small perturbation $\epsilon$ around the equlibrium point. One replaces $r=r_0+\epsilon$ into the equation of motion and approximates to first order in $\epsilon$. Then, one must see what kind of solutions has the differential equation, convergent ($r_0$ is stable) or divergent ($r_0$ is unstable).

  2. Calculating the second deriative of the potential, then checking if it is positive ($r_0$ is stable) or negative ($r_0$ is unstable).

Is one of these tests more general (maybe better) than the other? or both of them should work at the same level?


Particularly, I'm having trouble with the Lagrangian system with kinetic energy $T=\frac{1}{2}(\dot{r}^2+r^2\dot{\theta}^2)$ and $V=\beta r$, then:

$$\mathscr{L}=\frac{1}{2}\dot{r}^2+\frac{1}{2}\dot{\theta}^2 r^2-\beta r,$$

which seem to yield different results for each of these tests ($\alpha$ and $\beta$ are positive constants, the system has constant angular velocity $\dot{\theta}=\omega$).

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  • $\begingroup$ Those tests should absolutely give you the same result... can you give more detail on why you say they don't? $\endgroup$
    – Andrew
    Commented Jan 14, 2017 at 0:24
  • $\begingroup$ Could you explain how you got that potential from that lagrangian? $\endgroup$ Commented Jan 14, 2017 at 0:45
  • $\begingroup$ Sorry, just realized I had to be more explicit with the Lagrangian system, I updated the post. $\endgroup$
    – rsaavedra
    Commented Jan 14, 2017 at 1:12
  • $\begingroup$ I don't see why the system must have constant angular velocity. It probably has constant angular momentum, but that is not the same thing. That might account for different methods giving answers that are all wrong, but different from each other. $\endgroup$
    – alephzero
    Commented Jan 14, 2017 at 1:57

1 Answer 1

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For a system as simple as you have, it makes no difference which method you use.

The "perturbation" method is however slightly more general as it is applicable to systems of differential equations which may be linear but still coupled. Examples of such systems include predator-prey problems, and also systems for which the natural manifold is not the plane. The most common examples of the latter would be angular momentum systems, and more generally Lipkin-Meshkov-Glick types of models. There are also systems for which the potential (or the effective potential) is difficult to isolate, such as beads on rotating hoops of various shapes. In those cases the "perturbation" approach, when done with appropriate care, is foolproof (albeit typically more tedious than using then 2nd derivative test when this test is easily available).

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