Effective potential energy minimum

[LagrangeEuler]

Effective potential energy is defined by
$$U^*(\rho)=\frac{L^2}{2m\rho^2}+U(\rho)$$
in many problems I found that particle will have stable circular orbit if $$U^*(\rho)$$ has minimum.

1. Why is that a case? Why circle? Why not ellipse for example?

2. Is this condition equivalent with
$$\frac{f'(\rho)}{f(\rho)}+\frac{3}{\rho}>0$$?

 

[vanhees71]

Around a minimum $\rho_0$ you (very often) have
$$U(\rho)=U_0(\rho_0)+\frac{m \omega^2}{2}$$
with $m \omega^2=U''(\rho)>0$. This implies that if the particle is close to the minimum $\rho_0$, it behaves as in a harmonic-oscillator potential. A particular solution is of course $\rho(t)=\rho_0=\text{const}$. Then you have a circular orbit. Otherwise the particle oscillates around such an orbit.

 

[LagrangeEuler]

And when this is not the case? I am really trying to understand this but I have difficulties.

 

[vanhees71]

What do you mean? If you have a twice continuously differentiable function in some interval, the necessary condition for a (local) extremum is that its first derivative vanishes at the corresponding point. To have a minimum, it's sufficient (but not necessary) that the 2nd derivative is positive.

The criterion fails, if the function's 2nd derivative is also 0 at this point. Then it's either a saddle point or an extremum, depending on whether the first non-vanishing derivative is odd or even, respectively.

E.g. the function $f(x)=x^3$ has $f'(x)=3 x^2$, $f''(x)=6x$, $f'''(x)=6$. The 1st derivative vanishes at $x=0$ and there also $f''(0)=0$ but $f'''(0) \neq 0$. Obviously the graph of the function has a saddle-point, i.e., a tangent with vanishing slope but no extremum in $x=0$.

In the same way, you find that $f(x)=x^4$ has a tangent of vanishing slope at $x=0$ with $f''(0)=0$ but also $f'''(0)=0$ and $f^{(4)}(0)=24>0$, i.e., a minimum at $x=0$.

 

[PJC01]

1. The reason why a stable circular orbit is preferred in many cases is because it represents the lowest energy state for the particle. In other words, the particle will experience the least amount of resistance or force in a circular orbit, compared to other types of orbits such as elliptical. This is due to the fact that a circular orbit has a constant distance between the particle and the center of the orbit, resulting in a constant effective potential energy. On the other hand, an elliptical orbit has varying distances between the particle and the center, resulting in varying effective potential energy. This can lead to instability and the particle may eventually leave the orbit.

2. The condition \frac{f'(\rho)}{f(\rho)}+\frac{3}{\rho}>0 is not equivalent to the condition for a stable circular orbit. This condition is known as the Rayleigh criterion and is used to determine the stability of an orbit. It states that if the second derivative of the effective potential energy is positive, then the orbit is stable. However, this does not necessarily mean that the orbit will be circular. It only determines the stability of the orbit, not its shape. Therefore, the condition for a stable circular orbit, which considers the minimum of the effective potential energy, is not equivalent to the Rayleigh criterion.