# Effective potential energy minimum

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
$$\frac{f'(\rho)}{f(\rho)}+\frac{3}{\rho}>0$$?

## Answers and Replies

vanhees71
Science Advisor
Gold Member
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.

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

vanhees71
Science Advisor
Gold Member
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##.