# Effective potential energy minimum

1. Jan 12, 2015

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

2. Jan 13, 2015

### 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.

3. Jan 13, 2015

### LagrangeEuler

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

4. Jan 13, 2015

### 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 $x=0$.