I Effective potential

1. Feb 23, 2017

gau55

In Lagrangian and Hamiltonian mechanics it's common to define part of the kinetic energy as the "effective potential energy" but i am unclear on which expression we define this from, if we look at the lagrangian and identify the part of the kinetic energy thats dependant only on the the generalised coordinate and not velocity to be part of the effective potential, we get a different sign for one of the terms in V(eff) than if we do the same thing when looking at the lagrangian. From my experience in scleronimic systems where the hamiltonian is the total energy and is constant it is correct to identify the effective potential from the hamiltonian, whereas when energy is not conserved we choose to identify the effective potential from the lagrangian. They are always different in sign and from this i conclude that in the first case it is only correct to take it out of the hamiltonian and its wrong to do it the other way, and vice versa in the second case i described. My question is what is the difference that causes all of this and how is this choice the result of the need for the negative spacial derivative of the effective potential to equate to the effective force felr by the mass

2. Feb 24, 2017

stevendaryl

Staff Emeritus
Well, the most common case is treating angular momentum as producing a modification of the potential.

Many classical situations are described by an energy equation:

$KE + PE = E$

where $KE$ is the kinetic energy, and $PE$ is the potential energy. In polar coordinates $r, \theta$, the kinetic energy for motion in a plane is given by:

$KE = \frac{m}{2} \dot{r}^2 + \frac{L^2}{2mr^2}$

where $L$ is the angular momentum. The angular momentum term is the kinetic energy due to angular velocity and the other term is due to radial velocity.

I don't think that there is anything deeper about "effective potentials" than just pretending that $\frac{L^2}{2mr^2}$ is a kind of repulsive potential energy:

$PE_{eff} = PE + \frac{L^2}{2mr^2}$

Then the energy equation becomes (since $L$ is a constant) an effective one-dimensional equation:

$KE_{r} + PE_{eff} = E$

Writing it out explicitly:

$KE_{r} = \frac{1}{2} m \dot{r}^2$
$PE_{eff} = V(r) + \frac{L^2}{2mr^2}$

where $V(r)$ is the real potential.

Treating the angular momentum term as part of the potential, instead of kinetic energy, is exactly analogous to treating "centrifugal force" as part of the force, rather than part of the acceleration.