Hamiltonian for spherically symmetric potential

[Vitor Pimenta]

Homework Statement

A particle of mass m moves in a "central potential" , V(r), where r denotes the radial displacement of the particle from a fixed origin.
From Hamilton´s equations, obtain a "one-dimensional" equation for [tex]{\dot p_r}[/tex], in the form [tex]{{\dot p}_r} = - \frac{\partial }{{\partial r}}\left[ {{V_{eff}}\left( r \right)} \right][/tex], where [tex]{V_{eff}}\left( r \right)[/tex] denotes an "effective" potential that is a funcion of r only.

Homework Equations

Hamiltonian: [tex]H = \frac{{{p_r}^2}}{{2m}} + \frac{{{L^2}}}{{2m{r^2}}} + V\left( r \right)[/tex] , where L is the angular momentum with respect to the origin, which is a constant of the motion.

[tex]\frac{{\partial H}}{{\partial r}} = - {\dot p_r}[/tex]

The Attempt at a Solution

[tex]\begin{array}{l}
{{\dot p}_r} = - \frac{{\partial H}}{{\partial r}} = - \frac{\partial }{{\partial r}}\left[ {\frac{{{L^2}}}{{2m{r^2}}} + V\left( r \right)} \right]\\
\therefore {V_{eff}}\left( r \right) = \frac{{{L^2}}}{{2m{r^2}}} + V\left( r \right)
\end{array}[/tex]

The problem is that it doesn´t make sense to me that the effective potential is different than the normal one ( V(r) ). Besides, the force acting on the particle shouldn´t have a dependence on its angular momentum L.

 

[Vitor Pimenta]

It seems that [tex]\frac{{{L^2}}}{{2m{r^2}}}[/tex] has the form of a potential energy the centripetal force could produce ...

 

[hilbert2]

The Attempt at a Solution

[tex]\begin{array}{l}
{{\dot p}_r} = - \frac{{\partial H}}{{\partial r}} = - \frac{\partial }{{\partial r}}\left[ {\frac{{{L^2}}}{{2m{r^2}}} + V\left( r \right)} \right]\\
\therefore {V_{eff}}\left( r \right) = \frac{{{L^2}}}{{2m{r^2}}} + V\left( r \right)
\end{array}[/tex]

The problem is that it doesn´t make sense to me that the effective potential is different than the normal one ( V(r) ). Besides, the force acting on the particle shouldn´t have a dependence on its angular momentum L.

The effective potential is different from the normal potential because there's an apparent "centrifugal" effect on the radial motion. The effective potential should not contain the operator ##L^{2}##, you should replace it with its eigenvalue for orbital quantum number ##l##.

 

[Vitor Pimenta]

hilbert2, thanks for the reply !

I wonder what reference frame the Hamiltonian is about, since it includes the effect of a "false" force (centrifugal). Also, L^2 is not an operator, but a scalar number (which is a constant of the motion), so what was that about replacing it for a quantum number, since we´re not considering quantum mechanics ?

 

[Orodruin]

hilbert2, thanks for the reply !

I wonder what reference frame the Hamiltonian is about, since it includes the effect of a "false" force (centrifugal). Also, L^2 is not an operator, but a scalar number (which is a constant of the motion), so what was that about replacing it for a quantum number, since we´re not considering quantum mechanics ?

It is not about any particular reference frame, you can use whatever coordinates you wish (you may even mix spatial and momentum variables as long as your transformation is canonical).

It is not a fictitious effect, your coordinate system is curvilinear and you should expect the motion in one direction (in this case parametrised through the angular momentum, which is a constant of motion) to affect the motion in the other coordinates. Thus, in a rotating coordinate system, it would be the effect of the centrifugal force, while in a fixed coordinate system it is an effect of moving in curvilinear coordinates. The result is of course going to be the same.

Besides, the force acting on the particle shouldn´t have a dependence on its angular momentum L.

This is the effective potential. It tells you how to consider motion in one direction only, in the fixed frame it tells you (for example) the force needed to keep the body moving on a circular path - this does depend on the angular momentum.

 

[Vitor Pimenta]

I get it somewhat, but this needs further exploration by me (yay, more fun incoming)
Thanks again for all the help