# Classical Limit and Angular frequency

1. Sep 12, 2009

### keniwas

1. The problem statement, all variables and given/known data
Show that in classical mechanics, for circular motion, in an arbitrary central potential
V (r), one has w = dE/dL where E is the energy, L is the angular momentum, and w is the
angular frequency of motion around the orbit.

This is the basis for a problem on understanding the correspondence between classical and quantum mechanis in the limit of large circular orbits (i.e. akin to Bohr's atom).

2. Relevant equations
Total Energy of the system in Circular Motion
$$E=\frac{1}{2}m(\dot{r}^2+\frac{L^2}{mr^2})+V(\vec{r})$$
Where
$$L=r^2\omega m$$

3. The attempt at a solution
Once I got the energy in terms of angular momentum, I simply took the derivative with respect to L, however this leaves me with the angular velocity $$\omega$$ and an additional term from the potential (which I can't find any argument as to why it should be zero in this case...)
$$dE/dL = 0+\frac{L}{mr^2}+\frac{dV}{dL}=\omega+\frac{dV}{dL}$$

I am clearly missing somthing important in the problem... is my form for the total energy of the system wrong? or is there somthing special about the central potential I am missing?
Any input is greatly appreciated.

Last edited: Sep 12, 2009
2. Sep 12, 2009

### elduderino

I think, simply, as its a circular motion, the r has to be constant making the potential constant. Which makes the derivative zero.

PS--Be a bit more clear in your problem, as I think the $$\dot{r}^2$$ is different from the other "r" in the system (though it wouldnt matter for the problem). are you considering a point mass?

3. Sep 12, 2009

### keniwas

It would be a point mass (or as close as you can get to one as an electron). The $$\dot{r}^2$$ would be the velocity in the radial direction (i.e. a change in the radius of your circular orbit).

I thought something similar to the logic regarding a fixed circular orbit, however if I change my angular momentum will that not nessecarily correspond to a change in the potential of the system? Since a change in angular momentum would change the effective radius?

4. Sep 12, 2009

### gabbagabbahey

For a circular orbit, the radius must be constant, so $\dot{r}=0$.

Your potential is given in the functional form $V=V(r)$ suggesting it depends only only the radius, which is constant for a circular orbit

$$\frac{\partial V}{\partial L}=\frac{\partial r}{\partial L}\frac{\partial V}{\partial r}=(0)\frac{\partial V}{\partial r}$$

5. Sep 12, 2009

### keniwas

Thank you for the help!