Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In my teacher's notes there is a discussion of the Hamiltonian for a central force field with potential $V(r)$.

The Hamiltonian is formulated in spherical polar coordinates: $$H=\frac{p_r^2}{2m}+\frac{p_\phi^2}{2mr^2\sin^2\theta}+\frac{p_\theta^2}{2mr^2}+V(r)$$

Then the conservation of $L_z$ is trivial, because $\phi$ is cyclic. However it is asserted, as if it were evident, that $L^2$ is also conserved. One consequence is that the Hamiltonian can be written in the form $H=\frac{p_r^2}{2m}+\frac{L^2}{2mr^2}+V(r)$.

Now, I know that for any radial force it is trivial to prove that $\vec L$ is conserved, but how can you prove that $L^2$ is conserved just from looking at the previous Hamiltonian? Moreover, does it suffices to prove that $L^2$ and $L_z$ are conserved in order to find that all the three components of $\vec L$ are conserved?

I'm a bit confused, so any help is appreciated

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

By comparing the two expressions for $H$ you can infer an expression for $L^2$, which you need to prove, by transforming the standard expression for $L^2$ to spherical polar coordinates.

Conservation of $L^2$ and $L_z$ is enough to infer the spectrum, but one cannot deduce from it conservation of $L$.

share|improve this answer
    
OK, I have found the expression of $L^2$, even if the derivation is a bit messy $L^2 = m^2 r^4 (\dot \theta^2 + \dot \phi^2 \sin^2\theta ) = p_\theta^2 +p_\phi^2 / \sin^2\theta$. But why is $L^2$ conserved? And what's "the spectrum"? –  Ralph Jul 11 '12 at 18:07
1  
To see conservation, work out $\frac{d}{dt} L^2$. This is easiest if you know Poisson brackets. -- The spectrum - thats the set of labels you put on the spherical function in the separation of variables. As I don't know the context of your question, it is not easy to know what needs explanation. –  Arnold Neumaier Jul 11 '12 at 19:13
    
Thanks, so the idea was to find that $[L^2,H]=0$. I am just an undergraduate studying classical mechanics for the first time, so there are many techniques I am unfamiliar with –  Ralph Jul 11 '12 at 21:08
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.