I wondering about the interpretation for the energy difference between the Hamiltonian and the total mechanical energy for systems where the Hamiltonian is conserved, but it is not equal to the total mechanical energy.
For example, consider a bead (mass $m$) on a frictionless hoop (radius $R$) in the presence of gravity. The hoop is spun around an axis parallel to the gravitational acceleration at constant angular speed ($\omega$). This is the typical set up for this problem.
The total energy for this system (using $\phi$ to denote the angle from the bottom of the hoop) is:
$$
E = \frac{p_{\phi}^2}{2mR^2} + \frac{1}{2} mR^2 \omega^2 \sin^2 \phi + mgR (1- \cos \phi)\tag{1}
$$
where
$$p_{\phi} = mR^2\dot{\phi}\tag{2}.$$
The Hamiltonian is:
$$
H = \frac{p_{\phi}^2}{2mR^2} – \frac{1}{2} mR^2 \omega^2 \sin^2 \phi + mgR (1- \cos \phi).\tag{3}
$$
So the difference between the total mechanical energy and the Hamiltonian is:
$$
E-H = mR^2 \omega^2 \sin^2 \phi\tag{4}
$$
which is twice the rotational kinetic energy, I think. I'm just trying to get a handle on what this difference means. Any help is appreciated.
Best Answer
I think this theorem might help to you.
Assume that $L=T-U$ is lagrangian of the system. $T$ is kinetic energy that presented as a quadratic form of $\dot{q}$: $T=\frac{1}{2}\sum a_{ij}\dot{q_i}\dot{q_j}$, $a_{ij}=a_{ji}(q,t)$; $U=U(q)$.
Theorem: Under these assumptions Hamiltonian $H$ is total energy of system $H=T+U$
Proof of theorem: Using Euler's theorem on homogeneous functions $\frac{\partial f}{\partial x}x=2f$. Then we have: $H=p\dot{q}-L=\frac{\partial L}{\partial \dot{q}}\dot{q}-(T-U)=2T-(T-U)=T+U $ $\blacksquare$
So if you have system with these assumptions you can say that Hamiltonian and the total energy are the same thing.
I know for sure that if the potential energy depends on the velocity, the energy will be different from Hamiltonian.
You can get more about this in Arnold's Mathematical Methods of Classical Mechanics