3
$\begingroup$

In classical mechanics, the Hamiltonian is well defined by the Lagrangian. Whereas, energy is a very ambiguous term. We just say $E=T+U$, and usually it equals to Hamiltonian. Does there exist a way that, by just looking at the Lagrangian mathematically, we immediately know the relationship between the Hamiltonian and the energy of the system?

And if we have a system, the Hamiltonian of which does not equal to energy, what is the physical meaning of that difference?

$\endgroup$
3
5
$\begingroup$

There are some technical conditions (on the type of constraints in your system) but operationally $H$ is the total energy when $\sum_i \dot q_i p_i$ is $2\times$ the kinetic energy. Then clearly $$ H=\sum_i \dot q_i p_i-L=2T-(T-U)=T+U\, . $$

If this is not the case, $H$ may be conserved but it's just not $E$. This occurs in a wide variety of systems, such as the flyball governor, and systems where some external agent maintains a constant rate of rotation (v.g. beads on rotating wires of various shapes.) The dynamics is still constrained to remain on curves or surfaces of constant $H$, but there is usually no physical interpretation to this conserved quantity.

The simplest cases where $H$ is the energy are natural systems, for which the kinetic energy is quadratic in the velocities $$ T=\sum_{ij} m_{ij}\dot q_i\dot q_j $$ and there is no explicit time-dependence on $t$ in the Lagrangian.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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