1
$\begingroup$

Is there any way to compute the Legendre Transformation of a Hamiltonian which is linear in momentum, for example, a crazy Hamiltonian like

$$H(q,p) = \alpha p q + m\omega q^2 .$$

This function is convex (and also concave) in $p$, which is a sufficient condition for the Legendre transformation to work (as far as I know). However if I try to find $\dot{q}(p)$ , which I would normally then invert and sub into $p \dot{q} -H(q,p) $, I get stuck because $$\frac{\partial H}{\partial p} = \dot{q} = \alpha q.$$

$\endgroup$
1
$\begingroup$

No, when we consider the action principle $S=\int\! dt(p \dot{q} -H(q,p))$ for OP's Hamiltonian, then $p$ acts as a Lagrange multiplier that imposes the EOM $\dot{q} \approx \alpha q$. This EOM does not have a regular Lagrangian formulation, i.e. a formulation without the use of other variables than $q$.

| cite | improve this answer | |
$\endgroup$

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.