2017-05-18 13:01:25 8 Comments

Can both Lagrangian and Hamiltonian formalisms lead to different solutions?

I have a simple system described by the Lagrangian \begin{equation} L(\eta,\dot{\eta},\theta,\dot{\theta})=\eta\dot{\theta}+2\theta^2. \end{equation} The equations of motion are obtained from Euler-Lagrange eq.: \begin{eqnarray} 4\theta-\dot{\eta}=0\; \mathrm{and}\; \dot{\theta}=0, \end{eqnarray} yielding the solution $\eta(t)=4\theta_0t+\eta_0$ where $\eta_0$ and $\theta_0$ are constants.

But when I obtain one of the equations of motion from the Hamiltonian (via Legendre transformation), \begin{equation} H=\left(\frac{\partial L}{\partial\dot\eta}\right)\dot\eta+\left(\frac{\partial L}{\partial\dot\theta}\right)\dot\theta - L =-2\theta^2, \end{equation} \begin{equation} \dot\eta=\frac{\partial H}{\partial p_\eta}=0, \end{equation} the situation is surprisingly different from the Lagrangian approach because $\eta$ is now a constant!

Can someone give a proper explanation for this inconsistency? Am I doing something wrong here?

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## 1 comments

## @J.G. 2017-05-18 13:21:57

The problem here is that, because there exist constraints of the form $f(q,\,p)=0$, the phase space coordinates of the usual Hamiltonian formulation aren't independent. I'm not sure how you encountered this Lagrangian, but this issue is a common hiccup in electromagnetism and (if you'll pardon a more obscure example) BRST quantisation. The good news is you can still form a Hamiltonian description equivalent to the Lagrangian one. The trick is to append suitable terms to the "naïve" Hamiltonian, as explained here, and as a result the Poisson brackets are upgraded to what are called Dirac brackets.

For your problem the full Hamiltonian is $H=-2\theta^2+c_1 p_\eta+c_2( p_\theta-\eta)$, where the $c_i$ remain to be computed as functions of undifferentiated phase space coordinates. In fact $c_1=\frac{\partial H}{\partial p_\eta}=\dot{\eta}=4\theta$ while $c_2=\frac{\partial H}{\partial p_\theta}=\dot{\theta}=0$, so $H=-2\theta^2+4\theta p_\eta$. You can verify this gives you the right equations of motion.

## @donnydm 2017-05-18 14:39:36

Thanks! It's actually a part of perturbed NLS Lagrangian after an attempt of finding solution using variational method. Are the Dirac's procedure covers all Lagrangian of 'irregular' forms?

## @J.G. 2017-05-18 16:07:28

@donnydm Given any Lagrangian with consistent equations of motion, this technique will ensure an equivalent Hamiltonian formulation, although if there are higher-order derivatives you also need a trick due to Ostrogradski. (An example of an inconsistent Lagrangian is $L=q$.)