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I am unable to understand how to put the equation of the simple pendulum in the generalized coordinates and generalized momenta in order to check if it is or not a Hamiltonian System.

Having

$$E_T = E_k + E_u = \frac{1}{2}ml^2\dot\theta^2 + mgl(1-cos\theta)$$

How can I found what are the $p$ and $q$ for $H(q,p)$ in order to check that the following holds, i.e. the system is a Hamiltonian system.

$$\frac{dq}{dt}=\frac{\partial H}{\partial p}~~~~~~~~~~~~~~\frac{dp}{dt}=\frac{-\partial H}{\partial q}$$

BRabbit27
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1 Answers1

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The Lagrangian is $${\cal L}=\frac{1}{2}ml^2\dot{\theta}^2-mgl(1-\cos\theta).$$ The conjugate momentum is $$p_\theta=\frac{\partial{\cal L}}{\partial\dot{\theta}}=ml^2\dot{\theta}$$ and so the Hamiltonian is $${\cal H}=\sum_q \dot{q}p_q-{\cal L}=\frac{1}{2}ml^2\dot{\theta}^2+mgl(1-\cos\theta)=\frac{p_\theta^2}{2ml^2}+mgl(1-\cos\theta).$$

Jonathan
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  • What does $p_\theta$ mean? what does $p_q$ mean? – BRabbit27 Jan 10 '14 at 09:41
  • Could you develop more after you write $\cal H = \sum - \cal L$? – BRabbit27 Jan 10 '14 at 09:49
  • As stated, $p_\theta$ is the conjugate momentum to $\theta$. What kind of development are you looking for? It's a substitution of the second equation into the third equation. link – Jonathan Jan 10 '14 at 16:37
  • I am just trying to prove the system is a Hamiltonian system, but if I take the partial derivatives I don't get $\dot q = \partial H / \partial p$ and $\dot p = \partial H / \partial q$, I guess I am doing something wrong. – BRabbit27 Jan 10 '14 at 21:01
  • In the Hamiltonian you wrote, what would be $q$, would it be $\theta$? If so, I'm getting $\partial H / \partial p = \dot \theta$ and $dq/dt = mgl sin\theta \cdot \dot \theta$ which are not the same, therefore not a Hamiltonian, but of course it is, so I'm missing something. – BRabbit27 Jan 11 '14 at 11:25
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    Yes, the coordinate is $\theta$. $\partial H/\partial p=\dot\theta$ doesn't make sense: you need to express $H$ as a function of the coordinate and the momentum, without reference to the time derivative of the coordinate. – Jonathan Jan 11 '14 at 16:24