12
$\begingroup$

Not sure if this is too much physics to be here...

Consider $$H:\mathbb{R}^{2N+1}\rightarrow\mathbb{R}$$ of class $C^2$, let $H(x,y,z)$ such that $x\in\mathbb{R}^N$, $y\in\mathbb{R}^N$ and $z\in\mathbb{R}$. Let $\varphi$ be the flow associate with the Hamiltonian system $$\dot{x}_i=-\frac{\partial H}{\partial y_i}$$ $$\dot{y}_i=\frac{\partial H}{\partial x_i}$$ $$\dot{z}=1$$ I have to prove that if $\eta$ is a 1-form given by $\eta=\sum_{i=1}^Nx_i \ dy_i-H \ dz$ and $c$ is a closed curve in $\mathbb{R}^{2N+1}$, then for all $s$ we have $$\int_{\varphi(s,c)}\eta=\int_c\eta.$$

Thank you in advance!

$\endgroup$
  • $\begingroup$ How is this a Hamiltonian system? As presented down here the phase space is not even-dimensional. What is the symplectic form associated to it? $\endgroup$ – Novo Jul 29 '13 at 14:54
  • $\begingroup$ I agree with @Novo that this system is not Hamiltonian in usual sense. As an advice: since $\varphi(s, \dot)$ is a diffeomorphism, you can do the change of variables in l.h.s. integral, returning back to the parameterisation of curve $c$. $\endgroup$ – Evgeny Jul 29 '13 at 15:00
  • $\begingroup$ Ok, but if it isn't a Hamiltonian system, then what is? And how can I solve it? And thank you for the observations. $\endgroup$ – diff_math Jul 29 '13 at 15:39
  • 1
    $\begingroup$ The system is a time-dependent Hamiltonian system. The $x-y$ subspace has a symplectic form. You can see this in the form of the equations for the $\dot{x}$ and $\dot{y}$. $\endgroup$ – Robert Lewis Jul 29 '13 at 18:12
  • 1
    $\begingroup$ What is $\varphi(s,c)$? $\endgroup$ – user7530 Jul 29 '13 at 18:14
3
$\begingroup$

By Stokes $\int_c\eta = \int_D d \eta$ for any disc $D$ with $\partial D = c$ (which exists because all curves are contractible in $\mathbb{R}^{2n+1}$), and then also $\int_{\phi(s,c)} \eta = \int_{\phi(s,D)} d \eta$

Thus it suffices to show that Lie drivative of $d \eta$ is $0$.

By Cartan's magic formula

$$L_X d \eta = d (i_X (d\eta)) = d (i_X \omega -i_X (dH \wedge dz))$$

We have $i_X dz =1$.

We then compute $i_X \omega = dH - \frac{\partial H}{\partial z} dz$

and thus $d H = i_X \omega + \frac{\partial H}{\partial z} dz$ which contracted with $X$ again gives

$i_X dH = i_X (\frac{\partial H}{\partial z} dz)= \frac{\partial H}{\partial z}.$

Plugging this in we get

$$L_X d \eta=d (i_X \omega -i_X (dH \wedge dz)) = d( dH - \frac{\partial H}{\partial z} dz + (i_X dH) \wedge dz - d H \wedge (i_X dz))= d (0)=0.$$

$\endgroup$

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.