# Finding energy as a function of Symplectic area?

## Homework Statement

Find the energy E of the harmonic oscillator (H(x,p)=p2/2m+mω2x2) as a function of the system's symplectic area.

## Homework Equations

Canonical equations and A=$\int p dx$ (over one period)

## The Attempt at a Solution

From Hamilton's equations I get :

$\dot{x}=\partial H/ \partial p$ and $\dot{p}=- \partial H/ \partial x$

So

$dot{x}=p/m$ and $\dot{p}=-2m\omega2x$

$x(t)=pt/m ; p(t)=-2m \omega 2xt$

Then I integrate

$\int pdx = \int p d(pt/m)$

But I'm not sure how to handle the d(pt/m) term. If I do chain rule (in time) I get something like

$d(pt/m)=1/m (p+\dot{p}))dt$

and I'm not really sure what the answer is if I do it in x. Since I dunno what dp/dx is. (other than m*dv/dx=m*dx'/dx ... but I'm not sure what good that does me).

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## Answers and Replies

TSny
Homework Helper
Gold Member
Think about the shape of the orbit (for a given energy E) on a p vs x plot. There is a simple geometric formula for the area enclosed in the orbit,##\int{pdx}##.

[EDIT: Or, if you really want to carry out the integration, use E = p2/2m + (1/2)2x2 to find p as a function of x for fixed E.]

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