Finding energy as a function of Symplectic area?

  • #1

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=[itex]\int p dx[/itex] (over one period)


The Attempt at a Solution


From Hamilton's equations I get :

[itex]\dot{x}=\partial H/ \partial p[/itex] and [itex]\dot{p}=- \partial H/ \partial x[/itex]

So

[itex]dot{x}=p/m[/itex] and [itex]\dot{p}=-2m\omega2x[/itex]

[itex]x(t)=pt/m ; p(t)=-2m \omega 2xt[/itex]

Then I integrate

[itex]\int pdx = \int p d(pt/m)[/itex]

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

[itex]d(pt/m)=1/m (p+\dot{p}))dt [/itex]

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).
 
Last edited:

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
13,099
3,416
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.]
 
Last edited:

Related Threads on Finding energy as a function of Symplectic area?

Replies
5
Views
5K
Replies
0
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
7
Views
7K
Replies
7
Views
5K
Replies
3
Views
2K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
3
Views
3K
Top