A general problem in (q,p) -> (Q,P) for Hamiltonian

  • Thread starter cwasdqwe
  • Start date
  • #1
31
3
Hello everyone, this is my first thread. Hope to be helpful here, as well as to find some help! :D


Homework Statement



Given a Hamiltonian [itex]H(q,p)[/itex](known) and given a transformation of coordinates [itex](q,p)\rightarrow (Q,P)[/itex]:

a) Show that it is a canonic transformation
b) Solve the equations of motion
c) Is this a simmetry of H? Find the constants of motion.

Homework Equations



(See attempted solution)

The Attempt at a Solution



Just need to know if I'm wrong in some of these, and in that case being sure I'm doing right

a) By using the Poisson brackets, they must verify all of these

[itex]\left\{Q,P\right\}=1[/itex]
[itex]\left\{ P,Q \right\}=-1[/itex]
[itex]\left\{ Q,Q \right\}=0[/itex]
[itex]\left\{ P,P \right\}=0[/itex]


b) I tried by using the chain rule in the Hamiltonian

[itex]\frac{\partial H}{\partial p}=\frac{\partial H}{\partial P}\frac{\partial P}{\partial p} + \frac{\partial H}{\partial Q}\frac{\partial Q}{\partial p}[/itex]

[itex]\frac{\partial H}{\partial q}=\frac{\partial H}{\partial P}\frac{\partial P}{\partial q} + \frac{\partial H}{\partial Q}\frac{\partial Q}{\partial q}[/itex]

then I have a system of two equations and the two [itex]\frac{\partial H}{\partial Q}[/itex] and [itex]\frac{\partial H}{\partial P}[/itex], which will give me the equation of motion by using Hamilton's:

[itex]-\frac{\partial H}{\partial Q}= \dot{P}[/itex]
[itex]-\frac{\partial H}{\partial P}= \dot{Q}[/itex]


c) This point I haven't got clear. I've tried to relate it with 1, knowing that a quantity f is a constant of motion if the Poisson Bracket [itex]\left\{ H,f \right\}=0[/itex], i.e., if commutes with the Hamiltonian, but yet I've not it clear.

Thank you very much!
 
Last edited:

Answers and Replies

Related Threads on A general problem in (q,p) -> (Q,P) for Hamiltonian

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
2
Views
4K
Replies
6
Views
10K
Replies
6
Views
1K
  • Last Post
Replies
6
Views
2K
Replies
3
Views
712
  • Last Post
Replies
5
Views
5K
  • Last Post
Replies
4
Views
3K
Top