Hamiltonian Function thru new Variables Q,P -- Show that Q is cyclic

AI Thread Summary
The discussion centers on deriving the Hamiltonian function using new variables Q and P for a particle in a gravitational field. The user successfully determined the constant A as -1/2m^2g but struggles to express the Hamiltonian without the variable Q. Participants suggest that the user can eliminate Q by correctly formulating the Hamiltonian in terms of P and Q. A sign error in the user's calculation is pointed out, indicating the need for correction. The conversation emphasizes the importance of accurately expressing the Hamiltonian to achieve the desired cyclic nature of Q.
ardaoymakas
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Homework Statement
3b) Print out the Hamilton function using the new variables Q and P. Show that by choosing the appropriate constant A, the variable Q becomes cyclic and therefore the Hamilton function can be written down without Q.
Relevant Equations
H = (p^2/2m) + mgq
q = P - AQ^2 , p = - Q
I took the derviative of the Hamiltonian function with respect to Q and assumed that it was equal to 0 in order to find the Konstant A. I did find the Konstant A as -1/2m^2g but I still cant write the Hamiltonian equation without having the Q as a variable. Can someone please help?

Translation:
The Hamilton function for a particle moving vertically in a homogeneous gravitational field with gravitational constant g is given by
----
We introduced new variables Q and P. The variables q and p can be expressed by Q and P using the following transformation formulas:
-----
a)Evaluate the Poisson bracket {Q ,P}q,p. Is the transformation canonical?

b)Print out the Hamilton function through the new variables Q and P. Show that by choosing a suitable constant A, the variable Q becomes cyclic and therefore the Hamilton function can be written down without Q.
Screenshot 2024-01-18 at 23.57.59.png
 
Last edited:
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Welcome to PF!
But posting attachments as pdf that have to be downloaded will hurt your response rate, particularly from those using iPads etc. Can you imbed the image?
 
haruspex said:
Welcome to PF!
But posting attachments as pdf that have to be downloaded will hurt your response rate, particularly from those using iPads etc. Can you imbed the image?
Hey,
Thanks! Is this better?
 
Much better, thanks.
ardaoymakas said:
I did find the Konstant A as -1/2m^2g but I still cant write the Hamiltonian equation without having the Q as a variable.
I've never worked with Hamiltonians, so not usually able to answer such a question, but it seems to me it is trivial to find the A which eliminates Q (and yes, it is the A you found). Just write out H in terms of P and Q. What does the Q term look like?

Edit: I see you have a sign error.
 
haruspex said:
Much better, thanks.

I've never worked with Hamiltonians, so not usually able to answer such a question, but it seems to me it is trivial to find the A which eliminates Q (and yes, it is the A you found). Just write out H in terms of P and Q. What does the Q term look like?

Edit: I see you have a sign error.
When I write it in terms of Q I get H = (-Q)^2 / 2m) + mgP + (Q^2)/2m which isnt the formula without Q
 
ardaoymakas said:
When I write it in terms of Q I get H = (-Q)^2 / 2m) + mgP + (Q^2)/2m which isnt the formula without Q
Did you note my edit? It says your expression for A has the wrong sign.
 

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