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

[ardaoymakas]

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:
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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.

 

[haruspex]

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?

 

[ardaoymakas]

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?

 

[haruspex]

Much better, thanks.

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.

 

[ardaoymakas]

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

 

[haruspex]

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.