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

  • #1
ardaoymakas
3
0
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:
Physics news on Phys.org
  • #2
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?
 
  • Like
Likes MatinSAR
  • #3
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?
 
  • #4
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.
 
  • Like
Likes MatinSAR
  • #5
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
 
  • #6
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.
 
  • Like
Likes MatinSAR

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

What is a Hamiltonian function?

A Hamiltonian function, often denoted as H, is a function used in classical mechanics to describe the total energy of a system. It is expressed in terms of generalized coordinates and momenta, and it plays a central role in Hamiltonian mechanics, providing a reformulation of classical mechanics that is equivalent to Newtonian mechanics.

What does it mean for a variable to be cyclic in the context of Hamiltonian mechanics?

A variable is considered cyclic (or ignorable) in Hamiltonian mechanics if the Hamiltonian function does not explicitly depend on that variable. This implies that the corresponding conjugate momentum is conserved, leading to a simplification in solving the equations of motion.

How can we identify if a variable Q is cyclic in the Hamiltonian function?

To determine if a variable Q is cyclic, one must inspect the Hamiltonian function H(Q, P, t). If H does not explicitly depend on Q, then Q is a cyclic variable. This means that the partial derivative of H with respect to Q is zero, indicating that Q does not influence the Hamiltonian directly.

What is the significance of identifying a cyclic variable in a Hamiltonian system?

Identifying a cyclic variable in a Hamiltonian system is significant because it indicates a conserved quantity. According to Noether's theorem, the conservation of the conjugate momentum P associated with the cyclic variable simplifies the problem, reducing the number of variables and equations needed to describe the system's dynamics.

Can you provide a simple example where a variable is shown to be cyclic in a Hamiltonian system?

Consider a simple Hamiltonian H = (P_x^2 + P_y^2) / 2m + V(x), where P_x and P_y are the momenta conjugate to the coordinates x and y, respectively, and V(x) is a potential that depends only on x. Here, y does not appear in the Hamiltonian, making it a cyclic variable. Consequently, the conjugate momentum P_y is conserved, simplifying the analysis of the system's motion.

Similar threads

Change in Hamiltonian under Non-Canonical transformation
Replies
1
Views
918
Hamiltonian mechanics - the independence of p and q
Replies
6
Views
2K
A Recover Hamilton equation from 2-form defined on phase space
Replies
3
Views
521
I Exploring the Flexibility of Coordinates in Euler-Lagrange Equations
Replies
5
Views
1K
Given a set of equations, show if it is a Hamiltonian system
Replies
9
Views
1K
Canonical transformations of a quantized Hamiltonian?
Replies
3
Views
1K
Exploring Effects of Adding Derivatives to Lagrangian on Hamiltonian Eqns.
Replies
7
Views
5K
Hamiltonian conjugate dynamic variables
Replies
3
Views
1K
I Canonical transformations and generating functions
Replies
6
Views
1K
Is any Hamiltonian system integrable?
Replies
1
Views
2K

Hot Threads

Recent Insights

Back
Top