I How do canonical transformations relate to Hamiltonians?

  • I
  • Thread starter Thread starter dyn
  • Start date Start date
  • Tags Tags
    Transformations
AI Thread Summary
Canonical transformations relate Hamiltonians through a numerical equivalence rather than identical functional forms. The Hamiltonian for a harmonic oscillator, H(p, q), and the transformed Hamiltonian K(P, Q) can yield the same numerical results when evaluated at corresponding values, despite their different expressions. The distinction arises from the use of different coordinate systems, such as Cartesian versus polar coordinates. K and H being equal in value does not imply they share the same functional form, which only occurs under specific symmetries. Understanding this relationship is crucial for grasping the broader concept of canonical transformations in Hamiltonian mechanics.
dyn
Messages
774
Reaction score
63
Hi
The Hamiltonian for a harmonic oscillator is H = 1/(2m) ( p2+m2ω2q2). A canonical transformation is then made to a new Hamiltonian K( P , Q )

It is said that K ( P , Q ) = H ( p , q ) but K ( P , Q ) = ωP ( cos2Q +sin2Q ) = ωP

I don't understand how K ( P , Q ) = H ( p , q ) when they have different forms ? I thought if K = H then they must have the same form but H is a sum of 2 squares but K just equals ωP

Thanks
 
Physics news on Phys.org
dyn said:
I don't understand how K ( P , Q ) = H ( p , q ) when they have different forms ?
They have different forms for different parameters, i.e. (P,Q) and (p,q). One uses Cartesian coordinates. Another uses polar coordinates.
 
Last edited:
Thank you. I think i might be getting confused with symmetries , so let me see if i have got this right.

K ( P , Q ) = H ( p , q ) means that if i evaluate H at a certain value of p and q and then evaluate K at the transformed values of P and Q i get the same numerical answer ? There is no implication that K and H have the same functional form ?

If K and H had the exact same functional form then i could write H ( P , Q ) = H ( p , q ) and this occurs when the canonical transformation is a symmetry ?

Is that right ? Thanks
 
K and H does not have the same function form. Q does not appear in Hamiltonian K, which is called cyclic coordinate. Hamilton equation of motion for conjugate momentum is
\dot{P}=0
The generating function of transformation is
\displaystyle W_{1}(q,Q)={\frac {1}{2}}m\omega q^{2}\operatorname {cot} {Q}
 
Last edited:
dyn said:
Thank you. I think i might be getting confused with symmetries , so let me see if i have got this right.

K ( P , Q ) = H ( p , q ) means that if i evaluate H at a certain value of p and q and then evaluate K at the transformed values of P and Q i get the same numerical answer ? There is no implication that K and H have the same functional form ?

If K and H had the exact same functional form then i could write H ( P , Q ) = H ( p , q ) and this occurs when the canonical transformation is a symmetry ?

Is that right ? Thanks
This post is a general question. It is not specific to the harmonic oscillator. I am just trying to find out if i understand the concept in general terms ?
 
Thread 'Is 'Velocity of Transport' a Recognized Term in English Mechanics Literature?'
Here are two fragments from Banach's monograph in Mechanics I have never seen the term <<velocity of transport>> in English texts. Actually I have never seen this term being named somehow in English. This term has a name in Russian books. I looked through the original Banach's text in Polish and there is a Polish name for this term. It is a little bit surprising that the Polish name differs from the Russian one and also differs from this English translation. My question is: Is there...
This has been discussed many times on PF, and will likely come up again, so the video might come handy. Previous threads: https://www.physicsforums.com/threads/is-a-treadmill-incline-just-a-marketing-gimmick.937725/ https://www.physicsforums.com/threads/work-done-running-on-an-inclined-treadmill.927825/ https://www.physicsforums.com/threads/how-do-we-calculate-the-energy-we-used-to-do-something.1052162/
Hi there, im studying nanoscience at the university in Basel. Today I looked at the topic of intertial and non-inertial reference frames and the existence of fictitious forces. I understand that you call forces real in physics if they appear in interplay. Meaning that a force is real when there is the "actio" partner to the "reactio" partner. If this condition is not satisfied the force is not real. I also understand that if you specifically look at non-inertial reference frames you can...
Back
Top