Can Canonical Transformations Solve Velocity Dependent Potential Problems?

[QuantumDefect]

Hello,
I've looked through a couple books on this subject and found the basic theory but none actually apply it to a problem. I was wondering if someone would be so kind as to maybe do a practice problem for me? The reason I say this is because I have a homework problem and have solved for the hamiltonian and the canonical equations however, I would like to find a new set of co-ordinates in which the momenta might be constants. In my problem I have a co-ordinate velocity dependent potential and would like to find the transformations in which the hamiltonian is a constant. Is this possible? I have just gotten into the Hamiltonian formalism and am extremely excited, more so when I first learned about lagrangian mechanics. Thank you so much.

 

[Physics Monkey]

Hi QuantumDefect,

These methods are really cool, aren't they? I could do a problem for you, something simple like the harmonic oscillator. Such a problem is almost certainly worked in greater detail in your book, but if you're actually interested I could step you through it.

Unfortunately, there is no good way to just guess or figure out a canonical transformation that leaves the Hamiltonian independent of the coordinates. Is it possible? Sure. Is it easy to find? Almost never. A general approach starts from the Hamilton-Jacobi equation. The general approach says that the solution to the HJ equation (a non-linear differential equation) is the desired generator of the transformation. This method has the advantage of casting the problem in a form (differential equation) familiar to most physicists, but the problem is equally insoluable from an analytic point of view.

Without some details, I probably couldn't say much more. Feel free to post your problem and we can talk about it.

 

[QuantumDefect]

Thanks Physics Monkey!
Thats why none of the books took it into greater depth! I'll read up more on the Hamilton-Jacobi equation and if I have any questions, I'll come back. However, you answered my question and I am extremely grateful.Many thanks,

~QuantumDefect