Sunday, 17 May 2015

hamiltonian formalism - Finding action-angle variables


Given a 1 d.o.f Hamiltonian H(q,p) what is the general procedure for finding action angle variables (I,θ)?


I have read the Wikipedia page on action angle variables and canonical transforms but have difficulty applying the general methods to specific problems. Can someone explain the method to me using a simple general example?



Answer



In local coordinates the canonical transformation to action angle coordinates (q,p)(Q,P) can be related by, Pi=12πpidqi     and     Qi=Pipidqi

For Example:


Consider the one dimensional harmonic oscillator with the following Hamiltonian H=12m[p2+m2ω2q2]. Rearrange this for p and take the hypersurface H=E. p=±2mEm2ω2q2

Then use the above equation to compute P. P=12π2mEm2ω2q2dq
The integral is now over 0 to 2π which is easier to handle. This works out as, 12π2π0cos2Q dQ2Eω=Eω
Therefore we have used the quoted formula to compute the action variable for the harmonic oscillator.


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