Hamilton-jacobi and action angle

  • Thread starter subny
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hi

i am a MS physics student. We are following the goldstein book for classical mech and have exams next week.

Does anyone know of any good (easy - "how to do it" procedure explained) notes or pdf hand outs on the Hamilton Jacobi eq and the action angle variables.

I do not have time to study too deeply - I only want to get the hang of it and know how to do the sums.

If you know any such problem solving oriented notes pls post the links here or in my inbox.

Alternatively you can write a coupel of lines here yourself explaining how to go about the sums -


EG - for lagrangian -


1 - find the degrees of freedoms -
2 - for each one choose a generalized coordinate/variable.
3 - find out the PE and the KE
4 - The Lagrangian then is L = KE -PE
5 - solve the EL eq for each generalized coordinate.


Something very simplistic like this - Thanks a lot.
 

Answers and Replies

  • #2
vanhees71
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I don't know, what you mean by "sums". The concept of the Hamilton-Jacobi Partial Differential Equation (HJPDE) is to look for new canonical coordinates ##(Q_i,P_i)## such that the motion is described by ##(Q_i,P_i)=\text{const}.## (let me warn you that in almost all cases there is no solution, but only for "integrable systems", i.e., systems where you have sufficient symmetry so that you have enough conserved quantities according to Noether's theorem). One can show that this is equivalent to the demand that the new Hamiltonian vanishes:
$$H'(Q,P,t)=0.$$

To that end you look for the corresponding generator of the canonical transformation. We choose the generator in the form ##g(q,P,t)##. Then
$$p=\partial_q g, \quad H'(Q,P,t)=H(q,p,t)+\partial_t g \stackrel{!}{=}0.$$
This leads to the HJPDE,
$$H \left (q,\frac{\partial g}{\partial p},t \right ) + \frac{\partial g}{\partial t}=0.$$
If ##H## is not explicitly time dependent, we know that the energy is conserved. Then
$$\frac{\partial g}{\partial t}=-H=-E \; \Rightarrow \; g(P,q,t)=-E t + S(P,q).$$
The HJPDE then simplifies to
$$H \left (q,\frac{\partial g}{\partial q} \right )=E.$$
Let's look at the harmonic oscillator as the most simple example. The Hamiltonian is
$$H=\frac{p^2}{2m}+\frac{m \omega^2}{2} q^2.$$
We can apply the simplified version, because energy is conserved since ##H## doesn't depend explicitly on time. Thus ##E## is the new canonical momentum, and we have the HJPDE
$$\frac{1}{2m} \left (\frac{\partial S}{\partial q} \right )^2 + \frac{m \omega^2}{2} q^2=E.$$
We can easily find a solution, because it's just an integral:
$$\frac{\partial S}{\partial q}=\sqrt{2 m E-m^2 \omega^2 q^2}$$
with the solution
$$S(q,E)=\int \mathrm{d} q \sqrt{2 m E-m^2 \omega^2 q^2}.$$
We don't need to consider an ##E##-dependent integration constant, because this only adds another ##E##-dependent constant to the constant ##Q##, which we get from
$$g=-E t+S(q,E) \; \Rightarrow \; Q=\frac{\partial g}{\partial E}=-t+\frac{1}{\omega} \arcsin \left (\sqrt{\frac{m \omega^2}{2E}} q \right ).$$
Resolved to ##q##:
$$q=\sqrt{\frac{2E}{m \omega^2}} \sin[\omega(t+Q)].$$
Of course, ##E## and ##Q## are determined by the initial conditions ##q(t_0)=q_0## and ##v(t_0)=\dot{q}(t_0)=v_0##.
 

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