I have an equation of potential given:

$$U = U_0\tan^2( \alpha(t)q)$$

I need to find a motion rules for that potential in terms of action-angle variables. Using the fact that Hamiltonian is equal to energy of the system I write:

$$p = \pm \sqrt{2m(E-U_0\tan^2( \alpha(t)q)}$$

By definition the action variable is defined as

$$ I = \frac {1} {2\pi} \oint_{\gamma(E)}pdq \, ,$$

so I receive the following

$$I=\frac {1} {2\pi} \oint_{\gamma(E)}\pm \sqrt{2m(E-U_0\tan^2( \alpha(t)q)} \, dq \, .$$

This integral is not easy to take so did I maybe missed somewhere? Also I'm not sure about integral limits: should that be

$$q_1=\frac {1} {a(t)} \arctan(\sqrt{\frac EU_0}); q_2=0$$


$$q_1=\frac {1} {a(t)} \arctan(\sqrt{\frac EU_0}); q_2=\frac {1} {a(t)} \arctan(\sqrt{\frac EU_0}) + \pi$$

because period of $\tan^2(a(t)q)$ is $\pi$

or even

$$q_1=\frac {1} {a(t)} \arctan(\sqrt{\frac EU_0}); q_2=-\frac {1} {a(t)} \arctan(\sqrt{\frac EU_0})$$

because on phase space q is actually different only by a sign (+ or -)?

I've attached images of my solutions on paper:



  • $\begingroup$ I don't get the $\alpha(t)$ part but see physics.stackexchange.com/questions/327121/… and math.stackexchange.com/questions/2131086/… $\endgroup$ Feb 2, 2020 at 22:36
  • $\begingroup$ a(t) is just a some function depends of t. Yes I saw that answers but could you please confirm that I'm right with chosen limits? -1/a(t) arctan(); +1/a(t) arctan() $\endgroup$
    – fairafair
    Feb 3, 2020 at 7:48
  • $\begingroup$ Sorry but this is not a check-my-work site, especially as the answer is already elsewhere and can be adapted with minimal effort. $\endgroup$ Feb 3, 2020 at 7:49

1 Answer 1


Let me remark first, that a general form time-dependent hamiltonain with one degree of freedom $$H(p, \, q , \,t )$$ is unlikely to be integrable and can have a rather complicated dynamics. The time dependence of $\alpha(t)$ makes the given hamiltonain time-dependent.

Usually, when transforming hamiltonain systems, one applies generating functions. In this case, we have a time-dependent potential, which means that the hamiltonain $$H(p,\,q,\,t) = \frac{p^2}{2m} \, + \, U_0\,\tan^2\Big(\alpha(t)\, q\Big)$$ so we would look for time dependent generating function that moves from original $(p, \, q, \, t)$ coordinates to action-angle coordinates on each spacial slice $\{(p, q) \, \in \, \mathbb{R}^2\} \times\{t\}$. However, the computations for this particular example seem to be fairly heavy.

Step 1: Solve the indefinite integral $$G(k,\, x) = \int_{0}^{x} \, \sqrt{k - \tan^2(w)}\, dw$$ Step 2: Form the function \begin{align} F\big(E, \,q,\, t\big) &= \sqrt{2mU_0}\int_{0}^{q} \, \sqrt{\frac{E}{U_0} - \tan^2\Big(\alpha(t)\,u\Big)}\, du \\ &= \frac{\sqrt{2mU_0}}{\alpha(t)}\int_{0}^{\alpha(t)q} \,\sqrt{\frac{E}{U_0} - \tan^2\big(w\big)}\, dw \\ &= \frac{\sqrt{2mU_0}}{\alpha(t)}\,\, G\left(\frac{E}{U_0}, \, \alpha(t)q\right) \end{align} Step 3: Calculate the action variable \begin{align} I = I(E,\, t) &= \frac{1}{2\pi} \oint_{H_t = E} \, p\, dq = \frac{1}{2\pi} \, 4\, \sqrt{2mU_0} \int_{0}^{\frac{1}{\alpha(t)}\arctan\left(\sqrt{\frac{E}{U_0}}\right)}\, \sqrt{\frac{E}{U_0} - \tan^2\Big(\alpha(t)\,u\Big)}\, du \\ &= \frac{2\sqrt{2mU_0}}{\pi\, \alpha(t)} \int_{0}^{\arctan\left(\sqrt{\frac{E}{U_0}}\right)}\,\sqrt{\frac{E}{U_0} - \tan^2\big(w\big)}\, dw\\ &= \frac{2\sqrt{2mU_0}}{\pi\, \alpha(t)} \,\, G\left(\,\frac{E}{U_0}, \,\, \arctan\left(\,\sqrt{\frac{E}{U_0}}\,\right)\,\right) \end{align}
Step 4: Solve the equation $I = I(E, \, t)$ for $E$ to find the function $$E = E(I,\, t)$$ Step 5: Define the generating function $$S(I, \, q, \, t) = F\Big(E(I, \, t), \, q, \, t\Big)$$ Step 6: Calculate the angle variable $$\varphi = \frac{\partial S}{\partial I}\big(I, \, q, \, t\big)$$ Step 7: Solve the equation $\varphi = \frac{\partial S}{\partial I}\big(I, \, q, \, t\big)$ for $q$ to find the function $$q = q(I, \, \varphi, \, t)$$ Step 8: In order to find the hamiltonain function $\tilde{H}\big(I, \, \varphi, \, t\big)$ with respect to the action-angle variables $(I, \varphi)$, recall the Hamilotn-Jacobi equation: $$\frac{\partial S}{\partial t}\big(I, \, q, \, t\big) \, + \, H\left(\frac{\partial S}{\partial q}\big(I, \, q, \, t\big), \, q, \, t\right) = \tilde{H}\left(I, \, \frac{\partial S}{\partial I}\big(I, \, q, \, t\big), \, t\right)$$ By construction, we have that
$$ H\left(\frac{\partial S}{\partial q}\big(I, \, q, \, t\big), \, q, \, t\right) = E(I, \, t)$$ so the hamiltonian in action-angle coordinates should be $$\tilde{H}\left(I, \, \varphi, \, t\right) = \frac{\partial S}{\partial t}\big(I, \, q, \, t\big)\Big{|}_{q = q(I,\, \varphi, \, t)} \, + \, E(I, \, t)$$ Step 9: The equations of motion in the new action-angle variables are \begin{align} \frac{d I}{dt} &= \frac{\partial\tilde{H}}{\partial \varphi}\big(I, \, \varphi, \, t\big)\\ \frac{d \varphi}{dt} &= - \, \frac{\partial\tilde{H}}{\partial I}\big(I, \, \varphi, \, t\big) \end{align}

  • $\begingroup$ Thank you! But does it mean that I have to take a double integral of action variable (for q and for t)? Because by definition dI/dt should be equal to zero while in that equations it will be const*da(t)/dt? I've found some equations from lecture on my textbook (imgur.com/a/UvbFJXt) which seems to be related to that example but couldn't understand them much because I don't remember what was said for that =( p.s. you have a little misspelling in Hamiltonian (p^2/1m) $\endgroup$
    – fairafair
    Feb 4, 2020 at 20:54
  • 1
    $\begingroup$ No, you do not have to take double integral of the action variable. The only time you need to take integral is when computing the generating function, which in this example, is equivalent to finding the anti-derivative $G$. Your notes show the same philosophy, it is just that there instead of $\alpha$ the parameter $\lambda$ is used and the discussion is on hamiltonain systems with several degrees of freedom (your example is with one degree) and the generating function is separable, i.e. it is the sum of separate generating functions (in your example this is automatically the case). $\endgroup$ Feb 4, 2020 at 21:23
  • 1
    $\begingroup$ I have written the generating function directly as $S(I, q, t)$ but you can also think of it as $\tilde{S}\big(I,\, q,\, \alpha(t)\big)$ and when calculating $\frac{\partial S}{\partial t}$ one can use the chain rule $\frac{\partial \tilde{S}}{\partial \alpha}\, \frac{\partial \alpha}{\partial t}$. $\endgroup$ Feb 4, 2020 at 21:26
  • 1
    $\begingroup$ @fairafair After you calculate the integral in step 1 and find the function $G$, you just have to plug a bunch of stuff: in step 2 it's $k = E/U_0$ and $x = \alpha(t)q$ (and there is a factor in front of $G$), in step 3 it's $q = \frac{1}{\alpha(t)} \arctan\Big(\sqrt{E/U_0}\Big)$, in step 4 and 5 you replace $E$ by $I$ and obtain the generating function $S$. From that moment on, it is mostly differentiation and (again) substitution. $\endgroup$ Feb 4, 2020 at 21:40
  • $\begingroup$ @fairafair I just realized that in your notes, what I read as $\lambda$ is actually $\alpha$, so in your notes and in this example it is always $\alpha = \alpha(t)$. $\endgroup$ Feb 4, 2020 at 22:42

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