Course:PHYS350/Action-Angle Variables

Requirements

A powerful technique for systems that have as many constants of the motion as degrees of freedom is called the action-angle variables. Integrable systems in $n-$dimensions have $n$ constants of the motion that we will call $F_{i}$. We would like to transform the Hamiltonian from $p_{i}$ and $q_{i}$ to a new set of momenta and coordinates, $P_{i}$ and $Q_{i}$ such that

1. The equations of motion are preserved
2. The new momenta $P_{i}$ are functions of the constants of the motion $F_{i}$.
3. The new coordinates $Q_{i}$ are ignorable. They don't appear in the Hamiltonian.

With the Lagrangian technique we were free to use different coordinates to simplify the equations of motion. In the Hamiltonian picture we can define the coordinates and the momenta. Let's do an example. Let's say that we have a particular set of $p_{i}(t)$ and $q_{i}(t)$ that satisfy the equations of motion, then we must have

$\delta \int Ldt=\delta \int \left(\sum _{i}p_{i}{\dot {q}}_{i}-H(p_{i},q_{i},t)\right)dt=0$

from the principle of least action. If we want new set of coordinates and momenta that satisfy the equations of motion we have

$\delta \int Ldt=\delta \int \left(\sum _{i}P_{i}{\dot {Q}}_{i}-H(p_{i},q_{i},t)\right)dt=0.$

The New Variables

For both of these to be true in general it is sufficient (but not necessary) that we have

$\int \sum _{i}p_{i}{\dot {q}}_{i}dt=\int \sum _{i}P_{i}{\dot {Q}}_{i}dt$

so the transformation preserves the area. For motion that is periodic in the coordinates, we can take the integrals over a period of the motion in one of the coordinates. To make life easy for ourselves, we would like for new coordinate ${\dot {Q}}_{i}$ to increase linearly in time. For example it could go from 0 to $2\pi$ in one oscillation, so ${\dot {Q}}_{i}=2\pi /T_{i}$ where $T_{i}$ is the time for the particle to oscillation in the coordinate once.

Momenta

To achieve this we could pick

$P_{i}={\frac {1}{2\pi }}\oint p_{i}dq_{i}$

which is clearly a constant of the motion so we have

$\int _{t}^{t+T_{i}}P_{i}{\dot {Q}}_{i}dt=P_{i}{\frac {2\pi }{T_{i}}}T_{i}=2\pi {\frac {1}{2\pi }}\oint p_{i}dq_{i}=\int _{t}^{t+T_{i}}p_{i}{\dot {q}}_{i}dt$

and the principle of least action is preseved coordinate by coordinate. We can use Hamilton's equations to determine

${\dot {Q}}_{i}={\frac {\partial H}{\partial P_{i}}}$

and system is solved at least formally.

We may like to know the old coordinates in terms of the new ones. We have

$\int _{t_{0}}^{t}p_{i}{\dot {q}}_{i}dt=\int _{t_{0}}^{t}P_{i}{\dot {Q}}_{i}dt$

$\int _{q_{i,0}}^{q_{i}}p_{i}dq'_{i}=\int _{0}^{Q_{i}}P_{i}dQ_{i}'=\int _{0}^{P_{i}}Q_{i}dP_{i}'$

Let's take the derivative of both sides with respect to $P_{i}$ to yield

$Q_{i}={\frac {\partial }{\partial P_{i}}}\int _{q_{i,0}}^{q_{i}}p_{i}dq'_{i}={\frac {\partial }{\partial P_{i}}}\int ^{q_{i}}p_{i}dq'_{i}$

where the final equation holds because the initial value of $q_{i}$ is arbitrary.

Some Examples

Harmonic Oscillator

For a harmonic oscillator we have

$H={\frac {1}{2}}m{\dot {x}}^{2}+{\frac {1}{2}}kx^{2}={\frac {p^{2}}{2m}}+{\frac {1}{2}}kx^{2}.$

The energy is clearly conserved. Let's take the energy equal to ${\frac {1}{2}}kx_{0}^{2}$ where $\pm x_{0}$ are the bounds of the motion, so we have

${\frac {1}{2}}kx_{0}^{2}={\frac {p^{2}}{2m}}+{\frac {1}{2}}kx^{2}.$

What is the area of the path through phase space as $x$ goes from $x_{0}$ to $-x_{0}$ and back again? The equation above clearly defines an ellipse in phase space

$1={\frac {p^{2}}{kmx_{0}^{2}}}+{\frac {x^{2}}{x_{0}^{2}}}$

of area $\pi {\sqrt {km}}x_{0}^{2}$, so

$P={\frac {\sqrt {km}}{2}}x_{0}^{2}={\sqrt {\frac {m}{k}}}H={\sqrt {\frac {m}{k}}}\left({\frac {p^{2}}{2m}}+{\frac {kx^{2}}{2}}\right)$

and

$H={\sqrt {\frac {k}{m}}}P$

and

${\dot {Q}}={\sqrt {\frac {k}{m}}}\equiv \omega .$

We have found the period of the oscillation without really solving any differential equations and only some basic geometry and algebra. We didn't get any information about the shape of the oscillation $x(t)$ but possibly we weren't interested in that in the first place.

We can find $x(t)$ geometrically as well

$1={\frac {p^{2}}{kmx_{0}^{2}}}+{\frac {x^{2}}{x_{0}^{2}}}$

that can be solved with

$p={\sqrt {km}}x_{0}\cos Q,x=x_{0}\sin Q$

where

$Q={\sqrt {k/m}}t+\beta$.

We can also do this using the integral relationship; we have

$Q={\frac {\partial }{\partial P}}\int ^{q}pdq={\frac {\partial }{\partial P}}\int ^{q}{\sqrt {2m\omega P-kmx^{2}}}dx$

        $=\int ^{q}{\frac {m\omega }{\sqrt {2m\omega P-kmx^{2}}}}dx=\int ^{q}{\sqrt {\frac {m\omega }{2P}}}{\frac {1}{\sqrt {1-{\frac {k}{2\omega P}}x^{2}}}}dx$

        $={\sqrt {\frac {m\omega }{2P}}}{\sqrt {\frac {2\omega P}{k}}}\sin ^{-1}{\sqrt {\frac {k}{2\omega P}}}x=\sin ^{-1}{\frac {x}{x_{0}}}$

so

$x=x_{0}\sin Q\,$

as before. We can obtain the value of $p$ in terms of $Q$ using the conservation of energy.

Gravity

Let's write out the Hamilonian for a particle in a gravitational field of a point mass. Let's use polar coordinates to get

$H={\frac {1}{2}}mr^{2}{\dot {\theta }}^{2}+{\frac {1}{2}}m{\dot {r}}^{2}-{\frac {km}{r}}={\frac {p_{\theta }^{2}}{2mr^{2}}}+{\frac {p_{r}^{2}}{2m}}-{\frac {km}{r}}.$

$p_{\theta }$ is clearly a constant of the motion so we have

$P_{\theta }={\frac {1}{2\pi }}\oint p_{\theta }d\theta ={\frac {1}{2\pi }}\int _{0}^{2\pi }p_{\theta }d\theta =p_{\theta }.$

We get this simple result because $\theta$ is already an angular coordinate (it is periodic in $2\pi$) and it is cyclic.

The Hamiltonian is clearly constant; let's take its value to be $E$ to give

$E={\frac {P_{\theta }^{2}}{2mr^{2}}}+{\frac {p_{r}^{2}}{2m}}-{\frac {km}{r}}.$

Let's solve this for $p_{r}^{2}$ to yield

$p_{r}^{2}=2mE+{\frac {2km^{2}}{r}}-{\frac {P_{\theta }^{2}}{r^{2}}}$

so

$P_{r}={\frac {1}{2\pi }}\oint {\frac {\sqrt {2mEr^{2}+2km^{2}r-P_{\theta }^{2}}}{r}}dr$

      $={\frac {\sqrt {-2mE}}{2\pi }}\oint {\frac {\sqrt {-r^{2}-{\frac {km}{E}}r+{\frac {P_{\theta }^{2}}{2mE}}}}{r}}dr$

      $={\frac {\sqrt {-2mE}}{\pi }}\int _{a}^{b}{\frac {\sqrt {(r-a)(b-r)}}{r}}dr$

where $a$ and $b$ are the minimum and maximum radii respectively. Continuing we have

      $={\frac {\sqrt {-2mE}}{\pi }}\left[{\frac {1}{2}}\pi \left(a+b\right)-\pi {\sqrt {ab}}\right]$

We don't know $a$ and $b$ but we don't have to. Let's multiply out the two terms to give

$(r-a)(b-r)=-r^{2}+\left(a+b\right)r-ab$

so

$P_{r}={\frac {\sqrt {-2mE}}{\pi }}\left[{\frac {1}{2}}\pi \left(-{\frac {km}{E}}\right)-\pi {\frac {|P_{\theta }|}{\sqrt {-2mE}}}\right]$

      $=-|P_{\theta }|+{\frac {1}{2}}km{\sqrt {\frac {2m}{-E}}}$

Let's solve for $H=E$ in terms of $P_{r}$ and $P_{\theta }$ to yield

$H=E=-{\frac {k^{2}m^{3}}{2\left(P_{r}+|P_{\theta }|\right)^{2}}}$

so we find that

${\dot {Q}}_{r}={\dot {Q}}_{\theta }\operatorname {sgn} P_{\theta }={\frac {k^{2}m^{3}}{\left(P_{r}+|P_{\theta }|\right)^{3}}}={\frac {1}{k}}\left(-{\frac {2E}{m}}\right)^{3/2}=\left({\frac {r_{\rm {min}}+r_{\rm {max}}}{2}}\right)^{-3/2}{\sqrt {k}}$

so the orbits must be closed and we recovered Kepler's third law. We don't know the shape but we do know that the two non-linear frequencies are equal to each other.

Canonical Transformations

The action-angle variables are specific case of a general class of canonical transformations. Canonical transformations are changes in the coordinates and momenta that preserve the equations of motion. Specifcally, we have

$\delta \int Ldt=\delta \int \left(\sum _{i}p_{i}{\dot {q}}_{i}-H(p_{i},q_{i},t)\right)dt=0$

from the principle of least action. If we want new set of coordinates and momenta that satisfy the equations of motion we have

$\delta \int Ldt=\delta \int \left(\sum _{i}P_{i}{\dot {Q}}_{i}-K(P_{i},Q_{i},t)\right)dt=0.$

The most general set of transformations that satisfy these requirements are called canonical transformations. We can satisfy these if

$\lambda \left(\sum _{i}p_{i}{\dot {q}}_{i}-H\right)=\sum P_{i}{\dot {Q}}_{i}-K+{\frac {dF}{dt}}.$