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Phase space

The emphasis for this week in my non-linear physics class was on phase space. Analysis of dynamical systems in phase space (or configuration/state space) converts a numerical problem into a geometric one. Even if you aren’t able to solve the underlying differential equation, the geometric picture helps you make predictions on the behavior of the system. From control engineering, an analogy with the Root Locus method was drawn.

pend0.png

The picture above shows the phase space for the pendulum equation:

\theta^\prime^\prime + \sin \theta = 0

Each point on the phase space acts as an initial condition for the system. As I had written earlier, solving this equation is not trivial. If instead we look at the phase plane, we can determine three distinct regions of operations:

  • fixed points at the origin and at the intersection of the two red lines. These correspond to steady states, but aren’t very interesting.
  • closed orbits around the origin. These correspond to periodic motions.
  • trajectories apart from orbits and fixed points.

Physically, we can reason that for a high initial velocity (higher up on y-axis), the period tends to infinity. The position (x-axis) also increases constantly.

First Integral

\theta^\prime^\prime + \sin \theta = 0

\theta^\prime \theta^\prime^\prime + \theta^\prime \sin \theta = 0

[ \frac{1}{2} (\theta^\prime)^2]^\prime + [ -\cos \theta]^\prime = 0

[ \frac{1}{2} (\theta^\prime)^2 -\cos \theta]^\prime = 0

which means that

\frac{1}{2} (\theta^\prime)^2 -\cos \theta = \epsilon

where \epsilon is a constant. These indicate level curves in the phase plane, corresponding to constant “energy.”

For non-dissipative systems, flows in phase space have constant volume. If you were to pick a circle centered around one of the periodic trajectories, representing an ensemble of trajectories, then the area (in 2D) of the circle remains constant. Ofcourse, if the trajectories are not circle (as they are near the origin), the circle we initially start with gets deformed. This can be seen in the following figure:

pendn.png

Computing the area of the circle is simple, but computing the area of the arbitrary deformed polygon is a little harder. This is the focus of the next section.

Area of a polygon

Consider a polygon of N edges with vertices (x_i,y_i). The polygon should also be closed so that (x_0,y_0) is the same vertex as (x_N,y_N). The area is given by

A = \frac{1}{2} \Sigma_{i=0}^{N-1} ( x_i y_{i+1} - x_{i+1} y_i)

This method works by projecting a line from each vertex to a horizontal line (usually the x-axis) below polygon. The area enclosed between two such line segments is the sum of the area of the triangle in the polygon and the area of the trapezoid outside the polygon. As we loop through the vertices of the polygon, the area outside the polygon eventually cancels out. This is a form of the Stokes Theorem.

Possibly related:

This entry was posted on Sunday, January 14th, 2007 at 3:13 pm and is filed under Physics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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