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4. Geometrical meaning of the parameter $\alpha$

It is instructive to have a visual picture of the map ${\bf r}\mapsto{\bf r}^\prime$ of the sphere $S^2$ implemented by the operator $P({\bf n},\alpha ).$ To this end let us assume the vector ${\bf n}$ is pointing North, i.e. ${\bf n}=(0,0,1)$. Then the result ${\bf r}^\prime$ of applying the operator $P({\bf n},\alpha )$ to a point ${\bf r}$ on the sphere is on the same longitude as the original point ${\bf r}$, but its latitude $\theta$ changes - it moves towards the North Pole along its meridian, the new latitude being given by the formula:


\begin{displaymath} \theta^\prime = \arccos\left (\frac{(1 - \alpha^2)\cos(\thet... ...pha\cos(\theta))}{1 + \alpha^2 + 2\alpha\cos(\theta)}\right ) \end{displaymath} (20)

Figure: The amount of shift $\theta-\theta^\prime (\theta ), $ as a function of $\theta$ for different values of $\alpha .$

Remark: Here $\theta$ is not exactly the "geographical latitude". It is zero at the "North Pole" ($\sigma_3=+1$), $90$ degrees at the equator, and $180$ degrees at the "South Pole" ($\sigma_3=-1$).

Each map $P({\bf n},\alpha )$ maps the sphere onto itself in an injective way. For ${\bf n}=(0,0,1)$ the map is easy to picture. All points of the sphere move towards the North Pole along their meridians, except of the two fixed points: North and South Pole. All of the Northern hemisphere, and a strip below the equator, shrinks, while the other part, near the South Pole, stretches. The amount of stretching can be found by plotting the function $\theta-\theta^\prime (\theta )$ - it has a maximum at $\theta$ corresponding to $z=-\alpha$. Thus the parameter $\alpha$ gets a simple interpretation: it is the value of $z$ coordinate for which shrinking of meridians is replaced by stretching - an equilibrium point. This point is always on the southern hemisphere. For $\alpha$ close to zero, where the map is close to the identity map, the equilibrium point is close to the equator. Then, as $\alpha$ approaches the value of $1.0$, corresponding to the sharp projection operator, the equilibrium latitude gets closer and closer to the South Pole. In the limit of $\alpha=1$ all of the sphere shrinks to the North Pole, only the South Pole remains where it was.

next up previous
Next: 5. Quantum Fractals and Up: 4. Notes Previous: 3. Importance of fuzziness

2002-04-11