Mast Kalandar

bandar's colander of random jamun aur aam

Tue, 21 Oct 2008

Going for a spin


geometry, math [link] [comments ()] [raw]

When a sphere (globe or ball) is turned about its centre a number of times, there is at least one point that starts and ends at the same place. This a consequence of the algebraic statment that a 3x3 orthogonal matrix with determinant has 1 as an eigenvector with eigenvalue 1. One of my students recently asked me for a geometric proof and I came up with the following construction. (Of which I am evidently quite proud!)

We begin by making a more precise statement of the above proposition. Let a, b, c, ..., z be a succession of points on the sphere and A, B, C, ..., Z be a sucession of rotations; where A is a rotation about the axis through a, B a rotation about the axis through b and so on. We will show how to find a point on the sphere so that the combined effect of these rotations is a rotation about the axis through that point.

It is enough to show how this can be done for two successive rotations since the rest of the argument/construction follows by induction/recursion.

So we restrict ourselves to the simpler case of two points a and b on the sphere and two rotations A and B respectively about the axes through these points.

Let me introduce some elementary notions and results. The term "great circle" is used for the circle obtained by cutting a sphere with a plane passing through the centre of the sphere; any two points on the sphere lie on a (common) great circle. The term "antipode of a point p" denotes a point q that is diametrically opposite to the given point p; the axis through a point also passes through its antipode. Two points on the sphere that are not antipodal lie on a unique great circle, whereas the antipode of p lies on every great circle through p.

Consider the point c which goes to the point b under the rotation A and cb be the great circle containing these two points. Let d denote a point on bc that lies halfway between b and c. There are two such points which are antipodes; pick any one. Let ad be the great circle containing a and d.

Similarly, let e be the point which a goes to under the rotation B and ae be the great circle containing these two points. Let f denote a point on ae that lies halfway between a and e; let bf denote the great circle which contains b and f.

A point g (and its antipode) where the great circle ad meets the great circle bf is fixed under the combination of the two rotations. In fact, the result of the rotation A followed by the rotation B is a rotation G about the axis through the point g.

The proof that this construction works relies on the following description of a rotation X of a sphere about the axis through a point x on the sphere. Let y be any other point and z be its image under the rotation. As above let t be the point halfway between y and z on the great circle containing these two points. The rotation X is obtained as a succession of two reflections; first a reflection in the plane containing the x, t and the centre of the sphere and second a reflection in the plane containing x, z and the centre of the sphere. One can see this by noting that the composite of two reflections is indeed a rotation; moreover, a rotation that fixes x is uniquely determined by what it does to y.

We use the above description to write the composite of A and B as succession of four reflections through planes as follows:

the plane containing d, a and the centre of the sphere; the plane containing b, a and the centre of the sphere; the plane containing a, b and the centre of the sphere; the plane containing f, b and the centre of the sphere

The two reflections in the middle cancel each other out. We are then left with the composite of two reflections. Clearly, the points that lie on the fixed plane of each of these reflections are fixed by this composite rotation. This intersection is an axis of the sphere through the point g.


Mon, 20 Oct 2008

Xen on lenny x86_64

Sun, 05 Oct 2008

Will I be at FOSS.in 2008?

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