The motion of tectonic plates can be expressed as a rotation on the sphere

By a displacement of a sphere we mean a rigid movement of its surface in
which the center of the sphere remains fixed, and the distance between any two
points remains the same. That is, two points A and B move to points A'=f(A) and
B'=f(B), where the *one-to-one* mapping f() preserves the distance between
the points. The movement of crustal plates in plate tectonics is an example of
such displacements.

A nontrivial displacement of the sphere has the remarkable property that one, and only one, pair of diametrically opposite points has the same position before and after the displacement. Any displacement can, therefore, be regarded as a rotation about an axis passing through its two invariant points. A general movement of the sphere is a succession of such displacements, for each of which there may be different pairs of unmoved points.

The proof of such a fundamental property ought to be simple and satisfying, removing all doubt of its truth. If you do not already know one, you might like to look away at this point and see if you can come up with one, before reading the proof given below.

First, we agree that any displacement can be specified by considering two points A and B and giving their positions before and after the movement, under the sole restriction that the distance between them does not change. Let us say that A is carried into A', and B into B'. Construct the perpendicular bisectors of AA' and BB' (on the sphere) and extend them. They will meet at two opposite points C and C', because great circles always will. In triangles ABC and A'B'C the three sides are equal, each to each, because of the properties of the perpendicular bisector, so the triangles are congruent, and angles ACB and A'CB' are equal. Point A has rotated along a small circle around C through an angle ACA' = ACB + BCA', while point B has rotated through an angle BCB' = BCA' + A'CB'. Since the same angle BCA' has been added to equals, the sums are equal, and ACA' = BCB'. Therefore, the movement of these two points is a a rotation of the sphere about an axis passing through C. The movement of any other point on the sphere can be obtained in the same way. Only pooints C and its diametrically opposite point C' remain fixed. This was what was to be proved.

The same theorem holds in a plane, if we allow a linear displacement to be a rotation about a point at infinity. For a small displacement, our theorem locates the instantaneous center of rotation, a very useful concept in plane kinematics that makes it easy to find the velocity at any point. When considering motion, one must remember that the center of rotation C is not, in general, a fixed point, but may itself be moving. The existence of a center of rotation is a consequence of the rigidity of the plane or the spherical lamina, that merely emphasizes that all you can do with a rigid body is rotate it. The theorem seems more remarkable to me when applied to a sphere, where it might seem that a general motion might have to involve rotations about different axes. The theorem shows that this is not true, and a single rotation will suffice.

Crustal plates are assumed to be rigid, so they will move like the laminas discussed above. The pole of the displacement is called the *pole of spreading*, and together with the angle of rotation specifies the movement. The *transform faults* that occur at right angles to the spreading centres when the centres are displaced are on small circles whose centre is the pole of spreading.

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Composed by J. B. Calvert

Created 16 May 1999

Last revised 8 April 2001