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 Exercise [11.05] 
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Joined: 12 Jan 2010, 18:55
Posts: 2
Post Exercise [11.05]
The solution to this exercise is nicely depicted in Figure 41.4 of reference [1], which is attached.

Misner,Thorne,Wheeler Fig.41.4.GIF
Misner,Thorne,Wheeler Fig.41.4.GIF [ 6.47 KiB | Viewed 1484 times ]

In this picture, a rotation is first performed around the rotation axis pointing in z-direction with an angle Theta_1, followed by a rotation around the axis pointing in approximate y-direction with an angle Theta_2.

In analogy to Fig.11.4 of Penrose's book, a triangle can be constructed on the unit sphere showing the relation of the two consecutive rotations and the resulting composite rotation. The vertices of this triangle will be those where the three rotation axes intersect the unit sphere.

One side of the triangle is given by the two vertices corresponding to the first and second rotations. The other two sides, and thus vertex "3" are defined by the angles (-1/2 Theta_1) and (+1/2 Theta_2) at vertex "1" and "2", respectively.

The third vertex, "3", then corresponds to the rotation axis of the composite rotation, and the rotation angle of the composite rotation is two times the angle between the grand arcs given by vertices "1"-"3" and "2"-"3" (1/2 Theta_net in the picture).

That all this is correct, can easily be seen by using the insight gained from Exercise [11.6]. From there we know, that any 3D-rotation can be decomposed into two consecutive reflections on two planes, where the intersection of those planes is the rotation axis and the angle between the planes is half the rotation angle. Note, that the two planes can be freely rotated about this axis, it is only the angle between them that matters.

The sides "1"-"3" and "1"-"2" of the triangle on the unit sphere define just these two mirror planes for the first rotation and the sides "1"-"2" and "2"-3" for the second rotation. The possibility to rotate these planes freely around the rotation axes is used here to make the second mirror plane of the first rotation and the first mirror plane of the second rotation coincide. When composing the two rotations, or four consecutive reflections, the second and the third reflection, both "1"-"2" cancel each other, and only the reflections "1"-"3" and "2"-"3" remain.

[1] C. W.Misner, K.S.Thorne, J.A.Wheeler, Gravitation (Freeman 1973)

12 Jan 2010, 20:13

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [11.05]
No doubt the above is correct, but doesn't really give an intuitive picture of why this works.

Look at the following diagram:

11_05.png [ 14.77 KiB | Viewed 1252 times ]

Imagine that it's drawn on the surface of a sphere, so that vertices v1, v2, and v3 each represent axes of rotation through the sphere's center. The black triangle is formed by connecting them up (the edges are actually grand arcs). The blue triangles A, B, and C are formed by reflecting the black triangle along each of its edges in turn (technically, reflecting in the plane of the grand arc).

Now note that rotation Theta1 (around v1) takes A to B, Theta2 takes B to C, and Theta3 takes A to C directly... i.e. it's the result of composing rotations Theta1 and Theta2. But looking at the diagram, it should also be clear that the angles of the black triangle are half of Theta1, Theta2 and Theta3, respectively (after adjusting the sign of Theta3 appropriately).

So, I think that this diagram is probably what Prof. Penrose meant by the "dual" of figure 11.4b, as it intuitively shows how half-angles-of-rotation can be used to obtain the composite rotation (axis and angle) resulting from the composition of two other rotations.

Last edited by deant on 15 Jul 2010, 19:47, edited 4 times in total.

12 Jul 2010, 08:58

Joined: 12 Jan 2010, 18:55
Posts: 2
Post Re: Exercise [11.05]
very nice, I think the case is closed!

12 Jul 2010, 09:50
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