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Dynamics of Coin Tossing II

It appears that the instability that we described in the last article is indeed new; at least not known to experts such as Persi Diaconis and Richard Mongomery. At their encouragement I describe in more detail the effect of small pertrubations on a coin. We can look at this as a problem in Riemannian geometry or one in Classical Mechanics. Each brings its own insights and informs the other. Let us begin with the geometric point of view.

The Riemannian Submersion S^3\to S^2

In the orthogonal basis used above,

<e_1,e_1>=<e_2,e_2>=1, <e_3,e_3>=a^2

where we choose units such that

Q_1=1, Q_3=a^2. Since e_3 is a Killing vector whose orbits are circle, we can quotient by it to get a manifold S^2=SU(2)/SO(2). We work in the covering space SU(2)=S^3 of the orthogonal group for convenience.

There is an induced metric on S^2; it sectional curvature is given by O’Neill’s formula for Riemannian submersions ( Petersen p. 58)

S(e_1,e_2)=R_{1212}+3\left|{1\over 2}[e_1,e_2]\right|^2=1-{3\over 4}a^2+{3\over 4}a^2=1.

Thus the induced metric on the sphere is the standard one. In fact S^3 in this point of view is the circle bundle of S^2. But it is not the circle of unit tangent vectors; their length is a.

Geodesics on S^3 do not go project to geodesics in the base. The bundle U(1)\to SU(2)\to S^2 carries a natural connection; the horizontal vectors are orthogonal to the vertical vectors. Note that this connection is independent of a: the horizontal component is the same no matter what the radius of the fiber is. Thus the induced curves on the sphere are not geodesics, but are the same as what one would get for an isotropic rigid body.

The fact that a may not be one only affects the lift of these curves to the total space; that is where the negative sectional curvature can come in. When a>1 we are `stretching’ the vectors in the vertical direction. If this stretching is too large, it leads to an instability.

More coming…

Classical Mechanics

Landau Lifshitz sections 33 and 35 is a standard discussion of the rigid body.

The Lagrangian is, in Euler co-ordinates,

L={1\over 2}Q_1[\dot\phi^2+\sin^2\theta+\dot\theta^2]+{1\over 2}Q_3[\dot\phi\cos\theta+\dot\psi]^2.

This leads to the hamiltonian

H={1\over 2Q_1}\left[p_\theta^2 +{p_\phi^2-2p_\phi p_\psi\cos\theta +p_\psi^2\over \sin^2\theta}\right]+{p_\psi^2\over 2Q_1}[1-\eps],\ \eps=1-{I_1\over I_3}.

p_\phi and p_\psi are conserved. So is the total angular momentum:

L^2=p_\theta^2 +{p_\phi^2-2p_\phi p_\psi\cos\theta +p_\psi^2\over \sin^2\theta}.

The equation for \theta is not affected by the fact that Q_1\neq Q_3. This is what we saw geometrically also.

More coming …

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