San José State University

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Thayer Watkins
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The Classical and Quantum Mechanics
of a Thin Ring Spinning About Two Axes

The Classical Analysis

Consider a thin ring of mass M and radius R. (Here thin means that it is a line.) The linear mass density of the ring is ρ=M/(2πR). The moment of inertia I1 for spinning about an axis perpendicular to the plane of the ring which passes through the ring center is MR². There is another spin of a ring which is like the flipping of a coin. This is a rotation about an axis which is a diameter of the ring. The moment of inertia for this type of spinning is given by

I2 = ∫(R·cos(θ))²ρRdθ = ρR³∫cos²(θ)dθ

where the integration is from 0 to 2π. The integral of cos²(θ) from 0 to π is ½ so the integral from 0 to 2π is 1 and hence

I2 = ρR³ = (2πRρ)R²/(2π) = (1/(2π))MR².

This flipping rotation is of particular interest for the structure of nuclei. Experimental measurement indicate that that nuclei have at least approximately spherical shapes. If a circular band of nucleons rotates in this flipping fashion the dynamic appearance of the nucleus would be that of a sphere.

A spin can also take place perpendicular to the diameter considered above. The moment of inertia about that axis is same as that of I2. It unnecessarily complicates the analysis and will not be included now.

The angular momentum of the ring spinning at an angular rate of ω about the axis perpendicular to its plane is L=MR(Rω)=MR²ω. This means that

ω = L/(MR²)
and hence
½I1ω² = ½MR²(L²/(MR²)² = L²/(2MR²)

For the spin about a diameter the angular momentum Λ is found as follows.

Λ = ∫ (R·cos(θ))(R·cos(θ)Ω)ρdθ = R²ρΩ∫cos²(θ)dθ
with the integration being from 0 to 2π
which reduces to
Λ = R²ρΩ = R²(M/(2πR)Ω = MRΩ/(2π)
and hence
Ω = 2πΛ/(MR)

Thus

½I2Ω² = ½((1/(2π))MR²)(2πΛ/(MR))² = Λ²/(4πMR)

The kinetic energy of the spinning ring is then

K = L²/(2MR²) + Λ²/(4πMR)

and there is no potential energy. This means that the ring will continue spinning at the rates ω and Ω indefinitely. These rates can have any real values. This is the classical solution.

The Quantum Mechanical Analysis

To get the quantum mechanical solution the energy is expressed as

E = ½I1(dθ/dt)² + ½I2(dφ/dt)²

The momentum associated with θ is (∂E/∂θ) which is I1(dθ/dt). Let this be denoted as pθ. Thus (dθ/dt)=pθ/I1. Likewise the momentum associated with φ is pφ=(dφ/dφ)/I2. Therefore the Hamiltonian function for the the spinning ring is

H = pθ/(2I1) + pφ/(2I2)

and the Hamiltonian operator is

H^ = −h²(∂²/∂θ²) −h²(∂²/∂φ²)
and the time-independent
Schrödinger equation is
h²(∂²ψ/∂θ²) −h²(∂²ψ/∂φ²) = Eψ

If it is assumed that the wave function ψ is of the form Θ(θ)Φ(φ) then above equation becomes

h²Θ"(θ)Φ(φ) −h²Θ(θ)Φ"(φ) = EΘ"(θ)Φ(φ)
which upon division
by Θ(θ)Φ(φ) gives
h²Θ"/Θ −h²Φ"/Φ = E

This equation can be expressed as

−Θ"/Θ = Φ"/Φ + E/h²

The LHS is independent of φ and the RHS independent of θ. Therefore their common value must be a constant, say k². This means that

Θ"(θ) + k²Θ(θ) = 0

The solution is

Θ(θ) = A·cos(k(θ+θ0)

where A and θ0 are constants. By proper choice of the coordinate system θ0 can be made equal to zero. Then k(2π) must be an integral multiple of 2π. Therefore k must be an integer.

The other equation is

Φ"/Φ + E/h² = k²
or, equivalently
Φ" + (E/h² − k²)Φ = 0

From the previous case it is found that the coefficient of Φ must be a squared integer; i.e.,

(E/h² − k²) = q²
or, equivalently
E/h² = k² + q²

Thus the energy of the spinning ring is quantized such that it is equal to a multiple of of the sum of squared integers. The wave function then has the form

ψ(θ, φ) = cos(kθ)cos(qφ)
for 0≤θ≤2π
and 0≤φ≤2π

This means that the probability density function P(θ, φ) is given by

P(θ, φ) = cos²(kθ)cos²(qφ)

When a second rotation axis is included the quantization condition for energy involves the sum of the squares of three integers.

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