Date sön 11 juni 2017

Executive summary This question is about covering a sphere with great circles so that the resulting density is uniform and that if we for each circle associate a normal vector, then the magnitude of the sum of them would have an upper limit of 1/2. This has a bearing in a classical model of the electron that yields, where Mills (see link below) classically caclulats the correct exitation energy for hydrogene.

a Question The question is defined in the following link Question.

Theorem Mills example yields the maximal total angular momentum and when the total momentum equals the electrons spin then the hydrogene ionisation energy can be classically calculated within almost 1/1000 relative error

Beeing a maximum needs some physical interpretation. One could be that these solutions are the solutions with the least information in them e.g. the spin is ℏ/2 due to the laws of thermodynamics.

Detailed Proof First of all the angular momentum $$L$$ for a solid uniform loop of mass $$m$$, radius $$r$$ and moving at speed $$v$$ is

$$L = \int d(m_e \hat r \times \hat v) = dm_e v r \hat n,$$

where $$\hat n$$ is a normal to the loop. Assume that we have a set of such loops with all the same magnitude and so that when they are added together form a uniform density on the spherical shell of radius $$r$$. Then Conjecture A, in the linked quesiton will lead you to conclude that if we assume that the universal law is that we reach the maximum then the overall angular momentum is

$$L_{tot} = \frac{1}{2} m_e v r \hat z$$

, where we fixated the direction of $$L_{tot}$$ towards $$\hat z$$. Now it is not un unreasonable assumption that all these loop form the intrinsic spin of the electron, which is in the $$\hat z$$ direction of the standard setup: $$\hbar / 2$$ e.g. we conclude that

$$\hbar/2 = |L_{tot}| = \frac{1}{2} m_e v r.$$

We can then extract the velocity as a function of $$r$$ according to

$$v = \frac{\hbar}{m_e r}.$$

This is an important constraint because with it we can now reach conclusions regarding force balances.

So we conclude that the density of loop normals are uniform on the sphere and hence for each point on the loop we have the repelling centripetal force:

$$F_c = \frac{m_e v^2}{r} dSdl = \frac{m_e (\frac{\hbar}{m_e r})^2}{r} dSdl = \frac{\hbar^2}{m_e r^3}dSdl$$

And the attractive electrostatic force:

$$F_e = \frac{e^2}{4\pi \epsilon_0 r^2} dSdl$$

The forces should be in balance and we get,

$$\frac{\hbar^2}{m_e r^3} = \frac{e^2}{4\pi \epsilon_0 r^2}.$$

For which we can solve for $$r$$ and get,

$$r = \frac{4\pi \epsilon_0 \hbar^2}{m_e e^2}$$

That is we come back to the historical Bohr radius,

$$r = a_0$$

Now when we can calulate the ionisation energy of the system which is the electrival energy plus the kinetic energy so, the change of electrical energy for the system is

$$V = \int_{|\hat R|>r} \phi(R) d\hat R = -\frac{e^2}{4\pi \epsilon_0 r} = -27.212 e V$$

The kinetic energy is,

$$T = \frac{1}{2} m v^2 = \frac{1}{2} m_e \frac{\hbar^2}{m_e^2 r^2} = \frac{1}{2} m_e \frac{\hbar^2}{m_e^2 \frac{4\pi \epsilon_0 \hbar^2}{m_e e^2} r} = \frac{e^2}{8\pi \epsilon_0 r} = - \frac{1}{2} V = 13.606 e V$$

With this we get the ionisation energy as

$$E_{ionisation} = -(V + T) = -V + \frac{1}{2}V = - \frac{1}{2}V = 13.606 e V$$

And measured ionisation energy is

$$E_{measured} = 13.59 eV$$

A relative error of about 1/1000.

The author behind these idea, Randell Mills, is controversial. But here is a link to his work: GUTCP, it's rather huge.