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

Sketch of a proof Now we can generalize the theorem in the link above to assume that the sphere has a radious r and that the total weight of it is m and that each loop moves solidly with a velocity v. If we then assume that each loop has angular momentum ℏ then the theorem in the link wonders if the maximum total angular momentum atainable, and attained in Mills example is ℏ / 2 - the spinn of the electron. To each loop add the charge density so that the total charge is e and because of the uniformity of G, the sum of it would kill the charge field nicely. Also one can setup the force balance for each geodesics or loop between the electrostatic central force and the centrifugal force. Then by using the add hoc mvr=ℏ you can solve for v and get an equation in the radius and constants. So now the radius can be solved for. With the radius and velocity at hand one can now easily calculate the change in electrostatic energy and kinetic energy of the system as the electron is removed and you will get a figure of an ionization energy that is correct to almost three full digits or a relative error of a couple of 1/1000. This is actually amazing if the add hoc part can be motivated. Now the z component of the spin of the electron is ℏ/2 and put that as the total angular momentum e.g. it should represent electron spin, you get the magic add hoc equation above. The problem is if there is infinite many constructions of a type G covering with different total angular momentum, than the special selection Mils did, these calculations could be arbitrary and an act of tuning. Proving the underlying theorem in the link above would increase the value of this derivation in my eyes.

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 centrifugal 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.


Comments

comments powered by Disqus

~ Follow me ~