Helium Energy levels

The question is if we can find an alternative derivation of the ground state for the hydrogen atom then Bohr and especially Mills GUTCP. Now Mills indicate that this is due to a tilt of \(\theta = \pi/3\) radians. The idea here now is to look at two circles that are tilted and consider the forces as if they where two conducting lines that intersects. The force on a line segment between two parallel lines are

$$ dF = \frac{\mu_0}{2\pi}\frac{I_1 I_2}{r(l)} d l. $$

Assume that without interaction, \(|I_1|=|I_2|=I\). Now assume that they are counter clockwise and that they are tilted \(\theta\) radians with opposite equal currents, then (assuming in the same spherical shell, they annihilates each other. This means the what's left of \(I_1\) after the annihilation is

$$ |I_1| = (1-\cos(\theta))I $$

Now for a tilt we also need to scale \(dl\) with \(\cos(\theta)dl\). This is a singularity calculation e.g. everything happens at \(l=0\) and in a limiting argument with two loops of almost the same radius we do not need to scale \(r(l)\). It all depends on getting balance close to the singularity. Not that the sign is symmetric around the intersection point with total force equal to zero (but with a torque). Anyhow in all at \(l=0\) we have the infinitesimal force

$$ F = \cos(\theta)(1-\cos(\theta))I^2/r(0)\,dl $$

Note that we have an extreme point for \(\cos(\theta) = 1/2\) or at the angle \(\theta = \pi/3\). So this is a possible motivation that we basically minimizes the needed \(I\). The same angle is used by Mill's GUTCP and as this little work is inspired by his work and basically is a search for some alternative angles compared to GUTCP.

The next observation is

$$ \sin(\theta) = \sqrt{\sin(\theta)^2} = \sqrt{1-\cos(\theta)^2} = \sqrt{(1-\cos(\theta))(1+\cos(\theta))} = \sqrt{1/2(1+1/2)} = \sqrt{s(1+s)}, $$

with \(s=1/2\).

This give us some clues as this quantity is used by Mills. Anyhow consider the fact that the two electrons in Mills model for hydrogen consists of loops with normal's pointing to one upper half sphere and a another one to a lower half sphere tilted \(\theta\) between the upper half sphere and a lower half sphere. We note that one principle in Mills theory are that two loops cannot have the same normal as that would imply a loop where \(mvr=\hbar\) is not satisfied (superposition). So we must annihilate one. The fraction to annihilate can be calculated through the usual integral in spherical coordinates and essentially lead to,

$$ \int_{0}^{\theta_0}\cos(\theta)d\theta = \sin(\theta). $$

Which means that sin(\theta) fraction of the kinetic energies / momentum must be removed it is hence natural to consider removing

$$ \sin(\theta) K(r) $$

This is almost the correction Mills uses. But here is another fact, the momentum for a loop, \(mvr\) is constant, Also the fraction of removed loops are independent of the radius and if also the kinetic energy shall be independent then as our total kinetic energy is proportional to "Point Force" \(\times\) "Area" or

$$ (mv^2/r) \cdot r^2 $$

and hence as \(r\) scales as \(r/Z\) for the one electron (remember we start with two non interacting electrons and then remove kinetic energy from the annihilation), we then motivate in order to make the fraction of removed forces<Area invariant, the correction as

$$ \frac{C}{Z}\frac{\hbar^2}{m_e r^3}\sqrt{s(s+1)} $$

The motivation I can find behind this is that in order to remove a loop we need a fixed force that is a fraction of the overall loops independent of the radious also the force balance at a loop crossing in the dual loop analysis indicates as well at a uniform scale. But I agree that this is not an exact argument and of cause cen be flawed.

Remains to argue for why \(C=1\). Note the solution of the radius (see Mills) is then of the form

$$ (Z-1)r(Z) = a_0\Big( 1 - C\frac{\sin(\theta)}{Z} \Big ) $$

And if we take the implicit derivative around \(Z=1\) we get

$$ r(1) = C a_0 \sin(\theta) $$

At \(\theta=\pi/2\) we have \(\sin(\theta) = 1\) But the removed fraction of loops is all of the second half sphere and we should have \(r(1)=a_0\) which imply \(C=1\).

This is of cause a sketch and a proper deduction should spend a lot more on details but as this is an inspirational post and I'm not an academic, you have to take this for what it is.