# Helium Energy levels II

I will start by first do a retake of getting the energy levels of Helium. The idea is that for helium we have two spherical shells at the same radius. Each of them has a B-field where the component of the vector that gives and internal field is the radial component and has the form,

$$B_{\theta} = A \cos(\theta)$$

Now the idea is to superimpose two such field with a angle $$\alpha$$ between them. This means that the addition of the fields are (the other electron has a minus because it is coming from a reversed loop),

$$B_{\theta} = B_{\theta} - B_{\theta+\alpha} = A(\cos(\theta)-\cos(\theta+\alpha)) = A(\cos(\theta)(1-\cos(\alpha)) + \sin(\theta)\sin(\alpha)).$$

Now the $$\sin(\theta)$$ disappears as we can match that with the value at the antipodal, e.g. at $$-\theta$$ and the net contribution is the term,

$$A \cos(\theta)(1-\cos(\alpha)).$$

Now when we tilt 0 radians, we do not have any overlapping loops and if we tilt $$\pi/2$$ radians every loop is duplicated, thus motivating that the fraction remaining for force balance is $$cos(\theta)$$. Hence the overall force, internal to the shell is governed by an integral of,

$$B \cos(\theta)\cos(\alpha)(1-\cos(\alpha)).$$

As this is the force and we have a factor in front that only depends on $$\alpha$$ this factor remains when we calculate the potential related to this force. And as the force is attractive, we end up at the maximum of this factor minimizing the energy at $$cos(\alpha)=1/2$$ leading to the angle $$\pi/3$$.

The radial component at the surface are of a form $$B\sin(\theta)$$ and the field "between" the two shells that lead to a radial component proportional to,

$$\sin(t)/2 - \sin(t+a) = \sin(t)/2 - \sin(t)\cos(a) - \cos(t)\sin(a) = \cos(t)\sin(a).$$

The idea now is that the electrostatic interaction matches the magnetic interaction and both should be scaled by $$\sin(\alpha)$$ as we vary the angle $$\alpha$$.

So let's analyze the electrostatic interaction of two shells, we have the term,

$$1/|x-y| = \Sum_{n=0} P_n(cos(\theta))r_1^{n}/r_2^{n+1}.$$

With $$P_n$$ being the Legendre polynomials. Now as we will integrate over a uniform sphere all terms, but $$1/r_2$$ remains and the rest cancels. Now if we did not have an interaction we would have the Kinetic energy matching of the form for a certain radius if

$$Av^2 = ZE/r_2.$$

Hence the electric interaction term would be of the magnitude (keeping the momentum constant:

$$E/r_2 \sin(\alpha) = A v^2 / Z \sin(\alpha).$$

Now we stretch things and say that this shows that we should add this term as the interaction term to force balance and we get a match with the measured ground energy levels of Helium and similar 2 electron ions.

The astute reader would recognize that we have magnetic energy + electric energy and in this argument we should have a factor of two. But as we can't in the argument specify a single electron the interactin term needs to be distributed on both electrons giving a half each. In all we end upp with Mills equation that is quite good even if we skip adding the electrostatic term.