but this would ignore the presence of the positive end of the weak field produce by a \(T^{+}\) particle on a layer just below. (This is likely the path of an electron in an unpaired orbit.) This weak field associated with the other orbit of the pair, together with the proton captured by the weak field of the current orbit, share the electron. If we superimpose the attraction from the proton and the weak field,
The electron spin around the proton is displaced towards the center of the other orbit in the pair.
The resultant spin of the negative particle generates a \(T\) / \(B\) field that is the oscillatory component of the photon emitted. The photon is thus emitted perpendicular to the plane of the electron spin.(???) This plane is perpendicular to the line joining the proton and the center of the other orbit of the pair. The center of the spin is along this line too. And,
\(tan(\theta)=\cfrac{d}{r}\)
where \(d\) is the orbital separation between the paired orbits and \(r\), the orbital radius.
We have encountered this before, in that case, two similar particles behaving like waves are attracted to each other, in the post "It's All Fluorescence Outside, Inside" dated 29 Jul 2015 and post "High Frequency EMW From Laser" dated 31 Jul 2015, etc
Note: Previously we have also taken the direction normal to the plane of spin as the direction of radiation. A loop of wire sends off EMW in the plane of its loop. It is the same for photons? How do a dipole leave energy behind and send off such energy? In both cases, whether photons are emitted parallel or perpendicular to the plane of the spin, the photons emitted still form a cone with apex angle \(2\theta\), given a particular viewing direction.
