No Nucleus Needed

\(\psi\) wraps around a sphere.  A certain amount of energy is associated with \(\psi\) motion around the sphere.  If \(h\) amount of energy is in the system, going once per second around the sphere, then travelling at light speed \(c\) around a radius of \(a_{\psi\,n}\), the amount of potential energy associated with this motion is,

\(E_n=h.\cfrac{c}{2\pi a_{\psi\,n}}\)

but,

\(2\pi a_{\psi\,n}=m\lambda_m\)

where \(m\) is the number of wavelength, \(\lambda_m\) of \(\psi\) around the circle.

With \(m=1\),

\(E_n=h.\cfrac{c}{\lambda_m}=h.f\)

which is Planck's relation given \(m=1\).  When \(\psi\) is a half wave in time and a half wave in space space,

\(m=\cfrac{1}{2}\)

\(E_n=h.\cfrac{2c}{\lambda_m}=2h.f=h^{'}.f\)

with a mysterious factor of \(2\).

When \(m=2, \,3...\)

\(E_n=h.\cfrac{c}{m\lambda_m}=\cfrac{h}{m}.f=h_m.f\)

\(h_m\) changes with \(m\) because \(c=f\lambda\).

And,

\(h_{m_1}.f\gt h_{m_2}.f\)

when \(m_2\gt m_1\).

When there is more wavelengths around the circular path, stored energy decreases, which suggest another way \(\psi\) can absorb and release packets of energy; by changing \(m\).  Also, given \(a_{\psi\,n}\) at fixed value, \(f\) increases with \(m\) but \(h_m\) decreases, so, \(E_n=h_m.f\) remains a constant.  \(f\) increases with decreasing \(a_{\psi}\), with increasing \(E_n\).  Catastrophe when the number of particles with small \(a_{\psi}\) is also large.

\(E_n\) is independent of \(m\).  \(h\) is defined as one wavelength going around once per second along the circular path, then a corresponding increase in the number of wavelengths, \(m\) increases the frequency by \(m\) but this increase cancels with the decrease in \(h\) by the reciprocal of \(m\).

With these issues with \(m\) aside, when two particles coalesce,

\(a_{\psi\,n_1}\rightarrow a_{\psi\,n_2}\) 

the bigger particle has less potential energy associated with its \(\psi\) in circular motion,  The amount of energy released due to this drop in potential is.

\(\Delta E=E_{n_1}-E_{n_2}=h\cfrac{c}{2\pi a_{\psi\,n_1}}-h\cfrac{c}{2\pi a_{\psi\,n_2}}\)

\(\Delta E=hf_{n_1}-hf_{n_2}=h\left(f_{n_1}-f_{n_2}\right)\)

\(f_{n_1}\gt f_{n_2}\)

using \(h\) to denote all cases of \(m\), as discussed above, we have an expression similar to the discrete energy level transitions expression for electrons around a nucleus,  In this case, however it is just \(\psi\) alone going around a circular path and energy level decreases with increasing \(a_{\psi\,n}\)

No nucleus needed.