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\(\newcommand{\lambdabar}{{\mkern0.75mu\mathchar ’26\mkern -9.75mu\lambda}}\)
Since any circular orbit has an advance, it is assumed there is an unobserved (hidden) binding energy of the electron to the proton in this classical approach. The classical Coulomb potential energy is:
\(
\begin{equation}\label{eqCoulombEnergy}\tag{1}
E_{C}=\frac{e^2}{4\pi\epsilon r}
\end{equation}
\)
Using the value for the advance, if one simply takes the ratio of the advance to the circumference of the orbit it is:
\(
\begin{equation}\label{eqRatio}\tag{2}
ratio = \frac{adv}{2 \pi r}=\frac{\left( \frac{adv}{2\pi} \right) }{r}
\end{equation}
\)
Equation \ref{eqRatio} is the advance of the orbit in radians. Use equation \ref{eqRatio} and multiply it by the Coulomb energy, a derived total energy is (classical orbit plus the extra energy due to the orbital advance):
\(
\begin{equation}\tag{3}
E_T(r)= E_C(r)+E_A(r)=\frac{e^2}{4\pi\epsilon r}+ \left( \frac{e^2}{4\pi\epsilon r} \right) \left(\frac{\left( \frac{adv}{2\pi} \right) }{r} \right)
\end{equation}
\)
where r is the radial distance between the proton and electron, and the additional energy due to the advance is:
\(
\begin{equation}\label{eq999}\tag{4}
E_A(r)=\left( \frac{e^2}{4\pi\epsilon r} \right) \left(\frac{\left( \frac{adv}{2\pi} \right) }{r} \right)
\end{equation}
\)
Considering photon energy with wavelength r is:
\(
\begin{equation}\label{eqEgammaRT}\tag{5}
E_{\gamma}(r)=\frac{h c}{r}
\end{equation}
\)
Setting equation \ref{eqEgammaRT} equal to \(E_T(r)\) with K equaling the value of the proportionality between \(E_{\gamma}(r)\) and \(E_T(r)\) (Note: if they are equal, the value of K will be one):
\(
\begin{equation}\label{q1928}\tag{6}
K \left(\frac{h c}{r}\right)=\frac{e^2}{4\pi\epsilon r}+ \left( \frac{e^2}{4\pi\epsilon r} \right) \left(\frac{\left( \frac{adv}{2\pi} \right) }{r} \right)
\end{equation}
\)
solving for K:
\(
\begin{equation} \label{q1234}\tag{7}
k= \left( \frac{\left(\frac{e^2}{4\pi\epsilon r} \right)}{ \left( \frac{h c}{r} \right) } \right)+
\left( \frac{ \left( \left( \frac{e^2}{4\pi\epsilon r} \right) \left(\frac{\left( \frac{adv}{2\pi} \right) }{r} \right) \right) }{ \left( \frac{h c}{r} \right)} \right)
\end{equation}
\)
Reducing and rearranging equation \ref{q1234}:
\(
\begin{equation}\tag{8}
K= \frac{1}{2 \pi}\left( \frac{e^2}{2 \epsilon_0 h c} + \frac{e^2}{2 \epsilon_0 h c} \left( \frac{adv}{2\pi r} \right) \right)
\end{equation}
\)
From equation \ref{q1928}, the photon energy equivalent to \(E_{T}(r)\) as a function of \(\lambda\) is:
\(
\begin{equation} \label{q4321}\tag{9}
E_{\gamma}(\lambda) =
\overbrace{
\underbrace{
\frac{E_C(\lambda)}{K(\lambda)}
}_\text{Coulomb/K}
+
\underbrace{
\frac{E_A(\lambda)}{K(\lambda)}
}_\text{Advance/k}
}^\text{Total Binding Energy}
\end{equation}
\)
When one investigates \(\lambda=\lambda_{ebar}\) (the reduced Compton wavelength) equation \ref{q4321} generates the following values:
\(
\begin{align}\label{qList}
\lambda &= \lambda_{ebar} \mbox{(Where }\lambda_{ebar} \mbox{ is the reduced Compton wavelength for an electron) }\\
Advance &= 2 \pi r_e \mbox{(Where }r_e \mbox{ is the classical electron radius)} \\
Ratio &= \alpha \mbox{ (The Fine Structure Constant) }\\
r &=\frac{r_e}{\alpha} \\
E_{\gamma}(\lambda_{ebar}) &= 2 \pi m_e c^2 \\
E_C(r) &=\alpha m_e c^2 \\
E_A(r) &= \alpha^2 m_e c^2 \\
\end{align}
\)
If an investigator examined the energy from \(E_A(r)\) it is logical to imagine they would consider it a Coulomb based energy. As such, from a classical viewpoint the energy would relate to equation \ref{eqCoulombEnergy}. To remove ambiguity the new Coulomb based equation representing an investigator’s evaluation of \(E_A(r)\) will be marked by a prime symbol. Relating equation \ref{eqCoulombEnergy} to equation \ref{eq999}:
\(
\begin{equation}\label{q3465}\tag{10}
E’_{C}(r’)=\frac{e^2}{4\pi\epsilon r’}=E_A(r)=\left( \frac{e^2}{4\pi\epsilon r} \right) \left(\frac{\left( \frac{adv}{2\pi} \right) }{r} \right)
\end{equation}
\)
The value of \(E_A(r)\) when \(\lambda=\lambda_{ebar}\) is:
\(
\begin{equation}\label{qt1}\tag{11}
E_A(r)= \alpha^2 m_e c^2
\end{equation}
\)
This energy in equation \ref{qt1} is equivalent to the binding energy of the electron to the proton in the Hydrogen atom (27.21 eV). Visually the figure below displays a Coulomb binding energy as predicted by the model consisting of two parts: the classical Coulomb binding energy \(E_C\) (shown in blue) and an additional binding energy due to the electron’s orbital advance \(E_A\) (shown in red).

Recall from the model, the advance of the electron is an invariant distance \(r_e\). If one steps the radial magnitude by integer multiples of \(r_e\), discrete energy levels are generated. Using equation \ref{q3465} and solving for \(r{‘}\) (when \(\lambda=\lambda_{ebar}\) ):
\(
\begin{equation}\label{qrprime}\tag{12}
r{‘}=\frac{\left( \frac{r_e}{\alpha}\right)^2}{r_e}
\end{equation}
\)
Discretize \(r_e\) by adding $n$ to the equation where \(n=\{1,2,3,..n\}\):
\(
\begin{equation}\tag{13}
r{‘}=\frac{\left( \frac{n r_e}{\alpha}\right)^2}{r_e}
\end{equation}
\)
One can now ascertain the conversion of \(E_{C}^{‘}(r{‘})\) into a step function of \(r_e\) (the gravitational induced effect of the electron orbital advance) by:
\(
\begin{equation}\tag{14}
\frac{e^2}{4\pi\epsilon r{‘}}=\frac{e^2}{4\pi\epsilon r} \rightarrow r{‘}=\frac{\left( \frac{n r_e}{\alpha} \right)^2}{r_e}=\frac{n^2r_e}{\alpha^2}
\end{equation}
\)
Thus the energy in equation \ref{q3465} has discrete allowable energy magnitudes via the following equation:
\(
\begin{equation}\label{eqCoulombEnergyB}\tag{15}
E_{C}^{‘}(n)=\frac{e^2}{4\pi\epsilon \frac{n^2r_e}{\alpha^2}}=\frac{\alpha^2 e^2}{4 \pi \epsilon_0 n^2 r_e}
\end{equation}
\)
Interpreting equation \ref{eqCoulombEnergyB} the total binding energy of the Hydrogen atom may only take on discrete values due to the advance of the electron (a gravitational effect). Thus, the following binding energies are allowed for the electron-proton:
\(
\begin{align}\tag{16}
E_{C}^{‘}(1)&=\frac{\alpha^2 e^2}{4 \pi \epsilon_0 (1^2) r_e} \\
E_{C}^{‘}(2)&=\frac{\alpha^2 e^2}{4 \pi \epsilon_0 (2^2) r_e} \\
E_{C}^{‘}(3)&=\frac{\alpha^2 e^2}{4 \pi \epsilon_0 (3^2) r_e} \\
…\\
E_{C}^{‘}(n)&=\frac{\alpha^2 e^2}{4 \pi \epsilon_0 n^2 r_e}
\end{align}
\)
According the Virial theorem the average kinetic energy of an orbiting particle will have half the average potential energy of the bound system. Thus, to remove an electron from an n state energy, \(\frac{1}{2} E_{C}^{‘}(n)\) is required. Listed below are the values for the first three states.
\(
\begin{align}\tag{17}
\left( \frac{1}{2} \right) E_{C}^{‘}(1) = 13.6\; eV \\
\left( \frac{1}{2} \right) E_{C}^{‘}(2) = 3.40\; eV \\
\left( \frac{1}{2} \right) E_{C}^{‘}(3) = 1.51\; eV \\
\end{align}
\)
If one considers the difference between energy levels, the differential energies match the Hydrogen spectral emissions:
\(
\begin{align}\tag{18}
\left( \frac{1}{2} \right) E_{C}^{‘}(1) -\left( \frac{1}{2} \right) E_{C}^{‘}(2) = 10.20\; eV\\
\left( \frac{1}{2} \right) E_{C}^{‘}(1) -\left( \frac{1}{2} \right) E_{C}^{‘}(3) = 12.09\; eV\\
\end{align}
\)
These are representative of the Lyman series. Other series may be derived in a similar manner.

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