\(\Delta E_{2\rightarrow 1}=h.f_{\psi\,c}\left(\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1n_2}}-\cfrac { n_{ o2 }^{ 2 }-n_{ o1 }^{ 2 } }{ (n_{o1 }n_{ o2 })^{ 2 } }\right)\)
from the post "Particles In Orbits" dated 18 Oct 2016, with \(n_i=n_{oi}=i\),
\(\Delta E_{2\rightarrow 1}=h.f_{\psi\,c}\left(\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1n_2}}-\cfrac { n_{ 2 }^{ 2 }-n_{ 1 }^{ 2 } }{ (n_{1 }n_{ 2 })^{ 2 } }\right)\)
is ARBITRARY. It is the result of comparing the leading constant to Planck relation \(E=h.f\)
\(\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1n_2}}\)
is due to the difference in potential energy associated with the spin of \(\psi\) with \(n_1\) and \(n_2\), at \(a_{\psi\,n_1}\) and \(a_{\psi\,n_2}\), respectively, and the term,
\(\cfrac { n_{ 2 }^{ 2 }-n_{ 1 }^{ 2 } }{ (n_{1 }n_{ 2 })^{ 2 } }\)
is the change in energy as momentum changes due to a change in \(n\), assuming that the particles after coalescence/separation are still at light speed.
Let's formulate the change in energy as per Bohr model due to a quantized change in momentum. Since \(n\) is an integer, the change in momentum is equally spaced and so if \(L_{n1}\) is the momentum of the
\(KE_{n1}=\cfrac{L_{n1}^2}{2m_1}\)
\(KE_{n2}=\cfrac{L_{n2}^2}{2m_2}\)
the change in \(KE\) is
\(\Delta KE=KE_{n2}-KE_{n1}=-\Delta PE\)
\(\Delta PE=\cfrac{L_{n1}^2}{2m_1}-\cfrac{L_{n2}^2}{2m_2}\)
since we know that the change in momentum is entirely due to a change in \(n\). For a particle spinning at light speed \(c\),
\(L_{ n2 }=\cfrac{2}{5}m_{ 2 }a^2_{ \psi \, \, n2 }.\cfrac{c}{2\pi a_{\psi\,\,n2}}=\cfrac{c}{5\pi}m_{ 2 }a_{ \psi \, \, n2 }\)
\( L_{ n1 }=\cfrac{2}{5}m_{ 1 }a^2_{ \psi \, \, n1 }.\cfrac{c}{2\pi a_{\psi\,\,n1}}=\cfrac{c}{5\pi}m_{ 1 }a_{ \psi \, \, n1 }\)
where \(I_{ni}=\cfrac{2}{5}m_i.a^2_{\psi\,\,ni}\) is the moment of inertia of a sphere of radius \(a_{\psi\,\,ni}\). So,
\(\Delta PE=\cfrac { m_{ 1 }\left( a_{ \psi \, \, n1 }c \right) ^{ 2 } }{ 50 \pi^2} -\cfrac { m_{ 2 }\left( a_{ \psi \, \, n2 }c \right) ^{ 2 } }{ 50\pi^2 } =\cfrac { c^{ 2 } }{50\pi^2 } \left( m_{ 1 }a^{ 2 }_{ \psi \, \, n1 }-m_{ 2 }a^{ 2 }_{ \psi \, \, n2 } \right) \)
As particle of radius \(a_{ \psi \, \, n1 }\) is made up of \(n_1\) basic particle of radius \(a_{\psi\,c}\), \(n=1\)
\(\cfrac { 1 }{ n_{ 1 } } =\left( \cfrac { a_{ \psi \, \, c } }{ a_{ \psi \, \, n1 } } \right) ^{ 3 }\), \( a_{ \psi \, \, n2 }=\sqrt [ 3 ]{ n_{ 2 } } a_{ \psi \, \, c }\)
and particle of radius \(a_{ \psi \, \, n2 }\) is made up of \(n_2\) basic particle of radius \(a_{\psi\,c}\), \(n=1\)
\(\cfrac { 1 }{ n_{ 2 } } =\left( \cfrac { a_{ \psi \, \, c } }{ a_{ \psi \, \, n2 } } \right) ^{ 3 }\), \( a_{ \psi \, \, n1 }=\sqrt [ 3 ]{ n_{ 1 } } a_{ \psi \, \, c }\)
\( m_{ 1 }=n_{ 1 }m_{ c }\)
\( m_{ 2 }=n_{ 2 }m_{ c }\)
So after substituting for \(m_{i}\),
\( \Delta PE=\cfrac { m_{ c }c^{ 2 } }{ 50\pi^2 } \left( n_{ 1 }a^{ 2 }_{ \psi \, \, n1 }-n_{ 2 }a^{ 2 }_{ \psi \, \, n2 } \right) \)
And after substituting for \(a_{\psi\,i}\),
\(\Delta PE=\cfrac { m_{ c }c^{ 2 } }{50\pi^2 } \left( n_{ 1 }\left( \sqrt [ 3 ]{ n_{ 1 } } a_{ \psi \, \, c } \right) ^{ 2 }-n_{ 2 }\left( \sqrt [ 3 ]{ n_{ 2 } } a_{ \psi \, \, c } \right) ^{ 2 } \right) \)
\( \Delta PE=\cfrac { m_{ c }c^{ 2 } }{ 50\pi^2 } a^{ 2 }_{ \psi \, \, c }\left( n_{ 1 }^{ 5/3 }-n_{ 2 }^{ 5/3 } \right) \)
\(L_{ c }=\cfrac{2}{5}m_{ c }a^2_{ \psi \, c }.\cfrac{c}{2\pi a_{ \psi \, c }}\)
\( \Delta PE=\cfrac { L_c^{ 2 } }{ 2m_c } \left( n_{ 1 }^{ 5/3 }-n_{ 2 }^{ 5/3 } \right) \)
What happened to \(\cfrac{1}{n^2_1}-\cfrac{1}{n^2_2}\)? The above expression is the change in \(KE\) required for the specified change in momentum, it is not the associated change in energy level of the particle in its field.
Consider again,
\(L_{ n2 }=\cfrac{2}{5}m_{ 2 }a^2_{ \psi \, \, n2 }.\cfrac{c}{2\pi a_{\psi\,\,n2}}=\cfrac{c}{5\pi}m_{ 2 }a_{ \psi \, \, n2 }\)
\(\cfrac{L_{ n2 }}{L_{ n1 }}=\cfrac{m_2}{m_1}.\cfrac{a_{ \psi \, \, n2 }}{a_{ \psi \, \, n1 }}\)
\(\cfrac{L_{ n2 }}{L_{ n1 }}=\cfrac{n_2}{n_1}.\sqrt[3]{\cfrac{ n_2 }{ n_1 }}=\left(\cfrac{n_2}{n_1}\right)^{4/3}\)
Which is not \(\cfrac{L_{n1}}{L_{n2}}=\cfrac{n_1}{n_2}\) as in Bohr model of
\(L=n\hbar\)
And if we follow through the derivation for the energy difference between two energy levels, \(n_1\) and \(n_2\) we have,
\(E_{\small{B}}=R_{\small{E}}\left(\cfrac{1}{\left(n_1^{4/3}\right)^2}-\cfrac{1}{\left(n_2^{4/3}\right)^2}\right)=R_{\small{E}}\left(\cfrac{1}{n_1^{2.667}}-\cfrac{1}{n_2^{2.667}}\right)\)
this does not effect the first term, \(\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1n_2}}=\cfrac{1}{\sqrt[3]{n_1}}-\cfrac{1}{\sqrt[3]{n_2}}\) associated with the decrease in potential energy of the system in circular motion, as \(\psi\) move from \(n_1\rightarrow n_2\).
\(E_{\small{B}}\) derived above is the potential energy change in the field (\(E\propto\frac{1}{r}\)) that accompanies a change in the angular momentum of \(\psi\) as the energy level transition \(n_1\rightarrow n_2\) occurs. \(E_{\small{B}}\) is recovered from the amount of potential energy associated with the system in circular motion, released.
\(C=\cfrac { \sqrt [ 3 ]{ n_{ 2 } } -\sqrt [ 3 ]{ n_{ 1 } } }{ \sqrt [ 3 ]{ n_{ 1 }n_{ 2 } } } -\cfrac { n_{ 2 }^{ 2.667 }-n_{ 1 }^{ 2.667 } }{ (n_{ 1 }n_{ 2 })^{ 2.667 } }\)
Does a system in circular motion have potential energy \(h.f\) in addition to the system \(KE\) as quantified by its angular momentum? The work done field against the force \(-\frac{\partial\psi}{\partial r}\) as \(\psi\) moves away from the center along a radial line, is strictly not \(\propto\frac{1}{r^2}\) but, the Newtonian \(F\),
\(F\propto -\int{F_{\rho} dr}\)
\(F\propto -ln(cosh(r))\)
\(E\propto -\int{ln(cosh(r))} dr\)
and things get very difficult.
A series plot of ((x)^(1/3)-(a)^(1/3))/(a*x)^(1/3)-((x)^(2.667)-(a)^(2.667))/(a*x)^2.667 for a =1 to 6 is given below,
where the plots for \(n_1=2\) and \(n_1=3\) is almost coincidental.
After adjusting for \(2\rightarrow 2.667\), the plots are essentially the same except that the overlapping plots are not \(n_1=2\) and \(n_1=5\) but \(n_1=2\) and \(n_1=3\).
It seems that \(n_1=2\) and \(n_1=3\) are degenerate, instead.
The important points are: a pair of degenerate plots (\(n_1=2\) and \(n_1=3\)) very close together occurs naturally, that the plot for \(n_1=1\) is an absorption line and the rest of the plots \(n_1\ge 2\) are the emission background.
Good night.
Note: The potential energy of a ball in circular motion held by a spring, is the energy stored in the spring as the spring stretches to provide for a centripetal force. It is clear in this case, that the \(KE\) of the ball, (equivalently stated as angular momentum) is a separate issue from the potential energy stored in the spring. The energy required to increase the ball's \(KE_1\rightarrow KE_2\) is not the same as the energy required to stretch the spring further as the ball accelerates from \(KE_1\rightarrow KE_2\).
It could be that \(\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1n_2}}\) defines the energy change needed at the perimeter, \(a_{\psi\,\,n2}\) and \(-\cfrac { n_{ 2 }^{ 2.667 }-n_{ 1 }^{ 2.667 } }{ (n_{ 1 }n_{ 2 })^{ 2.667 } }\) is the change in energy along a radial line from \(n_1\) to \(n_2\).
Note: Positive is emission line, and negative is absorption line
