Adjusting for Bohr - Angular Momentum

The expression,

$\Delta E_{2\rightarrow 1}=h.f_{\psi\,c}\left(\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1n_2}}-\cfrac { n_{ o2 }^{ 2 }-n_{ o1 }^{ 2 } }{ (n_{o1 }n_{ o2 })^{ 2 } }\right)$

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 $E=h.f$

The term,

$\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1n_2}}$

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 $a_{\psi\,c}$, $n=1$

$\cfrac { 1 }{ n_{ 2 } } =\left( \cfrac { a_{ \psi \, \, c } }{ a_{ \psi \, \, n2 } }  \right) ^{ 3 }$,   $a_{ \psi \, \, n1 }=\sqrt [ 3 ]{ n_{ 1 } } a_{ \psi \, \, c }$

Since we have assumed that the volume of the particles are conserved and their density is a constant,

$m_{ 1 }=n_{ 1 }m_{ c }$

$m_{ 2 }=n_{ 2 }m_{ c }$

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