\(\cfrac { 1 }{ \varepsilon _{ o }}\oint{ \cfrac { 1 }{ 4\pi a^2_{\psi} }}\cfrac{d\,q_{\psi}}{d\,t}\,d\,A_l= \oint_{2\pi a_{\psi}}{\cfrac{d\,(2 \pi mc)}{d\,t}}\,d\,\lambda= \oint_{2\pi a_{\psi}}{F_\psi}\,d\,\lambda\)
If, we formulate instead,
\(\oint { \cfrac { d\, }{ d\, t } \left\{ \cfrac { q_{ \psi } }{ 4\pi \varepsilon _{ o }a^{ 2 }_{ \psi } } \right\} } \, d\, A_{ l }=\oint _{ 2\pi a_{ \psi } }{ \cfrac { d\, (2\pi mc) }{ d\, t } } \, d\, \lambda \, =\oint _{ 2\pi a_{ \psi } }{ F_{ \psi } } \, d\, \lambda \)
\(E_{\psi}=\cfrac { q_{ \psi} }{ 4\pi \varepsilon _{ o }a^{ 2 }_{ \psi } }\)
\(\cfrac { d\, }{ d\, t } \oint { E_{\psi} } \, d\, A_{ l }=\oint _{ 2\pi a_{ \psi } }{ F_{ \psi } } \, d\, \lambda \)
This expression is without any appendages because if the \(2\pi\) factor missing from the circular momentum term is the only reason
\(\varepsilon _{ o }\ne\mu_o\),
then,
\(\varepsilon _{ o }=\mu_o=\cfrac{1}{c}\)
after we made the correction, which would set,
\(Z_o=1=\sqrt{\cfrac{\mu_o}{\varepsilon_o}}\)
that space present equal resistance to \(B\) and \(E\) field. With such a change to \(\varepsilon_o\), the numerical value of \(q\) changes also.
\(\cfrac { 1 }{ \varepsilon _{ o }}\oint{ \cfrac { 1 }{ 4\pi a^2_{\psi} }}\cfrac{d\,q_{\psi}}{d\,t}\,d\,A_l= \oint_{2\pi a_{\psi}}{\cfrac{d\,(2 \pi mc)}{d\,t}}\,d\,\lambda\)
\(\cfrac { 1 }{ 4\pi\varepsilon _{ o }}\oint{ \cfrac { 1 }{ 2\pi a^2_{\psi} }}\cfrac{d\,q_{\psi}}{d\,t}\,d\,A_l= \oint_{a_{\psi}}{\cfrac{d\,( mc)}{d\,t}}\,d\,\lambda= \oint_{2\pi a_{\psi}}{F_\psi}\,d\,\lambda\)
without the correction to circular momentum.
Since,
\(J_{\psi}=\cfrac { 1 }{ 2\pi a^2_{\psi} }\cfrac{d\,q_{\psi}}{d\,t}\)
\(\cfrac { 1 }{ 4\pi\varepsilon _{ o }}\oint{ J_{\psi}}\,d\,A_l= \oint_{2\pi a_{\psi}}{F_\psi}\,d\,\lambda\)
And then we make the adjustment to \(\varepsilon _{ o }\), \(4\pi\varepsilon _{ o }\rightarrow\varepsilon _{ new }\) to correct for the mssing \(2\pi\) factor for circular momentum.
\(\cfrac { 1 }{\varepsilon _{ new }}\oint{ J_{\psi}}\,d\,A_l= \oint_{2\pi a_{\psi}}{F_\psi}\,d\,\lambda\)
Since, \(\mu_{new}=\varepsilon_{new}=\cfrac{1}{c}\)
\(\cfrac { 1 }{\mu_{ new }}\oint{ J_{\psi}}\,d\,A_l= \oint_{2\pi a_{\psi}}{F_\psi}\,d\,\lambda\)
Why not simply make the adjustment \(mc\rightarrow2\pi mc\), and obtain,
\(\cfrac { 1 }{ 2\varepsilon _{ o }}\oint{ J_{\psi}}\,d\,A_l= \oint_{2\pi a_{\psi}}{F_\psi}\,d\,\lambda\)
as in the post "Maxwell And Particles"???
This expression is wrong because it is still using the old definitions of \(\mu_o\) and \(\varepsilon_o\).
