\(h=\cfrac { 1 }{ 8\pi \varepsilon _{ o } } q=2\pi a_{ \psi \, }mc\)
where \(q\) is the dipole charge.
\(h=\cfrac { 1 }{ 2 \pi\varepsilon _{ o }} \cfrac { q }{ 4\pi a^2_{\psi} }\pi a^2_{\psi}= \oint_{2\pi a_{\psi}}{mc}\,d\,\lambda\)
where the closed looped integral is around \(2\pi a_{\psi}\).
\(h=\cfrac { 1 }{ 2 \pi\varepsilon _{ o }}\oint{ \cfrac { q }{ 4\pi a^2_{\psi} }}\,d\,A_l= \oint_{2\pi a_{\psi}}{mc}\,d\,\lambda\)
where the area integral is over the area, \(A_{a_{\psi}}=\pi a^2_{\psi}\) surrounded by \(2\pi a_{\psi}\).
Differentiating with respect to time,
\(\cfrac{d\,h}{d\,t}=\cfrac { 1 }{ 2 \pi\varepsilon _{ o }}\oint{ \cfrac { 1 }{ 4\pi a^2_{\psi} }}\cfrac{d\,q}{d\,t}\,d\,A_l= \oint_{2\pi a_{\psi}}{\cfrac{d\,mc}{d\,t}}\,d\,\lambda\) --- (*)
Compare this with Ampere's Circuital Law, (The fourth Maxwell's Equation with time varying component set to zero),
\(\mu_o\oint{J}\,d\,A_l=\oint_{2\pi l}{B} \,dl\)
In the first place, the factor of \(\cfrac{1}{2\pi}\) in expression (*) suggests that for a body in circular motion at speed \(c\), its momentum is not,
\(p=mc\) but \(p=2\pi mc\)
Consider the centripetal force, \(F\) that accounts for circular motion,
\(p=\int { F } dt=\int { \cfrac { mv^{ 2 } }{ r } } dt\)
\(\because \cfrac{d\,p}{d\,t}=F\)
Since,
\( \theta =\omega t\) and \( v=rw\)
\( p=\int { \cfrac { m\left( r\omega \right) ^{ 2 } }{ r } } \cfrac { 1 }{ \omega } d\theta =\int { mr\omega } d\theta \)
Over one period \(\theta=0\rightarrow\theta=2\pi\),
\( p=2\pi mv\)
And we got it all wrong. This is important! We have instead,
\(\cfrac{d\,h}{d\,t}=\cfrac { 1 }{ \varepsilon _{ o }}\oint{ \cfrac { 1 }{ 4\pi a^2_{\psi} }}\cfrac{d\,q}{d\,t}\,d\,A_l= \oint_{2\pi a_{\psi}}{\cfrac{d\,(2 \pi mc)}{d\,t}}\,d\,\lambda\)
Secondly, for a photon \(m=0\), but a change in its \(\psi\) is momentum (from the posts "Bouncy Balls, Sticky Balls, Transfer Of Momentum" and "de Broglie Per Unit Volume"), so
\(\cfrac{d\,(2\pi mc)}{d\,t}=F_\psi\)
is a force that acts on \(\psi\). Depending on the nature of the photon it may be \(\psi\) along \(t_T\),\(t_g\) or \(t_c\) and one other space dimension, for which the photon is defined. \(B\) is perpendicular to \(J\) in Maxwell's equation. The direction of
\(\cfrac{d\,q}{d\,t}\) is also perpendicular to \(F_\psi\)
The nature of \(F_\psi\) depends on the nature of \(q\). \(q\) is thus qualified as, \(q_{t_T}\), \(q_{tg}\) and \(q_{tc}\) depending on the dimensions between which the photon is oscillating. In all cases, these include a space dimension, the other being a time dimension, \(t_T\),\(t_g\) or \(t_c\). So a photon in motion exert a force around it, this force, \(F_\psi\) is a force in space (ie. \(F=ma\)) and also along one of the time axes \(t_T\),\(t_g\) or \(t_c\). Previously, in Maxwell's Equation, the effect of the force in the time dimension in which the charge is oscillating has been ignored. So expression (*) in the case of a photon, ph(\(t_c\),\(t_g\)) is,
\(\cfrac { 1 }{ 2\varepsilon _{ o }}\oint{ \cfrac { 1 }{ 2\pi a^2_{\psi} }}\cfrac{d\,q_{t_g}}{d\,t}\,d\,A_l= \oint_{2\pi a_{\psi}}{\cfrac{d\,(2 \pi mc)}{d\,t}}\,d\,\lambda= \oint_{2\pi a_{\psi}}{F_\psi}\,d\,\lambda\)
\(F_{\psi}\) then acts on \(t_g\) and a space dimension.
This equation for particles in motion through an area \(A_l\) exert a force field, effecting similar particles, around a loop \(2\pi a_{\psi}\) is just a restatement of Ampere's Law but extended to all particles. It may be possible that all of Maxwell's equations can be similarly extended.
Let,
\(J_{\psi}=\cfrac { 1 }{ 2\pi a^2_{\psi} }\cfrac{d\,q_{t_T}}{d\,t}\)
\(\cfrac { 1 }{ 2\varepsilon _{ o }}\oint{ J_{\psi}}\,d\,A_l= \oint_{2\pi a_{\psi}}{F_\psi}\,d\,\lambda\)
This however is using the old definition of \(\mu_o\) and \(\varepsilon_o\).
This however is using the old definition of \(\mu_o\) and \(\varepsilon_o\).
\(p=2\pi mc=m2\pi c\)
makes \(\mu_o\) and \(\varepsilon_o\) equivalent but orthogonal.
Consider,
\(\varepsilon_{new}=i\mu_{new}\)
\(\sigma ^2=(\cfrac{1}{\sqrt{2}}\varepsilon_{new})^2+(\cfrac{1}{\sqrt{2}}\mu_{new})^2\) --- (**)
\(\sigma ^2=\varepsilon^2_{new}=\varepsilon_{new}\mu_{new}\)
When \(c\) is not in circular motion,
\(c=\cfrac{1}{\sqrt{\mu_{new}\varepsilon_{new}}}=\cfrac{1}{\sigma}\)
as such \(\sigma=\varepsilon_{new}=\cfrac{1}{c}\) is the resistance to \(c\).
When \(c\) is in circular motion, where \(\sigma\) has been resolved into two components as in (**). Along one of this component,
\(\varepsilon_o=\cfrac{1}{\sqrt{2}}\varepsilon_{new}\)
\(\mu_o\) wrongfully incoperated a factor of \(\cfrac{1}{2\pi}\) when momentum was taken to be \(mv\) and not \(2\pi mv\). So,
\(2\pi \mu_o=\cfrac{1}{\sqrt{2}}\mu_{new}\)
And so,
\(\mu_o\varepsilon_o=\cfrac{1}{4\pi}\mu_{new}\varepsilon_{new}\)
This means in the expression for \(c\),
\(c=\cfrac{1}{\sqrt{\mu_{new}\varepsilon_{new}}}=\cfrac{1}{\sqrt{4\pi\mu_o\varepsilon_o}}\)
\(\mu_o\varepsilon_o\) has to be adjusted by a factor of \(4\pi\) because the \(2\pi\) factor needed when momentum is circular was not accounted for. (\(2\varepsilon_o\rightarrow\varepsilon_o\) and \(2\pi\mu_o\rightarrow\mu_o\))
\(\mu_o\) is defined as \(4\pi \times10^{-7}\) NA-2. This is wrong for failing to account for the fact that in circular motion, the momentum is \(2\pi mv\) and not \(mv\).
It is correct and consistent to set
\(\mu_o=\varepsilon_o=\cfrac{1}{c}\)
at the same time, all momentum \(p\) in circular motion is replaced with \(2\pi p\), ie.
\(p_{cir}=2\pi p\)
where \(p_{cir}\) is the momentum in circular motion. Equivalently, all velocity \(v\) in circular motion is replaced with \(2\pi v\).
\(v_{cir}=2\pi v\)
Equivalently,
\(p_{cir}=2\pi p=2\pi mv=m2\pi v=mv_{cir}\)
Correcta, Errata...
Note: This derivation for Ampere's Law applicable to all particles is not rigorous, but indicative none the less.
