Maxwell, Planck And Particles

From the post "Just Lots of Colors, Retro Disco" Jun2014, where the photon is modeled as a dipole, and the posts "de Broglie Per Unit Volume" and "de Broglie Per Person",

$h=\cfrac { 1 }{ 8\pi \varepsilon _{ o } } q=2\pi a_{ \psi \,  }mc$

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 $\varepsilon_o$.

More importantly, the correct momentum in circular motion,

$p=2\pi mc=m2\pi c$

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.