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=18πεoq=2πaψmc
where
q is the dipole charge.
h=12πεoq4πa2ψπa2ψ=∮2πaψmcdλ
where the closed looped integral is around
2πaψ.
h=12πεo∮q4πa2ψdAl=∮2πaψmcdλ
where the area integral is over the area,
Aaψ=πa2ψ surrounded by
2πaψ.
Differentiating with respect to time,
dhdt=12πεo∮14πa2ψdqdtdAl=∮2πaψdmcdtdλ --- (*)
Compare this with Ampere's Circuital Law, (The fourth Maxwell's Equation with time varying component set to zero),
μo∮JdAl=∮2πlBdl
In the first place, the factor of
12π in expression (*) suggests that for a body in circular motion at speed
c, its momentum is not,
p=mc but
p=2πmc
Consider the centripetal force,
F that accounts for circular motion,
p=∫Fdt=∫mv2rdt
∵dpdt=F
Since,
θ=ωt and
v=rw
p=∫m(rω)2r1ωdθ=∫mrωdθ
Over one period
θ=0→θ=2π,
p=2πmv
And we got it all wrong.
This is important! We have instead,
dhdt=1εo∮14πa2ψdqdtdAl=∮2πaψd(2πmc)dtdλ
Secondly, for a photon
m=0, but a change in its
ψ is momentum (from the posts "Bouncy Balls, Sticky Balls, Transfer Of Momentum" and "de Broglie Per Unit Volume"), so
d(2πmc)dt=Fψ
is a force that acts on
ψ. Depending on the nature of the photon it may be
ψ along
tT,
tg or
tc and one other space dimension, for which the photon is defined.
B is perpendicular to
J in Maxwell's equation. The direction of
dqdt is also perpendicular to
Fψ
The nature of
Fψ depends on the nature of
q.
q is thus qualified as,
qtT,
qtg and
qtc depending on the dimensions between which the photon is oscillating. In all cases, these include a space dimension, the other being a time dimension,
tT,
tg or
tc. So a photon in motion exert a force around it, this force,
Fψ is a force in space (ie.
F=ma) and also along one of the time axes
tT,
tg or
tc. 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(
tc,
tg) is,
12εo∮12πa2ψdqtgdtdAl=∮2πaψd(2πmc)dtdλ=∮2πaψFψdλ
Fψ then acts on
tg and a space dimension.
This equation for particles in motion through an area
Al exert a force field, effecting similar particles, around a loop
2πaψ 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,
Jψ=12πa2ψdqtTdt
12εo∮JψdAl=∮2πaψFψdλ
This however is using the old definition of μo and εo.
More importantly, the correct momentum in circular motion,
p=2πmc=m2πc
makes
μo and
εo equivalent but orthogonal.
Consider,
εnew=iμnew
σ2=(1√2εnew)2+(1√2μnew)2 --- (**)
σ2=ε2new=εnewμnew
When
c is not in circular motion,
c=1√μnewεnew=1σ
as such
σ=εnew=1c is the resistance to
c.
When
c is in circular motion, where
σ has been resolved into two components as in (**). Along one of this component,
εo=1√2εnew
μo wrongfully incoperated a factor of
12π when momentum was taken to be
mv and not
2πmv. So,
2πμo=1√2μnew
And so,
μoεo=14πμnewεnew
This means in the expression for
c,
c=1√μnewεnew=1√4πμoεo
μoεo has to be adjusted by a factor of
4π because the
2π factor needed when momentum is circular was not accounted for. (
2εo→εo and
2πμo→μo)
μo is defined as
4π×10−7 NA
-2. This is wrong for failing to account for the fact that in circular motion, the momentum is
2πmv and not
mv.
It is correct and consistent to set
μo=εo=1c
at the same time, all momentum
p in circular motion is replaced with
2πp, ie.
pcir=2πp
where
pcir is the momentum in circular motion. Equivalently, all velocity
v in circular motion is replaced with
2πv.
vcir=2πv
Equivalently,
pcir=2πp=2πmv=m2πv=mvcir
Correcta, Errata...
Note: This derivation for Ampere's Law applicable to all particles is not rigorous, but indicative none the less.