Differential form of Faradays Law (4-23)
04_23 Faradays_Law_for_rest_circuit integral form E,B
/$ ∮[•∂(l)′: Ei(r - v*t,t))
= -1/c*∮[dArea′: ∂[∂(t): Bi(r - v*t,t)]•n′)
/* This is an interesting one ...
I want to differentiate the LHS by •dl′, the RHS by *da′
This First attempt is a BOTCH, completely unnecessary and ineffective, but
of interest to see what happens
1. First try "reversing the entanglement"
Faradays Law (4-2) itself creates the "entanglement"
Equate, using solid angle &
Ω = non-planar solid angle
area A subtended by solid angle
Ω = A/r^2 = 2*π*(1 - cosΘ) = 4*π*sin^2(Θ/2/s")
??planar radians dθ = ds/r where s = segment of circle??
solid steradian dΩ = dA/r
R = radial distance (or displacement for non-symmetry) from the
charge q to the circular loop, and circular "cap" area
For a perfect sphere around charge, and considering [E,B] flux through
a circular loop and area at constant distance y from charge
(Note - this DOESNT have to be a great circle through the center of
the charge), relating dΩ and dy : (wanda.fiu.edu)
"Figure 2: Geometry of the cross section and the solid angle"
let Ra = the radius of the circular loop defined by [y,Ω]
(vs R = the radius (distance) from the charge to the circular loop )
In this web-page (wanda.fiu.edu) dΩ is defined in terms of a dθ for annular
rings, going from the center of the circular loop to its perimeter :
/$ dΩ_annular = 2*π*R*sinθ*R/R^2*dθ = 2*π*sinθ*dθ
/* 1a. But rather than using this definition (rather than a pie slice of
constant solid angle and planar angle), and [being lazy, wanting to be different,
and not having to deal with radiation problems of spherical symmetry]
(as opposed to constant radiation spherically) Ill use one with
constant dA, dl
/$ Ω_full_sphere = 4*π steradians
A_surface_section = 4*π*Rs^2*Ω_circle/Ω_full_sphere
/* Define θ as angle CCW from some arbitrary starting point on the circular loop
/$ ∂(l) = ((y*dz)^2 + dy^2)^0.5
∂(Area) := 4*π*y^2*dz/
/* Should be the circular area - not "chordal"
/$ |y| = constant, dy/dz = 0
∂(l) = |y|*dz
/* AH HAH! - perhaps [θ,φ] as defined here are the sense of Lucas's use
in (4-15)(4-17) !!??!!
1b. Wait - go back to the "standard" annular formula, as that provides
a differential solid angle, which in turn is a basis for Area of the
spherical surface "cap" :
/$ Ω_full_sphere = 4*π steradians
dΩ_annular = 2*π*sinθ*dθ
dA_annular = A_sphere*dΩ_cap/Ω_full_sphere
/* and
/$ A_cap = A_sphere*Ω_cap/Ω_full_sphere
∂(l) = ((y*dz)^2 + dy^2)^0.5
∂(Area) := 4*π*y^2*dz/
/* NO WORKEE! - I NEED the other form to relate dl&dA, BUT, this can be
used to calculate A_cap, L_cap
1.b.i
let : θf be the angle from the center of the cap to its
outer (circular) edge
where :
/$ A_cap = A_sphere/Ω_full_sphere*Ω_cap
= 4*π*R^2 / (4*π) * ∫[dθ,0 to θf: 2*π*sinθ*dθ)
∫[dθ,0 to θf: 2*π*sinθ*dθ)
= diff(0 to θf: - 2*π*cos(θ))
= -2*π*(cos(θf) - 1)
= 2*π*(1 - cos(θf))
/* check for θf = π, 2*π*(1 - cos(θf)) = 2*π*(1 - (-1)) = 4*π (correct!)
so :
/$ A_cap = 2*π*(1 - cos(θf))*R^2
1.β.ii
/$ let : L_cap = length around outer edge of cap, which is a circle
where : R_cap_edge = R*sinθf
so : L_cap = 2*π*R*sinθf
/* 1.a.i. Going back to "pie-slices"
Now, for pie-slices of cap, going from φ = 0 to 2*π
From symmetry, I argue that :
/$ : A_pie/A_cap = φ_pie/φ_cap = φ/2/π
so : dA_pie/A_cap = ∂(φ)/2/π
also : L_pie/L_cap = φ_pie/φ_cap = φ/2/π
also : dL_pie/L_cap = ∂(φ)/2/π
/* i.e. same "ratio formulae" for [A_pie, L_pie] and [dA_pie, dL_pie]
which means its easy to jointly integrate?
1.c Reminder of original objective : get the Differential form of
Faradays Law (4-23)
04_23 Faradays_Law_for_rest_circuit integral form E,B
/$ ∮[•∂(l)′: Ei(r - v*t,t)) = -1/c*∮[dArea′: ∂[∂(t): Bi(r - v*t,t)]•n′)
/* Substituting for dl and da (1.a.i. above using [dL,dA])
/$ ∮[•(L_cap/2/π*∂(φ))′: Ei(r - v*t,t))
= -1/c* ∮[*(A_cap/2/π*∂(φ))′: ∂[∂(t): Bi(r - v*t,t)]•n′)
/* extracting constants
/$ L_cap/2/π *∮[•∂(φ)′: Ei(r - v*t,t))
= -1/c*A_cap/2/π *∮[*∂(φ)′: ∂[∂(t): Bi(r - v*t,t)]•n′)
/* Now both have the same basis of integration. Multiply each side by
2*π and gathering terms :
/$ ∮[•∂(φ)′: Ei(r - v*t,t))
= -1*A_cap/L_cap/c* ∮[*∂(φ)′: ∂[∂(t): Bi(r - v*t,t)]•n′)
where
A_cap/L_cap
= [2*π*(1 - cos(θf))*R^2] / [2*π*R*sinθf]
= (1 - cos(θf))/sinθf*R
= (1 - cos(θf))/sinθf*R
/* Rearranging (I have a dot product "leftover" - I should have used a unit vector
for the dotProd)
/$ 0 = ∮[•∂(φ)′: Ei(r - v*t,t)]
+ ∂[∂(t): Bi(r - v*t,t)]•n′ * (1 - cos(θf))/sinθf*R/c
)
/* differentiating both sides by dφ′, (for this [R,θf] are constant) :
/$ 0 = Ei(r - v*t,t)
+ ∂[∂(t): Bi(r - v*t,t)•n′] * (1 - cos(θf))/sinθf*R/c
/* finally :
/$ Ei(r - v*t,t) = ∂[∂(t): Bi(r - v*t,t)•n′] *(-1)(1 - cos(θf))/sinθf*R/c
/* Nuts - 1.c was NOT successful (I need to look at it again later)
HOWEVER, my result makes sense in that for the same total B in the cap,
independent of R (Bi with 1/R^2 but area goes up that much),
Ei will go down with 1/R.
2. Do as Lucas suggests for (4-27) p69h0.8, and convert ONLY the
line integral to a surface integral via Stokes theorem.
Kreyszig1972 p364h0.5 Stkes theorem :
/$ ∬[∂(Area): (∇v)n) = ∮(ds: v*t)
/* from which (using n=n′,∇=∇′) :
/$ ∮[•∂(l)′: Ei(r - v*t,t))
= ∬[∂(Area): (∇′v)n′)
= ∬[∂(Area): (∇′Ei(r - v*t,t))n′)
/* Note that ∇′Ei(r - v*t,t)) is aligned (+,-) with Bi
(03Feb2016 as the gradient of E is radially out from the particle!),
and is like the Bi integral in (4-23). Placing into (4-23) :
/$ ∬[*∂(Area): (∇′Ei(r - v*t,t))n′) = -1/c*∬[dArea′: ∂[∂(t): Bi(r - v*t,t)]•n′)
/* Differentiating both sides by dA :
/$ (∇′Ei(r - v*t,t))n′ = -1/c*∂[∂(t): Bi(r - v*t,t)]•n′
/* One can remove the common •n′ :
/$ ∇′Ei(r - v*t,t) = -1/c*∂[∂(t): Bi(r - v*t,t)]
/*++++++++++++++++++++++++++++++++++++++
/*add_eqn "Question
04_27rev1
Faradays_law_differential_form
/$ ∇´Ei(r - v*t,t) = -1/c*∂[∂(t): Bi(r - v*t,t)]
/% ∇′EIodv(POIo,t) = -1/c*∂[∂(t): BIodv(POIo,t)]
/* OK - Note that this is for v=0 which seems inconsistent with "(r - v*t,t)", and I still unsure that Ive properly defined ∇′
Wasnt Stokes theorm already used in Faraday defn for (4-2)(4-18)?
If so, might this be a circular argument somehow?
(?nothing wrong with that - as it would at least "close the loop" and show consistency).
29May2016 - This is OK => 14Sep2015 p69h0.95 Lucas comment that ET = E0+Ei is obscure to me at present.
13Sep2015 My approach with 1.c was NOT successful (I need to look at it again later)