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)