paradox of 11_20_13 RESOLVED
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A brief working note by Phil dated 11.20.13, from his transmission lines files, tied to Appendix B of a lines write-up. He solves the 2D Poisson equation for Az inside a wire with uniform current density, using the 2D propagator ln(1/R) and an angular integral from Appendix B. The apparent paradox, that a solution of the form α+βr² seems to satisfy the homogeneous Laplace equation, is resolved by noting that for axial symmetry the only atoms are A + B ln r.
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Paradox of the Day PhL 11.20.13
Resolution: The atoms for the 2D Laplace in polars are ejnθ [ r±n ] and A + B lnr. If situation is axially symmetric, then only atoms are A + B lnr. Thus, r2 is not an atom in this case. Duh.
This is in lines Appendix B Version 2. Suppose we start with
22DAz(r) = -μ2Jc(r) r < a Jc(r) = Jc = I/(πa2) = uniform current density in wire
so we then have
22DAz(r) = -μ2 I/(πa2) r<a
Think of this as just some constant source
22DAz(r) = Kθ(r<a)
The particular integral of this equation I think should be
Az(r) = (1/2π) ∫dV' Kθ(r<a) ln(1/R) = (K/2π) !Syntax Error, Idr' r' !Syntax Error, Idθ' ln(1/R)
since (1/2π) ln(1/R) is the 2D propagator. Rewrite as
Az(r) = - (K/4π) !Syntax Error, Idr' r' !Syntax Error, Idθ' ln(R2)
From my App B picture I know that
R2 = r2 + r'2 -2rr'cos(θ-θ')
I then show in App B that
!Syntax Error, Idθ' ln[r2 + r'2 -2rr'cos(θ-θ')] = 2!Syntax Error, Idθ' ln[r2 + r'2 -2rr'cos(x)]
= 2 Q(r',r) = 4π
Thus, we get
Az(r) = - (K/4π) !Syntax Error, Idr' r' 4π = -K !Syntax Error, Idr' r'
= -K !Syntax Error, Idr' r' [ ln(r')θ(r'>r) + ln(r)θ(r'<r) [
= -K !Syntax Error, I θ(r'>r) dr' r' ln(r') - K ln(r) !Syntax Error, I θ(r'<r) dr' r'
= -K !Syntax Error, Idr' r' ln(r') - K ln(r) !Syntax Error, I dr' r'
= -K [!Syntax Error, Idr' r' ln(r') + ln(r) !Syntax Error, I dr' r' ] = -K * joe
Let's hand this to our friend Maple
Thus we get a solution
Az(r) = -K (1/4) [ 2a2lna + (r2-a2) ]
Now here is the paradox: This solution has the form Az(r) = α + βr2 . Both terms are atoms for the 2D Laplace equation, and therefore
22DAz(r) = 22D[α + βr2] = 0
But the paradox is that the starting equation was
22DAz(r) = Kθ(r<a)
We can do this manually since
22DAz(r) = r-1∂r(r ∂r Az(r)) = r-1∂r(r ∂r [α + βr2])
= r-1∂r(r ∂r [βr2]) = β r-1∂r(r ∂r r2)]) = β r-1∂r(r 2r)]) = 2β r-1∂r(r2) = 2βr-1(2r) = 4β
Oops! Huh? Well OK, for n = 0 the only atoms are C + D log r, my bad, Paradox resolved.