paradox of the day 2_18_14 REVIEWED
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A short dated note (2.18.14) by Phil comparing two of his own documents on bipolar coordinates and line conductors. He traces a factor of 2 in the transverse potential solution of the Laplace problem, comparing ln(s2/s1) with ln(s2^2/s1^2), and checks it against an MIT web text on line charges. He concludes that K1 = -2ξ1 and K2 = -2ξ2, so the constants K1 and K2 are not equal to ξ1 and ξ2.
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Paradox of the Day PhL 2.18.14
This concerns a factor of two confusion relating bipolar doc and lines doc. I kept thinking K1 = ξ1 but that we not correct. In fact, K1 = -2ξ1 and this is now all made clear in bipolar doc.
1. In my new bipolar document, I show that
ξ = (1/2) ln() = ln () (4.5)
and my potential solution is then simply
φt(ξ) = ξ = ln ()
and the problem this is the solution to is this
t2φt(x,y) = 0 φt(C1) = ξ1 φt(C2) = ξ2 (5.4.3)
because this matches the two boundary conditions ξ1 on the right and ξ2 on the left.
2. In lines doc (6.2.1) I claim that
s2/s1 = e-B => ln(s2/s1) = - B . (6.2.1)
and therefore
B = ln ()
Therefore, it must be true that B = ξ and these are the exact same variable.
3. In lines doc, my "scaling boundary condition" is this:
φt(x) ≈ ln(s22/s12) // limiting form as point x = (x,y) moves far from the conductors
(5.3.13)
where φt is a solution to this transverse problem,
t2φt(x,y) = 0 φt(C1) = K1 φt(C2) = K2 K1- K2 = K (5.4.3)
Now I certainly can identify K1 = ξ1 and K2 = ξ2 so I have this problem:
t2φt(x,y) = 0 φt(C1) = ξ1 φt(C2) = ξ2 ξ1- ξ2 = K (5.4.3)
4. Here then is the paradox. In the bipolar doc, I claim that the solution to (5.4.3) is this
φt = ξ = ln ()
whereas in my lines doc, I claim instead that the solution is this
φt(x) ≈ ln(s22/s12) // limiting form as point x = (x,y) moves far from the conductors
(5.3.13)
Somewhere there is a factor of 2 error going on!!!
Web data. Site http://web.mit.edu/6.013_book/www/chapter4/4.6.html says that
where λi is the charge per unit length and where r1 = s1 etc.
My lines doc solution to this problem is this:
φ(x,y,z) = q(z) φt(x,y) (5.1.1)
φt(x) ≈ ln(s22/s12)
so my lines doc solution is then
φ(x,y,z) = q(z) ln(s22/s12) = q(z) ln(s2/s1)
and this AGREES with the web quote.
5. In lines doc I have: " We then have from (6.1.2) and (6.2.1), "
φt(x) = ln(s22/s12) = 2 ln(s2/s1)
φt(C1) = 2 ln(s2/s1)|C1 = -2B1
φt(C2) = 2 ln(s2/s1)|C2 = -2B2 . (6.3.1)
It seems strange that I have those factors of 2 sitting there. So look back to the quoted equations:
φt(x) = ln(s22/s12) . for all values of r, close and far (6.1.2)
s2/s1 = e-B => ln(s2/s1) = - B . (6.2.1)
So the factors of 2 are just coming from φt(x) = 2 ln(s2/s1). So I basically have
φt(C1) = K1 = -2B1 = -2ξ1
φt(C2) = K2 = -2B2 = -2ξ2
When I wrote things up, I just "made up" the names K1 and K2. I made no claim they were equal to anything in particular. I later found out that K1 = -2ξ1 , for example.