6. the jigger problem
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Dated 2.28.10, this note by Phil treats the "jigger problem": the upper half of an oblate hyperboloid whose neck at ζ = 0 is filled with metal, so V = 0 there. It sets up the series solution in Legendre-type functions pnm, qnm, imposes the neck condition using Gamma-function values, and fixes the coefficients with Gauss's Law on a pillbox. The result matches the full bloid solution except that a factor 2 becomes 4.
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The Jigger Problem PhL 2.28.10
The problem here is to find the Green's Function for an oblate hyperboloid upper half where the neck is filled with metal, so the upper half looks like the upper half of a metal "jigger". There is no "lower bloid" to worry about in the jigger problem.
1. The Smythian form and picture 1
2. Condition V = 0 at the neck which is ζ = 0 1
Status after doing all the neck matches: 2
3. Apply Gauss's Law to pillbox around point charge q. 2
4. Final result and comparison with the full bloid solution: 3
1. The Smythian form and picture
If we follow the text of bloid attempt 4, we find that the Smythian form must be this:
Vo(ζ,ξ) = Σnm [Bnm pnm(ζ0) + Cnm qnm(ζ0)] qnm(ζ) Pnm(ξ) cos(mφ)
Vi(ζ,ξ) = Σnm [Bnm pnm(ζ) + Cnm qnm(ζ) ] qnm(ζ0) Pnm(ξ) cos(mφ) //upper ζ ≥ 0
This is what it has to look like to match at the boundary between Vo and Vi . We can write this as:
V(ζ,ξ) = Σnm [Bnm pnm(ζ<) + Cnm qnm(ζ<) ] qnm(ζ>) Pnm(ξ) cos(mφ) //upper ζ ≥ 0
2. Condition V = 0 at the neck which is ζ = 0. We have at the neck:
0 = Vi(0+,ξ) = Σnm [Bnm pnm(0+) + Cnm qnm(0+) ] qnm(ζ0) Pnm(ξ) cos(mφ) //upper
and from leg prop doc we have
pnm(0±) = 2m+1 / [ Γ(1/2 - n/2 - m/2) Γ(1 + n/2 - m/2) ]
qnm(0±) = (-1)m 2m-1 [ Γ(1/2 + n/2+ m/2) / Γ(1 + n/2- m/2)]
so our condition at the neck is this:
Bnm pnm(0+) + Cnm qnm(0+) => Cnm = – [ pnm(0+)/ qnm(0+)] Bnm
[ pnm(0)/ qnm(0)] =
= 2m+1 / [ Γ(1/2 - n/2 - m/2) Γ(1 + n/2 - m/2) ] / (-1)m 2m-1 [ Γ(1/2 + n/2+ m/2) / Γ(1 + n/2- m/2)]
= (-1)m 22 / [ Γ(1/2 - n/2 - m/2) Γ(1 + n/2 - m/2) ] * [Γ(1 + n/2- m/2)/ Γ(1/2 + n/2+ m/2)]
= (-1)m 22 / [ Γ(1/2 - n/2 - m/2) ] * [1/ Γ(1/2 + n/2+ m/2)]
= 4 (-1)m / [Γ(1/2 + n/2+ m/2) Γ(1/2 - n/2 - m/2)] // which agrees with Attempt 4
Let's define
g3(n,m) = 1/ [ Γ(1/2 + n/2+ m/2) Γ(1/2 - n/2 - m/2) ]
so
[ pnm(0)/ qnm(0)] = 4 (-1)m g3(n,m)
Our relation is then:
Cnm = – [ pnm(0+)/ qnm(0+)] Bnm = – 4 (-1)m g3(n,m) Bnm
Status after doing all the neck matches:
V(ζ,ξ) = Σnm [Bnm pnm(ζ<) + Cnm qnm(ζ<) ] qnm(ζ>) Pnm(ξ) cos(mφ) //upper ζ ≥ 0
with condition:
Cnm = – 4 (-1)m g3(n,m) Bnm
g3(n,m) = 1/ [ Γ(1/2 + n/2+ m/2) Γ(1/2 - n/2 - m/2) ]
3. Apply Gauss's Law to pillbox around point charge q.
This section is exactly word for word as in the bloid solution, where we apply only to the upper bloid. The conclusion is that no condition is put on Cnm and that
Bnm = (2-δm,0) (q/[4πc1ε]) Pnm(ξ0) sec[(n+m)π/2] f(n,-m)/ Knm(ξ1)
4. Final result and comparison with the full bloid solution:
V(ζ,ξ) = Σnm [Bnm pnm(ζ<) + Cnm qnm(ζ<) ] qnm(ζ>) Pnm(ξ) cos(mφ) //upper ζ ≥ 0
with condition:
Bnm = (2-δm,0) (q/[4πc1ε]) Pnm(ξ0) sec[(n+m)π/2] f(n,-m)/ Knm(ξ1)
Cnm = – 4 (-1)m g3(n,m) Bnm
g3(n,m) = 1/ [ Γ(1/2 + n/2+ m/2) Γ(1/2 - n/2 - m/2) ]
so we write our jigger solution as
V(ζ,ξ) = Σnm Bnm [pnm(ζ<) – 4 (-1)m g3(n,m) qnm(ζ<)] qnm(ζ>) Pnm(ξ) cos(mφ) //upper ζ ≥ 0
We can now compare this to our full bloid result:
V(ζ,ξ) = Σnm Bnm [pnm(ζ<) – 2(-1)m g3(n,m) qnm(ζ<)] qnm(ζ>) Pnm(ξ) cos(mφ) //upper ζ ≥ 0
V(ζ,ξ) = Σnm Bnm [ 2(-1)m g3(n,m) qnm(ζ<)] qnm(ζ>) Pnm(ξ) cos(mφ) //lower ζ ≤ 0
So our jigger solution is the same as the upper bloid solution with a factor 2 → 4 !!