Re: Spheres vs Toroids

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Antonio Carlos M. de Queiroz" <acmq-at-compuland-dot-com.br> 

Tesla list wrote:

 > Original poster: "Godfrey Loudner" <ggreen-at-gwtc-dot-net>
 >
 > Hello Antonio
 >
 > The messy part of the exact formula is
 > f(x) = Sum[Q(n-1/2,x)/P(n-1/2,x), {n, 1, infinity}]
 > where x = (D-d)/d.

In Snow's "Circular 544" there is a recurrence formula that calculates
The toroidal functions Q(n+1/2) and P(n+1/2) from the corresponding
functions of order n-1/2 and n-1/2-1. It's just the regular formula
for Legendre functions. The first terms, or orders -1/2 and 1/2
can be calculated from elliptic functions, and so the sum is very
easy to evaluate precisely without messy numerical integrations.
The sum converges quickly. I just stop adding terms when the last term
added is smaller than a given fraction of the sum (1e-6).

 > Set f(x,k) = Sum[Q(n-1/2,x)/P(n-1/2,x), {n, 1, k}], i.e.,
 > the sum of the first k terms. By playing around with integral
 > inequalities, one can show the below.
 >
 > absolutevalue[f(x)-f(x,k)] < e whenever x > 1 and
 >
 > k > ln[Pi(2x)^(1/2)/e(x-1)(x^2-1)^(1/4)]/ln[x].
 >
 > This can be slightly improved if one admits having
 > Q(-1/2,x) into the estimate. The problem with my
 > estimates is that they may over shoot the number of
 > terms necessary to achieve a given accuracy. I don't
 > know how optimal the estimates are.

I admit that my simple criterion could fail in a series that
doesn't converge. But this series converges, except for the
case when the toroid hole is closed. I will test your rule.

 > Let's take your D = 90cm and d = 30cm, then x = 2.
 > Now with e = 1/10^m with accuracy to m places in
 > mind.
 >
 > place value accuracy     number of terms
 >           1                 6 or more
 >           2                 9 or more
 >           3                13 or more
 >           4                16 or more
 >           5                19 or more
 >           6                23 or more
 >
 >  >From my notes, Mathematica gives 40.56477807 pf

I get: 40.5647768934 pF, with 7 terms in the series, stopping with the
first term that is smaller than 1/1e6 of the sum. The terms are:

Argument: 2.0000000000
P-0.5:    0.9012862994 Q-0.5:    1.6566381702
P+0.5:    1.3291381622 Q+0.5:    0.2240142928
P+1.5:    3.2439396660 Q+1.5:    0.0451587242
P+2.5:    9.5831240340 Q+2.5:    0.0100993416
P+3.5:   30.5393254980 Q+3.5:    0.0023700825
P+4.5:  101.1307275221 Q+4.5:    0.0005719164
P+5.5:  342.7613792184 Q+5.5:    0.0001405377

With error set to 1e-10, with 11 terms I get 40.5647780687 pF
1e-15 results in 15 terms and gives the same first 12 digits.
Our calculations agree.

 > However with x very close to 1 from above, my estimate
 > indicates that an astronomical number of terms might be
 > required. If D = 2d, we have a torus with no hole and x = 1.
 > I have tried to compute the limit of the formula as
 > x approaches 1 from above, but without success so far.

This would be interesting, but the formula is good enough for practical
calculations in this case:
With relative error of 1e-12:
D=1, d=0.49: 49 terms,    48.2356246774 pF
D=1, d=0.499: 147 terms,  48.3978426230 pF
D=1, d=0.4999: 441 terms, 48.4139700271 pF
To verify, with error of 1e-15:
D=1, d=0.49: 61 terms,    48.2356246775 pF
D=1, d=0.499: 186 terms,  48.3978426232 pF
D=1, d=0.4999: 563 terms, 48.4139700279 pF
And finally, still with 1e-15:
D=1, d=0.4999999: 15796 terms, 48.4157591228 pF
D=1, d=0.499999999: 142566 terms, 48.4157608907 pF
D=1, d=0.49999999999: 1197801 terms, 48.4157603364 pF
The computation time was only perceptible (~1s) in this last case.
D=1, d=0.4999999999999: 9457225 terms, 48.4157146119 pF
Numerical error starts to be apparent, as the capacitance decreased.
But by this point, there is no space for an electron in the hole...

Do you have some idea about how to calculate the electric field at
the toroid surface?

Antonio Carlos M. de Queiroz