31 T. H. Havelock. 
doublet at the point (0, 0, —f), let the direction cosines of its axis be (l, m, 1). 
By the same process as for the horizontal doublet in (16) and (17), the surface 
values of ¢ and 0¢/dz are found to be 
ee am d6 i (il cos 8+ im sin 6 — n) sec? 6 Ort 
Tw 
bee 0 K — Ky Sec” 0 + ip sec 0 
i Ay sof siiands manna a 
with Q = ens/ tie (eens ety sino), 
With the same notation as before, 
F, = ky (ul — n (k — ky sec? 6)} D sec? 0, 
F, = — ky {un sec 0 +1 (K — ky sec? 8) cos 6} D sec? 0, 
Fy = —kym (k — ky sec? 0) D sin 0 sec? 0, 
F, =—«yumD sin 0 sec? 0, 
G, = k[wk, 1 sec? 0 — nk (K — Ky sec? 8) + p? sec? 0}] D, 
G, = — x[I {« (« — Ky sec? 0) + p? sec? 0} cos 0+ pny sec? 0] D, 
G; = — xm {k (K — ky sec? 8) + p? sec? 6} D sin 8, 
G, =— wm«,xD sin 0 sec? 0, 
D = (M/r) e“/{(x — xy sec? 0)? + pu? sec? 6}. (28) 
We find that & FG simplifies very much even if we take the expressions as they 
stand ; since we are only concerned ultimately with uw zero, we could further 
simplify the work by omitting superfluous terms. The expression for R 
reduces to the limit of an integral of the same type as in (23), and the result is 
7/2 é 
1 16rpxp tM] (2 cos? 6 + m? sin? @ + n2) e~ *xoF 8°°8 sec’ 8d0 
0 
<= Oper iO o [p [Ko (@) (1 4 =| K, (a)} 
)Ke @x (1 a a a =) K, (@)} | (29) 
als 
where « = ky f = gf/c?. 
6. The only further stage to which we need carry the calculation is for two 
doublets in any positions: M at the point (h, &, — f) with its axis in the direction 
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