190 
MR. ARTHUR SCHUSTER ON THE 
where X and Y have the values given in (l), and for p we must substitute the 
expression (18), remembering, however, that we must ultimately restore the factor p 0 . 
We may also temporarily omit the factor C in the expression for the electric forces. 
To find the current function R we must eliminate S, but owing to the fact that p 
contains 8 and X, this does not seem to be possible directly. The difficulty must be 
turned by eliminating in the first place R, and if S is then found, R may be 
determined from the first of the above equations. It may occur to the reader that R 
might be more directly obtained if the resistivity were introduced instead of the 
conductivity. This is true, but the results are less valuable, as may be seen from the 
fact that, as suggested above, the Fourier expansion may have to be applied to the 
conducting power in so far as it depends on the position of the sun. If the resistivity 
were introduced as the variable, the high and possibly infinite values which the factor 
would take when the conductivity sinks low or vanishes would present difficulties 
much greater than those met with by keeping the conductivity as the variable 
quantity. 
The elimination of R requires in the first instance the reduction of 
i.e. of 
p (4s x sill 0+ 4r Y) +X sin 8 ^ + Y 
\cIl 7 cl\ / ciu c/A. 
Introducing the values of X and Y, the operation reduces to 
(1 +y' cos 8 + y sin 8 cos X) sin 8—^—y' sin 8 cos 8 ^ 
c/X c/X 
-y cos 8 ( cos 6 cos X —sin 8 sin X 
\ c/X dO 
d\Jj 
The part independent of y and y is 
— sind^= — cr sin dip/ cos (crX — a). 
The part depending on y f is 
— 2y f sin 8 cos 8 ~ — 
c/X 
— 2 y'ar sill 8 COS 8xjj n a COS (crX — a) 
-Vo- sin ( n ~ °~ + 1) Vn+i + ( n + q ') Vi 
2n+\ 
We are left with the part dependent on y, the factor of which is 
cos X (cos 2 8— sin 2 8) + sin 8 cos 8 sin X ( ~. 
c/X d8 
