FIGURE OF EQUILIBRIUM OF A ROTATING MASS OF LIQUID. 271 
S and T is measured inwards. If then we still choose to measure outwards, as I 
shall do, our formula l^ecomes 
cIv 
■y,, — ^ = ItttS or 0 , 
according as ^ does or does not cut the double layer. 
It may be Avell to remark that v being proportional to rS, 'C dvjdii is small compared 
with ItttS. It is also important to notice that the term ItttS is independent of the 
form of the surface, and that dvkln will be the same to the first order of small 
c[uantities for a slightly deformed ellipsoid as for the ellipsoid itself 
We have now to apply these results to our problem. 
The position of a point in the region R may be defined by the distance measured 
inwards from J along one of the curves orthogonal to J. The surface of the pear as 
defined in this way is given by e, a function of 6 and <f). Any jjoint on a curve may 
then be defined by se, where s is a proper fraction. If s is the same at every point 
the surface s is a deformed ellipsoid ; s = 1 gives the pear and s = 0 the ellipsoid J. 
If c/o- is an element of area of J, the corresponding element on the surface s will be 
(1 — Xes') da. The value of X will be determined hereafter, and it is only necessary 
to remark that it is positive because the areas must decrease as we travel inwards. 
Let s and s + ds be two adjacent surfaces ; then the mass of negative density 
enclosed between them in the tube of which (1 — \es) da and (1 — Xe (,s -f ds))da 
are the ends is — pe(I — 'Kes) da ds. If this element of mass be regarded as 
surface density on s, that surface density is clearly — pe ds. If the same element of 
mass were carried along the orthogonal tube and deposited as surface density on J, 
that surface density would be — pe (1 — Xes). The sum for all values of s of all such 
transportals would constitute the condensation — C already considered. 
The double system D consists of tlie volume density — p in R, and the positive 
condensation + C on J, the total mass being zero. 
Let 2 , a proper fraction, define a surface between J and the jDear. Consider one of 
the orthogonal curves, and let Lq be the potential of D at the point P where the 
curve leaves J and V. the potential at the point Q where it cuts 2 . Then I require 
to find Fq — Tb. 
Since s denotes a surface intermediate between J and the pear, -- ( Vq — Vdj ds is 
the excess of the potential at P above that at Q of surface density — pe ds on s and 
surface density + pe (1 — Xes) ds on J. Such a system is a double layer, but there is 
a finite distance between the two surfaces, and the form of p ( Vq — V^) will clearly 
iio 
be difierent according as 2 is greater or less than s. 
The arc e.s may be equally divided by a large number of surfaces, and we may take 
t to define any one of them. Now we may clothe each intermediate surface t with 
equal and opposite surface densities i pe [1 — Xe (s — ?)] dt. 
