e film has two surfaces the
surface-tension of the film is double the tension of the surface of the
liquid of which it is formed.

To determine the relation between the surface-tension and the pressure
which balances it when the form of the surface is not spherical, let us
consider the following case:--

[Illustration: Fig. 9.]

Non-spherical soap bubble

Let fig. 9 represent a section through the axis Cc of a soap-bubble in
the form of a figure of revolution bounded by two circular disks AB and
ab, and having the meridian section APa. Let PQ be an imaginary section
normal to the axis. Let the radius of this section PR be y, and let PT,
the tangent at P, make an angle a with the axis.

Let us consider the stresses which are exerted across this imaginary
section by the lower part on the upper part. If the internal pressure
exceeds the external pressure by p, there is in the first place a force
[pi]y^2p acting upwards arising from the pressure p over the area of the
section. In the next place, there is the surface-tension acting
downwards, but at an angle a with the vertical, across the circular
section of the bubble itself, whose circumference is 2[pi]y, and the
downward force is therefore 2[pi]yT cos a.

Now these forces are balanced by the external force which acts on the
disk ACB, which we may call F. Hence equating the forces which act on
the portion included between ACB and PRQ

[pi]y^2p - 2[pi]yT cos[alpha] = -F (9).

If we make CR=z, and suppose z to vary, the shape of the bubble of
course remaining the same, the values of y and of a will change, but the
other quantities will be constant. In studying these variations we may
if we please take as our independent variable the length s of the
meridian section AP reckoned from A. Differentiating equation 9 with
respect to s we obtain, after dividing by 2[pi] as a common
factor,

dy dy d[alpha]
py -- - T cos[alpha] -- + Ty sin[alpha] -------- = 0 (10).
ds ds ds

Now

dy
-- = sin[alpha] (11).
ds

The radius of curvature of the meridian section is

ds
R1 = - -------- (12).
d[alpha]

The radius of curvature of a normal section of the surface at right
angles to the meridian section is equal to the part of the normal cut
off by the axis, which is

R2 = PN = y/cos[alpha] (13).

Hence dividing equation 10 by $y sin \alp