nequal, both of thick crust and of hotter nucleus beneath also, whether
the latter be _now_ liquid or not. Were the contraction, lineal or
cubical, for equal decrements or losses of heat, or in equal
times--equal both in the material of the solidified crust and in that of
the hotter nucleus--there could be no such tangential pressures as are
here referred to, at any epoch of the earth's cooling. But in accordance
with the facts of experimental physics, we know that the co-efficient of
contraction for all bodies is greater as their actual temperature is
higher, and this both in their solid and liquid states.

Hence for equal decrements of heat, or by the cooling in equal times,
the hotter nucleus contracts more than does its envelope of solid
matter.

The result is now, as at all periods since the signs changed of the
tangential forces thus brought into play--_i.e._, since they became
tangential _pressures_--that the nucleus tends to shrink away as it
were from beneath the crust, and to leave the latter, unsupported or but
partially supported, as a spheroidal dome above it.

Now what happens? If the hollow spheroidal shell were strong enough to
sustain, as a spheric dome, the tangential thrust of its own weight and
the attraction of the nucleus, the shell would be left behind altogether
by the nucleus, and the latter might be conceived as an independent
globe revolving, centrally or excentrically, within a shell outside of
it. This, however, is not what happens.

The question then arises, Can the solid shell support the tangential
thrust to which it would be thus exposed? By the application to this
problem of an elegant theorem of Lagrange, I have proved that it cannot
possibly do so, no matter what may be its thickness nor what its
material, even were we to assume the latter not merely of the hardest
and most resistant rocks we know anything of, but even were it of
tempered cast-steel, the most resistant substance (unless possibly
iridio-osmium exceed it) that we know anything about. Lagrange has shown
that if P be the normal pressure upon any flexible plate curved in both
directions, the radii of these principal curvatures being r' and r'',
and T the tangential thrust at the point of application and due to the
force P, then:

P = T (1/r' + 1/r'')

When the surface is spherical, or may be viewed as such, r' = r'' and

P = 2T/r or, T = P x r/2

In the present case P is for a unit square (taken relative