DYNAMICAL PRINCIPLES TO PHYSICAL PHENOMENA. 
487 
If the rate of dilatation be uniform and equal to fi, the dilatation for 6 degrees is fid, 
the stress required to produce this dilatation is icfid, where k is the bulk modulus. 
Now, we can show by Hamilton’s principle that there will be a stress of this amount 
if, in the expression for the positional kinetic energy, there is the term 
where v 0 is a constant volume. 
Since 
j t<fi6 dv, 
J v n 
fi = (- — 
p \v de 
K — “ V 
p constant 
dp^ 
dv 
9 constant 
p being the pressure to which the surface of the solid or liquid is exposed, 
/3k = — 
dv dp 
dd dv 
dp 
Te 
v constant 
so that the term in the positional kinetic energy may be written 
6 (j y) dv. 
\ClC7 /u constant 
(15) 
Thus a dynamical system with this term in the expression for the positional kinetic 
energy will behave like the solid or liquid so far as expansion by heat goes. 
We may add to this the term 
M CyO, 
where M is the mass of the solid or liquid, and c x a constant, as this term will not give 
rise to any stresses tending to strain the body. Unless this term has different values 
for the different states in which the body can exist, the temperature being kept 
constant, it will disappear from the variation equation. 
We can now solve the problem of finding the density of the saturated vapour of a 
liquid at any temperature. 
Let us suppose that we have a mass N of the liquid and its vapour in a closed 
space : let £ be the mass of the vapour, N — £ that of the liquid, u\, w z , the mean 
intrinsic potential energies of unit mass of the vapour and liquid respectively, Q the 
volume occupied by the vapour. Then, if there is no energy due to surface-tension, 
electrification, and so on, the value of L for the vapour, using the same notation as 
before, is 
f(A0 + R01og® ! )-fi!> 1 , 
for the liquid 
(N — f)c,0 + (' e | dv — (N — f) w 2 . 
