
: 
‘a 
, 


Electric Origin of Rigidity and Consequences. 419 
molecule must contain Zand to form the doublet which 
gives cohesion. This is the most important characteristic of 
the molecules in metals, that each atom contains in itself 
both Zand. This conception is fundamental in the present 
icy y; especially in section 5. 
Equation (4) becomes 
s 
Nae 
= S 


23 OS a 
tee ee ss. CO) 
K (m/p)? 
This shows the electric origin of rigidity, but it needs 
interpreting. Consider the electric doublets in a metal at 
absolute zero so arranged that their charges form alternate 
plane laminze of positive and negative electricity of surface- 
density ¢. The average electric force between the laminz 
must be daro/K; and according to Maxwell’s view the energy 
in unit volume of the dielectric is $¢.470/K. To make our 
laminar distribution correspond with the charges of the 
electric doublet we must have c=e/(m/p)#; and, moreover, 
to give the properties of a homogeneous body we should use 
three laminar distributions at right angles to one another, 
such that each would give one- third of the energy due to the 
doublets in the actual body which they represent. ‘Thus, 
then, the electrostatic energy per unit ‘volume due to the 
components of the doublets in any one direction can be 
written 

27 & (mip)! 21 &(m/p)s (6) 
(mfp)? ~ K * (njfp) ” 
This becomes identical with (5) if s=(in/o)2, a result which 
would make it appear that in the monatomic molecules of 
metals the two electrons $ and 4 are on the surface of the 
atom at the ends of a diameter. With this result it is pos- 
sible, by means of the values for M*/ for the metals given 
in Table xxix. of “ Further Studies” and by the equation 
m?l=e?s*/KK, to obtain values of K the mean dielectric 
capacity of the metals, but we cannot at present follow this 
side track. We must now show how the rigidity at absolute 
zero and the electrostatic energy of the 
A € doublets resolved in any one direction are 
identical. 
Consider two electrons % and » of amount 
e at distance r apart in the positions A and 
B. Subject them toa small shear of amount 
w indicated by the displacements AC and 
D B BD each equal to wr/2 at right angles to AB. 
The electrical work done is d(2/rK) = 
—@drf[iP?K. But dr=w?r/2, and therefore the work done 
2G 2 
