426 Mr. W. Sutherland on the 
of the J electrons with X and of the » electrons against X, 
so as to give a current proportional to X. The cases in 
which the ¥ electron of one doublet is brought near to the 
like electron of another by the oscillations, or the ) electron 
is brought near to its like, need no further consideration as 
there is repulsion. But when the ¢ of one doublet approaches 
the ) of another, the cases in which break up of the doublets 
occurs must most frequently happen when % is moving for- 
ward with X and backward. While the near J and b of 
two doublets are forming a new doublet, the remote p and J 
of these two doublets are released as free electrons at the end 
of their vibrations when they are at rest. Therefore they 
begin at once to move in the field of X. Now, in the theory 
of Riecke and Drude and Thomson, these free electrons are 
supposed to travel a certain free path under the force X and 
thus maintain thecurrent. This isa possible mode of genesis 
of the current, but it seems to me more: probable that, as 
these free electrons will rush together to form new doublets, 
the passing on of % forward and of 5 backward at a favour- 
able conjunction of circumstances is the real source of current. 
As the other action is dependent on this one, its effects may 
be merged in those of its cause. The result of the oscillatory 
process, » then, i is to send forward a certain fraction of the Z 
electrons and backward the same fraction of tne b electrons. 
Denote this fraction by @ and the number of each sort of 
electrons per unit volume by Q, then we have $Q electrons 
per unit volume moving forward with velocity v, which is 
the mean velocity of vibration determined sbove, the same 
number moving backwards; so the total current across unit 
area 1s 
26Qev= Qs X/Coe't, . . . 1 1 Q) 
So for the electric conductivity y we have 
y= bQe's' Cot... °. 
But Cw?/2 is the electrokinetic energy of our metallic 
stion or atom plus doublet. The electrokinetic energy per 
unit volume is by section 3, 
NG?/T*= (0/7)? 4e*s?/R Ga) t: 
But Q=1/(2a)’, so ! 
Co?/2 = (6/T)*fes/K(2a)’}, and y=K@T"/26’r. . (12) 
We have still.to determine ¢ and tr. We have found that 
vis half the period of rotation of the doublet. We have 
pictured the rotation as movement on a curve like a screw 
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