Full text: The quantum and its interpretation

MAGNETIC TUBES 
135 
x. 1] 
identical closed trajectories in a molecular magnetic field, the 
number of lines of force cut by the radii vectores at each revolu 
tion is one and the same universal constant.” In other words, 
all the electron-resonators are transversed by a like tube of 
magnetic force. The product of the induction flux and the 
charge is equal to Planck’s constant. Bernoulli’s treatment is 
perhaps not altogether satisfactory since he makes the assump 
tion—which must be wide of the mark—“ that the molecular 
magnetic field may be regarded as uniform.” 
We consider first the number of magnetic tubes threading 
through a circular orbit in which a point charge e is travelling 
with large velocity. Such a charge making v revolutions per 
second may be regarded as equivalent to a current i = ev. Let 
L denote the self-inductance of the equivalent current circuit, 
then the electrokinetic energy is given by 
T = |L« 2 10 :1 
But by definition the product Li is the number of magnetic 
tubes threading through the circuit, i.e. 
Li = N TO 10 : 2 
Substituting this value in the expression for T we get 
T = IN,/*' ..... 10 : 3 
Now according to the quantum theory the steady motion of the 
system must be such that the kinetic energy obeys the relation 
2 $Tdt — nh 10 : 4 
where n is an integer and the integration is to be extended over 
a complete period of the motion. Although the quantum theory 
has been expressed by different investigators in many different 
ways, in this simple case where the energy can be expressed by 
one single term, all forms of the quantum theory lead to the 
same condition. This fact has been pointed out by W. Wilson. 
We now make a further assumption, namely, that as all mass 
may be regarded as electromagnetic mass the kinetic energy of 
the moving charge may be identified as electrokinetic energy, 
that is, we assume that the value for T given in equation (3) 
may be employed in equation (4). Substituting the value for T 
given in (3) we get 
$>N m evdt — nh 10 : 5 
If in this motion Maxwellian principles are suspended there is 
no radiation, and the kinetic energy may be assumed constant. 
Consequently N m is constant, for both e and v are constant. 
Since (pvdt = 1, we find
	        
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