dw D i One Or(4) 
za + Art — wt Grit) + 4 — Or Dld + S (6) (4 OP ref — — wt ]}] — 
Owe Owe 
2 Ore Or) 
— ()? 74) we | ——- — we —— }} ; 
Oe Oue 
Taking «=4, this formula yields the instant z®, for w= 1, 
and all terms vanish except the first. 
So we see that for a= 1, 2, 3 it gives the simultaneous displacements. 
We can simplify considerably. Writing 
0 d ; 
= (c) we = 
dze _ der 
> Pen inet «2 Or g 
(c) en OE qe) 
we get for the simultaneous displacements : 
dor Owe dod 
veeel 2 — = (0) os + LEWoe—, (62) 
da\4) Oue 
For ~w=4 we have s4)=0. 
6.3. Let us inquire what will be the polarization of matter, viz. 
the electrical moment per unit of volume. The electrical moment of 
one atom being es*, where e is the charge of an electron, the answer 
is, in a first approximation, that the polarization has components 
Ne st, (Gale): 
Proceeding more carefully, we must take some closed surface, 
a sphere, say, sum up the electrical moments of the atoms within 
and divide by the volume. But what about the border atoms, which 
are intersected by the sphere? Must we leave them out, or must we 
reckon them as lying within the sphere? The difference will be of 
second order only, but it does make a difference. 
A similar question has been raised by Lorentz in his Theory of 
Hlectrons (note 53). Lorunrz decides himself to leave out the inter- 
sected atoms, and this is certainly right when we restrict ourselves 
to the first order terms, neglecting 6%. But here we retain 6’. 
Fortunately, our calculus leads us to the answer: it will show a 
correction to be made to the same effect as establishing the rule: 
the atoms are to be reckoned as lying within the surface, whenever 
more than half of the line joining nucleus and electron lies 
within the surface. This is a quite satisfactory rule. 
Thus the polarization is: 
0 Nest sc 
dze 
6.4. The magnetic momentum of an atom has the components 
Nest — } 2 (Cc) (6.3) 
