328 
(9) 
F=n 0 2 — n 2 — -f- o) e 2 N/m 
(10) 
k — an unknown constant. 
From (1) and (2) F and 2 k n can be found. Writing for brevity 
di) 
,>2 — / 
(v 2 — a 2 — l) 2 + 4 v 9 -S- 
(12) 
2vx / . ., 
(• v 2 x 2 — iy + 4 v* y* — 9 W /!> 
we obtain: 
(13) 
£ 
<; I g 
1 
K 
1! 
6s 
(14) 
e 2 N 
2kn= 4n — q (v, x). 
m J ’ 
§ 12. As 
let us assume 
an example of the application of formula (13), § 11., 
that the term Fis of the form (4), § 11. We have then 
(1) 
»0 2 — n '‘ = Î n ~ i 1 + 3 f( v : *)} 
whence, substituting from (11), § 11., we obtain 
, 2 x + . e*N _ I _ 
(v 2 — x 2 -\-2){v 2 — x' 2 — l)-\-4v 2 x 2 3 m 'n 0 2 — n 2 ' 
Like equation (4), § 6., this formula includes Lorentz' theorem as 
a particular case; but resting as it does on the assumption of a 
mono-electronic substance it is less general than the cited equation. 
§ 13» Let us suppose we are dealing with a mono-electronic 
substance in a part of the spectrum far removed from the absorp¬ 
tion band of the body. We shall then have by formula (13), §11, 
(1) ( v 2 — 1)F = 4 Jie?N/m. 
If this be applied on the opposite sides of the absorption band it 
appears that 
(2) in the region of small n : F 0, v 2 >> 1 
(3) in the region of large n : L<0, v 2 < 1. 
